Nonequilibrium Thermodynamics, Second Edition: Transport and Rate Processes in Physical, Chemical and Biological Systems

Nonequilibrium Thermodynamics, 2nd Edition Transport and Rate Processes in Physical, Chemical and Biological Systems by...

Author: Yasar Demirel

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Nonequilibrium Thermodynamics, 2nd Edition Transport and Rate Processes in Physical, Chemical and Biological Systems

by Yasar Demirel

• ISBN: 0444530797 • Publisher: Elsevier Science & Technology Books • Pub. Date: September 2007

PREFACE* Natural phenomena consist of simultaneously occurring transport processes and chemical reactions. These processes may interact with each other and lead to instabilities, fluctuations, and evolutionary systems. The objective of this book is to explore the unifying role of thermodynamics in natural phenomena. Nonequilibrium thermodynamics is based on the entropy production character of nonequilibrium and irreversible processes and provides a link between classical thermodynamics and transport and rate processes. In 1850, Clausius introduced the concept of noncompensated heat as a measure of irreversibility. In 1911, Jaumann formulated the rate of entropy production as the product of flows and thermodynamic forces reflecting the second law of thermodynamics, stated as, "A finite amount of organization may be purchased at the expense of a greater amount of disorganization in a series of interrelated spontaneous actions." Every process in nonequilibrium conditions operates with thermodynamic forces, such as pressure, temperature, or chemical potential gradients, which cause flows. After the publication of Onsager's reciprocal relations in 1931, Casimir, Meixner, Prigogine, and De Groot made early attempts to formulate a general theory for irreversible processes. Nonequilibrium thermodynamics mainly formulates the rate of entropy production and the relations between the conjugate flows and the forces in a system. The rate of entropy production can be derived using the general balance equations of mass, momentum, energy, entropy, and the Gibbs relation based on local equilibrium. The rate of entropy production equation identifies the independent forces and flows that are related by phenomenological equations containing the proportionality constants called phenomenological coefficients. A matrix of these coefficients is symmetric, according to Onsager's reciprocal relations. The theory treating near global equilibrium phenomena is called linear nonequilibrium thermodynamics, and phenomenological equations linearly relate forces and flows. This edition updates and expands most of the chapters of the first edition by focusing on the balance equations of mass, momentum, energy, and entropy together with the Gibbs equation for coupled processes of physical, chemical, and biological systems. Every chapter contains example problems and practice problems to be solved. Chapter 1 briefly describes basic elements of classical thermodynamics, such as irreversibility, equilibrium state, thermodynamic principles, the Gibbs equation, and the phase equilibria. Chapter 2 briefly introduces transport processes and chemical reactions. Here, nonequilibrium systems (with thermodynamic branch), transport coefficients, and some well-known examples of coupled phenomena are briefly introduced. Chapter 3 discusses the general balance equations, which are used in the Gibbs equation to derive the rate of entropy production based on local equilibrium. Chapter 4 focuses on thermodynamic analysis of transport processes, chemical reactions, and power-generating systems. It also introduces the concept of exergy, the equipartition principle, and pinch analysis with various example problems to underline the contribution of thermodynamic analysis toward creating optimum designs and assessing the performance of existing designs. Chapter 5 introduces the thermodynamic optimum approach in the design and optimization of various processes. Here, the concept of thermoeconomics is emphasized. Under proper financial, normative, environmental, and technical constraints, thermoeconomics can minimize the overall cost by identifying the thermodynamically optimum design and operation. Chapter 6 summarizes the diffusion in nonelectrolyte and electrolyte systems and explores some related applications. Chapter 7 describes coupled heat and mass transfer and the level of coupling without chemical reaction. Chapter 8 briefly summarizes chemical reactions and coupled phenomena. Chapter 9 combines transport processes and chemical reactions; it focuses on the dynamic balance equations consisting of coupled flows as well as the coupling between chemical reactions of scalar character with the heat and mass flows of vectorial character. Chapter 10 briefly describes passive, facilitated, and active transport through membranes. Chapter 11 introduces various applications of thermodynamics in biological systems as well as energy conversion and coupling phenomena in bioenergetics. Based on the nonequilibrium thermodynamic approach, the stability aspects of various transport processes and chemical reactions are covered in Chapter 12. Stability analysis cannot predict the behavior of evolutionary processes. However, it can predict when a system will become unstable. Chapter 13 briefly describes some organized structures maintained with the outside supply of energy and matter. Some biological systems are good examples of maintaining dissipative organized structures. Chapter 14 summarizes some of the other thermodynamic approaches, such as extended nonequilibrium thermodynamics. The appendix supplies some data needed in the example and practice problems. *An instructor resource containing the Solution Manual can be obtained from the author - [email protected] http ://books.elsevier.com/Demirel/thermodynamics/

or from

xviii

Preface

This book introduces the theory of nonequilibrium thermodynamics and its use in simultaneously occurring transport processes and chemical reactions of physical, chemical, and biological systems. Thus, it provides a unified approach in describing natural phenomena and would be effective in senior and graduate education in chemical, mechanical, system, biomedical, tissue, biological, and biological systems engineering, as well as physical, biophysical, biological, chemical, and biochemical sciences. All through the first and second edition, the work of many people who contributed to both the theory and applications of thermodynamics, transport processes, and chemical reactions has been visited and revisited. I acknowledge and greatly appreciate all these people. I am also thankful for the help I received from Dr. Yelizaveta P. Renfro and the production team of Macmillan India Ltd. for editing and reviewing the chapters. Y. Demirel, 2007

PREFACE TO FIRST EDITION Classical thermodynamics is based on a limited number of natural laws, which have led to a vast number of equations describing macroscopic behavior of various types of systems. However, classical thermodynamics is mainly limited to energy conversion in equilibrium, and particularly applied to reversible and closed systems. Beside the equilibrium, there are instabilities, fluctuations, and evolutionary processes. The objective of this book is to bring out and emphasize the unifying role of thermodynamics in transport phenomena, chemical reactions, and coupled processes in physical and biological systems by using the nonequilibrium thermodynamic approach. The development of nonequilibrium thermodynamics is based on the entropy-generating character of irreversible processes to provide a link between the classical thermodynamics, and the transport and rate processes. In 1850, Clausius introduced the concept of noncompensated heat as a measure of irreversibility. In 1911, Jaumann introduced the concepts of the entropy production and the entropy flux. Donnan and Guggenheim in 1934 related the coupled natural processes to the second law of thermodynamics stated as "A finite amount of organization may be purchased at the expense of a greater amount of disorganization in a series of interrelated spontaneous action." After the publication of Onsager's reciprocal relations in 1931, Casimir, Meixner, Prigogine and De Groot made early attempts of a macroscopic and general theory for irreversible processes. Irreversible processes cause entropy generation because of net thermodynamic forces and flows within the system. Nonequilibrium thermodynamics is mainly concerned with the analysis of entropy generation and the study of the relations between the conjugate flows and the forces. Using the general balance equations of mass, momentum, energy, entropy, and the Gibbs relation, entropy generation or the dissipation function can be derived. The rate of entropy generation or the dissipation function identifies the flows and forces that are related by the phenomenological equations. These equations contain the proportionality constants called the phenomenological coefficients. This coefficient matrix is symmetric according to the Onsager's reciprocal relations. Such an analysis is necessary to understand the complex, coupled transport and the global behavior of physical and biological systems. The theory treating near-equilibrium phenomena is called the linear nonequilibrium thermodynamics. It is based on the local equilibrium assumption in the system and phenomenological equations that linearly relate forces and flows of the processes of interest. Application of classical thermodynamics to nonequilibrium systems is valid for systems not too far from equilibrium. This condition does not prove excessively restrictive as many systems and phenomena can be found within the vicinity of equilibrium. Therefore equations for property changes between equilibrium states, such as the Gibbs relationship, can be utilized to express the entropy generation in nonequilibrium systems in terms of variables that are used in the transport and rate processes. The second law analysis determines the thermodynamic optimality of a physical process by determining the rate of entropy generation due to the irreversible process in the system for a required task. Some processes may have forces operating far away from equilibrium where the linear phenomenological equations are no longer applicable. Such a domain of irreversible phenomena, such as some chemical reactions, periodic oscillations, and bifurcation, is examined by extended nonequilibrium thermodynamics. Extending the methods of thermodynamics to treat the linear and nonlinear phenomena, and such dissipative structures are attracting scientists from various disciplines. This book introduces the theory of nonequilibrium thermodynamics and its use in transport and rate processes of physical and biological systems. The first chapter briefly presents the equilibrium thermodynamics. In the second chapter, the transport and rate processes have been summarized. The rest of the book covers the theory of nonequilibrium thermodynamics, dissipation function, and various applications based on linear nonequilibrium thermodynamics. Extended nonequilibrium thermodynamics is briefly covered. All the parts of the book can be used for senior- and graduate-level teaching in engineering and science. All through this book, the work of many people who contributed to both theory and the applications of nonequilibrium thermodynamics has been visited and revisited. All those whose work has contributed in preparing this book are acknowledged, and greatly appreciated. Y. Demirel, 2002

LIST OF SYMBOLS a

A A* B,C Be Br

C,c

Ct,,Cv CSs D Da Ds,e DD, e DT, e e eX

E Ex

f fi

F g G zXG~ zXI-tr h H I J

4 4 k

kB K

ke L Le Lik Lqr Lsr

Jr m

M n

N

Nr Nu P Pe

activity, interfacial area Helmholtz free energy, chemical affinity, area nondimensional affinity virial coefficients, Bejan number Brinkman number concentration, cost heat capacities at constant-pressure and constant-volume, respectively reactant concentration at surface diffusion coefficient, diameter Damk6hler number effective diffusion coefficient for the substrate S coupling coefficient related to the Dufour effect coupling coefficient related to the Soret effect energy specific exergy activation energy of the chemical reaction, energy exergy fugacity of pure component fugacity of species in a mixture force acceleration of gravity Gibbs free energy Gibbs free energy change on reaction reaction enthalpy enthalpy (specific) enthalpy, Henry's law constant electric current diffusive mass flux conduction heat flux W m 2 reaction velocity thermal conductivity Boltzmann constant chemical equilibrium constant resistance coefficients effective thermal conductivity first-order reaction rate constant characteristic half thickness Lewis number phenomenological coefficients (conductance) coupling coefficient between chemical reaction and heat flow coupling coefficient between chemical reaction and mass flow volumetric reaction rate mass molecular weight number of moles, number of components number of moles number of independent reactions Nusselt number pressure, volumetric rate of entropy production Peclet number

xxii Pr

Pi

er qo q

Q R Re S

S Sc prod

t 7"

rr U V

V W

W

Ws X

X Y Z

Z

List of symbols

Prandtl number partial pressure of species i reduced pressure degree of coupling between processes i and j heat flow vector volumetric flow gas constant Reynolds number specific entropy entropy Schmidt number entropy production time temperature reduced temperature internal energy velocity vector volume mass fraction work shaft work liquid mole fraction, ratio of forces thermodynamic force vapor-phase mole fraction dimensionless distance compressibility factor

Greek letters

~e

#

3' "Yi

S S~j E:

Y q~ KT

/x P

0 A P II O" T 60

eigenvalues effective thermal diffusivity coefficient of thermal expansion, thermicity group specific heat ratio activity coefficient small deviations Kronecker delta function dimensionless coefficient related to Soret effect, extent of a chemical reaction, elasticity coefficient generalized Thiele modulus, dimensionless local entropy generation rate Arrhenius group, dimensionless efficiency, shear viscosity dimensionless temperature isothermal compressibility viscosity, chemical potential stoichiometric coefficient, kinematic viscosity dimensionless concentration controlling parameters density osmotic pressure dissipation function, dissipated available energy dimensionless parameter dimensionless time accentric factor, dimensionless parameter related to Dufour effect

===

List of symbols

Subscripts av az b c

d D e eq EOS f L liq m rain mix opt P prod q r

rev R s

sat S T th V vap

average azeotrope backward, bulk critical, charging discharging Dufour effective equilibrium state equation of state forward, formation liquid liquid melting minimum mixing or mixture optimum packed production heat reaction, reduced property reversible residual properties surface, isentropic property along a two-phase coexistence line Soret thermal diffusivity thermodynamic vapor vapor

Superscripts E R

sat

0 o

excess reduced saturation rate pseudo reference (initial) conditions standard conditions

XXlll

Table of Contents

Chapter 1 - Fundamentals of Equilibrium Thermodynamics Chapter 2 - Transport and Rate Processes Chapter 3- Fundamentals of Nonequilibrium Thermodynamics Chapter 4 - Using the Second Law: Thermodynamic Analysis Chapter 5 - Thermoeconomics Chapter 6 - Diffusion Chapter 7 - Heat and Mass Transfer Chapter 8 - Chemical Reactions Chapter 9 - Coupled Systems of Chemical Reactions And Transport Processes Chapter 10 - Membrane Transport Chapter 11 - Thermodynamics and Biological Systems Chapter 12 - Stability Analysis Chapter 13 - Organized Structures Chapter 14 - Nonequilibrium Thermodynamics Approaches

1 FUNDAMENTALS OF EQUILIBRIUM THERMODYNAMICS 1.1

INTRODUCTION

The name thermodynamics stems from the Greek words therme (heat) and dynamis (power). Thermodynamics is a science of energy in which, temperature as it is related to the average of molecular motion is an important concept. Guggenheim defines thermodynamics as that part of physics concerned with any equilibrium property's dependence on temperature. Thermodynamics also formulates the average changes taking place among large numbers of molecules; therefore, it is a macroscopic science. Thermodynamics first emerged as a science after the construction and operation of steam engines in 1697 by Thomas Savery and in 1712 by Thomas Newcomen in England. Later, Carnot, Rankine, Clausius, Kelvin, Gibbs, and many others developed formulations of thermodynamic principles for describing the conservation and conversion of energy. The theoretical formulation of classical thermodynamics is a set of natural laws governing the behavior of macroscopic systems; these laws lead to a large number of equations and axioms that are exact, based entirely on logic, and attached to well-defined constraints. As natural phenomena are far from reversible, adiabatic, isothermal, equilibrium, or ideal, the engineer must exercise a pragmatic approach in applying the principles of thermodynamics to real systems. Some new texts have attempted to present thermodynamic principles and formulations in ways applicable to students and engineers, which would allow the formulations to be applied in modeling, designing, and describing some natural and complex phenomena. Principles of thermodynamics find applications in all branches of engineering and the sciences. Besides that, thermodynamics may present methods and "generalized correlations" for the estimation of physical and chemical properties when there are no experimental data available. Such estimations are often necessary in the simulation and design of various processes. This chapter briefly covers some of the basic definitions, principles of thermodynamics, entropy production, the Gibbs equation, phase equilibria, equations of state, and thermodynamic potentials.

1.2 1.2.1

BASIC DEFINITIONS Systems

A thermodynamic system is a part of the physical universe with a specified boundary for observation. A system contains a substance with a large amount of molecules or atoms, and is formed by a geometrical volume of macroscopic dimensions subjected to controlled experimental conditions. An ideal thermodynamic system is a model system with simplifications to represent a real system that can be described by the theoretical thermodynamics approach. A simple system is a single state system with no internal boundaries, and is not subject to external force fields or inertial forces. A composite system, however, has at least two simple systems separated by a barrier restrictive to one form of energy or matter. The boundary of the volume separates the system from its surroundings. A system may be taken through a complete cycle of states, in which its final state is the same as its original state. In a closed system, material content is fixed and internal mass changes only due to a chemical reaction. Closed systems exchange energy only in the form of heat or work with their surroundings. In an open system, material and energy content are variable, and the systems freely exchange mass and energy with their surroundings. Isolated systems cannot exchange energy and matter. A system surrounded by an insulating boundary is called a thermally insulated system. A system and its surroundings are considered the universe.

2

1. Fundamentals of equilibrium thermodynamics

The properties of a system based on the behavior of molecules are related to the microscopic state, which is the main concern of statistical thermodynamics. In contrast, classical thermodynamics formulate the macroscopic state, which is related to the average behavior of large groups of molecules leading to the definitions of macroscopic properties such as temperature and pressure.

1.2.2 Processes Energy through its conversion and degradation can cause physical and chemical processes to occur. A process takes place in a system. Any process within an adiabatic system is known as an adiabatic process. A process that takes place with only an infinitesimal change in the macroscopic properties of a system is called an infinitesimalprocess. The classification of processes according to Planck considers three independent infinitesimal processes. They are natural processes, unnatural processes, and reversible processes. Natural processes actually occur and always proceed in a direction toward equilibrium. Unnatural processes are those that proceed in a direction away from equilibrium that never occurs. A reversible process is a case between natural and unnatural processes and proceeds in either direction through a continuous series of equilibrium states. Guggenheim provides the following simple example: consider the evaporation of a liquid at an equilibrium pressure Peq. If P < Peq, a natural evaporation takes place. However, when P > Peq, the evaporation is unnatural. I f P - P e q - 6 , where 6 > 0, evaporation takes place, and in the limit ~ ~ 0 the process becomes reversible. Variables in a system that remain constant with time are in a steady-state process, while those that change with time are in an unsteady-state process or in a transient process. At steady state, a system exchanges energy or matter at a constant rate.

1.2.3 Thermodynamic Properties Thermodynamic properties or coordinates are derived from the statistical averaging of the observable microscopic coordinates of motion. If a thermodynamic property is a state function, its change is independent of the path between the initial and final states, and depends on only the properties of the initial and final states of the system. The infinitesimal change of a state function is an exact differential. Properties like mass m and volume V are defined by the system as a whole. Such properties are additive, and are called extensive properties. Separation of the total change for a species into the external and internal parts may be generalized to any extensive property All extensive properties are homogeneous functions of the first order in the mass of the system. For example, doubling the mass of a system at constant composition doubles the internal energy. The pressure P and temperature T define the values at each point of the system and are therefore called intensive properties, some of which can be expressed as derivatives of extensive properties, such as the temperature T -- ( O U/OS)v,N i. IfXdenotes any extensive property (not necessarily a thermodynamic property) of a phase, we may derive intensive properties denoted by X/and called as partial properties as follows:

, (i 4: j)

(1.1)

For any partial property, we have dX = Ei (OX/Oni)dni = "ElXidl'li at constant T and P. The Euler theorem shows that X = ]~i Yini.

1.2.4 Energy Energy may be transferred in the form of heat or work through the boundary of a system. However, the conversion of work to heat or heat to work should be used with caution, as they are not numerically equal to each other. In a complete cycle of steady-state process, internal energy change is zero and hence the work done on the system is converted to heat ([work[ = Iheatl) by the system. The mechanical work of expansion or compression proceeds with the observable motion of the coordinates of the particles of matter. Chemical work, on the other hand, proceeds with changes in internal energy due to changes in the chemical composition (mass action). Potential energy is the capacity for mechanical work related to the position of a body, while Idnetic energy is the capacity for mechanical work related to the motion of a body. Potential and kinetic energies are external, while sensible heat and latent heat are internal energies.

1.2 Basicdefinitions

3

The conservation o f mass in an open system states that the change in the total mass is equal to the mass exchanged with the surroundings. In a system, we may consider two changes to the mass of species j: the internal change dimj and the external change with the surrounding demi (1.2)

dmj = dem j + dim j

During a process, energy can be transferred and converted from one form to another, while the total energy remains constant. This is known as the conservation o f energy principle.

1.2.5

Entropy

Entropy change is determined by the following equation: dS-

6qrev

T

(1.3)

where 6qrev is the reversible heat flow. When a fluid system changes from state A to state B by an irreversible process, then the change of its entropy is to be AS = SB - SA. Some important properties of entropy are: • • • •

• • • • •

Entropy is a state function and an extensive property. The determination of entropy requires the measured enthalpy and the use of relation T(OS/OT)p = (OH/OT)p = Cp. For a single phase, dS >- q/T, the inequality is for a natural change, while the equality is for a reversible change. The change of entropy is expressed as, dS = deS + diS where deS (des = q/T) is the change due to the interaction of a system with its surroundings, and dis is the increase due to a natural change, such as a chemical reaction, within the system and is always positive for irreversible changes (dis > 0) and zero at equilibrium (diS = 0). The entropy of a system is the sum of the entropies of all changes within the system. The entropy change of ice melting at 273.15 K is A S m - - ~ m / T = 21.99 J/(mol K). The entropy change of water vaporization at 373.15 K is 2~Sv = zXttv/T = 108.95 J/(mol K). The entropy of an insulated closed system remains constant in any reversible change, increases in any natural change, and reaches a maximum at equilibrium. Entropy remains constant for any reversible adiabatic change so that dS = O. For any complete cycle, the change of entropy is zero.

Some research topics in entropy are the determination of entropy changes in the mixing of very similar species (e.g., isotopes), in very cold systems (i.e., T ~ 0), and in highly dispersed systems.

1.2.6

Changes in Enthalpy, Entropy, and Volume in Terms of Temperature and Pressure

Some general property relations in terms of temperature and pressure are

clH :

-~

dr+

-~

dP : Cp d r + V - T --~

P

dS _

T

//

_0_~ S

dV =

dT +

OV

/

dP

(1.4)

T

--~ r dP - C P - - ~ -

dT +

-Of p

dP = ~ Vd T - K VdP

P

(1.6)

T

We can determine some of the partial derivatives above by using the following Maxwell relations, thermal expansion/3, and isothermal compressibility K:

/

0--fi)r - -

/

-~-

p

--/3V

After dividing d H = TdS + VdP by dP at constant temperature and using the Maxwell relations, we have

(1.7)

4

1. Fundamentals of equilibrium thermodynamics Table 1.1 Molar heat capacities for some gas compounds at T = 298.15 K and P = 1 atm

Species

Cp (J/mol)

Cv (J/mol)

y - Cp/Cv

(5/2)R (7/2)R 20.79 37.11 28.82 29.36 29.12

(3/2)R (5/2)R 12.47 28.46 20.44 20.95 20.74

5/3 7/5 1.6672 1.3039 1.4099 1.4014 1.4040

Ideal monoatomic gas Ideal diatomic gas Noble gases CO2 H2 O2 N2

Source: Kondepudi and Prigogine (1999).

= (1- fiT) V

_p(a~p) =(KP-flT) V T

(1.8)

(1.9)

T

P

and

--v

,,

v

(1.10)

For ideal gases, we have dH = Cp dT

(1.11)

dP ds:c --di-T _ R 7-

(1 12)

Table 1.1 shows the heat capacities at constant pressure and volume for some gas compounds at 298.15 K and 1 atm. Tables B3-B5 list the heat capacities of various compounds as function of temperatures. With small changes in T and P, and constant values of/3 and K, integration of Eq. (1.6) yields

In V2= ~(T2-T1)-K(P2-Pi) Vm

(1.13)

Equation (1.13) is a practical representation of a compressible fluid. For liquids, we have dH = Cp d r + (1 - fl T) VdP dT dS = Cp - 7 -

~ VdP

(1.14)

(1.15)

For incompressible fluids, thermal expansion fl and isothermal compressibility K are practically zero, and Eqs. (1.14) and (1.15) become dH = Cp dT + VdP

dS = Cp

dT

T

Table 1.2 shows expansivity and isothermal compressibility for some compounds at 298.15 K at 1 atm.

(1.16)

(1.17)

1.2 Basic definitions

5

Table 1.2 Expansivityand isothermalcompressibilityof some liquids and solids Species

/3 x 104 ( l / K )

Water Benzene Ethanol Tetrachloromethane Mercury Copper Lead Iron

KX

2.1 12.4 11.2 12.4 1.8 0.501 0.861 0.354

106 (1/atm) 49.6 92.1 76.8 90.5 38.7 0.735 2.21 0.597

Soume: Kondepudi and Prigogine (1999).

1.2.7

Change of Internal Energy and Entropy in Terms of Temperature and Volume

Some general property relations in terms of temperature and volume are dU-

OU d r + - ~ ~"

-~

dS-

-~

--~

dr+

r

V

dV

(1.18)

dV

(1.19)

T

From the following Maxwell relations, thermal expansion/3, isothermal compressibility K, and other relations, we have -~,

-T

~

~,

T

v

(1.20)

(1.211

After dividing d U - T d S - PdV by dV at constant temperature and using the Maxwell relations, we have

/, - P : r

7

77

(1.22)

Therefore, Eqs. (1.18) and (1.19) become (1.23)

dT dS-C,.--~+

(OP)dV=C - ~ ,"

dT+~dV --T- K

(1.24)

"~(~ dT dT [3 --2- - ~ VdP = C v ~ + -- d V 1 T K

(1.25)

where

f-

K

From Eqs. (1.15) and (1.24), we have dS

6

1. Fundamentals of equilibrium thermodynamics

Equation (1.25) yields a relation between Cp and Cv for real gases (1.26) For incompressible fluids, Eq. (1.26) reduces to Cp = Cv.

1.2.8

Chemical Potential

For ideal systems, the chemical potential is expressed by [&j = ]j,jo ( T , P ) + R T lnxj

(1.27)

where xj is the mole fraction of species j. The chemical potential can also be defined in terms of the concentration of species j, cj = N / V [&j -- ~j~jo ( T , V ) + R T lncj

(1.28)

For nonideal systems, we use activity in place of concentration, and Eq. (1.27) becomes [Ubj = [lbjo ( T , P ) + R T In ~ j X j

(1.29)

In the presence of external fields, the potential energy is included in the chemical potential. When the external field is an electric field, we get electrochemical potential/29 - = ]&jo ( T , P ) + R T ]&j

In "~jXj + FzjqJ

(1.30)

where F is the Faraday, or electric charge per mole (F = 96,500 C/mol), zj the valence of the species j, and ~ the electric potential.

1.2.9

Absolute Activity

The absolute activity is related to the molar chemical potential by

a=exp( ) The absolute activity is often used in phase equilibrium. For example, the equilibrium condition for the distribution of species i between vapor and liquid phases is a yap = a] iq

1.3

(1.32)

REVERSIBLE AND IRREVERSIBLE PROCESSES

Consider the equations that describe time-dependent physical processes; if these equations are invariant with regard to the algebraic sign of the time, the process is called a reversible process; otherwise it is called an irreversible process. Reversible processes are macroscopic processes that occur in the vicinity of global equilibrium. We can reverse the reversible process at any stage by a slight change in an external parameter. Guggenheim describes the reversible process and reversible change as follows: a process in a system interacting with its surroundings is a reversible process if the system and its surroundings are in equilibrium throughout the process. If equilibrium is not established, the result is a reversible change. For example, if heat flows from one system in equilibrium to another system in equilibrium, then a reversible change occurs; it is not a reversible process unless these two systems are at the same temperature. In a reversible process, it would be possible to perform a second process in at least one way to restore the system and its environment to their respective original states, except in case of differential changes higher than the first order. A reversible process proceeds with infinitesimal driving forces (i.e., gradients) within the system. Hence, for a linear transport system, reversible change occurs slowly on the scale of macroscopic relaxation

1.3

Reversible and irreversible processes

7

times, and dissipative effects cannot be present. Time appears only through its arithmetic value in the equations for reversible processes. For example, the equation describing the propagation of waves in a nonabsorbing medium is 1

02bl --

(1.3 3)

V2U

2 c o Ot 2

where Co is the velocity of propagation and u is the amplitude of the wave. Equation (1.33) is invariant in the substitution of t for (-t); hence, the propagation of waves is a reversible process. For a simple reversible chemical reaction, if one path is preferred for the backward reaction, the same path must also be preferred for the reverse reaction. This is called the principle of microscopic reversibility. Time can be measured by reversible, periodic phenomena, such as the oscillations of a pendulum. However, the direction of time cannot be determined by such phenomena; it is related to the unidirectional increase of entropy in all natural processes. Some ideal processes may be reversible and proceed in forward and backward directions. The Fourier equation 10T

(1.34)

-- V2T

c~ Ot

is not invariant with respect to time, and it describes an irreversible process. The term a is the thermal diffusivity. Irreversibility is a consequence of the dynamics of collisions in which the transfer of mass, energy, and momentum takes place. Hydrodynamics specifies a number of nonequilibrium states by the mass density, velocity, and energy density of the fluid. Hydrodynamic equations thus comprise a wide range of relaxation processes, such as heat flow, diffusion, or viscous dissipation, which are all irreversible.

1.3.1

Arrow of Time

Entropy in an isolated system increases dS/dt > 0 until it reaches equilibrium dS/dt = 0, and displays a direction of change leading to the thermodynamic arrow of time. The phenomenological approach favoring the retarded potential over the solution to the Maxwell field equation is called the time arrow of radiation. These two arrows of time lead to the Einstein-Ritz controversy: Einstein believed that irreversibility is based on probability considerations, while Ritz believed that an initial condition and thus causality is the basis of irreversibility. Causality and probability may be two aspects of the same principle since the arrow of time has a global nature. The term irreversibility has two different uses and has been applied to different "arrows of time." Although these arrows are not related, they seem to be connected to the intuitive notion of causality. Mostly, the word irreversibility refers to the direction of the time evolution of a system. Irreversibility is also used to describe noninvariance of the changes with respect to the nonlinear time reversal transformation. For changes that generate space-time symmetry transformations, irreversibility implies the impossibility to create a state that evolves backward in time. Therefore, irreversibility is time asymmetry due to a preferred direction of time evolution.

1.3.2

Dissipative Processes

All natural processes proceed toward an equilibrium state and dissipate their driving power; phenomenological relations, such as Fourier's law of heat conduction, can express them. Real physical processes progress with dissipative phenomena, such as mechanical or electrical friction, viscosity, and turbulence. These dissipative phenomena internally generate heat, and decrease the amount of energy available for work. In an isolated composite system, the change in the internal energy of a subsystem equals the change in heat d U - 6q -

OU

dS

(1.35)

OS

This phenomenon is associated with the level of entropy production due to the irreversibility of the process. Entropy is not conserved; it is the extensive parameter of heat.

1.3.3

Some Properties of Reversible Processes

• A reversible process can be reversed at any point by external conditions, retrace its path, and restores the original state of a system and its surroundings. • Equations that describe time-dependent processes are invariant with regard to the algebraic sign of the time.

8

1. Fundamentalsof equilibrium thermodynamics

• • • • • •

Processes are differentially removed from equilibrium. Reversible processes traverse a succession of equilibrium states. Reversible processes are characterized by frictionless flow. Flow (heat or mass) occurs when the net driving force is only differential in size. A reversible process represents a limit to the performance of actual processes. For cyclic processes, clS = 0

1.3.4 • • • • • • • • • • •

(1.36)

Some Properties of Irreversible Processes

Irreversible processes are actual processes carried out in finite time with real substances. Equations that describe time-dependent processes are not invariant with regard to the algebraic sign of the time. No infinitesimal change in external conditions can reverse process direction. In irreversible processes, heat transfer occurs through a finite temperature difference. In irreversible processes, mass transfer occurs through a finite chemical potential difference. An example of an irreversible process is a spontaneous chemical reaction or electrochemical reaction. Irreversible processes are characterized by the flow of fluids with friction, and sliding friction between any two matters. An example of an irreversible process is electric current flow through a conductor with a resistance. An example of an irreversible process is magnetization or polarization with hysteresis. An example of an irreversible process is inelastic deformation. In Agtota 1 > 0, no single process is possible for which the total entropy decreases. d s > 6Cl ~ dS = O T'

1.4

(1.37)

EQUILIBRIUM

If a physical system is isolated, its state changes irreversibly to a time-invariant state in which no physical or chemical change occurs, and a state of equilibrium is reached in a finite time. Some conditions of equilibrium are: (i) for a system thermally insulated with an infinitesimal change at constant volume: dS = O, d V = O, d U = 0, (ii) for a system thermally insulated with an infinitesimal change at constant pressure: dS - O, dP = O, d H - 0, (iii) for a system thermally insulated with an infinitesimal change at constant volume and temperature: dA = O, d V = O, d T = 0, and (iv) for a system thermally insulated with an infinitesimal change at constant pressure and temperature: d G = O, d T = O, dP = O. At equilibrium, all the irreversible processes vanish, and temperature, pressure, and chemical potentials become uniform; this means that no thermodynamic force exists in the system. No perturbation will cause a change in a neutral equilibrium. Any two phases in hydrostatic equilibrium must have the same pressure; in thermal equilibrium, any two phases must have the same temperature. If two phases are in equilibrium with respect to any species, then the chemical potential of that species must have the same value in these phases. Consider an elementary general chemical reaction vsS = vpP

(1.38)

where ~'s and Vp are the stoichiometric coefficients. The condition for chemical equilibrium is Vs/Xs = Vp~p

(1.39)

where/tz i is the chemical potential of a species i. This chemical equilibrium condition is equivalent to the vanishing affinity A defined by A=-~_vi~

i -0

(1.40)

1.4 Equilibrium

9

A system may be in a stable, metastable, unstable, or neutral equilibrium state. In a stable system, a perturbation causes small departures from the original conditions, which are restorable. In an unstable equilibrium, even a small perturbation causes large irreversible changes. A metastable system may be stable or unstable according to the level and direction of perturbation. All thermodynamic equilibria are stable or metastable, but not unstable. This means that all natural processes evolve toward an equilibrium state, which is a global attractor. The emergence of macroscopic reversibility from microscopic irreversibilities is referred to as dynamic equilibrium with the mechanisms of cancellation of the opposite molecular processes. An extremum principle minimizes or maximizes a fundamental equation subject to certain constraints. For example, the principle of maximum entropy (dS)u- 0 and, (d2S)u < 0, and the principle of minimum internal energy (dU)s = 0 and (d2U)s> 0, are the fundamental principles of equilibrium, and can be associated with thermodynamic stability. The conditions of equilibrium can be established in terms of extensive parameters U and S, or in terms of intensive parameters. Consider a composite system with two simple subsystems of A and B having a single species. Then the condition of equilibrium is (1.41)

d U -- (T A - TB )dS A - ( P A -- PB)dVA + (/d'A --/XB)dNA = 0

Hence the thermal, mechanical, and chemical equilibrium conditions in terms of the intensive properties are

(1.42)

rA - rB, PA - PB, ~A -- "B

since dSA, dVA, and d~A are the infinitesimal changes in independent variables. Similarly, the equilibrium conditions are expressed in terms of entropy

dS_( 1

~ rA

TB

(

PA rA

PB =0, rB

/XA rA

1]dUA_(PA

TA

TB )

PB

)

/d'A

]d'B

TA

TB

dNA= 0

(1.43)

and the equilibrium conditions become

1

1

ra

TB

=0,

/XB = 0 rB

(1.44)

Example 1.1 Equilibrium in subsystems Consider a closed isolated cylinder with two subsystems of 1 and 2 containing air with an equal volume of 1 L and equal temperatures of 298.15 K. There is a fixed piston at the boundary of the subsystems, which have different pressures of P1 = 2 atm and P2 = 1 atm. Estimate the temperature, volume, and pressure of the subsystems when the piston is released. Assume that the piston is impermeable to air, freely movable, and heat conducting. Solution: Assume that the air is an ideal gas with constant heat capacity. The initial states (i) of the subsystems are PliVli - nlRTli ' f 2iV2i - n2RT2i

(1.45)

8~ v~ = nlR r~r, Pay Vat - n2Rr2s

(1.46)

In the final state (f), we have

where the total volume is constant, and we have

Vt~ +V2;-2V1,-2L According to the first law and constant heat capacity Cv, we have (UI / - U l i ) + ( U 2 t , - L i

2, )-- 0, n i t v (Tlf - T l i ) +

n2C v ( T 2 f - T 2 i )

= 0

(1.47)

10

1. Fundamentals of equilibrium thermodynamics

Since Tlf-- T2f, Tli = T2i, and nli ::fi n2i , we have T V = Tli This is a natural result as the internal energy of an ideal gas depends on the temperature only, and the system is isothermal at the initial and final conditions. From Eqs. (1.45) and (1.46), we have

eli - nli - Vlf - 2 P2i n2i V2f

(1.48)

With a total volume of 2 L, the final volumes and pressures become 4

vv . 7L,.

1.5

.

2

3

. 7L, s.

2 atm

THE FUNDAMENTAL EQUATIONS

The f u n d a m e n t a l equations relate all extensive properties of a thermodynamic system, and hence contain all the thermodynamic information on the system. For example, the fundamental equation in terms of entropy is (1.49)

S = S ( U , ... , X j , . . . )

The extensive properties of U and X are the canonical variables. The fundamental equation in terms of internal energy U is U = U (S, ... , Xj , . . .)

(1.50)

For the entropy and internal energy, the canonical variables consist of extensive parameters. For a simple system, the extensive properties are S, U, and V, and the fundamental equations define a fundamental surface of entropy S = S(U,V) in the Gibbs space of S, U, and V. Differential forms of the fundamental equations contain the intensive thermodynamic properties. For example, dS and d U are

dS-~ (O~g) X

d U - + ' ~i (O~Xi) dXi " U,Xj=/=Xi

d U = --~ x dS

"3I-

"

-~i

S , X j @X i

(1.51)

(1.52)

Here, the first-order partial derivatives are the intensive properties T,/, and Y. In terms of the intensive properties, Eqs. ( 1.51) and ( 1.52) become d s = l d u + ~_ Ii dXi T i

(1.53)

dU

(1.54)

= TdS +

~_, Yi dXi

i The first term on the right side of Eq. (1.53) or Eq. (1.54) represents heat associated with the thermodynamic temperature T, and the remaining terms are the work terms. The pairs of intensive and extensive properties, such as 1/T and U, or Ii and X,., are the conjugate properties.

1.6

1.6

The thermodynamiclaws

11

THE THERMODYNAMIC LAWS

A set of thermodynamic laws governing the behavior of macroscopic systems lead to a large amount of equations and axioms that are exact, based entirely on logic, and attached to well-defined constraints. These laws are summarized in the following sections.

1.6.1

The Zeroth Law of Thermodynamics

Two systems in thermal contact eventually arrive at a state of thermal equilibrium. Temperature, as a universal function of the state and the internal energy, uniquely defines the thermal equilibrium. If system 1 is in equilibrium with system 2, and if system 2 is in equilibrium with system 3, then system 1 is in equilibrium with system 3. This is called the zeroth law of thermodynamics and implies the construction of a universal temperature scale (stated first by Joseph Black in the eighteenth century, and named much later by Guggenheim). If a system is in thermal equilibrium, it is assumed that the energy is distributed uniquely over the volume. Once the energy of the system increases, the temperature of the system also increases (dU/dT> 0).

1.6.2

The First Law of Thermodynamics

A change in a state function accompanying the transition of a system from one state to another depends only on the initial and final states and not on the path between these states. If the system returns to its original state, the integral of the change is zero au

- o

(~.55)

Such systems are called cyclic processes. The Poincare statement of the first law states that in a cyclic process, the work done by the system equals the heat received by it. According to the first law of thermodynamics, the state function of internal energy Uin a closed system is equal to the sum of the heat received by the system 6q and the mechanical work 6 Wperformed on the system by the surroundings dU = 6q + 6 W

(1.56)

Heat and work always refer to the system, and the sign convention for q and W chosen specifies which direction of energy transfer to the system is positive. The sign convention adapted here assumes that heat transferred into the system from the surroundings is positive, while work transferred into the system (work done on the system) at which energy is transferred into the system from the surroundings is positive. For compression work we have 6 W = - P d V as the compression leads to - d V and positive work. The signs of heat and work referring to the surroundings would then be opposite qsur = --q and W~ur= - W . Changes of heat and work depend on the path of a change and are not state functions. Therefore, it is not possible to define a function q or W that depends only on the initial and final states. When we consider an open system, we have a flow of energy due to heat transfer and to exchange of matter. The conservation of energy states that the total energy is conserved in any change of state. The total energy is the sum of the internal, kinetic, and potential energy of the system, heat, and type of work (such as electrical, mechanical, or chemical work). In general, the term 6 W represents all different forms of work. Work is the product of an intensive variable and a differential of an extensive variable. For example, if the system is displaced by a distance dl under a force F, it performs the work o f - F d l . I f - d N i moles of substance i with the chemical potential/x i flow from the system to its surroundings, the chemical work o f - t z d N , occurs. Thus the total work becomes

6 W = - P d V + Fdl + Ode + ~ tx i dN i + ' "

(1.57)

i=1

Here, - P d V refers to the sign convention recommending that work done on the system is positive as the compression work leads to - d V and positive work. For an open system, an additional contribution to the energy due to the exchange of matter dUm occurs dU - 6q + 6 W

+ dU m

(1.58)

12

1. Fundamentals of equilibrium thermodynamics

For systems with chemical reactions, the total energy may be considered a function of T, V, and N/: U = U(T, V, Ni). The total differential of U is

dU=(O~T) dT+(O-~V ) dV+~(O~Ni) dNi=6q+6W+dUm V, Ni

T ,Ni

i

(1.59)

V ,T,Ni, ~

The exact form of the function U(T, V,N/) for a certain system is obtained empirically.

Example 1.2 Relationships between the molar heat capacities Cp and

Cv

The first law of thermodynamics leads to a relation between the molar heat capacities. The change in internal energy expressed in volume and temperature U = U(T, V) is

dU=(O-~T ) dT+(O-~V ) d V = 6 q - p d V V

(1.60)

~.

The heat effects are

OU dT + P+ - ~ 6q= - ~ v

r

dV

(1.61)

At constant volume, the heat capacity is (1.62) V

V

and at constant pressure, the heat capacity is

G=U

+P+

(1.63)

The difference between Eqs. (1.63) and (1.62) is

C p - C v = P+ --~ T

--O-T p

(1.64)

The fight side of Eq. (1.64) shows the energy effect due to the expansion of volume at a constant pressure process. For a mixture of ideal gases, the internal energy is a function of temperature only, and hence Eq. (1.64) and PV = RT yields

Cp - Cv = R

(1.65)

The gas constant R is common to all gases and determined by the product of the Boltzmann constant and the Avogadro number R=(1.3805×10 -23) J/K (6.0225×1023) mo1-1 =8.3143 J/(mol K) In contrast, for a fluid, the enthalpy and entropy may be expressed as a function of temperature and pressure and in terms of thermal expansion/3 = (0 V/OT)p/V and isothermal compressibility K = -- (0 V/OP)r/V as follows:

dH = Cp d T + VdP = Cp d T + (1- [3T) VdP

(1.66)

dT dS = Cp --T- - ~ VdP

(1.67)

1.6

The thermodynamiclaws

13

For the internal energy and entropy, from Eqs. (1.23) and (1.24) we also have dU - CvdT +[-~ T-

P]dV

d S - Q ~dT + --[3d V T K

By comparing Eqs. (1.67) and (1.25), we have dT dT dS = Ci, ---2-- ~ VdP = Q ~ + - - d V 1

T

(1.68)

K

or

Cp-Cv=T/3V

~-f v

K

(1.69)

-~ p

At constant volume, Eq. (1.69) yields (1.70) V

For an incompressible fluid/3 = 0, and Cp = Cv = C. The second partial derivatives of the state functions at constant volume and entropy are V

,

=asV,

-

=KsV

(1.71)

I"

where Cv is the heat capacity at constant volume, as is the adiabatic thermal expansion, and Ks is the compressibility. The second partial derivatives of the state functions at constant pressure and temperature are OH

= Cp , P

-~

= ~T ' P

= apV P

-

= KrV

(1.72)

T

Here, Cp is the heat capacity at constant pressure, at, is the isobaric thermal expansion, and Kp is the isothermal compressibility. Table 1.1 shows the molar heat capacities of some gas compounds.

1.6.3

The Second Law of Thermodynamics

The work of Carnot, published in 1824, and later the work of Clausius (1850) and Kelvin (1851), advanced the formulation of the properties of entropy and temperature and the second law. Clausius introduced the word entropy in 1865. The first law expresses the qualitative equivalence of heat and work as well as the conservation of energy. The second law is a qualitative statement on the accessibility of energy and the direction of progress of real processes. For example, the efficiency of a reversible engine is a function of temperature only, and efficiency cannot exceed unity. These statements are the results of the first and second laws, and can be used to define an absolute scale of temperature that is independent of any material properties used to measure it. A quantitative description of the second law emerges by determining entropy and entropy production in irreversible processes. If a system is in equilibrium, then all the forces X~ are fully known from external parameters ai, so that the first law is (Sq - d U - (5 W = d U - ~ X i (a i )da i

(1.73)

Equation (1.73) is a Pfaffian equation and a; is an independent variable. Caratheodory's theory states that starting from a known original state, there may be other states that cannot be reached by an adiabatic process along the path 6q = 0. This shows the existence of an integrating factor for 6q; hence we have

14

1. Fundamentalsof equilibrium thermodynamics A d S = dr/

(1.74)

where dr/is a total differential of the variables a i. Therefore, r/must be a state function known as entropy S, and the integrating factor is the reciprocal of absolute temperature T, so that d S - ~qrev

(1.75)

T Equation (1.75) is a mathematical statement of the second law of thermodynamics for reversible processes. The introduction of the integrating factor for 6q causes the thermal energy to be split into an extensive factor S and an intensive factor T. Introducing Eq. (1.75) into Eq. (1.56) yields the combined first and second laws dU = TdS + ~ W

(1.76)

Every system is associated with an energy and entropy. When a system changes from one state to another, the total energy remains constant. However, the total entropy is not conserved, and increases in irreversible processes while remaining unchanged in reversible processes. The notion of entropy is not a directly intuitive concept. We can relate the entropy of an irreversible process to the external and internal properties, regardless of the energy content of the system. We can attain the same distribution of internal parameters imposed both reversibly and irreversibly by a set of external parameters. These different paths result in different work and energy changes in the system. However, we assume that a set of local parameters determines the entropy, and we can devise an ideal process that would reversibly bring the system to any configuration of the irreversible process. For example, diffusion of a substance is a nonequilibrium process, and the local concentration profile is necessary to define the system. We may apply reversibly a centrifugal field to the system to maintain the same concentration profile in a state of equilibrium. The energy applied reversibly to the centrifugal field is different from an irreversible diffusion process. Thus, the thermodynamic states of an irreversible diffusion process and the corresponding equilibrium system are different. Entropy may be computed as the corresponding entropy of the real system.

Example 1.3 Entropy and distribution of probability Entropy is a state function. Its foundation is macroscopic and directly related to macroscopic changes. Such changes are mostly irreversible and time asymmetric. Contrary to this, the laws of classical and quantum mechanics are time symmetric, so that a change between states 1 and 2 is reversible. On the other hand, macroscopic and microscopic changes are related in a way that, for example, an irreversible change of heat flow is a direct consequence of the collision of particles that is described by the laws of mechanics. Boltzmann showed that the entropy of a macroscopic state is proportional to the number of configurations 1~ of microscopic states a system can have S = k In f~

(1.77)

where k is the Boltzmann constant (k = 1.3805 × 10 -23 J/K). Consider a system with two chambers containing a total number of particles n. The total number of possibilities for distributing the particles between the two chambers is f~, which is the total number of distinct microstates with nl number of particles in chamber 1 and n 2 number of particles in chamber 2 f~ = (nl + n2)! nl! n2!

(1.78)

Equation (1.77) shows that disorganization and randomness increase entropy, while organization and ordering decrease it, and equilibrium states have the maximum value of D. In the above system, 1~ reaches its maximum value when n I = n 2. In parallel, the increase in entropy corresponds to the increase in the number of microscopic states or states with higher probability. The concept of entropy as a measure of organized structures is attracting scientists from diverse fields such as physics, biology, and communication and information systems.

1.7

BALANCE EQUATIONS

Balance equations consist of conserved, such as mass and energy, and nonconserved, such as entropy, properties. The following sections summarize the general balance equations of mass, energy, and entropy.

1.7 Balance equations

1.7.1

15

Mass Balance

In an open system, mass flow rate for the flowing streams through the boundary is (1.79)

rh = yAp

where v is the average velocity, A the cross-sectional area, and p the density. Assuming that the flow is positive when it enters into the control volume and using Eq. (1.79), the mass balance is dm

)in -- - -

d m + E (rhi )out -- Z ( g h i dt i i

+- Z

dt

i

(vAP)i, out - Z (vAP)i, in = 0 i

(1.80)

This form of mass balance is also called the continuity equation. At steady state, the accumulation term (dm/dt) becomes zero, and we have

E (/~i)out -- E i

1.7.2

(vAP)i,out - Z (vAP)i,in

(ghi)in - - E

i

i

0

=

(1.81)

i

Energy Balance

Each unit of mass flow transports energy e at a rate, O = [U + (1/2)v 2 + zg]rh, where z is the elevation above a datum level, and g is the acceleration of gravity. However, considering the flow work of all the entering and leaving streams in terms of the product of the pressure and volumes of each stream, we have a total energy associated with a stream i defined by

I/ ,v2 / U+-

+zg

J E{

rh+(PV)rh

-

2

H+-

i

+zg 2

/] rh

=O i

(1.82)

i

Using Eq. (1.82), we have a compact form of energy balance

d(mU__~_~)+ E (ei)out -- ~ dt

i

( e i ) i n -- ~/-F- m

(1.83)

i

where q and W are heat transfer rate and shaft work. Equation (1.83) assumes that the center of mass of the control volume is stationary. If the kinetic and potential energy changes are small enough, then Eq. (1.831) reduces to

d(mU)) + ~_~(I;t~)out i

dt

~ (/;/~)in

-

-

0 + f¢l

(1.84)

i

Equation (1.84) is widely applicable to many thermal engineering systems. If a system is at steady state, then the accumulation term vanishes

Z (~'i)out-- Z (el)in -i

(1.85)

4 Jr- ~/1~

i

or

•

1.7.3

(( ,v2 !/

(/ ,v2 i)

/,out

2

(1.86) /,in

Entropy Balance

The rate form of entropy balance is ±(rhS) +

d(mS) dt

+

dSsulT dt

0 Xpr°d > --

(1.87)

16

1. Fundamentals of equilibrium thermodynamics

where A(rhS) is the net entropy change of the flowing streams, and Sprod is the rate of entropy production. The term dSsurr/dl is the entropy changes within the surroundings defined by

dSsurr ~ Oi dt

-.

~

(1.88)

With Eq. (1.88), and for a steady-state condition, Eq. (1.87) reduces to A ( g h S ) - t~.. -~/ ~0 qi __ Sprod "

(1.89)

Equation (1.89) determines the rate of entropy production due to irreversibility within a control volume. The concept of entropy production is elaborated further in the next section. 1.8

ENTROPY AND ENTROPY PRODUCTION

By using Eq. (1.76) with a pressure-volume work, we have

dS =

dU + PdV T

(1.90)

The entropy of a system is an extensive property, and it changes through the exchange of mass and energy• If a system consists of several processes, the total entropy change is equal to the sum of the entropy changes in each process. The total change of the entropy dS results from the flow of entropy due to exchanges with surroundings (deS) and from the changes inside the system (diS)

dS=deS+diS

(1.91)

The value of diS is zero when the change inside the system is reversible, and it is positive when the change is irreversible

diS = 0

(Reversible change)

(1.92)

diS > 0

(Irreversible change)

(1.93)

For an isolated system, there is no interaction with the surroundings so that

dS = diS > 0

(1.94)

The rate of entropy production Sprod is expressed by Sprod = d i S ~ 0

dt

(1.95)

Irreversible processes produce entropy in any isolated, open, or closed system and Eq. (1.94) holds. In every macroscopic region of the system, the entropy production of irreversible processes is positive. A macroscopic region contains enough molecules for microscopic fluctuations to be negligible. The second law of thermodynamics states that the sum of the entropy production of all processes for any system and its environment is positive. When interfacial phenomena are considered, the entropy production is based per unit of surface area. The entropy source strength • is the rate of entropy production per unit volume (I) -- Spr°d ~ 0

(1.96)

dV The product of the entropy source strength and the absolute temperature is called the dissipation function = T~ >_ 0

(1.97)

1.8 Entropyand entropy production

17

When the T is the environmental temperature, the dissipation function represents the energy dissipated to the environment. The entropy source strength and the dissipation function are not state functions, and they depend on the path between the given states.

Example 1.4 Entropy production and subsystems Equation (1.91) can be applied to various irreversible processes. Let us consider a system consisting of two closed subsystems of I and II, and maintained at uniform temperatures of T I and Tn, respectively. The total entropy dS is expressed as dS -

6I q

d S I + d S II

6n q

= U + ri-----f-

(1.98)

The interactions of heat in each subsystem are given by 6Iq = 6~q + 6~q,

(3IIq = 6~Iq + 6~Iq

(1.99)

Using Eq. (1.99) and the conservation of energy 6~Iq + 6~q = 0, Eq. (1.98) yields

dS=---fif-+-~

,

q

T II = d e S + d i S

TI

(1.100)

The entropy production per unit time is

dt

dt

TI

T II > 0

(1.101)

Equation (1.101) shows that the rate of entropy production is the product of flow (heat flux) (6~q/dt), and the thermodynamic force, (1/T I - 1/Tn). Equation (1.90) is the total differential of the entropy as a function of the variables U and V only. To generalize this relation, we also consider the changes in the amounts of species. Using the mole amounts for the species, we have a general expression for the change of entropy from the Gibbs relation ~j

dS = d U + PT d v - E ---T-

(1.102)

J

Equation (1.102) is one of the main relations for calculating entropy production.

Example 1.5 Entropy production in a chemical reaction in a closed system Derive an expression for the entropy production for a chemical reaction in a closed system. Solution: The reaction is a single elementary reaction and homogeneous. Entropy production due to a chemical reaction in a closed system is given by dS = -6q - + - - dAe

T

T

(1.103)

where A is the affinity of the chemical reaction, A = - E vjix2, s the extent of the reaction, and v/the stoichiometric coefficient of species j. Equation (1.103) shows that the entropy change contains two contributions: one is due to interactions with the surroundings deS = 6q/T, and the other is due to a change within the system diS = Ade/T. Therefore, the rate of entropy production can be expressed in terms of the rate of reaction Jr diS dt

1 - --AJ r > 0 T

(1.104)

18

1. Fundamentals of equilibrium thermodynamics

where

Jr--

d8

dt

Equation (1.104) is similar to Eq. (1.101) in relating the rate of entropy production to the product of the flows (here the rate of reaction) and the scalar thermodynamic force that is A/T. Equation (1.104) can be readily extended to several chemical reactions taking place inside the system

diS_ 1 t dt T E AkJrk >- 0

(1.105)

k=l

When the chemical reaction reaches equilibrium, affinity vanishes A = -Eufl~j = 0. The entropy generated per unit time and unit volume is called the rate of volumetric entropy production or the entropy source of density = A Jr ~> 0

(1.106)

VT

Example 1.6 Entropy production in mixing Consider the following two mixing processes: Mixing process 1: In a steady mixing process, 14.636 kg/s of saturated steam (stream 1) at 133.9 kPa is mixed with 15.0 kg/s of saturated steam (stream 2) at 476 kPa. Mixing process 2: In a steady mixing process, 11.949 kg/s of super heated steam (stream 1) at 773.15 K and 6000.0 kPa is mixed with 60.60 kg/s of saturated steam (stream 2) at 2319.8 kPa. Assume that the mixing processes are adiabatic. Determine and compare the rate of entropy production and the lost work for these two mixing processes producing a product of stream 3. Solution: Assume that the surroundings are at 298.15 K and the kinetic and potential energy changes are negligible. Mixing process 1: Available data from the steam tables in Appendix D are th1 = 14.636 kg/s,

/'h 2 --

15.0 kg/s

H 1 = 2688.3 kJ/kg, H 2 = 2745.4 kJ/kg S 1 = 7.2615kJ/(kgK), S2 = 6.8358 kJ/(kg K) To = 298.15K Assuming negligible kinetic and potential energy changes, mass, energy, and entropy balances yield 0 = rh3 - rh2 - rh1 ~ rh3 = 29.636 kg/s

0 = ( t h H ) 3 - ( t h H ) 2 - ( r n H ) l --~ H 3 =

[(thH)2 + (thH)l ]

= 2717.2 kJ/kg

th3

From the steam table, we read T3 = 401.15 K and $3 = 7.0462 kJ/(kg K). The entropy balance yields the rate of entropy production, Sprod Sprod -- (thS)3 - (thS)2 - ( t h S ) l = thl (83 - S1 ) +/J/2 ($3 - $2 ) = 0.004724 kW/K

1.8 Entropyand entropy production

19

The rate of work loss Elos~,becomes /~loss = ToSprod - 1.4083 k W

The dissipated work potential is small as we mix two saturated steams at relatively low temperature levels. Mixing process 2: Available data from the steam tables are rnI - 11.949 kg/s, rn2 = 60.60 kg/s P1 - 6000 kPa, P2 = 2319.8 kPa (saturated) H l = 3422.2 kJ/kg, H 2 = 2799.9 kJ/kg S, - 6.8818 kJ/(kg K), S 2 - 6.2817 kJ/(kg K) From the mass and energy balances, we have 0 = /~/3 -- #12 --/~1 ~ th3 = 7 2 . 5 4 9 kg/s

0 = (thH)3 - (thH)2 - (rhH)l

--, H 3 =

[(,n/-/)2 + (rh/4)~ ]

= 2902.4 kJ/kg

rh3 From the steam table we read T3 - 523.15 K, S 3 - 6.5454 kJ/(kg K), and P3 = 2000 kPa. The entropy balance yields the rate of entropy production, Sprod Sprod -- (?T/S)3 - (tI/8)2 - ( t~/S)l -- #/l (83 - S1 ) + 1~/2(33 - 32 ) = 11.9603 kW/K

The rate of work l o s s /~loss becomes /~loss -- ToSprod -- 3565.96 k W

These two simple examples show that mixing the saturated steam with the superheated steam in the second mixing process causes much greater entropy production and lost work potential than mixing two saturated steams in mixing process 1.

1.8.1 ThermodynamicCoupling When a system exchanges mass and energy with its surroundings, it also exchanges entropy, and moves away from the equilibrium state. As the system produces entropy due to irreversible processes taking place, the entropy flowing out of the system is greater than the entropy flowing into the system. As the system keeps exchanging entropy with the environment, it may reorganize itself and transform into a higher order state maintained by matter and energy being exchanged with the environment. Various irreversible processes inside the system may continue and interact with each other as the mass and energy exchange continue. These interactions are called thermodynamic couplings, which may allow a process to progress without its primary driving force or in a direction opposite to the one imposed by its own driving force. For example, in thermodiffusion, a species diffuses not because of a concentration gradient but because of a temperature gradient. Sometimes, a species may flow from a low to a high concentration region, which must be coupled with a compensating spontaneous process with positive and larger entropy production. The principles of thermodynamics allow the progress of a process without or against its primary driving force only if it is coupled with another process. This is consistent with the second law, which states that a finite amount of organization may be purchased at the expense of a greater amount of disorganization in a series of coupled spontaneous processes. Such coupled processes are of great importance in physical, chemical, and biological systems, such as the Bdnard instability and biological pumps of sodium and potassium ions.

20 1.9

1.

Fundamentals of equilibrium thermodynamics

THE GIBBS EQUATION

By introducing Eq. (1.57) into Eq. (1.76), we have

dU = T d S - PdV + Fdl + ~, de + ~ [LidN i -+-".

(1.107)

i=1

Equation (1.107) relates the total change in internal energy to the sum of the products of intensive variables T, P, F, /zi, ~, and the changes in extensive properties (capacities) of dS, dV, dl, dN,., and de. The Bronsted work principle states that the overall work A W performed by a system is the sum of the contributions due to the difference of extensive properties AK across a difference of conjugated potentials X,.,1-X,.,2

(1.108)

AW = ~ (Xi,I - Xi,2)AKi i=1

Equation (1.107) is more useful if it is integrated with the Pfaffian form; however, this is not a straightforward step, since intensive properties are functions of all the independent variables of the system. The Euler relation for U(S, V, l, e, Ni) is

+ ~ N. V,l,e,Ni

S,I,e,Ni

V,S,e,Ni

V,I,S ,Ni

i=1

OU

+""

(1.109)

S,V,I,e,Nj

Comparing Eq. (1.109) with Eq. (1.107) yields the definitions of intensive properties for the partial differentials

-~

V,/,e,/Vi : T,

- . ~ S,',e,Ni---P,

-~

V,S,e,Ni

(1.110) V,I,S,Ni

=

°2v i S,V,I,e,Nj

= I~i

The chemical potential/z indicates that the internal energy is a potential for the chemical work (or mass action) txidNi, and it is the driving force for a chemical reaction. The chemical potential cannot be measured directly, and its absolute values are related to a reference state. However, the change of chemical potential is of common interest. By introducing the definitions given in Eq. (1.110) into Eq. (1.107), we obtain the integrated form of the Gibbs equation

U = T S - P V + FI + ~te+ ~ I~iNi

(1.111)

i=l Differentiation of Eq. (1.111) yields

dV

=

rdS + S d r

-

P d V - VdP + Fdl + ldF + qJde + e&b + ~

P'i d N i

+ ~ Ni d]Dl,i

(1.112)

Comparison of Eq. (1.112) with Eq. (1.107) indicates that the following relation must be satisfied:

SclT- VdP + ldF + ed~p + ~_~N i d lz i = 0

(1.113)

Equation (1.113) is called the Gibbs-Duhem relation, which becomes particularly useful at isobaric and isothermal conditions, and when the force and electrical work are neglected, we have

~Nidld, i : ~Ni i=l j=l

( OIdfi I : 0 , tONj),

j:l,2,...,n,

i=/=j

(1.114)

1.9

21

The Gibbs equation

Equation (1.114) determines the changes in chemical potential with the addition of any substance into the system. From the Gibbs fundamental equationf(U,S,V,N), we have the three functions of S, V, and N, the respective differential relations, and the Euler equations given by dU + p ~_dV dN T iX-T-,

S = S(U, V, N ) - ~ dS = r

= s

1 V u-+P--iX-T T

V = V ( U , S , N ) - - , dV = - d U + T ~dS+ ix dN V=U--1 + s T#+ N_ P P ---fi-, P P

N

T _ P

(1.115)

(1.116)

and N = N(U,S,V)-,dN

- dU tx

T dS + p dV tx tx

N =U 1 IX

s T +v P tx ix

(1.117)

Using the molar-specific volume (v-- V/N) and molar-specific entropy (s = S/N), a simplified version of the Gibbs-Duhem relation results dix = - s d T + vdP

(1.118)

By partial differentiation, Eq. (1.118) can be transformed to a form called the thermal equation

( O~pp) = v = v ( T,P )

(1.119)

T

and the corresponding caloric equation is ( O~T)

= s = s(T,P)

(1.120)

P In Eq. (1.73), ifXk are the external variables to maintain the nonequilibrium distribution of the internal parameters of ~:kin a state of equilibrium, we have a potential energy o f - E~kXk.This energy is the additional work of the external parameters to maintain the distribution of internal parameters. The internal energy of equilibrium system Ueq is related to the internal energy of the nonequilibrium system U by

Ueq

=U--Z~kX

k

(1.121)

The irreversible work 6 Weq is related to the work necessary in reaching the same conditions reversibly

meq

- ~ m nt- ~ ~kdYk

(1.122)

In contrast, the entropy change in the corresponding reversible process is TdS = d geq

- • Weq

( 1.123)

By inserting Eqs. ( 1.121 ) and (1.122) into Eq. (1.123), we have rdS = dU - ~ X k d ~ . - ~ ~kdYk + 8 W + Z ~kdXk - dU - ~ W - ~ X k d ~

(1.124)

The entropy term TdS in Eq. (1.124) is the same for the irreversible process and the corresponding reversible process. Therefore, Eq. (1.124) represents the Gibbs equation for irreversible process. With the first law of thermodynamics, dU - 6 W = 6q, Eq. (1.124) becomes TdS = 6 q - ~_, XkdsCk

(1.125)

22

1. Fundamentals of equilibrium thermodynamics

For an adiabatic process 6q = 0, we have (1. 126)

TdS = m~__~X k d ~ k

Equation (1.126) represents the change of entropy for an irreversible process in an adiabatic system as a function of the internal and external parameters. This may be an important property to quantify the level of irreversibility of a change, and hence yields (i) a starting point to relate the economic implications of irreversibility in real processes, and (ii) an insight into the interference between two processes in a system.

1.10

EQUATIONS OF STATE

Equations of state relate intensive properties to extensive properties, and are obtained from the Euler equation as partial derivatives. In the entropy representation, we have the following equations of state:

-

,

V,l,e,Ni

-

,

T

(1.127)

-

V,l,e,Ni

T

V,l,e,Ni

In contrast to fundamental equations, equations of state do not contain all the information on a system, since the intensive properties are partial derivatives of the extensive ones. To recover all the information, all the equations of state are inserted into the Euler equation.

1.7 Heat c a p a c i t i e s for real gases For real gases, the internal energy U is not a function of temperature only because of molecular interactions, such as the collision of molecules, which depends on the distance between the molecules. Therefore, the change in volume affects the energy. The molecular forces have a short range. At low densities, the molecules are far apart from each other; hence the effect of interactions is small and negligible, and it vanishes as the volume approaches infinity. Perfect gas has low enough pressure or density so that all virial coefficients are ignored. A gas is slightly imperfect when all virial coefficients are ignored except the second virial coefficient B. Integration of the Helmholtz equation (OU/O V)T = T2[O(P/T/OT)]v yields Example

(1.128)

real'

A similar expression is also obtained from the Gibbs free energy equation when pressure approaches zero. If the partial differentiation inside the integral is determined from an equation of state (EOS), then Urealcan be calculated. For example, (P/T) from the van der Waals EOS is P m nR T V-nb

1 an 2

(1.129)

T V2

Therefore, from Eqs. (1.128) and (1.129), we get

rea

)

(1.130)

The second term on the fight side of Eq. (1.130) represents the energy of molecular interactions per unit volume. As the volume increases, the interactions get smaller, as is the case for gases at low density. If we use the Berthelot equation

P -

nRT

1 an 2

V - nb

T V2

(1.131)

23

1.10 Equationsof state

in Eq. (1.128), we obtain (1.132)

Ureal- Uidea1-- a 7 Equation (1.130) also yields a relation for the heat capacity of a real gas at constant volume

Cv,rea1 =

(0ereal)

=

(0eideal)

+

0 [IVT2 [ 0 (P)])

dV

(1.133)

After reorganizing Eq. (1.133), we have

Cv,real: Cv,idealq-Ivr~OT2)v dV

(1.134)

If the second partial derivative inside the integral is determined from an EOS, then the heat capacity of a real gas at constant volume can be calculated. For example, the integral in Eq. (1.134) vanishes for the van der Waals equation, and as Eq. (1.129) shows, pressure is a linear function of temperature. However, by using the Berthelot EOS, (Eq. 131), the heat capacity is obtained from Eq. (1.134).

Example 1.8 van der Waals isotherms Use the critical parameters for ammonia and estimate the van der Waals constants a and b. Plot the van der Waals isotherms for ammonia at T1 = 200 K, T2 = 406 K, and T3 = 550 K, when the volume changes from 0.04 to 0.25 L.

Solution: Use the van der Waals equation (Eq. 1.129) p-

nRT V-

nb

an

2

V2

with the parameters

(1.135)

Figure 1.1 shows the isotherms obtained from the MATHEMATICA code below. (*Ammonia*) Tc = 405,7; Pc = 11fi.8"0,9860 (*atm*); R = 0,082 l; (% atm/Mol K*) a = (27/64)*(R"2)*(Tc"2)/Pc

b = (1/8)*(R*Tc/Pc) p[V_,T_]: = (R*T/(V-b)-(a/VA2)); Plot[{p[V200],p[V406], p[-V,550]}, {V,0.04,0.2}, Frame ~ True, OridLines ~ Automatic, PlotStyle -, {Thickness[0,0085]}, FrameStyle --, Thickness[0.007], FrameLabel ~ {"V, L", "P, atm "}, RotateLabel ~ True, DefaultFont --. {"Times-Roman", 12}] a = 4.2043 b = 0.0374004

1. Fundamentals of equilibrium thermodynamics

24

3000

/

2000

a2 lOOO

0

~ .

.

.

0.05

.

0.075

, .

O. 1

.

O. 125

O. 15

O. 175

0.2

V,L

Figure 1.1. Isotherms obtained from van der Waals equation.

Example 1.9 Estimation of molar volume of a gas at high pressure (a) Estimate the molar volume of 100 mol of methane from the van der Waals equation at 310 K and 15 atm. (b) Plot pressure versus temperature when the temperature changes from T = 250 to 450 K and V = 75 L. Solution: Use the van der Waals equation nRT

an 2

V - nb

V2

p-

At the critical point, the vapor, liquid, and critical volumes are the same, ( V v - 3v v + 3 v g v - v? = 0

Vc)3 =

0,

and expansion gives

(1.136)

or in polynomial form, we have

V3-

b+

RT¢ ) V2 + - a- V - ~ =ab O Pc

(a) Tc = 190.6; Pc = 45.99*0.9869; (*Tc, in K, Pc in atm*)

R = 0.0821; (*L atm/Mol K*) a = (27/64)*R^2*(Tc^2)/Pc;(*L^2 atm mol^-2 *) b = (1/8)*R*Tc/Pc; (*L mol^-l*) n = 1 0 0 ; P - 15;T= 310; (*Solve for V*) Solve[{P-(n*R*T/(V-n*b)-(a*n^2)/V^2) = = 0},V] {{V~4.4772- 4.42488i},{V~4.4772 + 4.42488i},{V~ 165.029}} V ~ 165.029 L/mol

(b) (*methane*) n = 100;V= 75; PIT_]: = n*R*T/(V-n*b)- (a*n^2)/V^2 Plot[P[T], {T, 250,400}, Frame ~True, GridLines ~Automatic, FrameLabel~ {" T, K", "Patm"},

Pc

(1.137)

1.10 Equations of state

25

42.5 40 37.5

E

35 32.5

30 27.5 25 f - : . . . . ' . , . 260 280

~ ' 300

' ' ' 320 340 T,K

i... 360

, .... 380 400

Figure 1.2. Change of pressure with temperature for methane.

RotateLabel --,True, DefaultFont--, {"Times-Roman", 12}] Clear Figure 1.2 shows the change of pressure with temperature.

Example 1.10 Estimation of volume of a gas at high pressure using generic cubic equation of state Estimate the volume of n-butane at 15 bar and 400 K. Solution: A generic cubic equation of state to calculate the molar volume is

RT +b P

V=

a(r) V-b P (v + eb)(V + o-b)

(1.138)

The solution can be obtained by iterative methods or using a software package with an initial estimate from the ideal gas law. Using V = ZRT/P, equations for vapor and vapor-like root Zvap and for liquid and liquid-like root Zliq are obtained Zvap -- 13

Zva p ---1 + / ~ - q/3

(Zva p + e/3)(Zva p + o'/~) ' ( 1 -- Zliq + ~ )

Zliq =/3 + (Zli q + el3)(Zli q + o'~)

q13

(1.139)

where

/3-

~Pr Tr

,

q-

¢"~(Tr) , ~-~rr

Tr -

r ,/Dr Tc

=-

P Pc

Iterative calculations using the implicit Eqs. (1.139) start with an initial value of Zvap close to one and an initial value of Zli q close to zero. Once the value of compressibility is obtained, then volume is obtained from V = ZRT/P. Table 1.3 lists the parameters for the generic equation of state. (*n-butane Redlich-Kwong equation*) R = 83,14; (*bar cm3/Mol K*) T = 400 (*K*) P = 15; (*bar*)

26

1.

Fundamentals of equilibrium thermodynamics

Table 1.3 Parameters for the generic equation of state

z~

Equation of state

a(Tr)

o-

van der Waals Redlich-Kwong Soave-Redlich-Kwong Peng-Robinson

1 Tr-1/2 al a2

0 1 1 1 + ]2

0 0 0

1/8 0.08664 0.08664 0.07779

1 - ]2

27/64 0.42748 0.42748 0.45724

3/8 1/3 1/3 0.3074

al = [1 + (0.480 + 1.574o) -0.176o)2)(1 - Tr-1/2)] 2. a2 = [ 1 + (0.37464 + 1.54226o) - 0.26992o)2)(1 - Tr- 1/2)]2. o) (Omega) is the accentric factor. Smith et al. (2005)

Source:

Vo = R*T/P (*cm3*) Zo = 0.9; Tc = 425.1; Pc -- 37.96; T1 = T/Tc; Pr = P/Pc; bb = 0.08664*Pr/T1; q = 0.42748"(T1"(- 1/2))/(0.08664"T1); FindRoot[Z -- = 1+bb-q*bb*(Z-bb)/(Z*(Z+bb)), {Z, Zo}] Clear 2217.07 initial volume {Z~0.815911} V = R T Z / P = 1808.90 c m 3

1.10.1

Joule Thomson Coefficient

The Joule Thomson coefficient is the ratio of the temperature decrease to the pressure drop, and is expressed in terms of the thermal expansion coefficient and the heat capacity

OT)

(OH/OP)T _ _ V ( 1 - a T ) /, :-

(0H/0r)

(1.140)

-

Example 1.11 Entropy of a real gas Determine the entropy of a real gas. Solution: Using

(OAIOV)r = - P or (OG/OP)r= V, we have A (T,V,n)= A (T, Vo,n) -

pdV

(1.141)

The difference between real and ideal systems of the Helmholtz energies yields v

Areal

(T,V,n)-Aideal (T,V,n)=-I (Preal-Pideal)dV

(1.142)

By using the van der Waals EOS in Eq. (1.142), we obtain Areal--A i d e a l - a ( ~ ) - n R T l n (

V-nb

(1.143)

where Aidea 1 = gidea 1 -- TSidea 1 = gidea 1 - n T

s O + C v InT + R

In

Using Eqs. (1.143) and (1.128) in A r e a l - - g r e a l - TSreal, we can calculate the entropy of a real gas. For example, using the van der Waals EOS, we get

27

1.10 Equations of state

Sreal =

n[s o +C v l n T + R ln(V-nb)]n

(1.144)

The ideal gas form of S is

[

Sidea1 = n s o + Cv In T + R In

(1.145)

Comparison of Eq. (1.144) with Eq. (1.145) shows that the van der Waals gas entropy has ( V - nb) instead of V.

Example 1.12 Chemical potential of a real gas Similar to Eq. (1.142), the Gibbs free energy for a real gas is

Greal(T,P,n)-Gideal(T,P,n ) --

(1.146)

.foP (Vreal -- Videal)dP

Using the definition for chemical potential/.t = (OG/On)p,Tand Eq. (1.146), we get ~real

(T,P)-/~ideal (T,P) = So (/Areal --

(1.147)

Videal)dP

Using the compressibility factor Z, the volume of a real gas is Vreal= ZRT/P. Therefore, the chemical potential in terms of Z in Eq. (1.147) is

~real(T,P)-Id, ideal(T,P)= R T f ~ ( Z - 1

dP

(1.148)

The chemical potential can also be expressed in terms of fugacityf (1.149)

/Zreal (T,P)-/Zideal (T,P) = R T l n ( f ) where ]Zidea 1(T, P) = / z 0 (T, P0 ) 4- R TIn P

and

In

-

P

From various EOSs, it is possible to determine the compressibility and hence the chemical potential of a real gas. For example, using the virial equation, which is valid for gases at low density, we have

Z-

PV - 1+ B'(T)P +C'(T)P 2 +... RT

(1.151)

By using Eq. (1.151) in Eq. (1.148) and ignoring the term with p2, we have #rea] (T, P) = #ideal(T, P) + RTB'(T)P +...

(1.152)

In terms of the virial coefficient B(=B'RT), the approximation in Eq. (1.152) becomes /Zreal (T, P) = ~ideal

(T, P) 4- BP +...

(1.153)

Similarly, by using the van der Waals EOS and the chemical potential from the Helmholtz energy/z(V,P) = (OA/On)v,v, we obtain (C

1 ]RT-T I.t(c,T) = (U o - 2ac)+| -__~v+ 1-bc ~R where c = n/V

[

so + C v l n T - R l n

c

(1.154)

28

1.

1.10.2

Chemical

Affinity

Fundamentals of equilibrium thermodynamics

of Real

Gases

The affinity of a chemical reaction is A = -~pi[.£i, where b,i is the stoichiometric coefficient, which is negative for reactants and positive for products. For a real gas mixture, the affinity is determined by P

Areal

-~ Aidea' -- Z

VifO (V/'real -

Vi'ideal )dP

(1.15 5)

i

Using the relation of RT/P for an ideal gas and an appropriate EOS for the volume of a real gas, the affinity for a real gas may be determined. 1.10.3

The

Clapeyron

Equation

In the case of a pure compound, vaporization and melting processes need latent heat to be supplied at certain temperatures. Still, under isothermal and isobaric conditions, various phases of a compound can coexist at phase equilibrium. A compound can be at various phases under suitable pressure and temperature. Phase diagrams (see Figures 1.3-1.5) show these different phases. For example, using a P versus T diagram (Figure 1.3), one can find the melting and vaporization temperatures at certain pressures. The saturation curve shows the temperature and pressure at which two phases are in equilibrium with each other. A Txy diagram for a binary mixture (Figure 1.6) can display the boiling temperatures for changing mixture compositions at a specified pressure. Chemical potentials of a species in vapor and liquid phases are equal at phase equilibrium. Entropy production due to the irreversible processes at equilibrium must be zero, and hence the affinity of liquid-vapor conversion must vanish for each compound

(1.156)

A=/Xva p ( T , P ) - t Z l i q ( T , P ) = 0, and /Zvap(T,P ) = ~liq(T,P)

The Clapeyron equation relates pressure to temperature, and hence boiling or melting points can be calculated with changing pressure. By using Eq. (1.156), we can equate the Gibbs-Duhem equation for two phases Carbon Dioxide: Temperature - Pressure Diagram 10000.0 ............................... i

t ................................

......................... i

........................... t

1000.0

........................................ ~ .................................

! .................................................... I

i

I........

......

i Liquid L

100.0

¢¢t

L_ ............................................~........................................

el

i

lO.O iii il .............................................; .................................. !

ii

iiiiiiiiiii iiii illI

....................................

............... I............................... t......................i............................... I~

i

i

i Vapori

1.0 ....~......

......i ..... i ....... i

0.1

.....

.................................... ~..............................

i ........................................ I .........................I ..........................I .................................... I,,

CO~Ta~ V1.0

Drawn

......

i

q.~

-100

-90

-60

-70

-6O

-50

-40

-30

-20

-10

0

10

20

30

40

Temperature. °C Figure 1.3. Temperature pressure diagram for carbon dioxide (with permission from Chemicalogic Corporation).

1.10

29

Equations of state

Carbon Dioxide: Pressure - Entlmlpy Diagram I .000

i

#

i ~'~i 1

/ ~'tl<, ., ~

,//

\;~ ,

100 1.-

f

li

m

*L

as~

i~ i~ '

il it

II

0..

!! 10 i

5

.....

i

i

?:i

.....

1

,

,:

•

................

ChemicaLogic Corporation

;7

Drawn with COlTab TM i A Spreadsheet Add-in !or the Thermodynamic and Transpolt Properties ~ Carbon Dioxide

www.c hemicalogic.com . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Enthalpy, kJ/kg i ......................

...........................

,

............................................

.....

-500

-400

-300

-200

-100

0

100

200

Figure 1.4. Pressure-enthalpy diagram for carbon dioxide (with permission from Chemicalogic Corporation).

( - S d T + VdP )vap - ( - S d T

(1.157)

+ VdP)liq

Using the molar enthalpy of transition ~-/transition = T(Svap - Sliq), we get the Clapeyron equation dP dT

_

Svap - Sliq __ aHtransition

Vvap--

~liq

(1.158)

TA V

Considering a liquid-vapor transition and Vvap >> Vliq, we get the approximate Clapeyron equation L~d--/vap

dP dT

(1.159)

TVva p

If we use the ideal gas equation to approximate the volume of vapor as Vvap - R T / P , Clapeyron equation d In P

_

AHvap

dT

we obtain the Clausius-

(1.160)

RT 2

Equation (1.160) is also valid for a solid-vapor equilibrium. Assuming that the latent heat of vaporization remains constant between small temperature intervals of AT, then integration of Eq. (1.160) yields

In -PIl

-

"vap/l 11) R

~2 - -~-1

(1.161)

30

1. Fundamentals of equilibrium thermodynamics Pressure-Enthalpy Diagram for Water and Steam Based on the IAPWS.95 Formulation for General and Scientific Use 10000

1000

! O0 L-¢¢l .¢}

¢; lO c~ t... 13.

0.1

0.01 0

500

1000

1500

2000

Drawn wltrl StoamTab Duo V3.0.

2500

3000

Enthalpy. kJlkg

3500

4000

4500

5000

copyright @ 1998 ChomlcaLoglc Corporation.

Figure 1.5. Pressure-enthalpy diagram for water (with permission from Chemicalogic Corporation).

In Eq. (1.161), temperature T is the boiling point at pressure P. Table 1.4 shows enthalpies and temperature of transition for some species. 1.10.4

Fugacity

To avoid some possible difficulties in determining chemical potentials, Lewis proposed a new property called the fugacity f . At low pressure and concentration, the fugacity is a well-behaved function. The fugacity function can define phase equilibrium and chemical equilibrium. For an ideal gas, the fugacity of a species in an ideal gas mixture is equal to its partial pressure. As the pressure decreases to zero, pure substances or mixtures of species approach an ideal state, and we have C

lim J ' = 1 (pure species i) P

(1.162)

P-*O

lim ~

= 1 (species i in mixtures)

P---~O yi P

(1.163)

where f represents the fugacity of species i in a mixture, and not a partial property, and Yi is the gas phase mole fraction of species i. f depends on composition, temperature, and pressure aln~) OP

_

T,Yi

RT

(1.164)

1.10 Equationsof state

31

388

384 N

Saturated liqui .......... Saturated vapor

380 -

376

372

368 ()

i

i

i

i

0.2

0.4

0.6

0.8

xj, Yl

Figure 1.6. Temperature composition diagram for n-heptane(1)-toluene(2) binary system at 1 atm by Aspen Plus. Table 1.4 Enthalpies and temperatures of phase change of some compounds at P = 1 atm

Compound

Tm (K)

~¢-/m (kJ/mol)

Tb (K)

2if-/v (kJ/mol)

H,

14.01 54.36 63.15 217.0 195.4 161.2 250.3 273.15 175.2 156

0.12 0.444 0.719 8.33 5.652 4.39 2.5 6.008 3.16 4.6

20.28 90.18 77.35 194.6 239.7 319.4 350 373.15 337.2 351.4

0.46 6.82 5.586 25.23 23.35 26.74 30.0 40.656 35.27 38.56

_

Or

N~ CO, NH3 CS~ CC14 H~O CH3OH C2HsOH _

Source: Kondepudi and Prigogine (1999).

a In f ] aT

-_Hi-H~

i°

(1.165)

RT 2

P,Yi

where ~ and H i are the partial volume and enthalpy, respectively. H / i s the value of enthalpy of species i in an ideal gas state at the same pressure and temperature as the real gas mixture. For a mixture, the departure function of Gibbs energy which is the difference between the partial Gibbs energy in the real state and in an ideal gas state, is related to the fugacity function by A

Gi - G i - Ixi - ~i,ideal -- R T In ~ Yi P

(1.166)

From Eq. (1.166), we define the fugacity coefficient of species i in a mixture by ~i-

~

(1.167)

YiP

(~i is a measure of deviation from the ideal gas mixture behavior, and unity in an ideal gas mixture. The fugacity coefficient may be determined from compressibility factor data at constant temperature and composition ln~i-Io(Zi-1) where Z i - a ( n Z ) / a n i .

dP

(1.168)

32

1.10.5

1. Fundamentals of equilibrium thermodynamics

Excess Thermodynamic Properties

An excess property is the difference between the actual property value of a solution and the ideal solution value at the same composition, temperature, and pressure. Therefore, excess properties represent the nonideal behavior of liquid mixtures. The major thermodynamic properties for ideal mixtures are aideal ~-~ Z xiGi nt-R T Z xi In x i i i

(1.169)

Sideal = 2 x i S i - R Z x i l n x i i i

(1.170)

Hideal -- Z i

xiHi

(1.171)

Videal : Z xiVi i

(1.172)

Therefore, the excess properties are

(1.173)

sE -- S - ( Z

lnxi]

- RZ

H E = H-~.xiH i

V E =V-~_xiV i

(1.174)

(1.175)

i

(1.176)

i

For a solution, Eq. (1.166) yields the partial excess Gibbs energy for species i, G/e

Gie = G i - G~ = R T In ~// xifi

(1.177)

where Xi is the mole fraction of solution. The dimensionless ratio in Eq. (1.177) defines the activity coefficient, ")/i

')/i -

(1.178)

xifi or

[°(ncE/Rr)] In ")/i -

Oni

P,T

(1.179)

,nj

The activity coefficient is a measure of the deviation of liquid solutions from ideal behavior, and unity in ideal solutions. We have the definitions of excess properties of Gibbs energy, volume, and enthalpy, which are experimentally measurable Gie = R T In ")/i or G E = RT~_~ x i In '}/i

(1.180)

33

1.10 Equations of state

V~:-RT(O(GE/RT)) OP

(1.181) T,x

H~:-_RT2(O(GE/RT)J OT P,x

(1.182)

Therefore, the excess entropy becomes H E _ G E S E --

(1.183)

From the Gibbs-Duhem equation at constant temperature and pressure, we have

~., xid In Yi i

0

(1.184)

The effect of temperature and pressure on the activity coefficients are obtained from ViE _ (O(lnyi))

RT

OP

(1.185)

T,x,

Hf _ _( O(lnyi) ) RT 2 OT p,xi 1.10.6

(1.186)

The Phase Rule

For a nonreacting equilibrium system with n species and p phases, the number of independent phase equilibrium equations is ( p - 1)n. The number of phase-rule variables is 2 + ( n - 1)p, consisting of intensive variables of temperature pressure and (n - 1) compositions for each phase. The difference between the phase-rule variables and the number of independent phase equilibrium equations is the degrees of freedom of the system, F

F - 2 +(n-1)p-(p-1)n= 2 - p + n

(1.187)

Therefore, we can fix F number of variables to determine the phase equilibrium fully. For a closed system, the number of variables will include the p number of extensive variables of the masses of the phases 2 + (n - 1)p +p. The number of independent phase equilibrium equations will include the material balance equations for each n species, and we have (p - 1)n + n. Therefore, the degrees of freedom of a closed system is F- 2+(n-1)p+p-[(p-1)n+n]

=2

(1.1883

So, the equilibrium state of a closed system with specified masses of species is fully determined by fixing two independent variables. This rule is called Duhem's theorem.

1.10.7

Fluid Phase Equilibrium

Using the fugacity function, we can define a multicomponent vapor-liquid and liquid-liquid equilibrium by

.~;,vap --.!i,liq

.~,,iql- J~i,liq2 (i =

1,2, ... ,n)

(1.189)

where liql and liq2 represent the liquid phases in equilibrium. In terms of activity and the fugacity coefficients, the vapor-liquid equilibrium from Eq. (1.189) becomes

YiChiP - xi Yif

(1.190)

34

1.

Fundamentals of equilibrium thermodynamics

Table 1.5 Antoine constant for some species In psat= A - B/(T+ C) where Pis in kPa and Tis in °C

Species Acetone Acetic acid Acetonitrile Benzene 1-Butanol iso-Butanol Carbon tetrachloride Chlorobenzene Chloroform Dichloromethane Diethyl ether 1,4-Dioxane

Ethanol Ethylbenzene Methanol Methyl acetate Methyl ethyl ketone Phenol 1-Propanol 2-Propanol

Toluene Water

A

B

C

Temperature range (°C)

14.3145

2756.22

228.060

- 2 6 to 77

15.0717 14.8950

3580.80 3413.10

224.650 250.523

24 to 142 - 2 7 to 81

13.7819 15.3144 14.6047 14.0572 13.8635 13.7324 13.9891 14.0735 15.0967 16.8958 13.9726 16.5785 14.2456 14.1334 14.4387 16.1154 16.6796 13.9320 16.3872

2726.81 3212.43 2740.95 2914.23 3174.78 2548.74 2463.93 2511.29 3579.78 3795.17 3259.93 3638.27 2662.78 2838.24 3507.80 3483.67 3640.20 3056.96 3885.70

217.572 182.739 166.670 232.148 211.700 218.552 223.240 231.200 240.337 230.918 212.300 239.500 219.690 218.690 175.400 205.807 219.610 217.625 230.170

6 to 37 to 30 to - 14 to 29 to - 2 3 to - 38 to - 4 3 to 20 to 3 to 33 to - 11 to - 2 3 to - 8 to 80 to 20 to 8 to 13 to 0 to

104 138 128 101 159 84 60 55 105 96 163 83 78 103 208 116 100 136 200

Source: B.E. Poling, J.M. Prausnitz and J.E O'Connell, The Properties of Gases and Liquids, 5th ed., Appendix A, McGraw-Hill, New York (2001).

If the vapor phase is close to an ideal phase, then Eq. (1.190) yields the modified Raoult's law Yi P = xi'Yi p/sat

(1.191)

where psat is the saturation pressure of pure components. The Antoine equation may be used to estimate the saturated vapor pressure B lnP = A - ~

T+C

(1.192)

where A, B, and C are the Antoine constants. Table 1.5 lists these constants for some species. When the pressure is low, the vapor phase may be assumed to be ideal, while the liquid phase nonideality is represented by the activity coefficient Yi for species i Yi- PYi/(xiPsat) • Raoult's law represents the vapor-liquid phase equilibrium if both phases are ideal PYi = xi Psat

(1.193)

For vapor-liquid equilibrium representation, temperature-composition, or pressure-composition diagrams are widely used to estimate the temperature or pressure of saturated liquid and saturated vapor phases. A temperature composition diagram for the n-heptane(1)-toluene(2) system at 1 atm is shown in Figure 1.6. A pure-species pressure-temperature diagram displays the change of pressure with temperature at different states. It also shows the critical point and triple point where the solid, liquid, and solid phases are at equilibrium, as shown in Figure 1.3. Another important phase diagram is the pressure-enthalpy diagram (Figures 1.4 and 1.5) where the enthalpy values are displayed at various pressures.

1.10.8

Henry's Law

Consider a very dilute gas solute in a liquid phase. Henry's law relates the mole fraction of the solute i in the gas phase to the mole fraction of the solute i in the liquid phase Hi

Yi = --P xi

(1.194)

1.10

35

E q u a t i o n s of state

Table 1.6 Change of Henry's law constants with temperature for some species in water T(°C)

CO2

H2S

CO

0 5 10 15 20 25 30 35 40

728 876 1040 1220 1420 1640 1860 2090 2330

26800 31500 36700 42300 48300 54500 60900 67600 74500

35200 39600 44200 48900 53600 58000 62000 65900 69600

H; is in atm/mol fraction. Source: R.H. Perry and D.W. Green (Ed.), Perty's Chemical Engineers ' Handbook, 7th ed., McGraw-Hill, New York (1997).

where P is the total pressure, and Hi is the Henry's law constant, which depends on temperature. Experimental values of Hg are tabulated for many gas solutes. Table 1.6 lists Henry's law constants of various gases at different temperatures. Values of Hi are correlated by an Arrhenius type of relation H i - H 0 exp - ~ - ~

(1.195)

1.13 Henry's law constant Determine the saturation concentration of carbon dioxide (CO2) in water at 1atm and 25°C. From Eq. (1.194), we have

Example

Yco2 P =

Hco2Xco2

From Table 1.6, we read the value of Hi for carbon dioxide at 25°C • Hco2 = 1640 atm/mol. The mole fraction of carbon dioxide in air is about 0.000314. Therefore, the partial pressure of carbon dioxide is YCO2 P =

0.000314 (1 atm) = 0.000314 atm

The mole fraction of carbon dioxide at the liquid water surface is obtained from Henry's law

XCO 2 --

Pco2 _ 0.000314 = 1.91 x 10 - 7 mol C O 2 Hco ~ 1640 mol solution

For 1 m 3 of dilute solution of carbon dioxide, the number of moles of water will be m3)(1 × 103 kg H20/m3)/(0.018 kg/mol) = 5.56 × 104 moles

nil20 -- (1

Since the total number of moles of the solution and the water are almost the same, the moles of carbon dioxide in 1m 3 of solution is

nco ~ - 1.91 × 10 - 7 -

mol CO 2 (5.56 X 10 4 mol solution) mol solution = 1.064 × 10 - 2 mol/m 3 of carbon dioxide

The saturation concentration is (1.064 × 10 .2 mol/m 3)(0.044 kg/mol)= 0.4684 × 10 - 3 kg CO2/m 3

1.10.9

Activity Coefficient Models

We can estimate the activity coefficients by using the excess Gibbs energy models. Based on the local composition concept, the Wilson, NRTL, and UNIQUAC models for excess Gibbs energy provide relations for activity coefficient

36

1. Fundamentals of equilibrium thermodynamics Table 1.7 Temperature-dependent parameters of the NRTL model

System Methyl acetate(1)-benzene(2) Methanol(1)-ethyl acetate (2) Ethanol(1)-ethyl acetate (2) Ethyl formate(1)-ethanol(2) 1,4-Dioxane(1)-acetonitrile(2) Ethyl formate(I)- 1-propanol(2) Ethanol(1)-cyclohexane(2) Ethanol(1)-toluene(2) Acetonitrile(1)-benzene(2) Water(1)-butyl glycol(2) 2-Propanol(1)-n-heptane(2) Benzene(1)-n-heptane(2) Ethanol(1)-acetone(2) Acetone(1)-water(2) Isobutyric acid(1)-n-heptane(2) Butanol(1)-n-hexane(2)

Tmin -

c1

c2

c3

Tmax (°C)

(cal/mol) (cal/(molK))

25-50 25-55 25-55 25-45 40 25-50 5-65 25-60 45-70 5-85 30-60 25-50 25-100 5-200 25-45 15-59

-124.44 568.81 617.11 357.15 -151.06 383.90 1772.50 1459.60 229.62 802.63 1829.80 -172.11 260.36 653.56 1109.80 1536.00

C6 X 102 (l/K)

c4

(cal/mol) (cal/(mol K))

1.84 -2.14 -2.52 -1.55 0.10 -3.30 -1.19 -5.97 1.18 3.73 -8.35 -4.58 -4.07 9.27 3.34 0.53

386.82 355.10 378.16 563.42 635.70 565.59 956.01 591.59 487.52 654.50 864.35 771.67 398.00 657.75 508.12 605.00

2.36 0.98 2.28 0.62 0.92 1.16 0.50 0.09 2.88 1.81 0.69 3.81 1.61 1.71 0.83 1.55

0.1507 0.5931 0.5999 0.5173 0.6960 0.5307 0.4546 0.4808 0.6402 0.1000 0.4874 0.3639 0.4199 0.6725 0.6758 0.5381

0.5800 -0.1185 -0.1243 -0.3895 0.0817 -0.2120 0.0574 0.0020 -1.0850 0.4269 0.0945 -0.1181 -0.2058 0.0825 0,0938 0.0890

c2(T- 273.15) (cal/mol). c4(T- 273.15) (cal/mol). oz= C5+ c6(T- 273.15). Source:Y. Demirel and H. Geceg6rmez, FluidPhaseEquilibria,65 (1991) 111. a21 = c 1+ a12 = c3 +

estimations. Besides these, the group contribution method UNIFAC also predicts the activity coefficients. For example, for a binary solution, the N R T L model for excess Gibbs energy is

GE--xIx2RT(

G217-21 + G127-12 I X1 "[" x2G21

(1.196)

x2 at- XlG12

Using Eq. (1.183), we determine the activity coefficients

ln~h = x 2

G21

7"21

-[-

x 1 -[- x2G21

G12

In 3/2 : x2 7"12

X2 + x1G12

G127"12 )2

(1.197)

(x 2 -q-XlG12

jr.

G217"21

(1.198)

(x 1 -k-xzG21) 2

a12/RT, 7"21 : a21/RT. Here, a, a12 , and a21 are the binary parameters estimated from experimental v a p o r - l i q u i d equilibrium data. The adjustable energy parameters, a12 and a21, are usually a s s u m e d to be independent o f composition and temperature. However, w h e n the parameters are temperature dependent, prediction ability o f the N R T L m o d e l enhances. Table 1.7 tabulates the t e m p e r a t u r e - d e p e n d e n t parameters o f the N R T L m o d e l for some binary liquid mixtures. For a m u l t i c o m p o n e n t solution, the Wilson equation is where G12 = exp(-ceT"12), G21 "-- exp(-otT"21), and "/'12 --

(~k : 1 i f / :

In a/i : 1 - In ~

k

XkAik ) -- ~m Xmhmi Z k xk hmk

hik = (VklVi ) exp(-aik/RT)

k)

(1.199)

(1.200)

37

1.10 Equations of state

-2oo1\ _,oo

! \

_~oo_. ©

! .... ......

.... i . . . . . . . .

.......\ ,

_~oo . . . . . . . . .

! ..............

-.-~ ,.,

/

...... i ......

/--~ !--/ ..... /

-900 ()

, 0.2

i 0.4

i 0.6

i 0.8

Xl

Figure 1.7. Gibbs energy of mixing for 1-propanol(1)-water(2) by the Aspen Plus simulator using the NRTL model. where Vi and Vk are the molar volumes of pure liquids i and k at the solution temperature, and aik is the adjustable energy interaction parameter, which is mostly temperature dependent. For a ternary system, we need the adjustable energy parameters of the three pairs A12, A21" A~I, A13; and A23 , A32.

1.10.10

Mixing Functions

Upon mixing of pure components, the thermodynamic properties change as shown by 2~. In Eqs. (1.201)-(1.204), we have the property change of mixing Gmi x

-

AG

- G-

xiG i - G u + RT~_. x i lnx i

~ i

AS - S - ~

(1.201)

i

x i S i - S E - R~__~ x i i

In x i

(1.202)

i

A H - H - y_, x i H i - H E

(1.203)

i

A V - V - y_~ x i V i - V e

(1.204)

i

Figure 1.7 displays the Gibbs energy of mixing for 1-propanol(1)-water(2) at 50°C and 1 atm obtained from the Aspen Plus simulator using the NRTL model.

Example 1.14 Estimation of partial excess properties The heat of mixing (excess enthalpy) for a binary mixture is H F~ _ xl x2 (2axl + ax2 )

Here, a is the parameter in J/mol. Derive relations for H1E and H e in terms of xl. The partial properties are dH e

H ( - HE + ( 1 - x 1 ) ~

(1.205)

dXl

H~ - H E - x 1

dH e dxl

(1.206)

38

1. Fundamentals of equilibrium thermodynamics

W i t h x 2 = 1 - Xl, the equation for H E

becomes HE

:

ax 1 _ ax 3

The differentiation is dH E

= a - 3ax21

dXl

Therefore, Eqs. (1.205) and (1.206) yield partial molar excess enthalpies H E = a x 1 - a x 3 + ( 1 - x 1) ( a - - 3 a x e ) = a - 3ax21 + 2 a x 3

H E = a x 1 - ax31 - x 1( a - 3ax21) = 2ax31

and H E = Xl H E + x2 H E

1.10.11 Azeotropes Azeotrope mixtures reach a point at which liquid and vapor compositions become the same at a certain temperature and pressure. Some azeotropes show a maximum boiling temperature, while others show a minimum boiling temperature. Table 1.8 shows some examples of binary and ternary azeotropes. Azeotrope mixtures cannot be separated into their pure species by a single distillation column. For a binary vapor-liquid system, the Gibbs-Duhem relations are SliqdT-

VliqdP + XldlJc 1 + ( 1 -

(1.207)

Xl) d / z 2 = 0

Table 1.8 Some binary and ternary azeotropic data at 1 atm

Binary species Water( 1)-chloroform Ethanol Ethyl acetate n-Butanol Nitromethane Acetonitrile Pyridine Methanol( 1)-acetonitrile Acrylonitrile

Toluene Ethyl acetate Ethanol(1)-acrylonitrile Ethyl acetate

Benzene Hexane

Ternary species Water(1)-chloroform(2)Methanol Ethanol Acetone Water(1)-ethanol(2)Benzene Hexane

x~

Tb (K)

A/-/v (J/mol)

0.160 0.096 0.312 0.753 0.511 0.307 0.768 0.231 0.724 0.883 0.684 0.445 0.462 0.448 0.332

329.2 351.3 343.5 365.8 356.7 349.6 367.1 336.6 334.5 348.1 335.4 343.9 344.9 341.4 331.8

32155.3 40683.6 34821.8 41464.4 39643.5 36358.3 41514.4 36717.8 38605.1 39778.0 37992.7 38065.2 37557.3 37647.5 34691.9

X1

X2

Tb (K)

AHv (J/mol)

0.066 0.129 0.163

0.698 0.795 0.353

325.4 328.4 333.5

34088.7 32954.9 32303.9

0.233 0.112

0.228 0.274

338.0 329.5

36847.4 34658.3

Source: Y. Demirel, Thermochimica Acta, 339 (1999) 79.

39

1.10 Equationsof state

(1.208)

SvapdT- VvapdP + yldtx 1 + ( 1 - Y l ) d/x2 = 0 If we subtract Eq. (1.207) from Eq. (1.208), and divide by dXl, we have

dT _

(Svap - Sliq ) dx 1

de --1-( x1 _ Yl ) ( d ~1 dx 1

(Vvap -- ~liq ) dx----~

(1.209)

dtZ2dXl) - - 0

At the azeotropic point (x, = Yl) and at constant temperature, we get

dP (Vvap

-

Vliq ) dx 1

(1.210)

= 0

Equation ( 1 . 2 1 0 ) means that dP/dxl = 0, and at a specified temperature and the total vapor pressure of a binary liquid mixture is a minimum or a maximum at the composition of the azeotropic mixture. A similar analysis of constant pressure systems indicates that, at a specified pressure and temperature of a binary liquid, mixture is a minimum or a maximum at the composition of the azeotropic mixture. If we simultaneously vary the temperature and pressure for an azeotropic mixture, we have the Clapeyron relation

dP __ (Svap - Sli q ) dT

(1.211)

(Vva p - Vii q )

At atmospheric pressure, the n-butanol-water system exhibits a minimum boiling azeotrope and partial miscibility, and hence a binary heterogeneous azeotrope. Figure 1.8 shows the Tyx and Pyx phase diagrams for 1-propanol(1)water(2) azeotropic mixture obtained from the Aspen Plus simulator using the NRTL activity coefficient model. Example 1.15 Binary liquid mixture phase diagrams Prepare phase diagrams for acetone(1)-water(2) mixture using Raoult's law: (a) Temperature compositions diagram, Tyx, at P -- 1 atm. (b) Pressure compositions diagram, Pyx, at 50°C. The Antoine constants are Acetone (1): A - 14.3145, B = 2756.22, C = 228.06 Water (2): A = 16.3827,

B = 388.570,

C = 230.17

(c) By using Raoult's law, we assume that the vapor and liquid phases are ideal or close to ideal. For the Tyx diagram, we obtain the saturation temperatures at the specified pressure P = 101.33 kPa from the Antoine equation (Eq. (1.192)) ~.sat =

B~i

-- C i

(1.212)

A/. - l n P

104

T-x --O--T-y __

100 ,- _~

21 19 17

96 L) o

15 92 13 88 84

11 0.2

0.4 ( .6 3'1,xl (a)

0.8

9

0

0.2

0.4

0.6

0.8

1

Yl, Xl

(b)

Figure 1.8. Phase diagrams for 1-propanol(1)-water(2) by the Aspen Plus simulator using the NRTL model.

40

1. Fundamentals of equilibrium thermodynamics

At P = 101.33 kPa, we have T~at = 56.38°C and T~at-- 100.0°C. For ideal mixtures, the total pressure is

P = Z xiPi sat

(1.213)

i

For a binary mixture, this equation yields P -- X1P1sat nt- (1 - x 1) P2sat

(1.214)

From this equation, we estimate the liquid composition x 1 p - p 2 at

m

X1

elsat _ P2sat

(1.215)

From Raoult's law, Eq. (1.193), we estimate the vapor composition yl XlP1sat Yl = ~ P By changing the temperature between T~at and T2sat, we estimate the saturation pressures p~at and p~at from the Antoine equation at each temperature. From Eq. (2.215), we estimate the equilibrium compositions at each temperature and prepare Table 1.9. Figure 1.9 displays the Tyx diagram with the saturated liquid and saturated vapor lines plotted from Table 1.9. (d) To prepare the Pyx diagram, we estimate the saturation pressures p~at and p~at at the specified temperature T = 50°C from the Antoine equation p~at __ 82.072 kPa and p~at __ 12.326 kPa at T = 50°C Next, we specify the Xl between 0 and 1, and estimate the total pressure P and Yl from Eq. (1.193) to prepare the total pressure and equilibrium compositions shown in Table 1.10. In Figure 1.9, we can compare both the Tyx and Pyx diagrams obtained from Raoult's law and the NRTL model using the Aspen Plus simulator. As we see, ideal behavior does not represent the actual behavior of the acetone-water mixture, and hence we should take into account the nonideal behavior of the liquid phase by using an activity coefficient model.

Example 1.16 Estimation of fugacity coefficients from virial equation Derive a relation to estimate the fugacity coefficients by the virial equation

P V - 1 + BP __ I _+_( BP¢ ] P~

Rr

Rr

(1.216)

t, Rrc )-gr

Table 1.9 Temperature compositions data estimated from Raoult's law for

acetone(1 )-water(2) at 101.33 kPa T (°C)

p~at (kPa)

p~at (kPa)

xl

Yl

56.038 60.038 64.038 68.038 72.038 76.038 80.038 84.038 88.038 92.038 96.038 99.999

101.33 115.951 132.195 150.187 170.051 191.919 215.923 242.201 270.894 302.145 336.099 372.529

16.539 19.959 23.962 28.624 34.032 40.276 47.456 55.681 65.066 75.736 87.823 101.33

1 0.84768 0.71482 0.59808 0.49476 0.40261 0.31978 0.24473 0.17618 0.11304 0.0544 0

1 0.9699 0.9325 0.8864 0.8303 0.7625 0.6814 0.5849 0.4710 0.3370 0.1804 0

41

1.10 Equations of state

where B is the second virial coefficient. The gamma-phi relation for vapor-liquid equilibrium is Yi dPi P - xi Yi

p/sat

(1.217)

This equation results from J~/,vap - J~i,liq"The term {hi is ~i - ~

"

exp

[

- V/'liq(PRT- Pi

)

(1.218)

90

104

I M

sat]

74

--'-Pxl ...... i .... -*-P-Y / ! i

8O 70

....

¢

i

6O

<

50 40

64

30

1

2O 54 0

0~.2

0.4

' 0.6

10 0.8

I

0

0.2

Yl,Xj (a)

90

I

.

.

94

.

.

"~

.

~

--O-- T-y

r 1-

7O

89 84 ~,.

.

6O

_

79 74

50

<

4O

69 64 59 54

0.8

(b)

104

99

I

0.4 0.6 Yl,Xl

...... j

3O

~P-x

2O 0

0~.2

t 0.4

0.6

0.8

10

1

l

0

0.2

YW,xT (c)

0.4 0.6 Yl, xi (d)

0.8

1

Figure 1.9. Phase equilibrium diagrams for acetone(1)-water(2) mixture estimated from the Raoult's law and the NRTL model using the Aspen Plus simulator.

Table 1.10 Pressure compositions estimated from Raoult's law for acetone(1)-water(2) mixture at T - 50°C xl

P (kPa)

vl

xl

P (kPa)

Yl

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45

12.326 15.813 19.300 22.787 26.275 29.762 33.249 36.737 40.224 43.711

0 0.2595 0.4252 0.5402 0.6247 0.6893 0. 7405 {}.7819 0.8161 {).8449

0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95

47.199 50.686 54.173 57.660 61.148 64.635 68.122 71.610 75.097 78.584

0.8694 0.8905 0.9089 0.9251 0.9395 0.9523 0.9638 0.9741 0.9835 0.9921

1

82.072

1

42

1. Fundamentals of equilibrium thermodynamics

Here, the exponential term is the Poynting correction factor, which may be negligible at low to moderate pressures. Disregarding the Poynting factor, Eq. (1.218) becomes (1.219)

t~sat For moderate pressures, the virial equation allows for the estimation of the fugacity coefficient 4,

1 ] gii +-~_,~.~Yjyk(2aji-6jk)

= exp~T

~ i ~ sat

and ~sat = e x p ~

j k

RT

(1.220)

For a binary gas mixture, we have

[

q~ = exp - ~ where 6ji = 2Bji- Bjj- Bii, ~ii -- ~jj = 0, and ~ji = ~ij. From Eqs. (1.219) and (1.220), we have

(~i = exp

[Bii(P-Pisat)+(1/2)PERT E 'YJYk(2t~ji ) ] j

6jk

(1.221)

For a binary mixture of species 1 and 2, Eq. (1.221) becomes

dPl =exp[Bll(P-PlSat)+py2612 ] '

=exp[ B22(P-P~at)+Py?~12 RT

(1.222)

Therefore, the gamma-phi relation of a binary vapor-liquid equilibrium for species 1 in Eq. (1.217) becomes

Yl =

P1sat K I X1 -

P

a/1 exp[(Bll ( P - PlSat) +

(1.223)

py2~12 )/RT 1 x1

Values of the pure-species virial coefficients Bkk come from the Pitzer correlations

BPc - B ° + RL

ogB1

(1.224)

where B ° = 0.083-(0.422/T ]6) and B 1 = 0.139-(0.172/T42). The cross coefficients Bik are obtained from

Oij - RVcij (O 0 Jr-¢,,oij.O1) Pcij

(1.225)

The mixing rules are

O)i Jr" O)j w°=

Tcij = (TciTcj )1/2 (1Pcij=

(1.226)

2

RZcijTaJ Vco

k,j)

(1.227)

(1.228)

43

1.10 Equations of state

=

Zci + Zcj

(1.229) 3

' cVij =

" cj

(1.230)

2

where k• is an interaction parameter characteristic of i - j molecular pair. k/j = 0 for i =j.

Example 1.17 Heterogeneous azeotrope The heterogeneous mixture n-butanol(1)-water(2) exhibits a minimum boiling azeotrope. The temperature composition data at 1 atm is shown in Table 1.11. (a) Calculate the activity coefficient for n-butanol and water. (b) Prepare TO' and yx diagrams and estimate the compositions of the two liquid phases in the immiscible region. Solution: Assume that the pressure is low and the vapor phase is ideal. The vapor-liquid phase equilibrium is represented by the modified Raoult's law (Eq. 1.191) PYi = Xi yiPsat

The experimental activity coefficients y and the activities are Yi = (Pyi/xi Psat) and ai = 'yixi. The Antoine equation with the following constants determines the vapor pressure in kPa with the temperature in °C B lnP= A - ~

T+C

Water A B C

n-Butanol

16.3872 3885.70 230.17

15.3144 3212.43 182.74

Figure 1.10 shows the temperature composition and vapor to liquid composition diagrams. Table 1.12 shows the estimated activity coefficients. The azeotropic composition of n-butanol at atmospheric pressure is about 0.247. The region of immiscibility is about Xl - 0.02-0.44.

1.10.12

Osmotic Equilibrium

Consider two liquid mixtures of a and/3 separated by a membrane permeable to the solvent of species 1 and impermeable to all other species in either mixture. The equilibrium condition for species 1 requires that the pressure of the solvent must be the same in either mixture. Therefore, the solute species in either mixture would not be in equilibrium. Also, there is no hydrostatic equilibrium established between the mixtures, and the difference of pressure is Table 1.11 Heterogeneous isotropic data for n-butanol(1)-water(2) system T(°C) 100 95.8 95.4 92.8 92.8 92.7 92.7 92.7 93.0

Yl

xl

T(°C)

Yl

xl

0.0 0.150 0.161 0.237 0.240 0.246 0.246 0.246 0.250

0.0 0.008 0.009 0.019 0.020 0.098 0.099 0.247 0.454

93.0 96.3 96.6 100.8 106.4 106.8 110.9 117.5

0.247 0.334 0.340 0.444 0.598 0.612 0.747 1.00

0.450 0.697 0.709 0.819 0.903 0.908 0.950 1.00

Source: T.E. Smith and R.F. Bonner, Ind. Eng. Chem., 41 (1949) 2867.

44

1.

Fundamentals of equilibrium thermodynamics

120

1

115

0.8 110

0.6

? 105 ~

100

0.4

95 0.2 90 (~

85 0

0.2

0.4 0.6 Xl, Y2

0.8

0

0.2

0.4

0.6

0.8

1

Xl

(a)

(b)

Figure 1.10. Phase diagrams for butanol(1)-water(2) partially miscible mixture.

Table 1.12 Activity coefficient calculations for butanol(1)-water(2) system at 1 atm T (°C)

Yl

xl

P1 (kPa)

P2 (kPa)

100 95.8 95.4 92.8 92.8 92.7 92.7 92.7 93 93 96.3 96.6 100.8 106.4 106.8 110.9

0 0.15 0.161 0.237 0.24 0.246 0.246 0.246 0.25 0.247 0.334 0.34 0.444 0.598 0.612 0.747

0 0.008 0.009 0.019 0.02 0.098 0.099 0.247 0.454 0.45 0.697 0.709 0.819 0.903 0.908 0.95

52.06 43.86 43.14 38.68 38.68 38.52 38.52 38.52 39.01 39.01 44.78 45.33 53.75 66.94 67.98 79.37

101.31 87.06 85.79 77.93 77.93 77.64 77.64 77.64 78.52 78.52 88.66 89.64 104.24 126.72 128.47 147.57

117.5

1

1

3'1

43.3052 42.0074 32.6638 31.4235 6.6012 6.5345 2.6191 1.4298 1.4252 1.0841 1.0716 1.0217 1.0021 1.0044 1.0036

]/2

]/lXl

]/2X2

0.9970 0.9997 1.0110 1.0080 1.0906 1.0918 1.3064 1.7722 1.7664 2.5113 2.5632 2.9853 3.3129 3.3254 3.4734

0.3464 0.3781 0.6206 0.6285 0.6469 0.6469 0.6469 0.6491 0.6414 0.7556 0.7598 0.8368 0.9049 0.9120 0.9534

0.9890 0.9907 0.9918 0.9879 0.9837 0.9837 0.9837 0.9676 0.9715 0.7609 0.7459 0.5403 0.3214 0.3059 0.1737

balanced by a force produced by the membrane. This kind of equilibrium is called the osmotic equilibrium of the solvent species 1. We can determine the pressure difference P ~ - P t ~ (called the osmotic pressure II that must be applied to the mixture to bring it into a certain equilibrium condition) necessary to maintain osmotic pressure by integrating the relation between P~ and Pt~ 0In fl _ 0 In a 1 _ V1 OP

OP

(1.231)

RT

where J] and a 1 are the fugacity and activity of species 1. The osmotic pressure is obtained as

IIV1_ ln(fl ° ] Z)

(1.232)

when the compressibility and the pressure effects are negligible. For example, for an ideal solution, Eq. (1.232) becomes HidealV1 = - I n x 1

RT

(1.233)

45

1.10 Equationsof state

For the solution sufficiently dilute in solute 1, we have Hideal V1

RT

= - l n ( x I ) ~ - ln(1 -

x 2

)

~

X2

(1.234)

and H - x2RT

(1.235)

This equation is called the van't Hoff equation. If the solute molecule is a strong electrolyte and completely dissociating m ions, then Eq. (1.233) becomes

nv~_

-- - ln(x~-y1 ) RT

(1.236)

Equation (1.236) yields H = mx2RT

V~

(1.237)

Osmotic pressure may be considerable for even highly dilute solutions; for example, for an aqueous nonelectrolyte solution with x2 - 0.001 at 200 K, we have

H - x2RT - 0"001(82"06cm3 atm/(molK)) V1

18.02cm 3

(1300 K) = 1.36 atm

The value of osmotic coefficient & is the ratio of the osmotic pressure to the one at the ideal mixture condition 4) = H/l-]idea 1.

1.10.13

Generalized Correlations for Gases

Compressibility factor (Z): The following Lee-Kesler correlation can be used to estimate the compressibility factor Z - Z ° + ~oZl

(1.238)

The parameters Z ° and Z ~ are listed as function of Tr and Pr in Tables F l-F4. The accentric factor 6o is tabulated in Table B 1. The definition for the residual property is XR -X-Xideal, where X and Yidea 1 are the actual and ideal gas properties, respectively. Residual volume ( VR) is

VR - V -

RT

Videa I -- /~-(,Z -

l)

(1.239)

Residual enthalpy (HR): The following correlation can be used to estimate the residual enthalpy HR -- H ° + w HIR RT~. RTC RT~

(1.240)

The parameters H ° and H~ are listed as function of Tr and Pr in Tables F5-F8. The critical constants are listed in Table B 1. Residual entropy (SR): The following correlation can be used to estimate the residual entropy: SR -- so + w S~ R R R

(1.241)

46

1. Fundamentals of equilibrium thermodynamics

The parameters ~ and S 1 are listed as function of Tr and Pr in Tables F9-F12. The residual Gibbs energy can be obtained by its definition and in terms Of HR and SR:GR = HR-- TSR. Fugacity coefficient (4~): The Lee-Kesler generalized correlation for fugacity coefficient is ln4~ = ln4~° + o)ln~b1

(1.242)

The parameters ~b° and ~1 are listed as function of Tr and Pr in Tables F 13-F 16.

1.11

THERMODYNAMIC POTENTIALS

At equilibrium, the extensive properties U, S, V, Ni, and the linear combination of them are functions of state. Such combinations are the Helmholtz free energy, the Gibbs free energy, and enthalpy, and are called the thermodynamic potentials. Table 1.13 provides a summary of the thermodynamic potentials and their differential changes. The thermodynamic potentials are extensive properties, while the ordinary potentials are the derivative of the thermodynamic potentials and intensive properties. A thermodynamic potential reaches an extremum value toward equilibrium under various conditions. The Helmholtz free energy A is particularly useful for systems at constant volume and temperature. Combining Eq. (1.76) and Eq. (1.244) at constant temperature yields -(dA)r = 6W

(1.255)

The total reversible work performed by a system is equal to the decrease in the Helmholtz free energy. The Gibbs free energy is especially suitable for isothermal and isobaric systems; from Eqs. (1.76) and (1.248), we have

-(dG)r,p = 6 W - PdV

(1.256)

Thus, the decrease in the Gibbs free energy is the useful work that is equal to the total work minus the pressure-volume work. Since PdV is usually negligible for a condensed phase and living tissues, the use of thermodynamic potential G is common in such systems. For a closed system under isobaric conditions and using Eq. (1.76), we have

Table 1.13 Thermodynamic potentials for closed systems and their changes between equilibrium states for a

homogeneous fluid with constant composition Thermodynamic potentials 1. Helmholtz free energy, A Definition Change With Eq. (1.107) Chemical potential

dA = d U - TdS - S d T dA = - S d T P d V + F d l + Ode + ~_, IzidNi

(1.243) (1.244) (1.245)

txi =

(1.246)

A = U - TS

T, V, l, e, N j

2. Gibbs free energy, G Definition Change With Eq. (1.107) Chemical potential

G = U - TS + P V d G = d U - TdS - S d T

+ P d V + VdP

dG=-SdT+VdP+Fdl+Ode+~_,txidNi

txi =

(1.247) (1.248) (1.249) (1.250)

T,P,l,e, N j

3. Enthalpy, H Definition Change With Eq. (1.107) Chemical potential

d H = d U + P d V + VdP

(1.251) (1.252)

d H = TdS + VdP + F d l + O d e + ~_,IzidNi

(1.253)

H = U + PV

~i =

(1.254) S,P,l,e,N 2

1.12

47

Cross relations

( d H ) p = (•q)p

(1.257)

The enthalpy is the same as the heat exchanged with the surroundings. The Gibbs free energy can be related to enthalpy G : H

-

(1.258)

TS

Hess's law referring to the heat evolved in a chemical reaction is conveniently formulated in terms of enthalpy.

1.12

CROSS RELATIONS

Maxwell first noted the cross relations based on a property of the total differentials of the state functions. The cross differentiations of a total differential of the state function are equal to each other. Table 1.14 summarizes the total differentials and the corresponding Maxwell relations. The Maxwell relations may be used to construct important thermodynamic equations of states. The cross relations can be seen in a reversible change of a rectangular rubber sheet subjected to two perpendicular forces Fx and F,, under isothermal conditions. If the extent of stretching in both directions of x and y are Ax and Ay, we have F x = M 1l a x + M 1 2 A y

(1.267)

F~, -- M 2 1 ~ x + M 2 2 A y

(1.268)

The ordinary elastic moduli in the x and y directions are denoted by Mll and M22

(1.269)

Table 1.14 Total differentials of the state functions and the Maxwell relations for closed systems of a homogeneous fluid of constant composition Total differentials/the Maxwell relations The internal energy: U = U (5', V)

(i

OU s dV = T d S - PdV d U = -OL,' ~ l dS + --~

(1.259)

(1.260) The Helmholtz free energy: A = A(77, V)

dA = ( aA , dT + ( OA T dV = - S d T - PdV aS

(1.261)

,,262,

The Gibbs free energy: G = G (77,P)

dG= -#G ~ p dT + - ~ 7 dP = - S d T + VdP

-

~

(1.263)

T-- -Of p

(1.264)

( ~aH ] ,, dS +( 7OH ] dP = TdS + VdP # ,

(1.265)

Enthalpy: H = H (S. P) dH

:

~T

OV

48

1. Fundamentals of equilibrium thermodynamics

However, the cross moduli m12 and m21 relate the force in one direction to the stretch in the other direction, and we have

(1.270) From Eq. (1.110), we can express Fx and Fy as

Y

x

The cross relations impose that

oy Jx

),

(1.272)

Therefore, the matrix of the moduli becomes symmetric mll

m12

M21 M22

(1.273)

where m12 = m21. Similar matrices may occur in nonequilibrium thermodynamic descriptions of irreversible systems.

1.13

EXTREMUM PRINCIPLES

Equilibrium thermodynamics has various extremum principles. At various conditions, a thermodynamic potential will approach an extremum value as the system approaches equilibrium. For an isolated or closed system, we may consider the following extremum principles: • The entropy of an isolated system reaches the maximum possible value at equilibrium. diS > 0 at constant U and V

(1.274)

dU = 6q- PdV = TdeS- PdV

(1.275)

• For a closed system

Since the total entropy change dS = deS + diS, we have d U = T d S - P d V - TdiS d U = - T d i S <- 0 at constant S and V

(1.276)

For entropy to remain constant, we keep T, V, and dN,. constant, and the entropy produced, diS, has to be removed from the system. The decrease in energy is generally due to the conversion of mechanical energy into heat. • The Helmholtz free energy reaches a minimum possible value at equilibrium. dA = - T d i S <- 0 at constant T and V

(1.277)

• The Gibbs free energy reaches a minimum possible value at equilibrium. dG = - T d i S <- 0 at constant P and T

(1.278)

• The enthalpy reaches a minimum possible value at equilibrium. d H = - T d i S <- 0 at constant S and P

(1.279)

49

Problems

PROBLEMS 1.1

Derive equations for the critical constants from the van der Waals equation of state. nRT

p-

V-

an 2

nb

V2

1.2

(a) Use the critical parameters for n-pentane and estimate the van der Waals constants a and b. (b) Plot van der Waals isotherms for pentane at T1 - 400 K, 7'2 - 460 K, and T3 - 600 K, when the volume changes from 0.085 to 3.5 L.

1.3

(a) Use the critical parameters for CO2 and estimate the Redlich-Kwong constants a and b at T - 3 0 0 K. (b) Plot van der Waals isotherms for CO2 at T l - 325 K, T 2 - 175 K, and 7'3- 120 K, when the volume changes from 0.037 to 0.45 L.

p -

1.4

RT

a ( T / T c ) -1/2

v-b

v(v+b)

(a) Using the critical parameters for CO2, estimate the van der Waals constants. (b) Plot the pressure estimated from van der Waals equation in a tank with a volume V = 30 L at a temperature of 310 K if the amount changes between 10 and 450 mol. p

m

nRT V-

an 2

nb

V2

1.5

(a) Estimate the molar volume of 200 tool of hydrogen from van der Waals equation at 400 K and 20 atm. (b) Plot pressure versus temperature when the temperature changes from T = 250 to 450 K at V = 50 L.

1.6

Estimate the saturated molar volume of propylene at 10 bar and 400 K.

1.7

Estimate the saturated vapor and saturated liquid molar volume of n-butane at 355 K and 9.45 bar using the Soave-Redlich-Kwong equation of state.

1.8

Estimate the saturated vapor and liquid molar volumes of CO2 at 274K and 35.6 bar using the PengRobinson equation of state.

1.9

Estimate the volume of methyl chloride at 300 K and 4 bar using: (a) Virial equation. (b) Redlich-Kwong equation: Generic and compressibility factor equations.

1.10

A well-insulated 30-m 3 tank is used to store exhaust steam. The tank contains 0.01 m 3 of liquid water at 30°C in equilibrium with the water vapor. Determine the amount of wet-exhaust steam, in kg, from a turbine at 1 atm at the end of an adiabatic filling process. The wet steam has the quality of 90%, and the final pressure within the tank is I atm. Assume that heat transfer between the liquid water and the steam is negligible. Exhaust steam 1()1.3 kPa

Liquid + vapor Storage tank

50

1.11

1.

Fundamentals of equilibrium thermodynamics

Initially the mixing tank shown below has 100 kg of water at 25°C. Later two other water inlet streams 1 and 2 add water and outlet stream 3 discharges water. The water in the tank is well mixed and the temperature remains uniform and equal to the temperature of outlet stream 3. Stream 1 has a flow rate of 20 kg/h and is at 60°C, while stream 2 has a flow rate of 15 kg/h and is at 40°C. The outlet stream has a flow rate of 35 kg/h. Determine the time-dependent temperature of the water in the mixing tank.

Mixingtank t~ ~3 1.12

Carbon dioxide is being withdrawn from a tank at a rate of 0.5 kg/min. The tank has a volume of 0.2 m 3, and initially contains the gas at 12 bar and 300 K. The tank does not exchange heat with the environment. The heat capacity of the carbon dioxide is 37.14 J/(mol K), and remains constant. Determine: (a) The temperature and pressure after 10 min. (b) The temperature and pressure change with time.

1.13

A 2-kW heater is used to heat a room with dimensions 3.0 m x 4.5 m x 4.0 m. Heat loss from the room is negligible and the pressure is always atmospheric. The air in the room may be assumed to be an ideal gas, and the heat capacity is 29 J/(mol K). Initially the room temperature is 290 K. Determine the rate of temperature increase in the room.

1.14

A steam at 900 psia and 700°F is throttled in a valve to 55 psia at a rate of 20 lb/min at steady state. Determine the entropy production due to expansion of the steam.

1.15

A compressor is compressing air from 1 atm and 300 K to 12 atm. Assume that the compression is adiabatic. Determine the entropy production if the work needed for compressing 25 mol/s of air for the compressor is 85 kW.

1.16

A steam turbine consumes 4000 kg/h steam at 540 psia and 800°E The exhausted steam is at 20 psia. Turbine operation is adiabatic. (a) If the expansion does not produce entropy, determine the exit temperature of the steam and the work produced by the turbine. (b) Determine the entropy production if the turbine efficiency is 80%. (c) As seen below, the use of a throttling valve reduces the pressure to 200 psia from 540 psia. Estimate the temperature outside the valve (state 2) and the temperature of the exhausted steam from the turbine.

1

~ ~

2

Steam Valve 1

@W i3 1.17

Steam at 8200 kPa and 823.15 K (state 1) is being expanded to 30 kPa in a continuous operation. Determine the final temperature (state 2), entropy produced, and work produced per kg of steam for the following operations: (a) Adiabatic expansion through a valve. (b) Adiabatic expansion through a turbine. (c) Isothermal expansion through a turbine.

1.18

Consider an ideal gas at temperature TI and pressure P1 used as a working fluid in an adiabatic turbine operated at steady state. The exit pressure of the gas is fixed at P2. Prove that in a reversible operation (entropy production = 0): (a) The exit temperature is at a minimum. (b) The work produced by the turbine is at a maximum.

51

Problems

1.19

Consider a steam power generation system. The system burns fuel at 1273.15 K, and cooling water is available at 290 K. The steam produced by the boiler is at 8200 kPa and 823.15 K. The condenser produces a saturated liquid at 30 kPa. The turbine and pump operate reversibly and adiabatically (entropy production = 0). For every kg steam produced in the boiler determine: (a) The net work produced. (b) The heat discharged in the condenser. (c) The heat absorbed in the boiler. (d) Actual and ideal thermal efficiencies.

~ QH 4 Boiler

J

mp

Condenser

]-"

I-

2

QC 1.20

Derive the entropy departure equation below.

[ 1.21

Prove that the following is correct: (a) (OH/OV)r- 0 if (OH/OP)r = O. (b) The partial derivative (OS/OV)p of a fluid has the same sign as its thermal expansion coefficient c~ = (1/V)(OV/OT)p, and is inversely proportional to o~.

1.22

Derive the following Maxwell relations for open systems

(~VJs,n----(~SS)I,n'

{~nT)s,~,-(-~)v,n '

and{~n)s,v--(~V)s, ~

1.23

For an isothermal fluid flow described by the Redlich-Kwong equation of state, develop expressions in terms of the initial temperature and the initial and final volumes for the changes in internal energy, enthalpy, entropy, and the Gibbs free energy.

1.24

Determine the standard heats and free energies of a reaction at 298.15 K for the reactions below: (a) N2(g) + 3Hz(g)- 2NH3(g) (b) CaCO3(s ) - CaO(s) + C02(g )

1.25

Determine the minimum amount of work to separate 1 mol of an equimolar binary mixture of isomers into its pure components at the same temperature and pressure.

1.26

Determine the heat and work needed to reversibly and isothermally separate an equimolar binary mixture into its pure species if the excess Gibbs free energy for the mixture is

G E - Axlx 2 where A is (a) independent of temperature, and (b) dependent on temperature. 1.27

Calculate the minimum work for separating air into pure oxygen and nitrogen in a continuous operation at 298.15 K. The air at inlet is at 1 bar and contains 79% nitrogen.

52 1.28

1.

Fundamentals of equilibrium thermodynamics

The following table shows the isothermal vapor-liquid equilibrium data for acetone(1)-methanol(2) at 55°C. P (kPa)

xl

Yl

P (kPa)

xl

Yl

68.728 72.278 75.279 77.524 78.951 82.528 86.762 90.088 93.206 95.017 96.365

0.0000 0.0287 0.0570 0.0858 0.1046 0.1452 0.2173 0.2787 0.3579 0.4050 0.4480

0.0000 0.0647 0.1295 0.1848 0.2190 0.2694 0.3633 0.4184 0.4779 0.5135 0.5512

97.646 98.462 99.811 99.950 100.278 100.467 100.999 101.059 99.877 99.799 96.885

0.5052 0.5432 0.6332 0.6605 0.6945 0.7327 0.7752 0.7922 0.9080 0.9448 1.000

0.5844 0.6174 0.6772 0.6926 0.7124 0.7383 0.7729 0.7836 0.8959 0.9336 1.000

Source: D.C. Freshwater and K.A. Pike, J. Chem. Eng. Data, 12 (1967) 179.

(a) Calculate the activity coefficient for acetone and methanol. (b) Prepare Pxy and yx diagrams and estimate the composition and pressure of the azeotrope. 1.29

Carbon dioxide C O 2 at 198.15 K and 1 atm has a heat capacity Cv = 28.46 J/mol. Calculate the heat capacity Cp at 298.15 K and 15 bar using the van der Waals EOS p

m

RT V-b

a V2

1.30

Prepare phase diagrams for acetone(1)-methanol(2) mixture using Raoult's law: (a) Temperature compositions diagram, Tyx, at P = 1 atmosphere. (b) Pressure compositions diagram, Pyx, at 40°C.

1.31

Prepare phase diagrams for acetone(1)-l-propanol(2) mixture using Raoult's law: (a) Temperature compositions diagram, Tyx, at P = 1 atm. (b) Pressure compositions diagram, Pyx, at 50°C.

1.32

Plot the Gibbs energy of mixing versus mole fraction of acetone for acetone(1)-water(2) mixture using the NRTL model at 1 atm and 50°C.

REFERENCES A. Bohm, H.-D. Doebner and P. Kielanowski (Eds), Irreversibility and Causality, Springer-Verlag, Berlin (1998). E.A. Guggenheim, Thermodynamics. An Advanced Treatment for Chemists and Physicists, North Holland, Amsterdam (1967). A. Katchalsky and EE Curran, Nonequilibrium Thermodynamics in Biophysics, Harvard University Press, Cambridge (1967). S.I. Sandler, Chemical, Biochemical, and Engineering Thermodynamics, 4th ed., Wiley, New York (2006). J.M. Smith, H.C. Van Ness and M.M. Abbott, Introduction to Chemical Engineering Thermodynamics, 7th ed., McGraw-Hill, New York (2005). J.W Tester and M. Modell, Thermodynamics and Its Applications, 3rd ed. Prentice Hall, Upper Saddle River, NJ (1997).

REFERENCES FOR FURTHER READING B.G. Kyle, Chemical and Process Thermodynamics, 3rd ed., Prentice Hall, Upper Saddle River (1999). D. Kondepudi and I. Prigogine, Modern Thermodynamics, From Heat Engines to Dissipative Structures, Wiley, New York (1999). N.W. Tschoegl, Fundamentals of Equilibrium and Steady-State Thermodynamics, Elsevier, Amsterdam (2000).

2 TRANSPORT AND RATE PROCESSES 2.1

INTRODUCTION

Using a limited number of principles, classical thermodynamics describes systems in mechanical, thermal, and chemical equilibria. Classical thermodynamics provides a set of extremum principles and mainly targets systems at equilibrium. According to the extremum principles, the entropy of an isolated system attains a maximum value while its free energy reaches a minimum at equilibrium. Classical thermodynamics plays an important role in science and engineering fields, such as physical chemistry and biochemistry. However, many real processes, such as physical and biological processes, occur in nonequilibrium, irreversible, and open systems. For such systems, the laws of classical thermodynamics provide a set of inequalities describing only the direction of change. Consequently, we can use kinetic equations and statistical models to describe the transport of matter, energy, and electricity, as well as biological systems, such as muscle contractions and complex coupled reaction cycles. There is a close connection between molecular mass, momentum, and energy transport, which can be explained in terms of a molecular theory for low-density monatomic gases. Equations of continuity, motion, and energy can all be derived from the Boltzmann equation, producing expressions for the flows and transport properties. Similar kinetic theories are also available for polyatomic gases, monatomic liquids, and polymeric liquids. In this chapter, we briefly summarize nonequilibrium systems, the kinetic theory, transport phenomena, and chemical reactions.

2.2

NONEQUILIBRIUM SYSTEMS

A system reaches the thermodynamic equilibrium state when it is left for a long time with no external disturbances. At equilibrium the internal properties are fully determined by the external properties. This makes it easy to describe such systems; for example, if the temperature is not uniform within the system, heat is exchanged with the immediate surrounding until the system reaches a thermal equilibrium, at which the total internal energy U and entropy S are completely specified by the temperature, volume, and number of moles. Therefore, at equilibrium there are no thermodynamic forces operating within the system (Figure 2.1). Equilibrium systems are stable. For small deviations, the system can spontaneously return to the state of equilibrium. Equilibrium correlations result from short-range intermolecular interactions. Existence of the extremum principles is a characteristic property of equilibrium thermodynamics. However, natural systems consist of flows caused by unbalanced driving forces, and hence the description of such systems requires a larger number of properties in space and time. Such systems are away from the equilibrium state, and are called nonequilibrium systems; they can exchange energy and matter with the environment, and have finite driving forces (Figure 2.1 ). The formalism of nonequilibrium thermodynamics can describe such systems in a qualitative and quantitative manner by replacing the inequalities of classical thermodynamics with equalities. Constant driving forces cause steady flows, which leads to a stationary state. For example, a constant temperature difference applied to a metal bar will induce a heat flow that will cause a change in all local temperatures. After a while, a constant distribution of temperature will be attained and the heat flow will become steady. The steady state flow and constant distribution of forces characterizing a system form the ultimate state of irreversible systems corresponding to the states of equilibrium in classical thermodynamics. All living systems are open systems capable of attaining a stable stationary state, at which the variables do not change with time. As dS/dt -- 0, entropy produced by steady-state flow is equal to the entropy given off to the surroundings. Only open systems capable of exchanging entropy with their environment can reach a steady state. Prigogine pointed out that steady nonequilibrium systems produce entropy at a minimal rate. Biological systems progress toward increasing size and complexity, and they do not decay toward an equilibrium state. Still, all living organisms produce less entropy by maintaining a stationary state.

54

2.

Transport and rate processes

J=0

J=0 A/~ =0 AT=0 AP=0

Equilibrium System

A#~0 AT@0 AP~0

J~0

J~:0

Nonequilibrium System Figure 2.1. Equilibrium and nonequilibrium systems.

State Linear region ~ ~ . . _ _ Nonlinear region axc

ax

Figure 2.2. Thermodynamic branch. AXc indicates the critical distance from equilibrium state.

2.2.1

Thermodynamic Branch

States away from global equilibrium are called the thermodynamic branch (Figure 2.2). Systems not far from global equilibrium may be extrapolated around equilibrium state. For systems near equilibrium, linear phenomenological equations may represent the transport and rate processes. The linear nonequilibrium thermodynamics theory determines the dissipation function or the rate of entropy production to describe such systems in the vicinity of equilibrium. This theory is particularly useful to describe coupled phenomena, and quantify the level of coupling in physical, chemical, and biological systems without detailed process mechanisms. In some systems, the distance from equilibrium reaches a critical point, after which the states in the thermodynamic branch become metastable or unstable. This region is the nonlinear region where the linear phenomenological equations are not valid. We observe bifurcations and multiple solutions in this region.

2.2.2

Local Equilibrium

In nonequilibrium systems, the intensive properties of temperature, pressure, and chemical potential are not uniform. However, they all are defined locally in an elemental volume with a sufficient number of molecules for the principles of thermodynamics to be applicable. For example, in a region k, we can define the densities of thermodynamic properties such as energy and entropy at local temperature. The energy density, the entropy density, and the amount of matter are expressed by uk(T, Nk), sk(T, Nk), and Nk, respectively. The total energy U, the total entropy S, and the total number of moles N of the system are determined by the following volume integrals: U = ~v ['" ukdV'

S = ~v [''skdV'

N = fv NkdV

(2.1)

Since the temperature is not uniform for the whole system, the total entropy is not a function of the other extensive properties of U, V, and N. However, with the local temperature, the entropy of a nonequilibrium system is defined in terms of an entropy density, sk.

2.3

2.2.3

Kineticapproach

55

Dissipative Structures

When the distance from global equilibrium is beyond a critical point, the states in the thermodynamic branch become metastable or unstable (Figure 2.2). Prigogine called these states dissipative structures. Dissipative structures may be highly organized and ordered states, and need a constant supply of mass and energy from the outside. This part of the thermodynamic branch corresponds to the nonlinear region, in which the relations between the general flows and forces are nonlinear. In this region, chemical reactions and transport processes are characterized by the local potentials in which each macroscopic variable is described by an average quantity and fluctuating quantity. The compensation function and generalized hydrodynamics are an integral part of the extended theory of irreversible processes.

2.2.4

NonequilibriumThermodynamics and Design:Thermodynamic Analysis

Linear nonequilibrium thermodynamics can help to design thermodynamically optimum processes. This trend is called thermodynamic analysis, in which the rate of entropy production and dissipated energy is calculated. The entropy production approach is especially important in terms of process optimality, as each process contributing to the entropy production can be identified and estimated separately. Through the minimization of excessive irreversibilities by modifying the operating conditions or the design, a thermodynamic optimum can be attained for a process with a required task. The trade-offbetween the various contributions of entropy production, equipartition of entropy production, or the uniform driving force may be used as a criterion for designing a thermodynamically optimum process. The distribution of the volumetric entropy production rate may identify the regions within the system where excessive irreversibilities occur. Thermodynamic analysis has been applied extensively in thermal engineering, separation by distillation, and chemical reactor design, which are all energy intensive processes.

2.3

KINETIC APPROACH

Statistical mechanics can provide phenomenological descriptions of nonequilibrium processes. An alternative approach based on kinetic theory is favorable especially in describing the transport and rate phenomena. A kinetic theory of nonequilibrium systems has been developed for dilute monatomic gases at low pressure. Substantial progress has also been achieved in extending the theory to dense gases, real gases, and liquids. A rigorous kinetic theory of monatomic gases at low density was developed early in the 20th century by Chapman in England and independently by Enskog in Sweden. Initially, the kinetic theory was limited to low density, nonreacting systems of simple, spherical molecules with no internal degrees of freedom. Typically, kinetic approaches start with the Boltzmann equation for the velocity distribution function of each component in a multicomponent system, and the time evolution of the distribution function is obtained by solving the governing kinetic equations with a set of initial conditions. Evolution of the velocity distribution function with time is calculated with an external force acting on a molecule and using an intermolecular potential energy function, such as the Lennard-Jones potential. The conservation laws appear in kinetic theory as a result of collision phenomena for the mass, momentum, and kinetic energy of molecules. The conservation relations together with the equations of mass, heat, and momentum yield the equations of change describing the hydrodynamic fields of velocity, temperature, and concentration for reacting and nonreacting systems. The kinetic theory also provides us with the expressions for transport coefficients of momentum, energy, and mass. For a gas mixture at rest, the velocity distribution function is given by the Maxwell-Boltzmann distribution function obtained from an equilibrium statistical mechanism. For nonequilibrium systems in the vicinity of equilibrium, the Maxwell-Boltzmann distribution function is multiplied by a correction factor, and the transport equations are represented as a linear function of forces, such as the concentration, velocity, and temperature gradients. Transport equations yield the flows representing the molecular transport of momentum, energy, and mass with the transport coefficients of the kinematic viscosity, u, the thermal diffusivity, c~, and Fick's diffusivity, Dij , respectively. The kinetic theory leads to the definitions of the temperature, pressure, internal energy, heat flow density, diffusion flows, entropy flow, and entropy source in terms of definite integrals of the distribution function with respect to the molecular velocities. The classical phenomenological expressions for the entropy flow and entropy source (the product of flows and forces) follow from the approximate solution of the Boltzmann kinetic equation. This corresponds to the linear nonequilibrium thermodynamics approach of irreversible processes, and to Onsager's symmetry relations with the assumption of local equilibrium. If the collisions of molecules produce a chemical reaction, the Boltzmann equation is modified in obtaining the equations of change; these problems are addressed and analyzed in the context of quantum theory, reaction paths, saddle points, and chemical kinetics. Mass, momentum, and energy are conserved even in collisions, which produce a chemical reaction.

56

2.3.1

2.

Transport and rate processes

Boltzmann's H-Theorem

We can describe irreversibility by using the kinetic theory relationships in maximum entropy formalism, and obtain kinetic equations for both dilute and dense fluids. A derivation of the second law, which states that the entropy production must be positive in any irreversible process, appears within the framework of the kinetic theory. This is known as Boltzmann's H-theorem. Both conservation laws and transport coefficient expressions can be obtained via the generalized maximum entropy approach. Thermodynamic and kinetic approaches can be used to determine the values of transport coefficients in mixtures and in the experimental validation of Onsager's reciprocal relations.

2.3.2

Microscopic and Macroscopic Domains

The kinetic equations serve as a bridge between the microscopic domain and the behavior of macroscopic irreversible processes through the description of hydrodynamics in terms of intermolecular collisions. Hydrodynamics can specify a large number of nonequilibrium states by a small number of reproducible properties such as the mass, density, velocity, and energy density of a fluid conserved during the collision of molecules. Therefore, the hydrodynamic equations can describe a wide range of relaxation processes of nonequilibrium states to equilibrium state. We call such processes decay processes represented by phenomenological equations, such as Fourier's law of heat conduction. The decay rates are determined by the transport coefficients. Reliable transport coefficients provide microscopic and macroscopic information, and validate the results of molecular dynamics.

2.4

TRANSPORT PHENOMENA

The majority of systems in physics, chemistry, and biology consist of open, irreversible processes. Besides equilibrium states, stationary states are also of great interest. In stationary states, the flows of mass and energy between a system and its environment do not change with time, allowing technological processes to be carried out on a continuous basis. There exist a number of linear phenomenological laws describing irreversible processes in the form of proportionalities between the flow J/and the conjugate driving force Xk

Ji = LikXk

(2.2)

where Lik is the constant called the phenomenological coefficient and can be related to transport coefficients such as thermal conductivity or the rate constant for a chemical reaction. The validity of Eq. (2.2) should be determined experimentally for a particular process. For example, the equation will be valid for an electric conductor if the conductor obeys Ohm's law. This linear representation implies that the driving force that indicates the distance from equilibrium should not be too large for the system to be characterized in the linear region of the thermodynamic branch (Figure 2.2).

2.4.1

Momentum Transfer

In Figure 2.3a, we have a fluid between two large parallel plates separated by a distance H. This system is initially at rest; however, at time t = 0, the lower plate is set in motion by a constant force F in the positive x-direction at a constant velocity v. As time proceeds, the fluid gains momentum, and achieves a linear steady-state velocity profile. Newton's law of viscosity relates the shear stress to the velocity gradient in a Newtonian fluid; for a one-dimensional flow we have

ryx = - / z ~

d~x dy

(2.3)

where ~ is the viscosity. The shear stress is a tensor with magnitude, direction, and orientation. For example, the shear stress, ~'yx,is the force in the x-direction on a unit area normal to the y-direction. The viscosity is a measure of fluid's resistance to the deformation rate. The concept of viscosity requires a study of the microscopic motion and collision of fluid molecules. Equation (2.3) describes the resistance to flow of all gases and all liquids of the Newt°nianfluids" Most common fluids, such as water, air, and kerosene, are Newtonian fluids at normal temperature and pressure. Equation (2.3), however, cannot describe polymeric liquids, suspensions, pastes, slurries, and other complex fluids, which are referred as the non-Newtonianfluids. In non-Newtonian fluids the shear stress is not directly proportional to the rate of deformation or the shear rate, and may be represented by the power law model in terms of the flow behavior

2.4

Transportphenomena

57'

initially

t=O vx, TI, wi = 0

small t ~ b . _

v v

Y

' ~vx(y)

i

large t v

(a)

small t

Y

L.x

~

) large t

(b) T]

T0

small t

)

y

large t (c) Wi = 0

Wi = WiO

Figure 2.3. Steady-state transport of (a) momentum, (b) heat, and (c) mass.

index. Some non-Newtonian fluids are classified as thixotropic fluids, which show a decrease in viscosity with time, and rheopectic fluids, which show an increase in viscosity with time. In the vicinity close to the moving solid surface, the fluid gains a certain amount of x-momentum, and transfers it to the adjacent layer of fluid, so that the fluid develops a motion in the x-direction. Hence, x-momentum is being transmitted through the fluid in the y-direction, and ~,,x represents the flow of momentum in the x-direction; this interpretation is consistent with the molecular phenomena of momentum transport and the kinetic theory of fluids, and with heat and mass transport. In gases, the momentum is transported by the motion and collision of molecules; in liquids, the transport is mainly due to the motion of pairs of molecules bonded with intermolecular forces. Momentum flows "downhill" from a region of high velocity to a region of low velocity, and the velocity gradient therefore is a driving force. Often kinematic viscosity, u, which is the viscosity divided by the density of the fluid, is used

-

/.t

(2.4)

P

Table 2.1 shows the units of quantities associated with the momentum flow. In Eq. (2.3), we have a simple steady-state shearing flow with the velocity function ofy alone. In more complicated flows, we need the velocity components in three directions and with time, and in Cartesian coordinates we have vx - vx(x, y , z , t ) ,

v,, - v~,(x, y , z , t ) ,

v: - vz(x, y , z , t )

(2.5)

58

2.

Transport and rate processes

Table 2.1 Units of quantities related to Eq. (2.3) Quantity

SI

c.g.s.

British

ryx Vx

Pa = N/m 2

dyn/cm 2

lbf/ft 2

m/s

cm/s

ft/s

/, u

Pa s mZ/s

P = Poise = g/(cm s) cmZ/s

lbm/(ft s)

ft2/s

In three directions, there will be nine stress components to. The viscous forces appear only when there are velocity gradients within the fluid. The forces per unit area (molecular stresses) acting on the body, rr, both by the thermodynamic pressure and by the viscous stresses are given by ~rx = P~x + r x ,

~ry = P ~ y + r y , a'rz = PS~ + r z

(2.6)

Here 8x is the unit vector in the x-direction. The components of forces are scalars and defined by (2.7)

rrO. = P6ij + rij

where i andj may be x, y, or z, and 6ij is the Kronecker delta, which is 1 if i =j, and 0 if i 4=j. The following stresses are called normal stresses: rrxx = P + rxx;

try, = P + ry,;

rrzz = P + rzz

(2.8)

The remaining six quantities are called shear stresses. They have two subscripts associated with the coordinates, and are referred to as the components of the molecular momentum flow tensor, or the components of the molecular stress tensor, as they are associated with molecular motion. Usually, the viscous stress tensor, 'r, and the molecular stress tensor, ~, are simply referred to as stress tensors. For a Newtonian fluid, we may express the stresses in terms of velocity gradients and viscosities in Cartesian coordinates as follows:

-%

%

OVy "Jr"OVx ) ~ Ox Oy

(2.9)

OVz + OVy ) Oz

(2.10)

Ovx Ovz ) Zzx = Zxz = i~ lv Oz + Ox

(2.11)

2 Ovx rrxx = - P - - ~ / z V . v + 2 / Z ~ 0 x

(2.12)

2 Ovy try, = - P - - ~ / z V . v + 2 ~ ~

(2.13)

2 Ovz rrzz = - P - - ~ / z V . v + 2 ~ ~ 0 z

(2.14)

yz = z ~ = IX Oy

ay

In vector-tensor notation, Eqs. (2.9)-(2.11) become (2.15) where 8 is the unit tensor with components 6/j, Vv the velocity gradient tensor with components (O/Oxi)vj, (Vv) f the transpose of the velocity gradient tensor with components (O/Oxj)vi, and (V. v) the divergence of the velocity vector. The generalization in Eq. (2.15) involves the viscosity, ~, and the dilatational viscosity, K, to characterize a fluid. Usually it is not necessary to know the value of K in fluid mechanics problems. For gases we often assume it to be close

2.4

59

Transportphenomena

to the values of an ideal monatomic gas, for which K is practically zero. We also assume that liquids are incompressible fluids, (V.v) = 0, and K is negligible.

2.4.2

Combined Momentum Flow

A combined momentum flow tensor is defined when a convective effect exits (2.16)

05 = av + pvv = pg + "r + pvv

Equation (2.1 6) consists of two contributions: the molecular momentum flow tensor, av, and the convective momentum flow tensor, pvv. The term pg represents the pressure effect, while the contribution -r, for a Newtonian fluid, is related to the velocity gradient linearly through the viscosity. The convective momentum flow tensor pvv contains the density and the products of the velocity components. A component of the combined momentum flow tensor of x-momentum across a surface normal to the x-direction is dpxx = 7r~.x + pVxVx = p + , r x x + p v x v x

(2.17)

Similarly, the combined flow of y-momentum across a surface normal to the x-direction is 49~, = rC,, + pv~v,, = Z~y + pv~vy

(2.18)

The first index is the direction of transport, and the second is the component of momentum.

Example 2.1 Estimation of m o m e n t u m flow Estimate the steady-state momentum flow, zyz, in lbf/ft 2 when the lower plate velocity in Figure 2.3 is 0.5 ft/s in the positive x-direction. The plate separation is: y = 0.001 ft, and the fluid viscosity is 1.46 x 10-5 lbf s/ft 2. Solution: Assume a linear velocity profile. Approximation of the velocity profile: dv~ .-. A v ~ _ - 0 . 5 -

dy

0.001

Ay

-500s

-1

Momentum flow from Eq. (2.3): dv~ Zyx = / x m - - _ 1 . 4 6 × 1 0 - 5 ( _ 5 0 0 ) = 7.30X 10 -3 lbf/ft 2

ay

2.4.3

Estimation of Viscosity of Gases at Low Density

Viscosities of fluids change over many orders of magnitude. The viscosity is a strong function of temperature and increases with temperature for gaseous systems at low density, while the viscosity usually decreases with increasing temperature in liquids. Extensive data on viscosities of pure gases and liquids are available. Table 2.2 shows some experimental values of viscosities for fluids. When experimental data are not available, we can estimate the viscosity for gases at low density with negligible interactions form the following relation obtained from the kinetic theory of gases: /z -

2

x/'n'mKT

37r

7Td 2

(2.19)

where d and m are the diameter and mass of the spherical molecules, respectively, K the Boltzmann constant, and 7rd 2 is the collision cross-section. Equation (2.1 9) predicts the viscosity as a function of temperature without the effect of pressure up to - 10 atm. Experimental data indicate that viscosity increases more rapidly than predicted. To describe the temperature dependency, it is necessary to replace the rigid sphere model with another model that can represent the attractive and repulsive forces more accurately. It is also necessary to use the Boltzmann equation to obtain the molecular velocity distribution in nonequilibrium systems more accurately.

60

2.

Transport and rate processes

Table 2.2 Viscosities of some gases and liquids at atmospheric pressure

Substance

T (K)

/z (mPa s)

Gases Water

Air

/-Butane, i-C4H10 Methane, CH 4 Carbon dioxide, CO2 Nitrogen, N 2 Oxygen, O2 Hydrogen, H 2 Mercury, Hg Ethanol, CzHsOH Acetone, (CH3)2CO Benzene, C6H 6 Glycerol

273 288 298 313 333 353 373 273 293 313 353 373 296 293 293 293 293 300 293 273 323 273 298 293 298

0.0121 0.01716 0.01813 0.01908 0.02087 0.02173 0.0076 0.0109 0.0146 0.0175 0.0204 0.0089

Liquids 1.787 1.140 0.890 0.653 0.463 0.3548 0.2821

1.552 1.786 0.694 0.283 0.224 0.649 934.0

Source: Bird et al. (2002); R.C. Hardy and R.L. Cottington, J. Res. Natl. Bur. Standards, 42 (1949) 573; J.E Swidells, J.R. Coe, Jr. and T.B. Godfrey, J Res. Natl. Bur. Standards, 48 (1952) 1; Tables of Thermal Properties of Gases, National Bureau of Standards Circular, 464 (1955); N.A. Lange, Handbook of Chemistry, 15th ed., McGraw-Hill, New York (1999); H.L. Johnston and K.E. McKloskey, J. Phys. Chem., 44 (1940) 1038; CRC Handbook of Chemistry and Physics, CRC Press, Boca Raton, FL (1999); LandoltBornstein Zehwerte und Funktionen, Springer (1969).

The viscosity of a pure monatomic gas of molecular weight M may be expressed in terms of the Lennard-Jones parameters by = 2 . 6 6 9 3 × 10 -5

x/MT 0-2 ~'-~

(2.20)

where T is expressed in K, o-, the collision diameter, in A, and/z in g/(cm s). The dimensionless quantity 12~ is called the collision integral for viscosity; it describes the deviation from rigid sphere behavior, and varies slightly with the dimensionless temperature KT/e. Viscosity is a complex function of temperature, and Eq. (2.20) can predict the effect of temperature on viscosity satisfactorily. Equation (2.20) also predicts the viscosity of polyatomic gases satisfactorily. Tables B1 and B2 list the Lennard-Jones (6-12) potential parameters, critical properties of various species, and collision integrals.

2.4.4

Effect of Pressure and Temperature on Viscosity of Gases

The corresponding state correlation, which is based on "the principle of corresponding states" is widely used for correlating thermodynamic data. Figure 2.4, based on this approach, displays the effects of the pressure and temperature on viscosity by relating the reduced viscosity,/~r =/Z/~c, to the reduced temperature, Tr = T/Tc, and the reduced pressure, Pr = P/Pc. The viscosity of a gas approaches a low-density limit at --~1 atm pressure, and increases with increasing temperature. If critical P - V - T data are available, we can estimate the critical viscosity using the following empirical relations:

tx c =61.6(MTc) 1/2 (Vc) -2/3

or

/% =

7.70M1/Z(Pc)Z/3(Tc) -1/6

(2.21)

2.4

Transportphenomena

.

\ lo

~ \

8

.

.

.

\

~\\

7

:~

.

61

\.

\~,\

6

'~k

4

I~~

3

\ X

Dense gas

Xpr =p

~\

X

Pc

Two-phase

it

2

~egio~

"-""

IL\\5\\ \

ta

:~,"" ~ '

,.I

0.9 ~ 0.8 0.7 0.6 0.4

Critical ~ point

k 1., / ~ . ~ 5~//' ~/ "'~~0- ~~ ~ " Low density limit

1

Pr =

0.2.

-'~

/ 0.3

/

/"

0.2 / 0.4 0.5 0.6

0.8 1.0 0.2 0.3 0.4 0.5 0.6 Reduced temperature Tr = T/T c

0.8 10

Figure 2.4. Change of reduced viscosity as a function of reduced temperature and reduced pressure [O.A. Hougen, K.M. Watson and R.A. Ragatz, Chemical Process Principles Charts, 2nd ed., Wiley, New York (1960)].

Here/Xc is in/~P (micropoises), Pc in atm, Tc in K, and Vc in cm3/mol. Figure 2.4 can also be used to approximate viscosities of mixtures with the pseudocritical properties defined by

P~-

YiPci'

Tc=

i=1

yiTci'

,

I~c

i=l

Yil~ci

(2.22)

i=1

Here Yi is the mole fraction of species i in a mixture. We may use Figure 2.4 for fluid mixtures, with pseudocritical properties instead of critical properties. This procedure yields reasonable accuracy for mixtures of chemically similar substances. The following semiempirical relation also predicts the viscosity of a gas mixture within an average deviation of--~2%: #mix

-i=1

Yi E j Y; ~ij

(2.23)

where the dimensionless quantity ~0 is

(2.24)

62

2.

Transportand rate processes

Here n is the number of chemical species in a mixture, Yi the mole fraction of species i, ~i the viscosity of species i at the system temperature and pressure, and Mi the molecular weight of species i. Mainly, the dependence of viscosities on composition is nonlinear for mixtures of gases. Many additional empirical equations are available for estimating viscosities of gases and gas mixtures at low and high densities (Reid et al., 1987) as well as for liquids, suspensions, and emulsions (Bird et al., 2002).

Example 2.2 Estimation of viscosity at specified temperature and pressure Estimate the viscosity of carbon dioxide and hydrogen at T = 320 K and P = 36 atm. Solution: Assume pure species of carbon dioxide and hydrogen. Supply the critical parameters and estimate the reduced pressure and temperature to use in Figure 2.4 for reading the reduced viscosities approximately. Use Eq. (2.21) to estimate the critical viscosity: /'Lc-- 7"70M1/2 (Pc)2/3 (Tc )-1/6 Species

Tc (K)

Pc (atm)

Tr

Pr

Approximate (Figure 2.4)

/Xc (Eq. 2.21) (/xP)

0.5 2.75

343.42 33.35

/x --

~r~c

(/xP)

~r CO 2

H2

304.2 33.3

72.8 12.8

1.05 9.61

0.49 2.81

171.71 91.71

At the same temperature and pressure, the viscosity of carbon dioxide is considerably greater than the viscosity of hydrogen. No data are available for comparison.

Example 2.3 Estimation of viscosity of gas mixtures at low density Estimate the viscosity of the following gas mixture at 293 K and 1 atm using the data given in the following table: Species

Yi

M

Air Carbon dioxide

0.5 0.5

28.97 44.01

/x(mPa s) 0.0181 0.0146

Solution: Assume pure species of carbon dioxide and hydrogen at low density. The viscosities for gas mixtures/-/'mix at low densities may be estimated from Eq. (2.23): /.Lmix __ £ i=1

Yi I'l'i E j Yj f~ij

where x i is the mole fraction of species i and/.Li the viscosity of the pure species i. The coefficients ~0 are obtained from Eq. (2.24)" I

j

Mi/Mj

Mj/Mi

~.ti/i~j

*ij

EI.=lYJf~ij

1

1 2 1 2

1 0.658 1.519 1

1 1.519 0.658 1

1 1.2397 0.8066 1

1 0.656 0.383 1

0.828

2

/'Lmix ___£ i=1

0.691

Yit.Li __ 0.5(0.0181) + 0.5(0.0146) = 0.0109 + 0.0105 = 0.0214 mPas E j YJdPiJ 0.828 0.691

No data are available for comparison.

2.4

2.4.5

63

Transportphenomena

Estimation of Viscosity of Pure Liquids

The kinetic theory of gases is far more advanced than that of liquids partly because of complex interactions among the molecules of liquids. We may estimate the viscosity of a pure liquid from the following relation based on the Eyring rate theory:

Nhexp(3"8Tb] /x - --~--

(2.25)

T

where N is Avogadro's number, h the Planck's constant (=6.62 × 10-27 erg s), V the molar volume, and Tb the boiling point of liquid at 1 atm. Using the available nomograph to estimate the liquid viscosity (Bird et al., 2002; Griskey, 2002) is also recommended.

2.4.6

Heat Transfer

Consider a solid slab of area A located between two large parallel plates that are a distance H apart, as shown in Figure 2.3b. Initially, the solid material is at a temperature To. At time t = 0, the lower plate is brought to a higher temperature T1, and maintained at this temperature. A linear steady-state temperature distribution is developed based on a constant rate of heat flow q through the slab. For a small temperature difference AT = T1 - To, the heat flow is proportional to the temperature decrease over distance H. When the slab thickness approaches zero, the onedimensional form of Fourier's law relates the heat flow and temperature gradient

dT

q,, - - k ~

(2.26)

dy

where k is the thermal conductivity, which is assumed to be independent of direction. If the temperature varies in all three directions, we get the three-dimensional form of Fourier's law in vector form

q = -kVT

(2.27)

This equation is applicable to an isotropic medium only, so the heat is conducted with the same thermal conductivity k in all directions. The thermal conductivity k is a property of a conducting medium, and is mainly a function of temperature. High pressure affects the thermal conductivity in a gas medium. For a multicomponent system of n components, heat flow is

q - - k V T + ~ H i J i + qD

(2.28)

i=l

where H i is the partial molar enthalpy of species i, Ji the molar mass flow vector, and qD the heat flow vector induced by the concentration gradient (Dufour effect).

2.4.7

Combined Energy Flow

When there is a convective heat flow effect, we may define a combined energy flow e by e =

]

2 pv2 + pH v + [~-'v] + q

(2.29)

The combined energy flow vector has three contributions: the convective energy flow, the rate of work done by molecular mechanisms, and the rate of heat flow by molecular mechanisms. After combining Eqs. (2.28) and (2.29), we have

o

v. i=1

Since the contributions [(½)pp2]V and [~-"v] are usually small, Eq. (2.30) reduces to

e - - k V T + ~_, HiJ i + PHiV+qD i=l

(2.31)

64

2.

Transport and rate processes

Table 2.3 Units of quantities related to heat flow Quantity qy k Cp ce

sI

c.g.s.

British

W/m 2 W/(m K) J/(kg K) m2/s

cal/(cm 2 s) cal/(cm s°C) cal/(g°C) cm2/s

Btu/(ft 2 h) Btu/(ft h °F) Btu/(lbm°F)

ft2/s

With the molar concentration, we have n

e = - k V T + ~ H i J i nti=1

ciHiv + qD

(2.32)

i=1

After using the total molar mass flow Ni, w e get

e = - k V T + ~ H i N i + qD

(2.33)

i=1

Equation (2.26) for heat conduction and Eq. (2.3) for momentum transfer are similar, and the flow is proportional to the negative of the gradient of a macroscopic variable; the coefficient of proportionality is a physical property characteristic of the medium and dependent on the temperature and pressure. In a three-dimensional transport, Eqs. (2.27) and (2.15) differ because the heat flow is a vector with three components, and the momentum flow ~"is a second-order tensor with nine components. 2.4.8

Thermal Diffusivity

Besides thermal conductivity, k, thermal diffusivity, c~, is also widely used, and is defined by k

(2.34)

Here Cp is the heat capacity at constant pressure. Thermal diffusivity has the same units as kinematic viscosity, and they play similar roles in the equations of change for momentum and energy. The dimensionless ratio P r - v _ Cptx c~ k

(2.35)

is the Prandtl number Pr, which shows the relative ease of molecular momentum and energy transport in the hydrodynamic and thermal boundary layers, respectively. The Prandtl number for gases is near unity, and hence the magnitudes of energy and momentum transfer by diffusion are comparable. For liquid metals, Pr << 1 and thermal diffusion is much greater than momentum diffusion; the opposite is true of liquids for which Pr >> 1. Table 2.3 shows the units that are commonly used for thermal conductivity and heat transport.

Example 2.4 Estimation of heat flow through a composite wall with constant thermal conductivities A pipe with an outside diameter of 10 cm and a length of 110 m is carrying hot fluid. The pipe is insulated with 0.5 cm thick silica foam and 10 cm thick fiberglass. The pipe wall is at 120°C and the outside surface of the fiberglass is at 30°C. Estimate the heat flow in the radial direction of the pipe. The thermal conductivities of silica foam and fiberglass are 0.055 and 0.0485 W/(m K), respectively. Solution: Assume radial heat flow only, and the thermal conductivities of the insulation layers are constant. For heat flow in the radial direction of a pipe only, Eq. (2.26) is

dT qr = -kA ~ dr

(2.36)

2.4

Transportphenomena

65

// q

q

q

r4

Temperature profile

Temperature profile (a)

(b)

Figure 2.5. Insulation layers around (a) a pipe and (b) a slab.

where A (=27rrL) is the surface area of the pipe with a length L normal to the heat flow qr in the radial direction. Inserting the area relation and integrating Eq. (2.36) between an inner radius, ri, and outer radius, r o, we find (2.37)

qr = 27rkL(Ti - T° )

ln(ro/~) If we have two layers of insulation with thermal conductivities of ka and kb around a pipe (Figure 2.5), then Eq. (2.37) becomes qr =

2 7rL(T i - T o ) overall

(2.38)

ln((r2/t] )/k a ) -~ ln((r3/r2)/k b )

Heat flow for a composite slab (Figure 2.5) is

qr

(Ti -- To ) overall ( ~ ( a / k a M) + (~¥'b/kb N) -t- (z~kXc/kcA)

(2.39)

Equation (2.38) is used in the solution of the problem qr =

2rr(110)(120- 30) l n ( ( 0 . 0 5 5 / 0 . 0 5 ) / 0 . 0 5 5 ) + ln((O. 105/0.055)/0.0485)

= 4126.8W

Another example of one-dimensional heat flow is the radial heat flow through the wall of a hollow sphere. Starting with Eq. (2.36) dT

qr = - k A

dr

and the area normal to the radial heat flow ofA = 47rr 2, we have qr -- - k 4 7 r r

2 dT

(2.40)

dr

By integrating this equation with the boundary conditions: r-

ri,

r-ro,

T-

Ti

T-To

we get qr = - 4 rrk

~-Vo (1/ri ) + (1/ro)

(2.41)

66

2.

Transport and rate processes

For one-dimensional steady-state heat conduction in the x-direction, we have a general relation for the temperature profile obtained from (V2T = 0) -- xs dx

= 0

(2.42)

where s is the shape factor: s = 0 for a rectangular shape, 1 for a cylindrical shape, and 2 for a spherical shape. Using the boundary conditions in Eq. (2.42), the temperature profiles are Rectangular shape T(x)-T2-Tlx+T L

1,

a t x = 0 , T = T 1 and x = L , T = T

(2.43)

2

Cylindrical shape Ti . - .T° l n r ,

T ( r ) . T i.

ln(ro/r~)

rl

a t r = r i , T = T i and r = r o , T = T O

(2.44)

Spherical shape

T ( r ) = Ti -

(2.45)

a t r = r i , T = T i and r = r o , T = T o

1/ri _ 1/ro

The temperature profiles are linear for a rectangular shape, logarithmic for a cylindrical shape, and hyperbolic for a spherical shape. The thermal resistances for various shapes in estimating the one-dimensional heat flow with constant thermal conductivity are Geometry Thermal resistance

Rectangular

Cylindrical

Spherical

L/Ak

ln( (ro/ri)/27rLk)

(ro/ri)/47rrirok

Example 2.5 Estimation of heat flow with temperature-dependent thermal conductivity The temperatures at the surfaces of a 0.2 ft thick rectangular box are 40°F and 120°E The box is filled with air. The thermal conductivity of the air is a linear ftmction of temperature: k = k0(1 + aT) with k = 0.0140 Btu/(ft h °F) at T = 32°F and k = 0.0183 Btu/(ft h °F) at T = 212°E Estimate the heat flow and the temperature profile in the air when the resistances at walls are negligible. Solution: Assume that the resistances at the walls are negligible, and the system is at steady state and has one-dimensional heat flow. k = 0.0140Btu/(fth°F)

atT = 32°F

k = 0.0183Btu/(fth°F)

atT = 212°F

Using the k values above, we find the parameters: k0 = 0.0134 and a = 0.001719, which are valid between 32 and 212°E The heat flow is q A

-k -

dT

dx

---,

q ix=2 A x=O d x

:

-k

f T2=120°F o a T~=32°F (1 + a T ) d T :

-k o

I

a T2

T +-~

ll20Of

:-6.669Btu/(ft 132°F

2 h)

2.4

Transportphenomena

67

The temperature profile is obtained from

q - - k dT ~ - -qx = - k o ( T + a2T) A dx A 2 D

q _ _ k ( T 1-T2) A L where k is an average value for the thermal conductivity estimated as 120°F

T2 - T 1

r 2 -1"1

2

= 0.0151Btu/(ft h °F) 132°F

Using the value of k in the heat flow equation above, we determine the temperature profile

aT2 + T + 0.0151(T1 - T2) 7x = 0 2 This quadratic equation will have two solutions and the solution with the positive sign must be chosen to satisfy the boundary conditions - 1 + x/1 - 4(a/2)[0.015 I(T1 - T2)x/L] T

2.4.9

m

Estimation of Thermal Conductivity

Thermal conductivity can vary from - 0 . 0 1 W/(m K) for gases to --~1000 W/(mK) for pure metals. Tables 2.4-2.6 show some experimental values of thermal conductivities. When available, experimental values should be used in calculations; otherwise, several empirical relations may provide satisfactory predictions. Table 2.4 Thermal conductivities, heat capacities, and Prandtl numbers of some gases and liquids Substance Gases at 1 atm pressure Hydrogen, H2 Oxygen, 02

Carbon dioxide, CO~_ Methane, CH 4 NO

r (K)

k (W/(m K))

Cp (kJ/(kg K))

Pr

100 300 100 300 200 300 200 300 200 300

0.06799 0.1779 0.00904 0.02657 0.0095 0.01665 0.02184 0.03427 0.01778 0.02590

11.192 14.316 0.910 0.920 0.734 0.846 2.087 2.227 1.015 0.997

0.682 0.720 0.764 0.716 0.783 0.758 0.721 0.701 0.781 0.742

300 350 400 250 300 350 250 350 250 300 350 200 300

0.6089 0.6622 0.6848 0.1808 0.1676 0.1544 0.1092 0.0893 0.1478 0.1274 0.1071 0.1461 0.1153

Liquids at their saturation pressures Water, H20

Ethanol, C2HsOH

Carbon tetrachloride, CC14 Diethyl ether, (C2Hs)20

1-Pentene, CsH~0

4.183 4.193 4.262 2.120 2.454 2.984 0.8617 0.9518 2.197 2.379 2.721 1.948 1.907

6.02 2.35 1.35 35.8 15.2 8.67 16.0 5.13 5.68 4.13 3.53 8.26 3.72

Source: Bird et al. (2002); Data Compilation of Pure Compound Properties, Design Institute for Physical Property Data, AIChE, New York, NY (2000).

68

2.

Transport and rate processes

Table 2.5 Thermal conductivities, heat capacities, and Prandtl numbers of some liquid metals at atmospheric pressure

Metal

T (K)

k (W/(m K))

Cp (kJ/(kg K))

Pr

Mercury, Hg

273.2 372.2 977.2 366.2 644.2 422.2 700.2

8.20 10.50 15.1 86.2 72.8 45.2 39.3

0.140 0.137 0.0146 0.0138 0.0130 0.795 0.753

0.0288 0.0162 0.013 0.011 0.0051 0.0066 0.0034

Lead, Pb

Sodium, Na Potassium, K

Source: Bird et al. (2002); Liquid Metals Handbook, 2nd ed., U.S. Government Printing Office, Washington, DC (1952); E.R.G. Eckert and R.M. Drake, Jr., Heat and Mass Transfer, 2nd ed., McGraw-Hill, New York (1959).

Table 2.6 Thermal conductivities of some solids

Solid

T (K)

k (W/(m K))

Aluminum, A1

273.2 373.2 273.2 291.2 373.2 273.2 373.2 291.2 373.2 273.2 373.2

202.5 205.9 387.7 384.1 379.9 55.4 51.9 46.9 44.8 418.8 411.9

Copper, Cu

Cast iron, Fe Steel Silver

Source: Bird et al. (2002); Griskey (2002); Reactor Handbook, Vol. 2, Atomic Energy Commission AECD-3646, U.S. Government Printing Office, Washington, DC (1955).

2.4.10

Effect of Temperature and Pressure onThermal Conductivity

In the corresponding states approach, the reduced thermal conductivity, kr = k/kc, is plotted as a function of the reduced temperature and the reduced pressure, as shown in Figure 2.6, which is based on a limited amount of experimental data for monatomic substances, and may be used for rough estimates for polyatomic substances. Figure 2.6 shows that the thermal conductivity of a gas approaches a limiting function of T at --~1 atm pressure. Thermal conductivities of gases at low density increase with increasing temperature, while they decrease with increasing temperature for most liquids. This correlation may change in polar and associated liquids. For example, water exhibits a maximum in the curve of k versus T. The corresponding states provide a global view of the behavior of the thermal conductivity of fluids.

Example 2.6 Estimation of thermal conductivity at specified temperature and pressure Estimate the thermal conductivity of methane at T = 380 K and P = 30 atm. At 300 K and 1 atm, the thermal conductivity is

0.03427 W/(m K). Solution: Assume Figure 2.6 can be used for polyatomic gases. Use the critical parameters and estimate the reduced pressure and temperature to use from Figure 2.6 for reading the reduced thermal conductivity approximately. For methane Tc = 190.7 K and Pc = 45.8 atm. The reduced conditions are Tr = 300/190.7 = 1.57 and Pr = 1/45.8 = 0.022. Using these reduced properties in Figure 2.6, we read the approximate reduced thermal conductivity as kr = 0.5. Estimate the critical thermal conductivity kc = 0.03427/0.5 = 0.0684 W/(m K).

2.4

Transportphenomena

69

10 .

.

.

.

8

7

-40~.,,

3o b-

I

10~

3

~

......

¢o

!1 I \ \ , ' ~ ' - - - ~ ~ "

\ l.s rf

07 0.6

i '1 0 ~ ~ v

o.,

o.3

0.2

o.1 S 0.3

~~d~/q ~"

0.4

0.6

0.8 1.0

2

3

4

5

6 7 8 910

Reducedtemperature,T r = T I T c Figure 2.6. Change of reduced thermal conductivity with reduced temperature and reduced pressure for monatomic substances [O.A. Hougen, K.M. Watson and R.A. Ragatz, Chemical Process Principles Charts, 2nd ed., Wiley, New York (1960)].

At the specified temperature and pressure the reduced conditions are Tr _ 380 _ 2.0 and Pr - 30 - 0 . 6 5 190.7 45.8 With the reduced parameters, Figure 2.6 yields kr = 0.66 (approximately). Therefore, the predicted value is k = kckr = 0.0684(0.66) = 0.0451W/(m K). The predicted values for polyatomic gases may not be satisfactory as Figure 2.6 is based on a limited set of data from monatomic gases only.

2.4.11

Thermal Conductivity of Gases at Low Density

The thermal conductivities of dilute monatomic gases are well understood. The thermal conductivity of a dilute gas composed of rigid spheres of diameter d is expressed as k -

2 x/vTmnT 3vr

-

-

7rd2

C~

( m o n a t o m i c gas)

(2.46)

where m is the mass of a molecule and K the Boltzmann constant. Equation (2.46) predicts that k is independent of pressure, and the prediction is satisfactory up to --- 10 atm for most gases. The predicted temperature dependence is weak.

70

2.

Transport and rate processes

The Chapman-Enskog formula for monatomic gases at low density and temperature T produces better predictions, and is given by k = 1.9891 × 10 . 4 x/-T/M

O-2~-~k

(2.47)

where k is in cal/(cm s K), T in K, o- in A~, and the collision integral for thermal conductivity, Ok, is identical to that for viscosity, 1~,. The values of collision integrals are given for the Lennard-Jones intermolecular potential as a function of the dimensionless temperature KT/s in Table B2. A simple semiempirical equation for polyatomic gases at low densities is given by k=

+-~R

(2.48)

when Cp is in cal/(mol K), R in cal/(mol K), and ~ is the viscosity in g/(cm s). This equation is called the Euckenformula, and it can provide a simple method of estimating the Prandtl number for nonpolar polyatomic gases at low density P r - Cptx_

k

Cp Cp + (5/4)R

(2.49)

The thermal conductivities for gas mixtures kmix at low densities are

kmix --- ~ i:1

xiki ZjXj~ij

(2.50)

where x i is the mole fraction of species i and ki the thermal conductivities of the pure gases. The values of ~/j are identical to those appearing in Eq. (2.24) in the viscosity equation.

Example 2.7 Estimation of thermal conductivity of monatomic gases Estimate the thermal conductivity of helium at T = 400 K and P = 1 atm and compare with the experimental value of 4.41 cal/(cm s K) (W.K. Saxena, S.C. Saxena, J. Phys. DAppl. Phys., 1 (1968) 1341). Solution: Assume that the pressure is low. Use Eq. (2.47): k = 1.9891 × 10 .4 ,~-f/M or2[~k where k is in cal/(cm s K) and T in K. Read the Lennard-Jones parameters for helium from Table B 1 and B2" o" = 2.576, -e = 10.2 K, and M = 4.003 k At T = 400 K, we have TK/e = 400/10.2 = 39.21. From Table B2, we have the approximate value for collision integral, 12k = 0.673. From Eq. (2.47)" k = 1.9891 × 10 . 4

x/400/4.003 = 4.45×10 .4 cal/(cms K) 2.5762(0.673)

Comparing this result with the experimental value of 4.41 × 10-4 cal/(cm s K) yields a deviation of 0.9%, which is low.

2.4

71

Transportphenomena

Example 2.8 Estimation of thermal conductivity of polyatomic gases Estimate the thermal conductivity of NO at 200 and 300 K at low pressure. Use the data in the following table" T (K)

Cp (cal/(mol K))

200 300

7.283 7.154

Solution:

Assumptions" The pressure is low. Read the Lennard-Jones parameters for NO from Table B 1 and B2: o- = 3.47, -e __ 119 K, M = 30.01 k With these parameters, we calculate the values of TK/e and the collision integrals, and use them in the following equation (Eq. 2.20) for estimating the viscosity" /z = 2.6693x10 -5 ~ / M T _ 2.6693×10_ 5 ~/30.01(200) = 1"363X10- 4 g/(cms) O"2~-~ 3.472 (1.260) A semiempirical Eucken equation is used to estimate the thermal conductivity of polyatomic gas with R = 1.987 cal/(mol K). k = Cp +-~R

=(7.283+2.484) 1"363×10-4 = 0.0185 W/(m K) 30.01

The results are shown in the following table" T (K)

200 300

Cp,ex p (Table 2.4) (cal/(mol K))

TK/e(K)

7.283 7.154

1.68 2.52

~ - ~ 1.260 1.093

/~ X 104 (g/(cm s))

kest (W/(m K))

kexp(Table 2.4) (W/(m K))

Deviation (%)

1.363 1.924

0.0185 0.0258

0.01778 0.0259

4.0 0.0

Example 2.9 Estimation of thermal conductivity of gas mixtures at low density Estimate the thermal conductivity of the following gas mixture at 293 K and 1 atm using the data given in the following table: Species

Yi

M

k (W/(m K))

/x (mPa s)

Air Carbon dioxide

0.5 0.5

28.97 44.01

0.026 0.01601

0.0181 0.0146

Solution: Assume that the system is at low density. Thermal conductivities for gas mixtures kmi×at low densities may be estimated from Eq. (2.50)" kmi x -- ~ i=1

yiki EjYj*ij

where x i is the mole fraction of species i and k i the thermal conductivities of the pure gases. The coefficients ~ij are identical to those appearing in Eq. (2.24) in the viscosity equation

,

(,+ i 'J21 / /TM ]2

72

2.

Transport and rate processes

i

j

Mi/Mj

Mj/M i

t.zi/[Lj

1

1 2 1 2

1 0.658 1.519 1

1 1.519 0.658 1

1 1.2397 0.8066 1

2

~1 £jYj% Yiki

k m i x = "=

0.5(0.026) + 0.5(0.0160) 0.806

0.701

f~iJ" 1 0.613 0.403 1

22j=lYj~ij 0.806 0.701

= 0.0161+0.0114 = 0.0275 W/(mK)

No result is available for comparison at these conditions.

2.4.12

Estimation of Thermal Conductivity of Pure Liquids

Based on the theory of Bridgrnan, the following expression can be used to estimate the thermal conductivity of a pure liquid: N)

k - 2.80 7-

/3

KVs

(2.51)

where N is Avogadro's number, K the isothermal compressibility [K =-1/V(O V/OP)T], Vthe molar volume, and vs the sonic velocity defined by

/ Cp

OP

lj2 (2.52)

U V where p is the density and the quantity (OP/Op)Tmay be obtained from isothermal compressibility data.

Example 2.10 Estimation of thermal conductivity of pure liquids Estimate the thermal conductivity of water at 300 K and 1 atm. The density of liquid water at 300 K is 995.7kg/m 3, and its isothermal compressibility u = -1/p(Op/OP)T is 0.4477 × 10-9 m2/N (= pa-1). Solution: Assume that Cp = Cu for water. Estimate

OP _ 1 _1 = 2.24 × 106 m 2/S 2 Op p(1/p)(Op/OP) 995.7(0.4477 × 10-9) The speed of sound is

Cp OP Vs=-~v-~p r

= 1497.7 m/s

Using the Avogadro number, the Boltzmann constant, and the molar volume of water (M/p = 0.01807 m3/kg), we have k = 2.80(6"023 × 1026 kg mol 2/3 1.3805 × 10.23 J/K (1497.7 m/s) = 0.599 W/(m K) 0.01807 m3/kg The experimental value of k at 300 K is 0.608 W/(m K) (Table 2.4), and the deviation is --~1.4%.

2.4 Transportphenomena 2.4.13

73

Mass Transfer

We consider a thin, horizontal, fused-silica plate of area A and thickness H. Initially, both horizontal surfaces of the plate are in contact with air (Figure 2.3c). We assume that the air is completely soluble in silica. At time t - 0, the air below the plate is replaced by pure helium, which is appreciably soluble in silica. The helium slowly penetrates into the plate by molecular motion, and eventually appears in the air above the plate. This molecular transport of one substance relative to another is known as d![jFusion. The air above the plate is removed rapidly, so that there is no measurable helium concentration there. In this system, the index i shows helium andj shows silica, and the concentrations are given by the mass fractions w; and w/, respectively. Eventually, the concentration profile tends toward a straight line with increasing t, and we have w ; - w~0 at the bottom surface, and w i = 0 at the top surface of the plate. At steady-state diffusion, the molar flow vector that is the flow rate of helium per unit area j~ is proportional to the concentration gradient in terms of the mass fraction V w i (2.53)

Ji = - P D ! j V w i

where p and D~9 are the density and the diffusivity coefficient of the silica-helium system, respectively. Similarly, we relate the molar flow vector to the concentration gradient by (2.54)

Ji = - c D ! j V x i

Equations (2.53) and (2.54) are called Fick's first law of diffusion, and indicate that mass flows from a high to a low concentration region. It is valid for any binary fluid or solid solution, provided that j; is defined as the mass flow relative to the mixture mass average velocity v defined by v-

1 P ~_~piv i = ~ . w , v i i

(2.55)

i

and in general we have

(2.56)

Ji - - P W i (Vi -- V)

Here the term ( v i - v) is called the diffusion velocity. The mass flow ]j is defined analogously. As the two chemical species interdiffuse, there is a shifting of the center of mass in the y-direction if the molecular weights of components i andj differ. The flows Ji and jj are measured with respect to the motion of the center of mass, and Ji + J j - 0. Molar average velocity is 1

VM

--'--ECiVi --EXiVi C

t•

(2.57)

i

and in general we have Ji

-- Ci(Vi

(2.58)

--VM)

We can use the molecular mass flow vector j; and the convective mass flow vector pv to define the combined mass flow vector ni, and combined molar flow vector Ni n; - ji + piv

(2.59)

Ni = Ji + civ

(2.60)

Equation (2.59) is the mass flow in terms of mass average velocity v, and using the diffusion velocity (v;- v), we obtain Ji - - P i ( V i

- - V) - - n i - pi v - - - p D ( / V w

Ji = ci(vi - VM) = Ni - civ = - c D i j V x i

For pair i-j there is just one diffusivity coefficient D!i = D/;.

i

(2.61)

(2.62)

74

2.

Transportand rateprocesses

Other type of mass flow is the one that is relative to the plane of no net volume flow. When JV,A and JV,B are the vectorial molar mass flows of species A and B relative to the plane of no net volume flow, we have

JV,AVA + Jv,BVB = 0

(2.63)

where VAand VB are the partial molar volumes of species A and B in the mixture. The mass flows are

vB

JV,A----~-JA

and

VA

JV,B=--V--JB=0

(2.64)

where V is the molar volume of the mixture. When the partial molar volumes are equal to each other, then JV, A = JA"

Example 2.11 Mass flow across a stagnant film Consider that a liquid A in a tube with a large diameter is evaporating into a stagnant gas B. Derive the relations for mass flow and the concentration profile. Assume that the liquid level is maintained at y = Yl. Solution: Assume that at the liquid-gas interface, the concentration of A is XA1, which is the gas-phase concentration of component A corresponding to equilibrium with the liquid at the interface. The mole fraction XA1 is the vapor pressure of A divided by the total pressure provided that A and B form an ideal gas mixture and that the solubility of gas B in liquid A is negligible. A stream of gas mixture A-B of concentration XA2 flows slowly past the top of the tube, to maintain the mole fraction of A at XA2.The entire system is kept at constant temperature and pressure. There is a net flow of gas upward from the gas-liquid interface. The transport process is in the y-direction and at steady state with no convective mass transfer, and the reaction source is

-- VNAy =

0

or

dNAy - 0 dy

(2.65)

For a molecular mass transport of component A, we have the mass flow in the y-direction

NAy=JAy+XANAy or NAy--

Jay

(2.66)

1 m XA

Substituting Eq. (2.66) in Eq. (2.65) and using Fick's law of diffusion, we find

d(CDAB dXA) -- 0 dy 1 - x A dy

(2.67)

For an ideal gas mixture c = P/RT, so that at constant temperature and pressure c must be constant. Furthermore, for a binary gas system, DAB is very nearly independent of the composition. Therefore, the product cDAB is constant, and Eq. (2.67) reduces to

d( 1

dXA) = 0 dy 1--x A dy

(2.68)

This is a second-order differential equation for the concentration profile expressed in terms of mole fraction of component A. With the boundary conditions Y = Y l , XA -- XA1 Y = Y2, XA = XA2 integration of Eq. (2.68) yields

--ln(1--XA)=IlY+I 2

(2.69)

2.4

75

Transportphenomena

After determining the integration constants Il and I2, the final solution becomes

2.4.14

1--XA

1-- XA2 exp( Y -- Y] )

I--XA1

1--XA1

(2.70)

Y2 --Yl

Estimation of Diffusivities

In general, diffusivity depends on pressure, temperature, and composition. With respect to the mobility of molecules, the diffusion coefficients are generally higher for gases and lower for solids. The diffusivities of gases at low densities are almost independent of concentration, increase with temperature, and vary inversely with pressure. Liquid and solid diffusivities are strongly concentration dependent and generally increase with temperature. Tables 2.7 and 2.8 show some of the experimental binary diffusivities for gas and liquid systems.

Table 2.7 Molecular diffusivities of some gases at atmospheric pressure

Gas pair

(i-j)

Air-ammonia Air-carbon dioxide Air-ethanol Air-n-octane Air-water Air-chlorine Air-benzene Air-naphthalene Air-hydrogen Carbon dioxide-benzene Carbon dioxide-ethanol Carbon dioxide-methanol Carbon dioxide-water

Carbon dioxide-carbon monoxide Hydrogen-ammonia

Hydrogen-benzene Hydrogen-ethane Hydrogen-water Oxygen-ammonia Oxygen-benzene Oxygen-ethylene Ammonia-hydrogen

Propane-iso-butane Propane-n-butane

T (K)

D~/(cm2/s)

273 273 317.3 298 313 298 298 313 273 298 298 298 318 273 298.6 298 307.2 328.6 273.2 273 293 358 273 311.3 273 293 293 296 311.3 293 263 298 298 378.2 378.2 437.7

0.198 0.136 0.177 0.132 0.145 0.0602 0.260 0.288 0.124 0.0962 0.0611 0.410 0.0715 0.0693 0.105 0.164 0.198 0.257 0.139 0.745 0.849 1.093 0.317 0.404 0.439 0.850 0.253 0.094 0.101 0.182 0.58 0.233 0.0439 0.0823 0.0768 0.107

Source: Bird et al. (2002); Griskey (2002); J.O. Hirschfelder, C.E Curtiss, and R.B. Bird, Molecular Theo~ of Gases and Liquids, 2rid ed., Wiley, New York (1964); Reid et al. (1987); S. Gotoh, M. Manner, J.P. Sorensen and WE. Stewart, J Chem. Eng. Data, 19 (1974) 169; Data Compilationof Pure CompoundProperties,Design Institute for Physical Property Data, AIChE, New York, NY (2000).

76 ..

2.

Transport and rate processes

Table 2.8 Diffusivities in some liquid mixtures Liquid pair; solute A (concentration in g mol/1)-solute B

T (K)

DAB × 105 (cm2/s)

Ammonia (3.5)-water Ammonia (1.0)-water Ethanol (3.75)-water Ethanol (2.0)-water Ethanol (0.05)-water Chloroform (2.0)-ethanol Sodium chloride (0.05)-water Sodium chloride (0.2)-water Sodium chloride (1.0)-water

278 288 283 289 283 293 291 291 291

1.24 1.77 0.50 0.90 0.83 1.25 1.26 1.21 1.24

A-B

T (K)

XA

DAB × 105 (cm2/s)

Chlorobenzene-bromobenzene

283.3 283.3 283.3 283.3 313.1 313.1

0.0332 0.2642 0.5122 0.7617 0.0332 0.2642

1.007 1.069 1.146 1.226 1.584 1.691

Water-n-butanol

303.2 303.2 303.2

0.131 0.222 0.358

1.240 0.920 0.560

298.15 298.15 298.15 298.15 298.15

0.026 0.266 0.408 0.680 0.880

1.076 0.368 0.405 0.743 1.047

Ethanol-water

Source: P.A. Johnson and A.L. Babb, Chem. Rev., 56 (1956) 387; P.W.M. Rutten, Diffusion in Liquids, Delft University Press, Delft, The Netherlands (1992); A. Vignes, I.E.C. Fundam., 5 (1966) 189; M.T. Tyn and W.E Calus, J. Chem. Eng. Data, 20 (1975) 310.

The molecular theory yields the self-diffusivity of component i at low density

Dii -

2 4'n'miKT 1 3rr

~

-

-

,rrd2

(2.71)

p

Equation (2.71) can be compared with Eq. (2.46) for the thermal conductivity of gases, and with Eq. (2.19) for the viscosity. For binary gas mixtures at low pressure, D O• is inversely proportional to the pressure, increases with increasing temperature, and is almost independent of the composition for a given gas pair. For an ideal gas law P = c R T , and the Chapman-Enskog kinetic theory yields the binary diffusivity for systems at low density

D 0. = 0.0018583 ~/T3 (1/Mi + 1~My )

(2.72)

2 Po'ijaDij

where 1

O"U = "~(0" i + O ' j ) ,

~

~

K

=

Si

--

,

12ij = f

KT

K

Here D,y in cm2/s, o-0.in A, T in K, and P in atm. The dimensionless quantity DDij is the collision integral for diffusion, and is a function of the dimensionless temperature KT/eo..The parameters o-• and e~ are those appearing in the LennardJones potential between molecules i andj.

2.4

77

Transportphenomena

Example 2.12 Estimation of diffusivity in a gas mixture at low density Estimate the diffusivity of benzene in air at 25°C and 1 atm. Solution: Assume that the system pressure is low. Use the Chapman-Enskog equation for the binary diffusivity at low density (Eq. 2.72):

Oij

=0.0018583

~/T3 (1/A4i -+-l / J~,[j) pcr2 f~~.:

The data and critical properties are

Species Benzene (A) Air (B)

M

(2-(A)

e/K (K)

Tc(K)

Pc (atm)

78.11 28.97

5.443 3.617

387 97

562.6 132

48.6 36.4

CrAB = ~(CrA + crB) = 4.53A, eAB = K KT

298

eAB

193.7

= 193.7

-1.538

The approximate collision of integral f~0 =flKT/eij)= 1.187 DAB = 0.0018583 X/2983(1/28"97 + 1/78.11i = 0.0854 cm2/s (1)(4.53)21.187 From Table 2.7, experimental diffusivity is = 0.0962 cm2/s, and deviation is --- 11.2%.

2.4.15

Effect of Temperature and Pressure on Diffusivity

At low pressures, we use the following expression developed from a combination of kinetic theory and corresponding states approach to estimate the effect of temperature on diffusivity:

T D~/=a

, /

1 + 1 MC ~

(PciPcj)I/3 (TciTcj ) 5/12 P

(2.73)

Here D~j is in cm2/s, P in atm, and T in K, and the constants a = 2.745 × 10-4 and b - 1.823 for nonpolar gas pairs, excluding helium and hydrogen, and a = 3.64 × 10-4 and b = 2.334 for pairs consisting of water and a nonpolar gas. Equation (2.73) predicts the data at atmospheric pressure within an average deviation of less than 10%. A corresponding states plot of the self-diffusivity DAA,, which is the interdiffusion of labeled molecules of A at the low-pressure limit, is shown in Figure 2.7. The reduced self-diffusivity that is CDAA, at pressure P and T divided by CDAA,at the critical point is plotted as a function of the reduced pressure and reduced temperature. Figure 2.7 shows that CDAA,increases sharply with increasing temperature, especially for liquids. The values of CDAA, decrease toward a low-pressure limit at each temperature. We can use the following empirical relation for estimating the critical self-diffusivity between i and labeled species i* ( cDii. )c:

)1/2

(cD.,)c

- 2.96 X IO-6 ~

Mi

+~

Mi *

Pc2/3

Tic/6

(2.74)

78

2.

Transportand rate processes

/ /

-¢

2

1.5

II ~o

~ 1.0

L~ •~. 0.8

~ £~ow-pressure

~

limit

....

~"pr

/y;t7

Vapor / , ~ 1 ~ ~

= 10

p'~pr=5

"" .

.

.

.

0.6

"~

Two- I phase I 0.4 region#

/

t I

/Sat-u'rated liquid 0.2

i t

0.6

6.8 1.0 1.5 2 3 Reduced temperature, T r = T / T c

4

5

Figure 2.7. Change of reduced self-diffusivity with reduced temperature and reduced pressure [J.J. van koef and E.G.D. Cohen, Physica A, 156 (1989) 522; B.I. Lee and M.G. Kesler, AlChE J., 23 (1975) 510].

where c is in mol/cm 3, Dii in cm2/s, Tc in K, and Pc in atm. For binary diffusion of chemically dissimilar species at low pressure, we use (cDij)c -- 2.96 X 10 -6

~

+~

Mi

Mj.

)1/2(PciPcj)1/3 (TciTcj )1/12

(2.75)

Here the c-multiplied diffusion coefficients are used because their dependence on pressure and temperature is simpler, and they are frequently used in mass transfer calculations. We may calculate the diffusion coefficient at a specified temperature and a specified pressure from a known value by using the following relation:

a~ ( T2 ' P2 ) -- Dij ( TI ' P1) ~

~-Tll)

~-~D(T2)

For pressures below 25 atm and away from critical values, Eq. (2.76) yields reliable corrections. If gas species 1 is diffusing through a mixture of known composition, then the diffusion coefficient obtained by

(2.76)

Ol_mixture is

1

Ol-mixture

(Y2/D12)+(Y3/D13)+...+(Yn/Dln)

(2.77)

where Y2 is the mole fraction of species i in the mixture estimated on a component-i-free basis, and obtained from * _

Y

2

Y2 -- Zj=z yj

=

Y2

Y2 + Y3 + " "

+ Yn

(2.78)

2.4 Transportphenomena

79

Example 2.13 Estimation of diffusivity in a gas mixture at low pressure Estimate the diffusivity of carbon dioxide in benzene at 318 K and 1 atm. Solution: Assume that the pressure is low. Use Eq. (2.73) to find the binary diffusivity at low pressure:

~-'I'-~

Dij-a x/TcTT~/

M,

1

1/2

)1/3

(PciPcj (TciTcj ) 5/12

M,

P

The critical properties are Species Carbon dioxide (A) Benzene (B)

M

Tc (K)

Pc (atm)

44.01 78.11

304.2 562.6

72.8 48.6

1/2

(TcATcB)5/12=

151.55, ( 1 • ~,M A

1 MB

= 0.188,

( P c a e c B ) 1/3 - -

15.2

with a = 2.745 x 10 -4 and b = 1.823 for nonpolar gas mixture a (T/4TciTc/)

b=

2 . 6 1 6 × 1 0 -4 .

With P = 1 atm DAB = (2.616 × l0 --4 )(0.188)(15.23)(151.55) = 0.0739cm2/s Experimental diffusivity from Table 2.7 is DAB = 0.0715 cmZ/s, and deviation is 3.44%.

Example 2.14 Estimation of diffusivity in a gas mixture of isotopes Estimate the diffusivity of C 1 4 O in ordinary CO at 225 K and 172.5 atm. The measured value of DAA, -- 0.109 cm2/s at T = 194.7 K and P = 1 atm. Solution:

Assumptions- Assume that the critical value (DAA,)c obtained at P = 1 atm pressure can be used. For CO the critical properties are Tc = 133 K and Pc = 34.5 atm. As we have a measured value, use Figure 2.7 with the reduced temperature Tr = 194.7/133 = 1.46 and the reduced pressure Pr = 1/34.5 = 0.116 to read the approximate value of (CDAA,)r = 1.4. The value of concentration is c = critical value is

(CDAA*)C =

P/RT =

CDAA (CDAA*)r

6.25 x 10-5 mol/cm 3 (R = 82.05 atm cm3/(mol K)). Therefore, the

_ (6.25×10-5)(0.109)

= 4 . 7 6 3 × 1 0 .6 mol/(cms)

1.43

At the required reduced temperature and pressure" Tr = 225/133 - 1.7 and Pr = 172.5/34.5 = 5. If the approximate reduced value o f ( C D A # ) r - 1.5 mol/(cm s); the predicted value is CDAA* = (CDAa*)c (cDaA)r- (4.763 × 10-6)(1.5)= 7.145 × 10 -6 cm2/s The predictions may not be satisfactory as the critical value is estimated at a low pressure of 1 atm. The critical value may also obtained from Eq. (2.74)

(cDii*)c

= 2.96 X 10 -6

/

1

Jr" 1

Mi Mi*

/¸j2

pc~/3 _ 2.96 × 10 -6 ~ - J r Tl/6c, 28.01

~

3(}.01

1/2 34.52/3 - 3 . 6 4 8 × 1 0 -6 mol/(cms)

1331/6

80

2.

Transport and rate processes

Therefore, the predicted value is CDAA* -- (CDAA*)c

(cDAA*)r-- (3.648

× 10-6)(1.5) -- 5.472 × 10 -6 mol/(cms)

Example 2.15 Estimation of diffusivity in a gas mixture Estimate the diffusivity of an air-carbon dioxide mixture at 200 K and 103 atm. The measured value of DAB = 0.177 cm2/s at T = 317.3 K and P = 1 atm. Solution: Assume that the critical value (DAA.)C obtained at P = 1 atm pressure can be used. The critical properties are Species Carbon dioxide (A) Air (B)

M

Tc (K)

Pc (atm)

44.01 28.97

304.2 132

72.8 36.4

As we have a measured value, use Figure 2.7 with the reduced temperature T 317.3 Tr = ~/TcATc------~ = 4304,2(132 ) = 1.58 and the reduced pressure P

1 Pr = 4PcAPc------~ = 436.4(72.9 ) = 0.019 We can use Figure 2.7 to read the approximate reduced value of (CDAB)r = 1.65. The value of concentration: c = P/RT= 3.84 × 10-Smol/cm 3 (R = 82.05 atmcm3/(mol K)). Therefore, the critical value is CDAA*

(CDAA*)c =

_ (3.84 ×10-5)(0.177)

(CDAA*)r

= 4.119 × 10 -6 mol/(cm s)

1.65

At the required reduced temperature and pressure

T~

200 x/304.2(132 ) = 1.0, P~

103 ,]36.4(72.9) = 2.0

From Figure 2.7 we read the approximate reduced value of (CDAB)r = 0.87 mol/(cm s). The predicted value is CDAB -- (CDAB)c (CDAB)r -- (4.110 × 10-6)(0.87) -- 3.570 X 10 -6 cm2/s

The predictions may not be satisfactory as the critical value is estimated at a low pressure of 1 atm. The critical value may also obtained from Eq. (2.74) / 1/2

(°DAB)c = 2.96 X 10 -6 _ ~ 1 nt" _ ~ 1 MA

MB

)1/3

(PcA PcB

= 4.05 x 10 -6 mol/(cm s)

(TcATcB) 1/12

Therefore, the predicted value is

CDAB=

(CDAB)c (CDAB)r = (4.05 X 10-6)(0.87) = 3.52 X 10 -6 mol/(cm s)

2.4

81

Transportphenomena

Example 2.16 Estimation of diffusivity of a component through a gas mixture Estimate the diffusivity of carbon dioxide (CO2) through a gas mixture of benzene and methane with the known mole fractions given in the following table. The mixture is at 300 K and 2 atm. Welty et al. (1984)

Yi

T (K)

0.15

318

0.55 0.30

318 273

Species Carbon dioxide (1) Benzene (2) Hydrogen (3)

D1j (cm2/s)

P (atm)

0.0715 0.550

1 1

Solution:

Assume that Eq. (2.76) is used in temperature and pressure corrections of the experimental diffusivities by ignoring the collision integral correction since the temperatures are close to each other:

Dij ( T2 ' P2 ) = Dij ( TI ' P1) -~2

--~1)

Therefore, we have 3 0 0 ~ 3/2

D12 (300 K,2 atm) - 0.0715

D13(300K,2atm ) - 0 . 5 5

(,)(3oo] ~ ~,~]

= 0.0327 cm2/s

-0.316cmZ/s

We need to calculate the compositions of benzene and hydrogen on a carbon dioxide-free basis:

Ybenzene

=(

)

0.55 - 0.647 1 -- 0.15

and

Yhydrogen =

/ 0.3 ) 1 -- 0.15

= 0.353

Using these estimations in Eq. (2.77), we have Dl_mixtur e --

= 0.323 cm 2/s (0.647/0.0327) + (0.353/0.316)

No experimental values are available for comparison.

2.4.16

Diffusion in Liquids

Even the binary system diffusivities in liquid mixtures are composition dependent. Therefore, in multicomponent liquid mixtures with n components, predictions of the diffusion coefficients relating flows to concentration gradients are empirical. The diffusion coefficient of dilute species i in a multicomponent liquid mixture, Dim , may be estimated by Perkins and Geankoplis equation 1

Dim -

(2.79)

0.8 ~ , x i DijlX °8 (i 4=j ) /J'm .j=l

where XJ is the mole fraction of species j, Dim the effective diffusion coefficient for a dilute species i in the mixture in cmZ/s, D O.the infinite dilution binary diffusion coefficient of species i in species j, and ~m and/XJ the viscosities of mixture and pure species j in cP, respectively. A modified version of the Vignes equation may be used to represent the composition effect on the liquid diffusion coefficient I~

D!j = _ [ ( D ° #

0

j)x/+(D./i/~i

)x i

]

(2.80i

82

2.

Transport and rate processes

where D ° is the infinite diffusion of species i in solventj and F the thermodynamic factor

FI'naAI l'naB 0 In XA r,P

(2.81)

0 In x B I T,P

where a is the activity of species A. The value of F is close to unity for gases away from the critical conditions, while it is a necessary correction for liquids. For diffusion in liquids, we mainly rely on empirical expressions. For example, the Wilke-Chang equation predicts the diffusivity for dilute mixtures of species of i inj by

DO.=7.4×10 -8 TX/~~.

(2.82)

Here V~is the molar volume of the solute i in cm3/(mol) at its normal boiling point (for some values of V~,see Tables 2.9a and b),/z the viscosity of solvent in/xP, q9 an association parameter for solvent, and T the absolute temperature in K. Recommended values of the association parameters are 2.6 for water, 1.9 for methanol, 1.5 for ethanol, and 1.0 for benzene, ether, heptane, and other unassociated solvents. Equation (2.82) should be used only for dilute nondissociating solutions. Table 2.9a Atomic volumes at the normal boiling point Element

Atomic volume (cm3/(mol))

C H O (except as below) O in aldehydes, ketones O in methyl esters O in methyl ethers O in ethyl ethers O in ethyl esters O in higher esters O in higher ethers

14.8 3.7 7.4 7.4 9.1 9.9 9.9 9.9 11.0 11.0

O in acids (-OH) O joined to S, P, N N doubly bonded N in primary amines N in secondary amines

12.0 8.3 15.6 10.5 12.0

Element

Atomic volume (cm3/(mol))

Br C1 C1 in RCHC1R C1 in RC1 (terminal)

27.0 21.6 24.6 21.6

F I S P Ring, three-membered as in ethylene oxide Ring, four-membered Ring, five-membered Ring, six-membered Naphthalene ring Anthracene ring C1 in RCHC1R

8.7 37.0 25.6 27 -6 -8.5 -11.5 -15 -30 -47.5 24.6

Source: G.L. Bas, The Molecular Volumesof Liquid Chemical Compounds, David McKay Co., New York (1915).

Table 2.9b Molar volumes at the normal boiling point Compound

Hydrogen, H 2 Oxygen, 0 2

Nitrogen, N2 Bromine, Br2 Chlorine, C12 Iodine, I2 Carbon monoxide, CO Carbon dioxide, COz Carbonyl sulfide, COS

Molar volume (cm3/(mol)) 14.3 25.6 31.2 53.2 48.4 71.5 30.7 34.0 51.5

Compound

Sulfur dioxide, SO 2 Nitric oxide, NO Nitrous oxide, N20 Ammonia, NH3 Water, H20 Hydrogen sulfide, HzS Air

Molar volume (cm3/(mol)) 44.8 23.6 36.4 25.8 18.9 32.9 29.9

Source: G.L. Bas, The Molecular VolumesofLiquid ChemicalCompounds,David McKay Co., New York (1915).

2.4

83

Transportphenomena

The Stokes-Einstein equation can also be used to estimate the diffusivity of binary liquid mixtures

Dii = ~

KT

(2.83)

4~Tr/xj

where r/. is the radius of diffusing species i and ~j the viscosity of solventj. Another relation is by Tyn and Calus D//= 8.93 × 10 -8

V/

(Vo-) i

T

(2.84)

where Vs.is the molar volume of solvent at the normal boiling point in cm3/mol, and o/and crj the surface tensions in dyn/cm = 10.3 N/m 2. This relation should not be used for diffusion in viscous solvents for which the viscosities are above 20-30 cP. Binary and multicomponent diffusions are different in nature; in binary diffusion flow, species i is always proportional to the negative of the concentration gradient of species i. In multicomponent diffusion, however, other situations may arise: (i) in reverse diffusion, a species diffuses in a direction opposite to the direction imposed by its own concentration gradient; (ii) in osmotic diffusion, a species diffuses although its concentration gradient is zero, and (iii) in the case of a diffusion barrier, a species does not diffuse even though its concentration gradient is nonzero. The Wilke-Chang equation may be modified for a mixed solvent case

Dim =

7.4 ×10 -8 Tx/-qtM

(2.85)

~mV/0"6

where 0M = ~ x/0j Mj (i 4=j) /=1

Example 2.17 Estimation of diffusivity in a dilute liquid mixture Estimate the diffusivity of ethanol in a dilute solution of ethanol-water at 25°C. The density of ethanol is 0.79 g/cm 3 and the viscosity of water at 25°C is 0.95 cP. Solution: Assume that the mixture is a dilute solution. Use the Wilke-Chang equation:

Dii = 7.4×10_8 T ~ j M / •

~ V/0.6

The molar volume of ethanol at its boiling point can be estimated from the Rackett equation

v = VcZ~1-r~>')~

(2.863

The critical properties for ethanol are (Table B 1) Tc = 513.9K, Vc = 167cm3/(mol), Tb = 351.4K(boilingatlatm), and Tr = Tb/Tc =0.6838 From the Rackett equation we have V = 59.8 cm3/(mol). The association parameter for solvent water is 4; = 2.6. Using the Wilke-Chang equation, we find D e w - 7.4× 10-8 [ 298x/216(18) ] 0.95(59.8)o. 6 =1.364×10 -s cm2/s The experimental value (Table 2.8) at 298.15 K and Xe = 0.026 is 1.076 × 10.5 cm2/s, so that the estimation has ---26% relative error.

84

2.4.17

2.

Transportand rate processes

Diffusion in Electrolyte Solutions

On dissociation of a salt, ions start to diffuse in a solution. Without an electric potential effect, however, the diffusion of a single salt is treated as molecular diffusion. For dilute solutions of a salt, the Nernst-Haskell equation is used to estimate the diffusivity coefficient DO = RT(1/n+ + 1/n_ )

(2.87)

F 2 (1/A° + 1/A°)

where R is the gas constant (8.314 J/(mol K)), 1/h ° and 1/h ° the limiting ionic conductances (at zero concentration) ((A/cm 2) (V/cm) (g equiv/cm3)), n + and n_ the valences of cation and anion, respectively, and F the Faraday constant (96,500 C/(g equiv)). Table 2.10 lists the values of limiting ionic conductances at 298 K for some ionic species. The values of 1/h ° and 1/h ° at a temperature T other than 298 K need a correction term of T/(334~w), where/~w is the viscosity of water at T in cP. The mutual diffusivities of some inorganic salts in aqueous solutions are tabulated in Table 2.11.

2.4.18

Diffusion in Colloidal Suspensions

Consider a dilute suspension of spherical particles A in a stationary liquid B. If the spheres are sufficiently small, yet large with respect to the molecules of stationary liquid, the collisions between the spheres and the liquid molecules B lead to a random motion of the spheres. This motion is called the Brownian motion. Dilute diffusion of suspended spherical colloid particles is related to the temperature and the friction coefficient ~:by KT DA B --

KT

67r/ZBRA

(2.88)

where R A is the radius of the sphere. The reciprocal of the friction coefficient, sc, is called the mobility.

2.4.19

Diffusion in Polymers

For a dilute solution of polymer A in a low molecular weight solvent B, the polymer molecules are modeled as beadspring chains. Resistance in the motion of beads is characterized by a friction coefficient ~. As the number of beads is proportional to the polymer molecular weight M, we have

Table 2.10 Limiting ionic conductances in water at 298 K ((A/cm 2) (V/cm) (g equiv/cm3)) Anion

h°

Cation

h°

OHCIBrINO~ CLO~ HCO~ HCOj CH3COj CICHzCO~ CNCHzCO~ CH3CHzCO ~CH3(CH2)2CO2 C6H5CO J HC20~ (1/2)C2 O2(1/2)SO42(1/3)Fe(CN) 3(1/4)Fe(CN) 4-

197.6 76.3 78.3 76.8 71.4 68.0 44.5 54.6 40.9 39.8 41.8 35.8 32.6 32.3 40.2 74.2 80 101 111

H+ Li + Na + K+ NH~Ag + Ti + (1/2)Mg 2+ (1/2)Ca 2+ (1/2)Sr 2+ (1/2)Ba 2+ ( 1/2)Cu 2+ (1/2)Zn 2+ (1/3)La 3+ ( 1/3)Co(NH3)63+

349.8 38.7 50.1 73.5 73.4 61.9 74.7 53.1 59.5 50.5 63.6 54 53 69.5 102

Source: H.S. Hamed and B.B. Owen, The Physical Chemistry of Electtvlytic Solutions, ACS Monogr. 95 (1950).

2.4

Transportphenomena

85

1 DAB

(2.89)

-- .__

x/M

Tables 2.12-2.14 show some values of diffusion coefficients in solids and polymers. In a flow of dilute solution of polymers, the diffusivity tensor is anisotropic and depends on the velocity gradient. The Maxwell-Stefan equation may predict the diffusion in multicomponent mixtures of polymers. Phenomenological systems show that in relatively slow processes, the conjugate flow J is largely determined by frictional forces, and is linearly related to the conjugate force X J-

(2.90)

LX

where the coefficient L is a proportionality factor, which is not necessarily constant but independent of both J and X.

Table 2.11 Mutual diffusion coefficients of some inorganic salts in aqueous solutions Solute

T (°C)

Concentration (mol/1)

DAB × 105 (cm2/s)

HC1

12

0.1

2.29

H2SO 4 HNO3

20 20

NH4C1 H3PO 4 HgC12 CuSO4 AgNO3 CoC1, MgSO4 Ca(OH)2 Ca(NO3), LiC1 NaOH NaC1

20 20 18 14 15 20 10 20 14 18 l5 18

0.25 0.05 0.25 1.0 0.25 0.25 0.4 0.17 0.1 0.4 0.2 0.14 0.05 0.05 0.4

1.63 2.62 2.59 1.64 0.89 0.92 0.39 1.28 1.0 0.39 1.6 0.85 1.12 1.49 1.17

18

0.8 2.0 0.01

1.19 1.23 2.20

0.1 1.8 0.4 0.8 2.0 0.046 0.2

2.15 2.19 1.46 1.49 1.58 1.49 1.43

KOH

KC1

18

KBr KNO 3

15 18

Source: International Critical Tables, McGraw-Hill, New York (1926-1930).

Table 2.12 Diffusivities in some solids A-B

7"(K)

DAB (cm2/s)

He-SiO2

293.2

2.4-5.5 × l0 ~o

H2-SiO2

773.2

0.6-2.1 × 10 s

H2-Ni Bi-Pb Hg-Pb

358.2 438.2 293.2 293.2

Sb-Ag A1-Cu

293.2 293.2

3.5 x 10 2~ 1.3 × 10 3o

Cd-Cu

293.2

2.7 × 10 ~5

Source: R.M. Barrer, Diffusion in and Through Solids, Macmillan, New York (1941 ).

1.16 × 10.5 × 1.1 × 2.5 ×

10-s 10 8 10 ~' 10 ~5

86

2.

Transport and rate processes

Table 2.13 Diffusion coefficients in some solid polymers: DAB X 106 (cm2/s)

Polymer

He

Polyethylene terephthalate (glassy crystalline) Polycarbonate (Lexan) Polyethylene, density 0.964 Polyethylene, density 0.914 Polystyrene Butyl rubber Polychloroprene (neoprene) Natural rubber Silicone rubber, 10% filler (extrapolated) Polypropylene, isotactic Polypropylene, atactic Polyethyl methacrylate Butadiene-acrylonitrile (Perbunan) Polybutadiene Polyvinyl acetate (glassy)

H2

0 2

1.7

0.0036 0.64

3.07 6.8 10.4 5.93 21.6 53.4 19.5 41.6 44.1 11.7 9.52

CO 2

4.36 1.52 4.31 10.2 67.1 2.12 5.7

0.00054

0.021 0.170 0.46 0.11 0.81 0.43 1.58 17.0

0.0048 0.124 0.372 0.058 0.058 0.27 1.10

0.11 0.43 1.5 0.051

0.030 0.19 1.05

4.5 9.6 2.10

CH 4

0.00017

0.057 0.193

0.89

0.0019

Source: J. Crank and G.S. Park (Eds), Diffusion in Polymers, Academic Press, New York (1968); D.W. van Krevelen, Properties of Polymers, 3rd ed., Elsevier, Amsterdam (1990). Table 2.14 Diffusivities in some molten and thermally softened polymers

Gas

Diffusivities × 106 (cm2/s) Polyethylene

Helium Argon Krypton Monochlorodifluoromethane Methane Nitrogen Carbon dioxide

Polypropylene

17.09 9.19

10.51 7.40

4.16 5.50 6.04 5.69

4.02 3.51 4.25

Polyisobutylene

Polystyrene

12.96 5.18 7.30 2.00 2.04 3.37

0.42 0.348 0.39

Source: R.G. Griskey and P.L. Durill,AIChE J, 12 (1966) 1147; R.G. Griskey and EL. Durill,AIChE J., 15 (1967) 106; R.G. Griskey, Mod. Plast., 54 (1977) 158; J.L. Lundberg, M.B. Wilk and M.J. Huyett,J Appl. Phys., 1131 (1960); J.L. Lundberg, M.B. Wilk, and M.J. Huyett, J Polym. Sci., 57 (1962) 275; J.L. Lundberg, M.B. Wilk, and M.J. Huyett, Ind. Eng. Chem. Fundam., 2 (1963) 37; J.L. Lundberg, E.J. Mooney, and C.E. Rodgers, J Polym. Sci.,A-2(7) (1969) 947; D.M. Newitt and K.E.J. Weale, Chem. Soc. (Lond.), 1541 (1948).

2.5

THE MAXWELL-STEFAN EQUATIONS

For multicomponent diffusion in gases at low density, the Maxwell-Stefan equations provide satisfactory approximations when species i diffuses in a h o m o g e n e o u s mixture

xi:n i XiX ('i

(2.91)

where xi = ci/c, c is the mixture concentration, and D~j the binary diffusivities of species i in j, and there are (1/2)n(n - 1) o f binary diffusivities required for n-component system. For gases at low and high densities, liquids, and polymers, the M a x w e l l - S t e f a n equation can be used with diffusivities called the M a x w e l l - S t e f a n diffusivities, which can be related to the Fick diffusivities through the t h e r m o d y n a m i c correction factor for nonideal systems. In a simple limiting case, a dilute species i diffuses in a homogeneous mixture, Nj --~ 0, and Eq. (2.91) in y-direction becomes

dx--L = - J i £ xj dy j=l cD~.

(i 4: j )

(2.92)

2. 7

87

Electric charge flow

If we define a diffusivity of species i in a mixture by D i m - -Ji/(dxi/dy), we have

xj -/=1D~/

Dim -

(i 4: j)

(2.93)

These simple relations are applied to some ternary systems (see Chapter 6).

2.6

TRANSPORT COEFFICIENTS

The one-dimensional flows of momentum, energy, and mass at constant densities are

d ~xv = - u - - ( P v v ) dx d qx = - o ~ - - ( p C p T ) dx d

JAr =--DAB -7--(PA) ax

(2.94)

(Cp = constant)

(2.95)

(2.96)

Therefore, momentum transport occurs because of a gradient in momentum concentration, energy transport is due to a gradient in energy concentration, and mass transport is the result of a gradient in mass concentration. These three transport processes show analogies in their formulations. However, these analogies do not apply in two- and three-dimensional transport processes, since ~-is a tensor quantity with nine components, while Ja and q are vectors with three components. The mass diffusivity Dij, the thermal diffusivity c~ = k/pCp, and the momentum diffusivity or kinematic viscosity ~' = Ix~p, all have dimensions of (length)Z/time, and are called the transport coefficients. The ratios of these quantities yield the dimensionless groups of the Prandtl number, Pr, the Schmidt number, Sc, and the Lewis number, Le P r - u _ Cpix

Sc-

Le-

(2.97)

u D~i ce

D!/

pDi/ -

k ~ pCpO(j

(2.98) (2.99)

These dimensionless groups of fluid properties play important roles in dimensionless modeling equations of transport processes, and for systems where simultaneous transport processes occur. The close interrelations among mass, momentum, and energy transport can be explained in terms of a molecular theory of monatomic gases at low density. The continuity, motion, and energy equations can all be derived from the Boltzmann equation for the velocity distribution function, from which the molecular expressions for the flows and transport properties are produced. Similar derivations are also available for polyatomic gases, monatomic liquids, and polymeric liquids. For monatomic liquids, the expressions for the momentum and heat flows include contributions associated with forces between two molecules. For polymers, additional forces within the polymer chain should be taken into account.

2.7

ELECTRIC CHARGE FLOW

According to the Ohm's law, the flow of electricity that is the current I is directly proportional to electric potential difference or applied voltage AO 1_±4'

(2.100)

88

2.

Transport and rate processes

where R is the resistance of the medium to the current. The value of resistance is influenced by the medium configuration, and for many materials is independent of current. When an electric field is applied, the free electrons experience acceleration in a direction opposite to that of the field, and the flow of charge is called the electric current. Quantum mechanics predicts that there is no interaction between an accelerating electron and atoms in a perfect crystal lattice. Since the current reaches a constant value after the electric field is applied, there exist frictional forces, which counter the acceleration. The frictional forces are the result of the scattering of electrons by imperfections in the crystal due to impurity atoms, dislocations, and vacancies. Thermal vibrations of the atoms may also cause frictional forces. The frictional forces cause the resistance that may be described by the drift velocity Vd and the mobility of an electron, /Ze. The drift velocity represents the average electron velocity, while the electron mobility indicates the frequency of scattering phenomena, and has units of mZ/(v s). Thus, we have (2.101)

Vd = ].Zee

where e is the electric field intensity, and is defined as the voltage difference between two points divided by the distance I separating them A6 1

(2 102)

The conductivity of most materials may be expressed in terms of number of free electrons n per unit volume, absolute magnitude of the electric charge of an electron (]e I = 1.6 × 10-19 C) in Coulomb, and the mobility of electrons ~=nlel~

(2.103)

e

The electric conductivity specifies the electric character of the material. Solid materials, in three groups of conductors, semiconductors, and insulators, exhibit a wide range of electric conductivities. Metals have conductivities on the order of 107 (12 m) -1, insulators have conductivities ranging between 10-l° and 10-2° (1~ m) -1, and the conductivities of semiconductors range from 10-6 to 104 (12 m) -1. The resistivity p is independent of specimen geometry and related to resistance R by RA

p-

(2.104)

1 where A is the cross-sectional area normal to the direction of the current. For most metals and their alloys, resistivity increases with temperature due to the increase in thermal vibration and other irregularities, such as plastic deformations, which serve as electron-scattering centers. Resistivity also changes with composition for alloys. Electric conductivity is defined as the reciprocal of resistivity in ~ m -

1

(2.105)

P Using Eqs. (2.104) and (2.105), Ohm's law becomes Je = O'e

(2.106)

where Je is the current flow (density) that is the current per unit of specimen area I/A. Table 2.15 shows the units of electric parameters. Table 2.15 Units and symbols of electric parameters Quantity Electric charge (C) Electric potential (V) Electric field strength (V/m) Electric current (A) Resistance (~) Resistivity (~1 m) Conductivity (12 m) -1 Current flux (density) Electron mobility

Symbol

SI

Q #s e I R p oJe /ze

C kg m2/(s 2 C) kg 1TI/(S2 C) C/s kg mZ/(s C 2) kg m3/(s C 2) s C2/(kg m 3) C/(s m 2) m 2 (V s) = s C/kg

2.9

2.8

Chemical reactions

89

THE RELAXATION THEORY

For describing transport phenomena, we use one of the constitutive equations of Newton's law of viscosity, Fourier's law, or Fick's law; each one of these relates flow to conjugate thermodynamic driving forces. The conservation laws for momentum, heat, and mass transfer lead to parabolic equations of change, which suggest that the velocity of the propagation of an external disturbance such as a thermal disturbance at any point in the transfer medium is infinite. This can be seen in Figure 2.3, when the surface of the semi-infinite solid material suddenly is brought to T1 from initial uniform temperature To. The solution of temperature profile shows that at time t = 0, T = To, but for t > 0 the material temperature is T(y, t) everywhere, implying that the change in surface temperature is felt everywhere in the material. This phenomenon is explained by the hypothesis of heat flow relaxation, which states that Fourier's law is an approximation of a more exact equation called the Maxwell-Cattaneo equation

0q Ot

q = -kVT-rq

(2.107)

where rq is the relaxation time of heat flow. Analogous equations for the irreversible flows of momentum and mass can also be expressed. For example, for mass transfer, an identical equation to Eq. (2.107) is obtained from the nonstationary version of the Maxwell-Stefan equation, and for momentum diffusion, a similar equation is obtained from the Maxwell equation for viscoelastic fluids. The relaxation time for heat transfer is around rq = 10-12 s for metals, 10-9 S for gases at normal conditions, and 10-~ to 10-13 s for typical liquids. Relaxation times can be greater in rarefied gases, viscoelastic liquids, capillary porous bodies (% = 10-4 s), liquid helium ( % - 4.7 × 10-3 s), turbulent flows (rq = 10-3 to 103 s), and dispersed systems. Combining the thermal energy conservation with Eq. (2.107) yields --c~

V2T -

iJt

~

(2.108)

c2 0 t 2

where c~ is the thermal diffusivity and Co the propagation speed of the internal wave

o-i

Tq

(2.109)

Equation (2.108) is a hyperbolic type, and its solution for the semi-infinite solid medium suggests that two regions exist in a solid; the first region is called the disturbed region where the heat diffusion occurs, and the second region is undisturbed. Fourier's law of heat transfer predicts heat diffusion everywhere in the medium. However, as soon as the surface temperature changes, the wall heat flow q(0, 0) does not start instantaneously, but rather grows gradually. The heat flow rate depends on the current relaxation time and not the relaxation in state. For example, chemical reaction phenomena may illustrate the state relaxation, and heat and viscous stress relaxations and also current relaxation in electric circuits associated with a change in the magnetic energy may illustrate the current relaxation. The wall heat flow reaches a maximum and decreases in time, since temperature gradient at the wall decreases. Therefore, the Fourier's and Fick's laws are inappropriate for describing short-time effects, which may be theoretically important although the relaxation times are typically very small. No exact general criterion is available when it is necessary to include the relaxation terms in the equations of change; however, relaxation terms are necessary for viscoelastic fluids, dispersed systems, rarefied gases, capillary porous mediums, and helium, in which the frequency of the fast variable transients may be comparable to the reciprocal of the longest relaxation time.

2.9

CHEMICAL REACTIONS

Chemical reaction rate depends on the collisions of molecules, per second per unit volume. Since the number of collisions of a species is proportional to its concentration, the chemical reaction rate is proportional to the product of concentrations (mass action law). Thus, for a single homogeneous elementary chemical reaction ulA(g) + u2B(g ) = u3C(g ) + u4D(g)

(2.110)

90

2.

Transport and rate processes

the flow of reaction (velocity) Jr refers to the difference between the forward rate Jrf = kfCAC B and backward rate 4b = kbCcCD J r ~-~ J r f -- Jrb m_ kfCACB -- kbCcCD

(2.111)

where kf and kb are the forward and backward reactions rate constants, respectively. The ratios of mole changes of reacting and produced species are related to the extent of reaction e defined by -dN----~A = Pl

dN-------~B- d N c - dND - d e 92

P3

(2.112)

94

where 9i are the stoichiometric coefficients, which are positive for products and negative for reactants. For a single homogeneous reaction, a generalized reaction rate Jr is a scalar value, and can be expressed in terms of the extent of reaction Jr -

dE

(2.113)

dt

The affinity A is (2.114)

A=-~_~vi~i

where/.L i is the chemical potential of component i. For the chemical reaction system given in Eq. (2.110), the affinity becomes A = P l e A + P2/-~B --(P3/-~C + 94/-~D )

(2.115)

At constant temperature and pressure, the affinity of the chemical reaction is the negative of the change of the Gibbs free energy

116 T,P

If the value of A is greater than zero, the reaction moves from left to fight; if it is smaller than zero, the reaction proceeds from right to left; when A = 0, no reaction takes place.

2.10

COUPLED PROCESSES

When two or more processes occur simultaneously in a system, they may couple (interfere) and cause new induced effects. In a multicomponent system, there are five main driving forces that cause the transport of mass with respect to the mean fluid motion: (i) concentration gradient, (ii) pressure gradient, (iii) temperature gradient, (iv) electric field or an electric potential gradient, and (v) external forces affecting the various chemical substances unequally, such as magnetic effects. In a multicomponent fluid, we have flows of momentum, energy, mass, and electric current, each resulting from an associated thermodynamic driving force. There may be a contribution to each flow stemming from each driving force in the system. This is the result of coupling that can occur between flow-force pairs, which are tensors of equal order or differ by 2 in order. For example, momentum flow is a tensor of order 2, while mass flow or heat flow is a vectorial process. Table 2.16 tabulates some coupled processes, which are briefly discussed below.

2.10.1

Electrokinetic Effects

In 1808, Rous, a colloid chemist, observed that imposing an electric potential difference across a porous wet clay led not only to the expected flow of electricity, but also to a flow of water. He later applied hydrostatic pressure to the clay and observed a flow of electricity. This experiment displayed the electrokinetic effect and demonstrated the existence of coupled phenomena where a flow may be induced by forces other than its own driving force. Therefore, the electric current is evidently caused by the electromotive force, but it may also be induced by the hydrostatic pressure. When two

2.10

91

Coupled processes

Table 2.16 Direct and coupled transport phenomena Flow

Thermodynamic force

Heat

Thermal

Chemical

Electric

Hydraulic

Thermal conduction,

Chemical osmosis,

Peltier effect,

Thermal filtration

Jco = -ci(rKV Tllh, where

Jq = -k~V T

Fluid

Thermal osmosis,

Jq = Lqe( ~ / T ) ,

K is the hydraulic coefficient, Hh the osmotic pressure head (IIh = II/pg), (r the coefficient of osmotic efficiency Dufour effect

where E is the electric field

Electric osmosis

Advection, Jadv =-ciKVh

Diffusion, Ji = -D~Vc,

Electrophoresis

Hyper filtration,

JTq = CikTV T, where kr is the thermoosmotic permeability (m=/(K s)) Thermal diffusion (Soret effect),

Solute

Jco = - c icrKV h,

JTD -De sc iV T,

where h is the hydraulic head

=

Current

where s is the Soret coefficient Seebeck (Thompson effect)

Diffusion and membrane potential

Electric conduction

Rouss effect: thermokinetic effect

chambers containing electyrolytes are separated by a porous wall, an applied potential generates a pressure difference called the electrosmotic pressure. Also, mass flow may generate an electric current called the streaming current. Gradients of electric potential and pressure govern the behavior of ionic systems, selective membranes, and ultracentrifuges. In electrokinetic phenomena, induced dipoles can cause separations, such as dielectrophoresis and magnetophoresis, which may be especially important in specialized separations. Diffusion potential is the interference between diffusion and electric conduction in an anisotropic crystal where heat conduction occurs in one direction caused by a temperature gradient in another direction.

2.10.2

Thermoelectric Effects

In a nonisothermal system, an electric current (flow) may be coupled with a heat flow; this effect is known as the thermoelectric effect. There are two reciprocal phenomena of thermoelectricity arising from the interference of heat and electric conductions: the first is called the Peltier effect. This effect is known as the evolution or the absorption of heat at junctions of metals resulting from the flow of an electric current. The other is the thermoelectric force resulting from the maintenance of the junctions made of two different metals at different temperatures. This is called the Seebeck effect. Temperature measurements by thermocouples are based on the Seebeck effect.

2.10.3

Multicomponent Mass Flow

In a binary mixture, diffusion coefficients are equal to each other for dissimilar molecules, and Fick's law can determine the molecular mass flows in an isotropic medium at isothermal and isobaric conditions. In a multicomponent diffusion, however, various interactions among the molecules may arise. Some of these interactions are (i) diffusion flows may vanish despite the nonvanishing driving force, which is known as the mass transfer barrier, (ii) diffusion of a component in a direction opposite to that indicated by its driving force leads to a phenomenon called the reverse mass flow, and (iii) diffusion of a component in the absence of its driving force, which is called the osmotic mass flow.

2.10.4

Coupled Heat and Mass Flows

Another well-known example is the coupling between mass flow and heat flow. As a result, an induced effect known as thermal diffusion (Soret effect) may occur because of the temperature gradient. This indicates that a mass flow of component A may occur without the concentration gradient of component A. Dufour effect is an induced heat flow caused by the concentration gradient. These effects represent examples of couplings between two vectorial flows. The cross-phenomenological coefficients relate the Dufour and Soret effects. In order to describe the coupling effects, the thermal diffusion ratio is introduced besides the transport coefficients of thermal conductivity and diffusivity.

92

2.

Transport and rate processes

Table2.17 Excursion into history (Straub, 1997) 1687 1736 1749 1750 1755 1736-1819 1811 1821-1828 1822 1822-1838 1824 1842 1847 1848 1850 1852 1864 1865 1872 1873 1872-1957 1901 1905 1931 1941

2.10.5

Viscosity Mass point Hydrodynamic pressure Newton's basic law of motion Ideal equation of motion Steam engine Propagation of heat in solids Viscous equation of motion Equation of heat conduction Navier-Stokes equation Camot cycle, reversibility Conservation of energy Conservation of force Absolute temperature First law of thermodynamics Second law of thermodynamics Electromagnetic field Entropy Entropy (statistical) Equation of state for real fluids Thermodynamic affinity Statistical mechanics Mass-momentum-energy relations Reciprocal relations Dissipative structures

Newton Euler

James Watt Baron-Jaseph Fourier Cauchy Fourier Navier-Saint-Venant Sadi Carnot Robert Mayer Hermann von Helmholtz Kelvin Rudolf Clausius Kelvin Maxwell Rudolf Clausius Boltzmann Van der Waals De Donder Gibbs Einstein Lars Onsager Prigogine

Coupled Chemical Reaction and Transport Processes

The coupling between chemical reactions and transport in biological membranes, such as the sodium and potassium pumps, is known as active transport, in which the metabolic reactions cause the transport of substances against the direction imposed by their thermodynamic force of mainly electrochemical potential gradients.

2.10.6

Coupled Phenomena and Thermodynamics

Using the formulation of nonequilibrium thermodynamics, the cross-effects are mathematically described by the addition of new terms into the phenomenological laws. For thermal diffusion, for example, a term proportional to the temperature gradient is added to the right hand side of Fick's law. These relations need certain phenomenological coefficients, which can be related to the transport coefficients such as thermal conductivity, ordinary and thermal diffusion coefficients, and Dufour coefficient. Therefore, the measured transport coefficients will help determine the phenomenological coefficients. Coupled phenomena are experimentally verified, and they are part of a comprehensive theory of nonequilibrium thermodynamics and phenomenological approach. Studies on the coupled processes in biological, chemical, and physical systems are attracting scientists from various fields. Table 2.17 shows some of the historical developments in thermodynamics, transport phenomena, and rate process.

PROBLEMS 2.1

Estimate the viscosity and thermal conductivity of carbon monoxide using the Chapman-Enskog model at 1 atm and 250, 300, and 400 K and compare with the experimental (Welty et al. (1984)) values in the table below:

T (K) 250 300 400

/z × 105 (Pa s)

k × 102 (W/(m K))

1.5408 1.7854 2.2201

2.1432 2.5240 3.2253

93

Problems

2.2

Estimate the viscosity and thermal conductivity of hydrogen using the Chapman-Enskog model at 1 atm and 250, 300, 400, 600, and 800 K, and compare with the experimental values (Welty et al. (1984)) in the table below: T (K)

# × 10(' (Pa s)

k (W/(m K))

7.919 8.963 10.864 14.285 17.40

0.1561 0.182 0.228 0.315 0.384

250 300 400 600 800

2.3

Estimate the viscosity and thermal conductivity of carbon dioxide using the Chapman-Enskog model at 1 atm and 250, 300, and 400 K and compare with the experimental values (Welty et al. (1984)) in the table below: T (K)

# × 105 (Pa s)

k (W/(m K))

1.2590 1.4948 1.9318

1.2891 1.6572 2.4604

250 300 400

2.4

(a) Estimate the thermal conductivity of oxygen at T = 350 K and P = 1 atm using Eq. (2.48). At 350 K and 1 atm, the thermal conductivity is 0.0307 W/(m K) and Cp = 7.586 cal/(mol K) (Welty et al. (1984)). (b) Estimate the thermal conductivity at the same conditions using Eq. (2.47) and compare the result with the result from part (a).

2.5

Estimate the thermal conductivity of 02 at T = 300 K and P = 30 atm. The thermal conductivity of oxygen at 300 K and 1 atm is 0.02657 W/(mK) (Welty et al. (1984)).

2.6

Estimate the viscosity and thermal conductivity of helium using the Chapman-Enskog model at 1 atm and 422.2,477.8, and 533.3 K and compare with the experimental values (Welty et al. (1984)) in the table below: 7' (K)

# × 105 (Pa s)

k (W/(m K))

422.2 477.8 533.3

2.5299 2.7532 2.9466

0.1834 0.1973 0.2111

2.7

Estimate the viscosity of carbon monoxide at T - 4 0 0 K and P = 35 atm using the reduced viscosity chart in Figure 2.4.

2.8

Estimate the viscosity of a mixture described in the table below at T - 330 K and P = 25 atm using the reduced viscosity chart in Figure 2.4: Species

Mole fractions y;

N. H~ CO Air CO2

0.20 0.10 O.20 0.25 0.25

_

2.9

Estimate the viscosity of the following gas mixture at 293 K and 1 atm using the data (Welty et al. (1984)) given in the following table: Species Air Carbon dioxide

Yi

M

k (W/(m K))

/x (mPa s)

0.75 0.25

28.97 44.01

0.02624 0.01601

0.0181 0.0146

94

2.10

2.

Transport and rate processes

Estimate the radial direction heat flow through a hollow sphere described in the following figure: qout

r=~ T=

2.11

Estimate the thermal conductivity of helium at T = 500 K and P = 1 atm and compare with the experimental value of 5.07 cal/(cm s K) (W.K. Saxena and S.C. Saxena, J. Phys. D Appl. Phys., 1 (1968) 1341).

2.12

Estimate the thermal conductivity of the following table:

0 2at 200 and 300 K at low pressure. Use the data (Welty et al. (1984)) in T (K)

Cp(cal/(molK))

200 300

2.13

2.14

6.971 7.039

Estimate the thermal conductivity of the following gas mixture at 293 K and 1 atm using the data (Welty et al. (1984)) given in the following table: Species

Yi

M

k (W/(mK))

/z (mPa s)

Air Carbon dioxide

0.5 0.5

28.97 44.01

0.02624 0.01601

0.0181 0.0146

Two large vessels containing binary mixtures of gases A and B are connected by a truncated conical duct, which is 2 ft in length and has internal diameters at the ends of 8 and 4 in., respectively. One vessel contains 80mo1% of gas A, and the other 30mo1% of A. The pressure is 1 atm and temperature is 32°E The diffusivity under these conditions is 0.702 ft2/h. Disregarding the convection effects: (a) Calculate the rate of transfer of A. (b) Compare the results that would be obtained if the conical duct was replaced with a circular duct with a diameter of 6 in. Cone ~r z ~ - - - ~ o

Cylinder ~ l r ~ z

...~ut ...

2.15

Estimate the diffusivity of acetic acid in a dilute solution of acetic acid-water at 12.5°C. The density of acetic acid at its boiling point is 0.973 g/cm 3 and the viscosity of water at 12.5°C is 1.22 cP.

2.16

Estimate the diffusivity of ethanol in a dilute solution of ethanol-water at 15°C. The density of ethanol is 0.79 g/cm 3 and the viscosity of water at 15°C is 1.2 cP.

2.17

Estimate the diffusivity of ethylbenzene in water in a dilute solution at 293 K. The density of ethylbenzene at its normal boiling point (at 1 atm) is 0.761 g/cm 3 and the viscosity of water at 293 K is 1.0 cP.

2.18

Estimate the diffusivity of acetic acid, i, in a mixed solvent with 21 mol% ethanol, e, and 79 mol% water, w, at 298 K. The viscosities of ethanol and water are/~e = 1.1 cP and/~w - 0.894 cP at 298 K.

95

Problems

2.19

Estimate the diffusivity of carbon monoxide in nitrogen at 373 and 194.7 K and 1 atm. The experimental values of diffusivities are DAB - - 0 . 3 1 8 at 373 K and D A B - 0.105 at 194.7 K.

2.20

Estimate the diffusivity of carbon dioxide in air at 276.2 and 317.3 K and 1 atm. The experimental values of diffusivities are (Griskey, 2002)

2.21

T (K)

DAB (cm2/s)

276.2 317.3

0.142 0.177

Estimate diffusivity of carbon dioxide A in sulfur dioxide B at 263 and 473 K and 1 atm. The experimental values of diffusivities are (Griskey, 2002) T (K)

DAB (cm2/s)

263 473

2.22

0.064 0.195

Estimate the diffusivity ofhexane in air at 294 and 328 K and 1 atm. The experimental values of diffusivities are Griskey (2002) T (K)

DAB (cm2/s)

294 328

2.23

2.24

0.080 0.093

Estimate the diffusivity of C1402 in ordinary CO 2 at 300K and 145atm. The measured value of DAA~ = 0.125 cmZ/s at T - 312.8 K and P - 1 atm (Griskey (2002)). Estimate the diffusivity of an air-hexane mixture at 390K and 66atm. The measured value of DAB -- 0.093 cm2/s at T = 328 K and P - 1 atm (Griskey (2002)).

2.25

Estimate the diffusivity DAB for the system carbon dioxide and dinitrogen oxide N20 at 273.2 K and 1 atm. The reported experimental diffusivity at these conditions is DAB = 0.096 cmZ/s (Bird et al. (2002)).

2.26

Estimate the diffusivity DAB for the system nitrogen N2 and ethane C2H 6 at 298.2 K and 1 atm. The reported experimental diffusivity at these conditions is DAB = 0.148 cmZ/s (Bird et al. (2002)).

2.27

Estimate the diffusivity DAB for the system air and hexane C6H14 at 328 K and 1 atm. The reported experimental diffusivity (Griskey, 2002) at these conditions is DAB -- 0.093 cmZ/s.

2.28

Estimate the diffusivity DAB for the system methane CH 4 and hydrogen H 2 at 298.2 K and 1 atm. The reported experimental diffusivity at these conditions is DAB = 0.726 cmZ/s (Bird et al. (2002)).

2.29

Estimate and compare the diffusivities DAB for the systems propane C3H8 and normal butane n-C4H10, and propane C3H8 and iso-butane i-C4Hj0 at 378.2 K and 1 atm. The reported experimental diffusivities (Bird et al. (2002)) at these conditions are DAB -- 0.0768 cm2/s and 0.0823 cruZ/s, respectively.

2.30

Estimate the diffusivity of carbon monoxide through a gas mixture of ethylene, hydrogen, and nitrogen with the known mole fractions given in the following table. The mixture is at 295 K and 2 atm. Species Carbon monoxide ( 1) Ethylene (2) Hydrogen (3) Nitrogen (4)

y~

T (K)

D!/(cm2/s)

P (atm)

0.05 0.20 0.30 0.45

273 273 288

0.151 0.651 0.192

1 1 1

96

2.

Transport and rate processes

REFERENCES R.B. Bird. WE. Stewart and E.N. Lightfoot, TransportPhenomena, 2nd ed., Wiley, New York (2002). R.G. Griskey, Transport Phenomena and Unit Operations. A CombinedApproach, Wiley, New York (2002). R.C. Reid, J.M. Prausnitz and B.E. Poling, The Properties of Gases and Liquids, 4th ed., McGraw-Hill, New York (1987). J.R. Welty, C.E. Wicks and R.R. Wilson, Fundamentals of Momentum, Heat and Mass Transfer, 3rd ed., Wiley, New York (1984).

REFERENCES FOR FURTHER READING W.M. Deen, Analysis of Transport Phenomena, Oxford University Press, New York (1998). I. Prigogine, Introduction to Thermodynamics of Irreversible Processes, Wiley, New York (1967). S. Sieniutycz and P. Salamon (Eds), Flow, Diffusion and Rate Processes, Advances in Thermodynamics, Vol. 6, Taylor & Francis, New York (1992). A.H.P. Skelland, Diffusional Mass Transfer, Krieger Publishing Company, Malabar (1985). J.M. Soler, J. Contam. Hydrol., 53 (2001) 63. D. Straub, Alternative Mathematical Theory of Non-equilibrium Phenomena, Academic Press, New York (1997).

3 FUNDAMENTALS OF NONEQUILIBRIUM THERMODYNAMICS 3.1

INTRODUCTION

Physical systems identified by permanently stable and reversible behavior are rare. Unstable phenomena occurring macroscopically result from inherent fluctuations of the respective state variables. Near global equilibrium, the fluctuations do not disturb the equilibrium; the trend toward equilibrium is distinguished by asymptotically vanishing dissipative contributions. In contrast, nonequilibrium states can amplify the fluctuations, and any local disturbances can even move the whole system into an unstable or metastable state. Equilibrium states can be classified by correlations having a typical average range of action of about 10-10 m. Correlations existing far from equilibrium can extend to macroscopic distances on the order of 1 cm or more. This feature is an important indication of the qualitative difference between equilibrium and nonequilibrium states. Kinetic and statistical models often require more detailed information than is available to describe nonequilibrium systems. Therefore, it may be advantageous to have a phenomenological approach with thermodynamic principles to describe natural processes. Such an approach is the formalism used in nonequilibrium thermodynamics to investigate physical, chemical, and biological systems with irreversible processes. In the formalism, the Gibbs equation is a key relation since it combines the first and second laws of thermodynamics. The Gibbs relation, combined with the general balance equations based on the local thermodynamic equilibrium, determines the rate of entropy production. Quantifying entropy production helps in analyzing the level of energy dissipation during a process, and in describing coupled phenomena. Phenomenological laws may describe many common irreversible processes with broken time symmetry. An irreversible phenomenological macroworld and a microworld determined by linear and reversible quantum laws should be related to each other. Prigogine and his colleagues attempted to unify the basic micro and macroscopic descriptions of matter. The first attempts to develop nonequilibrium thermodynamics theory occurred after the first observations of some coupled phenomena of thermal diffusion and thermoelectric. Later, Onsager developed the basic equations of the theory, and Casimir, Meixner, and Prigogine refined and developed the theory further. This chapter outlines the principles of nonequilibrium thermodynamics for systems not far from global equilibrium. In this region the transport and rate equations are expressed in linear forms, and the Onsager reciprocal relations are valid. Therefore, sometimes this region is called the linear or Onsager region and the formulations are based on linear nonequilibrium thermodynamics theory. In this region, instead of thermodynamic potentials and entropy, a new property called entropy production appears. The formulation of linear nonequilibrium thermodynamics has proven to be valid and useful for a wide range of transport and rate processes of physical, chemical, and biological systems.

3.2

LOCAL THERMODYNAMIC EQUILIBRIUM

A local thermodynamic state is determined as elementary volumes at individual points for a nonequilibrium system. These volumes are small such that the substance in them can be treated as homogeneous and contain a sufficient number of molecules for the phenomenological laws to be applicable. This local state shows microscopic reversibility that is the symmetry of all mechanical equations of motion of individual particles with respect to time. In the case of microscopic reversibility for a chemical system, when there are two alternative paths for a simple reversible reaction, and one of these paths is preferred for the backward reaction, the same path must also be preferred for the forward reaction. Onsager's derivation of the reciprocal rules is based on the assumption of microscopic reversibility.

98

3.

Fundamentalsof nonequilibrium thermodynamics

The reversibility of molecular behavior gives rise to a kind of symmetry in which the transport processes are coupled to each other. Although a thermodynamic system as a whole may not be in equilibrium, the local states may be in local thermodynamic equilibrium; all intensive thermodynamic variables become functions of position and time. The definition of energy and entropy in nonequilibrium systems can be expressed in terms of energy and entropy densities u(T~k) and s(T, Nk), which are the functions of the temperature field T(x) and the mole number density N(x); these densities can be measured. The total energy and entropy of the system is obtained by the following integrations

S = ~{s[T(x)],[Nk(x)]}dV

(3.1)

U = ~ {u[T(x)], [Nk(x)]}dV

(3.2)

From the internal energy density u(x) and entropy density s(x), we obtain the local variables of (Ou/Os)v,xk = T(x), --(Ou/OV)s,Uk = P, and (Os/ONk)u = -t~(x)/T(x). The densities in Eqs. (3.1) and (3.2) are dependent on the locally well-defined temperature. In a nonequilibrium system, therefore, the total entropy S is generally not a function of the total entropy U and the total volume V. Also, the classical thermodynamic equations such as the Gibbs and the Gibbs-Duhem equations

i=1

i=1

are valid in a multicomponent medium. For a large class of nonequilibrium systems, thermodynamic properties such as temperature, pressure, concentration, internal energy, and entropy are locally well-defined concepts. Prigogine expanded the molecular distribution function in an infinite series around the equilibrium molecular distribution functionj~ f = f0 + fl + f2 + "'"

(3.5)

Equation (3.5) is valid not only for an equilibrium system, but also for a nonequilibrium system that is characterized by the equilibrium distribution function of (J~ +J]) representing a nonequilibrium system sufficiently close to global equilibrium. Prigogine's analysis applies only to mixtures of monatomic gases, and is dependent on the ChapmanEnskog model. The domain of validity of the local equilibrium is not known in general from a microscopic perspective. The range of validity of the local thermodynamic equilibrium is determined only through experiments. Experiments show that the postulate of local thermodynamic equilibrium is valid for a wide range of macroscopic systems of common gases and liquids, and for most transport processes if the gradients of intensive thermodynamic functions are small and their local values vary slowly in comparison with the local state of the system. For chemical reactions, the reactive collision rates are relatively smaller than overall collision rates. The change in an intensive parameter is comparable to the molecular mean free path, and energy dissipation rapidly damps large deviations from global equilibrium. The local equilibrium concept is not valid in highly rarefied gases where collisions are too infrequent. The extension of equilibrium thermodynamics to nonequilibrium systems based on the local equilibrium assumption is a well-accepted practice in nonequilibrium thermodynamics.

3.3

THE SECOND LAW OF THERMODYNAMICS

Let us consider the system shown in Figure 3.1. In region I, an irreversible process occurs. Region II is isolated and contains region I. Equilibrium is attained everywhere in region II. One possibility is that region I is a closed system that can only exchange heat with region II. The first law relates the internal energy change dU I to a quantity of heat gained from region II, 6q I, and quantity of work 6 WI is performed on region I, so that we have dU I = ~qi + dW I. According to the second law, region I obeys the general inequality dS = ~qrev > -~q -

T

T

(3.6)

3.3

The second law of thermodynamics

T,P,p II

Figure 3.1.

99

@

A closed system with a subsystem I.

where dS, the entropy gain, is determined by 6qrev, the heat that would be absorbed in a reversible change. Since we are considering real (irreversible) processes, TdS is always greater than 6q, which is the actual heat absorbed. It follows that in practice, region I fails to absorb the maximum amount of heat, which theoretically might be transformed into work. Then for real processes, we have dS - 6q + (3q' T T

(3.7)

Here 6q is the actual uptake of heat and 6q' is the additional heat that would have been absorbed from region II for a reversible change, and hence it is a positive quantity. In an actual change of state, the entropy increment dS contains the quantity of entropy 6q'/T. This may be, for example, due to a mixing process or a chemical reaction within region I. The relation of Carnot-Clausius gives the change of the entropy of a closed system dS - des + d i S -

6eq + diS T

(3.8)

Equation (3.8) identifies the two contributions: (i) entropy inflow from the environment deS, and (ii) entropy change inside the volume under consideration diS. The value of diS in every macroscopic region of the system is positive for irreversible changes, while it is zero for reversible changes dis > 0 (irreversible changes)

and

diS = 0 (reversible changes)

(3.9)

The des may be due to a flow of internal energy, convection entropy flow transported along with the macroscopic flow of the substance as a whole, or the entropy flow caused by diffusion of the individual components. The quantity des may be positive, negative, or zero in a special case. For a closed, thermally homogeneous system de S _ 6eq _ dU + PdV T T

(3.10)

where 6eq is the elementary heat due to thermal interaction between the system and the environment. The concept of a macroscopic region describes any region containing enough molecules for microscopic fluctuations to be disregarded. The second law formulation given in Eq. (3.8) may be interpreted as a local formulation, while the second law formulation of classical thermodynamics may be interpreted as a global formulation. With the local formulation, we can analyze various irreversible processes and interactions between them within the same nonequilibrium system. Explicit calculation of diS as a function of the appropriate variables is essentially the basis of nonequilibrium thermodynamics.

3.3.1

Entropy Change of an Ideal Gas

For an ideal gas, we can obtain the change of entropy in terms of volume (or pressure), temperature, and the number of moles. For a unit mass of fluid undergoing a mechanically reversible process in a closed system, the first law yields dU = 6qrev - PdV

(3.11)

100

3.

Fundamentalsof nonequilibrium thermodynamics

On the other hand, the differential form of enthalpy is d H = d U + P d V + VdP

(3.12)

When we eliminate d U between these relations, we get (3.13)

6qrev = d H - VdP For an ideal gas, we have the following relations

d H = Cp,idealdT, V -

RT p

(3.14)

R dP

(3 15)

Dividing Eq. (3.13) by temperature, we get 6qrev --C

T

dT

~p'idealT -- 7

By integrating Eq. (3.15) between temperatures T1 and T2, and pressures P1 and P2, the change of entropy becomes

=

ideal

In terms of volume we have

AS = r2 Cu,ideal T

Equations (3.16) and (3.17) consist of state properties that are independent of the process path, and are general relations for an ideal gas.

3.3.2

Entropy Change of Phase Transformation

Heat exchanged during a phase change under constant temperature is another way to estimate the value deS = 6q/T. For a solid-liquid transformation after a flow of molar heat of melting ~ / m at a melting temperature Tm and at constant pressure, we have

ASm

_

~J-/m rm

(3.18)

Similarly, we can estimate the change of entropy for vaporization using the molar heat of vaporization AHv at boiling point Tb ASv - ~ / v

Vb

(3.19)

Table 3.1 displays the entropy changes of melting and vaporization for some pure substances. The entropy of vaporization is proportional to the ratio of the degree of randomness in the vapor and liquid phases. For a pure component, ASv consists of translational, rotational, and conformational motion of molecules. The translational effect is the largest contribution to the entropy of vaporization.

3.3

101

The second law of thermodynamics

Table 3.1 Enthalpies and temperatures of phase change of some compounds at P = 1 atm Compound

Tm (K)

M-/m (kJ/mol)

ASm (kJ/(mol K))

Tb (K)

2ff-/,~(kJ/mol)

ASv (kJ/(mol K))

H2 02 N2 CO2 NH3 CS2 eel 4 H20 CH3OH C2HsOH

14.01 54.36 63.15 217.0 195.4 161.2 250.3 273.15 175.2 156.0

0.12 0.444 0.719 8.33 5.652 4.39 2.5 6.008 3.16 4.6

0.009 0.008 0.011 0.038 0.029 0.027 0.010 0.022 0.018 0.029

20.28 90.18 77.35 194.6 239.7 319.4 350.0 373.1 337.2 351.4

0.46 6.82 5.586 25.23 23.35 26.74 30 40.656 35.27 38.56

0.023 0.076 0.072 0.130 0.097 0.084 0.086 0.109 0.105 0.110

Source:

G.W.C. Kaye and T.H. Laby (Eds.) Tables of Physical and Chemical Constants, Longman, London (1986).

We can estimate the change in the entropy of vaporization at azeotropic temperature when the heat flow is known at azeotropic pressure

ASaz -

AH v

(3.20)

raz

We may estimate the heat of vaporization for azeotropic mixtures from the Lee-Kesler correlation, with some suitable mixing rules A H v - R T c [6.096448-1.2886T r +1.0167T] + w ( 1 5 . 6 8 7 5 - 1 3 . 4 7 2 1 T

r +2.615T])]

(3.21)

where

--

--

n

i=1

,

(.0 ---

X i (.0 i ,

Tr

i=1

2i=lxiVci

rc

Here, xi is the mole fraction of component i, n is the number of components, Tci and Vci are the critical temperature and volume, and wi is the acentric coefficient for species i. In Eq. (3.21), Tb is the normal boiling point in Kelvin at atmospheric pressure, R = 1.987 cal/(mol K), and M-Iv is in cal/mol. Table 3.2 shows the entropy of vaporization of some binary and ternary azeotropic mixtures obtained from the Lee-Kesler correlation.

3.3.3

Entropy Change of Expansion of a Real Gas

For an irreversible expansion of a real gas at constant temperature due to a heat reservoir, the change of entropy flow is deS - 6 q / T , where 6q is the heat flow between the gas and the reservoir to maintain the constant temperature. The increase of entropy during the expansion is

dis_

Ap

dV =

Pgas -- episton

T

(3.22)

dV

T

where Ppiston is the pressure on the piston. Therefore, the change of entropy becomes

d S - d e S + d~S - 6q + T

The t e r m ( P g a s -

Ppiston) dV

Pgas -- Ppiston

dV

T

is the uncompensated heat as named by Clausius.

(3.23)

3.

102

Fundamentalsof nonequilibrium thermodynamics

Table 3.2 Entropy of vaporization for some binary and ternary azeotropic data at 1 atm Binary species

X1

Tb (K)

AHv (J/mol)

ASv (J/(mol K))

Water(l) Chloroform Ethanol Ethyl acetate n-Butanol Nitromethane Acetonitrile Pyridine

0.160 0.096 0.312 0.753 0.511 0.307 0.768

329.2 351.3 343.5 365.8 356.7 349.6 367.1

32155.3 40683.6 34821.8 41464.4 39643.5 36358.3 41514.4

97.68 115.81 101.37 113.35 111.14 104.00 113.09

Methanol(I) Acetonitrile Acrylonitrile Toluene Ethyl acetate

0.231 0.724 0.883 0.684

336.6 334.5 348.1 335.4

36717.8 38605.1 39778.0 37992.7

109.08 115.41 114.27 113.28

Ethanol(I) Acrylonitrile Ethyl acetate Benzene Hexane

0.445 0.462 0.448 0.332

343.9 344.9 341.4 331.8

38065.2 37557.3 37647.5 34691.9

110.69 108.89 110.27 104.56

Ternary species

xl

X2

Tb (K)

AHv (J/mol)

ASv (J/mol)

Water( 1)-chloroform(2 ) Methanol Ethanol Acetone

0.066 0.129 0.163

0.698 0.795 0.353

325.4 328.4 333.5

34088.7 32954.9 32303.9

104.76 100.35 96.86

Water(I)--ethanol(2) Benzene Hexane

0.233 0.112

0.228 0.274

338.0 329.5

36847.4 34658.3

109.02 105.18

Source: Y. Demirel, Thermochim. Acta, 339 (1999) 79.

Example 3.1 Total entropy change of an air flow in a nozzle Air enters a nozzle at 400 K and 60 m/s and leaves the nozzle at a velocity 346 m/s. The air inlet pressure is 300 kPa, while the pressure at the outlet is 100 kPa. Heat lost in the nozzle is 2.5 kJ/kg. Determine the total entropy change if the surroundings are at 300 K. Solution:

Assume that air is an ideal gas. The nozzle operates at steady state. The properties of air from Table E4" T1 = 400 K

Hi = 400.98 kJ/kg

S1 = 1.9919 kJ/(kg K)

Energy balance for a nozzle at steady-state conditions yields Eout =/~in

(3.24)

The enthalpy at the outlet is v22- v~ H2 = H1 - 0out -

~

/1 kJ/kg m2/s k 2 g) = 340"42 2 = 4 0 0 . 9 8 - 2 . 5 - 3462 k- 602J ( 1000

Therefore, the conditions at the outlet are: T2 = 340 K, $2 = 1.8279 kJ/(kg K)

3.3

103

The second law of thermodynamics

From the entropy balance, we have: Total entropy change = entropy changes of the air + entropy change of the surroundings AStota 1 = ASai r q-ASsurr Z~Sai r - - S 2 - S 1 -

Rln P2 = 1.8279-1.9919-(0.287 kJ/(kg K)) lnk, ~

= 0.151kJ/(kg K) (3.25)

ASsurr _ qsurr,in _ kJ/_______~s 2.5 = 0.00833 kJ/(kg K)

ro z~Stota 1 -

ASair +

3ooK ASsurr =

0.151 + 0.00833 - 0.1593 kJ/(kg K)

Example 3.2 Total entropy change in a polytropic compressing of methane We compress methane from an initial state at 100 kPa, 300 K, and 20 m 3 to 250 kPa and 400 K. The compression process is polytropic ( P V ~ constant). The average heat capacity of methane is Cp,av- 40.57 J/(mol K). Estimate the supplied work and the total entropy change if the surroundings are at 290 K.

Solution: Assume that the methane is an ideal gas. The heat capacity is constant. The number of moles of methane is n--

P1V1 _

Rr~

1 × 105 Pa(20 m 3) 8.314(m 3 Pa/(molK)) 300K

= 801.86 moles

The mass of methane is" m = n M W = 801.86 (16)(1/1000)= 12.829kg The heat capacities are: Cp = 40.57 J/(mol K) and Cv - C p - R = 40.57-8.314 = 32.256 J/(mol K) (for an ideal gas) and the ratio is

y = Cp Cv

_

1.257

The entropy change of methane is

'methan :

32 0 12

The volume of methane after the compression is

V2=v T2 P'

- 20 m 3 400 ( 1.0 × 10 s j 300 2.5×105 =10"66m3

For this polytropic process, we have

PV(~ - c°nstant --' ~

~,V2)

and

{25x,05! (20/ 1.0×105

=

i0.66

-+a=1.456

(3.26)

104

3.

Fundamentals of nonequilibrium thermodynamics

The surface work of the polytropic compression becomes

Win__I2vlPdV__tlRTl((P21(a-1)/a I a _ l ( ~-~l )

-1

/ -

(3.27)

1)/1.456

W.ln 801.861.456__1(8.314)(300) ([(1102"5;<105;< 105 ))(1.456__

- 1 / ( 10001kJJ ) = 1456.95 kJ

----

From the energy balance, we estimate the heat discharged to surroundings

qout-

(°z-T)nRTl((P2)(a-1)/~ (3'-1) a - 1 (~ P1 )

(1.456--1.257)801.86(8.314)(300)((2.5;<105) (1"456-1)/1"456 qout = 1.257--1 1.456--1 ii0;<105

1//11k / Oooj)

(3.28)

qout = 1120.83 kJ Since the surroundings undergo a reversible isothermal process, we have ASsurr _ qsurr _ 1120.8____~3= 3.865kJ/K TO 290K The total entropy change is ASto t -- ASmethan e nt- ASsurr -- 3.250 + 3.865 = 7.115 kJ/K

3.3.4

Rate of Entropy Production

Using Eq. (3.8), the local value of entropy production • is related to the rate of entropy increase within the system by a volume integral (3.29)

p - diS - I d p d V dt

v

where • represents the entropy production due to local changes, while P or diS/dt characterizes the overall behavior of the system and may be called the volumetric rate o f entropy production. The term • is also known as the entropy source strength.

3.3.5

Chemical Reactions

For a closed system, if the change of mole numbers dNk is due to irreversible chemical reactions, the entropy production is 1 diS = - = ~_, txk d N k > 0

(3.30)

The rate of entropy production P is p _ diS _ _ _1 ~/xk dNk dt T dt

>0

(3.31)

3.3

The second law of thermodynamics

105

where/Xk is the chemical potential that can be related to measurable quantities, such as P, T, and No. In terms of the affinity A, Eq. (3.31) becomes

diSdt - ~-~(@-) dek>-Odt

(3.32)

For a reaction B ~ 2D, the affinity is A - #.~ - 2~D, and de/dt is the velocity of the reaction. At thermodynamic equilibrium, the affinity A and the velocity of the reaction vanish. 3.3.6

Diffusion

When a mass diffusion occurs in a closed system from higher chemical potential/Z 2 to lower potential/xl, we have the entropy production expressed by

diS--(tz2-1"/Cl) de'>OT

(3.33)

where de = - d N 1 = dN2. Here the flow of mass from one region to another is accounted for by the extent of reaction de, although no real chemical reaction takes place. The rate of entropy production is

p _ d i S - _(/x2 - / X l ) de dt T -~ > 0

3.3.7

(3.34)

Electrical Conduction

The rate of entropy production due to electrical conduction is d i S _ i)1

dt

T

(3.35)

where the product ~I represents the heat generated because of potential difference ~ and current I. This heat is also called Ohmic heat per unit time. Here the flow is the electric current and the corresponding force is O/T; the linear phenomenological equation is expressed by

~b I - L~ -T

(3.36)

where Le is the phenomenological coefficient. From Ohm's law ~ = IR, where R is the resistance, and hence,

Le - T/R. 3.3.8

Electrochemical Reactions

The rate of entropy production due to electrochemical reactions is

diS dt

A de T dt

(3.37)

where ~/is the electrochemical affinity defined by z~[i -- Ai -nt-2 i F ( O l - ~2 ) = ~61 - ~2

(3.38)

Here z/is the electrovalency of ionic species i, F is the Faraday number, which is the electrical charge associated with 1 mol ion of a species with an electrovalency of 1, and ~1 is the electrical potential at position 1. The term/2 i is the electrochemical potential of species i, and defined by

~ci - tz i + ziF6

(3.39)

106

3.

Fundamentals of nonequilibrium thermodynamics

The level of electrical current is related to the extent of the electrochemical reaction by

d~

I = ziF--;7 = ziFJ r

(3.40)

tit

For an isolated system, deS = 0 and diS > 0. However, for an open system, we have

deS =

dU + PdV T

+ (deS)matter and diS > 0

(3.41)

Systems that exchange entropy with their surroundings may undergo spontaneous transformation to dissipative structures and self-organization. The forces that exist in irreversible processes create these organized states, which range from convection patterns of Bbnard cells to biological cycles.

3.3.9

Rate of Energy Dissipation

The loss of energy is directly proportional to the rate of entropy production because of irreversible processes in a system. The loss of energy may be estimated based on the temperature of the surroundings of the system To, and we have

,oss-- zT0(dis) k dt

= (kg) (K) (kJ/(kg s K)) = kW

(3.42)

As Eq. (3.42) indicates, the surrounding conditions represent a state where the process reaches equilibrium at which the thermodynamic driving forces vanish. The value of energy Eloss is the rate of energy dissipated to the surroundings.

Example 3.3 Energy dissipation in a nozzle Steam enters a nozzle at 30 psia and 300°F, and exits as a saturated vapor at 300°E The steam enters at a velocity of 1467 ft/s, and leaves at 75 ft/s. The nozzle has an exit area of 0.5 ft 2. Determine the rate of energy dissipation when the environmental temperature is TO= 500 R. Solution: Assume that there are no work interactions, the potential energy effects are negligible, and the nozzle operates at steady state. The properties of steam: State 1: Superheated steam P~ = 30psia

T~ = 760 R

H 1 = 1189.0 Btu/lb m

S1 = 1.7334 Btu/(lb m R)

State 2: Saturated vapor T2 = 760R

H 2 = l179.0Btu/lb m

S2 = 1.6351Btu/(lb m R)

V2 =6.466ft3/lbm

The energy balance for a nozzle at steady-state conditions yields

Eout -- Ein

qout By estimating the steam flow rate, we can determine the heat loss from the nozzle:

1 rh = ~ A2v2

1 6.466(ft3/lbm)

(0.5 ft 2)(75 ft/s) =

5.8 lbm/s

3.3

107

The second law of thermodynamics

Therefore, the heat loss is

qout m/H2 H, (/out = -5.8[1179.7-1189.0 +

752 - 14672 ( 1Btu/lbm 2

~ 25,037 ft2/s 2

) = 302.55 Btu/s

The entropy balance contains the nozzle and its surroundings, and we have deS

diS

dt

dt

dS

-

dt

-0

Rate of net entropy + Rate of entropy production = Rate of change of entropy

qou, ) -+-Spr°d - -

th(S 1 - S 2 ) - - ~ a

Sprod -- 5.8(1.6351-1.7334)-+

0

302.55 500

(3.43) - 0.03501Btu/(s R)

The energy dissipated is /~loss -- ToSprod = 500(0.03501)= 17.506Btu/s

(3.44)

Example 3.4 Energy dissipation in a compressor Air enters a compressor at 15 psia and 80°E and exits at 45 psia and 300°E The inlet air velocity is low, but increases to 250 ft/s at the outlet of the compressor. The power input to the compressor is 250 hp. The compressor is cooled at a rate of 30 Btu/s. Determine the rate of energy dissipation when the surroundings are at 540 R. Solution: Assume that there are no work interactions, the potential energy effects are negligible, and steady flow occurs in the compressor. The properties of air can be obtained from Table E4 of Appendix: State 1: P~=15psia

T1 = 5 4 0 R

H l=129.06Btu/lb m

Sl=0.60078Btu/(lb mR)

=45psia

T2 = 7 6 0 R

H 2=182.08Btu/lb m

S2=0.68312Btu/(lb mR)

State 2:

The energy balance for a compressor at steady-state conditions yields /~7out = Ein Win +/41 H 1 + T

-~/°ut--/41

Using the energy balance, we can estimate the mass flow rate

H2+

108

3.

Fundamentals of nonequilibrium thermodynamics

W i n - Oout

- th( n2 - nl + v2-) 2v2

(250 hp)

0.7068 Btu/s]_ ( 2502 ]-@ ) 30 Btu/s - rh 182.08-129.06 + 2

1Btu/lb m ) ) 25037 ft2/s 2

rh = 2.7 lbm/s

The entropy balance contains the compressor and its surroundings, and we have deS + diS _ dS

dt

dt

dt

-0

Rate of net entropy + Rate of entropy production - Rate of change of entropy \

AS-

t)out / +

-- 0

ro )

(

,out/ =

r0

By taking into account the variable heat capacity, we have - A S = rh(S 2 - S 1 - R l n ~ ) - A S = 2.71bm/s(0.68312-0.60078-(0.06855Btu/lbm) l n ( 4 5 ) ) = 0.019Btu/(s R)

Sprod -- -- A,~ + !)out = 0 . 0 1 9 + 30 = 0 . 0 7 4 5 Btu/(s R)

TO

540

The energy dissipated is /~loss -- ToSprod -- 540(0.0745)- 40.26 Btu/s

Example 3.5 Energy dissipation in an adiabatic mixer In a mixer, we mix a hot water (stream 1) at 1 atm and 90°C adiabatically with cold water (stream 2) at 15°C. The hot water flow rate is 60 kg/h. If the warm water (stream 3) leaves the mixer at 30°C, determine the rate of energy dissipation if the surroundings are at 300 K.

Solution: Assume that the kinetic and potential energy effects are negligible, and this is a steady process. The properties of water from the steam tables in Appendix D: Stream 1: Hot water P~ = 100kPa

T1 =90°C

H 1 = 376.9kJ/kg

S1 = 1.1925kJ/(kgK)

P2 = 100 kPa

T2 = 15°C

H 2 = 62.94 kJ/kg

S2 = 0.2243 kJ/(kgK)

T3 = 30°C

H 3 = 125.7 kJ/kg

S3 = 0.4365 kJ/(kgK)

Stream 2: Cold water

Stream 3: Warm water P3 = 100 kPa

3.3

109

The second law of thermodynamics

The mass, energy, and entropy balances for the adiabatic mixer are Mass balance: thout -- thin ~ thl -t- th2 - th3

(3.45)

Energy balance:/~out

--/~in ~ thl Hi + th2H2 = th3 H3

(3.46)

Entropy balance: Sin - gout -+" Sprod = 0 --~ thiS 1 -Jr-th2S2 - th3S3 -t- Sprod = 0

(3.47)

Combining the mass and energy balances, we estimate the flow rate of the cold water thl H1 + th2 H2 = (thl + th2 ) H3

th2 = th~( H13 - H3 H 2 ) = 60 kg/h( 1376"9-125"7 2 5 . 7 - 6 2 . 9 4 ) = 2401 "53 The mass flow rate of the warm water is: th~ =thl +th: = 60.0 + 240.153 = 300.153 kg/h The rate of entropy production for this adiabatic mixing process is Sprod = th3S 3 -- thiS 1 - t h 2 S 2

Sprod = 300.153(0.4365) -- 60.0(1.1925) -- 240.153(0.2243) = 5.6 kJ/(h K) The energy dissipated because of mixing is: /~loss - -

~)'~prod

--

300(5.6) = 1680.0 kJ/h

Example 3.6 Energy dissipation in a mixer In a mixer, we mix a saturated steam (stream 1) at 110°C with a superheated steam (stream 2) at 1000 kPa and 300°C. The saturated steam enters the mixer at a flow rate 1.5 kg/s. The product mixture (stream 3) from the mixer is at 350 kPa and 240°C. The mixer loses heat at a rate 2kW. Determine the rate of energy dissipation if the surroundings are at 300 K. Solution: Assume that the kinetic and potential energy effects are negligible, this is a steady process, and there are no work interactions. The properties of steam from the steam tables: Stream 1: Saturated steam T1 = 110°C

H l = 2691.3kJ/kg

S 1 = 7.2388kJ/(kgK)

Stream 2: Superheated steam P2 = 1 0 0 0 k P a

T2 - 3 0 0 ° C

H 2 =3052.1kJ/kg

S2 - 7 . 1 2 5 1 k J / ( k g K )

2945.7 kJ/kg

S 3 = 7.4045 kJ/(kg K)

Stream 3: Superheated steam P~ = 350 kPa

T3 = 240°C

H 3 =

The mass, energy, and entropy balances for the mixer are:

Mass balance: thout -- thin--~ thl -t-th2 = th3 Energy balance:/~out

--/~in thlHl + th2H2 = 0out + th3H3

Entropy balance: Sin - Sour + '~prod = 0---~ thiS1 -Jr-t h 2 S 2 - th3S3 - qout _jr_ " = 0 T0 Spr°d

110

3.

Fundamentalsof nonequilibrium thermodynamics

Combining the mass and energy balances, we estimate the flow rate of the super heated steam

Oout --/hlH1 -F/h2H2 - (/hi mr-/h2)H3 /h2 - qout - - / h l (Hi - H 3 ) = 2 kW - 1.5 kg/s (2691.3 - 2945.7)kJ/kg = 3.605 kg/s H 2 -H 3 3052.1-2945.7 The mass flow rate of the warm water is:/h3 =/hi nt-/h2-- 1.5 + 3.605 = 5.105 kg/h. The rate of entropy production for this adiabatic mixing process is

Sprod --/h3S3 - / h i S1 -/h2 $2 q- qout

r0

2 kJ/s

o

Sprod = 5.105kg/s(7.4045)- 1.5 kg/s (7.2388)- 3.605 kg/s (7.1251) + ~

300

- 1.262 kJ/(s K)

The energy dissipated because of mixing is

L'loss -- T0Sprod = 300(1.262)= 378.76 kW

E x a m p l e 3.7 E n e r g y dissipation in a t u r b i n e A superheated steam (stream 1) expands in a turbine from

5000 kPa and 325°C to 150 kPa and 200°C. The steam flow rate is 10.5 kg/s. If the turbine generates 1.1 MW of power, determine the rate of energy dissipation if the surroundings are at 300 K. Solution: Assume that the kinetic and potential energy effects are negligible, this is a steady process, and there are no work interactions. The properties of steam from the steam tables: Stream 1" Superheated steam P1 = 5000kPa

T1 = 325°C

H 1 = 3001.8kJ/kg

S1 = 6.3408kJ/(kgK)

T2 = 200°C

H 2 =2872.9kJ/kg

S2 = 7.6439kJ/(kgK)

Stream 2: Superheated steam P2 = 150kPa

The mass, energy, and entropy balances for the mixer are

Mass balance:/hout = thin

Energy balance:/~out -"/~in ~/hill1 = 0out -]- Wout "]-/h2H2 Entropy balance: Sin - Sour + Sprod "- 0--}/hiS1 -/hiS2 - qout ..]_Sprod -- 0

r0

We estimate the heat loss from the energy balance ~¢out = -/~out +/hi(H1 - H2) = - 1100 kJ/s + 10.5 kg/s (3001.8- 2872.9) = 253.45 kJ/s

3.3

The second law of thermodynamics

111

And the entropy production from the entropy balance is 253.45 -- 14.526 kW/K kg/s (7.6439 - 6.3408) + ~ 300

Sprod -- ghl ($2 - S1) + qout __ 10.5

r0

The amount of energy dissipated becomes: 300(14.526)= 4 3 5 8 . 0 5 k W

/~loss -- ToSprod --

3.3.10 Stationary State Intensive properties that specify the state of a substance are time independent in equilibrium systems and in nonequilibrium stationary states. Extensive properties specifying the state of a system with boundaries are also independent of time, and the boundaries are stationary in a particular coordinate system. Therefore, the stationary state of a substance at any point is related to the stationary state of the system. In a stationary state the total entropy does not change with time, and we have dS

deS diS + -0 dt dt

dt

(3.48)

The term deS/dt is the reversible entropy change in time as a result of an entropy flow between the system and its surroundings. On the other hand, diS/dt represents the rate of entropy production inside the system. Equation (3.48) shows that the entropy exchange with the surrounding must be negative at stationary state

deS dt

--~

diS

< 0

dt

(3.49)

Therefore, the total entropy produced within the system must be discharged across the boundary at stationary state. For a system at stationary state, boundary conditions do not change with time. Consequently, a nonequilibrium stationary state is not possible for an isolated system for which deS/dt = 0. Also, a steady state cannot be maintained in an adiabatic system in which irreversible processes are occurring, since the entropy produced cannot be discharged, as an adiabatic system cannot exchange heat with its surroundings. In equilibrium, all the terms in Eq. (3.48) vanish because of the absence of both entropy flow across the system boundaries and entropy production due to irreversible processes, and we have deS/dt = diS/dt = dS/dt = O. For the total entropy to be constant the entropy flowing out of the system must be equal to the entropy entering the system plus the entropy generated within the system: d i S + (J~,, in - J s out ) -- 0 dt

'

'

(3.50)

From Eqs. (3.48) and (3.50), we have des d[

-

(Js,in - Js,out) < 0

(3.51)

So the stationary state is maintained through the decrease in entropy exchanged between the system and its surrounding. Entropy change inside an elementary volume by irreversible phenomena is the local value of the sum of entropy increments. By the second law of thermodynamics, the entropy production diS is always positive for irreversible changes and zero for reversible changes.

3.3.11 EntropyChange in aTwo-Compartment System Assume that region I in Figure 3.1 is an open system than can exchange both matter and energy with region II. The total entropy change in the system is dStotal -- d S I + d S II

(3.52)

112

3.

Fundamentals of nonequilibrium thermodynamics

From Eq. (3.8) we have d S I = de S~ + d i S I

(3.53)

d S II - - d e S I I = - - de SI

(3.54)

so that

since the entropy change in region II is solely due to exchange with region I. From Eqs. (3.52) to (3.54), it is clear that d S t o t a 1 ~ di SI

(3.55)

Thus di SI represents the total increase of entropy in the environment due to processes taking place in region I. It can be shown that the entropy production is associated with a loss of free energy or the capacity to do work. At constant temperature and pressure, the Gibbs free energy G measures the maximum work capacity, and the changes of Gibbs free energy in each region are dG I = dU I + PdV I - TdS I

(3.56)

dGII - d U H + P d V II - T d S II

(3.57)

Summing these two equations, and remembering that the volume and internal energy are constant in an isolated system, we have d G t o t a 1 = d G ~ + d G ~ = _ TdStota 1

(3.58)

Since for any real process d S t o t a 1 is necessarily positive, the free energy of the entire system decreases. The rate of decrease of the Gibbs free energy is of interest.

Example 3.8 Entropy production in a composite system Consider a composite system consisting of subsystem I enclosed inside subsystem II. The whole system containing subsystem I and II is isolated. However, in subsystems I and II some irreversible processes may take place. Discuss the total entropy production in the whole system. The second law of classical thermodynamics predicts that: clS = d S I + d S II >~ O. From Eq. (3.9) we may have the following two possible phenomena for each subsystem: (a) We may have: di $I > 0

and

diSII> 0

This phenomenon shows that in every macroscopic region of the system, the entropy production is positive and hence both processes are irreversible, which leads to: d S = d S I + d S II >~ 0 (b) We may also have: di $I >> 0

and

diS II < 0

This phenomenon shows that decrease or absorption of entropy in subsystem II may be compensated by a larger entropy production in subsystem I. This is possible only if subsystems I and II are coupled by some suitable coupling mechanisms leading to: d S = d S I + d S II>- O. With thermodynamic coupling a process in subsystem II may progress in a direction contrary to that determined by its own thermodynamic force. Some biological reactions represent coupled reactions for which the total entropy production is positive.

3.4

BALANCE EQUATIONS AND ENTROPY PRODUCTION

Balance equations of extensive quantities describe a change in a system (except in rare gases and shock waves). These balance equations also contain intensive parameters specifying the local state of a continuous medium. Intensive parameters described by the macroscopic properties of the medium are based on the behavior of a large number of particles.

3.4 Balanceequations and entropy production

113

It is necessary to consider the mechanics of a continuous medium to determine the thermodynamic state of a fluid. The properties of a fluid can be determined that are at rest relative to a reference frame or moving along with the fluid. Every nonequilibrium intensive parameter in a fluid changes in time and in space.

3.4.1

Total Differential

Consider the temperature as a function of time and space T = T (t, x, y, z); the total differential of T is expressed as

OT OT dT - .OT dt . + . dx +. ~ d y 3t 3x 3v

+

OT

dz

(3.59)

Oz

Dividing the total differential by the time differential, we obtain the total time derivative of T

dT _ OT + OT dx + _OT _ dv. + -OT - - - dz dt Ot Ox dt Oy dt Oz dt

(3.60)

The partial time derivative of T, (OT/Ot), shows the time rate of change of temperature of a fluid at a fixed position at constant x, y, and z

dT

OT

dt

Ot

(3.61)

If the derivative in Eq. (3.61) vanishes, then the temperature field becomes stationary. The terms dx/dt, dy/dt, and

dz/dt are the components of the velocity of the observer relative to the velocity of the fluid.

3.4.2

Substantial Derivative

If the velocity of the observer is the same as the mass average velocity of the fluid v with components v~, v,,, and v__, then the rate of temperature change is given by

DT

-

Dt

3T

OT

+ ~'.. ~

i~t

Ox

OT

+ v,.-

- Oy

OT

+ v_ - - -

(3.62)

- Oz

or

DT

OT

Dt

Ot

+v.VT

(3.63)

The special operator, DT/Dt is the substantial time derivative, and represents the time rate of change if the observer moves with the substance. A scalar or a vector function expressed in terms of O/at can be converted into the substantial form; for a scalar function T = T(x, y, z, t), we have

DT

O(pT) + Ot

Dt

=p

l

OT

+

aT

-7--+ v r - - +

Ot

-

3x

Ox

+

33' i~T

v~, - - + v _ - -

0~'

-

-

3z

~T)

+T

Oz

(Op

--+ at

O(pv,.) 3x

" +

O(pv,,)O(pv:)) Oy

+

3z

(3.64)

The second term in the second line of Eq. (3.64) is the equation of continuity and vanishes, so that in vector form we have

p

DT Dt

-

O(pT) Ot

~-(V.p vT)

(3.65)

This equation is valid for every local quantity, which may be a scalar, an element of a vector, or an element of a tensor.

114

3. Fundamentalsof nonequilibrium thermodynamics

3.4.3 Conservation Equation An extensive quantity E for a fluid in volume V can be expressed in terms of the specific quantity e, and we have

E = ~ pe dV

(3.66)

V

The partial time derivative of E pertaining to the entire body is equal to the total differential

OE Ot

dE dt

d ; pe dV = ~ O(pe) dV dt v v Ot

(3.67)

In Eq. (3.67) the quantity pe is determined per unit volume when the observer is at rest. The amount of substance entering through an elementary surface area dA per unit time is p v. dA, where dA is a vector with magnitude dA and pointing in a direction normal to the surface. Along with the substance flow there is a convection flow (p ve), and the amount transported per unit time is - f ( p u e ) . dA. The conduction flow Je is a vector with the same direction as the flow, and the amount transported per unit time by means of conduction without a flow of substance is - fJe" dA. The rate of energy production inside the elementary volume of substance at a given point is ~e -

dE dVdt

(3.68)

For the entire volume at rest relative to the coordinate system, the balance equation per unit time is expressed as

dEdt -!O(pe)ot d V = - f

A

+f

dV

(3.69)

Using the Gauss-Ostrogradsky theorem, Eq. (3.69) can be written over the entire volume

dEdt_!O(pe)otdV =-;[~'(pev)]dV-;(V'Je)dVv v +;~ev dV

(3.70)

From Eq. (3.70), the local balance equation for a fixed observer becomes

O(pe) Ot

--V'(pev)-V'Je +~e

(3.71)

The local balance equation for properties subject to a conservation law is called the conservation equation, which is given for e as follows

O(pe) - -V.(pev)- V'J e Ot

(3.72)

If the system is in a stationary state, the extensive property E does not change with time dE/dt- 0, and we have V.(J e + pev) = 0

(3.73)

Equation (3.73) shows that the net amount of E exchanged through the boundary must be zero, and the divergence of the sum of the conduction and convection flows governed by a conservation law is equal to zero in the stationary state. For the values e = 1, Je = 0, and @e -- 0, Eq. (3.71) becomes

Op _ _ V.(p v) = - p ( V . v) - v. Vp Ot

(3.74)

3.4

Balance equations and entropy production

115

The local balance equations for an observer moving along with the fluid are expressed in substantial time derivative form. From Eq. (3.71), we can express the substantial time derivative of e by

p De

+~c

--V.J<

Dt

(3.75)

On the right side of this equation, the divergence of the convection flow of e , - V • (pev), vanishes since the observer (coordinate system) is moving along with the fluid. In terms of the conservation law, where the source term vanishes, Eq. (3.75) becomes De

p~

Dt

-- - V . J

e

(3.76)

Engineering systems mainly involve a single-phase fluid mixture with n components, subject to fluid friction, heat transfer, mass transfer, and a number of 1 chemical reactions. A local thermodynamic state of the fluid is specified by two intensive parameters, for example, velocity of the fluid and the chemical composition in terms of component mass fractions w;. For a unique description of the system, balance equations must be derived for the mass, momentum, energy, and entropy. The balance equations, considered on a per unit volume basis, can be written in terms of the partial time derivative with an observer at rest, and in terms of the substantial derivative with an observer moving along with the fluid. Later, the balance equations are used in the Gibbs relation to determine the rate of entropy production. The balance equations allow us to clearly identify the importance of the local thermodynamic equilibrium postulate in deriving the relationships for entropy production.

3.4.4

Average Velocity

The mass flow of component i, pivi, is a vector showing the flow of a component relative to a motionless coordinate system. On the other hand, diffusion flow shows the transport of a component relative to a coordinate system moving at the reference velocity vr. The diffusion flow relative to the center-of-mass velocity v (or mass average velocity) is Ji -- P i ( V i

-- V)

(3.77)

where (vi-v) is the diffusion velocity. Mass average velocity is

v-- ~,wivi

(3.78)

i=1

Here wi is the mass fraction of component i. We can express the molar diffusion flow 3;,M based on the molar average velocity VM Ji,M

= C i ( V i -- VM)

VM

-- ~

XiV i

(3.79)

(3.80)

i=1

or based on the volume average velocity vv

Ji,t" =ci(vi-Vv)

(3.81)

1 ,~ Vivi

(3.82)

Vv -- 7 i=1

where V; is the partial molar volume. The sum of diffusion flows of all components is zero ~J, -~ i-:1

Ji ,M = ~

i=1

Of the n diffusion flows, only n-1 of them are independent.

i=1

J~,~.... = 0

(3.83)

116

3.4.5

3. Fundamentalsof nonequilibrium thermodynamics

The Mass Balance Equations

The mass balance equation for component i is similar to the general form given in Eq. (3.71) after setting e = w; and Je = Ji" The amount of component produced or consumed inside a unit volume per unit time is the result of chemi-

cal reactions. The mass balance equation is l

Ot~. _ - V ' ( t ~ v ) - V'ji + M i ~_~vi;Jv 0t

(3.84)

j=l

Here Jrj is the chemical reaction rate per unit volume for reaction j, % the specific stoichiometric coefficient of species i in the chemical reaction j, and Mi the molecular mass of component i. From Eqs. (3.76) and (3.63), we can represent the mass balance in the substantial time derivative

Dp i Dt

Opi

t

-- - - + v'VPi = - - V ' ( P i V ) - - V ' J i + mi E vijJrj + v'VPi

Ot

j=l

(3.85)

Using Eq. (3.74), Eq. (3.85) becomes Dp i

z = -p~ (V" v ) - V. Ji + Mi ~_~voS~j Dt j=l

(3.86)

When an observer moves at the center-of-mass velocity of the fluid, the conservation equation from the substantial derivative of the density becomes

Dp _ Op + v. VO = -p(V" v) Dt Ot

(3.87)

The conservation of mass with the specific volume v = lip is

Ov -

Dt

v ( v . v)

(3.88)

Using Eq. (3.86), the balance equation for the amount of substance can also be written in terms of mass fraction wi

O(pw/)

l

= _ v . ( p w/v) - v . j~ +

M, ~ vi;Jr;

0l

(3.89)

j=l

With the substantial derivative, Eq. (3.89) becomes

p

D(wi ) Dt

l - - V . Ji + Mi ~ VoJrj j=l

(3.90)

In the stationary state d m / d t - 0, and we have V.(pv) = 0

(3.91)

3.4.6 The Momentum Balance Equations Fluid motion may be described by applying Newton's second law to a particle. The momentum flow of a substance pvv is equal to the product of the mass flow pv and the barycentric velocity. Newton's second law of motion states that the change in the momentum of a body is equal to the resultant of all forces, mass force F and surface force or, acting on that body. If Fi is the force exerted per unit mass of component i, we have

F = ~1/ ~ F i P i=1

=

~wiF i i=1

(3.92)

3.4 Balanceequations and entropy production

117

The mass forces may be the gravitational force, the force due to the rotational motion of a system, and the Lorentz force that is proportional to the vector product of the molecular velocity of component i and the magnetic field strength. The normal stress tensor cr produces a surface force. No shear stresses occur ("r = 0) in a fluid, which is in mechanical equilibrium. The time derivative of the momentum density is given by 0 - - ( p v) - - V . ( p vv) + V.o- + p F Ot

(3.93)

By taking into account the following relations cr = - P g + ~"

(3.94)

V . ( P a ) - VP

(3.95)

0 - - ( p v ) - - V . ( p v v ) - VP + V.~- + pF Ot

(3.96)

Equation (3.93) becomes

where ~ is the unit tensor. The terms on the right side represent the change of momentum due to the convection momentum flow V • (pvv), the pressure force VP, the viscous force V. ~-, and the mass force oF, respectively. The momentum balance equation for a coordinate system moving along with the fluid is given by Dv ~ :-VP+V.~'+pF P Dt

(3.97)

The left side of Eq. (3.97) contains the center-of-mass acceleration dv/dt. The state of mechanical equilibrium is characterized by vanishing acceleration dv/dt = 0. Usually, mechanical equilibrium is established faster than thermodynamic equilibrium, for example, in the initial state when diffusion is considered. In the case of diffusion in a closed system, the acceleration is very small, and the corresponding pressure gradient is negligible; the viscous part of the stress tensor also vanishes "r = 0. The momentum balance, Eq. (3.97), is limited to the momentum conservation equation VP-

,~ ,qF~ = pF

(3.98)

i=1 Hence, the pressure gradient is equal to the sum of the mass forces acting on the substance in a unit volume.

3.4.7

The Energy Balance Equations

The time variation of the total energy e per unit volume is subject to a law of conservation, and given in terms of convection flow p ev and conduction flow Jc

O(pe) ht

- -V'(pev)- V'J e

(3.99)

The total specific energy of a substance e is e - u + - 1 1/2 +ep 2

(3.100)

and consists of the specific internal energy u, the specific kinetic energy l~v2, and the specific potential energy, ep. The conduction flow of the total energy J~. consists of the conduction flow of the internal energy J, the potential energy flow ( ~ = lepiji) due to the diffusion of components, and the work of surface forces - v . ~r, per unit surface area, and is expressed as follows

J ~' - Ju + ~ epiJi - v'o" i--I

(3.101)

118

3. Fundamentalsof nonequilibriumthermodynamics

The divergence of the total energy flow becomes (3.102) By using the mass flow, j i - Pi (v~- v), the time variation of the potential energy of a unit volume of the fluid is given by

O(pep) _ V.(pepV+~epiJi)_pF.v_~ji.Fi+£epiMi~vO.jr Ot i=1 i=1 i=1 j=l where the mass force F i is associated with the specific potential energy

j

(3.103)

epi of component i by (3.104 )

Fi = - V epi

with the properties of conservative mass forces

Oepi Ot

-0,

OFi Ot

=0

(3.105)

The last term on the right side of Eq. (3.103) is zero if the potential energy is conserved for the chemical reactionj

epiM iv/j -- 0

(3.106)

i=1 The balance equation for the kinetic energy is obtained by scalar multiplication of the momentum balance, Eq. (3.97), and the mass average velocity, and is given by p

D(pv2/2) ~ =-V.(Pv)+V.(v.z)+P(V.v)-z'(Vv)+pv.F Dt

(3.107)

The time variation of the kinetic energy per unit volume (for a motionless reference frame) is O(pv2/2)

~=-v.(lpv2v)-V.(Pv)+V.(v.~

") + P ( V . v ) - z " ( V v ) + p v ' F

(3.108)

Ot

In Eqs. (3.107) and (3.108), the relation o- = - P g + zis used. In Eq. (3.108), the term - 7 ' . (1/2 pvZv) is the convection transport of kinetic energy, V. (Pv) is the work of the pressure, V. (v. z) is the work of the viscous forces, and pv. F is the work of the mass forces. Part of the kinetic energy P(V.v) is transformed reversibly into internal energy, and the part - z : (Vv) is transformed irreversibly and dissipated. Combining Eqs. (3.108) and (3.103) under the conditions of conservation of energy given in Eq. (3.75) yields the total change in kinetic and potential energies per unit volume Op((1/2)v 2 + Ot

ep)__

V . [ p ( 2 v 2 + e p ) V _ v . o . + ~ e p i J- if]f ' ( V v ) - £ j i ' F i i = l i=1

(3.109)

We find the rate of change of the internal energy for an observer at rest by subtracting Eq. (3.109) from the total energy conservation relation Eq. (3.99) and using Eq. (3.102)

O(pu) -

Ot

- V" (puv) - V ' J u - P(V" v) + ~'" (Vv) + £ Ji" F/ i=1

(3.110)

The term - V . (puv) is the divergence of the convection internal energy flow, - V . J, is the divergence of the conduction internal energy flow, - P ( V • v) is the reversible increment of internal energy due to volume work, - z " (Vv)

3.4

Balance equations and entropy production

11 £

is the irreversible increment of internal energy due to viscous dissipation, and -ZiG lJi" Fi is the transport of potential energy by diffusion flows. We can represent Eq. (3.110) in terms of the substantial derivative Du

p~

Dt

=-V.J,-

P(V.v) + ~- • (Vv) +

Ji'E

(3.111)

,=~

The internal energy balance equation for the fluid is based on the momentum balance equation. The assumption of local thermodynamic equilibrium will enable us to introduce the thermodynamic relationships linking intensive quantities in the state of equilibrium and to derive the internal energy balance equation on the basis of equilibrium partial quantities. By assuming that the diffusion is a slow phenomenon, -YJ/=lJi/P << Pv2, the change of the total energy of all components per unit volume becomes

--t0[ i=1 ~ bli 3-~t'? 3-epi

(3.112)

--~--~ p U+--2 3-ep

This form is based on the concept of local thermodynamic equilibrium. From Eq. (3.112), the convection flow of the total energy is ~

( 1 ~ ) ~ ( l v 2 1 ~ i=1Pi b/i 3-~1"? 3-epi V i -- i=1 bliJi +p u+-2 +pe V3- i=1 epiJi (3.113)

= p e v + ~ u i J i +~epiJi i--1

i--I

Equation (3.113) contains the convection flow of the total energy and energy changes due to the diffusion flows. If jq is the pure heat conduction without a flow of internal energy due to diffusion of the substance, the total energy conservation given in Eq. (3.99) becomes ,t

O(pe) --V.(pev)-V. Ot We may relate the terms j,,

Jq3-

uij i3-

i=1

(3.114)

epiJi-v.~

i=1

Jq, and j; by !

Ju -- Jq 3-

(3.115)

Uij i

i=1 The second term on the right represents the net flow of internal energy transported along with the diffusion of species i.

3.4.8

The Entropy Balance Equations

The entropy balance form of Eq. (3.71) is

O(ps) ~t

- - V . ( p s v ) - V.J, + ~

(3.116)

Equation (3.116) shows that the rate of change of the entropy per unit volume of substance is due to the convection entropy flow psv, the conduction entropy flow j,., and the entropy source strength @. The conduction entropy flow is

J, - --Jq+ ~ siji • T i=1

(3.117)

The conduction entropy flow consists of the heat flow lq and the diffusion flow Ji. The jq is reduced heat flow that is the difference between the change in energy and the change in enthalpy due to matter flow. With the • f!

• t!

120

3.

Fundamentals of nonequilibrium thermodynamics

substantial derivative and using Eq. (3.117), we obtain the entropy balance equation based on local thermodynamic equilibrium

["£ /

Ds -17. Jq P D--~= -T +

sij~ + *

(3.118)

i=l

Example 3.9 Conservation of energy Describe the change of energy in closed and open subsystems in a composite system. The change of energy has two parts; deE is the part exchanged with the surroundings, and diE is the part produced or consumed within the system

(3.119)

dE = deE --}-diE

(a) Closed subsystems: For a closed subsystem with a chemical reaction characterized by the extent of reaction e, the total differential of E with respect to the variables V, T, and e is dE dV + - ~ dE = --~ T,e

v,e

dT + -~e

T,V

(3.120)

de

The energy flow from the surroundings is equal to the sum of the changes in heat and the mechanical work (pressure work) (3.121)

dE = 8 q - PdV

Combination of Eqs. (3.120), (3.121), and (1.22) yields (3.122)

6q = qvdV + C,,dT - ~ r d e

with

(oE)

OE

Cv= - ~ v,e where Cv is the heat capacity at constant volume and specified value of e, qv is the heat effect of pressure work (compression) at constant T, and ZkHrthe heat of reaction. In terms of enthalpy, the energy conservation is dH = 6 q + VdP

(3.123)

The total differential of enthalpy in terms of P, T, and e leads to dH :

--~

dP + - ~ T,e

dT + ~ V,~

de T,V

(3.124) 6q = HidP + C ; d T - AHrde

Having used the following definitions and Eq. (1.8)

P ,e

T ,e

T,P

where Cp is the heat capacity at constant pressure and e, and Hi the specific molar enthalpy of species i. (b) Open subsystems: Eq. (3.121) for an open subsystem must account for the exchange of matter with the environment, and is modified as follows dE = 6 q ' - PdV (for - dV)

(3.125)

3.5

121

Entropyproduction equation

Here 6q' accounts for the heat flow due to heat transfer as well as mass transfer. The enthalpy also is modified as follows d H = 6 q ' + VdP ( f o r - dP )

(3.126)

Subsystem I and I! may both exchange matter and energy, and we have dHI = 6I q, _

drill

V IdP~

= 6IIq, _ VIIdpII

(3.127)

where 6~q ' is the total flow of energy by phase ! for a time interval of dt. By assuming equal pressures P = pi = pH, we have the following change of the total enthalpy (3.128)

dH = 6Iq ' + 6 n q ' - ( V I + V I I ) d p

Comparing this equation with the first law of thermodynamics dH = 6q - VdP (for a closed system and for - dP), we obtain (3.129)

6q = 6~q ' + 6Hq '

where 6Iq ' is the summation of ordinary heat flow from the surroundings 6~q plus the energy flow 6~q from subsystem II 6Iq ' : 6~q + 6~q'

6nq ' : 6~'q + 6[Iq '

(3.130)

The total heat flow from the surroundings is 6q = 6~q + 6~'q

(3.131)

6~q'+6~Iq'=O

(3.132)

Therefore, from Eq. (3.130), we have

Equation (3.132) suggests that the energy flows exchanged between subsystem I and II are equal with opposite signs.

3.5

ENTROPY PRODUCTION EQUATION

Assuming that the local thermodynamic equilibrium holds, we can write the Gibbs relation in terms of specific properties

(3.133)

Tds - du + Pdv - £ ~idw i i=l

Equation (3.133) can be applied to a fluid element moving with the mass average velocity v. After replacing the differential operators with substantial time derivative operators in Eq. (3.133), we have Ds P Dt

p Du -

T Dt

pP Dv t

T Dt

p ~

Dw i

2 . , tx; ~

T i=~

Dt

The individual terms on the right-hand side of Eq. (3.134) are substituted by Eq. (3.111)

Du

£

p ~ = - V. J . - P(V. v) + "r" (Vv) + ji.Fi Dt ~=1

(3 1 3 4 )

122

3.

Fundamentals of nonequilibrium thermodynamics

by Eq. (3.88)

Dv Dt

pP - :

= P ( V . v)

and by Eq. (3.90) n D(wi ) l pXltLi ----~i(~7. ji) + Z AjJrj

i=1

Dt

)=1

where the affinity A of a chemical reactionj is

Aj --

(3.135)

Mild, iVij

i=I After the substitutions of the relations above, Eq. (3.134) becomes V.J~ I 1£ 1~ 1 t ~i~ ~---T7"" (Vv) -1- T i=1 Ji "Fi at- 7 i=1 i&i (V'ji) - "-f )=lZAj Jrj

Os

Dt

(3.136)

Using the following transformations

-v.

+Tao.vr

Eq. (3.136) reduces to Os ~---

P Dt

~V°

Ju - Z i = l l & i J i

r

_

1 j 1 tzi 1 1 £ AjJrj (3.137) T 2 "'VT - - Ji" TV T - Fi + Y r " ( V v ) - ~)=1 T ~=1

Comparison of Eqs. (3.132) and (3.137) yields an expression for the conduction entropy flow

"Is = T

+

siJi = 7

Ju -

i=1

tziJi

(3.138)

i=1

Using the relation between the chemical potential and enthalpy given by t.£i = h i - zS i = u i + PV i - zS i

(3.139)

tt we can relate the second law heat flow Jq, the conduction energy flow au, and the pure heat flow Jq! as follows

n tP

Jq = Ju - 2 h i J i

i=1

!

= Jq

PviJi

(3.140)

i=1

Heat flow can be defined in various ways if diffusion occurs in multicomponent fluids. The concept of heat flow emerges from a macroscopic treatment of the energy balance or the entropy balance. The internal energy of a substance is related to the molecular kinetic energy and the potential energy of the intermolecular interactions. If a molecule travels without colliding with other molecules, the loss of kinetic energy is due to diffusion. If the kinetic energy loss is the result of molecular collisions, it is classified as heat conduction. However, changes in the potential energy of intermolecular interactions are not uniquely separable. There is an ambiguity in defining the heat flow for open systems. We may split u into a diffusive part and a conductive part in several ways and define various numbers of heat flows. In the molecular mechanism of energy transport, the energy

123

3.5 Entropyproduction equation

of the system is associated with the kinetic energy of the molecules and with the potential energy of their interactions. The kinetic energy changes in an elemental volume are easily separated. If a molecule leaves the volume, the kinetic energy loss may be due to diffusion. If the kinetic energy loss occurs because a molecule at the surface of the volume transfers energy by collision to a molecule outside the volume, then this loss may be called heat flow. However, the potential energy of molecular interactions is the sum of the potential energies of interactions for each molecular pair. When some molecules leave the volume and other molecules collide at the surface with molecules outside the volume, they produce a complicated change in the potential energy. These changes cannot be uniquely separated into the contributions of pure diffusion and of molecular collisions. From Eqs. (3.118), (3.137), and (3.138), the entropy source strength or the rate of local entropy production per unit volume • is defined by

+ - j . . v 71 - - f1 ,:, Ji" r v --~ - V i +---fT-(Vv)--f

AjJrj

(3.141)

Equation (3.14 l) shows that • results from a sum of the products of conjugate flows and forces

di) -- ~ JiXi

(3.142)

i---1

Equation (3.141) identifies the following flows and forces to be used in the phenomenological equations. • Heat transfer:

(3.143) • Mass transfer:

-;

,3,44

where V0xi/T) = V(txi/T)r + hiV(1/T) • Viscous dissipation:

1

X v = - ~-(Vv)

(3.145)

1

• Chemical reaction:

T

,--z

T

u!/ ( j - 1 , 2 .... ,l)

(3.146)

Equation (3.141), first derived by Jaumann (1911 ), determines the local rate of entropy production by summing four distinctive contributions as a result of the products of flows and forces: • Entropy production associated with heat transfer q= J , X q

(3.147)

~d = ~J~Xi

(3.148)

• Entropy production due to mass transfer

i=1

• Entropy production as a result of viscous dissipation of fluid

~v ='r : Xv

(3.149)

124

3.

Fundamentals of nonequilibrium thermodynamics

• Entropy production arising from chemical reactions

~c : Z J r j - -

j=l

(3.150)

T

Equation (3.141) consists of three sums of the products of tensors that are scalars with rank zero ~0, vectors with rank one ~1, and tensor with rank two (I)2 1

1 l

*o : -~ z (V "v) - -~ ~_~Jrj Aj >-0

(3.151)

j=l

(I)l = a u ' V

(1)

l ~ji.[Fi_T~(_~)]~ 0

(3.152)

1 -r" (Vv) 's > 0

(3.153)

+ T- i=1

where t • (Vv) = "r' • (Vv) 's + z(V.v). The tensor (Vv)' is the sum of a symmetric part (Vv) 's and antisymmetric part (Vv) 'a, and the double dot product of these is zero.

3.5.1

Rate of Entropy Production

The time derivative of entropy production is called the rate of entropy production, and can be calculated from the laws of the conservation of mass, energy, and momentum, and the second law of thermodynamics expressed as equality. If the local entropy production, ~, is integrated over the volume, it is called the volumetric rate of entropy production

p_diS dt - I ~ P d V =

v

IZJiXi dV v i

(3.154)

This integration enables one to determine the total entropy production. When phenomena at the interface between two phases are considered, the amount of entropy produced is taken per unit surface area. Nonequilibrium thermodynamics estimates the rate of entropy production for a process. This estimation is based on the positive and definite entropy due to irreversible processes and of Gibbs relation

TdS = dU + PdV - ~ [&idNi

(3.155)

Entropy depends explicitly only on energy, volume, and concentrations because the Gibbs relation is a fundamental relation and is valid even outside thermostatic equilibrium. For an isotropic medium, the dissipation function or entropy rate can be split into three nonnegative parts /~/0

*-*o

/~/1

+ ' 1 + ' 2 - ~_~JiXi + ~ J i ' X i i=1

i=1

/12

+~Ji

"X i

(3.156)

i=1

where no is the number of scalar, n 1 the number of vectorial, and n 2 the number of tensorial (rank two) thermodynamic forces. The choice of thermodynamic forces and flows must ensure that in the equilibrium state when the thermodynamic forces vanish (X/= 0), the entropy production must also be zero. In contrast to entropy, the rate of entropy production and the dissipation function are not state functions since they depend on the path taken between the given states. 3.5.2

Dissipation

Function

From the rate of entropy production and the absolute temperature, we derive the dissipation function ~ , which is also a positive quantity

= TOp - T Z X i J i >_ 0

(3.157)

3.5 Entropy production equation

125

The increment of the dissipation function can be split into two contributions d~=dx~+dj~

(3.158)

Where d x ~ = TY_.,iJidX i and d j ~ = TY__,iXidJ i , When the system is not far away from global equilibrium, and the linear phenomenological equations are valid, we have d x ~ = dj~ = d ~ / 2 , and a stationary state satisfies d~_
- ~_~ Jr/~" >- 0

(3.159)

i=1

Classical thermodynamics states that the change of entropy production as a result of the irreversible phenomena inside a closed adiabatic system is always positive. This principle allows for the entropy to decrease at some place in the systems as long as a larger increase in the entropy at another place compensates for this loss. The quantities ~ and • are scalars; they are the products of two scalars, the dot product of two vectors, or the double dot products of two tensors of rank two.

3.5.3

NonequilibriumThermodynamic Postulates

The field of linear nonequilibrium thermodynamics provides a new insight into the transport and rate processes, as well as the coupled processes in physical, chemical, electrochemical, and biological systems. Nonequilibrium thermodynamics identifies the conjugated flows and forces from the rate of entropy production or from the dissipation function, and establishes the phenomenological equations with these forces and flows. Onsager's reciprocal relations relate the phenomenological coefficients pertaining to interactions or coupling between the processes. When the phenomenological equations relate the conjugate forces and flows linearly, the phenomenological coefficients obey the Onsager reciprocal relations. Therefore, the reciprocal rules reduce the number of unknown coefficients, which are related to the conventional transport and rate coefficients. There exist a large number of"phenomenological laws"; for example, Fick's law relates to the flow of a substance and its concentration gradient, and the mass action law explores the reaction rate and chemical concentrations or affinities. When two or more of these phenomena occur simultaneously in a system, they may couple and induce new effects, such as facilitated and active transport in biological systems. In active transport, a substrate can flow against the direction imposed by its thermodynamic force. Without the coupling, such "uphill" transport would be in violation of the second law of thermodynamics. Therefore, dissipation due to either diffusion or chemical reaction can be negative only if these two processes couple and produce a positive total entropy production. The phenomenological coefficients are important in defining the coupled phenomena. For example, the coupled processes of heat and mass transport give rise to the Soret effect (which is the mass diffusion due to heat transfer), and the Dufour effect (which is the heat transport due to mass diffusion). We can identify the cross coefficients of the coupling between the mass diffusion (vectorial process) and chemical reaction (scalar process) in an anisotropic membrane wall. Therefore, the linear nonequilibrium thermodynamics theory provides a unifying approach to examining various processes usually studied under separate disciplines. The form of the expressions for the rate of entropy production does not uniquely determine the thermodynamic forces or generalized flows. For an open system, for example, we may define the energy flow in various ways. We may also define the diffusion in several alternative ways depending on the choice of reference average velocity. Thus, we may describe the flows and the forces in various ways. If such forces and flows, which are related by the phenomenological coefficients obeying the Onsager relations, are subjected to a linear transformation, then the dissipation function is not affected by that transformation. The linear nonequilibrium thermodynamics approach mainly is based on the following four postulates: (i) The quasi-equilibrium postulate states that systems are not far from equilibrium; the gradients, or the thermodynamic forces are not too large. Within the system, local thermodynamic equilibrium holds. (ii) All flows in the system are a linear function of all the forces involved; the proportionality constants in these equations are the phenomenological coefficients. (iii) The matrix of phenomenological coefficients is symmetric provided that the conjugate flows and forces are identified by the entropy production equation or the dissipation function. (iv) The Curie-Prigogine principle states that the coupling is restricted by a general symmetry principle. If' the tensorial order of the flows and forces differs by an odd number, coupling cannot occur.

126 3.5.4

3.

Fundamentals of nonequilibrium thermodynamics

Gradient of Chemical Potential at Constant Temperature

Chemical potential is a function of T, Ni, and P, and the total differential of chemical potential is

:lo.il

~,ONi )r,P d N i +~" OT

dtxi

Ni,P d T + k. OP

Ni ,T

(3.160)

dP

Using the Gibbs energy density g, the second partial term becomes the partial molar entropy of species i

co.i

o(o /

o(ox/

_ _

t. 0 T ) Ni ,p

_ _

0T

_ _

_ _

=

ONi -'~ Ni ,P

T,P

--Si

T,P

If we consider a system under mechanical equilibrium, dP = 0, and the third term in Eq. (3.160) vanishes. Equation (3.160) then becomes VI./fi "-- (VId~i )r,P - si V T or V[L i Jr- S i V T = (Vtz i

3.5.5

)r,P

(3.161)

Simultaneous Heat and Mass Transfer

Equation (3.152) represents the entropy production for vectorial processes of heat and mass transfer. In Eq. (3.152), the conduction energy flow can be replaced by the heat flow Jq using Eq. (3.140) and the total potential/x* comprising the chemical potential and the potential energy per unit mass of component i/x* =/zl + epi where Vepi = - F i. Using Eq. (3.161), the isothermal gradient of the total potential is (3.162)

V T l~i = V[L i @ S i V T + Vepi

From the thermodynamic force for mass transfer, we have V T + l~ep i = ~Tl.Li _ hi VTT •

TV

-

-- E" = V l&i -- I&i T

(3 163)

where /[£i -- hi - T s i. Using the relation above, we can rearrange Eq. (3.152) as follows

-T1 i=1 ji "VTtXi

,,

(I)1 =Jq'V

>-- 0

(3.164)

Since only the n-1 diffusion flows are independent, we have n--1 Ji "VT~; = Z Ji" VT (/J'~ -- ~ ; ) i=1 i=1

Introducing this equation into Eq. (3.164), we have ,, ~1 = a q ' V

1

-T/=1

j i . V T ( i ~ ; _ tZn) > 0

(3.165)

Therefore, the thermodynamic driving force of mass flow becomes ,

xi

=

(3.166)

127

3.6 Phenomenologicalequations

We relate the dissipation function to the rate of local entropy production using Eqs. (3.151)-(3.153) xIJ' = T(~ = r ( %

+ ~ 1 + ( I ) 2 ) = xt/'o + xI$1 + xI*2

(3.167)

If the dissipation function identifies the independent forces and flows, then we have l

(3.168)

*0 = 7 (V" v ) - E A / J r / > - 0 j=l n-I tt

*

*

~ l - - - J q . V l n T - ' ~ . . , ji'VT(l, Zi --~n)Z>O

(3.169)

i=1

q~2 - a"" (Vv) 's _> 0

(3.170)

We can modify Eq. (3.169) using the following transformation of Eq. (3.144)

TV

]&i

T

/&i V T 4- Vepi = V l~ ~ - tJ~--j-iV T T T

- Fi - V I i i - ~

(3.171)

and •

n-1

•

Ji" V/'ti = E Ji" t~7(/"Li --/{L; ) i=1

(3.172)

i=I

Therefore, Eq. (3.169) becomes n-I qff , - - J , "V T - E Ji . v(/x7 - / x ; ) > _ 0 i--1

(3.173)

As shown by Prigogine, for diffusion in mechanical equilibrium, any other average velocity may replace the centerof-mass velocity, and the dissipation function does not change. When diffusion flows are considered relative to various velocities, the thermodynamic forces remain the same and only the values of the phenomenological coefficients change. The formulation of linear nonequilibrium thermodynamics is based on the combination of the first and second laws of thermodynamics with the balance equations including the entropy balance. These equations allow additional effects and processes to be taken into account. The linear nonequilibrium thermodynamics approach is widely recognized as a useful phenomenological theory that describes the coupled transport without the need for the examination of the detailed coupling mechanisms of complex processes.

3.6

PHENOMENOLOGICAL EQUATIONS

In nonequilibrium systems, spontaneous decaying phenomena toward equilibrium take place. When systems are in the vicinity of global equilibrium, linear relations exist between flows J; and thermodynamic driving forces Xk Ji = LikXk

(3.174)

where the parameters Li~Tare called the p h e n o m e n o l o g i c a l coej.ficients. For example, Fourier's law relates heat flow to the temperature gradient, while Fick's law provides a relation between mass diffusion and concentration gradient. The temperature and concentration gradients are the thermodynamic forces. The Fourier and Fick laws consider a single force and a single flow, and are not capable of describing coupled heat and mass flows. Choice of a force Xi conjugate to a flow J; requires that the product JiXi has the dimension of entropy production. The validity of Eq. (3.174) should be determined experimentally for a certain type of process; for example, linear relations hold for an

128

3.

Fundamentals of nonequilibrium thermodynamics

electrical conductor that obeys Ohm's law. Fluctuations occurring in turbulent flow deviate relatively little from the local equilibrium state. If a nonequilibrium system consists of several flows caused by various forces, Eq. (3.174) may be generalized in the linear region of the thermodynamic branch (Figure 2.2), and we obtain

Ji = Z LikXk

(3.175)

k

These equations are called the phenomenological equations, which are capable of describing multiflow systems and the induced effects of the nonconjugate forces on a flow. Generally, any force X/can produce any flow Ji when the cross coefficients are nonzero. Equation (3.175) assumes that the induced flows are also a linear function of nonconjugated forces. For example, ionic diffusion in an aqueous solution may be related to concentration, temperature, and the imposed electromotive force. By introducing the linear phenomenological equations, into the entropy production relation, • = ~JX, we have

OP-- ~ LikXiX k ~ 0 i,k=l

(3.176)

This equation shows that the entropy production is a quadratic form in all the forces. In continuous systems, the base of reference for diffusion flow affects the values of transport coefficients and the entropy due to diffusion. Prigogine proved the invariance of entropy for an arbitrary base of reference if the system is in mechanical equilibrium and the divergence of viscous tensors vanishes. Equation (3.176) leads to a quadratic form

f~--~tiiX?-.[- ~ Lik[-Lki XiXk ~0 i=1 i,k=l 2

(i 4: k)

(3.177)

or the following matrix form [Lll L12 ...L1 n

*-

X1

~.LikXiXk-[X1X2...Xn]

IL~I...Lz.2..::.'.L.z" X2 >-0

i,k=l

LLml Lm2""Lnn

(3.178)

Xn

A necessary and sufficient condition for • >--0 is that all its principal minors be nonnegative

Lii Lik = LiiLkk -- LikLki >--0 L~ Lkk

(3.179)

If only a single force occurs, Eq. (3.178) becomes

dp= LiiX2 >_0

(3.180)

and then the phenomenological coefficients cannot be negative Lii > O. For a system at metastable equilibrium, we may have J / = 0, • = 0, and Xk 4: O. Otherwise, all forces and flows are independent, and the inequality sign holds in Eq. (3.180).

3.6.1

Flows

Mass flow, heat flow, and chemical reaction rate are some examples of the "flows" J/. The thermodynamic "forces" X/ of the chemical potential gradient, temperature gradients, and the chemical affinity cause the flows. The affinity A is

Aj =-~VijlX i

(j = 1,2,...,/)

(3.181)

i=1

where vii is the stoichiometric coefficient of the ith component in the flh reaction, n the number of components in the reaction, and 1 the number of reactions.

3.6

129

Phenomenological equations

The flows may have vectorial or scalar characters. Vectorial f l o w s are directed in space, such as mass, heat, and electric current. Scalar flows have no direction in space, such as those of chemical reactions. The other more complex flow is the viscous flow characterized by tensor properties. At equilibrium state, the thermodynamic forces become zero and hence the flows vanish

Ji,eq (Xi

(3.182)

= 0) = 0

As an example, the diffusion flow vector J; for component i is the number of moles per unit area A per unit time t in a specified direction, and defined by J; -

1 dN i A

(3.183)

dt

Considering a small area dA at any point x, y, z perpendicular to average velocity vi, in which v i is constant, the volume occupied by the particles passing dA in unit time will be v~dA. If the concentration per unit volume is ci then the total amount of the substance is

N i ~--cividA

(3.184)

The local flow, which the amount of substance passing in a unit area per unit time, is

Ji --CiVi

(3.185)

Generally, these three quantities 'Ji, Ci, and v i are the functions of the time and space coordinates. If the area dA is not perpendicular to the flow vector, we consider a unit vector i, perpendicular to dA, whose direction will specify the direction of the area (3.186)

d A = idA

The scalar product vi" dA gives the volume dV, which is multiplied by the local concentration ci to find differential flow dQi , which is the amount of the substance passing an area at any angle with the velocity vector vi dQi

=

Ji

" dA

(3.187)

= civ i • d A

For a volume enclosed by a surface area A, the total amount of species i leaving that volume is Q, = I J i .dA

(3.188)

A

The divergence of the flow Ji is

V.J;-

OJi x '

Ox

+

OJi

'Y +

Oy

OJi -

'~

(3.189)

Oz

Here J;.x, Ji,y, and Ji._- are the Cartesian coordinates of the vector Ji. As the volume V and the product Ji" dA are scalars, the divergence is also a scalar quantity. A positive divergence means a source of component i, while a negative divergence indicates a sink, and at points of div Ji = 0, there is no accumulation and no removal of material. Transformation of the surface integral of a flow into a volume integral of a divergence is

I V'JidV l"

- IJi "dA

(3.190)

A

The divergence of the mass flow vector pv is used in the continuity equation Op

0t

- - ( V . pv)

(3.191)

130

3.

Fundamentals of nonequilibrium thermodynamics

Similarly the local equivalent of the law of conservation of mass for an individual component i is OCi - -

Ot

-(V'Ji)

(3.192)

Equation (3.192) cannot describe a flow process for a reacting component. Another conserved property is the total energy, and in terms of local energy density Pe for each point in the system, we have 0Pe - -(V. je ) Ot

(3.193)

where Je is the energy flow. The total entropy of a system is related to the local entropy density Sv

S = f sv dV v

(3.194)

dS _ r os v dV Ot ~ Ot

(3.195)

The total entropy changes with time as follows

The entropy flow is, on the other hand, is the result of the exchange of entropy with the surroundings

deS - -j" Js" dA -- -j" (V.js) dV Ot A V

(3.196)

An irreversible process causes the entropy production • in any local element of a system, and the rate of total entropy production is

diS - f ~ dV dt v

(3.197)

From Eqs. (3.196) and (3.197), the total change in entropy becomes

dS_deS

diS

dt

dt

dt

(3.198)

Inserting Eqs. (3.195) to (3.197) into Eq. (3.198), we obtain

f OSv dV = -[. (V.js) dV + [. • dV Ot v v v

(3.199)

Therefore, for any local change, an irreversible process in a continuous system is described by

OSv - - V . js + ~ Ot

(3.200)

Equation (3.200) is the expression for a nonconservative change in local entropy density, and allows the determination of the entropy production from the total change in entropy and the evaluation of the dependence of • on flows and forces. Stationary state flow processes resemble equilibria in their invariance with time; partial time differentials of density, concentration, or temperature will vanish, although flows continue to occur in the system, and entropy is being produced. If the property is conserved, the divergence of the corresponding flow must vanish, and hence the steady flow of a conserved quantity is constant and source-flee. At equilibrium, all the steady-state flows become zero.

131

3.6 Phenomenologicalequations

At stationary state, the local entropy density must remain constant because of the condition Osv/Ot= 0. However, the divergence of entropy flow does not vanish, and we obtain q~ = V.j~

(3.201)

Equation (3.20 l) indicates that in a stationary state, the entropy produced at any point of a system must be removed by a flow of entropy at that point. In the state of equilibrium, all the flows including the flow of entropy production vanish, and we obtain the necessary and sufficient condition for equilibrium as • =0

(3.202)

Equation (3.200) may be useful in describing the state of a system. For example, the state of equilibrium can be achieved for an adiabatic system, since the entropy generated by irreversible processes cannot be exchanged with the surroundings.

3.6.2 Thermodynamic Forces For thermodynamic vectorial forces, such as a difference in chemical potential of component i, proper spatial characteristics must be assigned for the description of local processes. For this purpose, we consider all points of equal /.Li as the potential surface. For the two neighboring equipotential surfaces with chemical potentials/.L i and ].Li -+- d/xi, the change in IX~with number of moles N is Otxi/ON, which is the measure of the local density of equipotential surfaces. At any point on the potential surface, we construct a perpendicular unit vector with the direction corresponding to the direction of maximal change in/x~. With the unit vectors in the direction x, y, and z denoted by i, j, and k, respectively, the gradient of the field in Cartesian coordinates is grad/x; = i O]&i + i

Ox

Ol&i + k OIJbi Oy

Oz

(3.203)

A thermodynamic driving force occurs when a difference in potential exists, and its direction is the maximal decrease in ~;. Consequently, at the point x, y, z, the local force X causing the flow of component i is expressed by Xi =-grad/x i

(3.204)

For a single dimensional flow, Eq. (3.204) becomes

Ot~i Xi - ' - - - i ~ Ox

(3.205)

From the definition of the chemical potential, we have

Ol~i

O

Ox

Ox

ONi

=-~i

Ox

(3.206)

where -dG shows the free energy available to perform useful work, 6 W, and the differential of work with distance, 6 W/dx, is a force. Therefore, Xi is a force per mole of component i, causing a flow in the direction of the unit vector. The overall thermodynamic force that is the difference in chemical potential for the transport of the substance between regions 1 and 2 in discontinues systems is the integral of Eq. (3.191) 2 f Xidx l

20

-if

i'zi a OX

dx

-

i (/.Li,1 -/dq, 2 ) -= ~/.L i

(3.207)

I

Here, ~]J'i is a difference in potential, while Xi is a conventional force used in classical mechanics. Electric potential q~that causes a current at the point x, y, z lead to the definition of electric force Xe Xe =-grad where Xe is the force per unit charge, or the local intensity of the electric field.

(3.208)

132

3.

Fundamentals of nonequilibrium thermodynamics

When we consider the difference in electric potential between two points instead of local electric forces, the quantity of electromotive force AO is defined in a single direction by 2 dl//

a~b = - i I ~ clx = i (~1 -- ~2 ) ldx

(3.209)

Other types of forces of irreversible processes may be derived similarly. In general, the flows and forces are complicated nonlinear functions of one another. However, we can expand the nonlinear dependence of the flows J/and the forces Xi in a Taylor series about the equilibrium

-Ji = Ji,eq(Xj = 0)+ ~ ( OJi ] j=l~OXj )eq xj

1 ~ (02Ji] 2 ( 022 )eq xj Jr'""

"~- ~.T j = l

- X i "- Xi,eq(Jk --0)+ ~ ( Oxi ] 1 ~(02Xi,) 2 ~ 0~k)eq Jk +-~..k=l OJ2 )eq J~ +"" k=l

(3.210)

(3.211)

If we disregard the higher order terms, these expansions become linear relations, and we have the general type of linear phenomenological equations for irreversible phenomena

Ji = ~LikXk

(i, k= 1,2,...,n)

(3.212)

i=1 ?7

Xi = Z KikJ k

(3.213)

k=l

Equation (3.212) shows that any flow is caused by all the forces, whereas Eq. (3.213) shows that any force is the result of all the flows present in the system. The coefficients Lik and Kik are called the phenomenological coefficients. The coefficients Lik are the conductance coefficients and Kik the resistance coefficients. The straight coefficients with the same indices relate the conjugated forces and flows. The coefficients with i 4=k are the cross coefficients representing the coupling phenomena. According to the principle of Curie-Prigogine, vector and scalar flows are able to couple only in an anisotropic medium. This theory has important consequences in living cells.

3.7

ONSAGER'S RELATIONS

Onsager's reciprocal relations state that, provided a proper choice is made for the flows and forces, the matrix of phenomenological coefficients is symmetrical. These relations are proved to be an implication of the property of "microscopic reversibility", which is the symmetry of all mechanical equations of motion of individual particles with respect to time t. The Onsager reciprocal relations are the results of the global gauge symmetries of the Lagrangian, which is related to the entropy of the system considered. This means that the results in general are valid for an arbitrary process. The cross-phenomenological coefficients are defined as (3.214)

(3.215)

t,

)

Ji-o

3.8

Transformation of forces and flows

133

The phenomenological coefficients are not a function of the thermodynamic forces and flows; on the other hand, they can be functions of the parameters of the local state as well as the nature of a substance. The values of Lik must satisfy the conditions L;; > 0

1

Lii Lkk > ~ (Lik + Lki

(i = 1,2,...,n)

)2

(i =/:k; i, k = 1, 2,..., n)

(3.216)

(3.217)

or

Kii > 0 (i -

1 ): KiiKkk > -4 (Kik + K~q

1,2,...,n)

(i 4= k; i,k = 1,2 .... ,n)

The matrix of phenomenological coefficients Lki and Kk/are related by K = L-1 where L-~ is the inverse of the matrix L. In a general matrix form in terms of the conductance Lij and resistance Kij coefficients, Eq. (3.178) becomes - XTLX = jTKj

(3.218)

This relation suggests that the local rate of entropy production is a quadratic form in all forces and in all flows if the cross coefficients differ from zero.

3.8

TRANSFORMATION OF FORCES AND FLOWS

Consider a system in which the thermodynamic forces are independent, while the flows are linearly related

0 - £ YkJk

(3.219)

k=l

If the constants are nonvanishing, y~ :~ 0, then we have the flow for component n

j~ = _ ~ k=l

/

Yk Jk

(3.220)

Using Eq. (3.220) in the local entropy production equation, we have

=

Jk Xk = ~ Jk Xk -- Y_LX,, k=l

k=l

Yn

(3.221)

Equation (3.221) has n-1 independent forces ( X k - (vk/y~)Xn) and n - 1 independent flows of Jk, therefore, the phenomenological equations are

Ji - £ LikXk (i,k = 1,2,...,n)

(3.222)

k=l or

Ji - ~,~ L'ik Xk - Y__LX , k--1

Yn

(i,k = 1,2,...,n- 1)

(3.223)

134

3.

Fundamentals of nonequilibrium thermodynamics

Onsager's reciprocal rules lead to (L~ =

Li'k).Substituting Eq. (3.223) in Eq. (3.220), we have Jn - - - - £ L~. i,k=l

Xk-

YiYky2X,,)

(3.224)

The comparison of phenomenological coefficients yields

Yk

n-1

n-1

Lik--Ltik' Lin----ZEikk=l \-~n)' L"i=-ZL'ik~,TSn)k=l

Yk Lnn = ~" r,k k=l

Equations (3.224) and (3.225) show that the Onsager reciprocal relations remain valid From Eq. (3.219), we have

0 = ~ YkLik k=l

0=

~.~YiLik

YiYk

~--~n 2 )

Lik = Lki (i, k =

(3.225) 1,2, ..., n).

(3.226)

i=l

Here, only (2n-l) equations are independent. From Eq. (3.226), we have

0 = ~ yiYkLik

(3.227)

i,k--1 These results prove that the Onsager reciprocal relations remain valid when homogeneous relationships relate the flows to each other.

3.8.1 Two-Flow Systems For a two-flow system, we have the phenomenological equations in terms of the flows J1 = LllX1 + L12g2

(3.228)

J2 -- L21X1 + L22X2

(3.229)

L22 x1-qg

(3.230)

From these relations we can derive the forces L12

L21 Lll X2 =-]L----~J1 + - ~ J 2

(3.231)

We can also write the phenomenological equations in terms of the forces X 1 = K l l J 1 Jr-K12J 2

(3.232)

X 2 = K21J 1 + K22J 2

(3.233)

The following relations link the phenomenological coefficients of Lik to Kik __ L22. L12. L21. Lll Kll--~' K12 = ILl' K21 = ILl' K22 - I L l

where [LI is the determinant of the matrix: ILl =

LllLz2- LIzLzl

(3.234)

3.8

135

Transformation of forces and flows

From Eq. (3.234) and Onsager's relations, we have

L22 ZllZ22 -Z12L21

L22 Zl 1L22 -(/-.12) 2

zt2 ZllZ22 -L12L21

K12 --K21 --

=

Zl 1L22 -- (LI2) 2

/-,11 Zl 1L22 -- (L12) 2

Lll K22

L21

ZllZ22 -LI2L21

Example 3.10 Relationships between t h e conductance and resistance phenomenological coemcients F o r a three-flow system, to derive :the relationships between the conductance and resistance phenomenological coefficients, Consider the linear phenomenological equations relating forces to flows with resistance coefficients,

KllJ 1+ K i2J2 + K13J 3 :

::

:

.........

....

& = K21J1 :,::: K22J2 +K23J3

:

After applying the Onsager relations to the linear matrix solutions, we: have

.

.

.

(K12

J1-

.

.

::

D

.

: :

.

.

..

.

.

.

.

3:- K23K13)i ..... x~ D : .... ' .

.

.

.

.

.

. .

J2 = L21X1 + ~ 2 X 2 + ~3X3 .

.

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~ l & + ~2X2 + L33X3

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.....

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+(

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::. . . .... 3K 23-K[2K-K11K23 2 33 : . . . . . +K22K33 . . . where D - -g~ag22 +. .:2K12K1

::

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:::;

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Example 3.11 ~ a n s f o r m a t i o n of phenomenological equations: dependent flows Transform the t h e ~ o d y namic forces and flows w h e n the forces are independent, while the flows are ~linearly dependent in a two-flow system: 0 = y J l + J2. : : i The local entropy production is : ...... ..... . . . . . . . ............. ?

.

.

.

.

.

:

~

=

JIXI + J2X2

:

:i

:i(3.235)

136

3.

Fundamentalsof nonequilibrium thermodynamics

Using the linear relation between the flows, 0 = Y J1 +

J2,

Eq. (3.235) becomes

= J~& - yJ1X2 = Jl(Xl

- yx2)

(3.236)

The linear phenomenological equations are J l = LI~& +&2X2

(3.237)

J2 = L21X1 + L22X2

(3.238)

However, from Eq. (3.236), the modified phenomenological equation for J1 is J1 = L ~ I X ~

(3.239)

J2 = - y J i = - y / 4 ~(Xl - yx2 )

(3.240)

where X[ = X1 -yX2 Then, the second flow is

Using the linear relation between the flows in Eqs. (3.237) and (3.238), we have - YJ1 -- - y L l l X 1 - yL12 X2

(3.241)

J2 = 521Xl -I- Z22X 2

(3.242)

Subtracting Eq. (3.242) from Eq. (3.241), we get yLll -+-L21 = 0----~L21 -- - y L l l

522

yL12 nt- L22 -- 0---+ L12 = - - -

(3.243) (3.244)

Substituting Eq. (3.244) into Eq. (3.237), we find J1 -- L i l Y 1 - L22 X 2

(3.245)

Y

Comparing Eq. (3.245)with Eq. (3.239), we have J1 -- Z l l X l - L22 X2 J( "-/-JII(X1 "+-y X 2 )

Y Therefore, L22

L l l - y2

(3.246)

From Eqs. (3.243) and (3.244), we have L22 Ll2 =

y2L11 -

Y

- yL11

Y

(3.247)

Equation (3.247) shows that Onsager's reciprocal relations are satisfied in the phenomenological equations (Wisniewski et al., 1976).

3.8

Transformation of forces and flows

137

Example 3.12 Transformation of phenomenological equations: dependent forces Transform the thermodynamic forces and flows when the flows are independent, while the thermodynamic forces are linearly dependent in a two-flow system: 0 = yX~ + X2. The local entropy production is rip = J1X1 -+-Y2 X 2

Using the linear relation between the thermodynamic forces, 0 - yX1 + X2, the local entropy production becomes (# -- Yl X l - Y J2 X l "- X1 ( J1 - Y J2 )

(3.248)

The linear phenomenological equations in terms of the resistance coefficients are X1 = KllJ1 + K12J2

(3.249)

X 2 = K21J 1 + K22J 2

(3.250)

However, From Eq. (3.248), the modified phenomenological equation for the force Xa is X 1 = K ' J ' = K ' (Yl - Y J2 )

(3.251)

Then, the second thermodynamic force is

x2

(3.252)

= - y K ' ( J~ - y J 2 )

Using the linear relation between the forces in Eqs. (3.249) and (3.250), we have y X 1 = y K 11Jl + YK12J 2

(3.253)

- Y X 1 - K21J1 + K 2 2 J 2

(3.254)

Subtracting Eq. (3.254) from Eq. (3.253), we get y K l l + K21

= 0-+-

K21

-

Kll

Y )'Kle + K22 = 0 - * K12 = -

~2

(3.255)

(3.256)

Substituting Eq. (3.256) into Eq. (3.249), we find K.. XI = K11J1 - 77/-z J2 Y

(3.257)

Comparing Eq. (3.257) with Eq. (3.251), we have K22 Xt - K"lt - YK'J2 = KllJl -~J2 Y

and K11 -

K22

v2

and

K12 = K21

(3.258)

These results show that Onsager's relations are satisfied in Eqs. (3.249) and (3.250) since the dependency of the forces to the flows are linear (Wisniewski et al., 1976).

138

3.

Fundamentals of nonequilibrium thermodynamics

Example 3.13 Transformation of phenomenological equations: dependent flows and forces Transform the phenomenological equations when the flows and forces are linearly dependent: 0 = z J1 + J2 and 0 = yX1 + X2. The local entropy production is ~ = J1X1 +J2X2 = J1X1 + ( - z J 1 ) ( - Y X 1 ) = J 1 Y l ( l + z y )

(3.259)

The linear phenomenological equations are J1 = L11X1 "+-L12X2 -- L ' X '

(3.260)

J2 "- L21X1 + L22X2

(3.261)

1 =(l+zy) -T

(3.262)

where X' is defined by X'=(l+zy)X

=

The phenomenological coefficient is L ' = L~I

(3.263)

Using the linear flows and forces in Eqs. (3.260), (3.261), and (3.262), we have J1 = L' (1 + zy) 321

(3.264)

J2 = - z L ' (I + zy) X1

(3.265)

J1 - - ( L l l - yL12)X1

(3.266)

J2 -- (L21 - yL22) Yl

(3.267)

From Eqs. (3.260) and (3.261), we obtain

Comparing Eqs. (3.264) and (3.265) with Eqs. (3.266) and (3.267), we find ( L l l - yL12) X 1 = L ' ( l + z y ) X 1 - - + ( l + z y ) L ' = I_ql-YL12

(3.268)

(L21- yL22) X1 = - L ' z ( l + z y ) X 1 --+-(l + z y ) L ' z = L21-YL22

(3.269)

In a two-flow system, there are two degrees of freedom in choosing the phenomenological coefficients. With the linear relations of flows and forces, there is one degree of freedom that is L12 -- L21, and L22 is proportional to L' L22 = wL'

(3.270)

With Eq. (3.270), the relations in Eqs. (3.268) and (3.269) become L21 - yLze = - L ' z ( 1 + zy) L21- y(wL') = - L ' z ( 1 + zy) L21 = - L ' (z + zZy) + ywL'

(3.271)

L12 -- L21 -- - L ' (z + z2 y - yw) and

&l - y & 2 = L' ( 1 + z y ) LI~ = L'(l+zy)- yL'(z+z2Y- yw) Lll = L ' ( 1 - z Z y 2 +y2w)

(3.272)

3.10

139

Heat conduction

Since the local entropy production is positive if L i i > O , L' =L{1 > 0 , L' = L{I > 0 and Lii > 0 restrict w to positive values, and we have

2

and L l l L 2 2 - L 1 2 > O ,

the conditions

1 - z 2 y 2 + y2 w > 0 - . y 2 w > z 2y2 _ 1 z2y 2 1 w > . y2. . . yZ

2

~w>z

1 -- y2

(3.273)

Inequality (3.273) leads to w > z 2 > 0 (Wisniewski et al., 1976).

3.9

CHEMICAL REACTIONS

For an elementary step reaction, we may relate the flow Jr and the affinity A to the forward Jrf and backward Jrb r e a c t i o n rates as follows

Jr - Yrf -- drb

(3.274)

R T In drf Jrb

(3.275)

A

=

If we solve these equations together, we obtain the reaction (velocity) flow Jr - Yrf ( 1 - e -A/RT )

(3.276)

Close to the thermodynamic equilibrium, where A / R T << l, we can expand Eq. (3.276) as

Jr - drf,eq

A RT

(3.277)

On the other hand, we have the following linear phenomenological equation for chemical reaction i

l

A/

dri - E L!/

,j=l

"

T

(3.278)

We can compare these linear phenomenological equations with Eq. (3.277) to obtain the phenomenological coefficients

Lij

Jrt;eq,!j

(3.279)

-

'

R

Here, we have Jrf, eq - Jrb,eq" For an overall reaction with l number of intermediate reactions, the linear phenomenological law is valid, if every elementary reaction satisfies the condition A / R T << 1, and the intermediate reactions are fast and hence a steady state is reached.

3.10

HEAT CONDUCTION

The entropy production for a heat conduction process is (3.280)

140

3.

Fundamentals of nonequilibrium thermodynamics

where Jq is the heat flow (or generalized flow) and VT is the thermodynamic force. Equation (3.280) identifies the forces and flows. The phenomenological equation and the Fourier equation for the heat conduction is (3.281)

Jq = Lqqgq = - k V T

Therefore, we have - J q = -Lqq -~ VT = kVT

and (3.282)

Lqq -- kT 2

If the dissipation function • = T~ is used to identify the thermodynamic forces, then the phenomenological coefficient is !

(3.283)

Lqq -- kT

Example 3.14 Entropy production and dissipation function in heat conduction Consider one-dimensional heat conduction in an isotropic solid rod. The surface of the rod is insulated and the cross-sectional area is constant (Figure 3.2). Describe the entropy production and the dissipation function for the heat conduction in an isotropic rod. The entropy change of the rod element is dS -

q + 6q _ q

T + dT

T

..~ Tq + T6q - q T - q d T = 6 q _ qd__TT T2 T T2

(3.284)

where 6q is the actual uptake of heat and dT is the actual increase in temperature. From Eq. (3.8) we have dS =

+

Comparing this equation with Eq. (3.284), we find the entropy production term diS

qdT -

T2 > 0

-

(3.285)

The rate of entropy production is p_

diS dt

qdT __qa__TT T2dt T2 > 0

(3.286)

Here,i/represents the heat flow rate. The rate of entropy production per unit volume is the entropy source strength

dt d V

ATZdt------~x= -

T q

~

-~x > 0

T+dT q+6q

x

x+dx

Figure 3.2. Heat conduction in an isotropic rod.

(3.287)

3.11

141

Diffusion

where A is the area, Jq the heat flow Jq = q/Adt, and Xq = dT/dx the thermodynamic force due to the finite temperature difference AT. Three-dimensional heat conduction in an isotropic solid is

1

dP = - J q - ~ V r = Jq "V

{1 ! =--T.VlnT

(3.288)

In Eq. (3.288), Jq is the heat flow, V(1/T) is the inverse temperature gradient representing the thermodynamic force for heat conduction, and Jq/T- Js is the entropy flow.

3.11

DIFFUSION

The local entropy production for diffusion of several substances per unit volume is

+--

J,.V --T-

Based on the entropy production, linear phenomenological equations for an isothermal flow of substance i become Ji = - Z - - Lik ~ V/xk

(3.290)

1

It is clear from the Gibbs-Duhem equation that not all the forces V(/xz/T) are independent. For example, for a twosubstance system at constant pressure and temperature, we have 0 = C1V~I +C2V/Z 2

(3.291)

The condition for no volume flow corresponding to no change in volume due to diffusion is

0 = J1V1 + J2V2

(3.292)

where V; is the partial molar volume for substance i. Therefore, for a two-substance system Eq. (3.289) becomes

~=_1

j1_ T

j2 c2

V~l~=--7/"

1+ Y

Jl.V~l V2c2 T

(3.293)

where V/x1 = (Otxl/Oc~)Vcl. Then, the linear phenomenological equation is

(3.294) Comparing this equation with Fick's lawJ~ = -D~Vcl, we have

J1---~

L/

1+

V2c2

~

Ocl

/

Vc I = - D lVc 1

(3.295)

Therefore, the diffusion coefficient is related to the phenomenological coefficient by

-- I"

1+V2c2 } t OC1 )

(3.296)

142

3.

Fundamentals of nonequilibrium thermodynamics

For diffusion flow of substance 1 in a dilute solution, we have Da -

LllR

(3.297)

x1

since ~1 = / z ° ( P , T) + RTln (C1/C) = ~O(p,T) + RTln (Xl) , where c is the concentration of the solution.

3.12

VALIDITY OF LINEAR PHENOMENOLOGICAL EQUATIONS

If a system is not far from global equilibrium, linear phenomenological equations represent the transport and rate processes involving small thermodynamic driving forces. Consider a simple transport process of heat conduction. The rate of entropy production is

+=-U E>0

(3.298)

The corresponding linear relation between the heat flow and the thermodynamic force is

Jq= _ L q q ( d T )

(3.299)

Equation (3.299) is identical to Fourier's law of heat conduction, k = Lqq/T 2. The validity of Eq. (3.299) is the same as the validity of Fourier's law, and the equation is valid when the relative variation of temperature is small within the mean free path distance A in the case of gases

AOT << 1 T Ox

--~

(3.300)

Since this condition is satisfied for most systems, the linear phenomenological equations are satisfactory approximations for transport processes. For an elementary chemical reaction, the local entropy production and the linear phenomenological equation are A

Jr=Lrr--~

(I) = A J r > 0

T

(3.301)

Considering a homogeneous chemical reaction S - P, the corresponding affinity is A = ~s -/Xp

(3.302)

For a mixture of perfect gases, the chemical potential is/x =/x ° + RT In C. We can relate the chemical potentials to the chemical equilibrium constant and the affinity by

RT In K(T) : -~__,l,i tx° (T)

(3.303)

i

A

=

- ~ vi/x° i

-

R T ~ vi In Ci = RT In K(T) i

(3.304)

(Cp/Cs)

From the kinetic expression, we have

Jr=Jrf-Jrb=kfCs-kbCp=

kfCs ( 1 - 1) K-~sP

(3.305)

where the indices f and b refer to forward and backward reactions.

exp/

(3.306)

3.13

Curie-Prigogine principle

143

Equation (3.306) is a general and nonlinear relation between reaction flow and affinity. However, when the reaction is close to equilibrium, we have

~T]

<
(3.307)

When this condition is satisfied, Eq. (3.306) becomes

Jr - Jrf, eq A R T

(3.308)

Comparing Eq. (3.308) with Eq. (3.306), we have

L.,. -

Jrf, eq R

(3.309)

Since the condition in Eq. (3.307) is highly restrictive, the linear laws for chemical reactions are not always satisfactory.

Example 3.15 Gibbs energy and distance from global equilibrium Discuss the effect of the distance from global equilibrium for a chemical reaction system: R = P. For the chemical reaction considered, with the concentrations of [P] and [R], we have K-

[P]eq [R]eq

(at chemical equilibrium)

[P] Q = [R]

(at nonequilibrium)

(3.310)

(3.311)

The displacement from equilibrium may be defined by /3 - QK

(3.312)

The Gibbs free energy change is

A.=

)

(3.313)

As Figure 3.3 displays, the absolute values of AG increase as the values of/3 move further from unity. For example, when 13 = Q/K = 10, then AG = 5743.1J/mol. The value of AG is at a minimum at/3 = 1 or K - Q, corresponding to the chemical equilibrium. If the values of/3 < 1, the AG < 0, and such reactions occur spontaneously when necessary mechanisms exist.

3.13

CURIE-PRIGOGINE PRINCIPLE

This principle as originally stated by Curie in 1908, is "quantities whose tensorial characters differ by an odd number of ranks cannot interact (couple) in an isotropic medium." Consider a flow J; with tensorial rank m. The value of m is zero for a scalar, it is unity for a vector, and it is two for a dyadic. If a conjugate force ~. also has a tensorial rank m, than the coefficient L!/is a scalar, and is consistent with the isotropic character of the system. The coefficients L!/are determined by the isotropic medium; they need not vanish, and hence the flow J; and the force ~. can interact or couple. If a force ~. has a tensorial rank different from m by an even integer k, then Li/has a tensor at rank k. In this case, L~X Xj is a tensor product. Since a tensor coefficient Li/of even rank is also consistent with the isotropic character of the

3.

144

Fundamentals of nonequilibrium thermodynamics Distance from equilibrium

7000 6000 5000

IAG IJ/mol

K/Q

5743.108 1728.848 0 1011.311 1728.848 2285.412 2740.159

0.1 0.5 1 1.5 2 2.5 3

i ........................................

4000 3000

T= 300 K R = 8.314 J/(mol K) [AG] = RT In (K/Q)

2000 1000

0

0.5

1

1.5

2

2.5

3

3.5

4

K/Q Figure 3.3.

Distance from global equilibrium in a chemical reaction system: R = E

fluid system, the Lij is not zero, and hence Ji and Xj can interact. However, for a force Xj whose tensorial rank differs from m by an odd integer k*, L/j has a tensorial rank of k*. A tensor coefficient Lij of odd rank implies an anisotropic character for the system. Consequently, such a coefficient vanishes for an isotropic system, and Ji and Xj do not interact. For example, if k* is unity, then Lij would be a vector. By definition, an isotropic system cannot support a vector quantity associated with it. Therefore, the vectorial flows ] q and Ji c a n only be related to the vector forces. The scalar reaction rates can be functions of the scalar forces and the trace of the dyadic, but not the vector forces. We, therefore, observe coupling between the diffusion flow and heat flow. We also observe coupling between the chemical reactions. According to the Curie-Prigogineprinciple, vector and scalar quantities interact only in an anisotropic medium. This principle has important consequences in chemical reactions and transport processes taking place in living cells.

3.14

TIME VARIATION OF ENTROPY PRODUCTION

The rate of entropy production inside a given system of volume V is

P-~ ~O£)dV-- f ~ J i X i dV V

(3.314)

V i=1

Equation (3.314) shows the volumetric rate of entropy production. Both the flows and the forces may change with time, while they remain constant at the system boundaries at stationary state only. The time variation of P is

dt

--~ ) dV -Jrf

Vi

X i --~ dV =

dt

+

dt

(3.315)

The first term in Eq. (3.315) represents the variation of the rate of entropy production in terms of the variation of thermodynamic force

dXPdt - ! Ox~OtdV-f~Jiv i

dyidt dV<-O

(3.316)

The second term in Eq. (3.315) represents the time variation of the flow

djP _ ! aj~ dV = ~ ~ x ~i dV

dt

dt

v i

dt

(3.317)

3.14

145

Time variation of entropy production

There is no definite sign for Eq. (3.317). When the generalized flows are expressed by linear phenomenological equations with constant coefficients obeying to the Onsager relations

Ji - £ Lik Xa.

(3.318)

k=l

then Eq. (3.317) and the Onsager relations (Lia- - Lki) yield

d,

-~

d V - I £ (LikXi) O,

• i

f" i , k = l

OX k ) =a.--, J ~ - ~ dV~,k=l

f {,

(3.319)

1 dP dxP -<0 2 dt dt

Therefore, from Eqs. (3.315)and (3.319), we have

dP - 2 d__,,XP _ 2 dsP <-- 0 dt dt dt

(3.320)

At stationary state, the boundary conditions are time independent, and the rate of entropy production is at a minimum, leading to minimum energy dissipation. From Eq. (3.319), we have

C)XOP Oj OP_<0 at

(3.321)

dt

Example 3.16 Entropy production and the change of the rate of entropy production with time in heat conduction For heat conduction in an isotropic medium (Figure 3.2), derive a relationship for the rate of entropy production, the dissipation function, and the rate of entropy production change with time. The entropy source strength ~ is

1 VT=j q.V(1]

(I) = - J q

T-~

T

-

=

-

Jq ~

T VlnT o

Hence, the phenomenological equation for heat conduction is

Therefore, we have Lqq- kT 2 The rate of entropy production is obtained from the local value of entropy production or entropy source strength

f[1312

-

I ® d v - Ia~,.v T d v : L~ v 7 f"

dv_0

(3.322)

I"

The dissipation function for a reference temperature of T0 is

* - r 0 f , d v - r,,~a~.v 7 d r - r0kr~ v

dv _>0

(3.323)

The time variation of the rate of entropy production with respect to the variation of the thermodynamic force ( d r ) is

I/ill ![(3]

dx P 0 dt - IJq'-:07 V

dV -

c? 1 -~ -f

[/3]

0 1 J q ' d A - f -07 -T

(V'Jq)dV

(3.324)

146

3.

Fundamentals of nonequilibrium thermodynamics

where the surface integral is zero as the temperature does not change with time. The divergence of the heat flow is obtained using the first law of thermodynamics

dU = 6 q - PdV

(3.325)

For a solid dV = 0, and we have dU = Cv dT, so therefore

6q OU aT - V ' J q = p--~= p---~-= pC v dt

(3.326)

Combining Eqs. (3.324) and (3.326) and assuming constant phenomenological coefficient for small temperature gradients and for Cu > 0, we get

dt

dt

\ Ot ) dV <_0

(3.327)

This shows that the rate of entropy production decreases with time because of heat flow in an isotropic solid, and a minimum is reached in an equilibrium state

3.15

MINIMUM ENTROPY PRODUCTION

The entropy production rate is

=I V

I

V i=1

(3.328) V i,k=l

The value of P will be minimal if its variation is equal to zero

6P = 61 ~ LikXkX i dV = 0

(3.329)

V i,k =1

The variation considered in Eq. (3.329) may be subject to various constraints. For example, the flows J/may vary when the forces Xi remain constant. It is also possible that the thermodynamic force may change while the flow remains the same, or they both may change. For a set of linear phenomenological equations, consider the following potentials

1 ~ LikXkXi >_0

4'=-2

1~

KikJiJk >_0

(i,k=l,2,...,n)

(3.330)

q~ = -2 i,k

These potentials have the following properties o4,_1

OX~

L kXk =

2~

1&

)__,KikJk = Xi 2k

(3.331)

and

02______~~ _ OJi _ Lik = Lki - OJk

oxioxk oxk

_

02~

o& oxko& (3.332)

02q) OJiOJk

-- OYi - Kik -- K~. - OXk -

OJk

OJi

02q9 OJkOJi

Equations (3.331) and (3.332) indicate that the first derivatives of the potentials represent linear phenomenological equations, while the second derivatives are the Onsager reciprocal relations.

3.15 Minimumentropy production

147

For an elementary volume, minimum entropy productions under various constraints are 6 X ~ - 0 ,~,]~ 4= 0 6 ( 0 0 - q ~ ) x , - 0

(3.333)

6x~ ~ 0 6J,, - 0 a ( @ - 4,)j~ - 0

(3.334)

aN; 4: o a J, 4= o a i r - ( q , + ~)] = 0

(3.335)

For the whole system under consideration, we have 6 X i - O 63 i 4: 0 6 I ( ~ - q ~ ) x ,

dV - O

(3.336)

6X~ 4:0 6J, = 0 61 (alp- O ) j d V = 0

(3.337)

I"

6 X i 4:0 6Ji 4:0

6I[dP-(O+q~)]dV-O

(3.338)

Total entropy decreases to a minimum value as the system approaches the stationary state. When the system reaches the state of equilibrium, the entropy increases to a maximum value. At steady state, a system loses a minimal amount of available energy. The concept of least dissipation may be one of the physical principles underlying the evolution of life. Living systems are endowed with a series of regulating mechanisms that preserve the steady state and bring the organisms back to their unperturbated condition. The principle of minimum entropy is restricted to linear phenomenological equations obeying the Onsager relations. For rapid metabolic processes, the linear phenomenological equations may not hold, and general laws applicable to all possible rate phenomena are still to be developed. Attempts are being made to extend the range of validity of the variational principles to include stationary states away from global equilibrium. In equilibrium thermodynamics, systems tend to maximize the entropy or minimize the free energy. Prigogine demonstrated that in linear nonequilibrium thermodynamics, entropy production in stationary states is minimal.

Example 3.17 Minimum entropy production in a two-flow system Determine the conditions for minimum entropy production for a two-flow system. Assuming that the linear phenomenological equations hold for a two flow coupled system J1 - L 11X1 + L12X2

J2 = L21X1 + L22X2

(3.339)

we can express the entropy production by the conductance coefficients L 0

(I0 ~- J1Yl --t-J2 X2 -/-'11 X2 -'[-L22X2 -~-(L12 -'[-L21 ) Yl X2

(3.340)

Equation (3.340) yields a paraboloid-like change of dissipation with respect to forces)(1 and X2, as seen in Figure 3.4. The system tends to minimize the entropy and eventually reaches zero entropy production if there are no restrictions on the forces. On the other hand, if we externally fix the value of one of the forces, for example, X2 = X20, then the system will tend toward the stationary state characterized by the minimum entropy production at X2 = X20. The system will move along the parabola of Figure 3.4 and stop at point ~0. At the minimum, the derivative of with respect to )(2 is zero dOp dX2

-- 2L22X2 +(/'12 + L21)X1 = 0 (3.341)

If the Onsager relations are valid, L12 = L21, and then Eq. (3.341) becomes 0 = 2(L22X2 + LI2X1) = 2 J 2 = 0

(3.342)

148

3.

Fundamentals of nonequilibrium thermodynamics

0

=,.. v

X2o

X2

Figure 3.4. Representation of entropy production in terms of the forces in a two-flow system.

since J2 is the flow given by the phenomenological equations. Therefore, a stationary point with respect to mass flow characterizes the state of minimum entropy production, and minimum energy dissipation. If a system has n independent forces (X1, X2,..., X,), andj of them are held constant 0(1, X2,..., Xj = constant), then the flows with Jj.+ 1, Jj+2,..., J, disappear at the stationary state with minimum entropy represented by

0 (diS):0

OXkt, dt )

(k:j+l...,n)

(3.343)

Since (diS/dt > 0), the extremum in Eq. (3.343) is a minimum. Such a state is called a stationary state ofjth order.

Example 3.18 Minimum entropy production in an elementary chemical reaction system Consider a monomolecular reaction, for example, the following isomerization reaction. R< 1 ..X< 2 >B

(3.344)

In this open reaction system, the chemical potentials of reactant R and product B are maintained at a fixed value by an inflow of reactant R and an outflow of product B. The concentration of intermediate X is maintained at a nonequilibrium value, while the temperature is kept constant by the reaction exchanging heat with the environment. Determine the condition for minimum entropy production. The entropy production per unit volume is ~ = 4 Jrl + A2 Jr2 > 0 T ?-

(3.345)

Where A1 and A 2 are the affinities for reactions 1 and 2. The linear reaction flows with vanishing cross coefficients are A1 Jrl = L22 ~A2 Jrl = LI 1 --T--

(3.346)

As the chemical potentials ~R and/..LB are fixed by the flow conditions, we have a constant total affinity A A = (/xR - / x x) + (/xx - / x B) =/x R - / x B = 4 + A2

(3.347)

3.15

Minimumentropy production

149

At the stationary state, we have (3.348)

Jrl = Jr2 After inserting Eq. (3.347) into Eqs. (3.345) and (3.346), we get

,=

A1Jr, + ( A-AI~) Jr2~0 T T

A1 Jrl =/-11 7 -

Jr2 = L22

A-4

(3.349)

(3.350)

After combining Eqs. (3.349) and (3.350), we get

A2 ( A - A1)2 ~(A1) =/-11-~k- L22 T2

(3.351)

The entropy production is at a minimum with respect to the affinity of reaction 1

OdP(A1) =/-ql 2A1 2 ( A - A1) OA------~ - ~ - L22 T2 = 0

(3.352)

Al_ A2 /-ql ~ L22 T = J r l - J r 2 = 0

(3.353)

Therefore, we have

Equation (3.353) proves that with the linear reaction flows, the entropy production is minimized at nonequilibrium stationary state where the reaction velocities are equal to each other Jr1 - Jr2.

Example 3.19 Minimum energy dissipation in heat conduction Use the minimum entropy production principle to derive the relation for nonstationary heat conduction in an isotropic solid rod. For an isotropic rod, we have Os

p - - = - V. J, + ~ Ot p

Os Ot

-

p Ou T Ot

(entropy balance) (3.354)

10T T Ot

-

where ,Is = ,Iq/T. From Eq. (3.337), we have

6X i =/:0 6J i = 0 6 f (dO- O)j, dV = 0

(3.355)

V

where Lqq V

4,=-5With the Gauss-Ostrogradsky theorem, we have

6

p-~t-qt

dV+6 Ji

Js'dA=O A

(3.356)

150

3.

Fundamentalsof nonequilibrium thermodynamics

For the isotropic rod with constant temperatures at the boundaries, Eq. (3.356) yields

v

T

(3.357)

Ot

By using the absolute inverse temperature as the variable subject to change, Eq. (3.357) becomes

+ E qq lllll,llll v

T

o

Ot

(3.358)

This variational equation based on Eq. (3.355) is equivalent to a differential heat conduction equation in the following form p C v OT _ - V . Ot

I(1t] ZqqV

= V .

,33,9

V

(3.360)

The Lagrangian of the variational problem is Lq

p C v OT TOt

Lqq 2

A Euler-Lagrange equation for the -variational problem of 6Iv L q d V may be obtained by considering the differential heat conduction equation, and we have OLq

3

3.15.1

0

OLq

. ~ Oxi 0 [O(1/T)/Ox i ] = 0

O(1/T)

(3.361)

Entropy Production in an Electrical Circuit

In electrical circuits, electrical energy is converted into heat irreversibly in resistors and capacitors, and entropy is produced. When there is an electrical field, the change of energy is d U -- TdS - p d V + ~_, t£i dN i + Z Fzil[ti dNi i i

(3.362)

where F is the Faraday constant, and z i the ion number. The product FzidN i represents the amount of charge transferred. When we transfer the charge dI from a potential ~bl to a potential ~2, then the rate of entropy production is d i S _ _ ~2 - i~1 ~ F z i dNi _ _ i~2 - ~1 d I dt T i dt T dt

(3.363)

In Eq. (3.363), the difference (~2--~tl) is the voltage across the element, while dI/dt is the electric current. For a resistor, using the Ohm law V = (~2-~1) - IR, where R is the resistance, the rate of entropy production is diS - VI _ R I 2 dt

T

T

> 0

(3.364)

In Eq. (3.364),/U 2 is the Ohmic heat rate produced by a current through an element, such as a resistor. For a capacitor with capacitance C, the rate of entropy production is diS - V c I - Vc d I _

dt

r

r dt

C

-T

dVc

d-T

(3.365)

151

3.15 Minimumentropy production

where d V c - - d I / C is the voltage decrease when we transfer charge of dI. We can modify Eq. (3.365) as follows

dt

- -T dt

2

(3.366)

- T d t --~

where the terms (CV2/2) - (I2/(2C)) represent the electrostatic energy stored in a capacitor. Only for an ideal capacitor there is no entropy production and no energy dissipation. For an inductance, the rate of entropy production is

dt

-

~£

(--~) = -

r dt

-

_> 0

r

(3 367)

where the energy stored in an inductance (in the magnetic field) is LI2/2. The voltage is V - -LdI/dt. The phenomenological equations for resistance, capacitance, and inductance are as follows V V I - L R --z-, I - Lc -2, I 1

and

I =--L L

V

(3.368)

T

where LR, Lc, and LL are the phenomenological coefficients, which may be related to resistance. Using Ohm's law, we have

1 R

-

LR

, R-

T

T

and

T

R-

Lc

(3.369)

LL

Example 3.20 Minimum entropy production in electrical circuits Determine the conditions that minimize the entropy generation in electrical circuits with n elements connected in series. Assume that the voltage drop across the circuit is kept constant. The entropy production is diS _ V - --I dt T

(3.370)

where V is the voltage across the element (t)2 - ~1) and I the current passing through the element. The phenomenological law is

Vj Ij - L/J T

(3.371)

Since the voltage drop across the circuit is kept constant, we have

V = ~ V/

(3.372)

J The total entropy production for the n circuit elements is

p _ 4~s _ Vl I~ + -=v2 + . . . + mv,,i n d~ r i 12

(3.373)

T

]2

After combining Eqs. (3.373), (3.371), and (3.372), we get

( v - G) p-

dis _ Lll dt ~

+

L22

J +"" + Lnn

T2

(3.374)

152

3.

Fundamentalsof nonequilibrium thermodynamics

To minimize the rate of entropy production, we use OP/OVj with n - 1 independent values of Vj, which leads to /1 = / 2 . . . . .

In

(3.375)

Therefore, in a circuit element, the entropy production is minimized if the current through the n elements is the same. In an electrical circuit, the relaxation to the stationary state is very fast, and nonuniform values of/are not observed.

PROBLEMS 3.1

Derive the relationships between the conductance type ofphenomenological coefficients Lik and the resistance type of phenomenological coefficients K/y in a three-flow system.

3.2

Consider a monomolecular reaction in Example 3.18, and determine the condition for minimum entropy production when the rate of entropy production is expressed in terms of the concentration. In this open reaction system, the chemical potentials of reactant R and the product B are maintained at a fixed value by an inflow of reactant R and an outflow of product B. The concentration of intermediate X is maintained at a nonequilibrium value, while the temperature is kept constant by exchanging the heat of reaction with the environment. Determine the condition for minimum entropy production.

3.3

Consider the following sequence of reactions R< 1 ) X l (

2 .X2(

3 )...(

n-1 ) X n _ l (

n >p

Identify the states at which the entropy production will be minimal. 3.4

Consider the following synthesis reaction H 2 4- Br2 = 2HBr This results from the following intermediate reactions Br2 < 1 >2Br H2+Br< 2 ; H B r + H H + B r 2 ." 3 , H B r + B r The affinity of the net reaction is maintained at a constant value by the flows of H2 and Br2. One of the reactions is unconstrained. Show that the stationary state leads to minimal entropy production.

3.5

Consider one-dimensional heat conduction in a rod with a length of L. Obtain the function that minimizes the entropy production.

3.6

Consider an elementary reaction of A = B, and calculate the change of Gibbs free energy when/3 = Q/K changes from 0.1 to 10.

3.7

For a three-component diffusion system, derive the relations between the diffusion coefficients and the phenomenological coefficients under isothermal conditions.

3.8

Transform the thermodynamic forces and flows when the forces are independent, while the flows are linearly dependent in a two-flow system: 0 = J1 + YJ2.

3.9

Transform the thermodynamic forces and flows when the flows are independent, while the thermodynamic forces are linearly dependent in a two-flow system: 0 = X1 + yX2.

3.10

Transform the phenomenological equations when the flows and forces are linearly dependent and the forces are linearly dependent: 0 = J1 4- zJ2 and 0 =)(1 + yX2.

Problems

153

3.11

A steam enters a nozzle at 500 kPa and 220°C, and exits at 400 kPa and 175°C. The steam enters at a velocity of 200 m/s, and leaves at 50 m/s. The nozzle has an exit area of 0.2 m 2. Determine the rate of energy dissipation when the environmental temperature is T0 = 300 K.

3.12

A steam enters a nozzle at 4000 kPa and 425°C with a velocity of 50 m/s. It exits at 286.18 m/s. The nozzle is adiabatic and has an inlet area of 0.005 m 2. Determine the rate of energy dissipation if the surroundings are at To = 300 K.

3.13

A steam enters a nozzle at 3200 kPa and 300°C with a velocity of 20 m/s. It exits at 274.95 m/s. The nozzle is adiabatic and has an inlet area of 0.01 m 2. Determine the rate of energy dissipation if the surroundings are at T o - 300K.

3.14

A compressor receives air at 15 psia and 80°E The air exits at 40 psia and 300°E At the inlet the air velocity is low, but increases to 250 ft/s at the outlet of the compressor. The power input to the compressor is 350 hE The compressor is cooled at a rate of 200 Btu/s. Determine the rate of energy dissipation when the surroundings are at 540 R.

3.15

In a mixer, we mix a hot water at 1 atm and 80°C adiabatically with a cold-water stream at 25°C. The flow rate of the cold water is 20 kg/h. If the product leaves the mixer at 50°C, determine the rate of energy dissipation if the surroundings are at 295 K.

3.16

In a mixer, we mix a hot water at 1 atm and 86°C adiabatically with cold-water stream at 25°C. The hot water flow rate is 60 kg/h. If the warm water leaves the mixer at 35°C, determine the rate of energy dissipation if the surroundings are at 300 K.

3.17

In a mixer, we mix liquid water at 1 atm and 25°C with a superheated steam at 325 kPa and 200°C. The liquid water enters the mixer at a flow rate of 70 kg/h. The product mixture from the mixer is at 1 atm and 55°C. The mixer loses heat at a rate of 3000 kJ/h. Determine the rate of energy dissipation if the surroundings are at 300 K.

3.18

In a mixer, we mix liquid water at 1 atm and 20°C with a superheated steam at 1350 kPa and 300°C. The liquid water enters the mixer at a flow rate 70 kg/h. The product mixture from the mixer is at 1 atm and 55°C. The mixer loses heat at a rate of 3000 kJ/h. Determine the rate of energy dissipation if the surroundings are at 290 K.

3.19

Steam expands in a turbine from 6600 kPa and 300°C to a saturated vapor at 1 atm. The steam flow rate is 9.55 kg/s. If the turbine generates a power of 3 MW, determine the rate of energy dissipation if the surroundings are at 298.15 K.

3.20

Air enters a nozzle at 400 K and 60 m/s and leaves the nozzle at a velocity of 250 m/s. The air inlet pressure is 300 kPa and the air leaves the nozzle at 100 kPa. If the nozzle loses 2.2 kJ/kg, determine the total entropy change if the surroundings are at 290 K.

3.21

(a) At steady state, a 4-kW compressor is compressing air from 100 kPa and 300 K to 500 kPa and 450 K. The airflow rate is 0.02 kg/s. Estimate the rate of entropy change. (b) If the compression takes place isothermally by removing heat to the surroundings, estimate the rate of entropy change of air if the surroundings are at 290 K.

3.22

Derive the following isentropic relation tbr ideal gases with constant specific heats.

T2-(P2] (7..,/~/TI where 7 is the ratio of heat capacities at constant pressure to heat capacity at constant volume. 3.23

Refrigerant tetrafluoroethane (HFC-134a) enters the coils of the evaporator of a refrigerator as a saturated vapor liquid mixture at 240 kPa. The refrigerant absorbs 100 kJ of heat from the interior of the refrigerator maintained at 273.15 K, and leaves as saturated vapor at 240 kPa. Estimate the total entropy change.

3.24

Methane gas is compressed from an initial state at 100 kPa, 280 K, and 10 m 3 to 600 kPa and 400 K. The compression process is polytropic ( P V ~ = constant). The average heat capacity of methane is Cp,av= 40.57 J/(mol K). Estimate the total entropy change if the surroundings are at 300 K.

3.25

Hydrogen gas is compressed from an initial state at 100 kPa, 300 K, and 5 m 3 to 300 kPa and 370 K. The compression process is polytropic ( P V ~ = constant). The average heat capacity of hydrogen is Cp,av= 29.1 J/(mol K). Estimate the total entropy change if the surroundings are at 290 K.

154

3.

Fundamentals of nonequilibrium thermodynamics

REFERENCES Y.A. Cengel and M.A. Boles, Thermodynamics. An Engineering Approach, 4th ed., McGraw-Hill, New York (2002). S.R. De Groot, Thermodynamics of Irreversible Processes, North-Holland Publishing, Amsterdam (1966). G.A.J. Jaumann, Wien. Akad. Sitzungsberichte (Math-Nature Klasse), 129 (1911) 385. D. Kondepudi and I. Prigogine, Modern Thermodynamics From Heat Engines to Dissipative Structures, Wiley, New York (1999). I. Prigogine, Introduction to Thermodynamics of Irreversible Processes, Wiley, New York (1967). S. Wisniewski, B. Staniszewski and R. Szymanik, Thermodynamics of Nonequilibrium Processes, D. Reidel Publishing Company, Dordrecht (1976).

REFERENCES FOR FURTHER READING E Kock and H. Herwig, Int. J. Heat Fluid Flow, 26 (2005) 672. H.C. Ottinger, Beyond Equilibrium Thermodynamics, Wiley, New York (2005). A. Perez-Madrid, Physica A, 339 (2004) 339. A. P6rez-Madrid, J. Chem. Phys., 123 (2005) 204108-1. J.M. Rubi and A. Perez-Madrid, Physica A, 264 (1999) 492. J.M. Rubi and A. Perez-Madrid, Physica A, 298 (2001) 177. D.P. Ruelle, Proc. Natl. Assoc. Sci., 100 (2003) 3054. I. Santamaria-Holek, A. Perez-Madrid and J.M. Rubi, J. Chem. Phys., 120 (2004) 2818.

4 USING THE SECOND LAW: THERMODYNAMIC ANALYSIS 4.1

INTRODUCTION

We all widely utilize aspects of the first law of thermodynamics. The first law mainly deals with energy balance regardless of the quality of that part of the energy available to perform work. We define first law efficiency or thermal efficiency as the ratio of the work output to total rate of heat input, and this efficiency may not describe the best performance of a process. On the other hand, the second law brings out the quality of energy, and second law efficiency relates the actual performance to the best possible performance under the same conditions. For a process, reversible work is the maximum useful work output. If the operating conditions cause excessive entropy production, the system will not be capable of delivering the maximum useful output. In the last 30 or so years, thermodynamic analysis had become popular in evaluating the efficiency of systems. Thermodynamic analysis combines the first and second laws of thermodynamics, and makes use of second law analysis, exergy analysis, and pinch analysis. Second-law analysis can identify the sources and quantity of entropy production in various processes in a system. Exergy analysis describes the maximum available work when a form of energy is converted reversibly to a reference system in equilibrium with the environmental conditions; hence, it can relate the impact of energy utilization to environmental degradation. Pinch analysis aims for a better integration of a process with its utility in reducing energy cost. On the other hand, the equipartition principle states that a process would be optimum when the thermodynamic driving forces are uniformly distributed in space and time. Thermodynamic analysis aims at identifying, quantifying, and minimizing irreversibilities in a system. Such analysis is of considerable value when efficient energy conversion is important. This chapter discusses second law analysis, exergy analysis, and pinch technology, providing some examples.

4.2

SECOND-LAW ANALYSIS

The mathematical statement of the second law is associated with the definition of entropy S, dS = 6qrev/T. Entropy is a thermodynamic potential and a quantitative measure of irreversibility. For reversible processes, dS is an exact differential of the state function, and the result of the integration does not depend on the path of change or on how the change is carried out when both the initial and final states are at stable equilibrium. The entropy of a closed adiabatic system remains the same in a reversible process, and increases during an irreversible process. A system and its surrounding create an isolated composite system where the sum of the entropies of all reversible changes remains the same, and increases during irreversible processes. The product of thermodynamic forces and flows yields the rate of entropy production in an irreversible process. The Gouy-Stodola theorem states that the lost available energy (work) is directly proportional to the entropy production in a nonequilibrium phenomenon. Transport phenomena and chemical reactions are nonequilibrium phenomena and are irreversible processes. Thermodynamics, fluid mechanics, heat and mass transfer, kinetics, material properties, constraints, and geometry are required to establish the relationships between physical configuration and entropy production and to minimize entropy production. Generally, we may minimize entropy production through a set of modifications in design and operating conditions. The second law of thermodynamics is applicable to all physical, chemical, and biological processes, as well as to heat and work conversions. The second law can quantify the thermodynamic equivalence of heat to work through exergy

156

4.

Usingthe second law: Thermodynamic analysis

and availability analysis, and hence it can provide specific insights into the design, operating conditions, or the retrofitting of an existing process. Some concepts and properties of entropy are: • Processes follow certain directions and paths, that must yield positive entropy production. This principle might force chemical reactions to proceed without reaching completion. • Entropy production is a measure of dissipated useful energy and degradation of the performance of a process; the level of dissipation depends on the extent of irreversibilities. • Entropy is a nonconserved property; it is conserved during an ideal reversible process only. • A reversible adiabatic process is isentropic, meaning that a substance will have the same entropy values at the beginning and end of the process. Systems such as pumps, turbines, nozzles, and diffusers are nearly adiabatic operations and are more efficient when irreversibilities, such as friction, are reduced, and hence operated under isentropic conditions. • Isentropic efficiency of a turbine rh at steady state is defined as the ratio of the actual work output Wactof the turbine to the work output of isentropic operation Ws

mact 7~t- ms

(4.1)

• Isentropic efficiency of a compressor r/c is the ratio of isentropic work to actual work input

~c -

Ws Wact

(4.2)

• Entropy does not exist in various forms. Second-law analysis can determine the level of energy dissipation from the rate of entropy production in the system. The entropy production approach is especially important in terms of process optimality since it allows the entropy production of each process to be determined separately. The map of the volumetric entropy production rate identifies the regions within the system where excessive entropy production occurs due to irreversible processes. Minimizing of excessive irreversibilities allows a thermodynamic optimum to be achieved for a required task. Estimation of the trade-offs between the various contributions to the rate of entropy production may be helpful for attaining thermodynamically optimum design and operation.

4.2.1

Entropy Balance

In every nonequilibrium system, an entropy effect exists either within the system or through the boundary of the system. Entropy is an extensive property, and if a system consists of several parts, the total entropy is equal to the sum of the entropies of each part. Entropy balance is Change in total entropy = Total entropy i n - Total entropy out + Total entropy produced Entropy balance in the rate form is given by

ASsystem--(Sin- Sout) nt-Sprod

(4.3)

where A shows the net change within the system. The first term on the fight in Eq. (4.3) shows the rate of net entropy exchange between the system and its surroundings, which may be by heat and/or mass (Figure 4.1). The rate of entropy production cannot be negative; however, the changes in entropy of the system may be positive, negative, or zero. For a reversible process, the entropy production is zero, and the entropy change of a system is equal to the net entropy transfer. The entropy of an isolated system during an irreversible process always increases, which is called the increase of

entropy principle. Heat and mass flows can transfer entropy. Entropy transfer through the boundary represents the entropy gained or lost by a system during a process. No entropy is transferred by work. According to the first law of thermodynamics, there is no difference between heat and work. According to the second law, however, energy exchange accompanied by entropy transfer is the heat transfer, and energy exchange that is not accompanied by entropy transfer is the work.

4.2

157

Second-law analysis

Sin System Mass >_0 Mass

-j

Heat ]

~Sout

~

Figure 4.1. Mechanism of entropy transfer for a general system. The general entropy balance relations for a control volume are given in terms of the rate of entropy change due to the heat transfer, mass flow, and entropy production

/~'-~cv-- Z 0--Qi~r-£/0 _nLZ fhinSin- Z FhoutSout-{-Sprod

(4.4)

where q0 and To are the environment's heat and temperature. For a general steady-state flow process, the rate form becomes

'~prod- 2 FhoutSout-- Z fhinSin- - 00 - ~ 0i r0

(4.5)

r

Second-law analysis can account for the quality of energy. This may lead to possible improvements in energy converting processes, and the effective use of resources. Some second law guidelines are: • • • • • • • • • •

Avoid excessively large or small thermodynamic driving forces in processes. Minimize the mixing of streams with large differences in temperature, pressure, or chemical composition. Do not discharge heat at high temperature into the environment. Do not heat refrigerated streams with hot streams. When choosing streams for heat exchange, try to match streams where the final temperature of one stream is close to the initial temperature of the other. Extremely large or small amounts of flows may not be easy to manage efficiently. When exchanging heat between two streams, the exchange is more efficient if the heat capacities of the streams are similar; otherwise, consider splitting the stream with the larger flow heat capacity. Hot or cold sources with temperatures far from the ambient temperature are useful. Minimize the throttling of fluid flow, steam, or other gases. Use exergy balance or exergy loss calculations to evaluate the utilization of energy and as a guide for process modifications.

These guidelines may be useful in designing and optimizing the processes such as power plants, heat exchangers, and other thermal systems. Suitable trade-offs between the use of energy and capital may be required by identifying and eliminating design parameters and operating conditions that cause excessive entropy production.

4.2.2

Throttling

Sometimes, fluids flow through a restriction, such as an orifice, a valve, or a porous medium, and a pressure drop occurs adiabatically. If the changes in kinetic and potential energies are negligible, the flow process is a throttling process, which causes no change in enthalpy at the inlet and outlet, and we have M-/= 0. Some properties of throttling processes are: • • • •

In an ideal gas, enthalpy is a function of temperature only, and temperature remains constant. At moderate pressure and temperature, a throttling process causes a drop in temperature for most real gases. If wet steam is throttled to a considerably low pressure, the liquid evaporates and the steam becomes superheated. The throttling of a saturated liquid causes vaporization (or flashing) of some of the liquid, which produces saturated vapor and liquid.

158

4.

Usingthe second law: Thermodynamic analysis

Example 4.1 Lost work in throttling processes n-Butane gas with a flow rate of 25 mol/s is throttled from 15 bar and 450 K to 1 bar in a steady-state flow process. Determine the final temperature and the lost work. Assume that the surroundings are at 298.15 K.

Solution: Assume that kinetic and potential energy effects are negligible To = 298.15K,

R = 8.314 J/(mol K)

This example uses the Lee-Kesler generalized correlation for the reduced enthalpy estimations (see Tables F5-F8) in a throttling process. The reduced properties lead to enthalpies

82 -- N0,ideal + I

Cpdr + I4~

To

and H 1 = N0,idea 1 + I Cpdr + I-I~ To

(4.6)

By using the throttling property of AH = 0 and the above equation, we have z~-I = O = C p, av ( T2 - T1) + H R2 - H R

(4.7)

where Cp,avis the average heat capacity between 7'1 and T2. At the outlet conditions, the n-butane gas is ideal, and hence H2R = 0. Therefore, Eq. (4.7) becomes Te = T1 + HR

(4.8)

Cp, av The critical properties for n-butane are: ire = 425.1 K, Pc = 37.96 bar, and the acentric factor to = 0.2. The reduced properties and heat capacity are:

rr-

T1 - 1.058, P~ = P__I= 0.395,

Cp =

R(1.935 + 0.00369T)

Pc

Using the generalized correlation, we have

: RT Pr

_ Tr

dTr

-T r ~

=-2830.12J/mol

(4.9)

where B0=0.083_~=0.422 -0.3026 and B 1 =0.139 rrl.6

0.172 - 0.00326 T4-2

(4.10)

As T2 is not known, an initial value Cp is calculated at T1 = 450 yielding Cp = 151.38 J/(mol K). With the known value of Cp, Eq. (4.8) yields T2 = 431.6 K. As the temperature difference is not large, the average value of Cp at the average temperature value is: 450+431.6 Tav

= 440.8,

Cp,av =

151.38 J/(mol K)

Therefore, the value of temperature at the outlet is Tz=T1 +H1R

.....

Cp,av

450

2830.12 --=431.3K 151.38

4.2

159

Second-law analysis

The value of entropy change of throttling is

/

-Rln(P21-S R

iP, )

= 18.377 J/(mol K)

(4.11)

where dB : - R P r t, dTr +

dBldTr) =

-2.2858 J/(mol K)

(4.12)

For a flow rate of 25 mol/s and To = 298.15 K, the lost work or dissipated energy is Eloss =

nToAS =

136.97 kW

Example 4.2 Dissipated energy in an adiabatic compression In an adiabatic compression operation, air is compressed from 20°C and 101.32 kPa to 520 kPa with an efficiency of 0.7. The air flow rate is 22 mol/s. Assume that the air remains ideal gas during the compression. The surroundings are at 298.15 K. Determine the thermodynamic efficiency rhh and the rate of energy dissipated Eloss.

Solution: Assume that kinetic and potential energy are negligible, and the system is at steady state. Basis: 1 mol/s air To - 298.15 K R = 8.314 J/mol K Cp/R= 3.355 + 0.575 × 10 -3 Twhere Tin K andR = 8.314 J/molK (Table B3) n = 22 mol/s For an isentropic operation (AS = 0), the final temperature Tzs is y-1

(a)

where

-

Y 1

y

_

R

and

3'-

Cp

Cp Cv

As the value of Cp is temperature dependent, Eq. (a) needs iterations. A value greater than 293.15 K may be an initial temperature. After iterations, we have Tzs = 460.59 K An average heat capacity may be estimated from

C r,av =

v,

(b) = 29.695 J/(mol K)

The work required under isentropic operation Ws is

~H s - Ws - Cp,av(TZs -

T~) - 4972.17 J/tool

Actual work mact is Wact - Ws - 7 1 0 3 . 1 1 J/mol rtc

160

4.

Usingthe second law: Thermodynamic analysis

"With the value of actual work, we can estimate the actual temperature by M-/ T2 = T I + ~ Cp,av Eq. (c) requires iterations,

as

(c)

Cp,avis dependent on the value of T. After iteration, we have T2 = 531.0K

Using Eq. (b), we have the new value of Cp,av= 29.863 J/mol K. The change of entropy is

AS=Cp,av ln(-~12) - R l n ( @ l ) = 4.414 J/mol K The value of ideal work is Widea 1 --

A H - ToAS = 5867.74 J/mol

The thermodynamic efficiency becomes T/th -- Wideal -- 0 . 8 2 6 Wact

The dissipated energy is ~+loss= h ToAS = 28.95 kW

4.2.3

Heat and Fluid Flow

Researchers and engineers extensively utilized second law analysis in the field of heat and fluid flow. Bejan developed the basic approach, methodology, and applications. In two-dimensional Cartesian coordinates, the local rate of entropy production per unit volume in a convective heat transfer is Sprod d x

dy =

qx + (Oqx/OX)dx qy + (Oqy/Oy)dy qx qy r +(OT/Ox)dx dy+ T + (OT/Oy) dy d x - ~ Td y - - - d rx + s+--clx Ox

vx+

I I(

+

Ox

dx

p + - - d x dy Ox

(4.13)

I(

O(ps) dx dy Ot

o+ ay

The first four terms on the right of the above equation account for the entropy transfer due to heat transfer, the next four terms represent the entropy convected into and out of the system, and the last term represents the rate of entropy accumulation in the control volume. Dividing Eq. (4.13) by dxdy, the local rate of entropy production becomes

Spr°d--T -

Ox

Oy

---f-2 qx --~x + qy

+ P --O-/+Vx --~x+ Vy

+S

(4.14)

Here, the last term on the right vanishes based on the mass conservation principle

Dp Dt

~+pV.v=0

(4.15)

4.2

161

Second-law analysis

where D / D t is the substantial derivative• Therefore, in vectorial notation, we have the volumetric rate of entropy production •

1

1

(4.16)

ds

@rod -- ~ - V ' q - - ~ q ' V T

+ p dz-7

From the canonical relation du - Tds - P d ( l / p ) , we obtain p

ds dt

-

p du

P

T dt

p T dt

dp

(4.17)

The first law of thermodynamics expressed locally in the convection of a Newtonian fluid is dH --=-V.q-P(V.v)+r" P dt

Vv

(4.18)

Introducing the above equation into Eq. (4.17) and combining it with Eq. (4.16), we have the following equation for an incompressible flow" •

1

1

Sprod = - 7 2 ( q ' V T ) + r ( ' r

(4.19)

" Vv)

The term ('r • Vv) represents the conversion of mechanical energy into the viscous dissipation heating. This heat source can be considerably high in flows with large viscosity and large velocity gradients, such as high-speed flights, rapid extrusion, and lubrication.

•

--

+ 2 # *'["*'7

Oxj

)]2

-/.tO

Ox i

where/x is the viscosity and ® the viscous dissipation function (in S-2). When the index i takes on the values 1, 2, 3, the velocity components Vx, v,,, v_ and the rectangular coordinates xi become x, y, z. Using the Fourier law q = - k V T and Eq. (4.20), Eq. (4.19) becomes • Sprod =

k ( V T ) 2 -+--/x--~

(4.21)

For a two-dimensional Cartesian coordinate system, the above equation can be expressed as aT

Spr°d -- 7

0N

+ --

ov

+

tx

7

2

Ov x

k Ox )

+2

Ov v

o),

+

Ov x

oy

+

Ov),

/2}

(4.22)

Ox

The above equation shows that the local irreversibilities are due to heat and viscous effects. Entropy production is positive and finite as long as temperature and velocity gradients exist.

E x a m p l e 4.3 T h e r m o m e c h a n i c a l c o u p l i n g in a C o u e t t e flow b e t w e e n p a r a l l e l plates Couette flow provides the simplest model for the analysis of heat transfer for flow between two coaxial cylinders or parallel plates (Figure 4.2). The Couette flow is important in lubrication, polymer flows, and food processing. The tangential annular flow is a model for a journal and its beating in which one surface is stationary while the other is rotating, and the clearance between the surfaces is filled with a lubricant oil of high viscosity. For such a system, the viscous energy dissipation appears as a heat source term in the energy equation, which is necessary to predict the temperature distribution in the narrow gap of a Couette device. Heat transfer and friction in a Couette flow causes entropy production and loss of useful energy.

162

4.

Using the second law: Thermodynamic analysis

For planar Couette flow, the rate of entropy production of an incompressible Newtonian fluid is

•

, (,,)2 (4.23)

Figure 4.2 shows a Couette flow of a fluid of constant density p, viscosity/~, and thermal conductivity k between parallel plates. The bottom plate is at rest, while the top plate is moving at a constant velocity Vl. The upper and lower plates are at uniform temperatures T1 and T2, respectively. The equation of motion for fully developed flow in the x-direction is

-Tyy ~ Tyy = - ~

(4.24)

The boundary conditions are v = •1 at y = H, and v = 0 at y = 0. We can integrate Eq. (4.24) twice to obtain the dimensionless velocity profile U (Figure 4.3) U = v Y[1 + A(1- Y)] v1 where A =

(-dP/dx)HZ/2~Vl and

Y=

(4.25)

y/H.

V=vl,T=T1 H

0

v=O,T=T2

Figure 4.2. The plane Couette flow. Reprinted with permission from Elsevier, Y. Demirel, Int. J. Heat Mass Transfer, 43 (2000) 4205.

-2

l

0.5

0.4

7 Figure 4.3. Dimensionless velocity field Ufor the plane Couette flow for H = 0.005 m. Reprinted with permission from Elsevier, Y. Demirel, Int. J. Heat Mass Transfer, 43 (2000) 4205.

4.2

Second-law analysis

1 63

For the case of (-dP/dx) = 0, the velocity is linear across the fluid. For a negative pressure drop the velocity is positive, and for a pressure increase, the velocity can become negative leading to backflow. At the point of reversal dv/dy - 0 at y = 0. This occurs when -dP/dx - -2/2v1/H 2. The velocity gradient dv/dy from Eq. (4.25) is

dv dy

_

(-dP/dx)

(H

2y) + v l

-

2/2

H

(4.26)

The energy equation for laminar and hydrodynamically developed flow is

d2T - /2 ( dv ] 2 @2 k~ dy)

(4.27)

with the boundary conditions of T = T2 at y = 0, and T = T1 at y = H. Substitution of Eq. (4.26) into Eq. (4.27) yields the temperature distribution

E 1

0 - T - T2 _ (- dP/dy)Y [all4 (1 - y3) _ 2bH 3(1- y2) + 6cH 2 (1 - Y)] + Y 1+ ~ Br(1 - Y) T1 - T 2

12k(T1 - T 2)

1

(4.28)

where a=

(-dP/dy) /2

; b=

(-dP/dy)H /2

+

2v 1

; c=

(-dP/dy)H 2

m

+vl; B r =

4/2

/2v2 k(T1 -T2)

Here, Br is the Brinkman number, which is a ratio of the viscous heating to the heat conducted through the gap of Couette device. From Eq. (4.28), we obtain the temperature gradient

dT

dy

(-dP/dY)[aH3(l_4Y 3 ) - 2 b H 2 ( 1 - 3 y 2 ) + 6 c H ( 1 - 2 Y ) ] + T 1 - T 2 I I + I H ~ B r ( 1 - 2Y) ] 12k

(4.29)

Figure 4.4 shows the temperature profile for T2 = 300 K and - 2 . 0 < Br < 8.0. The rise of temperature in the middle part of the Couette device is considerably large for high values of Br. Inserting Eqs. (4.26) and (4.29) into Eq. (4.23) yields an expression for the volumetric entropy production rate for a Couette flow.

8 Br

6

-2

y 0 . 6 ~ 1 Figure 4.4. Dimensionless temperature profile 0 for the plane Couette flow for Vl = 5.0 m/s and H = 0.005 m. Reprinted with permission from Elsevier, Y. Demirel, Int. J. Heat Mass Transfer, 43 (2000) 4205.

164

4.

Using the second law: Thermodynamic analysis

Example 4.4 Thermomechanical coupling in a circular Couette flow For a circular Couette flow (Figure 4.5), the entropy production rate for an incompressible Newtonian fluid held between two coaxial cylinders is k (dT) 2 ~[ d (~_)] 2 Sprod= 7 ~ . ~ ) q-y r~r •

(4.30)

The circular Couette flow between concentric cylinders is in the 0-direction only, and satisfies Vr =Vz = O, Vo = vo(r), and T = T(r). The inner cylinder is stationary while the outer cylinder rotates with an angular velocity w. Assuming a steady and laminar flow without end effects, the velocity distribution is v o _ 1 (r 2 - r i 2)

(4.31)

Wro R(d- d) where R = r/ro. With Eq. (4.31), we determine the second term on the fight of Eq. (4.30) as follows:

4w2 :ri4 T F4(F2-/~2)

(4.32)

- ~

T r~r

With surface temperatures of To and Ti for the outer and inner cylinders, respectively, the temperature profile is given by

0

(in.

To-~i-(B+I)

n21n n

1-~nn)+B

1/

R2

(4.33)

with the dimensionless quantities of n _ _ _4~ _ /L/,W2r2 n = rl" B = Br )2 " and Br = ro ' (1-n 2 ' k(To - T i )

(4.34)

where Br is the Brinkman number for the annulus. Equation (4.33) satisfies the boundary conditions of 0 = 0 at R - n, and 0 - 1 at R - 1. The temperature gradient can be obtained from Eq. (4.3 3)

d T _ ( T O _ Ti ) dr

2Bro r3

+

B

B+I)

~ - ~ rn 2 Inn r Inn

(4.35)

Substituting Eqs. (4.32) and (4.35) into Eq. (4.30), we can determine the rate of entropy production for the tangential annular flow.

v=wro, T = T o

\

v=O,T=T~

Figure 4.5. The circular Couette flow. Reprinted with permission from Elsevier, Int. J. Heat Mass Transfer, 43 (2000) 4205.

4.2

1 65

Second-law analysis

The first terms of Eqs. (4.23) and (4.30) show the entropy production due to the heat transfer Sprod,kT, while the second terms show the entropy production due to the fluid friction Sprod,6p; hence, the rate of entropy production expression has the following basic form:

Sprod = Sprod,kT + Sprod,kao

(4.36)

The irreversibility distribution ratio is Be = Spr°d'AT

(4.37)

Sprod

and is called the Bejan number Be. Be = 1 is the limit at which all irreversibility is due to heat transfer only. Irreversibility due to heat transfer dominates when Be >> 1/2, while Be << 1/2 indicates that irreversibility due to friction dominates. The effects of A and Br on the irreversibility distributions are shown in Figure 4.6 for unused engine oil with k = 0.14 W/(m K), kinematic viscosity v = 0.839 × 10 -4 m2/s, and p = 864.04 kg/m. We disregard the temperature

Br

2

4

//

0.6

0.4 Be

o.

~~-/t

R0 • 98 ~ ' - ~ . ~ 1 (a)

w=

1048

s -I ; R e

= 5000

8 sr

0.75 Be

0.5

0.

00

W

R0.98"~.~ 1 (b)

w = 8390

s-'

; Re = 40000

Figure 4.6. The Bejan number Be for the circular Couette flow for T~= 300 K, ro = 0.02 m, and r, = 0.019 m. Reprinted with permission from Elsevier, Y. Demirel, Int. J. Heat Mass Transfer, 43 (2000) 4205.

166

4.

Usingthe second law: Thermodynamic analysis

dependence of viscosity, density, and thermal conductivity. Except for the simple Couette flow where A = 0, the Be peaks appear in the ~ d d l e part of a Couette device due to the development of maximum temperature in that region. An increase in Be indicates a competition between the irreversibilities caused by heat transfer and friction. At high Reynolds numbers, the distribution of Be is relatively more uniform than at lower Re. For a circular Couette device, the Reynolds number (Re = wrZh,) at the transition from laminar to turbulent flow is strongly dependent on the ratio of the gap to the radius of the outer cylinder, 1 - n . The critical Re reaches a value --~50,000 at 1 - n = 0.05. We may control the distribution of the irreversibility by manipulating various operational conditions such as the gap of the Couette device, the Brinkman number, and the boundary conditions.

Example 4.5 Entropy production in a flow through an annular packed bed The introduction of suitable packing

into a fluid flow passage considerably enhances wall-to-fluid heat transfer, and hence reduces the entropy production due to heat transfer but increases the entropy production due to fluid-flow friction. Heat transfer to a fluid flowing in an annulus has a technical importance because we can heat or cool either or both of the surfaces independently. Entropy production provides a new criterion in analyzing such processes. In terms of the velocity and temperature profiles, the local rate of entropy production per unit volume of an incompressible Newtonian fluid for a two=dimensional annular flow is

Spr°d -- T-2- L ~,0---~-j

ik-~z ~]

(4.38)

+ --T-- l~.-~z ~]

Here, k and ~ are the thermal conductivity and dynamic viscosity of the fluid, respectively. The terms v and T denote the velocity and temperature of the fluid. The first term on the fight side of Eq. (4.38) shows the entropy production due to finite temperature differences in axial z and radial r directions, while the second term shows the entropy production due to fluid friction. We may construct the entropy production profiles using Eq. (4.38) if we know the temperature and the velocity fields. Assuming fully developed velocity and temperature profiles for the control volume of an annular packed bed, the energy equation is r

r Or -~r

= --

(4.39)

Olek-~2 )

Here, C~eis the effective thermal diffusivity of the bed and Tb the bulk fluid temperature. We assume that the plug flow conditions (v = Vav) and essentially radially flat superficial velocity profiles prevail through the cross-section of the packed flow passage, and the axial thermal conduction is negligible. The uniform heat fluxes at each of the two surfaces provide the necessary boundary conditions with positive heat fluxes when the heat flows into the fluid a t r = ro , -

= qo = constant

(4.40)

at r r=i

= qi = constant

(4.41)

, --

Equation (4.39) can be directly integrated because the term dTb/dZ is constant. The linearity of the energy equation allows the use of the superposition method to build solutions for asymmetric heating by adding the two fundamental solutions: (1) the outer wall heated with the inner wall insulated and (2) the inner wall heated with the outer wall insulated

f = T - To =

(StZR' + 1) + 2keDH (R 2 - 1 - 2(r*

f2 =T-To =--ff (StZr*R'+I)+

2keDH (R2-(r*) 2 +21n

lnR)

-21nR)

(4.42)

(4.43)

4.2

167

Second-law analysis

where R= r • r~ --; r =--; ro ro

R'-

2r° ; Zro+r~

z ro-ri

St and h are the Stanton number and the heat transfer coefficient, respectively. By adding the fundamental solutions j] andj~, we can obtain the temperature profile for the annular packed bed with asymmetric heating T = To (1 + zA)

(4.44)

where

A = StZR'(r* + n) + n + 1 +

Nukfr2R'

[R 2 (r* + n) - 2r* (1 + nr*) In R +

r* (2 In r* -

(r*

)2 ) -

n]

2keDpDH T--

q/h

--

TO

Tw-Tb. To

n_qo qi

The hydraulic diameter of the annular bed is DH = 2(ro -- ri) and Dp is the packing diameter. Relations for the Nusselt number and the effective thermal conductivity ke for the annular packed bed are Nu - hDp - 5. 9Re0.44

(4.45)

k e - kf (0.6 + 0.157PrRe)

(4.46)

k~

The heat transfer parameters have been derived for an annular packed bed in the range 200 < Re < 800 and D = DH/D p = 6. The average fluid temperature is Tav =

2I~' rTdr '

(4.47)

j'~°rdr

The following simple energy balance yields the gradient (dT/dz)" qi(1 + n)[2zr(r o + ri)]dz = GCpzr(r2o - ri2)dT

(4.48)

or it can be obtained directly from the differentiation of Eq. (4.44) with respect to axial distance z as

dT

M

dz

2qiR' n + r* Pekf

(4.49)

D

where Pe is the Peclet number and G the mass flux. From Eq. (4.44), we obtain the temperature gradient in the radial direction

OT Or

[

r"

qiR'r° R(n+ r* keDu ) -~- (1 + nr*)

]

(4.50)

Inviscid flow behavior dP/p - - d (v~/2) relates the pressure to the velocity. Using the Bernoulli and Ergun equations, we obtain the velocity gradient

dv dz

[ C 1 (1 - 8 ) 2 nt- C 2 ( l -

3 2 13 Dpp

8)Re]/x (4.51)

168

4.

Usingthe second law: Thermodynamic analysis

The Reynolds number is Re = GDp/iX. The constants C 1 and C2 are

C1 = 130 and C2

D 0.335D +2.28

(4.52)

These constants take into account the effect of confining walls. By substituting Eqs. (4.43)-(4.51) into Eq. (4.38), we get the following expression for the volumetric entropy production for the packed annulus: •

_ kf

Sprod -- ~

[(

qiR'ro k~DH

R(n+

r*

r.

)--~-(l+nr*)

(n+

(4.53) + 2/'z T (130(1- e)2/z e 3+ Dpp C2 2 (1 - s)Re/z/2

Here, the terms on the right side of the above equation show the entropy production due to heat transfer and fluid friction, respectively; hence, the entropy production expression has the following basic form: Sprod = Sprod,ar + Sprod,ae • The volumetric entropy production rate is positive and finite as long as temperature or velocity gradients are present in the medium. The dimensionless entropy production profile is J - - Sprod

kfT2 q2

(4.54)

where q = qi + qo. In second law analysis of convective heat transfer, one of the useful parameters is the dimensionless temperature ~- -

q/h

To

Tw - Tb - ~

To

(4.55)

The other dimensionless parameters are the following irreversibility distribution ratios:

(4.56) Sprod, AT

Be = Spr°d'AT --

Sprod

1 (1 + 4')

(4.57)

where Be is the Bejan number. Be = 1 is the limit at which irreversibility due to heat transfer dominates, and Be = 0 is the opposite limit at which irreversibility due to fluid friction dominates. In Eq. (4.53), local entropy production has been expressed in terms of z, R, Z, and D, including the properties of the fluid p and Cp. The rate of entropy production over the cross-section S" may be calculated by the following integration: S " : frr° SprodFdF

(4.58)

Example 4.6 Entropy production in a packed duct flow Fluid flow and the wall-to-fluid heat transfer in a packed duct are of interest in fixed bed chemical reactors, packed separation columns, heat exchangers, and some heat storage systems. In this analysis, we take into account the wall effect on the velocity profile in the calculation of entropy production in a packed duct with the top wall heated and the bottom wall cooled (Figure 4.7). We assume

4.2

169

Second-law analysis

ql

Y=O

dS"

T+dT

X=O

[

~x

Y=I

1

~/=

Figure 4.7. Control volume of the rectangular packed bed with asymmetric heat effects. Reprinted with permission from Elsevier, Y. Demirel and R. Kahraman, Int. J. Heat Mass Transfer, 42 (1999) 2337.

that the difference of wall-to-fluid bulk temperature is small. Physical properties of the fluid are constant, and there is no axial conduction, or natural convection. An approximate expression for the velocity profile in terms of the average velocity in a packed bed with H/dp -> 5 and T = 293 K is

- / 3 { 1 - ( 1 - a3Y)e a''" - [ 1 - a 3 ( 1 - Y)le a'(1-Y) }

U -

(4.59)

Vav

where the term/3 denotes the deformation factor that depends on the ratio H/dp and Rep -1

]3-

(

2(1- m + a2a3 )

1+

"

a1

2a3 (1- m ) )

+ - - " al2

(4.60)

where

a,-

aH nH - 7 , av - e "~ a3 - -7-

ap

--

'

~l p

Y-'

y H

n = - 1083 + 201.6a 4 - 3737a °5 + 5399a 1/3 , l < Rep < 1000

a-

411 4-

, a4 --lnRep

+4, Rep-

t7

u avdp v

Figure 4.8 shows the dimensionless velocity profile. The energy equation for a fully established laminar flow is

(4.61) where % is the effective thermal diffusivity and ?~ the bulk air temperature

; ~v TdA

(4.62)

4.

170

Using the second law: Thermodynamic analysis

(a) Z.5 1.25 1

U

0.75 0.5 0.25 .

.

.

.

0

.

.

0.2

.

.

.

.

.

.

.

0.4

.

.

.

.

.

0.6

0.8

1

0.6

0.8

1

Y

(b) 1.5 u

1 0.5

0

0.2

0.4

y

Figure 4.8.

Velocity profiles in the packed bed for (a)

H/dp = 5, Rep = 309,/3 = 0.816; (b) H/dp = 10, Rep = 154,/3 = 0.871.

Reprinted with permission from Elsevier, Y. Demirel and R. Kahraman, Int. J. Heat Mass Transfer, 42 (1999) 2337.

The heat flux at the upper and lower surfaces specifies the temperature gradient at the wall, and the necessary boundary conditions .

.

. ql

.

y. .constant, .

H;

ke 0T

q2 = constant

(4.63)

The linearity of the energy equation suggests that the superposition method may be applied to build solutions by adding two fundamental solutions for the top and bottom walls. For a constant heat flux, a simple energy balance is

qdx= PVavHCpdT b

(4.64)

and yields the temperature gradient in the flow direction. Using Eqs. (4.59)-(4.64), we can determine the velocity and temperature profiles and derive the rate of entropy production in the packed bed

•

ke[(OTI2+(OT]2] tz(dvl2 0.,]2) ~-~XJ +-Tt,dy)

(4.65)

Spr°d = ~ -

Here, the first term on the fight shows entropy production due to heat transfer, and the second term shows entropy production due to fluid friction. Equation (4.65) relates the rate of entropy production to heat duty q, the Reynolds number, and the Stanton number St = Dimensionless entropy production N is

h/pvavCp.

N = Sprod (ql -- q2) 2

H/dp,

(4.66)

4.2

Second-law analysis

171

15

(a) lO

1.

0.5

0.2 0.4 0.6 0.8

15

(b)

5

1

o

-,q 0.4-"-.,.,.. Y

0.6

Figure 4.9. Distribution of entropy generation N, in the packed bed: (a) H/dp= 10, ql = 66W/m 2, r = 0.2, Rep= 154,/3=0.871" (b) H/dp = 20, ql = 66 W/m 2, r = 0.2, Rep = 774,/3 = 0.905. Reprinted with permission from Elsevier, Y. Demirel and R. Kahraman, Int. J. Heat Mass Transfer, 42 (1999) 2337.

The distribution of entropy production is calculated for air flow with constant properties of k = 0.026 W/(m K), = 1.84 x 10 -5 kg/(m s), p = 1.17 kg/m 3, and Pr = 0.7, and is shown in Figure 4.9 for an inlet air temperature of To = 297 K. When there is no packing in the flow passage, we can determine the temperature profiles similarly using the parabolic velocity profiles of v = 6Vav( Y - y2) with the superposition approach T* = To (1 - z0*) where X = x / H and ~* = ( h H / 2 k O [ ( r - 1)( y4 _ 2 y3 _ 2 Y + r)] + (1 - r)(StX + 1).

(4.67)

4.

1 72

Using the second law: Thermodynamic analysis 15 X

0.5

0,2 0,4 0.6 0.8

Figure 4.10. Distribution of entropy generation in the empty bed for ReH = 1641, St = 0.005, ql = 34 W/m 2, r = 0.2. Reprinted with permission from Elsevier, Y. Demirel and R. Kahraman, Int. J. Heat Mass Transfer, 42 (1999) 2337.

Using the velocity and temperature gradients, we obtain the dimensionless entropy production for the empty bed (Figure 4.10). Comparison of Figures 4.9 and 4.10 indicates that outside the wall region, the distribution of the rate of entropy production is uniform in the packed bed, which is the thermodynamic optimality criterion. The profile of entropy production shows a typical S shape in the empty bed. Usually, minimization of entropy production leads to an increase in equipment size. For example, a large heat transfer area for a heat exchanger may be necessary. This, however, is not in practice an economically optimal solution, and we should consider other possible optimal flow configurations.

Example 4.7 Heat and mass transfer Second-law analysis may play an important role in optimizing forced

convective heat and mass transfer. The analysis may provide engineers with an optimum Reynolds number and the best geometric configuration after minimizing the entropy production. This may be the result of a trade-off between the irreversibilities due to heat and mass transfer. A general expression for the rate of entropy production is (Kjelstrup and Hafskjold, 1996) .

Sprod

_

_

1

R

q(T _ T~ )dA + f j i -~i A

(

Ts

T~

T~+

,

Ci ~

'

/

dA+--

T~

(4.68)

where T~ is the free stream temperature, q the heat flow, A the surface area of the body, Jr,.the mass flow of species i per unit area, R the gas constant, Mi the molecular weight of species i, Ci,s the concentration of species i at the body surface, C;,~ the concentration of species i at free stream conditions, F the total drag force exerted on body, and v~ the free stream velocity. In Eq. (4.68), we assume that the temperature difference Ts - T~I and the concentration difference ICi,s- Ci,~l are small. The irreversibilities are due to heat transfer across a finite temperature difference, mass transfer across a finite difference in the chemical potential of species, and flow friction. The drag force on the plate is

=J'o

(4.69)

4.2

173

Second-law analysis

where Cf is the local friction coefficient. The temperature and concentration differences are related to the respective flows as T - T~.

_

Ji

q.

(4.70)

-£, ci, S - c i ~ , km

where h and km are the local heat and mass transfer coefficients, respectively. For laminar and turbulent flows, we need appropriate correlation equations for the friction coefficient, heat transfer coefficient, and mass transfer coefficient. For laminar flow in the ranges of 5 × 106 > Re > 103, and Pr and Sc > 0.5, we have the following relations for the coefficients" C r = 0.664Re -°s

(4.71)

h = 0.458 Re O.SprO.33 _k

(4.72)

X

(4.73)

k m = 0.458 Re°SSc 033 Di X For turbulent flow in the ranges of 5 x 10 s < Re < 107, 60 > Pr > 0.6, and 3000 > Sc > 0.6, we have Cf = 0.0592Re -°'2

(4.74)

h = 0.296 ReO.SprO.33 _k

(4.75)

X

(4.76)

k m =0.296Re°8Sc °33 Di X

Substituting Eqs. (4.69)-(4.76) into Eq. (4.68) and performing the integration, the entropy production rate can be obtained for a laminar flow on a flat plate

•

w

Sprod -- 1.456~- ReT.°5

7,2 +

R8c-O.]33

,, R/pr°33

j2

MiT~

M2iCi,~Di

k

I )

(4.77)

+0.664 Re °5 ~ u ~ v ~

We can obtain a similar expression for the entropy production for a turbulent flow from Eqs. (4.68) and (4.74)-(4.76) . W ReTO. s• Sprod - 28.15 7

[ q2 ) pr-0.33 JiqR ~ + MiT~

k

j/2RSc-°33 ] M2Ci,~Di

(4.78)

where W and L are the width and length of the plate, D the mass diffusivity coefficient, and Sc and Pr the Schmidt number and Prandtl number, respectively. By setting 0Sprod/ORe - 0, we determine the optimum Reynolds number, and hence the minimum entropy production at known values q and J;. For a laminar flow over a plate, we have "~ ' 033 2 2.19 [(1/L2)((q'/T~)+(JiqR/MiT~))(Pr-°33/k)+(JiRSc- • / L M i Ci,~Di) ] ReL,op t =

p=u=v2/T=

(4.79)

174

4.

Usingthe second law: Thermodynamic analysis

For a turbulent flow over a plate, the optimum Reynolds number becomes

(1/t2)[((q2/T2) + (JiqR/MiT~))(Pr-0"33/k)+ (j2 RSc-O.33/M2Ci ~Di)1}0.625 ReL,opt =62.69

2 p~v~v~/T~

'

(4.80)

For a cylindrical geometry, the rate of entropy production becomes Spr°d = 0"462 1LRel)°466 [( -~+MiT~ q2 JiqR]

Pr-°33 k

~ +

j2RSc-°'33 ] ~ D

i

(4.81)

+2.743Re~754(p~v~v2) .......

.....

For cylindrical geometry, the following empirical relations, which are valid for 40 < Re < 1000, are used: Cf = 5.484 ReD0246

(4.82)

h = 0.689 Re~466pr 0"33 k D

(4.83)

km= 0.689Re~466Sc0.33De

(4.84)

D

The expression for the optimum Reynolds number is

ReD,opt = 1.5 7

{ (1/L2)[((q2/T2) + (JiqR/MiT~))(Pr-°33/k) + (JiRLSc-°33/Mz Ci ~Di)]t

p~v~v2/T~

0.82 (4.85)

The optimal Reynolds number defines the operating conditions at which the cylindrical system performs a required heat and mass transport, and generates the minimum entropy. These expressions offer a thermodynamically optimum design. Some expressions for the entropy production in a multicomponent fluid take into account the coupling effects between heat and mass transfers. The resulting diffusion fluxes obey generalized StefanMaxwell relations including the effects of ordinary, forced, pressure, and thermal diffusion.

Example 4.8 Chemical reactions and reacting flows The extension of the theory of linear nonequilibfium thermodynamics to nonlinear systems can describe systems far from equilibrium, such as open chemical reactions. Some chemical reactions may include multiple stationary states, periodic and nonperiodic oscillations, chemical waves, and spatial patterns. The determination of entropy of stationary states in a continuously stirred tank reactor may provide insight into the thermodynamics of open nonlinear systems and the optimum operating conditions of multiphase combustion. These conditions may be achieved by minimizing entropy production and the lost available work, which may lead to the maximum net energy output per unit mass of the flow at the reactor exit. One of the ways to reduce energy costs in the chemical process industry is to increase process reversibility by increasing equipment size. Engineers have to make a trade-off between the energy and area costs. Other tradeoffs are possible between the system output and transfer area and the system output and energy consumption. The equipartition of forces principle suggests that the trade-offs would be optimum for those processes with uniformly distributed thermodynamic driving forces over the transfer area. The local entropy production of a reacting mixture in a system with gradients in temperature T, and chemical potentials/z i, is given by

• Spr°d---JqV

-'-f1 i JiVtxi,r- j Jrj T

(4.86)

4.2

175

Second-law analysis

Here, Jq is the total heat flow, Ji the mass flow of component i, and Jr/the reaction rate (flow) of reaction j. For chemical reactions, linear phenomenological equations are l

jr/ - - Z L i

J

•

AG/ •

"

(4.87)

T

With a homogeneous reaction and mechanical equilibrium (VP = 0), consider a reactor consisting of a large number of n subsystems with equal volumes and the same reaction taking place in all subsystems. We assume that the subsystems have a uniform composition and temperature. The reaction flow in subsystem k is Jrk and the driving force is A G~/T. The total system is a nonhomogeneous reactor with variations in temperature and composition

Spro&min £OPkVk £Lk(~kGk) 2 k k

VI(

(4.88)

~_~JkVk = - ~ Lk - 7 - - V k = constant

(4.89)

_._

~

~

For a specified total reaction rate, AGa.

The Cauchy-Lagrange method of constant multipliers yields

oZ.,%

O(AGk/T) + A O(AGk/T) = 2L k

+ ALk - 0

(4.90)

Thus, we obtain AG k - - ' ~ T 2

(4.91)

The above equation implies that for a given total reaction rate and a given total volume, entropy production is minimal when the driving force AG/Tis equal in all n subsystems. According to the linear duality theory, the results of the optimization will be the same if we maximize the total reaction speed for a given entropy production. Therefore, a thermodynamically efficient reactor has a uniform AG/T in all parts of the reactor volume. This result is independent of the local variations in the reaction rate. Another consequence of Eq. (4.91) is that if we arrange the n subsystems in time instead of in space, then the collection of subsystems constitutes the reaction path of a batch reactor where Vk is the volume of subsystem k. For a specified conversion and time, we should minimize the sum of.Jk(AGk/T)Vk. This minimization leads to results similar to Eq. (4.91), and supports the principle of equipartition of forces. Hence, for a given total conversion and reaction time, minimum entropy production results when the driving force AG/T is equal in all n time intervals. Similarly, the conversion is maximum for a given entropy production and reaction time when the driving forces are uniform. The Gibbs-Helmholtz relation is

~(1/T)

p

(4.92)

If N/-/is constant, integrating the above equation from an equilibrium temperature Teq to the optimal temperature Topt yields

~G°pt - AH ( 1 1) Topt Topt Teq

(4.93)

There is no constant of integration due to the boundary condition that both AG/T and A(1/T) are zero at equilibrium. However, N/-/will be temperature dependent most of the time. For example, in producing ammonia from hydrogen

176

4.

Usingthe second law: Thermodynamic analysis

and nitrogen, the goal is to maximize the output of ammonia at the exit. An approximately constant ATbetween the optimal path and the equilibrium temperature provides the optimal temperature profile, which reduces the exergy loss by --~60% in the reactor. The equipartition of forces principle for multiple, independent rate-controlled reactions and multiphase and coupled phenomena, such as reactive distillations, may lead to the improved use of energy and reduced costs (Sauar et al., 1997).

4.3

EQUIPARTITION PRINCIPLE

The equipartition of forces principle combines the nonequilibrium thermodynamics approach with the CauchyLagrange optimization procedure. The principle shows that the best trade-offs between entropy production and transfer area in transport processes are possible when the thermodynamic driving forces are uniform over the transfer area. For example, in a rate-controlled chemical reaction, the distribution of ~ G / T should be uniform through space and time in the reactor system (the term ~G is the change in the Gibbs energy for a reaction). For example, mathematical models show that a cascade drying process with uniform driving force across every stage yields a substantial decrease in energy consumption. Some options for achieving a thermodynamic optimum are to improve an existing design so the operation will be less irreversible and to distribute the irreversibilities uniformly over space and time. This approach relates the distribution of irreversibilities to the minimization of entropy production based on linear nonequilibrium thermodynamics. For a transport of single substance, the local rate of entropy production is (4.94)

= JX

where J is the local flow of the substance and X the conjugate driving force. Assuming that the linear phenomenological relations hold between the flow and force, we have (4.95)

J = LX

where L is a phenomenological coefficient assumed to be constant and positive. The total entropy production is the integral of • over the time and space variables P = [[*dVdt = L[[ XZdVdt

JJ

JJ

(4.96)

The total flow is the integral over time and space of the local flow J = L f l X d V d t = L VtXav

JJ

(4.97)

where Xav is the average driving force. The total entropy production from Eq. (4.96) is

Pav-- JXav

(4.98)

The difference between the general case and the average value is (Hohmann, 1971)

P-Pav -= L [ ~ y 2 d V d t - ( Y a v ) 2 V t ] = tVt[(X2)av - (Xav) 2 ]

(4.99)

The square bracket on the fight of the above equation is the difference between the mean square and the square mean of the force distribution, and is the variance of X. We therefore have P-Pay > 0 Vt

(4.100)

Pav (equipartioned) < P (arbitrary)

(4.101)

or

The entropy production Pav of a process with a uniform driving force is smaller than that of a nonuniform situation with the same size, and duration of the same average driving force, and the same overall load J. Equations (4.94) and (4.95) show that the local flow and the local rate of entropy production • will be constant when X is uniform.

4.3

1"/7

Equipartition principle

Therefore, the equipartition of forces is analogous to equipartition of flows or of entropy production. The relations of flow and the entropy production in matrix forms are a - [L]. X - Xs'J-

(4.102)

X T'[LI'X

(4.103)

where [L] is the symmetric matrix of phenomenological coefficients due to the Onsager reciprocal relations. The total entropy production P and the total flow J (specified duty) are

P- ff xf.lLl.X.dVdt

(4.104)

a - f l [L]. XdVdt - Vt[L]. Xav

(4.105)

where Nayis the average driving force vector, and elements of it are the averages of the individual driving forces. Using the average force, the total entropy production becomes P~v = J" XaSv The excess entropy production P

-

(4.106)

is obtained as

Pay

P - P a y - If( XT "[L]'X-XaTv "[L]'Xav)dVd'

(4.107)

The above equation can also be expressed as

If ( e l ' e - e f a v ' e a v ) d V d t - [ ( e s ' e ) a v - e a v 'we a v ] Vt

P-R~v -

(4.108)

where e = [R] • X, and [L] is the positive matrix and may be decomposed into a product of matrices [ L ] - [R] T .[R]

(4.109)

The quantity of excess entropy production is positive by the Cauchy-Schwartz inequality (similar to the inequality in Eq. (4.101)), indicating that P > PaySince the minimization of entropy production is not always an economic criterion, it is necessary to relate the overall production and distribution of entropy to the economic analysis by considering various processes with different structures and operating configurations. One optimum requires a uniformly distributed entropy production rate in a heat exchanger, mixer, or separator. Consider the example of countercurrent and cocurrent heat exchangers shown in Figure 4.11. Temperature profiles

Wl

1

_-

Z

=_

2

K

Z

2

J I .....

,,] __

~]

F-

1 [

I

1F 1

1

countercurre11[

]

COCurrent

Figure 4.11. Heat exchangers with countercurrent and cocurrent operations.

178

4.

Usingthe second law: Thermodynamic analysis

show that the driving force AT, or 1/AXT,is more uniformly distributed in the countercurrent than in the cocurrent flow operation. This is the basic thermodynamic reason why a countercurrent is better than a cocurrent operation. The duty of the exchangers depends on the flow rate and inlet and outlet temperatures T1 and T2 of cold streams. The duty is the amount of heat transferred from the hot fluid to cold fluid. The heat exchangers are identical except for the flow arrangements. The cocurrent exchanger will require a higher flow rate and/or higher temperature of hot fluid, and hence the operating cost will be higher than that of the countercurrent exchanger. Alternatively, the cocurrent exchanger will require a larger heat transfer area for a specified flow rate and inlet temperature of the hot fluid, which will require a greater initial investment. Therefore, the countercurrent exchanger may minimize either operating or investment costs compared with the cocurrent exchanger.

Example 4.9 Entropy production in separation process: Distillation Distillation columns generally operate far from their thermodynamically optimum conditions. In absorption, desorption, membrane separation, and rectification, the major irreversibility is due to mass transfer. The analysis of a sieve tray distillation column reveals that the irreversibility on the tray is mostly due to bubble-liquid interaction on the tray, and mass transfer is the largest contributor to the irreversibility. A distillation column requires a large amount of heat in the reboiler and discharges a similar amount of heat at the condenser, and hence resembles a heat engine that requires optimum operating conditions. Heat causes a required separation of the components of a feed stream into products. Second-law analysis may be an effective tool for identifying the possible improvements in distillation column design by determining the entropy production due to thermodynamic inefficiencies in a column. The lost work profiles may quantify the inefficiencies in terms of the pressure drop, mixing, heat and mass transfer, and coupling between heat and mass transfer. The thermodynamic optimization of a distillation column leads to uniform irreversibility distributions. Such an optimization may be achieved through a set of internal and external column modifications, such as altering the feed condition or feed stage location, and using intermediate exchangers to reduce irreversibility in sections with large driving forces and to increase irreversibility in sections with small driving forces. Nonequilibrium molecular dynamics simulations show that the assumption of local equilibrium in a column with heat and mass transfer is acceptable. The dissipation function in a binary distillation is (Ratkje et al., 1995; Sauar et al., 1997) 2

= TSproO=

-JqVlnT- EJiVIXi,v

(4.110)

1

On the other hand, the rate of entropy production is Sprod =

-Jq

VT

1 _2

T2

T

Z1 JiVl&i,T

(4.111)

At constant pressure V/.t;,r = V/xc is the concentration-dependent part of the chemical potential gradient. Through the Gibbs-Duhem equation, we can relate the chemical potentials of heavy "h" and light "1" components in the gas phase as follows: V/x~ =

Yl V/xf Yh

(4.112)

where Yl and Yh are the mole fractions in the gas phase of the light and heavy components. From Eqs. (4.110) and (4.112), we obtain Sprod -

where

-Jq --~--

Vtx--LT

(4.113)

Jd (in m3/(m2 h)) is the relative mass flux across the interface Jd

J1

Jh

Yl

Yh

(4.114)

4.3

Equipartition principle

179

The phenomenological equations that follow from Eq. (4.113) are

Jq = - Lqq -VT ~ - LqlY1 V/~ T

(4.115)

Jd = -- Llq ~VT - LllYl V/rx___L

(4.116)

where Lji are the local phenomenological coefficients, which can be determined from experiments. For isothermal conditions, the phenomenological coefficients for mass transfer are

T ) (4.117)

LI1 = _ Yl V/~l

~XT=0

Using the chemical force for mass transfer VIZl _

Yl T

Jd

Llq V T

Lli

L11 T2

(4.118)

we obtain an expression for the heat flow

Jq=- Lqq-Lql Lu )---~+--~1Jd

(4.119)

On the other hand, Fourier's law of heat conduction without mass transfer is

(Jq )J,/=o = -kVT

(4.120)

where the thermal conductivity k becomes /-qq) 1

k = Lqq - Lql LI---~ T2

(4.121)

The total rate of entropy production for a stage is '~prod -- f fI)vdV v

(4.122)

The entropy production rate is determined with quasi-steady-state calculations. The following constant gradients in the gas phase at each stage are used: VT-

AT Ax

(4.123)

7/xl -

A/x1 2ix

(4.124)

By assuming that Yl and T are approximately constant, and using Eq. (4.122), the entropy production for a stage becomes •

1 AT r

Spr°d -- -- r 2 z2LlcJ

Yl Akh f

J qdV - T A x j J ddV

(4.125)

180

4.

Using the second law: Thermodynamic analysis

where d V = dAdx, A is the contact area, and x the distance. The integrals are the heat and mass transfer per unit time in the mixture volume, respectively, and can be calculated using Eqs. (4.115) and (4.116). Equation (4.122) determines the rate of entropy production. The flow on a stage Jd can be calculated from the diffusion coefficients in the gas phase and from the energy balance. In an adiabatic distillation, the heat flow across the interface contains the latent heat and heat conducted away from the interface kAT

(Jh + J1)Hn = (Jh + J1)Hn-1 + JhAHh + JlZ~J]l - ~Ax

(4.126)

If the enthalpy of the mixtures at stages n and n - 1, Hn and Hn_ 1, are similar, and the temperatures of these stages are close to each other, we have Jl~/1

~ --Jh~/h

(4.127)

This means that the heat of vaporization is approximately equal to the heat of condensation for the mixture. From Eqs. (4.118) and (4.127), the flow on a stage is expressed by

Jd=Jl(I + l zxH1 Yh ~J-/h

(4.128)

Fick's law predicts the diffusion of the light component in the gas phase (4.129)

J1 = - D Acl kx

where D is the diffusion coefficient of the light component and AC1 the concentration difference of the light component across Ax. The concentration difference is ACl -- Cl,g - Cl,g

(4.130)

where the concentration in the gas phase at the total pressure PT is PT Cl,g = Yl R T

(4.131)

* PT

(4.132)

At the liquid-vapor interface, we have

Cl,g = Yl R T

where the mole fraction Yl is the inlet composition of the liquid. Inserting Eqs. (4.128) and (4.129) into Eq. (4.117), and assuming constant driving forces, we express the phenomenological coefficient of the mass transfer as follows: 1.11_

T

[ (11 1 1 D Aq

YlA/xl

-~

Yh AHh

(4.133)

m

The average coefficient/-ql is dx fill -- a 0( ~ ) 2

(4.134)

The phenomenological coefficients obtained from Eq. (4.133) may vary considerably from enriching section to stripping section.

4.3

181

Equipartifion principle

The chemical driving tbrce on a stage has inlet and outlet concentrations as boundary conditions. In the enriching section below the feed plate, a flow Jd from liquid to vapor occurs while in the stripping section, the direction of flow is from vapor to liquid. For the column with specified inlet and outlet compositions, the entropy production rates are

~1 = I L1','X(dAdx' ~2 = I I~',2x2dAdx V

(4.135)

V

where the net separation flow Jd,1, for example, for stage 1 is . j j ,l dA dx = [. I~ l,l X l dA clx V

(4.136)

V

where Xis the chemical force in Eqs. (4.135) and (4.136). A specified level of separation determines the boundary for the forces, and an increase in the force in one stage must lead to a reduction in another stage. The sum of the entropy production rates is

Sprod,1 -+-Spro&2 = I (Lll,1x2 + Lll,222) dAdX V

(4.137)

The total flow Jj is given by (4.138) V

V

V

An increase in the flow for a given entropy production rate and a reduction in the entropy production rate for a specified separation are desired; the yield y is defined as the benefit-cost ratio in an economic sense, and given by Jd,~_LllX; _ 1 Y--

~i

L,1X~

Xi

(4.139)

When the derivative ofy with respect to Xi is higher in one stage than in another, the ratio of flow to entropy production increases by increasing the driving force, and hence the entropy production rate is adjusted by increasing or reducing the force. We can maximize the separation output by redistributing the forces between the stages. Such a distribution results from the following differentiation: d(1/Xl_____~)= d(1/X2 ) dX1 dX2

(4.140)

X 1- X2

(4.141)

The above equation yields

The equality of forces is independent of the individual values of the phenomenological coefficients. This means that the variation of the entropy production rate along the column follows the variation of the phenomenological coefficient Lll. The reversible operation is possible when X1 and )(2 approach zero and y increases toward infinity. Therefore, the practical way to improve second law efficiency is to apply the relationship between dX1 and dX2. For a constant Jd, we obtain m

m

dX1 = - Lll'---LdX2

Lll,2

The above equation relates the driving forces of the two stages (Tondeur and Kvaalen, 1987).

(4.142)

4.

182

4.3.1

Usingthe second law: Thermodynamic analysis

Separation Work

In a distillation column, we supply heat at a higher temperature source in the reboiler, and then discharge at a lower temperature in the condenser (Figure 4.12). Assuming the column to be a reversible heat engine, the net work available from the thermal energy is (Ognisty, 1995)

4,43 where To is the ambient temperature. The temperature corrections describe the maximum fraction of theoretical work extracted from thermal energy at a particular ambient temperature. The minimum separation work Wminrequired for separation is the net change in availability A (A = H - ToS) - W m i n - - ~ d s = A p r o d - Afeed

(4.144)

The change of availability of separation is the difference between the work supplied by the heat and the work required for separation, which contains the work lost due to irreversibilities ~ d s -- Wheat -- Wts

(4.145)

where Wts is the total work necessary for the separation. Minimizing the work lost due to irreversibility will minimize the total heat needed for separation. Efficiency based on the second law of thermodynamics Thh is mmin

(4.146)

.

't~th --

dPdt Wmin + V :o

Heat and mass transfer in a distillation column are coupled, and if the temperature field or chemical force is specified in the column, the other force would be defined. Maximum second law efficiency results from minimizing the entropy production rate with respect to one of the forces. For example, if the contribution of mass transfer is dominant, we should try to minimize the change of the entropy production with respect to the chemical force. The main effects through which work is lost are pressure drops due to fluid flows, heat transfer between fluids with different temperatures, and mass transfer between streams that are not in equilibrium: • The work lost due to a high-pressure drop (as high as 10psi) is considerable at the condenser and reboiler. The pressure drop is relatively smaller through the trays (0.1 psi or less). A large pressure difference affects the distance

I

"

Tc

....,',',',] Condenser

Distillate

Feed

=,... v

l qR

Bottoms

TR I Reboiler Figure 4.12. Distillation column as a heat engine between reboiler and condenser.

4.3

Equipartition principle

183

from equilibrium and causes the large temperature difference and hence utility costs between the condenser and reboiler to increase. • The work lost due to heat transfer results from differences in temperature between the inlet streams of liquid and vapor on each tray, and is usually a large contributor to the total lost work. Enthalpy profiles display the heat transfer on each tray. Since the heat and mass transfer are coupled, any changes in heat transfer through heating and cooling modifications will change the internal mass balances. A modification in mass transfer will have similar effects on heat transfer properties in the column. If a cheap power source is available, intermediate exchangers may be feasible, although the number of trays will need to increase due to the operating line being closer to the equilibrium curve. Lowering the duties of the reboiler and condenser reduces the overall flow in the column, and results in a smaller diameter column design. Improving thermodynamic efficiency leads to a basic trend of taller and more slender columns. • Large amounts of lost work due to mixing and mass transfer mainly occur around the feed trays. The mixing may take place between streams with widely different compositions. Amounts of heavier components decrease above the feed tray and lighter components diminish below the feed tray. From the thermodynamic perspective, we may adjust the location of feed tray to counterbalance the lost work. Often, the feed location is determined at the minimum utility loads and tray count or simply by taking into account light- and heavy-key component compositions. The relative cost of the heating and cooling media may influence the location of the feed stage. If a very cold feed enters a stage, then a large amount of heat exchange is necessary below the feed stage to strip the light components. When the heat transfer rises considerably around the feed tray location, feed preconditioning may be useful to unload the top or bottom sections of columns. Preconditioning the feed is less expensive than interheating or intercooling. Heat profiles and heat transfer lost work plots can be used together to determine if feed preconditioning is necessary. • In many multicomponent mixture separations by distillation, components can display large concentration changes within the column, which cause considerable lost work. One of the ways of overcoming this obstacle is to remove the key components from the feed. As seen in Figure 4.13, light nonkey components can be removed by using an absorber, and the bottom products of the absorber provide the feed to the main distillation column. The heavy nonkey components are removed by using a prestripper, and the excess products of the stripper become the feed of the main distillation column. As seen in Figure 4.13, heat pumping, vapor compression,

Absorber

Main column ~- Liglhts Lights Light key

Feed

_l

-

Feed

Light bottoms ....

Bottoms

I_.

FHeavy bottoms (b)

Stripper

Main column (a)

t

-D fl ¸

C

-~D C

~,B ~B

(c)

(d)

(e)

Figure 4.13. Prefractionation arrangements: (a) removing light keys with absorber; (b) removing heavy keys with stripper; (c) heat pumping; (d) vapor recompression; (e) reboiler flashing. B: bottom product, D: distillate, V: valve.

184

4.

Usingthe second law: Thermodynamicanalysis

and reboiler flashing may also be useful. These modifications will reduce the load of the column to prevent bottlenecking and reduce the required number of stages. • For a sieve tray distillation column, we may calculate the entropy production for heat, mass, and momentum transfer accounting for the movement of a bubble through a moving liquid pool. Some variables are orifice diameter and weir height for the required separation characteristics of components in vapor-liquid phases. Mass transfer and work done against liquid during bubble growth and the drag on bubbles are the major causes of entropy production on a tray. The bubble-liquid interactions are the major contribution of irreversibility, while the effects of the interaction of the flowing liquid with the tray internals are negligible. Bubbles forming at sieve trays may cause entropy production, as most of the heat and mass transfer occurs before bubble detachment. Bubble growth after detachment is small, while viscous drag on bubbles also contributes to irreversibility. The effect of the weir height is more significant than the effect of the sieve-hole diameter. Increasing the weir height shows a monotonic increase in the entropy production, while an increase in sieve-hole diameter is associated with maximum entropy production. This diameter range mainly depends on the properties of the mixture on the tray.

4.4

EXERGY ANALYSIS

Mass and energy are never lost in any physical transformation process. Energy remains constant but changes its form during a process. To determine what is lost in resource transformation processes, we need to utilize the second law of thermodynamics, which states that a part of accessible work potential is always lost in any real process. A certain amount of the total energy is not available to do useful work. For example, the same amounts of total energy may have different capacities to cause a change because of the varying available energy. The available energy is a measure of a process's maximum capacity to cause a change. The capacity exists because the process is in a nonequilibrium state.

4.4.1

Exergy

Exergy is the maximum amount of work theoretically available by bringing a resource into equilibrium with its sur-

rounding through a reversible process. Therefore, exergy is a function of both the physical properties of a resource and its environment. In all real processes, exergy loss always accompanies exergy transfer. The maximum work output of any process occurs if the process proceeds reversibly toward equilibrium with the environment (dead state or reference state). The actual work output is much smaller due to process irreversibility. The work loss in a continuous process is the difference in the exergy before and after the process. At the dead state, both the system and its surroundings possess energy but no exergy, and hence there is no spontaneous change within the system or the surroundings. Available energy, A = H - ToS, or exergy is a measure of the departure from the ambient or dead state. As shown in Figure 4.14, in a heat exchanger, the temperature of the hot stream decreases, and its availability goes down, while the temperature of a cold stream increases, and availability increases. So, a heat exchanger transfers available energy from the hot stream to the cold stream, and some of the available energy is lost to allow the heat transfer processes to occur within a finite time and cost. Exergy is an extensive property and a thermodynamic potential. In contrast to energy, exergy is not conserved and decreases in irreversible processes. Exergy is a broadly useful concept both in engineering and in proper resource management for reducing environmental destruction. Exergy expresses simultaneously the quantity and quality of energy; quality is the ability to produce work under the conditions determined by the natural environment. If we discharge the waste product of the process into the environment, external exergy loss occurs due to the deviation of

A

~,

Hot

A

70

Hot

7o ,,..._ r

y X

Figure 4.14. Heat transfer above and at the ambient temperature To.

X

4.4

185

Exergyanalysis

thermal parameters and the chemical composition between the product and the components of the environment. The thermal state and chemical composition of the natural environment represent a reference level (dead state) for the calculation of exergy. Exergy is a unifying concept of many forms of energy, such as heat, mechanical work, and chemical energy. We can derive the exergy Ex relation from the energy and entropy balances for the composite system shown in Figure 4.15 Ex - (Et - Uo)+ P o ( V - Vo)- T o ( S - So)

(4.147)

where E t is the total energy (E t - U + KE + PE) and U, V, and S denote the internal energy, volume, and entropy of the system, respectively. The terms with subscript '0' are the values of the same properties when the system was at the dead state. The terms KE and PE are the kinetic and potential energies, respectively. Some properties of exergy are" • Exergy is measured with respect to the environment; therefore, it is attributed to the composite system. If the environment is a reference state with zero exergy, then exergy becomes a property of the system. • If the system is not at the dead state, then it will undergo a spontaneous change. • The value of exergy loss cannot be negative. • Exergy decreases due to irreversibilities in the system. If a system undergoes a spontaneous change to the dead state without a device to perform work, then exergy is completely lost. • Exergy is the minimum theoretical work input necessary to change the system from the dead state to the specified sate. Specific exergy ex based on a unit mass is given by V2 e x -- ( U -- lto ) + PO ( I .... Vo ) - To ( s - so ) + --£- + g z 2

(4.148)

The kinetic energy (v2/2) and potential energy (gz) are relative to the surroundings and contribute fully to the magnitude of exergy. Using Eq. (4.147), the change in exergy between two states of a closed system is EX2 -- E X l :

(E 2 -

E l ) + Po(V2 - V1 ) - To(S 2 - S1)

(4.149)

where Po and TOshow the pressure and temperature of the surroundings.

4.4.2

Environment

Since exergy is a measure of the departure of the state of the system from that of the environment, it relates the system to the environment. When a system is in thermal, mechanical, and chemical equilibriums with the environment, there are no processes taking place and the system is at the dead state. At dead state, the system has no motion and elevation relative to coordinates in the environment. Only after specifying the environment can we estimate a value for exergy. Specifying environment usually refers to some portion of a system's surroundings. For example, in estimating exergy values, the temperature and pressure of the environment are usually the standard state values, such as 298.15 K and

/

Surroundings TI,.PI, .~'.~" Heat j/~'" . . . . . -'oS-." Work

/ /

\

", system

/'

Figure 4.15. Combined system.

~"/

/

4.

186

Usingthe second law: Thermodynamic analysis

101.31 kPa. Sometimes, the standard state values are the average values of the ambient temperature and pressure of a location where the process under consideration takes place. The environment is composed of large numbers of common species within the Earth's atmosphere, ocean, and crust. The species exist naturally. They are in their stable forms and do not take part in any chemical or physical work interactions between different parts of the environment. We mainly assume that the intensive properties of the environment are unchanging, while the extensive properties can change because of interactions with other systems. Coordinates in the environment are at rest with respect to each other, and relative to these coordinates, we estimate kinetic and potential energies. In the natural environment, however, there are components of states differing in their composition or thermal parameters from thermodynamic equilibrium state. These components can undergo thermal and chemical processes. Therefore, they are natural resources with positive exergy. Only for commonly appearing components can a zero value of exergy be accepted. A correct definition of the reference level is essential for the calculation of external exergy losses. The most probable chemical interaction between the waste products and the environment occurs with the common components of the environment.

4.4.3

Exergy Balance

The decrease of exergy of a system during a process can be expressed as Change in the total exergy = Total exergy i n - Total exergy o u t - Total exergy loss The exergy balance consists of internal exergy losses. Irreversible processes may cause the distribution ofexergy losses within the volume, and the partition of exergy losses may help in understanding the thermodynamic performance of the system. The exergy balance of a closed system (Figure 4.16) between states 1 and 2 is Ex 2 - Ex 1 =

+ Sprod

¢5q) -

=

(4.150)

6q - W

--F

(4.151)

b

where W and q denote work and heat transferred between the system and its surroundings, respectively, Tb is the temperature on the system boundary, and Sprod shows the entropy production by internal irreversibilities. For deriving the exergy balance for heat and work streams, first we multiply the entropy balance by the temperature To and then subtract it from the energy balance, and we obtain

gx2

_ gx1

.__ ~2

1 - -~b 6q - [ W - Po (V2 - V1)] - ToSprod

(4.152)

The above equation is analogous with the entropy balance of the second law. The first term in this expression shows the exergy transfer accompanying heat when the temperature at the heat transfer medium is not constant.

EXin

t

w°k:i Mass

Heat

-'-

Heat "-I ---~

Mass Work

Exou t

Figure 4.16. Mechanism of exergy transfer for a general system.

187

4.4 Exergyanalysis

For a steady-state flow, the energy balance is

Z

(hH+Jt+~)-

(nH+4+Ws)= 0

2

out of system

(4.153)

into system

The exergy balance for a steady-state system shows the exergy loss

~_~

hEx + gt 1- T °

into system k

~s

-+"~4/~s - out ofZ system

[

hEx+gt 1 - - ~

+I/Vs =L'Xloss

(4.154)

where Ws is the shaft work. The rate of loss exergy Exloss represents the overall thermodynamic imperfections, and is directly proportional to the rate of entropy production due to irreversibilities in a process. As the exergy loss increases, the net heat duty has to increase for the process to occur. Consequently, smaller exergy loss means less waste heat or thermodynamic imperfections. At absolute temperature T, the exergy transfer accompanying heat transfer becomes

(l )q

15')

The exergy transfer accompanies work EXwork = W -- W~ur~(for boundary work)

(4.156)

EXwork = W (for other forms of work)

(4.157)

In accordance with the second law, the exergy loss is positive in an irreversible process and vanishes in a reversible process. The change in exergy of a system can be positive, negative, or zero. When the temperature of a process where heat transfer occurs is less than the temperature of the environment, the transfer of heat and exergy flows in opposite directions. Work and the accompanying exergy transfer can be in the same or opposite directions. For an isolated system, there is no transfer of exergy between the system and its surroundings, and hence the change of exergy is equal to exergy loss (4.158)

EXl°ss = T°Spr°d

This equation shows the decrease of exergy principle, which states that the exergy of an isolated system always decreases for irreversible processes, and remains constant for a reversible process. This is similar to the increase of entropy principle, and is a statement of the second law. Exergy balance can also be expressed in exergy rate form

dEx dt

To

-

/

dV

i ! ¸ ,--;-- o;

-

-

loss

(4.159)

If we consider the exergy of a change from a given reference state (where exergy is zero), the work attainable in a real process would be

W = E x - T0AStota1

(4.160)

If the total entropy change vanishes, as in a reversible process, exergy defines an upper limit to the work that is extractable from any process. If heat is transferred between two reservoirs with temperatures T and :To,the exergy becomes

-T°ASt°t~' = To T o -

= q 1-

(4.161)

The above equation is a generalization of the Carnot relation. The ratio between the exergy and the heat Ex/q is called the exergyfactor. When T < To, there is a lack of energy in the system; the value of Ex/q greatly increases for

188

4.

Usingthe second law: Thermodynamic analysis

low temperatures. When T approaches absolute zero, 0 K, then Ex/q approaches infinity; at higher temperatures, Ex/q moves closer to unity. Therefore, exergy reflects the quality of energy; heat or cold is more expensive and valuable when it is needed the most. Since the exergy depends on the state of the environment, waste heat carries a higher exergy in winter than in summer. The exergy of sensible heat with temperature T is expressed as (4.162) If a power generation unit discharges 2000 MW of waste heat at T = 310 K from cooling water at a local ambient temperature To = 300 K, the above equation shows that the waste exergy discharged with the slightly warmed water is ----33MW. This waste energy causes a temperature increase in the local environment, which could gradually change the local ecology. The exergy of light relates to the exergy power per unit area of black body radiation 6x

[ 1()4

~x-O 1+~

--~

(4.163)

where e=o-T 4 is the energy power emission per unit area and tr ~ 5.67 × 108 W/(K 4 m 2) the Stefan-Boltzmann constant. Since the Earth receives sunlight with/'Sun = 6000 K, for the environmental temperature /'Earth= 300 K the exergy factor becomes

[l+ll:oool4300] 093336000

(4.164)

The exergy of material substances can be calculated if the pressure P and the temperature T are constant and equal to ambient conditions P0 and To, and the exergy is Ex = ~ ni(ix i - tXio )

(4.165)

i

where/J-,i and ].1,io are the chemical potentials of substance i in its present state and in its environmental state, respectively, and ni the number of moles. The chemical potential/zi is defined in terms of concentration c [Zi -- ],1,0 nt- R T In c i

(4.166)

where/x ° is the standard state chemical potential. Substitution of Eq. (4.166) into Eq. (4.165) yields EX "-" E t'li (l'LOi -- ]'zOo) + RTO E gli In c--!-/ i i Cio

(4.167)

where/z ° and Cio are the chemical potential and concentrations of component i, respectively, at environmental conditions (dead state). For a single-component system, Eq. (4.167) becomes

Combustion reactions often cause extensive exergy loss. Exergy calculations show that the entropy production can cause the loss of considerable potential work due to a reaction. An electrochemical membrane reactor or a fuel cell could reduce exergy loss considerably. For pure components, the chemical exergy consists of the exergy that can be obtained by diffusing the components to their reference concentration ci0 with a partial pressure of Pio. For an ideal gas, we obtain Ex = n R T o In p.P~

(4.169)

4.4

189

Exergyanalysis

where Pi and Pi0 refer to the partial pressure of the gas in the emission and in the environment, respectively. Equation (4.169) shows that exergy use may have ecological and environmental effects.

4.4.4 Flow Exergy As flow processes are common in industry, exergy of the mass flow crossing the system boundary is important. The main components of exergy are kinetic exergy, potential exergy, physical exergy, and chemical exergy. We define the kinetic and potential exergies by the kinetic and potential energies calculated in relation to the environment. Physical exergy results from the deviation of temperature and pressure from the environmental values. Chemical exergy results from the deviation of the composition from the composition of the environment. In an open system,.flow exergy is exergy transfer due to mass flow and flow work, and the specific flow exergy exf is

exf -- rh

(h-ho)-To(S-So)+---~+gz

(4.170)

where h and s represent the specific enthalpy and entropy, respectively, at the inlet or outlet; h0 and So represent the respective values at the dead state. The flow work rate is rh(Pv), where 1/7 is the mass flow rate, P the presure, and v the specific volume at the inlet or exit. The exergy rate balance for a control volume is

dt

-

.

- ~

q/ -

- P°

dt

/

+ Zi lhexfi - Ze lheXfe - EXl°ss

(4.171)

where the first three terms on the right represent the rate of exergy transfer, and the last term is the rate of exergy loss. The term qj shows the heat transfer rate through the boundary where the instantaneous temperature is ~., while the term Wcv shows the energy transfer rate by work other than flow work. The terms mexfi and mexfe denote the exergy transfer rates accompanying mass flow and flow work at the inlet i and exit e, respectively. For a control volume at steady state, exergy rate balance becomes

0=

1--~/ •

q/

(4.172)

.

i

e

This equation shows that the rate of exergy transferred into the control volume must exceed the rate of exergy transferred out, and the difference is the exergy destroyed due to irreversibilities. Exergy concepts for some steady-state processes are: • Energy remains the same in the throttling valve, while exergy is destroyed because of the expansion of the fluid. • Exergy is destroyed by irreversibilities associated with pressure drops due to fluid friction and stream-to-stream heat transfer due to temperature differences. • In a steam power plant, exergy transfers are due to work, heat, and exergy loss within the control volume. • In a waste heat recovery system, we might reduce the heat transfer irreversibility by designing a heat recovery steam generator with a smaller stream-to-stream temperature difference, and/or reduce friction by designing a turbine with a higher efficiency. • A cost-effective design may result from a consideration of the trade-offs between possible reduction of exergy loss and potential increase in operating cost.

4.4.5 Exergetic(Second Law) Efficiency Energy supplied by the heat transfer, qin, is either utilized, %, or lost to the surroundings, q., and the thermal efficiency, r/, is qu

r/-

. qin

(4.173)

190 The exergetic efficiency,

4.

T]th,

Usingthe second law: Thermodynamic analysis

iS (exergy recovered)/(exergy supplied)

'Oth = 1"]

1-

To/ru )

1-

To~Tin

(4.174)

Generally, the value of exergetic efficiency is less than unity even when r/= 1. Exergy use would increase as the temperature of utilization of energy approaches the temperature of inlet energy. The rate of exergy loss accompanying the heat loss 01 is (1- To/T1)gll , and depends on the operating temperature. Exergetic efficiency expressions can take different forms, including the following for engineering various steady-state processes: • A turbine with adiabatic operation

Wt

T/t =

(4.175)

/h(exf, in -- eXf, out)

where Wt shows the work produced by the turbine. • A compressor or pump with work input 1//and adiabatic operating conditions

th(eXf, out -- exf, in) n~

-- --

(4.176)

(--rP)

• A heat exchanger not mixing at adiabatic conditions with both streams at temperatures above To rhc (eXf, o u t - exf, in ) c

(4.177)

"0th = th h (exf, in _ eXf, out )h

where rhc and rhh represent the mass flow rates of cold and hot streams, respectively. • An adiabatic mixer with streams 1 and 2 entering and stream 3 leaving the system rhh =

th2 (exf,3 - e x f , 2 )

(4.178)

rhl (exf, 1 - e x f , 3 )

• For an adiabatic chemical reaction at constant pressure, the enthalpy remains constant. The loss in exergy is given by the exergy of reactants exl and the exergy of the reaction products ex2 W 1 = e x 1 - e x 2 = T O (s1 - s 2 )

(4.179)

and "Oth - -

W, r0 (sl 1- ' = 1ex 1

$2 )

(4.180)

ex 1

For a combustion reaction taking place in a well-insulated chamber with no work produced, exergetic efficiency becomes Tith

=

1 Exl°ss Ex F

(4.181)

where E x F is the rate of exergy entering with the fuel and Exloss the exergy loss. The primary exergy load Exp,i is a fraction of the total primary exergy. The transformed exergy load Ext,i is the ratio of the transformed exergy to the total primary exergy. The relationship between the individual efficiencies '0i and the overall efficiency r/is r / = E [Exp,i'r/i - Ext,i ( 1 - 'qi)] i

(4.182)

191

4.4 Exergyanalysis The primary exergy loads have the constraint

Z Exp, i =

(4.183)

1

i

Equation (4.182) shows that by increasing a local efficiency r/i or decreasing a transformed exergy load Ext, i, the overall efficiency rt of a process may increase as long as this does not cause any opposite and larger effect through a change in other parameters. An effective way of improving the overall exergy is to increase the primary load of the units with the largest efficiencies at the expense of those with the lowest efficiencies. The exergy efficiency by the second law is Exout Tit h - -

(4.184)

Exin

Intrinsic efficiency T/in takes into account the transiting exergy Extr EXou t --

Extr _ Exp

rim . . . . . . . . . . . . . .

Exin - Extr

Exc

(4.185)

The transiting exergy is the part of the exergy entering a unit operation; it traverses without undergoing any transformation and exergy loss. The terms Exc and Exp are the exergies actually consumed and produced, respectively. In Eq. (4.182), intrinsic efficiency is used.

Example 4.10 Thermodynamic efficiency in a power plant A steam power plant produces 65 MW electricity with an efficiency of 70%. It uses steam (stream 1) at 8200 kPa and 550°C. The discharged stream (stream 2) is at 75 kPa. If the expansion in the turbine is adiabatic, and the surroundings are at 298.15 K, determine: (a) The maximum work output; (b) The thermodynamic efficiency. Solution: Assume that kinetic and potential energies are negligible, and the system is at steady state. (a) Basis" 1 kg/s steam with the properties from the steam tables (Appendix D) H 1 = 3517.8kJ/kg, S 1 - 6. 8648 kJ/(kg K) at ~ = 550°C, P1 = 8200kPa 82, V =

7.3554 kJ/(kg K), $2, L = 1.213 lkJ/(kgK) at P2 = 75 kPa(saturated steam)

He, v = 2663.0 kJ/kg, H2, c = 384.45 kJ/(kg K) at P2 = 75 kPa (saturated steam') T~t=- 0.7, To = 298.15 K, R = 8.314 J/(mol K) If the turbine operates at isentropic conditions, then we have S 2 = S 1 < 7.3554 kJ/(kg K) Therefore, the discharged steam is wet steam, and the vapor fraction Xs (the quality at isentropic operation) is 6.8646-1.2131 x~= 7.4570-1.2131

= 0.905

The discharged steam enthalpy at isentropic conditions Hzs is H2s = H2, L (1 - Xs) + H2, v x s = 384.45(1 - 0.905) + 2663(0.905) = 2446.8 kJ/kg The enthalpy change at isentropic conditions is ~ / s = H2s - H1 = - 1070.98 kJ/kg

192

4.

Usingthe second law: Thermodynamic analysis

The actual enthalpy change is AH = "qtAHs = - 7 4 9 . 6 7 The actual enthalpy of the discharged steam is H2 = H1 + AH = 2768.11 kJ/kg > H2,v Therefore, the final state is superheated steam at 75 kPa and between 125 and 15°C. Using the linear interpolation of the values from the steam table, we have T2 -- 144.88°C, S2 = 7.7254 kJ/(kg K) The steam rate is -65,000 rh = ~ = 86.70 kg/s H2- H1 The maximum work with AS = $2 - $1 is Wmax = t h ( ~ -

ToAS ) = - 8 7 , 2 5 3 . 3 2

kW

(b) The thermodynamic efficiency is

W Tit - - - Wmax

4.4.6

-65,000.0

= 0.744

-87,253.32

Exergy Analysis Procedure

The thermodynamic analysis of an existing operation consists of three parts. The first part mainly assesses the thermodynamic performance of the current operation. If found to be necessary, in the second part, targets for modifications are identified to reduce the cost of operation. The third part involves the assessment of the thermodynamics and economic effectiveness of the modifications. Exergy analysis can help in all three parts above. The main steps of exergy analysis are: • • • • • • • • • •

4.5

Define the system boundary of processes to be analyzed. Define all the assumptions and the reference conditions of temperature and pressure. Choose the thermodynamic methods for property and phase equilibrium estimations. Consider possible heat recovery and heat integration strategies for all the processes analyzed. Obtain a converged solution using a simulator for the mass and energy balances. Estimate the rate of exergy flows for material and heat streams crossing the system boundary. Determine the total exergy losses. Determine the thermodynamic efficiency. Use exergy loss profiles to identify the regions performing poorly. Identify improvements and modifications to reduce the cost of energy and operation.

APPLICATIONS OF EXERGY ANALYSIS

Example 4.11 Energy dissipation in countercurrent and cocurrent heat exchangers The two most commonly used heat exchangers are countercurrent and cocurrent at steady-state flow conditions as shown in Figure 4.17. Estimate the energy dissipated from these heat exchangers if the surroundings are at 290 K. Consider the data below: Cocurrent Countercurrent:

Thl = 365 K, Th2 = 315 K, Tcl= 280 K, To2 = 305 K Thl = 365K, Th2 -" 315K, Tcm = 280K, Tc2 = 355K

4.5 Applications of exergy analysis

193

Thl

Thl Th2

Tc2

Tc2

Y

Tcl J

mh2

Tcl

J

q

q

Cocurrent

Countercurrent

Figure 4.17. Temperature profiles for heat exchangers operated in cocurrent and countercurrent modes.

The temperature difference between the hot and cold streams is 10 K at the end of the heat exchange. Assuming that kinetic and potential energies are negligible, we apply the following general energy balance by disregarding work interactions"

fhhCph(Th2 -- Thl ) -Jr-thcCpc(%2

(4.186)

-%1 ) - 0

where rhh and rhc are the flow rates of hot and cold streams, which may be related to the temperature changes of hot and cold streams when Cph-- (/pc: ivhc - Thl -- Th2 /h h

(4.187)

Tc2 - Tel

The total rate of entropy change for the hot and cold streams is (4.188)

A S - ;hhASh Jr-/~/cASc

If the pressure changes of the streams are negligible, and the heat capacities of both streams are constant and equal to Cp, then we have total entropy change A , ~ -- th h

Cp

/

In Th2 -~ t~/c In rhl

th h

c2/

(4.189)

Tcl

Applying Eqs. (4.187) and (4.189) for cocurrent and countercurrent operations, we find: Cocurrent heat exchanger I: th c _ fhl -- Th2 _ 365 - 315

rhh

A~o I = fh h

Cp

Tce -Tcl

(

305-280

=2.0

(3,5

305

In Th2]/.,hl-4-"thh *hc ln.Tc I - rhhCp In 365 + 2.0 I n 280

/~Xloss,i --

~()AS

1 -

- 0.0237rhhC p

2 7 3 . 1 5 K ( A S t ) - 6.473rhhC p

Countercurrent heat exchanger I|" rhc _ Th~ - The

3 6 5 - 315

Jr//h

355-

To2 - Tcl

= 0.666

280

ASu=rhhCl,(lnTh2+rh--!ClnfC2j-rhhCp(ln315+O.6661n 355 Thl

mh

Tel

365

280

~+Xloss,iI - ToASI1 - 273.15 K ( A S . ) - 2.932rhhC p

= 0.0107thhC p

4.

194

Usingthe second law: Thermodynamic analysis

The ratio of exergy losses yields EXloss,iI ~ =

/~Xloss,I

2.932 rhhCp

=0.453

6.473rhhC p

This ratio shows that the rate of energy dissipated in the cocurrent heat exchanger is almost twice the dissipation in the countercurrent heat exchanger. Although the heat exchanged between the hot and cold streams is the same, the countercurrent operation is thermodynamically more efficient.

4.5.1

Power Generation

Steam power plants produce electricity with rather low thermal efficiency. An increase in efficiency leads to savings in fuel costs and minimizes environmental effects. The two basic approaches in increasing the thermal efficiency of a cycle are: (i) design a process that transfers heat to the working fluid at high temperature in the boiler and (ii) design a process that transfers heat to the working fluid at low temperature in the condenser. These may decrease the temperature differences, and hence the level of irreversibility.

Example 4.12 Exergy analysis of a power plant A steam power plant operates on a simple ideal Rankine cycle (see Figure 4.18). The turbine receives steam at 698.15 K and 4200 kPa, while the discharged steam is at 40 kPa. The mass flow rate of steam is 3.0 kg/s. In the boiler, heat is transferred into the steam from a source at 1500 K. In the condenser, heat is discharged to the surroundings at 298 K. Determine the energy dissipated at each state. Solution: Assume that the surroundings are at 298 K and the kinetic and potential energy changes are negligible rhs = 3.0kg/s, V1 = 0.001022m3/kg P3 = 4100kPa, P4 = 40kPa H 3 = 3272.3 kJ/kg, S3 = 6.845 kJ/(kg K) H4, v = 2636.9 kJ/kg, H 1 = H4, L = 317.65 kJ/kg Sa,v = 7.6709 kJ/(kg K), S1 =

S4,L,

84, L --

1.0261 kJ/(kg K)

T0 = 2 9 8 K , TH = 1 5 0 0 K , Tc = 2 9 8 K

~mm,=

Figure 4.18. The effects of superheating the steam to higher temperatures and reducing the condenser pressure on the ideal Rankine cycle.

4.5

195

Applications of exergy analysis

Wp,in = Vl(P2, p1)= 0.001022(4100,50) ( llkPam3 kJ)

=4.14kJ

H 2 = H 1 + mp,in = 321.79 kJ/kg

(4.190)

(4.191)

Because this is an isentropic process $3 = $4 and S1 = $2. We estimate the quality of the discharged wet steam (S3,v < S4,v) after passing through the turbine: 6.845-1.0262 X4s =

= 0.875

(4.192)

7.6709-1.0261

H 4 = 317.65(1-0.875)+ 2636.9× 0.875-2356.6kJ/kg

S4 = S 3 =1.0261 × ( 1 - 0 . 8 7 5 ) + 7.6709 × 0.875 = 6.8402 kJ/(kg s) qin = H3 - H2 = 2 9 5 0 . 5 1 kJ/kg qout = H4 - H1 - 2 0 3 8 . 9 5 kJ/kg

Wnet=qin -- qout = 911.56 kJ/kg

rl = 1- qout

=

0.309

(4.193)

(4.194)

qin

The exergy balance for 3.0 kg/s working fluid yields EXloss,12 = 0 (isentropic process)

EXloss,23=thToIS 3 - S 2 + (~3q4i4n3) .)5 8=k W T u

EXloss,3 4 = 0 (isentropic process)

•

/

qout) =914.76kW

EXloss,41=rhTo S 1 - S 4 nt-~C

The exergy losses or the work losses are 79% and 21% in the boiler and condenser, respectively. In a Rankine cycle, exergy losses are due to irreversibilities occurring during heat transfer with finite temperature differences in the boiler and condenser. In order to decrease exergy losses, the temperature differences should be made smaller. Regeneration may help to decrease the temperature differences.

4.5.2

Improving the Efficiency of Power Plants

Some modifications to improve the efficiency of steam power production are

1. Modification of operating conditions of the condenser and boiler: Lowering the operating pressure of the condenser reduces the temperature of the saturated steam within the condenser. Therefore, the heat flow from the condenser to the environment will be at a lower temperature. The lower limit of the condenser pressure is the saturation pressure at the temperature of the cooling medium. For example, if the cooling water is at 20°C, and the temperature difference for the heat transfer is 10°C, the steam temperature must be above 30°C, and the condenser pressure must be above 4.24 kPa, which is the saturation pressure of water at 30°C. At the same time, we have to control the quality of the discharged steam from the turbine: steam with a high level of liquid water and of low quality lowers the

196

4.

Usingthe second law: Thermodynamic analysis

efficiency of the turbine and may corrode the turbine blades. The level of superheating in the boiler controls the quality of the discharged steam. Superheating the steam to high temperature increases the temperature at which the heat flows into the boiler (decreasing the temperature difference between the heat source and boiler), and increases the turbine efficiency and quality of the discharged steam. Figure 4.18 shows the effects of superheating the steam to higher temperatures and reducing the condenser pressure of the ideal Rankine cycle on a T-S diagram. The gray area underneath 3-3' is the increase in the work due to superheating the steam to a higher temperature, while the area underneath 1-4 is the increase in the work due to reducing the condenser operating pressure. However, the area underneath 2-2' shows the heat input increase, which is considerably smaller. Increasing the operating pressure of the boiler increases the boiling point and the average temperature for heat flow into the boiler. The operating pressures may be as high as 30 MPa (4500psia) in many power plants. The temperature of a furnace (heat source) is --~1350°C, which makes the energy in the furnace very valuable. This energy is used to produce steam at --~200-250°C. This process is irreversible, and hence causes a large amount of work loss. 2. Reheating the steam" Reheating enables the expansion of the steam in various stages instead of a single expansion process. Mainly, reheating increases the steam quality to protect the material. In an ideal reheat Rankine cycle with two-stage expansion, for example, the steam is expanded to an intermediate pressure isentropically in the high-pressure turbine section, and sent to the boiler to be reheated. In the low-pressure turbine section, the reheated steam is expanded to the condenser operating pressure. The reheating increases the average temperature at which the steam is heated, and hence a single reheating may increase the cycle efficiency by --~4%. A double reheating is common. Increasing the number of reheatings to more than 2 may not be feasible, as the additional increase in efficiency may be marginal. The optimum intermediate pressure in the reheating is about one-fourth of the maximum cycle pressure, while the reheating temperature is close or equal to the turbine inlet temperature.

Example 4.13 Simple reheat Rankine cycle in a steam power plant A simple ideal reheat Rankine cycle is used in a steam power plant (see Figure 4.19). Steam enters the turbine at 9000 kPa and 823.15 K and leaves at 4350 kPa. The steam is reheated at constant pressure to 823.15 K. The discharged steam from the low-pressure turbine is at 10 kPa. The net power output of the turbine is 65 MW. Determine the thermal efficiency and the work loss at each unit. Solution: Assume that the surroundings are at 298 K, the kinetic and potential energy changes are negligible, and this is a steady process. V1= 0.00101 m3/kg

T

3

5

t t

/

s Figure 4.19. Simple reheat Rankine cycle.

4.5

197

Applications of exergy analysis

P3 =9000 kPa, H 3 =3509.8kJ/kg, S 3 =6.8143 kJ/(kg K) P6 = 10 kPa, H6, v = 2584.8 kJ/kg, H 1 = H6, L = 191.81 kJ/kg $6,v = 8.1511 kJ/(kg K), $6,L = 0.6493 kJ/(kg K) T4=698.15K , P4=4350kPa, H 4 =3268.5kJ/kg, S 3 = S 4 = 6 . 8 1 4 3 k J / ( k g K ) Ts=823.15K, Ps=4350kPa, H s - 3 5 5 5 . 2 k J / k g ,

Ss=S6=7.1915kJ/(kgK)

T 0 - 2 9 8 K , TH=1600K, T c = 2 9 8 K 1 kJ ) Wp,in - V1(Pz - P~) = 0.00101(9000 - 10) 1kPa m 3 = 9.08 kJ/kg

H 2 = H 1 + Wp,in = 2 0 0 . 8 9

kJ/kg

Because this is an isentropic process $3 = $4 and $1 = $2. We estimate the quality of the discharged wet steam (S3,v < S4,v) after passing through the turbine: 7.1915-0.6493 X6s =

8.1511-0.6493

=0.872

H6s =191.81(1-0.872)+ 2584.8×0.872 = 2278.69 kJ/kg The turbine work output is: Wout =

H6) = 1517.80 kJ/kg

(H 3 - H4) + (H 5 -

kJ/kg

q23,in = H 3 - H 2 = 3 3 0 8 . 9 1

- H4 = 286.70

kJ/kg

qout = H 6 - H1 = 2 0 8 6 . 8 9

kJ/kg

q45,in --

qin = ( H 3

H5

- H2 ) +

(Hs

- H4 )= 3595.61

Wnet = Wou t - Wp,in -

_Wnet

T~t m w qin

kJ/kg

1508.72 kJ/kg

-n 0 . 4 2

The thermodynamic analysis is: S1 = S 2 =0.6493 kJ/(kg K), S3 = S 4 =6.8143, S 5 = S 6 -7.1915 kJ/(kg K) The exergy balance with a basis of rh = 1kg/s of working fluid yields: EXlo~s,12 = 0 (isentropic process)

(-- q23,in) ] = 1 2 2 0 . 8 9 k W

r.

)

198

4.

Usingthe second law: Thermodynamicanalysis

Table 4.1 Distribution of exergy losses in Example 4.13

Process

&loss (kW)

Percent

1-2 2-3 3-4 4-5 5-6 6-1 Cycle

0 (isentropic) 1220.89 0 (isentropic) 59.00 0 (isentropic) 137.31 1417.20

0 86.1 0 4.2 0 9.7 100

°

EXloss,34 -- 0 (isentropic process)

JEXloss,45--l'hTo(S5-$4-+.

q45'in-------~)) (-- W T -- H59.00 k

EXloss,56 = 0 (isentropic process)

%ut)

/~Xl°ss'61--/'hT° S 1 - $ 6 +-~--c --137.31kW

Table 4.1 shows the distribution of exergy losses at each process. In this ideal reheat Rankine cycle, the steam from the first part of the high-pressure section is expanded and reheated in the boiler until it reaches the boiler exit temperature. The reheated steam is expanded through the turbine to the condenser conditions. The reheating decreases the moisture within the discharged steam. The exergy losses or work losses are 86.1% and 9.7% in the boiler and condenser, respectively. The exergy loss in the reheating step is low (4.2%).

Example 4.14 Actual reheat Rankine cycle in steam power generation A reheat Rankine cycle is used in a steam power plant (see Figure 4.20). Steam enters the high-pressure turbine at 9000 kPa and 823.15 K and leaves at 4350 kPa. The steam is reheated at constant pressure to 823.15 K. The steam enters the low-pressure turbine at 4350 kPa and 823.15 K. The discharged steam from the low-pressure turbine is at 10 kPa. The net power output of the turbine is 65 MW. The isentropic turbine efficiency is 80%. The pump efficiency is 95%. Determine: (a) The thermal efficiency; (b) The work loss at each unit. Solution: Assume that the surroundings are at 298 K, the kinetic and potential energy changes are negligible, and this is a steady process. The steam data are: V1 = 0.00101 m3/kg T3=823-15K, P3 =9000kPa, H3=3509.8kJ/kg, S3=6.8143kJ/(kgK) P6=10kPa, H6,v=2584.8kJ/kg,

Hl=H6,L=191.81kJ/kg

$6,v = 8.1511 kJ/(kg K), $6,L = 0.6493 kJ/(kg K) T4 = 698.15 K, P4=4350kPa, H4s=3268.5kJ/kg,

S3=S4s=6.8143kJ/(kgK)

4.5

Applications of exergy analysis

~/

3

199

5

2

J

Figure 4.20. Actual reheat Rankine cycle.

T s = 823.15K, P5 = 4350kPa, H s = 3555.2kJ/kg, $5= S6s= 7.1915 kJ/(kg K) To = 298K, TH = 1600K, Tc = 2 9 8 K (a) In this actual reheat Rankine cycle, the steam expands in two stages. In the first stage, the steam expands to 4350 kPa and is sent to the reboiler where it is reheated at constant pressure to 823.15 K. In the second stage, the steam expands and the discharged steam at 10 kPa is sent to the condenser. Therefore, the total heat input and total turbine work output become qin = q23 + q45 = (H3 - H2) + (H5 - H4) mou t = W34 + W56 =

(H 3 - H4) + (H 5 - H6)

lkJ Wp,in -

VI(P2 - P 1 ) = 0 . 0 0 1 0 1 ( 9 0 0 0 - 1 0 )

~

lkPam3

=9.55kJ/kg

H 2 - H 1 + Wp,in = 200.88 kJ/kg Because this is an isentropic process $3 = S4s and S~ = S2s, Ss,v < S6,v we estimate the quality of the discharged wet steam after passing through the turbine"

~t =

H3- H4

-+ H 4 - H 3 - Tit (H 3 - H4s ) = 3316.76kJ/kg

H 3 -H4s 7.1915-0.6493 X6s =

=0.872

8.1511 - 0.6493

H6s = 191.81(1- 0.872) + 2584.8 × 0 . 8 7 2 - 2278.7 kJ/kg H5- H6 37 t z

H 5 - H6s

---, H 6 - H 5 - l~t ( H 5 - H6s ) - 2533.99 kJ/kg < H6, v (= 2584.8 kJ/kg)

(4.195)

4.

200

Usingthe second law: Thermodynamic analysis

2354.0-191.81 =0.978 x6 = 2 5 8 4 . 8 - 1 9 1 . 8 1 S 6 = $6, L (1 -

x 6 ) -k- 86, V x 6 =

7.9918 kJ/(kg K)

(4.196)

The turbine work output is: Wout = ( H 3 - H 4 ) + ( H 5 - H 6 ) = 1214.24 kJ/kg q23,in = ( H 3 - H 2 ) = 3308.43 kJ/kg q45,in = ( H 5 - H 4 ) = 238.44 kJ/kg qin = (H3 - H 2 ) + ( H 5 - H 4 ) = 3546.87 k J / k g Wnet = Wout - Wp,in = 1 2 0 4 . 6 8 kJ/kg

_

Tit

_

Wnet

--0.339

qin

The rate of steam is rhs -

Ws Wnet

65,000

- ~

= 53.95 kg/s

1204.68

Table 4.2a shows the state properties of the actual reheat Rankine cycle.

Table 4.2a State properties of the actual reheat Rankine cycle in Example 4.14 State

T (K)

P (kPa)

H (kJ/kg)

S (kJ/(kg K))

1 2 2s 3 4 4s 5 6(x 6 = 0.978) 6s(X6s = 0.872)

319 321.27 321.15 823.15 718.94 698.15 823.15 319 319

10 9000 9000 9000 4350 4350 4350 10 10

191.81 201.36 200.98 3509.8 3316.76 3268.5 3555.2 2534.0 2278.7

0.6492 0.6776 0.6492 6.8143 6.8800 6.8143 7.1915 7.9918 7.1915

Table 4.2b Distribution of exergy losses in actual and ideal reheat Rankine cycles Process

Example 4.14 (actual reheat Rankine) Exloss (kJ/kg)

1-2 2-3

0.477 1212.54

Example 4.13 (ideal reheat Rankine)

Percent 0.03 72.4

Exloss (kJ/kg)

0

0

1220.89

86.1

3-4

19.57

1.2

4-5

48.41

2.9

5-6

238.50

14.3

0

6-1

154.10

9.2

137.31

Cycle

1673.55

100

Percent

0 59.00

1417.20

0 4.2 0 9.7 100

4.5

201

Applications of exergy analysis

(b) Exergy balance with a basis of 1 kg/s working fluid yields Exloss,12 = rh(Wp,a - Wp,s) - 9.5577 - 9.0799 = 0.477 kW

EXloss,23 = t//TO S 3 -- S 2 q-

( - qin,23) ] _ 1 2 1 2 . 5 4 k W

r.

)

Exloss,34 - & T 0 ( & -- ,5'3) -- 19.57 k W

•

{

EXloss,45 = th TO 5'5 -- S 4 Jr- ( - qin,45) ] _ 4 8 . 4 1 k W

rH

)

EXloss,56 - thT0 (S 6 - $5) - 2 3 8 . 5 0 k W

•

(

qout] -

EXloss,61 = th TO S 1 - S 6 + U

154.10 kW

Table 4.2a shows the state properties in actual reheat Rankine operations, while Table 4.2b compares the exergy losses of ideal reheat and actual reheat Rankine cycle operations• The total exergy loss increases from 1417.2 to 1673.55 kJ/kg in the actual operation• This shows an increase of 18.0% in the actual operation. In this actual reheat Rankine cycle, the steam expands in two stages. In the first stage, the steam expands to 4350 kPa and is sent to the reboiler where it is reheated at constant pressure to 823.15 K. In the second stage, the steam expands and the discharged steam at 10 kPa is sent to the condenser. The exergy losses or the work losses are 72.8% and 9.2% in the boiler and condenser, respectively. The exergy loss in the reheating step is low (2.9%). Reheating reduces the moisture in the turbine.

3. Regeneration: Increasing the boiler feed temperature by using expanding steam is possible in a regenerative cycle• Steam extracted at intermediate pressures from various parts of the turbine is used in countercurrent heat exchangers (closed heaters) to heat the feed. The steam leaves the condenser as a saturated liquid at the condenser operating pressure. The condensed water pressure is increased to the feed heater pressure through pumping. By adjusting the fraction of steam extracted from the turbine, one can produce a saturated liquid output at the heater operating pressure. After passing through the heater a pump increases the pressure of the water to the boiler pressure. In open feed heaters, the expanded steam from the turbine is mixed with water. Regeneration increases the efficiency and helps to deaerate the water and control the discharged steam flow rate. The efficiency will increase further if more heaters are used. As many as eight heaters may be used. However, the savings in boiler fuel costs corresponding to the increase in the efficiency of the cycle should exceed the cost of the heaters.

Example 4.15 Ideal regenerative Rankine cycle A steam power plant is using an ideal regenerative Rankine cycle (see Figure 4.21). Steam enters the high-pressure turbine at 8200 kPa and 773.15 K, and the condenser operates at 20 kPa. The steam is extracted from the turbine at 350 kPa to heat the feed water in an open heater. The water is a saturated liquid after passing through the feed water heater. The work output of the turbine is 70 MW. Determine the thermal efficiency and the work loss at each unit. Solution: Assume that the surroundings are at 285 K, the kinetic and potential energy changes are negligible, and this is a steady-state process. The steam data are: l~ - 0.001017 m3/kg =773.15K,

P~-8200kPa,

H 5-3396.4kJ/kg,

Ss-6.7124kJ/(kg K)

202

4.

Using the second law: Thermodynamic analysis

& T

5

2

qout S

Figure 4.21. Ideal regenerative Rankine cycle.

P~ = P7 = 20 kPa, H7, v = 2609.9 kJ/kg, /-/1 = H6, L = 252.45 kJ/kg $7,v = 7.9094 kJ/(kg K), $7,L = 0.8321 kJ/(kg K) P3=350kPa, H3,L=584.27kJ/kg, H3,v=2731.50kJ/kg V~=0.001079m3/kg,

S3,L=l.7273kJ/(kgK),

S3,v=6.9392kJ/(kgK )

To=285K, TH=1600K , Tc=285K In this ideal regenerative Rankine cycle, the steam extracted from the turbine heats the water from the condenser, and the water is pumped to the boiler. Sometimes, this occurs in several stages. The condensate from the feed water heaters is throttled to the next heater at lower pressure. The condensate of the final heater is flashed into the condenser

Wpl = V~( p 3 - p~)= 0.001017(350- 20) (

lkJ ) 1kPa m 3 = 0.335 kJ/kg

H 2 = H 1 + Wpl = 252.78 kJ/kg

Wp2 = V1(P3- P~)= 0.001079(8200- 350)( 1kPa lkJm 3 / = 8.47 kJ/kg H 4 = H 3 + mp2 = 592.74 kJ/kg Because this is an isentropic process, 5'5 = $6 = $7. We estimate the quality of the discharged wet steam at states 6 and 7: x6=

6.7124-1.7273 =0.956 6.9392-1.7273

4.5

Applications of exergy analysis

203

H 6 = 584.27(1- 0.956) + 2731.50 × 0.956 = 2638.06 kJ/kg 6.7124-0.8321 X7 =

=0.830

7.9094-0.8321

H 7 = 252.45(1 - 0.83) + 2609.9(0.83) = 2211.18 kJ/kg The fraction of steam extracted is estimated from the energy balance: t h 6 H 6 + th 2 H 2 = th 3 H 3

In terms of the mass fraction z = th6/th3, the energy balance becomes

zH 6 + (1 - z)H 2 = H 3 The mass fraction is z = H3 - H2 = 0.139 H 6- H 2

The turbine work output is: qin = H5 - H 4 - 2 8 0 3 . 6 6 kJ/kg

qout =(1 -- z)(H 7 - H~) = 1686.52 kJ/kg W n e t = qin -- qout = 1117.13 kJ/kg

Tit

-

-

1 - qout = 0.398

(4.197)

qin

The rate of steam is"

~hs m

Ws _ 70,000 Wnet

= 62.66 kg/s

1117.13

The exergy balance with a base of 1 kg/s yields EXloss,12 = 0

L'Xloss,23 = # / T O(S 3 - 82) =- 255.13 k W

EXloss,3 4 = 0

F__,Xloss,45-lhTo{S5-$4-+-(- k q i nW) )

- T9 2 1n. 3 5

gXloss,56 -- 0

•

EXloss,71 -/~/T0

i Sl -

87 nt- -q°ut/ ~ 0 = 10.63 k W

4.

204

Usingthe second law: Thermodynamic analysis

Table 4.3a State properties of the ideal regenerative Rankine cycle in Example 4.15

State

T (K)

1

319

2 2s 3 4 4s 5 6(x 6 = 0.978) 6s(X6s = 0.872)

321.27 321.15 823.15 718.94 698.15 823.15 319 319

P (kPa)

10 9000 9000 9000 4350 4350 4350 10 10

H (kJ/kg)

S (kJ/(kg K))

191.81 201.36 200.98 3509.8 3316.76 3268.5 3555.2 2534.0 2278.7

0.6492 0.6776 0.6492 6.8143 6.8800 6.8143 7.1915 7.9918 7.1915

Table 4.3b Distribution of exergy losses at each process in Example 4.15

Process

Ideal regenerative Rankine cycle ~bXloss(kW)

Percent

0 255.13 0 921.35 0 10.63 1187.12

0 21.5 0 77.6 0 0.9 100

1-2 2-3 3-4 4-5 5-6 7-1

Cycle

"x

{

/~Xloss,cycle = th TO / qout

rc

qin | = 1187.12 kW

rH)

Table 4.3a shows the state properties of the ideal regenerative Rankine cycle. Table 4.3b shows the distribution of exergy losses at each process. As seen from this table, the highest exergy loss occurs due to heat transfer in the boiler.

Example 4.16 Actual regenerative Rankine cycle A steam power plant is using an actual regenerative Rankine cycle (see Figure 4.22). Steam enters the high-pressure turbine at 11,000 kPa and 773.15 K, and the condenser operates at 10 kPa. The steam is extracted from the turbine at 475 kPa to heat the water in an open heater. The water is a saturated liquid after passing through the water heater. The work output of the turbine is 90 MW. The pump efficiency is 95% and the turbine efficiency is 75%. Determine the work loss at each unit. Solution: Assume that the surroundings are at 285 K, the kinetic and potential energy changes are negligible, and this is a steady-state process. The steam data are:

Ps=ll,000kPa, Hs=3362.2kJ/kg , S5=6.5432kJ/(kgK), Ts=773.15K Pl=P8=10kPa, H8,v=2584.8kJ/kg, Hs,L=191.83kJ/kg, Vl=0.00101m3/kg $7,v = 8.1511 kJ/(kg K), S7,t, = 0.6493 kJ/(kg K) P3=P6=475kPa, H3,L=631.29kJ/kg, H3,v=2745.30kJ/kg

4.5

205

Applications of exergy analysis

Boiler

Open Heater

@

3

I

I_.., 6 Condenser

2 P1

(a)

T

4

5

y

S

(b) Figure 4.22. (a) Schematic of actual regenerative Rankine cycle and (b) T-S diagram.

$3, L =1.8408kJ/(kgK),

$3,v =6.8365kJ/(kgK)

T/p--0.95, r/t=0.75 , To=285K , TIq=1700K , T c = 2 8 5 K In this actual regenerative Rankine cycle, the steam extracted from the turbine heats the water from the condenser, and the water is pumped to the boiler. Sometimes, this occurs in several stages. The condensate from the feed heaters is throttled to the next heater at lower pressure. The condensate of the final heater is flashed into the condenser

%1-~(P2- ~ ) -

0.00101(475-10) (lkJ j 0.95

1kPa m 3 = 0.494 kJ/kg

H 2 = H 1 +Wv~ =191.83+0.494=192.32kJ/kg

206

4.

Usingthe second law: Thermodynamic analysis

Table 4.4a State properties of the actual regenerative Rankine cycle in Example 4.16

State

T (K)

P (kPa)

H (kJ/kg)

1 2 2s 3 4 4s 5 6 6s (X6s = 0.941) 7 (x7 = 0.920) 7s (Xys - - 0.785)

319 319.5 319 423.07 428 428 773.15 457.83 423.07 319 319

10 475 475 475 11000 11000 11000 475 475 10 10

191.81 192.32 631.81 643.89 3362.20 2806.46 2621.21 2394.47 2071.90

S (kJ/(kg K)) 0.6493 0.6510 0.6493 1.8408 1.8451 1.8408 6.5432 7.0125 6.5432 7.5544 6.5432

0.001097(11,000-550)( lkJ / 0.95 1kPa m 3 = 12.08 kJ/kg

Wp2 = V1(P4 - P3) =

9 4 = 9 3 -Jr-mp2 = 6 3 1 . 2 9 +

12.08 = 643.37 kJ/kg

Because this is an isentropic process $5 = S6s = S7s, and we estimate the quality of the discharged wet steam at states 6s and 7s: X6s =

6.5432-1.8408 6.8365-1.8408

=0.941

H6s = 631.29(1- 0.941) + 2745.3(0.941) = 2621.18 kJ/kg From the turbine efficiency, we estimate the enthalpy of superheated vapor at state 6 H 5-H 6 Tit =

--~ 9 6 = 9 5 -- Tit ( 9 5 -- H6s ) = 2 8 0 6 . 4 6 k J / k g

H 5 - H6s

XTs =

6.5432-0.6493 8.1511-0.6493

=0.786

HTs =191.81(1-0.786) + 2584.8(0.786) = 2071.90 kJ/kg From the turbine efficiency, we estimate the enthalpy at state 7 H5- H7 Tit --

--+ H 7 = H 5 - Tit ( g 5 - H7s ) = 2394.48 kJ/kg

H 5 - H7s

Table 4.4a shows the state properties of the actual reheat regenerative Rankine cycle. The fraction of steam extracted is estimated from the energy balance m6H6 + m2H2 = rn3H3. In terms of the mass fraction z = rh6/rh3 , we have z H 6 + ( 1 - z ) I t 2 = H 3. The mass fraction is 2=

~H

3 -= H

2

H 6 -H 2

732.03-192.68

=

2776-192.68

The turbine work output based on 1 kg/s working fluid is" {)in = H 5 - H e = 2 7 1 8 . 3 0 k W

0.168

4.5 Applications of exergy analysis

207

Table 4.4b Distribution of exergy losses at each process in Example 4.16

Actual regenerative Rankine cycle

Process

1-2 34 4-5 5-7 7-1 Regeneration Cycle

Exl .... (kW)

Percent

0.0247 0.6043 883.86 262.24 195.22 34.29 1376.62

64.2 19.0 14.2 2.4 100

To = 285 K, TIt = 1700 K, and T~ = 285 K.

l)out = ( 1 -

z)(H 7 - H1) = 1832.34 kW

~'~net -- 6)in -- qout -- 8 8 5 . 9 6 k W

The thermal efficiency is ~t - 1 - c)°ut -

0.326

~/in T h e rate o f s t e a m is

rhs _ Ws _ 90,000 _ 101.57 kg/s Wnet 886.04 The exergy balance based on specified hot and cold sources yields EXloss,12 = ~/~/pla - ~'~/pls - 0.0247 kW .

.

°

EXloss,34 - Wp2 a - mp2 s =0.6043 kW

~2Xloss,45 - To [ th(Ss - S4) + (- qin) ) - 883.86 k W

TH

EXloss,57 = rnT 0 (zS 6 + (1 -

z)S 7 -

S 5) - 262.24 kW

qout ) - 195.22 kW EXloss,71 - TO /~h(S1 - $ 7 ) ( 1 - Z) + - ~ c

/~Xloss,regen -- fh ~)(S 3 - zS 6 - (1 - z)S2) - 34.29 kW

/~Xloss,cycl e __ ~ ) ( ~/out~c ~Hj--~)in 1376.62kW

Table 4.4b shows the distribution of exergy losses at each process based on T o - 285K, T H - 1700 K, and Tc - 285 K. The highest exergy loss occurs due to heat transfer in the boiler. The regeneration stage work loss is minimum.

208

4.

Usingthe second law: Thermodynamic analysis

Example 4.17 Ideal reheat regenerative cycle A steam power plant is using an ideal reheat regenerative Rankine cycle (see Figure 4.23). Steam enters the high-pressure turbine at 9000 kPa and 773.15 K and leaves at 850 kPa. The condenser operates at 10 kPa. Part of the steam is extracted from the turbine at 850 kPa to heat the water in an open heater, where the steam and liquid water from the condenser mix and direct contact heat transfer takes place. The rest of the steam is reheated to 723.15 K, and expanded in the low-pressure turbine section to the condenser pressure. The water is a saturated liquid after passing through the water heater and is at the heater pressure. The work output of the turbine is 75 MW. Determine the work loss at each unit. Solution: Assume negligible kinetic and potential energy changes, and that this is a steady-state process. The surroundings are at 285 K. In this ideal regenerative Rankine cycle, the steam extracted from the turbine heats the water from the condenser, and the water is pumped to the boiler. Sometimes, this occurs in several stages. The condensate from the heaters is throttled to the next heater at lower pressure. The condensate of the final heater is flashed into the condenser.

Boiler

)" Open Heater ~.~

3

Condenser 2

©

1

-

P2 (a)

T 5

7

(b) Figure 4.23. (a) Schematic of ideal reheat regenerative Rankine cycle and (b) T-S diagram.

4.5

209

Applications of exergy analysis

The steam data are: P 5 - 9000kPa, H 5 -3386.8kJ/kg, S 5 = 6.6600 kJ/(kg K), T5 - 7 7 3 . 1 5 K Pl=Pa-10kPa,

Ha,v-2584.8kJ/kg,

Hs,L=191.83kJ/kg, Vl=0.00101m3/kg

$8,v - 8.1511 kJ/(kg K), $8,L = 0.6493 kJ/(kg K) P3 - 8 5 0 kPa,

732.03 kJ/kg,

H3, L -

H3, V =

2769.90 kJ/kg

PT=850kPa, H7=3372.7kJ/kg, S7-7.696kJ/(kgK),

Tv-723.15kPa

P6=850kPa, ~,=450.0kPa, H6-2779.58kJ/kg T o - 2 8 5 K , TH=1600K, T c - 2 8 5 K Work and enthalpy estimations yield:

W p l _ V l ( ~ _ p 1 ) _ 0 . 0 0 1 0 1 ( 8 5 0 _ 1 0 ) ( 1kPa lkJm 3 ] - 0.848 kJ/kg

H 2 - H 1+ Wpl - 192.68 kJ/kg

W p 2 = V l ( P 3 - P 1 ) - O ' O 0 1 0 7 9 ( 9 0 0 0 - 8 5 0 ) [ 1kPalkJ1113) = 9.046 kJ/kg

H4

- H 3 + Wp2 -

741.07 kJ/kg

Because this is an isentropic process, $ 5 - $6 and $ 7 - $8, and we estimate the quality of the discharged wet steam at state 8: 7.696-0.6493 X8 =

8.1511-0.6493

= 0.940

H a = 191.83(1-0.94) + 2584.8(0.94) = 2439.63 kJ/kg The fraction of steam extracted is estimated from the energy balance m6H 6 + m2H 2 =m3H 3. In terms of the mass fraction z =/n6/rh 3 , the energy balance becomes zH6 + (1 - z)t12 - H3. The mass fraction is H~-H~ z

=

-

-

H~ - H 2

=

732.03-192.68 2779.58-192.68

=0.208

Table 4.5a shows the state properties of the ideal reheat regenerative Rankine cycle based on: T0 - 2 8 5 K ,

TH - 1 6 0 0 K ,

Tc - 2 8 5 K

The turbine work output with rh - 1kg/s working fluid is" qin --

rh[(H5 - H4 ) + (1 - z)(H v - H 6 ) ] - 3115.18 kW

4.

210

Usingthe second law: Thermodynamic analysis

Table 4.5a State properties of the ideal reheat regenerative Rankine cycle in Example 4.17 State

T (K)

P (kPa)

H (kJ/kg)

S (kJ/(kg K))

1 2 3 4 5 6 7 8(x8 = 0.94)

319 321.27 446.1 447.15 773.15 449.97 723.15 319

10 850 850 9000 9000 850 850 10

191.81 192.68 732.03 741.07 3386.80 2779.58 3372.7 2439.63

0.6493 0.6493 2.0705 2.0705 6.6600 6.6600 7.6960 7.6960

To = 285K, TH = 1600K, and Tc = 285K.

qreheat = t h ( g 7 - 9 6 ) = 593.12 k W

qout = rh(1 - z) (H 8 - H1) = 1779.14 k W ~rnet -- qin -- l)out -- 1 3 3 6 . 0 3 k W

The thermal efficiency is T~t "-- 1 - qout __ 0 . 4 2 8 qin

The rate of steam is

hS

.

Ws _ 75,000 . . . . Wnet

56.36 kg/s

1336.03

The exergy balance for each unit with rh = 1 kg/s working fluid is: /~Xloss,12 -= 0

Eyloss,26 = thT0 (83 -

zS6 - 0 - z)$2)=47.88 k W

EXloss,34 = 0

/~Xloss,45 = To ( t h ( S 5

_ 84 ) + (--TH)qin) ] =

785.68 k W

EXloss,56 = 0

EXl°ss,67= T° (th(ST - S6) + (- q67)) =195"82 EXloss,78 -- 0

•

/

qout/

EXloss,S 1 = T o rh(S 1 - $ 8 ) + - ~ - c

= 228.13kW

4.5

211

Applications of exergy analysis

Table 4.5b Distribution of exergy losses at each process in Example 4.17

Ideal reheat regenerative Rankine cycle

Process

Percent

/% .... (kW) 1-2 2-6 3-4 4-5 5-6 6-7 8-1

0 3.9 0 62.5 0 15.5 18.1 100

0 47.88 0 785.68 0 195.82 228.13

Total Cycle

1256.89 1256.89

The work loss in the reheat stage is: (-- £/reheat) _

EXloss,reheat -- TO /~/(S 7 - S 6 ) nt- _

r.

= 195.72 kW

The work loss in the regeneration stage is: EXloss,rehea t -- # / T 0 (S 3 - [ z S 6 + ( 1 -

z)S2] ) -

47.88 kW

The work loss for the whole cycle is

qin/

= 1256.7 k W

Table 4.5b shows the distribution of exergy losses at each process. The highest exergy loss occurs due to heat transfer in the boiler.

Example 4.18 Actual reheat regenerative Rankine cycle A steam power plant is using an actual reheat regenerative Rankine cycle (Figure 4.24). Steam enters the high-pressure turbine at 11,000 kPa and 773.15 K, and the condenser operates at 10 kPa. The steam is extracted from the turbine at 2000 kPa to heat the water in an open heater. The steam is extracted at 475 kPa for process heat. The water is a saturated liquid after passing through the water heater. The work output of the turbine is 90 MW. The turbine efficiency is 80%. The pumps operate isentropically. Determine the work loss at each unit if the surroundings are at 290 K. Solution: Assume negligible kinetic and potential energy changes, and that this is a steady-state process. The steam data are: V1 = 0.00101m3/kg

P1 - P9 = 10 kPa, H1, v - 2584.8 kJ/kg, H9, L - 191.83.45 kJ/kg Sl, v =8.1511 kJ/(kgK), $1,L - 0 . 6 4 9 3 kJ/(kgK)

P3-P6=475kPa,

H3, L = 6 3 1 . 2 9 k J / k g ,

H3,v=2745.30kJ/kg

$3, L = 1.8408 kJ/(kgK), $3, v = 6 . 8 3 6 5 k J / ( k g K )

212

4.

Usingthe second law: Thermodynamicanalysis

Boiler

) 8 ]_.d

Open Heater 3

Condenser

I

2 1

P2

~ P1

(a)

T

9s!

r

(b) Figure 4.24. (a) Schematic of actual reheat regenerative Rankine cycle and (b) T-S diagram.

Ts = 773.15 K, Ps=ll,000kPa,

Hs=3362.2kJ/kg ,

Ss=6.5432kJ/(kgK)

T~t=0.80 , T0=290K, TH=1700K, Tc=290K In this actual regenerative Rankine cycle, the steam extracted from the turbine heats the water from the condenser, and the water is pumped to the boiler. Sometimes, this occurs in several stages. The condensate from the water heaters is throttled to the next heater at lower pressure. The condensate of the final heater is flashed into the condenser Wpl = V1(p2- p1)= 0.00101 (475_ 10)( 1kPa lkJm 3 ) = 0.46 kJ/kg H 2 = H 1+ Wpl = 191.83 + 0.46 = 192.28 kJ/kg

Wp2=Vg(p4_P3)=O.O0109101,O00_475)( lkJm 3 ) = 11.48 kJ/kg 1kPa

4.5 Applications of exergy analysis

213

H 4 - H 3 +//Vp2 = 6 3 1 . 8 1 + 11.48 = 642.77 kJ/kg P6=2000kPa,

H7=3467.3kJ/kg,

S 7 = 7 . 4 3 2 3 k J / ( k g K ), T 7 = 7 7 3 . 1 5 K

From the turbine efficiency, we estimate the enthalpy at superheated vapor at state 6 _

H5

-

T ] t ~ - -

H6

H 6 = H 5 - r/t (H 5 - H6s ) = 2691.67 kJ/kg

H5 - H 6 s P7=2000kPa,

_

T]t~

HT-H8

Sss=S7=7.4323kJ/(kgK), T7 = 7 7 3 . 1 5 K ~ H s s = 3 0 4 5 . 5 4 k J / k g H 8 = H 7 --Tit ( H 7 - Hss) = 3129.89 kJ/kg ~ S s = 7.5951 kJ/(kg K)

H 7 -- Hss

At the condenser operating conditions of 10 kPa, we have $7 = Sss = S9s, and we estimate the quality of the discharged wet steam at state 9s"

X9s =

7.4323-0.6493 8.1511--0.6493

= 0.904

H9s = ( 1 - X9s)l 91.83 + (X9s)2584.8 = 2355.51 kJ/kg From the turbine efficiency, we estimate the enthalpy at state 9:

T~t =

H 7 -H

9

H 9 = H 7 --T~t ( H 7 - H 9 s ) = 2577.87 kJ/kg

H 7 - H9s

We also estimate the quality of the discharged wet steam at state 9:

x9 =

2577.87-191.83 2584.80-191.83

= 0.997

S 9 =(1 - x9)0.6493 + (x 9 ) 8 . 1 5 1 1 = 8.1287 kJ/(kg K) The fraction of steam extracted is estimated from the energy balance rh6H6 + rh2H2 = m 3 H 3. In terms o f the mass fraction z = rh6/rh3, we have zH6 + (1 - z)//2 = H3. The mass fraction is

z=

H3- H2 631 81 - 192.29 ~ = " =0.149 H8- H2 3129.89 - 192.29

Table 4.6a shows the state properties of the actual reheat regenerative Rankine cycle. The turbine work output based on rh = 1 kg/s working fluids" Olin = t h [ ( H 5 - H 4) + ( H 7 - H 6 ) ] = 3 0 7 7 . 3 3 k W

qout = lh(1 - z)(H 9 - HI) = 2029.05 kW

//Vnet - qin -- £/out -- 1048.28 k W

The thermal efficiency is

T~th - - 1 - - qout __ 0 . 3 4 0 £/m

4.

214

Usingthe second law: Thermodynamic analysis

The rate of steam is

rh s -

I'Vs

90,000 - ~ = 85.85 kg/s Wnet 1048.28

The exergy balance based on specified hot and cold sources yields

]E,Xloss,45 -- To (lh(S5 - S 4 -+-S 7 - &) +

EXloss,59

( - /k/ i4n )W ) = T1 0 5 1I. 0 2

=lhTo[zS 8 +(1- z)39 + 86 - 87 -

85] = 224.37 kW

JEXloss,91=rhTo((S1-Sg)(1-z)+q°ut) =184"53kwTC

EXloss,regen = thT0 (S 3 - z S

( /~Xloss,cycle = T0 [

\

8 -O-z)S2)=44.16kW

0out

rc

qin)

= 1504.09 kW

Table 4.6b shows the distribution of exergy losses at each process based on ~t ~-- 0 . 8 0 , T O ~- 290 K, TH = 1700 K, and Tc - 290 K. The table shows that the highest exergy loss occurs due to heat transfer in the boiler. The work loss in the regeneration stage is minimal.

Table 4.6a State properties of the actual reheat regenerative Rankine cycle in Example 4.18 State

T (K)

P (kPa)

H (kJ/kg)

S (kJ/(kg K))

1 2 3 4 5 6s 6 7 8s 8 9s(x7 = 0.904) 9(x 9 = 0.997)

319 319 423.07 423.07 773.15

10 475 475 11000 11000 2000 2000 2000 475 475 10 10

191.83 192.29 631.81 643.29 3362.20 2914.83 3004.30 3467.30 3045.54 3129.89 2355.51 2577.87

0.6493 0.6493 1.8408 1.8408 6.5432 6.5432 6.7003 7.6765 7.6765 7.5951 7.6765 8.1287

571.58 773.15

319 319

Table 4.6b Distribution of exergy losses at each process in Example 4.18 Process

4-5 5-9 9-1 Regeneration Total Cycle

Actual reheat regenerative Rankine cycle Exloss (kW)

Percent

1051.02 224.37 184.53 44.16 1504.09 1504.09

69.9 14.9 12.3 2.9 100 100

r/t = 0.80, To = 290 K, TH = 1700 K, and Tc = 290 K.

4.5

215

Applications of exergy analysis

4.5.3 Cogeneration The cogeneration process produces electric power and process heat from the same heat source. This may lead to the utilization of more available energy and the reduction of waste heat. The process heat in industrial plants usually needs steam at 500-700 kPa, and 150-200°C. The steam expanded in the turbine to the process pressure is used as process heat. By adjusting the steam rate, the steam leaves the process as a saturated liquid. The saturated liquid is pumped to the boiler. Cycles making use of cogeneration may be an integral part of large processes where the adjusted energy of the expanded steam from the turbine at intermediate pressure may be fully utilized in producing electricity and process heat simultaneously. The utilization factor for a cogeneration plant is the ratio of the energy used in producing power and process heat to the total energy input. The utilization factor is unity if the plant does not produce any power.

Example 4.19 Energy dissipation in a cogeneration plant A cogeneration plant is using steam at 8200 kPa and 773.15 K (see Figure 4.25). One-fourth of the steam is extracted at 700 kPa from the turbine for cogeneration.

I

Turbine

Boiler 1-z 7 1_. Process Heater

3;

@

4

I

Mixer I I

Condenser

2

1 P1

P2

(a)

rI 5

6]

4

(b)

Figure 4.25. (a) Schematic of ideal cogeneration plant and (b) T-S diagram.

216

4.

Using the second law: Thermodynamic analysis

The extracted steam is condensed and mixed with the water output of the condenser. The rest of the steam expands from 700 kPa to the condenser pressure of 10 kPa. The steam flow rate produced in the boiler is 60 kg/s. Determine the work loss at each unit. Solution: Assume that the kinetic and potential energy changes are negligible, and this is a steady-state process. The surroundings are at 290 K. The steam data are: P1 = P8 = 10 kPa, HI, v = 2584.8 kJ/kg, H1, L = 191.83 kJ/kg, V1 = 0.00101 m3/kg S1,v = 8.1511 kJ/(kg K), S1, L = 0.6493 kJ/(kg K) P3=P6=700kPa,

H 3 = 6 9 7 . 0 6 k J / k g , S3=l.9918kJ/(kgK),

z=0.25

V4 = 0.001027 m3/kg P 6 = 8 2 0 0 k P a , H 6 = 3 3 9 6 . 4 k J / k g , S6=6.7124kJ/(kgK), T 6 = 7 7 3 . 1 5 K T 0 = 2 9 0 K , TH=1700K, T c = 2 9 0 K In this cogeneration cycle, the steam extracted from the turbine is used as process heat. The liquid condensate from the process heat is combined with the output of the condenser

1kJ ] = 0.697 kJ/kg

Wpl = V1(P2 - P1 ) = 0.00101 (700 - 10) 1 kPa m 3

H 2 - H 1 +Wpl = 1 9 1 . 8 3 + 0 . 6 9 7 = 192.53 kJ/kg From the energy balance around the mixer, we have th3/th 6 =0.25 t h 4 H 4 = t h 2 H 2 + t h 3 H 3 , H 4 = th2H2 + t h 3 H 3 th 4

H4 =

45(192.53) + 15(697.06) 60

= 318.66 kJ/kg

T4 = 349.15K, V4 = 0.001027kg/m 3

Wpz=V4(Ps_p4)=O.O01027(8200_700) (

l k Jm 3 ) = 7.70 kJ/kg 1kPa

H 5 = H 4 + mp2 -- 326.36 kJ/kg--. S 4 = 1.0275 kJ/(kg K) Because this is an isentropic process, $6 = $7 = $8 = 6.7124 and P7 = 700 kPa ~ H7 = 2765.68 kJ/kg. We estimate the quality of the discharged wet steam at state 8: 6.7124-0.6493 X8 =

=0.808

8.151 l - 0.6493

H 8 = 191.83(1 - 0.808) + 2584.80.3(0.808) = 2125.35 kJ/kg

4.5

217

Applications of exergy analysis

Using the reference values for H 0 = 71.31 kJ/kg and So = 0.2533 kJ/(kg K) at 290 K, we estimate the exergy flows To(Si - So) and miAi, which are shown along with the other properties in Table 4.7a. The energy balance yields the fraction of steam extracted

A i = H i - 14o -

mtotal --- th6 (H6 - H7) + rh8 (H7 - Hs) = 66,634.44 kW mpi ~- ml~pl 4.

th4Wp2

--

493.51 kW

The net work output is: mnet- ~'total-

~/in

--

Z Wpi - 66,140.93 kW

ms(H6 - H5) = 184,202.24 kW

£/process = rh7 (m7 - H 3 ) = 3 1 , 0 2 9 . 3 0 k W

qout = rh8 (H8-H~) = 87,032.01 kW ~/net -- £/in -- £/out -- £/process --

66,140.93 k W

The exergy balance based on specified hot and cold sources is To = 290 K, Tu = 1700 K, and Tc = 290 K /~Xloss,mixer = rh 2 A 2 4. th 3 A 3 - th 4 A 4 - 7 4 0 . 8 0 k W

i

EXloss,boiler =rJ/5A5 - t h 6 A 6 4. 1---~HH ~/in = •

67,494.52kW

or /~Xloss,boiler --th5 ~0 (S 6

- Ss) 4. [ T-~) (- c)in) = 67,494.52 kW

Exloss,process=rhT(AT-A3) -

1 - - ~ cC-

qprocess=10, 4 9 4 . 6 9 k W

or

•

/ ol

Exloss,process = F/73T0 (33 - $7) 4- ~

@rocess -

10,494.69 kW

Table 4.7a State properties of the ideal cogeneration plant in Example 4.19 State

T (K)

P (kPa)

H (kJ/kg)

S (kJ/(kg K))

A (kJ/kg)

rnA (kW)

1 2 3 4 5 6 7 8 (x~ = 0.808)

319 319.15 423.07 349.50 351.05 773.15 457.83 423.07

10 700 700 700 8200 8200 700 10

191.83 192.52 697.06 318.66 326.36 3396.40 2765.68 2125.87

0.6493 0.6493 1.9918 1.0275 1.0275 6.7124 6.7124 6.7124

5.68 6.38 121.58 22.83 30.53 1451.95 821.23 181.42

255.60 286.96 1823.77 1369.93 1832.08 87117.06 12318.47 8164.15

218

4.

Usingthe second law: Thermodynamic analysis Table 4.7b Distribution of exergy losses at each process based on Example 4.19

Process

Mixer Boiler Process Expansion Condenser Total Cycle

Cogeneration plant Exloss (kW)

Percent

740.80 67494.52 10494.69 0 7908.55 86638.58 86638.58

0.8 77.9 12.1 0 9.1 100

To = 290 K, TH = 1700 K, and Tc = 290 K.

/~Xloss,condenser -- th 8 (A 8 - A1) -

1-

qout = 7908.55 kW

or

&loss,condenser -- th8 T0 (S1 - - 8 8 ) +

(~-c°)0out= 7908.55 kW

EXloss,mrbkne -- th6 & - / h 7 A7 - / h s A8 - Wtotal - 0 kW

EXloss,cycle__thTc To(qoutq

qprocess TC

qTH in) = 86,638.58k w

Table 4.7b shows the distribution of exergy losses at each process based on To = 290 K, TH = 1700 K, and Tc = 290 K. The table shows that the highest exergy loss occurs due to heat transfer in the boiler. The work loss in the regeneration stage is minimal.

Example 4.20 Energy dissipation in an actual cogeneration plant A cogeneration plant uses steam at 900 psia and 1000°F to produce power and process heat (see Figure 4.26). The steam flow rate from the boiler is 16 lb/s. The process requires steam at 70 psia at a rate of 3.2 lb/s supplied by the expanding steam in the turbine. The extracted steam is condensed and mixed with the water output of the condenser. The remaining steam expands from 70 psia to the condenser pressure of 3.2 psia. If the turbine operates with an efficiency of 80% and pumps with an efficiency of 85%, determine the work loss at each unit. Solution: Assume that the kinetic and potential energy changes are negligible, and this is a steady-state process. The surroundings are at 540 R. The steam data are: th 3 th 6 =161b/s, z = ~ = 0.2 th 6

P1 =P8 = 3.2 psia, H1,v =1123.6 Btu/lb, H1,L =111.95 BtuAb, 1/1= 0.01631 ft3Ab 5'1,v = 0.2051Btu/(lb R), $1,L = 1.8810 Btu/(lb R)

4.5

219

Applications of exergy analysis

Boiler

Process I~ Heater

I Condenser

/

3 ;

2

1

T

Mixer

4

~.~

~

-'= P1

P2 (a)

T

5

6 7

24

Ib)

Figure 4.26. (a) Schematic of actual cogeneration plant and (b) T-S diagram.

P3 = P6 = 70psia, H 3 = 272.74 Btu/lb, S 3 =0.4411Btu/(lb R) V4 - 0.0175 ft3/lb P6-900psia, H6-1508.5Btu/lb, S6=l.6662Btu/(lbR), T6=1000°F To = 540 R, TH = 3000 R, Tc = 540 R The pump efficiency is % = 0.85. The turbine efficiency is r h = 0.80. In this cogeneration cycle, the steam extracted from the turbine is used in process heat. The liquid condensate from the process heat is combined with the output of the condenser Wpl = V1( P 2 - P1)= 0.01631(70- 3.2) ( 1Btu ) % 0.85 5.4039psiaft 3 = 0.237Btu/lb

220

4.

Using the second law: Thermodynamic analysis

H 2 -- H 1 + mpl -- 111.95 + 0.237 = 112.18 BtuAb From the energy balance around the mixer, we have th3/th 6 =0.2: ivh4 H 4 = th 2 H 2 nt- th 3 H 3 , /-/4 = th2H 2 + th 3 H 3 rh4

He

0.8(112.18) + 0.2(272.74)

= 144.29 Btu/lb ~ S 4 =0.2573 Btu/(lb R)

T4 = 635.87R, V4 = 0.0010911b/ft 3

Wp2 = V4/°5 - P4 = 0 . 0 1 7 5 0 0 0 - 70) ( 1Btu ~p 0.85 5.4039 psia ft 3

= 3.16 BtttOb

H 5 = H 4 + Wp2 = 147.46 Btu/lb Because these are isentropic processes, $6 = S7s = S8s = 1.6662 Btu/(lb R) and P7 = 70 psia. We have H7s = 1211.75 Btu/lb. We estimate the quality of the discharged wet steam at state 8s is: 1.6662-0.2051 X8s

=0.871

1.8810-0.2051

H8s = 111.95(1-0.871)+ 1123.6(0.871) = 993.93 Btu/lb Using the turbine efficiency, we estimate the enthalpies at states 7 and 8 H 6 -H 7

'0t -- - - - + H H 6 -H7s

7 = H 6 -'qt (H6-H7s

) = 1271.10 Btu/lb

H8 = H6 - ~t ( H 6 - Hss) = 1096.85 BmAb The steam quality at state 8 is 1096.85-111.95 =0.973 xs = 1 1 2 3 . 6 0 - 1 1 1 . 9 5 Using the reference values for H0 = 38.0 Btu/lb and So = 0.0745 Btu/(lb R) at 70°E we estimate the exergy flows .4, = H i - H o - T o ( S i - S o ) a n d thi~, which are shown along with the other properties below. Table 4.8a shows the state properties of the cogeneration plant. The turbine work output is" !)process = th3 (H7 - H3) -- 3194.75 Btu/h mtotal = T/t [th6 ( 9 6 - H7s) + th8 (H7s - Hss)] = 5975.17 B t u ~ Z ~/rPi = ghlWp1 + th4Wp 2 = 53.63 Btu/h

The net work output is" Wnet -- mtotal - ~ ~rpi = 6975.17 Btu/h 0in = rh5 (H6 - Hs) = 16(1508.5-147.46) = 21,776.64 Btu/h C)out= rhs (H 8 - H1) = 16(0.8)(1096.85-111.95) = 12,606.7 Btu/h

4.5

Applications of exergy analysis

221

Table 4.8a State properties of the actual cogeneration plant in Example 4.20 State

T (R)

P (psia)

H (Btu/lb)

S (Btu/(lb R))

A (Btu/lb)

1 2 3 4 5 6 7s 7 8s (xs~ = 0.87I) 8 (xs = 0.973 )

603.67 603.90 763.6 635.87 638.17 1459.67 820.72 937.38 603.67 603.67

3.2 70 70 70 900 900 7(1 70 475 10

111.95 112.18 272.74 144.29 147.46 1508.5 1211.75 1271.10 993.93 1096.85

0.2051 0.2051 0.4411 0.2573 0.2573 1.6662 1.6662 1.7384 1.6662 1.8366

3.37 3.61 36.72 7.53 10.69 610.93 314.18 373.53 96.36 107.22

#~A(Btu/h) 43.21 46.25 117.52 120.57 171.16 9774.91 1005.38 1195.30 1233.52 1372.41

The exergy balance based on specified hot and cold sources is: To = 540 R, TH = 3000 R, and Tc = 540 R /~Xloss,mixer =/4/2A 2 + th3A 3 -/4/4A 4 = 4 3 . 2 0 Btu/h

/~Xloss,boiler = ~/5 1 5 - th6 A6

+ (1- TT---°H ) 0in = 8253.10Btu~

or •

EYl°ss'b°iler =/h5T°(S6 - $ 5 ) - +

~H ( - q i n ) =

8253.10Btu~

/~TXloss,process = th7 ( A 7 - A3) - (1 - T~-) C)process = 1077.78 B t u ~

or

/~TXloss,process-/h3T0 (S 3

--$7)+-(T@)qprocess= 1077.78Btu/h

/~TXloss,condenser-- /4/8( & -- A 1 ) -

1--'~CC qout

=1329.19Btu~

or

/~Xloss,condenser = th 8T0 (S1 - 38) + (T~-) qout = 1 3 2 9 . 1 9 B t u ~

EXloss,turbine = rh6A6 - th7 "47 - th8A8 -
•

( qout + £/process

qin/ = 11,881.67Btu~

Table 4.8b shows the distribution of exergy losses at each process based on To = 540 R, Tn = 3000R, and Tc = 540 R. The table shows that the highest exergy loss occurs due to heat transfer in the boiler. The process heat and condenser units cause relatively less work loss.

222

4.

Usingthe second law: Thermodynamic analysis

Table 4.8b Distributionof exergy losses at each process in Example 4.20

Process

Mixer Boiler Process Expansion Condenser Total Cycle

Cogenerationplant

L'Xloss(Btu/h)

Percent

43.20 8253.10 1077.78 1178.39 1329.19 11881.67 11881.67

0.3 69.5 9.1 9.9 11.7 100

To= 540R, TH= 3000R, and Tc = 540 R. 4.5.4

Geothermal Power Plants

(1) Dry steam power plants: The geothermal steam goes directly to a turbine, where it expands and produces power. The expanded steam is injected into the geothermal well. This was first done at Lardarello in Italy in 1904. The world's largest geothermal power is at The Geysers in northern California. (2) Flash steam power plants: Geothermal fluids above 360°F (182°C) can be flashed in a tank at low pressure causing some of the fluid to rapidly vaporize. The vapor then expands in a turbine. (3) Binary-cycle power plants: Moderate-temperature geothermal fluids between 85 and 170°C are common. A hot geothermal fluid and a suitable working fluid with a much lower boiling point than geothermal fluid pass through a heat exchanger. The vaporized working fluid drives the turbines. In such a closed-loop system, no working fluid is emitted into the atmosphere. The waste liquid from the flash steam power plants can also be used as an energy source. The working fluids may be isobutene, isopentane, or n-pentane.

Example 4.21 A steam power plant using a geothermal energy source A steam power plant is using a geothermal energy source (see Figure 4.27). The geothermal water is available at 220°C and 200 kg/s. The hot water goes through a flash drum. Steam from the flash drum enters the turbine at 550 kPa and 428.62 K. The condenser operates at 10 ~ a . The water is a saturated liquid after passing through the condenser. If the surroundings are at 290 K, determine: (a) The net work output; (b) The exergy losses at each unit; (c) The exergy efficiency of each unit. Solution: Assume that the kinetic and potential energy changes are negligible, and this is a steady-state process. The steam data are: ~ = 4 9 3 . 1 5 K , P~=2319.8kPa, H~=943.7kJ/kg, S~=2.517kJ/(kgK) Tz=428-62K, P2 =550kPa, H l = H 2 = 9 4 3 . 7 k J / k g /-/2,v = 2551.7 kJ/kg, H2, L =655.80 kJ/kg $2,v = 6.787 kJ/(kg K), $2,L = 1.897 kJ/(kg K) P3=550kPa, H3=2751.7kJ/kg, S3=6.787kJ/(kgK) P4=10kPa, X 4 = 0 . 9 6 ~ H 4 = 2 4 8 9 . 0 8 k J / k g , S4=7.8634kJ/(kgK) T0=290K, Tc=290K, TH=1600K

4.5

Applications of exergy analysis

223

Turbine o Q_

03

)(

Condenser 5

Valve

"-I M <er 7

Figure 4.27. Schematic of geothermal power plant.

(a) In this geothermal power plant, the hot water is flashed and steam is produced. This steam is used in the turbine. The rate of vapor is estimated from the quality at state 2. The fraction of steam after flashing is" 943.7-655.8

x2 =

2751.7-655.8

=0.159

S 2 =(1-0.159)1.897 +0.159 (6.787) =2.675 kJ/(kg K)

The steam flow rate is" th 3 -=- X2 (rhl) =

0.159(200) = 31.84 kg/s

From the mass balance around the flash drum, we have rh6 = rn1- t h 3 -168.15 kg/s P3 =550 kPa, H 3 -2751.7kJ/kg, S3 =6.787kJ/(kgK) The discharged steam has the quality of 0.96 P4 = 10 kPa, 2 4 = 0.96 ~ H 4 = 2489.08 kJ/kg, S4 = 7.8634 kJ/(kg K) From the flash drum at state 6, we have P6 =550 kPa, S6 - 1 . 8 9 7 k J / ( k g K ) ~ H 6 =655.80kJ/kg Using the reference states of To = 290K, we have H0 = 71.31 kJ/kg and So = 0.2533 kJ/(kg K). The availability is Ai = H i - Ho - T o ( S i - So). Table 4.9a shows the state properties and the energy and exergy values at each state. The turbine work output is ~/in -- fhl

(H1 - H 0 ) - 174,478.00 kW

6)out ---/~/3

(H4 - H s ) = 73,161.35 kW

mnet - th3 (H3 - H4 ) = 8363.71 k W

224

4.

Usingthe second law: Thermodynamic analysis

Table 4.9a State properties and the energy and exergy values at each state in Example 4.21 State

T (K)

P (kPa)

H (kJ/kg)

S (kJ/(kg K))

A (kJ/kg)

1 2 3 4 5 6

493.15 428.62 428.62 319 319 428.62

2319.8 550 550 10 10 550

943.70 943.70 2751.70 2489.08 191.83 655.80

2.5178 2.6756 6.7870 7.8634 0.6493 1.8970

215.68 169.90 785.61 210.82 5.68 107.82

Table 4.9b Distribution of exergy losses at each process in Example 4.21 Process

Geothermal power plant Exloss (kW)

Flash valve Turbine Condenser Plant

Percent

9156.32 9941.85 6533.35 34773.29

35.7 38.8 25.5

7/ 78.7 45.7 19.4

To = 290 K, Tc = 290 K, and TH = 1600 K.

(b) The thermal efficiency is __ ~rne t __ 8 3 6 3 . 7 1 '0th -- ~ qin 174,478

= 0.0479

(c) The exergy balance based on specified hot and cold sources yields /~Xloss,valve -- th 1 (A 1 - A2) =-

"Oth,valve

9156.32 kW

- A2 - 7 8 . 7 % A1

/~Xloss,turbine = t h 3 ( / 3 - 1 4 ) ~rne t

~/turbine l~th,turbin e =//13 (A3 - A4)

=9,941.85 kW

=45.7%

/)Xloss,condenser = th 3 (14 - A 5) - 6,

Wturbine -l~plant -- _ _

533.35 kW

19.4%

m~A~

/~Xloss,plant = thlA1 - mturbine = 4 3 , 1 3 7 . 0 - 8363.7 = 34,773.29 kW Table 4.9b shows the distribution of exergy losses at each process based on To = 290K, Tc = 290K, and TFI = 1600 K. The highest exergy loss occurs due to heat transfer in the boiler. The work loss in the regeneration stage is minimal.

4.5

225

Applications of exergy analysis

Table 4.10a Values of enthalpy and entropy for R-134a in Example 4.22

T (K)

P (Mpa)

H~ (kJ/kgt

Hv (kJ/kg)

SL (kJ/kg K)

Sv (kJ/(kg K))

273.15 293.15

0.293 0.571

50.02 77.26

247.23 258.35

0.1970 0.2924

0.9190 0.9102

Example 4.22 Exergy analysis of a refrigeration cycle One of the most common refrigerant is 1,1,1,2tetrafluoroethane (CF3CH2F) known as R-134a. R-134a has a boiling point temperature o f - 2 6 . 2 ° C and a latent heat of 216.8 kJ/kg at 1.013 bar. It is compatible with most existing refrigeration equipment. However, it is not compatible with the mineral oils used as lubricant in currently used systems. Another refrigerant is pentafluoroethane (C2HFs) known as R-125. In a R-134a refrigeration cycle, the superheated R-134a (state 1) enters a compressor at 263.15 K and 0.18 MPa. The R- 134a (state 2) leaves the compressor at 313.15 K and 0.6 MPa, and enters a condenser, where it is cooled by cooling water. The R-134a (state 3) leaves the condenser at 293.15 K and 0.571MPa as saturated liquid, and enters a throttling valve. The partially vaporized R-134a (state 4) leaves the valve at 0.293 MPa. The cycle of R-134a is completed when it passes through an evaporator to absorb heat from the matter to be refrigerated. The flow rate of R-134a is 0.2 kg/s. The total power input is 60 kW. The cooling water enters the condenser at 293.15 K and leaves at 298.15 K. The surroundings are at 290 K and evaporator is at 273 K. Determine the exergy loss of each unit. Solution: Assume that kinetic and potential energy changes are negligible, and the system is at steady state. From Tables E 1 to E3 the data for R- 134a are: H 1 = 242.06 kJ/kg, (superheated vapor)

$1 = 0.9362 kJ/(kg K)

at T1 = 263.15 K, P1 = 0.18 MPa

H2 = 278.09 kJ/kg

$2 = 0.9719 kJ/(kg K)

at 7,2 = 313.15 K, P2 = 0.6MPa

(superheated vapor) H3 = 77.26 kJ/kg (saturated liquid)

at 7'3 - 293.15 K, P3 = 0.571 MPa

H4 = 77.26 kJ/kg (Throttling process)

P4 = 0.293 MPa

T4 = 273.15 K

To = 290 K Win = 60 k W m - 0.2 kg/s Tevaporator = 273 K Tcondenser -- 290 K The throttling process (Stage 4) causes partial vaporization of the saturated liquid. The vapor part of the mixture is known as 'quality.' Using the enthalpy values at 0.293 MPa (Table 4.10a) the quality is

x4

=

77.26-50.02 =0.138 247.23- 50.02

Then the value of entropy is S4 = (1-0.138) S L +0.138 S v =0.2967 kJ/(kg K) For the cycle, the total enthalpy change is zero. At the compressor, outside energy Win is needed, and at the evaporator, heat transfer qin f r o m the matter to be cooled is used to evaporate the refrigerant R-134a. The heat absorbed within the evaporator from the contents of the refrigerator is qin = rh(H1 - H 4 ) -- 3 2 . 9 6 k W

The energy balance indicates that the total energy (Win + qin) is removed by the cooling water. For the cycle, the first law yields Win --t-~/in- qout - - 9 2 . 9 6 k W

226

4.

Usingthe second law: Thermodynamicanalysis

We can find the coefficient of Performance (COP) to estimate the efficiency of a refrigerator qin

COP =

mideal,in

-- H1 - H 4 = 4 . 5 7

H2-H1

where Wideal,in = t h ( H 2 -- g l ) = 7.20 kW

Enthalpy values of the cooling water from the steam table (Table D 1) are Hl,cw = 83.86 kJ/kg

at T I = 293.15K

H2,cw = 104.89 kJ/kg

at T2 = 298.15 K

Heat removed in the condenser qc is equal to

!)out

qc = rhwater(H2,cw - Hl,cw ) = 92-96 kW Hence mwater -- 4.42 kg/s The total work (exergy loss) loss is

total

Tevaporate r

/ /

(--~¢°ut) -- 58"94 k W !)in- 1 -Tcondenser ~

The exergy loss at each step describes the thermodynamic performance using the exergy of streams obtained from

4 =(Hi -

ros/)

Using the enthalpy and entropy data of each stream, we have A 1 = - 2 9 . 4 3 kJ/kg A2 = -3.76kJ/kg A 3 = - 7 . 5 3 kJ/kg A4 = -8.79kJ/kg 1. Compressor .

EXloss,comp --/'h(A 1 - 1 2 ) ' + Win "- 54.86 kW

2. Condenser

EXloss,cond = rh(/2 - A3)- ( 1 - ~ Tcondenser

(--0out) - 1.75 kW

3. Valve /~Xloss,valve =- th(A 3 - A 4 ) = 0.25 kW

Table 4.10b Energy dissipated in various units in Example 4.22

Process

A~Xloss(kW)

States

Percentage loss

Compressor Condenser Valve Evaporator

54.86 1.75 0.25 2.07 58.94

1-2 2-3 3-4 4-1

93.0 3.0 0.5 3.5 100

4.5

227

Appfications of exergy analysis

4. Evaporator

- A1)+(1-~ Tevaporato r

E'Yloss,eva p = t h ( A 4

qin --

2.076 k W

Table 4.10b displays the levels of energy dissipated in various units. Thermodynamic analysis identifies the performance of individual processes. Finding ways to improve the thermodynamic performance of individual steps is equally important.

Example 4.23 Analysis of the Claude process in liquefying natural gas We wish to partially liquefy natural gas in a Claude process shown in Figure 4.28. It is assumed that the natural gas is pure methane, which is compressed to 80 bar and precooled to 300 K. In the expander and throttle the methane is expanded to 1.325 bar. The methane after the first heat exchange at state 5 is at 80 bar and 250 K. Thirty percent of the first heat exchangers output is sent to the expander. Only 10% of the first heat exchange is liquefied. The expander efficiency is 0.8. Determine the work loss in the liquefaction section excluding compression and precooling. Solution: Assume that kinetic and potential energy changes are negligible, and this is a steady-state process. The whole operation is adiabatic, and the surroundings are at 290 K. It is economic to cool the compressed methane for liquefaction by the gas that does not liquefy in the throttling process. In the Claude process, the gas at an intermediate temperature splits into two parts. One of them enters the expander and exhausts as a saturated or slightly superheated vapor, and produces work. The remaining gas is further cooled in the second heat exchanger and throttled to liquefy. The portion that is not liquefied is combined with the output vapor of the expander and recycled into the compressor. The data for methane are: At P = 8 0 p s i a and T = 3 0 0 K : H 4 =l119.7kJ/kg, S4=9.158kJ/(kgK ) At P=l.325psia a n d T = 3 0 0 K : H 4 =lll9.7kJ/kg, S4 =9.158kJ/(kgK) thl2 - 0.3 /1;14

Z

F

-I~

Cooler

.

__~

.

.

.

Exch. 1

Exch. 2

7

I I

11

I

Win

I

15 I

out

I

1

,,

1

I

.

3

.

.

.

.

.

Figure 4.28. Schematic of the Claude process considered in Example 4.23.

..._1

228

4.

Usingthe second law: Thermodynamicanalysis

Mass and energy balances for the control volume and adiabatic expander yield /hi5 + m 9 = / h 4

(a)

~rout "--/h999 nt thl5H15 - / h 4 H 4 --/hi2 (H12 - / - / 5 )

(b)

The combination of (a) and (b) yields the portion y that is liquefied

y

-

-

-

/h9 -- z(912 - 9 5 ) + m

9 4 -H15 =0.1

H9 - H15

/'h4

At P=1.325 psia and T = 115K" H 9 =297.7kJ/kg, S9 =5.035 kJ/(kg K) (saturated liquid) At P = 1.325 psia and T = 115 K" H12 = 802.5 kJ/kg,

=9.436 kJ/(kg K) (saturated vapor)

812

At P = 80 psia and T = 250 K (state 5)" H 5 =964.4 kJ/kg, S5 = 8.590 kJ/(kg K) For isentropic expansion $5 =

S12s =

8.590 kJ/(kg K), and we have the quality at state 12" 8.590-5.035

Xl2s =

9.436-5.035

= 0.807--, 9 1 2

s --

705.46 kJ/kg

FromH12,L = 297.7 kJ/kg, H12,v = 802.5 kJ/kg From the turbine efficiency, we estimate the enthalpy at state 12

"l~t ~-0.8

H12 - H 5

912 = H 5 -- T~T(912 s -- 9 5 ) = 757.2 kJ/kg

H12 s - H 5

From Eq. (c), we have H15 = 1141.79 kJ/kg. The mass and energy balance for heat exchanger 1 is" th 4 (95 - 94)--/~h15 (914 - 915 )

/h 4 =?h 9 -k-thl5 ;

?h9

H5 - H 4

m4

1-y

y ......... --, H]4 = H15 +

= 969.23 kJ/kg (at 1.325 bar)

The mass and energy balance for heat exchanger 2 is" th7(H 7 - H5) = thl4 (/-/12 - H14 )

//'/4 =/h7 nt- ?hi2; gh4 --/h9 nt- ?hi4 --*H7 = H 5 - (1-- y)(H14 - H12 ) = 691.84 kJ/kg (at 1.325 bar) 1--z

P3=P6=60psia, H3, L =262.21Btu/lb, H3,v=l177.6Btu/lb $3,L =0.4273 Btu/(lb R), $3,v = 1.6440 Btu/(lb R)

'l~t = 0 . 8 0 ,

T0 =

290 K,

Tc = 290 K

(c)

4.5 Applications of exergy analysis

229

Table 4.11 Distribution of exergy losses at each unit in

the Claude process in Example 4.23 Process

Claude process Exi .... (kW)

HX1 HX2 Expander Throttling Total ~mt -~- /~'XI..... total

Percent

41.21 41.82 73.60 295.45 452.15 514.29

9.1 9.2 16.3 65.4 100

The basis is 1 kg/s methane; the work loss calculations are mout -Widea 1 =

rh[Hls (1 -

y) + H9y

rhz(H12 - H11) = 62.14 kW -

H4]-

rhTo[Sls(1- y) + S9y - $4] = 514.32 kW

Heat exchanger 1" EXloss,HX1 =/~/mT0 (S 5 - & + (815 - 814)(1 - )2)) --

41.21 kW

Heat exchanger 2: EXloss,HX2 = t h m T 0 ((S 7 - 86)(1 - z) -Jr-(814 - 813)(1 - y)) =

41.82 kW

Expander: EXloss,expande r -- t h s T 0 ((812 - S 11)7,) =

73.60 kW

Throttling: EXloss,throttl

e

~---thsT0 (&y - 5'10(1- y - z ) - $7(1- z)) = 295.45 kW

Distribution of exergy losses at each process is based on: To = 290 K and Tc = 290 K. Table 4.11 displays the results of the exergy analysis. As seen, the highest exergy loss occurs in the throttling process.

Example 4.24 Power plant analysis A steam power plant (Figure 4.29) uses natural gas to produce 0.12 MW power. A furnace completely burns the natural gas to carbon dioxide and water vapor with 40% excess air. The flue gas leaves the furnace at 500 K. The combustion heat supplied to a boiler produces steam at 10,000 kPa and 798.15 K, which is sent to a turbine. The turbine efficiency is 0.7. The discharged steam from the turbine is at 30 kPa, and is sent to a condenser. The condensed water is pumped to the boiler. The pump efficiency is 0.90. Determine: (a) The thermal efficiency of an ideal Rankine cycle; (b) The thermal efficiency of an actual cycle; (c) The work loss of boiler, turbine, condenser, and pump. Solution: Assume that kinetic and potential energy changes are negligible, and the system is at steady state. Assume that the natural gas is pure methane gas, and the surroundings are at 298.15 K.

230

4.

Usingthe second law: Thermodynamic analysis

~ Methane+Air

Flue gas t

1

Boiler a,-

r-

2

Pump ( k Condenser

'

/

Heat to surroundings f at 298.15 K Figure 4.29. Steam power plant considered in Example 4.24.

(a) The basis is 1 kg/s steam To = 298.15K,

Cp=4.19kJ/(kgK)

R = 8.314J/(molK),

Hz=3437.7kJ/kg,S2=6.6797kJ/(kgK) at

Tz=798.15K, P2=10,000kPa

$3,L =0.9441kJ/(kgK), $3,v = 7.7695kJ/(kgK) H3,L = 289.3 kJ/kg, //3, v = 2625.4 kJ/kg

at at

P3 = 30 kPa (saturated steam) P3 = 30 kPa (saturated steam)

~t =0.70, ~p =0.75, /3=0.000584K -1, V = 1020cm3/kg at T = 342.15K For the Rankine cycle, the operation is isentropic, and $2 = $3. However, $2 < S3,v, and the discharged steam from the turbine is wet steam. The quality of the wet steam X3s is

w

2 - - S3,L

= 0.84

X3s -- $3, V - 8 3 , L

The enthalpy of the wet steam H3s is H 3 s = 9 3 , L nt- X3s (H3, V - H 3 , L ) =

2252.4 kJ/kg

The enthalpy difference is AH s = H3s - H 2 = - 1185.3 kJ/kg With the turbine efficiency of 0.7, we have AH = 7 ~ t A g s = - 829.71 kJ/kg H 3 = H 2 + M-/= 2607.99 kJ/kg

4.5 Applications of exergy analysis

The steam in the final state is also wet steam with a quality of x3 2 6 0 7 . 9 9 - 289.3 x3 = 2 6 2 5 . 5 - 289.3

= 0.9925

With this quality, the entropy of the final state is S 3 = 7.7186kJ/(kgK) W~ = AH s = - 1185.3 kJ/kg The heat loss in the condenser is qc = H4 - H3s = - 1963.1 kJ/kg From the isentropic pump operation, we have Wp s __ z~_/p s _ V(P1 - e3) = 10.19 kJ/kg

106

So the enthalpy at point 1 is His = ~ - / p s nt- H 4 = 10.189 + 289.3 = 299.47 kJ/kg The heat required in the boiler becomes qBs = H2 - His = 3138.21 kJ/kg The net work for the Rankine cycle is Wnet, s =

AHp + AH s = - 1175.1 kJ/kg

The efficiency of the Rankine cycle is Wnet, s

r/s - - - 0.37 qBs (b) For the actual cycle operation, we have AHps Wp = ~Hp - - - 13.58 kJ/kg T/p

The temperature change of water in the pump is

Arp=AH -

V ( 1 - fiT) (P~ - P3)/106

Cp

= 1.296 K or 1.296°C

The entropy change of the water in the pump is ASp

=CpIn

T3 + ATp P1 - P 3 T3 - / 3 V 106 -- 0.00989 kJ/(kg K)

Therefore, we have the value $1 = $4 + AS = 0.9540 kJ/(kg K). The net work for the practical cycle is Wnet, act = A H p +

AH = 816.12 kJ/kg

231

232

4.

Using the second law: Thermodynamic analysis

qc, act = H 4 - H 3 qB,act = H 2 - / - / 1 = 3134.81 kJ/kg

Therefore, the efficiency of the actual cycle is Wnet,act glact -- - - - - 0 . 2 6 qB, act

Comparison of the two efficiencies shows that both operations have relatively low efficiencies, although the actual cycle is considerably less efficient than the Rankine cycle. The steam flow rate thsteam is W

/J/steam - - ~ - "

120.52

/-/3 -H2

kg/s

With this steam rate, the total heat rates in the condenser and boiler become qc = thsteam qc, act "- 3 7 7 , 8 2 0 . 1 k W

279,457.5 k W

qB = thsteamqB,act = -

(c) Table 4.12a summarizes the streams connecting the four units of the power plant, while Table 4.12b shows the average heat capacities for the components of the flue gas. The basis is 1 mol CH 4. The combustion reaction in the furnace is CH 4 ÷ 2 0 2 ---+CO 2 -Jr-2HzO(g ). The changes during combustion are: 1. Unmixing the air: - 0.0, ASai r =

-hR~__, Yi In Yi = - 52.9 J/K i

2. Standard reaction enthalpy and entropy values at 298.15 K: zkH ° = - 393,509 + 2(-241, 818) - (74,520) = - 802,625 J/(mol CH4)

Table 4.12a Various states in the cycle in Example 4.24 Point

State of steam

T (K)

P (kPa)

H (kJ/kg)

S (kJ/(kg K))

1 2 3 4

Subcooled liquid Superheated vapor Wet vapor (x = 0.992) Saturated liquid

342.77 798.15 342.27 342.27

10000 10000 30 30

302.88 3437.7 2607.99 289.30

0.9540 6.6797 7.7186 0.9441

Table 4.12b Average heat capacities for Example 4.24: Cp/R= A + BT+ CT 2 (Tin K) Species

A

B × 103

Cp,av (298.15-500 K) (J/(mol K))

CO2 H20 02 N2

5.457 3.47 3.699 3.28

1.045 1.45 0.506 0.593

48.83 33.66 32.42 29.23

4.5

233

Applications of exergy analysis

A G O - - 394, 359 + 2 ( - 228, 512) - ( 5 0 , 4 6 0 ) = - 8 0 1 , 0 4 3

J/CH

4

AH ° - A G O

AS 0 =

= 5.306 J / ( m o l C H 4 K) 298.15

3. M i x i n g o f g a s e s to f o r m the flue gases" A H - 0.0, A S a i r - -

-hR~_, Yi In Yi = - 9 4 . 1 4 J / K i

4. H e a t i n g f r o m 298.15 to 5 0 0 K a n d u s i n g Table 4.12b, w e e s t i m a t e a total a v e r a g e heat capacity. Table 4.12c s h o w s the c o m p o s i t i o n s o f the inlet a n d outlet s t r e a m s and the a v e r a g e heat capacities

zM-I = Z Cn,i (500 - 298.15) - 85, 0 9 1 . 7 4 J / ( m o l C H 4 )

AS = ~

500 - 2 2 0 . 0 2 J / ( m o l K) Cp~v; In ~ 298.15

For the total c h e m i c a l s y s t e m , w e have AH = ~

AH/=0

- 8 0 2 , 6 2 5 + 0 + 85, 0 9 1 . 7 4 = - 7 1 7 , 5 3 3 . 3 J / m o l

i AS = ~

AS i = - 52.9 - 5.3 + 94.14 + 220.0 = 253.8 J / ( m o l K) i

T h e e n e r g y b a l a n c e a r o u n d the b o i l e r / f u r n a c e unit yields the m o l a r flow rate o f m e t h a n e thsteam (H2 - H I ) -- tlmethane (AH), f/methane -- 526.55 m o l / s T h e ideal w o r k o f the c h e m i c a l s y s t e m is

Wideal -- f/methane (/~H •

Sprod,1 :

o

- ToAS ) - 418.05 M W

.

msteam

($2 - S1) -Jr-f/methane ( 0 . 2 5 5 9 ) - 690.09 k W / K ~ EXloss,furnace,boile r : 2 9 8 . 1 5 ( 6 0 9 . 0 9 ) - 2 0 5 . 7 4 M W •

o

Sprod,2

-- mstea m (S 3 -

S 2 ) - - 125.21 k W / K

--, Exturbin e --

298.15(125.21) = 37.33 M W

qc, r - 120.81 k W / K --,/~Xcondense r = 298.15(120.81) -- 36.02 M W Sprod,3 -- r}/steam(84 -- $3) -+- ~ _ 298.15 Sprod,4 - - mstea m ( S 1 - S 4) - 1.192 k W / K --, E x p u m p - 298.15(1.192) = 0.35 M W

Table 4.12c Compositions of inlet and outlet gas streams for the furnace in Example 4.24

Species

n (inlet mol)

CH 4

1 0 0 2 (1.3) = 2.6 2.6 (79/21 ) = 9.78 12.38 (air)

COt H~O O~ N~ Total _

_

Flue gas (mol)

Flue gasy (%)

nCt..... (J/(kg K))

0 1 2 0.6 9.78 3.38

0 7.4 14.9 4.4 73.1 100.0

48.33 67.32 19.46 285.94 421.56

234

4.

Using the second law: Thermodynamic analysis

Table 4.12d Distribution of lost work among the units of the power plant in Example 4.24

1 2 3 4 W Total

Units

Exlo~s (MW)

Furnace/boiler Turbine Condenser Pump Work outlet

205.74 37.33 36.02 0.35 100.0 379.45

Percent 54.22 9.83 9.49 0.09 26.35 100.0

Exergy losses are Exloss,i = ToSproa,i.Table 4.12d shows the distribution of lost work in various units of the power plant. The maximum lost work occurs in the boiler/furnace unit, while the minimum lost work occurs in the pump.

4.5.5

Exergy Analysis of Distillation Columns

The simulator packages such as Aspen Plus and Hysys may be useful in analyzing distillation column systems to improve recovery and separation capacity, and to decrease the rate of entropy production. For example, for the optimization of feed conditions and reflux, exergy analysis can be helpful. A complete exergy analysis, however, should include both an examination of the exergy losses related to economic and environmental costs and suggestions for modifications to reduce these costs. Otherwise, the analysis is only theoretical and less effective. A typical distillation column achieves separation using heat at a high temperature in the reboiler, and discharging it at a lower temperature in the condenser. Therefore, the column resembles a heat engine delivering separation work. Distillation is an irreversible process with thermodynamic losses due to heat and mass transfer, mixing, pressure drop, internal design, and column configurations (such as number of feeds and side products and side heating and cooling). Pinch analysis has been at the forefront of research in the efficient use of energy, and the heat integration of distillation with the rest of the process has attracted a great deal of attention. Exergy analysis is used in designing thermodynamically optimum columns and assessing the performance of existing columns (Rivero, 2004; Demirel, 2004, 2006a,b). Thermodynamic analysis of existing distillation column operations may be separated into the following stages: (i) the assessment of the status of exergy use, (ii) if necessary, the modification and improvement of the operation to reduce irreversibility and hence the cost of energy, and (iii) the assessment of the thermodynamic and economic effectiveness of the modifications and improvements (retrofits). Improvements and modifications to distillation column operations involve feed conditioning (heating or cooling), adjusting the reflux ratio, adjusting the feed stage location, using side heating and cooling, and reducing the entropy production by equally distributing the driving forces over the column volume. Total exergy Ex of a multicomponent material stream consists of physical, chemical, and mixing parts. Disregarding kinetic and potential exergy contributions, the rate of exergy of a stream is ~+x= H - ToS, Ex = hEx, where h is the molar flow rate of a stream and Ex the molar exergy. Similarly, H and ~ are the stream enthalpy and entropy rates, respectively, and are based on reference conditions. To is the environmental temperature usually assumed as 298.15 K. The entropy balance determines the rate of entropy production due to irreversibilities

out of system

into system

[ + sl:Sp r°d

where Ts is the stream temperature. Equation (4.198) may also be used to estimate the rate of exergy loss, which is directly proportional to the rate of entropy production according to the Gouy-Stodola theorem.

4.5.6

Exergy Loss Profiles

The Aspen Plus simulator with a suitable thermodynamic model generates the stage-exergy loss profiles of distillation columns from the exergy balance of stage i & l o s s , i -- E"x i + V 1 -t- j Ex. iL__ 1j E x V _ j ~ , x L _ ~ _

q(1--~-] \1i)

(4.199)

4.5

Applications of exergy analysis

235

v and Exi_u L are the rates of the exergy of vapor and liquid phases for stages i + 1 and i - 1, respectively. As where /~.xi+~ an extensive property, exergy loss can be used in the sensitivity analysis and optimization of distillation columns. The difference between the exergies of products and feed streams of a column determines the minimum amount of exergy for a required separation work (Ognisty, 1995) "

Ex~ep -

~ ~+x- ~ / ~ x products feeds

(4.200)

For a reversible operation, the net exergy from heat is

into system

~

-- out of system

where qR is the heat absorbed in the reboiler and £¢~ the heat discharged in the condenser. As the exergy loss increases, the net heat exergy has to increase to enable the column to achieve the required separation. Supplying more heat increases the temperature in the reboiler, and hence the net heat exergy. However, the temperature of the reboiler is limited to avoid product thermal degradation, while the temperature of the condenser is set by cooling utilities. The unavoidable part of the total exergy loss EXloss,un is

N/0/

E"q°~'u~ = /~l 7 ,

Ex'°ss'i

(4.202)

where N is the number of stages including the condenser and reboilers and 7,. the temperature of stage i. A quantified potential improvement indicator (PI) is P! =

EXloss

-

-

EXloss,un

(4.203)

Exloss The indicator PI represents exergy losses that are avoidable because of the configuration of that column and the transport rate limitations.

4.5.7 Exergy Efficiency The difference between the net exergy of heat and the exergy of separation represents the total exergy loss in the column

&loss -- &heat --/~Xsep

(4.204)

The total exergy losses consist of configuration limitations due to the design of the column and transportation limitations due to the states of streams. For L'A-~,> 0, the exergy efficiency of column i becomes r/;-

EXsep,i /~Xsep,i . £Vh~t,i EXlos~,i+ +~Xsep,i

(4 205)

The values of efficiencies help in assessing the effectiveness of exergy usage, as the value of (1 - r/) is directly related to the exergy loss. They can be used to compare the performance of columns with each other. Smaller exergy losses mean the utilization of a higher portion of available energy, and hence fewer thermodynamic imperfections in a column. The low exergetic efficiency is typical for distillation systems with close boiling mixtures and with high energy requirements in the reboiler. An alternative is to use reboiler-liquid flashing. A compressor is used to return the reboiled vapor to the bottom of the column. The required reboiler duty is somewhat larger than the required condenser duty, and so an auxiliary steam-heated reboiler is needed. Thus, a trade-off is made between the power used in the compressor and the large reduction in reboiler steam. The converged mass and heat balances and the exergy loss profiles produced by the Aspen Plus simulator can help in assessing the thermodynamic performance of distillation columns. The exergy values are estimated from the enthalpy and entropy of the streams generated by the simulator. In the following examples, the assessment studies illustrate the use of exergy in the separation sections of a methanol production plant, a 15-component two-column

236

4.

Using the second law: Thermodynamic analysis

Table 4.13 Conventional column operation for the separation of propylene and propane a Stream

Temperature (K)

Flow rate (kmol/h)

Enthalpy (kJ/mol)

Entropy (kJ/(mol K))

325.0 319.5 330.9

272.2 189.2 113.0

13338 12243 14687

-4.1683 - 13.8068 -2.3886

Feed Distillate Bottoms aSeider et al. (2004).

system, and a vinyl chloride monomer (VCM) production plant. The distillation columns in the methanol plant use process heat from the other parts of the plant. The operating parameters for all the examples are summarized in Table 4.13. The column stages are numbered starting from the condenser. The rate of entropy production, the stream exergy flows, the total exergy losses of the columns, the minimum values of separation exergy, and the thermodynamic efficiencies are estimated. The efficiencies are used to compare the performance of the individual columns in each case study, while the potential improvement indicator shows the possible savings in exergy losses. Example 4.25 Column Exergy efficiency Propylene-propane mixture is a close boiling mixture. A reflux ratio of 15.9 (close to minimum) and 200 equilibrium stages are necessary. Table 4.13 shows the enthalpy and entropies of the saturated feed and saturated products from the simulation results with the Redlich-Soave equation of state. The reboiler and condenser duties are 8274.72 and 8280.82 kW, respectively. The reference temperature is 294 K. The lost work LWis obtained from Eq. (4.198) as L m -- ToSprod -- 1902.58 k W .

Availability Ai = Hi - ToSi analysis yields l/~'rmin = E t/A- E t:/A= 140.81 kW out in

The thermodynamic efficiency Tith is mmin T/th

4.5.8

l/Vmin+ L l~

Equipartition Principle

In linear nonequilibrium thermodynamics, the local equilibrium holds, and we use linear relations between the generalized flows and thermodynamic forces with constant proportionality coefficients. The rate of entropy production formulated by the linear nonequilibrium thermodynamics approach describes the level of the dissipated power (work) in a system. Separation systems with uniform driving forces or uniform entropy production in space and time only will dissipate less of the available energy. A uniform entropy production rate corresponds to either minimum energy costs for a required separation and area investment or minimum investment for a specified energy cost, and leads to thermodynamically optimum design. One way of achieving thermodynamically optimum distillation system may be the use of heat integration. Stage-exergy calculations are used to prepare exergy profiles throughout column. The stage-exergy loss profiles indicate the distribution of stage irreversibility, and hence the distribution of driving forces in a column operation. Therefore, such profiles show the current level of the utilization of available energy, as well as the effects of operating conditions and design parameters on the efficiency of the operation. Based on the exergy loss profiles, modifications to the feed stage location and feed condition and the use of intermediate exchangers can be considered. For example, exergy loss due to mixing at the feed stage can be identified and reduced using an external modification to the prefractionator. In addition, locating the heat exchangers in the regions where the largest deviations from isoforce exist may lead to uniform driving forces over the internal stages of a binary distillation system. An isoforce operation of an adiabatic column where large reflux ratios are avoided is consistent with a minimum exergy loss. Distillation columns operating with close to uniform thermodynamic forces are analyzed for separating n-pentane from n-heptane (Table 4.14), and ethanol from water (Table 4.15). As Eq. (4.113) shows, chemical separation force (viVlzi/T) should be uniform throughout the column for thermodynamic optimum. Separation of ethanol from water

4.5

Applications of exergy analysis

237

Table 4.14 Reduction in entropy production in distillation column for n-pentane and n-heptane a

Operation

ql~(MW)

Adiabatic Isoforce Near optimum

2.37 1.89 1.90

q(,

Reduction in entropyproduction(%)

(MW)

0.704 0.732 0.797

13.56 13.33

~'DeKoeijerand Rivero (2003). Table 4.15 Comparison of various types of distillation operations for separation of ethanol-water mixture a

Operation

Total exergylosses (kJ/h)

Distillate flowrate (kg/h)

Distillate composition(%)

Adiabatic Isoforce Diabatic

44.23 14.24 15.89

0.969 0.974 0.973

87.53 87.13 87.22

%auar et al. (1997). shows that the largest exergy loss occurs in the stage with the largest composition differences, and the total exergy losses are 433.8 kJ/kg in an adiabatic distillation and 248.41 kJ/kg in a diabatic operation, corresponding to a 42% decrease. For a specified number of stages, the minimum distance between the operating and the equilibrium curves corresponds to optimum exergy usage. The analysis of a heat-integrated distillation column utilizing the heat pump principle revealed that the exergy loss is considerably lower than that of a conventional column. Table 4.15 compares the adiabatic, isoforce, and diabatic operations. However, in minimizing the exergy loss or the rate of entropy production, an operation with driving forces that are too small (pinch in separation) at any stage should be avoided. Equipartition principle is mainly used to investigate binary separations by distillation; it should be extended to multicomponent separations with nonideal mixtures and by accounting for the coupling between driving forces.

Example 4.26 Assessment of separation section of a methanol plant Figure 4.30 shows the separation section of a methanol plant using natural gas, carbon dioxide, and water as the basic feed streams, and which produces 62,000 kg/h and 99.95% pure methanol. The methanol synthesis takes place in a tube-cooled reactor with an exit temperature of 240°C. The reactor output contains small amounts of dimethylether, n-butanol, acetone, water, and hydrogen besides the main product of methanol. The output is flashed and the liquid stream 407 is fed to the separation section where the methanol is purified in two distillation columns. The first column has 51 stages, a partial condenser at the top, and a side condenser at stage 2. It receives a side heat stream of 15.299 MW at stage 51, and operates without a reboiler. The feed enters at stage 14. The second column has 95 stages and a total condenser. It receives a side heat stream of 18.9 MW at stage 95 as a part of a reboiler duty of 282.283 MW, and operates with a very high reflux ratio. The feed enters at stage 60. The methanol is a side product drawn from stage 4. Table 4.16 summarizes both of the column operations. Converged mass and energy balance data from Aspen Plus are the result of the thermodynamic methods of Redlich-Kwong-Soave (RKS) for vapor properties, and the activity coefficient model of NRTL and Henry components method for the equilibrium and liquid properties. Figures 4.31-4.33 display the exergy loss profiles together with the unavoidable portion of the exergy loss for columns 1 and 2, respectively. Figure 4.31 shows that the main exergy losses are 0.367 MW (44%) at the feed stage and 0.117 MW (14%) at the condenser of column 1. The large exergy losses are due to mass transfer and mixing occurring at the feed stage, where streams at different compositions and temperatures are mixed. Figure 4.32 shows that column 2 operates with rather large exergy losses at the feed stage and at stages close to the reboiler. The losses occur mainly because of the feed stage and heat transfer and pressure drops (as high as 8-10 psi) in the vicinity of the reboiler. The temperature-exergy profile in Figure 4.32 displays large amounts of exergy losses mainly due to the heat transfer between the temperatures 355 and 393 K, and contributes a large part of the total loss, as distillation columns utilize heat for separation. Figure 4.33 displays the effects of possible side heating and positioning the heating units at the bottom section. The side heating is effective only at appropriate temperature levels, and the side heating in Figure 4.33d may be a desirable retrofit as it reduces the exergy loss from 4.2 to 0.74 MW at stage 94. Also, the steam used in the side heating would be at lower temperatures and would cost less.

238

4.

Usingthe second law: Thermodynamic analysis OFFGAS HOFFGAS(OUTi

::;iii~

JDISTVAP

~

JDISTL'Q

I MKwATERI

.................. ~'..

I

' .

.

.

i

.

........

:il

Y

I QTOP(IN)~'""J QTOP J...... [ QREB(IN)~ ..........I QREB

Figure 4.30. Separation section of the methanol plant for Example 4.26.

Table 4.16 Operating parameters for columns 1 and 2 for Examples 4.26-4.28 Parameter

Example 4.26 Column 1

No. of stages Feed stage Feed temperature (°C) Reflux ratio Condenser duty (MW) Condenser temperature (°C) Side condenser stage Side condenser duty (MW) Reboiler duty (MW) Boilup rate (kmol/h) Bottoms rate (kmol/h) Reboiler temperature (°C) Bottom temperature (°C) Heat stream duty (MW) Heat stream stage Heat stream temperature (°C)

51 14 43.74 3.7 1.371 32.75 2 - 8.144 1551.28 2995.14 85.85 15.29 51 105.15

0~

Example 4.27

Example 4.28

Column2

Column 1

Column2

Column 1

Column2

95 60 85.84 188765 281.832 74.85

30 8 58.3 0.68 10.990 122.0

50 24 116.6 0.422 3.199 230.0

15 8 47.60 1.O82 6.295 -1.50

10 7 145.70 0.969 9.065 50.6

282.283 24890.7 1050.96 119.71 119.71 18.90 95 136

12.752

4.332

11.116

7.076

344.0

145.7

658.0

165.9

360

................................................................

I- - - ~ ° t ~

.................. i ............... ,--O--Unavoi

ji~

i ...........................

~

350

I ab,e, ........

340 ~ 330

2

......... ! .......... i .......... i ..........

~ 320

1

310 0

(a)

0.1

0.2

0.3

Exergy Loss, MW

300

0.4 (b)

',

0

0.1

0.2

0.3

0.4

Exergy Loss, MW

Figure 4.31. Exergy loss profiles of column 1 for Example 4.26: (a) stage-exergy loss and (b) temperature-exergy loss.

Table 4.17 shows that column 2 operates with efficiency as low as 4.1%, while the efficiency o f c o l u m n 1 is 50.6%. The exergy values for the whole separation system of two columns are also low and need to be improved. The individual values o f PI show that it is possible to reduce by 13.1% the total loss o f 0.834 M W in c o l u m n 1 and by 19.1% the total loss o f 27.813 M W in column 2. Despite the heat integration, the separation section o f the m e t h a n o l plant performs poorly in utilizing the exergy in the distillation columns.

4.5 100 90 80 70 60 50 40 30 20 10 0 0.00

400 .___L . . . . . . . . . .

---r

- . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . .

r . . . . . . . . . . . . . . . . . . . . .

..........

±

. . . . . . . . . . . . . . . . . . . . .

-___

f, ~--,,

---r

. . . . . . . . . .

r.

___1.

. . . . . . . . . .

• . . . . . . . . . . . . . . . . . . . . .

.........

(a)

-

-

370

©

360

J____

350

•

,

1.00 2.00 3.00 Exergy loss, MW

ii ~

1----

.............. r O-T°ta' i Unavoidable ,

-

380

.~___ ,,

. . . . . . . . . . . . . . . . . . . .

-

%1

l

, .....................

ii

390

-___

___, . . . . . . . . . .

---r

239

Appfications of exergy analysis

F-

i _~___Unavoidabii~ iii!

i

340

4.00

0.00

i

1.00 2.00 3.00 4.00 Exergy loss, MW

(b)

Figure 4.32. Exergy loss profile of column 2 for Example 4.26: (a) stage-exergy loss profiles and (b) temperature-exergy loss profiles.

6~_~________

~00

1

e

;~

U ~

50

5-1

0

(a)

i

0.5

1

1.5

2

2.5

3

3.5

4

4.5

Exergy Loss M W

e

'1

0.5

0

(b)

1

1.5

2

Exergy Loss M W

I00

l)

dl II

I

4

0

A

i

r 50!

" 50

(c)

J

05

1 15 ~=ergy Loss M W

2

25

0 (d)

05

1 15 ~xe,gy Loss M W

2

25

Figure 4.33 Stage-exergy profile of column 2 for Example 4.26: (a) without side heating, (b) single side heater at stage 87 with qsd--150MW, (c) two side heaters at stage 87 with qsd " - - 1 5 0 M W and at stage 90 with qsd " - 6 0 M W , and (d) stage 87 with qsd - - 180 MW and stage 92 with qsd - - 50 MW. Table 4.17 Results of exergy analysis for Examples 4.26-4.28 Examples

Sprod

/~Xsep

&heat

Exloss

]~x1 . . . . . .

(MW/K)

(MW)

(MW)

(MW)

(MW)

(%)

/~Xreduced (MW)

PI (%)

Example 4.26 Column 1 Column 2 Columns 1 + 2

0.00217 0.0890 0.0912

0.856 1.135 1.991

11.690 28.114 29.804

0.834 26.979 27.813

0.724 21.810 22.535

50.6 4.1 6.6

0.110 5.169 5.279

13.1 19.1 18.9

Example 4.27 Column 1 Column 2 Columns 1 + 2 + V a

0.00831 0.00337 0.00898

0.907 0.542 1.114

3.385 1.547 4.932

2.476 1.005 3.818

1.946 0.617 2.563

26.8 35.0 22.6

0.530 0.388 1.255

21.4 38.6 32.8

Example 4.28 Column 1 Column 2 Columns 1 + 2

0.00555 0.00411 0.00966

2.164 0.327 2.491

3.814 1.263 5.077

1.650 0.936 2.586

1.406 0.739 2.145

56.6 20.9 46.3

0.244 0.197 0.441

14.7 20.9 17.0

7"]

aValve.

system is a 15-component ~ t u r e containing a pseudocomponent, which is:specified ~ t h an ~ e f i c a n Petroleum institute (AP I)degree ofAPi = 48!6 0ra specific gravity o f 01786 (see Fibre 4i 34). ;The first column operates a t 13!26 atm with 30 stages: ~ e feed enters at stage 8, The top product is rich in propane and the m i x ~ e of isobutene

240

4.

Using the second law: Thermodynamic analysis

c~

~

Figure 4.34. Flow diagram of separation system for Example 4.27.

35~

1

450 •

1

425 25

+Total

400 ~ ~ "

-.-0.--

Unavoidable

o 20

~ 375

~15 ,

10

(a)

~L

325

5 0

t

300 0

0.1

0.2 0.3 Exergy Loss, MW

0.4

0

(b)

I i 0.2 0.3 0.1 Exergy Loss, MW

0.4

Figure 4.35. Exergy loss profiles of column C1 for Example 4.27: (a) stage-exergy loss and (b) temperature-exergy loss. 650

55 50 45 40 o r~

(a)

35 30 25 20 15 10 5 0

600 550

o

500 450 400 350

0

0.1 0.2 0.3 Exergy Loss, MW

0.4

0

(b)

0.1

0.2 0.3 Exergy Loss, MW

0.4

Figure 4.36. Exergy loss profiles of column C2 for Example 4.27: (a) stage-exergy loss and (b) temperature-exergy loss.

and n-butane. The n-butane is the light-key component with a recovery of 99.0%, while the 2-methyl-butane is the heavy-key component. The bottom product is sent to column 2 after the pressure is lowered to 5.1 atm through a valve. The second column operates at 5.1 atm with 50 stages, and the feed enters at stage 24. The top product is rich with the 2-methyl-butane and n-pentane mixture. The n-undecane is the light-key component with a 98% recovery, while the n-dodecane is the heavy-key component. The bottom product at 347.85°C is used to heat the feed of the first column. Table 4.16 shows the operating parameters for the columns. The converged simulation was done using the thermodynamic method of RKS to estimate the properties. The Peng-Robinson equation of state was used for predicting the equilibrium and liquid properties, and the vapor phase was assumed to be ideal. As seen in Figures 4.35 and 4.36, the stripping section operates with a high level of exergy losses in column 1. Table 4.17 shows that the total exergy losses in columns 1 and 2 are 2.476 and 1.005 MW, and the thermodynamic

4.5

241

Applications of exergy analysis

CRACK QUENCH COL1

COL2 PUMP

Figure 4.37. Flow diagram for Example 4.28: vinyl chloride production plant.

16 14 12 10 8 6 4 2 0

I o

nv

I

290

().3

Exergy Loss MW

.

.

.

.

- ~

270 ~

o'.l

(a)

430 410 . 390 E 370 350 ~" 330 310-

.

.

.

-o--

~

t,~,,,oi~t~,bJ~,

,

0

0.1

(b)

0.2

0.3

Exergy Loss MW

Figure 4.38. Exergy profiles of column 1 for Example 4.28 (a) stage-exergy loss and (b) temperature-exergy loss.

10 ~ ' x _

450

9

+

Total

430

8

navoidable

7

-

~

410

-

= 390 c~

5

~ 37O

4 3

~

350 330 310

1

() (a)

0.1

0.2

Exergy Loss MW

().3

0 (b)

0.;5

0'.1

0.15

012

0.25

Exergy Loss MW

Figure 4.39. Exergy loss profiles of column 2 for Example 4.28: (a) stage-exergy loss profiles and (b) temperature-exergy loss profile.

efficiencies are 26.8% and 35.0%, respectively. The individual values of PI are 21.4% and 38.6% for columns 1 and 2, respectively. The exergy values for both column operations include the valve, and 1.255 MW can be saved from the whole operation.

Example 4.28 Assessment of separation section of vinyl chloride monomer (VCM) plant VCM is produced by the pyrolysis of 1,2-dichloroethane (EDC) at -483°C and 26.5 atm with a conversion of 55%. The pure EDC is fed to the reactor with a flow rate of 909.1 kmol/h. The feed is at 21 °C and 26.5 atm. The reactor outlet is cooled to 47.6°C. The first column operates at 25 atm with 15 stages. The feed is introduced at stage 8. The top product is anhydrous

242

4.

Usingthe second law: Thermodynamic analysis

hydrochloric acid, which is recycled for use in oxy-chlorination. The bottom product consisting of vinyl chloride and unreacted EDC is fed to the second column. The second column operates at 7.8 atm with 10 stages (with a feed stage of 7). The top product is VCM, while the bottom product with unused EDC is recycled to the reactor. Table 4.16 lists the other operating parameters for the columns, while Figure 4.37 shows the process flow diagram. The converged simulation is obtained from the thermodynamic method of RKS to estimate the thermodynamic properties and the phase equilibrium. Especially the second column is practically a binary distillation column. As the exergy loss profiles in Figures 4.38 and 4.39 show both columns operate with rather large exergy losses throughout the columns. The total exergy losses are 1.650 and 0.936 MW for columns 1 and 2, respectively, as seen in Table 4.17. The total exergy losses are mainly due to mixing and heat transfer. The thermodynamic efficiencies are 56.6% and 21.0% for columns 1 and 2, respectively. Such low thermodynamic efficiencies are not unusual in industrial column operations (Seider et al., 2004). The values of PI are 14.7% and 20.9% for columns 1 and 2, respectively. A total of0.441MW can be saved from both the column operations. For such a large industrial VCM plant, the amount of fuel that can be recovered in the distillation operation is as high as 9.1 MW. For existing operations, process heat integration and reducing the cost of steam in the reboiler may be considered first among other modifications and optimizations for potential improvements in the use of energy.

4.5.9

Exergy Analysis of Refinery

A1-Muslim et al. (2003) and A1-Muslim and Dincer (2005) performed an exergy analysis of single- and two-stage crude oil distillation. The single-stage system consists of a crude heating furnace and a 27-tray atmospheric distillation column. The feed is introduced in tray 23. The two-stage system consists of a furnace, a 13-tray atmospheric distillation column, another furnace to heat the bottom product of the first unit, and a second distillation column with 14 trays. The feed is introduced in tray 12. Table 4.18a compares the exergy analysis of the systems, and shows considerable reduction in exergy losses in the two-stage system. Rivero (2002) reported exergy analysis for an existing refinery operation. Table 4.18b shows the considerable economic gains due to the reduction in exergy losses after the optimization studies.

Table 4.18a Exergy analysis for single- and two-stage crude oil distillation systems a System

Exergy input (MW)

Exergy output (MW)

Overall exergy loss (MW)

Overall exergy efficiency (%)

Column exergy losses (MW)

498.8 352.0 29.4

69.8 110.9 58.8

429.0 241.1 43.8

14.0 31.5 125

137.2 121.6 11.4

Single stage Two stage Difference (%)

aA1-Muslimet al. (2003).

Table 4.18b Exergy analysis and exergy loss reduction in a modified refinery a Unit

Combined distillation unit Naphtha HDS unit Naphtha reforming unit HDSc unit Catalytic cracking unit Visbreaking unit Utilities plant Total

Exergy loss before optimization (%)

Exergy loss after optimization (%)

Proposal investments (US$1000)

20.4 3.2 10.9 3.4 19.5 2.9 39.8 100.0

17.2 2.7 7.9 2.8 12.7 3.7 36.7 84.3

2822.3 1101.2 1204.8 834.0 7822.0 1000.0 660.6 15245.0

aR. Rivero,Energy Convers. Manage., 43 (2002) 1199. bNPV: net present value (only operating cost is taken into account). CHDS:hydrodesulfurization.

Payback time (months) 17 69 5 53 3 3 4 5

NPVb 10 years of investments 7.37 1.80 3.98 2.04 47.4 33.27 32.19 24.48

4.6

4.6

243

Chemical exergy

CHEMICAL EXERGY

The chemical exergy is the maximum theoretical work of a combined system composed of a combustion cell and its surroundings (see Figure 4.15). Fuel enters the combustion cell operating at steady state at temperature To and pressure P0; oxygen enters from the environment at T0 and partial pressure Yo~P0, where Yo~ is the mole fraction of oxygen. The fuel and oxygen undergo a combustion reaction completely and produce water vapor and carbon dioxide. The reaction for a hydrocarbon is given by _

C,,H h + a + ~ 02 - aCO 2 +

(4.206)

H20

Disregarding the kinetic and potential energy effects, the energy balance over the control volume for a steady-state operation is

[

( )

Wc - ~/v qc + hv + a + -4 ho~- - ahco~ -

hugo

]

(4.207)

where NF is the molar flow rate of the fuel and hF the molar enthalpy of the fuel. Equation (4.208) represents the work produced by the combined system of a combustion cell and its environment, and heat is transferred to the environment. In terms of the enthalpies of products and reactants hp and hR, respectively, Eq. (4.207) becomes q,: - W,: = NF(hp

(4.208)

- hR)

We need enthalpy of formation data, since some fuels are normally composed of several chemical species. The heating value of a fuel is the enthalpy of combustion; a lower heating value occurs when all the water is in vapor state. The entropy balance for the combustion cell is

O = ~ ~•

T~

+ NF sv + a+

-4

So~- -- aSco:

--

SH20

+

(4.209)

The entropies of the mixture components can be calculated using the appropriate partial pressures

s i (T, P,.) - s o (T) - R In

(4.210)

~ef

where s o (T) is the absolute entropy of component i at temperature T and Pref- Eliminating the heat transfer rate between Eqs. (4.207) and (4.209), we have W,:- NF [hE + [ a + b ] h o ~ - a h c % - [ b ] h u ~ o ] (4.211)

- T o N v s F + a +-4 So~ - ascot_ -

b SH20 "-I'-O

We can determine the specific enthalpies and the specific entropies from the temperature, pressure, and composition of the environment. Once we specify the environmental conditions, all enthalpy and entropy terms are fully defined regardless of the process within the control volume. The term T0~ depends on the nature of the process and the irreversibility. Chemical exergy, /~Xch, is ~'Xch-NFIh F. + ( a + b ) h o ~ - a h c o ~ - ( b ) h H ~ o ] b - To l sv + ( a + ~ ) So~_ - asco~ -- ( b j sH20 1

(4.212)

Chemical exergy leads to maximum theoretical work when there is no irreversibility. A similar equation in terms of the Gibbs function g = h - Ts of respective substances becomes

244

4.

Usingthe second law: Thermodynamic analysis

E'Xch = NF[gF

+(a+b)go2- ghco2- (b)gH20(g)]

+RTlnl

(4.213)

(YO2)a+b/4 ] (Yco2)a (yH20)b/2

The specific Gibbs functions are evaluated at temperature To and pressure P0 of the environment, and are given by

g(To,Po)= g~ +[H(To,Po)-H(Tref -

Pref)]

-[TS(To,Po)-TrefS(Tref -

Pref)]

(4.214)

where gO is the Gibbs function of formation at specified conditions. Standard chemical exergy values, in units ofkJ/kmol, are based on reference conditions To and P0, such as 298.15 K (536.67 R) and 1 atm, respectively, and consist of a set of reference substances with standard concentrations of gaseous, liquid, and solid components. The standard chemical exergy tables often simplify the application of exergy principles.

4.7

DEPLETION NUMBER

Resource depletion may cause environmental change. By reducing resource depletion, we can reduce ongoing environmental transformation. We may quantify the resource depletion by the DepletionnumberDp, which is a nondimensional indicator EXDpper unit consumption Exc EXDp

(4.215)

Dp- Exc

Biological systems have evolved to allow sustainable consumption to occur with little or no depletion. Therefore, the depletion number provides a measure of system progress or maturity, and is a useful basis for studying the evolution of industrial resource use patterns and the implementation of resource conservation strategies. The depletion number is a function of three indicators showing the level of implementation of resource conservation strategies: • The exergy cycling fraction ~ is a measure of recycling that accounts for both the throughput and quality change aspects of resource consumption and upgrading. • The exergy efficiency rl is a universal measure of process efficiency that accounts for the first and second law principles. • The renewable exergy fraction f~ is a measure of the extent to which resources supplied to an industrial system are derived from renewable sources. Industrial systems consume resources by supporting processes associated with supplying and removing resources. Therefore, the temporal and spatial boundary conditions are important in defining the universal relationships among the conservation strategies of renewable and nonrenewable sources. Boundary conditions will determine which resources and processes constitute an industrial system. Spatial boundary conditions are mainly geographical and resource-specific, while temporal boundary conditions define the scope of time for the exergy transfer and loss in processes. The definition of depletion number defined in Eq. (4.215) may be

JEXDp--1+ ]~XDs 1+ ~ 1-- E~Ru --1 - + L'xTV 1-- f~vu --1 -- . Dp Exc Exc ~/RU L'Xc ~VU

[

J I

J

(4.216)

where L'XDslis the exergy dissipation rate, ~Ru and ~/Ru the renewable exergy fraction and transfer efficiency for the recovered resource upgrade process, respectively, EXTv the exergy transfer rate to the nonrenewable source, and ~ v u and ~vu the renewed exergy fraction and transfer efficiency, respectively, for the nonrenewable resource upgrade process. Two structural constants aDs-C and av-c are

C~Ds_C =

/~XDsl EXc(1- ~)

ExTV

C%-c = L'xC(1 - gt)

(4.217)

(4.218)

4.9

245

Informationcapacity and exergy

With these definitions, Eq. (4.216) becomes

I {

Op- 1+~

-1 +d-g0

c~D~_c+C~v_c

~RU

[1,v Jl --1

(4.219)

~VU

The above equation expresses the depletion number as a function of a system's structural constants, the exergy efficiency and renewed exergy fraction of the individual resource upgrade processes, and the extent of resource cycling. The generalized depletion number may result from numerous consumption processes, such as incomplete cycling or partial upgrading, and the direct reuse of resources without upgrading. Recycling may reduce the need for resources and the exergy requirements of manufacturing processes. Generally, increasing resource cycling reduces depletion due to less exergy transfer from other sources. For example, producing 1 tonne aluminum from bauxite requires 27,400 MJ of exergy transfer, while converting the recycled aluminum to feedstock requires far less exergy transfer.

4.8

OPTIMIZATION PROBLEM

In general, the methods of modeling, analysis, and optimization in engineering begin with deciding on the system geometry, architecture, and components, and the manner in which the components are connected. Engineering analysis involves the mathematical description of the conceptualized system and its performance. Finally, optimization leads to the most favorable conditions for maximum performance (e.g., minimum entropy production or minimum cost). The search for an optimal design may be considerably challenging, since one may contemplate a very large number of geometric possibilities, and boundary and initial conditions in a fluid flow network. In practice, one may examine a number of alternative configurations, optimize their performance, and compare the optimized alternatives. Finally, one should select a suitable configuration with the least irreversibility or cost. In complex designs, the designer has to deal with an increasing number of degrees of freedom. The thermodynamic optima of individual processes may be robust and useful shortcuts in the optimization of larger and more complex systems. This approach is important in physical, chemical, and biological systems with well-adapted extensions and ongoing modifications leading to better and more complex designs. For complex engineering problems, either there is no a single optimum solution or it is difficult to find a global optimum. Complex problems may involve various processes coupled to each other with various sources of irreversibility. Engineering design is usually associated with the exact economic optimum leading to a global minimum. On the other hand, only simple problems, such as determining the minimum value of a parabolic equation, have single optimum solutions. The level of irreversibility introduces thermodynamic imperfections that eventually decrease the overall performance of the system. Some optimization procedures are: (i) In a heat exchanger network system, one can calculate the total annual cost for a possible combination of heat exchangers for required heating and cooling loads. The minimum total annual cost approach does not take into account the significant differences between solutions that have similar total annual costs. For example, we may find significant differences in control, operability, safety, and environmental impact in a complex network. Even for a simple network problem with nh hot streams and nc cold streams, the number of evaluations required for the global economic optimum would be (nh × nc)!, and calculating every possible design combination and permutation may not be the best way to optimize a design. (ii) Mathematical modeling can improve our understanding of a thermal process and is the key to a good process design. However, mathematical modeling usually deals with the optimization of the design parameters for a specified process. Improvements in a design are often a result of changes in the process, and mathematical modeling does not usually address such changes. The engineer should be able to make the final design decision after carefully considering the results of mathematical modeling.

4.9

INFORMATION CAPACITY AND EXERGY

Exergy also appears as information capacity; the free energy of the information that a system possesses is kT In/, where I is the information we have about the state of the system and k the Boltzmann constant. A relation between the exergy and information is Ex= k ln 2ToI

(4.220)

246

4.

Usingthe second law: Thermodynamic analysis

The transformation of information from one system to another is often almost entropy-free energy transfer, and the information capacity I in binary units is expressed as a function of the probability P

/

ln2

j=l

Pj ln Pj - Z pO lnpO j=l

/

(4.221)

where f~ is the number of possibilities, P0 the probability at equilibrium (i.e., no knowledge), and P the probability when we have some information about the system. Information here is used as a measure of order or structure. With small amount of exergy in the form of information, we can control processes converting large amounts of energy and matter. The exergy carried as information is a structural exergy. Living systems survive and evolve by transforming solar exergy into complex, highly ordered structures directed and controlled by the information of the genes. One generation transfers the information to the next generation by DNA replication. The superiority of biological systems relies on the difference in information transfer techniques between biological and physical systems. Information must be stored and transported safely. In biological systems, the information transfer takes place with a continuous debugging or control. The specific molecular structures and the unique positions of single atoms of DNA molecules make systems far more efficient than technological systems. This is because technological information systems work with macroscopic structures such as printing, orientation of magnetic particles on a medium, or transistors in an electric circuit. The information transfer efficiency of the most advanced microprocessor is still several times poorer than that of bacteria.

4.10

PINCH ANALYSIS

Pinch analysis yields optimum energy integration of a process and its utilities by using the principles of thermodynamics. The target of the analysis is the minimum theoretical energy required to supply a hot or cold utility for the overall process. The pinch analysis matches cold and hot process streams with a network of exchangers. Hot and cold streams can only exchange energy up to a minimum allowable temperature difference ~Tmi n. The minimum temperature level ATmin is called the pinch point or pinch condition. The pinch point defines the minimum driving force and hence the minimum entropy production allowed in a network. The pinch point separates the overall operating temperature region into two regions. We must supply the hot utility above the pinch and the cold utility below the pinch. The value of ~XTmin, sometimes called the approach temperature, is a key design variable for a heat exchanger network. It has an impact on lost work associated with heat transfer. The net lost work Exloss for a heat flow of q between high temperature 7'1 and low temperature T2 is

Exl°ss =

,oss

rl-r2) TzT1

qT°

(4.222) (4.223)

(4.224)

Here, To is the absolute temperature of the environment. For a given heat load, energy loss is directly related to the value of ATmin. When the temperature levels move into the cryogenic region in a process, ~Tmi n must decrease as the square of the temperature level to maintain the same rate of lost work. An increase in ~Tmi n causes higher energy and lower capital costs (a smaller heat exchanger area). For example, an increase of 5°C from a value of ATmin = 10°C decreases the heat exchanger area by 11%, and increases the required minimum energy by --~9%. To find the value of optimum ZXTmin,we plot the total annual cost against AT (see Figure 4.40) and search for an optimum z~Tmin where the total annual cost of energy and capital costs are minimized. The optimum value for ATmin is generally in the range of 3-40°C for heat exchanger networks, but is unique for each network and needs to be established for a process. If no cooling media are required below --~10°C, the optimum ATmin is often in the range of 10-40°C. An increase in ~Tmi n causes the energy costs to increase. After establishing a ATtain, we can estimate the minimum hot and cold utility requirements from the composite curves.

4.10 Pinchanalysis

247

Pinch analysis can optimize a whole plant operation containing not only heat transfer, but separation and reaction units as well. Some examples are heat-integrated crude oil distillation systems, the total process energy integration in retrofitting an ammonia plant with 44 hot and cold streams (Wang et al., 2003), the heat exchanger network of a nitric acid plant (Matijasevia and Othmaeia, 2002), and the combination of a chemical reactor network with a heat exchanger network (Lavric et al., 2003). Some of the advantages of pinch analysis over conventional designs are the ability to set energy cost and capital cost targets for a network and the ability to update the process flowcharting. However, some of the modifications suggested by pinch analysis may require substantial capital investments. The analysis will be successful if target temperatures and utilities are set on the basis of process objectives rather than on flowcharting. For example, a flowchart may mix two streams with different temperatures to prepare a feed stream. This may cause degradation of available energy or of the thermodynamic driving force. To prevent this, the temperatures of both streams should be increased to the process operating temperature. Also, heat recovery from special streams like two-phase streams should be completed in a single heat exchanger due to phase separation and large pressure drops. The destination of process streams also should be fully evaluated to avoid adverse effects of streams with hazardous chemicals. However, process integration would be more complete and meaningful if its goals include environmental protection, emission control, and reduction in depletion of natural energy, in addition to lowering the cost of energy. Moreover, software developed for process integration should be able to interact with other software to access a wide range of optimization and design models.

4.10.1

Composite Curves

The second law determines the direction of heat flow and prevents crossovers of the curves of the hot and cold stream temperatures. Temperature-enthalpy diagrams called composite curves represent the thermal characteristics of hot and cold streams and the amount of heat transferred (see Figure 4.41). The enthalpy change rate for each stream is

!

~

total cost

,

\

nergy cost N

~

capital cost

AT,nin

AT

Figure 4.40. Optimum .XTm,nfrom energy cost and capital cost changes.

Hot utility ...................... ..-., -'~

,- . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

i,~ v

hot composite curve

_.

.C

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

,

ATtain cold composite curve

"_ii

Cold utility y

q

Figure 4.41. Hot and cold composite curves.

248

4.

Usingthe second law: Thermodynamic analysis

q = A H = rnCpAT = MCAT

(4.225)

where M-/is the enthalpy change rate, m the mass flow rate, Cp the heat capacity, AT the temperature change in a stream, and MC the heat capacity rate rhCp. We add the enthalpy change rates over each temperature interval that includes one or more of the streams. This leads to hot and cold composite curves of the streams, which need heating and cooling, respectively. If rhCp is constant, q versus T would be a straight line 1

(4.226)

dT = ~ d q thC P

Each temperature is a fixed value on the vertical axis, and enthalpy change rates are relative quantities. We estimate the enthalpy changes rather than absolute enthalpies, and the horizontal location of a composite line on the diagram is arbitrarily fixed. The location of ATmin on the composite diagram is where the hot and cold curves most closely approach each other in temperature in a vertical direction. We move one of the two curves horizontally until the distance of the closest vertical approach matches the selected ATmin. The overshoot of the hot composite curve represents the minimum cold utility (qc,min) required, and the overshoot of the cold composite curve represents the minimum hot utility (qh,min) required for the process. Above the pinch, only the hot utility is required, while only the cold utility is required below the pinch. No heat should be transferred across the pinch. For estimating the minimum hot and cold utilities required, Linnhoff and Flower (1978a,b) developed the temperature interval method based on the work of Hohmann (1971). Similarly, grand composite curves show the variation of heat supply and demand in a system. These diagrams enable engineers to minimize the expensive utilities, network area, and number of heat exchanger units. Pinch analysis may also lead to optimum integration of distillation columns, evaporators, condensers, furnaces, and heat pumps by reducing the utility requirements. Pinch analysis is utilized widely in industry leading to considerable savings, as it identifies energy targets, minimum driving forces, and capital cost targets. Dhole and Linnhoff (1993, 1994) developed the vapor and liquid composite curves for representing combined heat and mass transfer loss in a column system. Some pinch techniques consist of minimizing pressure drop effects, water and wastewater, and plant emissions. Pinch analysis can specify the exchanged heat and mass between hot/rich and cold/lean streams based on the first and second laws. According to the second law of thermodynamics:

(heatJmasslostbyhotJr,chstreams) (heatJmassainedbycolleanstreams) 0 below the pinch point

below the pinch point

-

(4.227)

Pinch analysis can optimize the combined heat and mass exchanger network and chemical reactor systems with heat exchangers.

4.10.2

Heat Exchanger Network Synthesis and Pinch Analysis

Heat exchanger network synthesis targets energy recovery that will minimize the annualized cost of the equipment and utilities. The synthesis starts by identifying a set of process streams to be cooled and a set of process streams to be heated. We also need flow rates, heat capacities, and inlet and outlet temperatures for all the process streams. In addition, we need to identify the available utilities, their temperatures, and their costs per unit of heat provided or removed. An energy-efficient heat exchanger network will result in a trade-off between the energy recovered and the capital costs involved in this recovery. The need for process energy integration increases as processes become more complex. Heat exchanger network synthesis requires mass and heat balances often with the following assumptions: (1) use of single-pass shell-and-tube exchangers, (2) no phase change of process streams, (3) equal values of overall heat transfer coefficients for exchangers between two process streams, and between process and utility streams, (4) temperature-independent heat capacity of process streams, (5) constant minimum approach temperatures for exchangers between two process streams and between process and utility streams. More sophisticated techniques can solve problems with multiphase shell-and-tube exchangers, phase changes of process streams, and varying overall heat transfer coefficients. We may analyze a heat exchanger network as a single heat

4.10 Pinch analysis

249

EXs,ci

EXsq EXw,hj

Heat exchanger network

EXs,hj

Exwq

Exw,ci Figure 4.42. Heat exchanger network representation as a single operation.

transfer operation (see Figure 4.42) with its minimum overall heat supply and minimum heat removal requirements. We can determine these minimum requirements by conducting a pinch analysis. The exergy flows corresponding to the heat supplied to the process Exsq and the heat withdrawn Exwq are

EXsq-qh,min{1--T~--)

(4.228)

m,n/l

(4.229)

The terms qh,min and qc,minare the energy load targets, TH and Tc the temperatures of hot and cold utilities, and To the temperature of the environment. Assuming that Tc - T0, the term Exwq becomes zero and the intrinsic efficiency is expressed as

Z i Exw, ci Tim -- Z / ( E X s , h /

The primary

Exs,ci

(4.230)

-- EXw, h/ ) + Exsq

Exp and transformed exergy EX t loads are Exsq Exc

(4.231)

~-~ i EXs'hj -- EXw'hj

(4.232)

Exp-

EX t z

Ex c The term Exc is the exergy consumed by the overall processes. We may combine the pinch analysis with the exergy analysis to reduce the number of unit operations and thermodynamic parameters, leading to an optimum operation. The minimum number of heat exchangers needed is obtained by (4.233)

NHx - N h + N c + Nhu + Ncu - 1

where NHx is the number of heat exchangers, Nh and Nc the number of hot and cold streams, and Nhu and Ncu the numbers of hot and cold utilities, respectively. There is not a unique network for any but a two-stream heat exchange problem, and the design engineer needs both insight and creativity. After establishing the minimum number of heat exchangers, we identify the stream for each exchanger by a heat balance. For negligible heat gains or losses from the exchanger, the heat balance equation is

-+-

0 - q - ~XH - ( MCAT)hot stream (MCAT)cold

stream

(4.234)

250

4.

Usingthe second law: Thermodynamic analysis

Some guidelines for heat exchanger network synthesis are: 1. Do not transfer heat across the pinch point. For example, any process stream heat that is transferred from one side of the pinch point to the other side only increases the requirements for both utilities. The optimal network uses the least number of heat exchangers. 2. Do not use a hot utility below the pinch point, and a cold utility above the pinch point to avoid needing more utility. 3. No heat exchanger should have an approach temperature less than the specified ATmin. 4. On the cold composite curve, each stream that is to be heated must enter or leave an exchanger at the pinch point. On the hot composite curve, each stream that is to be cooled must enter or leave an exchanger at the pinch point. 5. Start the analysis of exchangers in the sink and source sections at the pinch point where all temperatures are fixed. 6. A point of discontinuity on a composite curve indicates the addition or removal of a stream. Added or removed streams must enter or leave an exchanger at the temperature where the discontinuity occurs. 7. If a discontinuity occurs in a stream curve within a utility section, it may be possible by means of the adjacent process section to meet the duty of the stream by leaving the curve at the discontinuity and still not violate the ATmin. Doing so reduces the required number of exchangers by 1 without changing the utility requirements and often is an economic choice. 8. If there are only two streams in a section, they both go to the one exchanger that is reserved for the section. 9. If there are three streams in a section, the stream with the largest change in enthalpy should be split across two exchangers to satisfy the heat duties for each of the other two streams. 10. If there are four streams in a section we need three heat exchangers. If three streams are either heated or cooled, then the fourth stream is split into three flows to satisfy the heat duties for the other three streams. 11. If there are more than four streams in a section, the use of a computer-based algorithm may be necessary. 12. Avoid loops in the heat integration network.

Example 4.29 Minimum utilities by composite curve method Table 4.19 shows the hot and cold process streams and their heat capacities for the process shown in Figure 4.43. (a) Construct the balanced composite curves for the process with ATmm= 20 and 10°C, and compare the amounts of hot and cold utilities needed. (b) Suggest a heat exchanger network system for ATmin - 20°C. Solution: Assume that heat capacities of hot and cold streams are constant. Figure 4.43 shows the process flowchart, while Table 4.19 displays the hot and cold streams with the heat capacity data. The two reactant streams, each at 25°C, are to be heated to 180°C and fed into a reactor. The reaction is endothermic, and the product stream leaves the reactor at 150°C. After further heating the reactant stream to 250°C, it becomes the feed to the distillation column to recover the products. The liquid distillate product is at 190°C and cooled to 25°C for storage. The bottom product is cooled from 260 to 50°C. A review of the streams in Table 4.19 shows that only stream 1 is to be heated from 25 to 150°C. Streams 1 and 2 are heated from 150 to 180°C. Stream 2 only is heated from 180 to 250°C. When there is more than one stream in an interval, we sum the heat capacity rate values for all the streams. The same procedure is followed for the streams to be cooled. We estimate the enthalpy change rates by selecting a baseline value for the enthalpy change rate at one stream temperature. As Table 4.20 shows, a starting enthalpy change rate is chosen as 10,000 kW at 25°C for the streams to be heated, while for the streams to be cooled, the base value chosen is 15,000 kW at 260°C. We will adjust these arbitrary values once we have a unique composite diagram based on a specified ATmin.We add the enthalpy change rates to the initial enthalpy change rate values to tabulate the enthalpy rate values with the corresponding temperatures as seen in Table 4.20. We plot the sets of temperature versus enthalpy rate values in Figure 4.44. This is a composite diagram for the heat integration problem. It is apparent from the figure that the closest vertical approach of the two curves occurs at an enthalpy change rate of 10,000 kW. This is the pinch point for the two composite curves, and occurs where the temperature of the streams that are to be heated is 120°C and the temperature of the streams that are to be cooled is --- 153°C. This ATmin of 33°C is simply a consequence of the starting enthalpy rates that were initially chosen. To achieve a mTmin o f 20°C, we move one of the curves horizontally to bring the two curves closer together. One way of doing this is to move the curve representing the streams that are to be cooled to the right, so that a

4.10

Pinch

analysis

251

Table 4.19 Hot and cold process stream conditions in Example 4.29 Ti,1 (°C)

Stream

Tout (°C)

MC = rhCt) (kW/°C)

C1 C2

Reactor feed Reactor effluent

25 150

180 250

40 55

H1 H2

Bottoms Distillate

260 190

50 25

35 25

REACTOR

q -- t h C p A T (kW) 6200 5500 11700 -7350 -4125 -11475

COLUMN

Figure 4.43. Process flowchart displaying hot and cold process streams.

Table 4.20 Initial temperature interval in Example 4.29 Stream

Initial temperature

Temperature interval t °C 1

MC (kW/°C)

q (kW)

Initial enthalpy selection

interval

T (°C)

C1 C1 and C2 C2

25 150 180

15(1 180 250

40 95 55

5000 2850 3850 11700

H1 H2 and H1 H2

260 190 50

19(1 5(1 25

35 60 25

-2450 -8400 -625 - 11475

q (kW)

25 150 180 250

10000 15000 17850 21700

260 190 50 25

15000 12550 4150 3525

temperature of 140°C is intercepted at an enthalpy rate of 10,000 kW. If the curves for the required heating and cooling utilities are included in the composite diagram, the diagram is called a balanced composite diagram, as shown in Figure 4.45. Figure 4.46 shows the balanced composite diagram for an approach temperature of 10°C. Table 4.21 shows the revised enthalpy rates for an approach temperature ATmin -- 20°C, while Table 4.22 shows the revised enthalpy rates for an approach temperature ATmi n -- 10°C. (a) Utility requirements

Minimum hot utility required: It is clear from the composite diagrams in Figures 4.45 and 4.46 that above a cold stream temperature of--~ 190°C, there is no hot stream curve above the cold stream curve. Since all heat transfer is vertical on a composite diagram, there is no process stream available to heat the cold stream from 190 to 260°C. Therefore, a hot utility provides this heat, which is called the minimum hot utility requirement, for the process in Figure 4.43 with a specified ATmi n o f 20°C.

252

4.

Using the second law: Thermodynamic analysis

280 260

/

240 220 200 180 160 140

o

120 100 80

/

60 40 20 0 0

5000

10000

15000

20000

25000

q, k W

Figure 4.44. Initial composite diagram in Example 4.29.

280 A

260 ' 240-

!

220 200 180 160 °

140 120

-

100

-

/

80-

60402000

5000

10000

15000

20000

25000

q, k W

Figure 4.45. Composite diagram with 20°C approach temperature.

From Figure 4.45 and Table 4.21, the minimum hot utility for a specified ATmi n of 20°C is qH = 3050 kW. From Figure 4.46 and Table 4.22, the minimum hot utility for a specified ATmi n of 10°C is qH -- 2450 kW. Minimum cold utility required: Similarly, below a hot stream temperature of-vg0°c, there is no cold process stream available to cool the hot process streams. Thus, a cold utility must be used to remove this heat. The corresponding heat is called the minimum cold utility requirement. Table 4.21 shows that the minimum cold utility is 2825 kW. From Figure 4.45 and Table 4.21, the minimum cold utility for a specified ATmi n o f 2 0 ° C is qc -- 2825 kW. From Figure 4.46 and Table 4.22, the minimum hot utility for a specified A T m i n of 10°C is qH -- 2225 kW. This simple analysis shows that the smaller approach temperature reduces the utilities needed.

(b) Heat exchanger network system For this process with a ATminof 20°C, various heat exchanger networks can be devised. Generally, there are four sections in most network problems (see Figure 4.47). These are identified from a balanced composite diagram by

4.10

253

Pinch analysis

280 260 -~

•

.

.

.

.

+~

I i

240 -

/

220 -

i i

4--

200 180 160 140 ............ 120 100 -

6o

i

40 -4

I

20 q o!

I

()

50()()

10000

15000

20000

25000

q, k W

Figure 4.46. Composite diagram with 10°C approach temperature.

Table 4.21 Revised enthalpy for an approach temperature of 20°C in Example 4.29 Stream

C1

Revised

Required

MC

temperature

temperature

(kW/°C)

interval

interval

25

15O

q (kW)

Revised enthalpy selection

q(kW)

40

5000

T(°C)

10000

25

15000

150

C 1 and C2

150

180

95

2850

17850

180

C2

180

_5()

55

3850

21700

250

11700 18650

260

H1

260

190

35

-2450

16200

190

H2 and H 1

190

50

60

- 8400

7800

50

50

25

25

- 625

7175

25

qcol&min

2825

kW

qhot ....ill

3050

kW 'T

H2

-11475

2~T

20

~(

drawing three vertical lines at the pinch point, the hot utility initially required, and the cold utility initially required. This divides the diagram into four distinct sections which are, moving from left to right, the cold utility section, the process exchange section below the pinch, the process exchange section above the pinch, and the hot utility section. We count the number of streams participating in the heat exchange, including any utility stream, in each of the four sections. Within any one section, we count each stream only once, and each stream is counted in every section in which it appears. The minimum number of heat exchangers required for a given composite diagram can be obtained from such a diagram. The number of exchangers required in a section is NHx -- N S - 1

(4.235)

The minimum number of heat exchangers required for the problem is determined with the aid of Figure 4.47 and Eq. (4.234). On the graph, vertical lines are drawn to divide the curves into four independent exchange sections. The cold utility section shows three streams and requires two exchangers. The process section below the pinch requires two exchangers since there are three streams. The process section above the pinch requires three exchangers

254

4.

Usingthe second law: Thermodynamic analysis

Table 4.22 Revised enthalpy for an approach temperature of 10°C in Example 4.29 Stream

Revised

Temperature interval

temperature

MC (kW/°C)

q (kW)

selection

interval

C1 C 1 and C2 C2

25 150 180

150 180 250

40 95 55

5000 2850 3850 11700

H1 H2 and H 1 H2

260 190 50

190 50 25

35 60 25

-2450 - 8400 -625 -11475

qcold,min qhot,min AT

2225 2450 10

kW kW °C

260 240

!

t,iiiiiiill

200220 ] - - 1 tii 80

[..,

[ Below pinch [ section '

"-H1 ........ :i'..........

COldsectionUtility,]iiiiii!iiiiill ...........

iiil

'::

Revised enthalpy q (kW)

T(°C)

10000

15000 17850 21700

25 150 180 250

19250 16800 8400 7775

260 190 50 25

C2

,4o'6°.......... iiiiiiiiiiiiii iii iii ,

..... ;

ection

100 ............ i ......... ~ ............. 80 .................. i..... HI&H~ -------/____/__ ..... ,............ ~................ 60 ........................ Above pinch ........ I~ section 40 ................................. i C1 20 ................. !_H__2_:_........................................................... 0 0

5000

10000 15000 q, kW

20000

25000

Figure 4.47. Sections and streams above and below the pinch for an approach temperature of 20°C.

since there are four streams. Finally, there are two streams in the hot utility section, and so, one exchanger is necessary. Therefore, the minimum total number of exchangers required is 8. From the pinch decomposition of hot and cold streams (Figure 4.48) and a heat exchanger temperature profile (Figure 4.49), we may consider the possible locations of the heat exchangers at the pinch. Energy balances for the hot and cold streams yield Thi -- Tho -- M Cq h

(4.236)

Tci - Tco - MC q c

(4.237)

4.10

255

Pinch analysis

MC, kW/°C Pinch HI, 260 °C

H1

170 °C

~-

35

H2 " ~-

25

170°C

H2, 190 °C 170°C

170°C

C1

C1, 25 °C :e 150 °C

40 150 °C

C2, 250 °C 55 150 °C

qhot,min = 3050 kW

qcold,min = 2825 kW

Figure 4.48. Pinch decomposition of the hot and cold streams.

Thi MC h Tho AT l

MC~

Tci Figure 4.49. Temperature profile for a countercurrent heat exchanger.

Subtracting these equations, we find MC c (Thi -- Tco ) -

(Tho -- Tci ) -- A T 2 - A T 1 = q

MC h )

MChMCc

(4.238)

Above the pinch, by arbitrarily stating that AT1 ~Tmin, we have =

MCc A T 2 - ATmi n =

q

- MCh /

MChMCc

(4.239)

For AT2>-ATmin, the above equation requires that MCc-> MC h for a feasible match since q > 0 and the heat capacities are positive. Therefore, we match the lower outlet temperature of the hot streams to the lower inlet temperature of the cold streams that satisfy MCc-> MC h. Below the pinch, considering AT2 ~Tmin, we have =

MC c AT 1 -ATmi n

=-q

MC h )

MChMCc

(4.240)

256

4.

Using the second law: Thermodynamic analysis

H2, 25 °C CW-"

H1, - 50 °C -" CW

L.. H2, 86.66 °C 8 '1-" CW 1

1541.5kW

I [

1283.1 kW

~

7

L.. H1, 86.66 °C I-" I "- CW I "-

Cold utility section

H1, 86.66 °C -" ..[, "-I Cla, 25 °C H2,_.. 86.66 °C I -" -I

HI, 170 °C 2916.9 kW Cla, 150 °C

2083"5kW 5

I~. H2, 170 °C I-"

Clb, 25 °C

Clb, 150 °C Below the pinch

H2, -" 170 °C I -" .._ C2, 150 °C

4

[-"

"-I

H1,170 °C

I

1202.95 kW 3

"-i

H1,204.37 °C ._

"C2, 159.1 °C

i_, HI' 204.37

I .._

C1,150 °C

i_. H2, 190 °C 500 kW

i

1947 kW

2

C2, 159.1 °C

° C

~1-" "C1,180.1 °C

I-

H1,260 °C -_

C2, 194.5 °C

Above the pinch

...................................................................................................

C2, 250 °C Steam

I ] ~1

14 C2, 194.54 °C 3050 kW

1

] I

~ Steam

Hot utility section

...................................................................................................

Figure 4.50. Stream matching in various sections shown in Figure 4.47.

For AT1 -----ATtain, the above equation requires that MC h -> MCc for a feasible match. Therefore, we match the higher outlet temperature of the cold streams to the higher inlet temperature of the hot streams that satisfy MC h -> MC c. If necessary, the streams may be split to satisfy the conditions. Starting with the cold utility section, and referring to Figure 4.47, we match streams to obtain the desired heat exchange as seen in Figure 4.50. For example, for heat exchanger 6, the following energy balance yields the outlet temperature of the hot streams T = 86.66°C: q = MCc] (AT) = (MChl + MCh2)(AT) = 40(150 - 25) = 60(170 - T) Figure 4.51 displays a possible heat exchanger network system with eight heat exchangers in total. As seen, the cold stream splits into two with Cla of 41.7% and Clb of 58.3% of the C1 stream for cooling H1 and H2 streams in the two heat exchangers. Heat duties of each exchanger are estimated from the energy balances. This arrangement may be modified according to the economic analysis of a particular plant. For example, the total number of heat exchangers in the cold utility section may be reduced to a single exchanger. Consequently, the number of heat exchangers may be reduced to 5, as Eq. (4.233) suggests. This aspect is elaborated in the next example.

4.10

257

Pinch analysis

C2(250 °C)

3050 kW 1

H2(190 °C)

1

C2( 194.54 °C)

"

. . . . . . . . . .

H 1(260 °C)

-~'' I

50O kW

1947 kW

C2( 150 °C) H1(204.37 o c ) ;

!

C2(1591 °C)

Cooling water

C 1( 180 °C)

H2( 170 °C)

1202.95 W I t

C1(150 °C)

Hl(170 °C)

~

"l o4,~

i °.~ 2083.5 kW

~,~kW

......i

~ Cla

H 1(86.66 °C)

'~

Clb

1283.1 kW

1'

t H2(86.66 °C)

/

1541.5 kW

i

t

I

C1(25 °C)

Figure 4.51. A heat exchanger network for Example 4.29.

Table 4.23 Cold and hot streams for Example 4.30 Streams

Ti, (°C)

Tout (°C)

C1 C2

20 120

135 200

H1 H2

150 180

45 80

MC = filCt~ (kW/°C) 37 33.5 36 40

q=

rhCpAT (kW)

4255 2680 6935 -3780 -4000 -7780

Example 4.30 Pinch analysis by temperature interval method and grand composite curve Table 4.23 shows hot and cold streams. (a) Construct the balanced grand composite curve for the process with ATmi n = 10°C, arid find the minimum hot and cold utilities needed. (b) Suggest a heat exchanger network system. Table 4.24 shows the adjusted temperature intervals of the hot and cold streams for an approach temperature of 10°C. Therefore, the heat transfer calculations account for the specified approach temperature, and hence the smallest deriving force for heat transfer. The adjusted values in the temperature interval start with the highest temperature. Table 4.24 also shows the heats of hot and cold streams based on the adjusted temperature intervals. For example, there is no hot stream available to heat the cold stream C2 above 180°C. Therefore, the heat of the hot stream is zero. On the other han& between 180 and 150°C only the hot stream H2 is used to heat the cold stream, and we estimate the heat available q -- MCh2 (AT) -- 4 0 ( 1 8 0 - 1 5 0 ) - - 1200 k W

258

4.

Usingthe second law: Thermodynamic analysis

Table 4.24 Temperature interval method for an approach temperature of 10°C Heats of hot streams

Temperature intervals

Heats of cold streams

Grand composite curve

(kW)

(°C)

(kW)

(kW)

Available

Cascaded

qu

~qu

H2

The t

H1

C2

C1

Tcold Required

Cascaded

qc

~qc

2OO

210

0

0

7 0

1005

180

170

1200

1005

6 1200

1005

150

2010

140

380

167.5 1580

135

145

1140

2177.5

4 2720

1057.5

130

120

3800

3235

3 6520

1850

80

70

5085

1260

1295 7780

45

35

0

6380

1 7780

555

30

20

220

180

Cascaded ~q

Adjusted Eq

0 -1005 0 195 0 212.5 0 82.5 0 1950 0 -35 0 -555 0

0

1005

200

-1005

0

180

-810

195

150

-597.5

407.5

135

-515

490

120

1435

2440

80

1400

2405

45

845

1850

20

~ I hot,rain= 1005 kW

....

q',. _ - ] .

,

200

6935

Net q = qu - qc

,/"

~Pinch

160

140

k

120 r~ o

100

80

60

/

40

)'

q,cold.min = 1850 kW

20 mm

0

m

~

~mm

---.

500

---

mm

m

i~

1000

~.

am

,mm

1500

-,-

,

2000

2500

3000

q, kW

Figure 4.52. Grand composite curve for Example 4.30.

T (°C)

4.10

259

Pinch analysis

C2, 200 °C

I .............

Steam

,,._1 "-I

I_.., C2, 170 °C

1005 kW

I" 1 I

~ Steam

180

170

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

H2b, 80 °C

I

1_.. H2b, 180 °C

.2

kW

I r

C2,120°C

"-I ...........................

H2a, _, 159.6 °C [ -"

I

C2,170°C

i4 H2a, 180 °C 474.1 kW 3

v

....

C1, 122.18 ~-C

H1, 45 °C -"

.... 3780 kW

.~J

-"

"-I

H1,150 °C

4 ..... ,IJ-" ]-"

C1, 20 °C

C1,122.8 °C

H2a, 80 °C CW

y

C1,135 °C

,

-"

1850 kW

~

H2a, 159.6 °C -"

5

~- CW

Figure 4.53. Matching the streams for heat exchangers.

Steam

. . . . . . . . . .C2(200 . . °C) 1005 kW

H2(180 °C)

C1(135 °C)

1 Steam __.__._j

.

C2(170 °C)

(b)

(a)

(0.419)

......

l

(0.581)

' W l C2(120 o c ) 1 H2b(80 °C) H2a(159.6 Cooling water

[ ... ; 1850 kW

oo,in 1' water

i

1(122.18 °C)

1

Hl(150 °C)

H2a(80 °C)

I C1(20 °C) H1(45 °C)

Figure 4.54. A heat exchanger network system for Example 4.30.

260

4.

Usingthe second law: Thermodynamic analysis

However, between 150 and 145°C, we consider both the hot streams H2 and HI, and we have q = (MCh2 4- M C h l ) ( A T )

=

(36 + 40)(150 - 145)= 380 kW

Similarly, we estimate the required heats for the cold streams. These estimations yield the hot cascaded heats (~q) for the cold composite curves. In Table 4.24, the last column displaying the adjusted cascaded heats yields the grand composite curve indicating a pinch point at 180°C on the hot side. The grand composite curve is the result of overall net heat flows, including the utilities needed. It allows the engineer to determine the amount and the type of utilities needed. Figure 4.52 displays the grand composite curve with the pinch point at which the adjusted cascaded heat becomes zero. The grand composite curve shows that we need a hot utility of 1005 kW and a cold utility of 1850 kW for a specified approach t e m p e r a t e of 10°C. Figure 4.53 shows a possible matching between the hot and cold streams starting from the hot utility above the pinch. Based on these matchings, Figure 4.54 shows a heat exchanger network system. As there are a total of four hot and cold streams and a total of two hot and cold utility streams, ~om Eq. (4.233) we learn that we need a minimum of five heat exchangers. In the network, hot stream H2 is split into two. H2a has 58.1% of the hot stream H2 and heats the cold stream C1, while H2b heats the cold stream C2.

4,10.3

Distillation Column Targets

A "practical near-minimum thermodynamic condition" targets a reversible column operation at minimum reflux with appropriate heat integration and hence negligible entropy production. To achieve this, heaters and coolers with appropriate duties would operate at each stage; the reflux ratio would be close to its minimum, and hence the operating line approaches the equilibrium curve. This would correspond to the distribution of reboiling and condensing loads throughout the column, and hence over the temperature range of the operation. The Aspen Plus column-targeting tool for thermal analysis and hydraulic analysis is helpful in identifying the targets for appropriate modifications in order to reduce utility and capital costs, improve energy efficiency, and decrease column bottlenecking. The column-targeting tool of Aspen Plus produces the enthalpy and the exergy loss profiles based on the practical near-minimum thermodynamic condition. The enthalpy estimations take into account the thermodynamic losses due to column design and operating conditions, such as pressure drop, multiple feed and side products, as well as side heat exchangers. The pinch point in distillation requires that there should be no side reboiling below the pinch and no side condensing above the pinch in heat-integrated columns.

4.10.4

Column Grand Composite Curve

To analyze the energy-saving potential of distillation columns, it is customary to construct the temperature-enthalpy and stage-enthalpy curves, called column grand composite curves. Column grand composite curves are based on the practical near-minimum thermodynamic condition approximation proposed by Dhole and Linnhoff (1993, 1994), and show the theoretical minimum heating and cooling duties within the temperature range. The stage-enthalpy calculations take into account losses or inefficiencies stemming from the actual column design, such as pressure drops, multiple side products, etc. Column grand composite curves display the net enthalpies for the actual and ideal operations at each stage, and the cold and hot heat utility requirements. Therefore, the area between the actual and the ideal operations in a column grand composite curve should be small for a thermodynamically efficient operation. Column grand composite curves are constructed by solving the mass and energy balances for a reversible column operation. Simulation packages are making column grand composite curves readily available even for multicomponent, complex distillation column operations such as crude oil distillation. These simulators enable the process engineer to assess the performance of an existing operation and explore the possibility of reducing utility costs by improving efficiency in energy usage. Column grand composite curves can identify targets for restructuring and modification, and may be helpful in suggesting retrofits. Some of the retrofits consist of feed conditioning (preheating or precooling), feed splitting, reflux adjustments, and adding side condensers and reboilers. These retrofits target a practical near-minimum thermodynamic loss. The stage-enthalpy and temperature-enthalpy profiles represent the theoretical minimum heating and cooling requirements over the stages or the temperature range. Using the equilibrium compositions of light L and heavy H key components obtained from a converged simulation, we estimate the minimum vapor and liquid flow rates leaving the same stage with the same temperatures from the following mass balances: Vmin = ~ ( D L 4- LminXL) YL

(4.241)

4.10

261

Pinch analysis

Lmin _

1 (VminY H _ D H ) xH

(4.242)

where x* and y* are the equilibrium mole fractions of liquid and vapor streams, Lmi n and Vminthe minimum amounts of the liquid and vapor streams, and D the distillate. The enthalpies for the minimum vapor and liquid flows are obtained from the molar flow ratios Hr,

-Hv

* ( Vmm) -7

(4.243)

(4.244)

HL,,~,,, - H L

where V* and L* are the molar flows of equilibrium, and Hv* and HL* the enthalpies of equilibrium vapor and liquid streams leaving the same stage, respectively. From the enthalpy balances at each stage, the net enthalpy deficits are obtained H d e f = HLmH~ -- Hl]~,,, ~ + H D

Hdef-HLm,,-Hi

+HD

(before t h e f e e d stage)

-- Hfeed

(after the feed stage)

(4.245) (4.246)

After adding the individual stage-enthalpy deficits to the condenser duty, the enthalpy values are cascaded, and plotted in column grand composite curves. This is called the top-down calculation procedure, which will be the same as the bottom-up calculations for a stage without any feed. At the feed stage, mass and energy balances differ from a stage without feed, and finite changes of composition and temperature disturb the reversible operation. For the two procedures to yield similar results, the enthalpy deficit at the feed stage becomes

Hdef, F --

qc + D H D + H L (xD - YF,) _ HV (XD -- XF,)

(4.247)

The values of YF and x F may be obtained from an adiabatic flash for a single phase feed, or from the constant relative volatility estimated with the converged compositions at the feed stage and feed quality. This procedure can be reformulated for multiple feeds and side products as well as different key components. A pinch point near the feed stage occurs for nearly all binary ideal mixtures. However, for nonideal multicomponent systems, the pinch point exists in rectifying and stripping sections. A horizontal distance between the column grand composite curve pinch point and the vertical axis represents excess heat, and therefore the scope for reduction in reflux ratio. For smaller reflux ratios, the column grand composite curve will move toward the vertical axis, and hence reduce the reboiler and condenser duties, which may be estimated by

qR -- qR,min -- q c -- qc,min -- D ( A H v )

R - XD, - YF, YF -- XF

(4.248)

where AHv is the heat of vaporization. The horizontal distance of the column grand composite curve from the temperature axis determines the targets for installing a side reboiler or side condenser at suitable temperatures (or stages). On the other hand, a sharp change in the enthalpy represents inappropriate feed conditioning, such as poor feed quality or nonoptimal temperature. For example, a sharp change on the reboiler side may be due to a subcooled feed, and a feed preheater can be installed. Feed conditioning is usually preferred to side condensing or reboiling, since the side heat exchangers are effective at suitable temperature levels only.

Example 4.31 Column grand composite curves in a distillation column with a five-component mixture The column has 14 stages with a feed stage of 7 and a reflux ratio of 8.87. The feed has five components of ethane, propane, n-butane, n-pentane, and n-hexane. Table 4.25 shows the configuration of the column. Thermodynamic

262

4.

Usingthe second law: Thermodynamic analysis

properties are estimated by the Peng-Robinson equation of state. The column has a condenser duty of 3395.336kW, and a reboiler duty of 3432.206 kW. The condenser and the reboiler temperatures are 319.3 and 400.2 K, respectively. The reference temperature (dead state temperature) To is assumed to be 300 K. The simulation results show the lost work = 531.37kW, Wmin= 117.49kW, and the exergetic efficiency is 18.1%. Figures 4.55 and 4.56 show the column grand composite curves (T-H and stage-enthalpy), which can be useful for identifying the targets for feed preparation and location, reflux ratio, and heat integration modifications. The column grand composite curves indicate distortions as significant projections around the feed stage location (stages 8 and 9 pinch point), indicating that the current feed stage is inappropriate. To compensate for the inappropriate feed stage location, extra local reflux may be needed. Besides that, a feed stage too high or too low in the column will display sharp enthalpy changes on the condenser and the reboiler, respectively. The sharp enthalpy changes on the grand composite curves may also indicate the need for an adjustment of the feed quality. A sharp enthalpy change on the reboiler side suggests that the feed is subcooled, and a preheater should be installed. The horizontal gap between the pinch point and the ordinate in Figure 4.55, which is ----200kW, indicates the possible reduction in heat duties by reducing the reflux ratio at the expense of increasing the number of stages to achieve the specified separation. Obviously, the increase in the capital cost for a taller column should be weighed against the

Table 4.25 Column configuration for a five-component distillation used in the Aspen Plus simulator with the Peng-Robinson model: qc = 3395.3367 kW; qR--3432.2069 kW; Number of stages = 14; Location of feed stage = 7; Reflux ratio = 8.87.

Flow (kmol/h) Pressure (Atm) Temperature (K) Vapor fraction Enthalpy (kJ/kmol) Entropy (kJ/kmolK) Compositions Ethane n-Propane n-Butane n-Pentane n-Hexane

450

I

=

Feed

Distillate

Bottom

453.59 17.01 380.37 0.2830 1.3457 x 10s -403.93

102.51 16.87 319.30 1.0 1.0284 x 10s -277.61

351.08 17.14 400.26 0.0 1.4346 X 10s -443.76

0.0299 0.1999 0.3699 0.3499 0.0499

0.1326 0.8450 0.0219 1.404 X 10 -4 2.507 X 10-8

4.642 X 10 -6 0.0116 0.4716 0.4521 0.0646

I

IdealProfile Actual Profile

400 ~ - . - - A

© [-,

350

0

500

1000

1500 2000 2500 Enthalpy Deficit, (kW)

3000

3500

Figure 4.55. Column 1 grand composite curve of temperature-enthalpy in Example 4.31 obtained from the simulations with the Aspen Plus RADFRAC block using the Peng-Robinson equation of state.

4.10

263

Pinch analysis

savings in utility costs. Figure 4.55 also shows that the reboiler side is relatively close to an ideal operation while the condenser side is far from an ideal operation. The significant area underneath the pinch point suggests the need for a side condenser at an appropriate temperature. The need for heat integration through side condensing or side reboiling could be quantified from the area between the ideal and actual enthalpy profiles after considering the capital cost increase due to the modification. However, an external modification of feed conditioning is usually preferred to an internal modification of heat integration.

Example 4.32 Column grand composite curves in methanol plant Table 4.16 describes the existing base case operations for columns 1 and 2 of the methanol plant obtained from the converged simulations using the RKS equation of state to estimate the vapor properties. The activity coefficient model, NRTL, and Henry components method are used for predicting the equilibrium and liquid properties. Column 1 has 51 stages, and operates with a partial condenser with a duty of 1.371 MW at the top, and a side condenser with a duty of 8.144 MW at stage 2. It has no reboiler; however, it receives a side heat stream with a duty of 15.299 MW at the last stage from Section 2 of the plant. The column grand composite curves in

16

. . . . .

i ................

i _..._..__4._.i

-"

IdealProfile Actual Profile

0

500

]000

1500 2 0 0 0 2 5 0 0 Enthalpy Deficit, (kW)

3000

3500

Figure 4.56. Column 1 grand composite curve of stage-enthalpy in Example 4.31 obtained from the simulations with the Aspen Plus RADFRAC block using the Peng-Robinson equation of state.

360 -

350 -

340

'~ 330i..., & E 320 -

Actual Profile

310 -

\\

---o--- Ideal Profile 300 -2.5

I 0

, 2.5

, I I 5 7.5 10 12.5 Enthalpy Deficit, MW

i 15

17.5

Figure 4.57. Temperature-enthalpy deficit curves for column 1 of the methanol plant in Example 4.32. Column 1 configuration is given in Table 4.16.

264

4.

Usingthe second law: Thermodynamic analysis 400

390 380 b'"

Ideal Profile

4b

370

"

Actual Profile

ql,

~

360

Ir

j

(

350 .......

-50

....

0

50 100 150 200 Enthalpy Deficit M W

250

300

Figure 4.58. Temperature-enthalpy deficit curves for column 2 of the methanol plant in Example 4.32. Column 1 configuration is given in Table 4.16

Figure 4.57 show that within the rectification section, there exists a significant area difference between the ideal and actual enthalpy profiles, which identifies the scope for side condensing. The extent of the change around the feed stage determines the approximate feed preheating duty required, as the feed at 43.74°C is highly subcooled. Thus, a new heat exchanger with a duty of 1.987 MW is used as the second retrofit for the column, and the feed temperature has increased to 65 from 43.74°C (Demirel, 2006a,b). Table 4.16 shows that column 2 has 95 stages, and a total condenser with a duty of 281.832 MW. It operates with a high reflux ratio, and receives a side heat stream of 18.9 MW at the last stage from Section 2 of the plant and produces two side streams. The second side product is drawn at stage 86 at 361.2 K. Figure 4.58 also shows a significant area difference between the ideal and the actual enthalpy profiles above the feed stage representing the pinch point, and hence suggests side reboiling at appropriate temperatures to decrease the difference. With the two side reboilers, the duty of the reboiler decreases to 52.3 from 282.3 MW (Demirel, 2006a,b).

PROBLEMS 4.1

Methane gas with a flow rate of 15 mol/s is flowing through a throttling valve from 25 bar and 450 K to 1 bar in a steady-state flow process. Determine the lost work if the surroundings are at 298.15 K.

4.2

Fifty moles per second of air is throttled from 100°C and 7 bar to a pressure of 1 bar. Assume that air is an ideal gas with Cp= (7/2)R. Determine the work loss if surroundings are at 298.15 K.

4.3

Ten kilograms per second of superheated steam at 400°C and 1100 kPa is throttled to 125 kPa adiabatically through a valve. Determine the work loss if the surroundings are at 298.15 K.

4.4

In a steady-state mixing process, 50.25kg/s of saturated steam (stream 1) at 501.15 K is mixed with 7.363 kg/s of saturated steam (stream 2) at 401.15 K. The mixer is well insulated and adiabatic. Determine the energy dissipation (work loss) if the surroundings are at 298.15 K.

4.5

In a steady-state mixing process, 50.0 kg/s of saturated steam (stream 1) at 501.15 K is mixed with 17.0 kg/s of saturated steam (stream 2) at 423.15 K. The product steam (stream 3) is at 473.15 K. Determine: (a) The rate of heat loss; (b) The work loss if the surroundings are at 298.15 K.

Problems

265

4.6

In a steady-state mixing process, 15 kmol/s of air (stream l) at 550 K and 2 atm is mixed with 40 kmol/s of air (stream 2) at 350 K and 1 atm. The product (stream 3) is at 300 K and 1 atm. Determine the work loss.

4.7

A steady flow adiabatic turbine receives steam at 650K and 8200kPa, and discharges it at 373.15K and 101.32 kPa. If the flow rate of the steam is 12 kg/s, determine: (a) The maximum work; (b) The work loss if the surroundings are at 298.15 K.

4.8

A hot exhaust gas is heating a boiler to produce superheated steam at 100 psia and 400°F. In the meantime, the exhaust gas is cooled from 2500 to 350°E Saturated liquid water (stream 1) at 14.7 psia enters the boiler with a flow rate of 200 lb/h. Superheated steam (stream 2) is used in a turbine, and discharged as saturated steam (stream 3) at 14.7 psia. Determine: (a) The molar flow rate of the exhaust gas needed; (b) The lost work in each unit. Assume that the surroundings are at 70°E and the heat capacity of the flue gas is Cp = 7.606 + 0.0006077T, where T is in Rankine @ is in Btu/(lbmol R).

4.9

Steam expands in a nozzle from inlet conditions of 500°1,250 psia, and a velocity of 260 ft/s to discharge conditions of 95 psia and a velocity 1500 ft/s. If the flow is at 10 lb/s and the process is at steady state and adiabatic, determine: (a) The outlet temperature; (b) The work loss.

4.10

Water at room temperature of 20°C is throttled from 6.5 to 1 atm in household use. Determine the work loss when 15 L of water is used every day. The surroundings are at 298.15 K. The thermal expansion coefficient ofthe liquid water is/3 =(1/V)(OV/OT)p=2.07 × 10-4K -1

4.11

A steam power generation unit produces 65,000 kW electricity with an efficiency of 70%. It uses a steam (stream 1) at 8200 kPa and 550°C. The discharged stream (stream 2) is at 75 kPa. If the expansion in the turbine is adiabatic, and the surroundings are at 298.15 K, determine: (a) The thermodynamic efficiency; (b) The work loss.

4.12

In an adiabatic compression operation, air is compressed from 25°C and 101.32 kPa to 450 kPa with an efficiency of 0.8. The air flow rate is 15 mol/s. The air is assumed to be an ideal gas. The surroundings are at 300 K. Determine: (a) The ideal work required; (b) The thermodynamic efficiency; (c) The energy dissipated.

4.13

A pump operating adiabatically is pumping water from 20.0°C and 2.337 kPa to a pressure of 9870 kPa. The water flow rate is 10 kg/s. The pump efficiency is 0.77. If the surroundings are at 298.15 K, determine: (a) The ideal work; (b) The thermodynamic efficiency; (c) The dissipated energy.

4.14

A steam power plant uses natural gas to produce 0.1 MW power. A furnace completely burns the natural gas to carbon dioxide and water vapor with 30% of excess air. The flue gas leaves the furnace at 500 K. The combustion heat supplied to a boiler produces steam at 10,000 kPa and 798.15 K, which is sent to a turbine. The turbine efficiency is 0.75. The discharged steam from the turbine is at 30 kPa, and sent to a condenser. The condensed water is pumped to the boiler. The pump efficiency is 0.75. Assume that the natural gas is pure methane gas, and the surroundings are at 298.15 K. Determine: (a) The thermal efficiency of an ideal Rankine cycle; (b) The thermal efficiency of an actual cycle: (c) The work loss of each unit of boiler, turbine, condenser, and pump.

266

4.

Usingthe second law: Thermodynamic analysis

4.15

A steam power plant uses natur3al gas to produce 0.12 MW power. A furnace completely burns the natural gas to CO2 and water vapor with --~25% of excess air. The flue gas leaves the furnace at 465 K. The combustion heat supplied to a boiler produces steam at 9000 kPa and 798.15 K, which is sent to a turbine. The turbine efficiency is 0.7. The discharged steam from the turbine is at 20 kPa, and sent to a condenser. The condensed water is pumped to the boiler. The pump efficiency is 0.70. Assume that the natural gas is pure methane gas, and the surroundings are at 298.15 K. Determine: (a) The thermal efficiency of a Rankine cycle; (b) The thermal efficiency of an actual cycle; (c) The work loss of each unit of boiler, turbine, condenser, and pump.

4.16

In a two-stage continuous compression process, methane (stream 1) enters the first compressor at 300 K and 1 bar. The methane (stream 2) leaves the second compressor at 300 K and 60 bar. The flow rate of methane is 0.5 kg/s. The total power input is 400 kW. The intercooler between the compressors uses cooling water. The surroundings are at 295 K. Determine the energy dissipated.

4.17

In a two-stage continuous compression process, methane (stream 1) enters the first compressor at 300 K and 1 bar. The methane (stream 2) leaves the second compressor at 350 K and 80 bar. The flow rate of methane is 0.6 kg/s. The total power input is 450 kW. The intercooler between the compressors uses cooling water. The cooling water enters the cooler at 295 K and leaves at 305 K. The surroundings are at 295 K. Determine: (a) The cooling water rate; (b) The work loss.

4.18

In a two-stage continuous compression process, air (stream 1) enters the first compressor at 300 K and 1 bar. The air (stream 2) leaves the second compressor at 300 K and 40 bar. The flow rate of air is 0.5 kg/s. The total power input is 350 kW. The intercooler between the compressors uses cooling water. The cooling water enters the cooler at 293.15 K and leaves at 295.15 K. The surroundings are at 298.15 K. Determine: (a) The cooling water rate; (b) The work loss.

4.19

In a three-stage continuous compression process, propylene (stream 1) enters the first compressor at 300 K and 1 bar. The propylene (stream 2) leaves the second compressor at 300 K and 20 bar, and enters the third. The propylene leaves the third compressor at 300 K and 40 bar. The flow rate of propylene is 0.45 kg/s. The total power input is 550 kW. The two intercoolers between the compressors use cooling water. The cooling water enters each cooler at 293.15 K and leaves at 293.15 K. The surroundings are at 298.15 K. Determine: (a) The cooling water rates; (b) The total work loss.

4.20

An ideal Otto cycle operates with a compression ratio of 8.293 ( Vmax/Vmin).Air is at 101.3 kPa and 280 K at the start of compression (state 1). During the constant volume heat addition process, 1000 kJ/kg of heat is transferred into the air from a source at 1900 K. Heat is discharged to the surroundings at 280 K. Determine: (a) The net work output; (b) The work loss at each state.

4.21

An ideal Otto cycle operates with a compression ratio of 9 ( Vmax/Vmin).Air is at 101.3 kPa and 295 K at the start of compression (state 1). During the constant volume heat addition process, 900 kJ/kg of heat is transferred into the air from a source at 1800 K. Heat is discharged to the surroundings at 295 K. Determine: (a) The net work output; (b) The work loss at each state.

4.22

A steam power plant operates on a simple ideal Rankine cycle. The turbine receives the steam at 698.15 K and 4200 kPa, while the discharged steam is at 30 kPa. The mass flow rate of steam is 27 kg/s. In the boiler, heat is transferred into the steam from a source at 1550 K. In the condenser, heat is discharged to the surroundings at 300 K. The condenser operates at 295 K. Determine the work loss at each state.

=

=

Problems

267

4.23

A simple ideal Rankine cycle is used in a steam power plant. Steam enters the turbine at 6600 kPa and 798.15 K. The net power output of the turbine is 35 kW. The discharged steam is at 10 kPa. Cooling water is used in the condenser at a rate of 750 kg/s. Determine: (a) The thermal efficiency; (b) The work loss at each unit.

4.24

A reheat Rankine cycle is used in a steam power plant. Steam enters the high-pressure turbine at 10,000 kPa and 823.15 K and leaves at 4350 kPa. The steam is reheated at constant pressure to 823.15 K. The steam enters the low-pressure turbine at 4350kPa and 823.15K. The discharged steam from the low-pressure turbine is at 10 kPa. The net power output of the turbine is 65 MW. The turbine efficiency is 82%. The pump efficiency is 96%. Cooling water is used in the condenser at a rate of 700 kg/s. Determine: (a) The thermal efficiency; (b) The work loss at each unit.

4.25

A steam power plant is using an ideal regenerative Rankine cycle. Steam enters the high-pressure turbine at 8600 kPa and 773.15 K, and the condenser operates at 30 kPa. The steam is extracted from the turbine at 350 kPa to heat the feedwater in an open heater. The water is a saturated liquid after leaving the feedwater heater. The work output of the turbine is 75 MW. Determine the thermal efficiency and the work loss at each unit.

4.26

A steam power plant operates on a regenerative cycle. Steam enters the turbine at 700 psia and 800°F and expands to 1 psia in the condenser. Part of the steam is extracted at 60 psia. The efficiencies of the turbine and pump are 0.80 and 0.95, respectively. The work output of the turbine is 4000 Btu/s. Determine the work loss at each unit. Assume that the surroundings are at 530 R, and the kinetic and potential energy changes are negligible.

4.27

A steam power plant is using an ideal reheat regenerative Rankine cycle. Steam enters the high-pressure turbine at 9400 kPa and 773.15 K and leaves at 850 kPa. The condenser operates at 15 kPa. A part of the steam is extracted from the turbine at 850 kPa to heat the water in an open heater, where the steam and liquid water from the condenser mix and direct contact heat transfer takes place. The rest of the steam is reheated to 723.15 K, and expanded in the low-pressure turbine section to the condenser condition. The water is a saturated liquid after leaving the water heater at the heater pressure. The work output of the turbine is 80 MW. Determine the work loss at each unit if the surroundings are at 300 K.

4.28

A steam power plant is using an actual regenerative Rankine cycle. Steam enters the high-pressure turbine at 11,000 kPa and 773.15 K, and the condenser operates at 10 kPa. The steam is extracted from the turbine at 475 kPa to heat the water in an open heater. The water is a saturated liquid after leaving the water heater. The work output of the turbine is 90 MW. The pump efficiency is 95% and the turbine efficiency is 75%. Determine the work loss at each unit if the surroundings are at 290 K.

4.29

A steam power plant is using an actual reheat regenerative Rankine cycle. Steam enters the high-pressure turbine at 10,800 kPa and 773.15 K, and the condenser operates at 15 kPa. The steam is extracted from the turbine at 2000 kPa to heat the water in an open heater. The steam is extracted at 475 kPa for process heat. The water is a saturated liquid after leaving the water heater. The work output of the turbine is 95 MW. The turbine efficiency is 82%. The pumps operate isentropically. Determine the work loss at each unit if the surroundings are at 295 K.

4.30

A steam power plant is using a geothermal energy source. The geothermal water is available at 220°C and 200 kg/s. The hot water goes through a flash drum. Steam from the flash drum enters the turbine at 550 kPa and 428.62 K. The condenser operates at 10 kPa. The water is a saturated liquid after leaving the condenser. Determine: (a) The net work output; (b) The exergy losses at each unit; (c) The exergy efficiency of each unit. Assume that the surroundings are at 290 K and the kinetic and potential energy changes are negligible.

4.31

A cogeneration plant is using steam at 5500 kPa and 748.15 K to produce power and process heat. The amount of process heat required is 10,000 kW. Twenty percent of the steam produced in the boiler is extracted

268

4.

Usingthe second law: Thermodynamic analysis

at 475 kPa from the turbine for cogeneration. The extracted steam is condensed and mixed with the water output of the condenser. The remaining steam expands from 5500 kPa to the condenser conditions. The condenser operates at 10 kPa. Determine the work loss at each unit. Assume that the surroundings are at 290 K. 4.32

A cogeneration plant is using steam at 900 psia and 1000°F to produce power and process heat. The boiler produces the steam at a rate of 16 lb/s. The process heat requires steam at 70 psia at a rate of 3.5 lb/s supplied by the expanding steam in the turbine. The extracted steam is condensed and mixed with the water output of the condenser. The remaining steam expands from 75 psia to the condenser conditions. The condenser operates at 3.5 psia. If the turbine operates with an efficiency of 78% and the pumps with 95% efficiency, determine the work loss at each unit. Assume that the surroundings are at 530 R and the kinetic and potential energy changes are negligible.

4.33

A cogeneration plant is using steam at 875 psia and 900°F to produce power and process heat. The turbine operates with an efficiency of 80%. The boiler produces the steam at a rate of 15 lbm/s. The process heat requires steam at 85 psia at a rate of 5 lbm/s supplied by the expanding steam in the turbine. The extracted steam is condensed and mixed with the water output of the condenser. The remaining steam expands from 85 psia to the condenser conditions. The condenser operates at 3.2 psia. The pump efficiency is 85%. Determine the work loss at each unit. Assume that the surroundings are at 540 R and the kinetic and potential energy changes are negligible.

4.34

In a pentafluoroethane (R-125) refrigeration cycle, the saturated R-125 (state 1) enters a compressor at 250 K and 3 bar. The R-125 (state 2) leaves the compressor at 320 K and 23.63 bar, and enters a condenser, where it is cooled by cooling water. The R-125 (state 3) leaves the condenser as saturated liquid at 310 K and 18.62 bar and enters a throttling valve. The partially vaporized R-125 (state 4) leaves the valve at 255 K and 3.668 bar. The cycle is completed by passing the R-125 through an evaporator to absorb heat from the matter to be refrigerated. The R-125 leaves the evaporator as saturated vapor. The evaporator temperature is 275.15 K. The flow rate of R-125 is 0.75 kg/s. The total power input is 60 kW. The cooling water enters the condenser at 293.15 K and leaves at 295.15 K. The surroundings are at 298.15 K. (a) Determine the total work loss. (b) Perform a work loss analysis for each unit.

4.35

In a tetrafluoroethane (R-134a) refrigeration cycle, the superheated R-134a (state 1) enters a compressor at 263.15 K and 0.16 MPa. The R-134a (state 2) leaves the compressor at 313.15 K and 0.8 MPa, and enters a condenser, where it is cooled by cooling water. The R-134a (state 3) leaves the condenser at 281.15 K and enters a throttling valve. The partially vaporized R-134a (state 4) leaves the valve at 0.30 MPa. The cycle is completed by passing the R-134 through an evaporator to absorb heat from the matter to be refrigerated. The R-134a (state 1) leaves the evaporator as superheated vapor. The flow rate of R-134a is 0.2 kg/s. The total power input is 95 kW. The cooling water enters the condenser at 293.15 K and leaves at 296.15 K. The surroundings are at 298.15 K and the evaporator is at 270 K. Perform a work loss analysis for each unit.

4.36

In a tetrafluoroethane (R-134a) refrigeration cycle, the superheated R-134a (state 1) enters a compressor at 253.15K and 0.14MPa. The R-134a (state 2) leaves the compressor at 303.15K and 0.5 MPa, and enters a condenser, where it is cooled by cooling water. The R-134a (state 3) leaves the condenser at 297.15 K and enters a throttling valve. The partially vaporized R-134a (state 4) leaves the valve at 0.32 MPa. The cycle is completed by passing the R-134 through an evaporator to absorb heat from the matter to be refrigerated. The R-134a (state 1) leaves the evaporator as superheated vapor. The flow rate of R-134a is 0.16 kg/s. The total power input is 750 kW. The cooling water enters the condenser at 293.15 K and leaves at 295.15 K. The surroundings are at 298.15 K and the evaporator is at 277 K. Perform a work loss analysis for each unit.

4.37

A refrigerator using tetrafluoroethane (R-134a) as refrigerant operates with a capacity of 100,000 Btu/h. The refrigerated space is at 15°E The evaporator and condenser operate with a 10°F temperature difference in their heat transfer. Cooling water enters the condenser at 70°E Therefore, the evaporator is at 5°F, and the condenser is at 80°F. Determine the work loss at each unit if the compressor efficiency is 85%. Assume that the surroundings are at 70°E

4.38

A refrigerator using tetrafluoroethane (R-134a) as refrigerant operates with a capacity of 140,000 Btu/h. The refrigerated space is kept at 10°F. The evaporator and condenser operate with a 10°F temperature difference

269

Problems

in their heat transfer. Cooling water enters the condenser at 65°E Therefore, the evaporator is at 0°E and the condenser is at 75°E Determine the work loss at each unit if the compressor efficiency is 80%. Assume that the surroundings are at 65°E 4.39

A refrigerator using tetrafluoroethane (R- 134a) as refrigerant operates with a capacity of 250 Btu/s. Cooling water enters the condenser at 70°E The evaporator is at 10°F, and the condenser is at 80°E The refrigerated space is at 20°E Determine the work loss at each unit if the compressor efficiency is 75%. Assume that the surroundings are at 70°E and the kinetic and potential energy changes are negligible.

4.40

A refrigerator using R-134a as refrigerant (tetrafluoroethane) operates with a capacity of 2500 kW. Cooling water enters the condenser at 280 K. Evaporator is at 271.92 K, and the condenser is at 299.87 K in Pr. 4.40. The refrigerated space is at 280 K. Determine the work loss at each unit if the compressor efficiency is 80%. Assume that the surroundings are at 290 K, and the kinetic and potential energy changes are negligible.

4.41

Natural gas is partially liquefied in a Claude process shown in Figure 4.28. It is assumed that the natural gas is pure methane, which is compressed to 80 bar and precooled to 298.15 K. In the expander and throttle, the methane is expanded to 1.325 bar. The methane after passing through the first heat exchanger at state 5 is at 80 bar and 250 K. Thirty percent of the first heat exchange is sent to the expander. Only 8% of the first heat exchange is liquefied. The expander efficiency is 0.78. Determine the work loss in the liquefaction section excluding compression and precooling. Assume that the whole operation is adiabatic, and the surroundings are at 295 K.

4.42

The table below shows the hot and cold process streams and their heat capacities. (a) Construct the balanced composite curves for the process with z~Tmi n -- 1 5 ° C , and estimate the hot and cold utilities needed. (b) Suggest a heat exchanger network system for ATmin - 15°C. Stream C1 C2 HI H2

4.43

Tout (°C)

25 150 260 t90

180 250 50 25

MC = rhCp (kW/°C) 40 55 35 25

The table below shows the hot and cold streams. (a) Construct the balanced grand composite curve for the process with hot and cold utilities needed. (b) Suggest a heat exchanger network system. Stream C1 C2 H1 H2

4.44

Ti,, (°C)

Ti,, (°C)

Tout (°C)

20 120 150 180

135 200 45 80

ATmin -- 2 0 ° C ,

and find the minimum

MC = rhC/, (kW/°C) 37 34 36 40

The table below shows the hot and cold process streams and their heat capacities. (a) Construct the balanced composite curves for the process with ATmin - 10°C, and estimate the hot and cold utilities needed. (b) Suggest a heat exchanger network system for ATmin = 10°C. Stream C1 C2 H1 H2

T,,, (°C)

Tout (°C)

20 t20 150 180

135 200 45 80

MC = rhCl, (kW/°C) 37 34 36 40

270 4.45

4.

Usingthe second law: Thermodynamic analysis

The table below shows the hot and cold streams. (a) Construct the balanced grand composite curve for the process with hot and cold utilities needed. (b) Suggest a heat exchanger network system. Stream C1 C2 H1 H2

4.46

Tin (°C)

Tout(°C)

20 120 150 180

135 200 45 80

ATmi n =

10°C, and find the minimum

MC = rhCp(kW/°C) 37 34 36 40

Use the Aspen Plus simulator with the following input summary to estimate the thermodynamic efficiency of the distillation column: General Simulation with English Units: E psi, lb/hr, lbmol/hr, Btu/hr, cuft/hr. Flow basis for input: Mole COMPONENTS C3 C3H8 / IC4 C4H10-2 / NC4 C4H 10-1 / IC5 C5H12-2 / NC5 C5H12-1 / NC6 C6H 14-1 FLOWSHEET BLOCK COLUMN1 IN = FEED OUT = DIST BOTTOM PROPERTIES PENG-ROB PROPERTIES NRTL-2 STREAM FEED SUBSTREAM MIXED PRES = 4.4 < a t m > VFRAC = 0. MOLE-FLOW = 100. MOLE-FLOW C3 5. / IC4 10. / NC4 30. / IC5 20. / NC5 15. / NC6 20. BLOCK COLUMN 1 DSTWU PARAM LIGHTKEY = NC4 RECOVL = 0.9908 HEAVYKEY = IC5 & RECOVH = 0.01124 PTOP = 4.4 < a t m > PBOT = 4.4 < a t m > RR = 1.8

4.47

Use the Aspen Plus simulator with the following input summary to estimate the thermodynamic efficiency of the distillation column: Input Summary: IN-UNITS ENG COMPONENTS C2H6 C2H6 / C3H8 C3H8 / C4H 10-1 C4H 10-1 / C5H12-1 C5H12-1 / C6H14-1 C6H14-1

271

Problems

FLOWSHEET B L O C K D1 IN = FEED OUT = DIS BOT PROPERTIES RK-SOAVE PROPERTIES CHAO-SEA STREAM FEED S U B S T R E A M MIXED TEMP = 225 PRES = 250 MOLE-FLOW C2H6 30 / C3H8 2 0 0 / C 4 H 1 0 - 1 370 / C5H12-1 & 350 / C6H14-1 50 BLOCK D 1 DSTWU PARAM L I G H T K E Y = C3H8 RECOVL = 0.955 HEAVYKEY = C4H10-1 PTOP = 248 PBOT = 252 RDV = 1 RR = - 1 . 7 5

4.48

& RECOVH = 0.0134

Thermal analysis of the Aspen Plus simulator produces column grand composite curves of temperatureenthalpy and stage-enthalpy curves for rigorous distillation column simulations. These types of calculations are performed for RADFRAC columns. Using the following input summary for a RADFRAC column, construct the temperature-enthalpy, stage-enthalpy curves, and the stage-exergy loss profiles and assess the thermodynamic performance of the column by estimating and plotting the unavoidable part of exergy loss: Input summary: General Simulation with English Units: E psi, lb/hr, lbmol/hr, Btu/hr, cuft/hr. Flow basis for input: Mole COMPONENTS C3 C3H8 / IC4 C4H10-2 / NC4 C4H 10-1 / IC5 C5H12-2 / NC5 C5H12-1 / NC6 C6H 14-1 FLOWSHEET B L O C K RADFRAC 1N = FEED OUT = DIST BOTTOM PROPERTIES PENG-ROB PROPERTIES NRTL-2 STREAM FEED S U B S T R E A M MIXED PRES = 4.4 < a t m > VFRAC = 0. M O L E - F L O W = 100. MOLE-FLOW C3 5. / IC4 10. / NC4 30. / IC5 20. / NC5 & 15. / NC6 20. B L O C K RADFRAC RADFRAC PARAM NSTAGE = 28 COL-CONFIG C O N D E N S E R = TOTAL FEEDS FEED 14 PRODUCTS DIST 1 L / BOTTOM 28 L P-SPEC 1 4.4 < a t m > / 2 4 4.4 < a t m > COL-SPECS D:F = 0.44 M O L E - R R = 1.8 T-EST 1 308. < K > / 2 8 3 6 7 . < K >

4.49

Hydraulic analysis of the Aspen Plus simulator produces "thermodynamic ideal minimum flow" and actual flow curves for rigorous distillation column simulations. These types of calculations are performed for RADFRAC columns. Using the input summary given in problem 4.48 construct the stage-flow curves. Assess the thermodynamic performance of the column.

272 4.50

4.

Usingthe second law: Thermodynamic analysis

Using the following input summary for a RADFRAC column, construct the temperature-enthalpy, stageenthalpy curves, and the stage-exergy loss profiles and assess the thermodynamic performance of the column by estimating and plotting the unavoidable part of exergy loss: Input summary: IN-UNITS MET COMPONENTS HEPTANE C7H 16-1 / OCTANE C8H 18-1 / N O N A N E C9H20-1 / D E C A N E C 10H22-1 / C15 C15H32 FLOWSHEET B L O C K C O L U M N IN = FEED OUT = DIST BTMS PROPERTIES SRK STREAM FEED IN-UNITS MET V O L U M E - F L O W = 'cum/hr' ENTHALPY-FLO - 'MMkcal/hr' & HEAT-TRANS-C - 'kcal/hr-sqm-K' PRESSURE = bar TEMPERATURE -- C & V O L U M E = cum DELTA-T = C HEAD = meter M O L E - D E N S I T Y = 'kmol/cum' & M A S S - D E N S I T Y = 'kg/cum' M O L E - E N T H A L P = 'kcal/mol' & M A S S - E N T H A L P = 'kcal/kg' HEAT = MMkcal M O L E - C O N C - 'mol/l' & PDROP-PER-HT = 'mbar/m' PDROP = bar S U B S T R E A M MIXED TEMP = 1 0 0 . P R E S = 2.4 MOLE-FLOW - 1000. MOLE-FRAC HEPTANE 0.2 / OCTANE 0.2 / N O N A N E 0.2 / & D E C A N E 0.2 / C15 0.2 B L O C K C O L U M N RADFRAC PARAM NSTAGE = 15 HYDRAULIC = YES COL-CONFIG C O N D E N S E R = PARTIAL-V KEY-SELECT = SPLIT-FRACTI FEEDS FEED 3 PRODUCTS BTMS 15 L / D I S T 1 V P-SPEC 1 200. COL-SPECS MOLE-D = 400. M O L E - R R = 7.5 TRAY-REPORT TRAY-OPTION = ALL-TRAYS FORMAT = C O M B I N E D REPORT STDVPROF TARGET HYDANAL

4.51

Using the following input summary for RADFRAC columns, construct the temperature-enthalpy, stageenthalpy curves, and the stage-exergy loss profiles for each columns and assess the thermodynamic performance of the columns: General Simulation with English Units: F, psi, lb/hr, lbmol/hr, Btu/hr, cuft/hr. Flow basis for input: Mole COMPONENTS EDC C2H4CL2-2 / HCL HCL / V C M C2H3CL FLOWSHEET B L O C K PUMP IN = RECYCLE OUT = RECYC1N B L O C K CRACK 1N = FEED R E C Y C I N OUT = REACOUT B L O C K Q U E N C H IN = REACOUT OUT - COOLOUT B L O C K COL 1 IN = COOLOUT OUT = HCLOUT V C M I N B L O C K COL2 IN = V C M I N OUT = V C M O U T RECYCLE

273

References

PROPERTIES RK-SOAVE STREAM FEED S U B S T R E A M M I X E D T E M P = 70 P R E S = 390 M O L E - F L O W E D C 2000 BLOCK QUENCH HEATER PARAM PRES 5 D E G S U B = 10 =

-

BLOCK COL 1 RADFRAC P A R A M N S T A G E = 15 COL-CONFIG CONDENSER = TOTAL FEEDS COOLOUT 7 ABOVE-STAGE P R O D U C T S H C L O U T 1 L ! V C M I N 15 L P - S P E C 1 367 C O L - S P E C S D : F = 0.354 M O L E - R R = 1.082 BLOCK COL2 RADFRAC P A R A M N S T A G E = 10 COL-CONFIG CONDENSER = TOTAL FEEDS VCMIN 6 ABOVE-STAGE P R O D U C T S V C M O U T 1 L / R E C Y C L E 10 L P - S P E C 1 115 C O L - S P E C S D:F = 0.550 M O L E - R R = 0.969 BLOCK CRACK RSTOIC P A R A M T E M P = 900 P R E S = 390 STOIC 1 MIXED EDC -1 /HCL 1 ' VCM 1 C O N V 1 M I X E D E D C 0.55 BLOCK PUMP PUMP P A R A M P R E S = 390 STREAM-REPOR MOLEFLOW MOLEFRAC CRACK

QUENCH COL1

PUMP

REFERENCES H. A1-Muslim and I. Dincer, lnt. J. Energy Res., 29 (2005) 637. H. A1-Muslim, I. Dincer and S.M. Zubair, J Energy Resouc Tech., 125 (2003) 199. A. Bejan, Entropy Production Minimization, CRC Press, Boca Raton (1996).

CO~

274

4.

Using the second law: Thermodynamic analysis

L.T. Biegler, I.E. Grossmann and A.W. Westerberg, Systematic Methods of Chemical Process Design, Prentice Hall, Upper Saddle River (1997). C.G. Carrington and Z.E Sun, Int. J. Heat Fluid Flow, 13 (1992) 65. Y.A. Cengel and M.A. Boles, Thermodynamics," An Engineering Approach, 4th ed., McGraw-Hill, New York (2002). G.M. De Koeijer and R. Rivero, Chem. Eng. Sci., 58 (2003) 1587. Y. Demirel, Int. J. Heat Mass Transfer, 43 (2000) 4205. Y. Demirel, Nonequilibrium Thermodynamics. Transport and Rate Processes in Physical and Biological Systems, Elsevier, Amsterdam (2002). Y. Demirel, Sep. Sci. Technol., 39 (2004) 3897. Y. Demirel, Sep. Sci. Technol., 41 (2006a) 791. Y. Demirel, Int. J. Exergy, 3 (2006b) 345. Y. Demirel and H.H. A1-Ali, Int. J. Heat Mass Transfer, 40 (1997) 1145. Y. Demirel and R. Kahraman, Int. J. Heat Mass Transfer, 42 (1999) 2337. Y. Demirel and R. Kahraman, Int. J. Heat Fluid Flow, 21 (2000) 442. V.R. Dhole and B. Linnhoff, Comp. Chem. Eng., 17 (1993) 549. V.R. Dhole and B. Linnhoff, Comp. Chem. Eng., 13 (1994) S 105. W.R. Dunbar, N. Lior and R.A. Gaggioli, J. Energy Resour. Tech., 114 (1992) 75. E.P. Gyftopoulos and G.P. Beretta, J. Energy Resour. Tech., 115 (1993) 208. E.C. Hohmann, Optimum networks for heat exchange, Ph.D. dissertation, University of Southern California, LA (1971). S. Kjelstrup and B. Hafskjold, Ind. Eng. Chem. Res., 35 (1996) 4203. S. Kjelstrup, E. Sauar, D. Bedeaux and H. van der Kooi, Ind. Eng. Chem. Res., 38 (1999) 3046. B. Linnhoffand J.R. Flower, AIChE J., 24 (1978a) 633. B. Linnhoff and J.R. Flower, AIChE J., 24 (1978b) 642. M.J. Moran and H.N. Shapiro, Fundamentals of Engineering Thermodynamics, 4th ed., Wiley, New York (2000). T.O. Ognisty, Chem. Eng. Prog., 2 (1995) 40. D. Poulikakos and J.M. Johnson, Energy, 14 (1989) 67. S.K. Ratkje, E. Sauar, E.M. Hansen, K.M. Lien and B. Hafskjold, Ind. Eng. Chem. Res., 34 (1995) 3001. R. Rivero, M. Garcia and J. Urquiza, Energy, 29 (2004) 467. R. Rivero, Energy Convers. Mgmt., 43 (2002) 1199. E. Sauar, K.S. Ratkje and K.M. Lien, Comp. Chem. Eng., 21(Suppl.) (1997) 29. W.D. Seider, J.D. Seader and D.R. Lewin, Product & Process Design Principles, 2nd ed., Wiley, New York (2004). J.M. Smith, H.C. Van Ness and M.M. Abbott, Introduction to Chemical Engineering Thermodynamics, 7th ed., McGraw-Hill, New York (2005). D. Tondeur and E. Kvaalen, Ind. Eng. Chem. Res., 26 (1987) 50. Y. Wang, J. Du, J. Wu, G. He, G. Kuang, X. Fan, P. Yao, S. Lu, P. Li, J. Tao, Y. Wan, Z. Kuang and T. Tian, Appl. Energy, 76 (2003) 467.

REFERENCES FOR FURTHER READING N. Kahraman and Y.A. Cengel, Energy Convers. Manage., 46 (2005) 2625. V. Lavric, D. Bactens, V. Plesu and J. De Ruyck, Appl. Therm. Eng., 23 (2003) 1837. L. Matijasevia and H. Othmaeia, Appl. Therm. Eng., 22 (2002) 477. M.A. Rosen, Int. J. Exergy, 3 (2006a) 202. M.A. Rosen, Int. J. Exergy, 3 (2006b) 219.

5 THERMOECONOMICS 5.1

INTRODUCTION

Thermodynamics principles describe the flow, conservation, and conversion of energy, and hence have implications for energy management and economics. The economies of processes always involve matter, energy, entropy, and information, and the consideration of economics leads to certain structures with minimum overall costs. Thermodynamic formulations impose directions and limits on the probability of processes; they also imply the use of scarce resources, and compare the efficiencies of conversion between different kinds of energies, which may be a necessary step in net energy analyses and energy policy discussions. Thermoeconomics combines thermodynamic principles with economic analysis. Therefore, it may bring some fundamental changes in the economic evolution, design, and maintenance of processes. Thermal systems involve significant work and/or heat interactions with their surroundings, and appear in almost every industrial plant. Consequently, the design of thermal systems requires the application of principles from thermodynamics, fluid mechanics, heat transfer, and engineering economics. Thermoeconomics usually concerns exergy and economics for optimizing the design and operating conditions of thermal systems. The optimization of subsystems individually does not guarantee an optimum for the overall system, and often various design variables must be considered and optimized simultaneously. In the optimization, the cost of the thermal energy source plays an important role; change in fuel cost from one year to another and from one place to another will eventually affect the overall design and hence the economic considerations. The U.S. Department of Energy web site, "Energy Savers for Industry Plant Managers and Engineers," offers a wide variety of energy saving possibilities, such as an energy management action plan. The process engineer should minimize the input cost of a process by reducing exergy loss due to thermodynamic imperfections. Taking such a perspective, thermodynamic analysis considers the interrelations among energy, economy, and ecology. Such considerations may have a positive impact on sustainable development and environmental protection. For example, a thermodynamic analysis of a solar desalination unit shows that the thermoeconomic evaluation of the system is closely related to a complete economic analysis of the possible improvements leading to a unit in which fewer irreversible processes occur.

5.2

THERMODYNAMIC COST

Thermoeconomics assigns costs to exergy-related variables by using the exergy cost theory and exergy cost balances, and mass, energy, exergy, and cost considerations can be unified by a single formulation. There are two main groups of thermoeconomic methods: (a) cost accounting methods and optimization methods, such as exergy cost theory for a rational price assessment, and (b) optimization by minimizing the overall cost, under a proper set of financial, environmental, and technical constraints, to identify the optimum design and operating conditions. Cost accounting methods use average costs as a basis for a rational price assessment, while optimization methods employ marginal costs in order to minimize the costs of the products of a system or a component. Extended exergy accounting considers nonenergetic costs, such as financial, labor, and environmental remediation costs, as functions of the technical and thermodynamic parameters of systems. Since exergy is a measure of thermodynamic work, available heat, and irreversibilities within the system, it is a true and rational basis for assigning monetary costs. Therefore, the exergy costing is the main aspect of thermoeconomics. The cost of fuel represents economic values for exergy loss in combining exergy and economic analysis allows for optimizing design and operation of thermal systems. Consider a heat exchanger; the average temperature difference ATlm between hot and cold streams is a measure of irreversibility, which vanishes as ATIm --' 0. The cost of fuel increases with increasing ATlm, while the capital cost decreases. As seen in Figure 5.1, the total cost consisting

276

5.

Thermoeconomics

~ ~ ~

Totalcost

Capitalcost~

[

[

I !

I I

ATtm(NTU)

Figure 5.1. Annual cost optimization.

of fuel and capital costs between points "a" and "b" would be optimum. For minimizing capital cost, the optimum would be toward point "b," and for minimizing fuel cost, the optimum approaches "a." Some concerns in thermoeconomics evaluations are: • It is difficult to estimate the cost precisely, since costs of fuel and equipment change with time and location. • The optimization of an individual process does not guarantee an overall optimum for the system due to interactions among various processes within the system. • For the whole system, often several design variables should be considered and optimized simultaneously.

5.2.1 Thermodynamic Analysis andThermoeconomics Thermodynamic analysis can lead to a better understanding of the system's overall performance, and eventually to identifying the sources of losses due to irreversibilities in each process in the system. This will not guarantee that economic and useful process modifications or operational changes would be undertaken; the relationship between energy efficiency and capital cost must be based on an analysis of the overall plant system, and sometimes improved energy efficiency will require more investment than is feasible. Mainly, thermodynamic analysis methods of pinch analysis, exergy analysis, second law analysis, and equipartition principles are combined to analyze process and energy systems. Process simulation packages, such as Aspen Plus and Hysys, may help in improving thermodynamic efficiency. These will enable engineers to modify existing systems or design new systems with complete objectives and targets, taking into consideration environmental concerns and natural resources.

5.2.2

Extended Exergy

To account for the environmental impact in a more systematic way, a resource-based quantifier, called "extended exergy" estimates the resource-based value of a commodity. Consider a separation process with outputs containing hot streams with various chemicals having conditions considerably different from environmental temperatures and concentrations. To achieve zero environmental impact, these streams must be brought to both thermal and chemical equilibriums with the surroundings: thus, the real (exergetic) cost of the zero impact would correspond to the extended exergy ideally required to bring the conditions of effluents to equilibrium conditions with the surroundings. If an acceptable level of pollutant or the "tolerable environmental impact limit" for a certain pollutant is specified, then the environmental cost may be quantified.

5.2.3

Exergy Cost

For any process or subsystem i, the specific cost of exergy c in US$ per kW per unit time for a stream is

C

--

Ex

(5 1)

where 0 and /~x are the cost rate and the rate of exergy transfer for a stream, respectively. However, the cost of a product and other exiting streams would include the fixed capital investment CFCI and the annual operating cost of

5.2

Thermodynamiccost

277

Exhaust gases Fuel I

Boiler

Air . . . . . . .

I

T .......

Work

|

Feedwater

Exhausted steam Figure 5.2. A two-unit system of a boiler and a turbine.

process Cop. This will be called the total cost of process C p - CFC I + Cop. Then the cost rate balance for a single process is (5.2)

For example, consider the exergy costing on a boiler and turbine system shown in Figure 5.2. The cost rate balance for the boiler (control volume 1) relates the total cost of producing high-pressure steam to the total cost of the entering streams plus the cost of the boiler CB, and from Eq. (5.2) we have (5.3)

CHp/~XHp + CLp/~XLp -- CF EX F -t- CA JE,X A -t- CW ]E,Xw "]'-d B

where the subscripts HP and LP denote the high- and low-pressure steams, respectively, while F, A, and w are the fuel, air, and water, respectively. All the cost estimations are based on exergy as a measure of the true values of work, heat, and other interactions between a system and its surroundings. By neglecting the costs of air and water, and assuming that the combustion products are discharged directly into the surroundings with negligible cost, Eqs. (5.1) and (5.3) yield the specific cost of high-pressure steam (product)

CHp -- CF EXHp

(5.4)

-4- L.,XH--~

The ratio (F__,XF/JE,XHp ) > 1 due to inevitable exergy loss is in the boiler, and hence Clip > Similarly, the cost rate balance for the turbine (control volume 2) is

CF.

CE~J/E -t- CLpL"XLp = CHpEXHp ---Hd T

(5.5)

where CE, Cop, and C'T are the specific costs of electricity, low-pressure steam, and the total cost of the turbine, respectively; WE and Excp the work produced by the turbine and the exergy transfer rate of low-pressure steam, respectively. Assuming that the specific costs of low- and high-pressure steams are the same (cLp -- Clip), we have

CE

Using the exergetic efficiency of turbine r h -

--

CHp

[EXHp--EXLp)'q ~'T WE

WE/(+CxHp--L'xLp), CE --

~/E

(5.6)

Eq. (5.6) becomes

Clip + C v

As r/t < 1, the specific cost of electricity (product) will be higher than that of high-pressure steam.

(5.7)

278

5.

Thermoeconomics

Example 5.1 Cost of power generation A turbine produces 30 MW of electricity per year. The average cost of the steam is US$ 0.017/(kW h) of exergy (fuel). The total cost of the unit (fixed capital investment and operating costs) is US$1.1 × 105. If the turbine exergetic efficiency increases from 84% to 89%, after an increase of 2% in the total cost of the unit, evaluate the change of the unit cost of electricity. Solution: Assume that heat transfer effects between the turbine and surroundings are negligible. Also, kinetic and potential energy effects are disregarded. From Eq. (5.7), we have CE(S4%) = CliP nt_ ~ ~t

~rE

__ 0.017 + 1-1X10_______~ 5 = US$ 0.0239/(kW h) 0.84 30×106

0.017 (1.02)1.1×105 = US$ 0.0228/(kW h) c E (89%) = ~ + 0.89 30×106 The reduction in the unit cost of electricity after the increase in efficiency is --~4.4%. This simple example shows the positive effect of exergetic efficiency on the unit cost of electricity.

Example 5.2 Cost of power and process steam generation In steam power generation, the boiler uses natural gas as fuel, which enters the boiler with an exergy rate of 110 MW (Figure 5.2). The steam exits the boiler at 6000 kPa and 673.15 K, and exhausts from the turbine at 700kPa and 433.15 K. The mass flow rate of steam is 32.5 kg/s. The unit cost of the fuel is $0.0144/kW-hr of exergy, and the specific cost of electricity is $0.055/kW-hr. The fixed capital and operating costs of the boiler and turbine are $1150/hr and $100/hr, respectively. The exhaust gases from the boiler are discharged into the surroundings with negligible cost. The environmental temperature is 298.15 K. Determine the cost rates of the steam produced by the reboiler (HP) and discharged steam (LP) from the turbine. Solution: Assume that heat transfer effects between the boiler and turbine and surroundings are negligible. Also, kinetic and potential energy effects are disregarded. The environmental temperature is 298.15 K. At this temperature, we have the reference values of enthalpy and entropy: H o = 2547.2kJ/mol and SO = 8.5580kJ/(molK)

rh = 32.5kg/s

The cost data: c E = $0.055/(kW h)

CF -- $ 0 . 0 1 1 4 4 / ( k W

C13=$1150/h

h)

/~XF

-

-

110,000 kW

CT = $ 1 1 0 / h

Enthalpy and entropy values after the boiler: P1 = 6000kPa

T1 = 673.15K

H 1 = 3177.2 kJ/kg

S1 = 6.5408 kJ/(mol K)

H 2 = 2798.2kJ/kg

S2 = 7.8279 kJ/(mol K)

After the turbine: P1 = 700kPa

T1 = 433.15K

The work produced: = rh(H 1 - H 2) = 32.5 (3177.2 - 2798.2) = 12317.50 kW The cost of electricity: CE = cEIY = $ 677.43/h

5.2

279

Thermodynamiccost

Using the reference values for enthalpy and entropy, the rate of exergy of stream leaving the boiler (1) and turbine (2) is: Ex 1 = rh[H 1 - H o - To (S1 - So)] = 40,021.42 kW /~TX2 = t h [ H 2 - H 0 - T 0 ( S 2 - S 0 )] =

15,232.08 kW

The cost rate balance for the boiler yields the specific cost of steam produced by the boiler:

C 1L'X 1 "-- C v L'N v Jr- C B --~ c 1 - - c F

~E'XF CB .... + .

Exl

Exl

= $0.0683/(kW h)

The cost of boiler steam: C 1 = ClEX1 = $2733.46 h -1 . The cost rate balance for the turbine, Eq. (5.5), yields the specific cost of steam exhausted by the turbine"

~'X1

c2 -- Cl ~7x2

C------LE+ CT = $0.1415/(kW h) ~'x2 Ex 2

The cost of exhausted steam: -C2 - - c2/~7x 2 - -

$2156.53

h-1

The exergetic efficiency may be increased by minimizing the throttling of large thermodynamic driving forces which are changes in pressure, temperature, and composition. The effect of exergetic efficiency would increase for larger steam mass flow rates. At lower temperature levels, friction losses would have more negative effects on the unit cost of electricity.

Example 5.3 Thermoeconomic consideration of a refrigeration system A refrigeration heat exchanger provides an opportunity to study the trade-off between the cost of availability loss and the capital cost of the exchanger. We may need to supply a refrigeration flow to the condenser of a distillation column that returns reflux as a condensate at a certain temperature. The refrigeration temperature must be less than the condensing temperature, and the temperature difference of the refrigerant and the condenser AT is an important parameter. A larger difference results in a smaller and hence less expensive condenser; however, at low temperature, the power required for a unit of refrigeration increases due to the higher fuel costs to operate the refrigeration compressor. This creates a typical optimization problem involving the value of AT and the annual cost of fuel and the condenser (see Figure 5.1). From the second law consideration, the annual cost of fuel CF is given by

C v = CFFwFbtyToq A T T2

(5.8)

where cv is the cost of unit of fuel, Fw the units of shaft work required by the refrigeration system to deliver a unit of availability, Fb the units of fuel fired in the plant boiler per unit of shaft work produced, ty the operating time per year, To the absolute temperature of the refrigeration system condenser, or of ambient, q the heat or refrigeration duty per unit time, and AT the temperature difference. The annual capital cost Ca of the heat exchanger, assuming that the exchanger is large and the cost is directly proportional to its area, is given by

ceFiq

Ca

etgA T

(5.9)

where c e is the purchase cost per unit of the heat exchanger area, Fi the installation cost factor, Pt the allowable payout time in years, and U the overall heat transfer coefficient. The differential of total annual cost of fuel and capital with respect to AT is set equal to zero, and is used to determine the optimum AT

280

5.

Thermoeconomics

)1/2 z~Topt -- T

Cefi

cFFwFbI-IyroPtU

(5.10)

The above equation shows that the optimum temperature ATopt is proportional to the temperature level at which the heat transfer occurs, which is well known for the refrigeration systems. The value of refrigeration increases as the temperature decreases; hence, the smaller AT are used as the refrigeration temperature decreases. For aboveambient systems, the larger AT should be used at higher temperatures, even though the value of heat increases with increasing temperature. In some cases, availability saved at a heat exchanger does not lead to a fuel saving, but may be accompanied by further availability savings elsewhere in the process.

5.2.4

Cumulative Exergy Consumption

Exergy analysis evaluates the level of irreversibility, and hence identifies the possibilities of improvements for a process. It may play a primary role in minimizing the consumption of natural resources within the context of ecological economy. Exergy analysis is, however, a thermodynamic approach, not an economic one. Still, the partition of production costs between the useful products of a complex process can be managed by means of exergy. All useful products out of an industrial production line are the results of a complicated network of interconnected processes, which need the supply of raw materials, fuels, and other energies extracted from natural resources. The quality of the natural resources can be evaluated and expressed by means of exergy. The analysis of cumulative exergy consumption provides an insight into the possibilities of improving the technological network of production. The following derivations are from Szargut (1990). The total consumption of natural resources involved in the production of a product can be expressed by the overall index of cumulative exergy consumption rj,

rj ~-

E;k%

(5.11)

where ~kj is the exergy consumption, expressed as ~j = To~kAS~., of the n natural resources (k = 1, 2, ..., n) for the product of j, and Pj the final product from an industrial plant. The terms To and ]~kASkj indicate the environmental temperature and sum of entropy changes over the consumption of sources for product j, respectively. As the value of rj is related to a unit of the product leaving the system, it depends on the assumption of the system boundary. Usually, the system of production processes is analyzed without considering employees and local levels of consumption, and hence rj can be used for comparison of production processes in various countries. The exergy consumption index for a certain resource k can be determined separately for a final product Pj-

(I) k rkj B PJ

(5.12)

Exergy consumption of resources that are renewable should also be considered. The overall cumulative exergy consumption index may help in assessing various energy utilization problems for a specified product, such as the relationships between the amounts of raw materials and the products, or the cost of raw materials and the alternative production technologies available. In analyzing the production of materials and energy flows, the values of t) can be compared with the exergy of the product. The ratio of the specific exergy of the product ex to rj is called the cumulative degree of thermodynamic perfection ~ for a certain production network T~ B

ex F

(5 13)

Here, r is the cumulative exergy consumption index for a specified product. For the manufacturing of major products, we have r t < 1. Sometimes, for a byproduct we may have rt > 1, if the exergy of the product is greater than the exergy of the substituted product of a certain process. It would not be useful to calculate r/for certain products, such as cars and airplanes, because their usefulness results mainly from their system features, not from the chemical composition of their components. However, the calculation of r may be beneficial for all kinds of products, because the values of r can be used to compare various design variables and production technologies.

5.2

281

Thermodynamiccost

Cumulative exergy consumption can be calculated by the balance equations; the rkj for the useful products equals the sum of cumulative exergy consumption of all raw materials and semifinished products in the production network. For the linkj of the network and for the natural resources k, the balance equations are

r~/) - -/:i!/ ) E w . G (')+
(5.14)

t

where k is the index of a natural resource, i andj the indices of technological network, 1 and t the indices of the production technologies of the products i and j, a!j the coefficient of the gross consumption of the semifinished product i per unit of the complex useful product containing a unit of the major productj, f j the coefficient of the byproduction of the useful product i, wit the fraction of the manufacturing technology t of the product i, and ~kj the immediate gross exergy consumption of the neutral resources k per unit of the useful product. Equation (5.14) describes a complex process producing more than one useful product. A complex process is usually related to a major product, which determines the capacity and location of production. A useful product substituting the major product is called the byproduct. The coefficient ofbyproduction is expressed in terms of the substitution ratio zi, which is the ratio of the unit of the major product i substituted by the unit of the byproduct u, and given by

(5.15)

ft :) - zi" : U.l

wheref,j is the coefficient of production of byproduct u per unit of the major productj. Cumulative exergy consumption for the byproduct u is given by .~J-i'ri- f,i/'r,

(5.16)

r, = r, zi,

(5.17)

From Eqs. (5.15) and (5.16), we obtain

5.2.5

Cumulative Degree of Thermodynamic Perfection

The cumulative degree of thermodynamic perfection for real byproducts u is defined by CX,_ -

flu

ex u

r,,

rizi u

_

ex u - - ~ i

bi zi u

T] i -

(5.18)

T]z,iu

where r/i is the cumulative degree of thermodynamic perfection for the major product i substituted by the product u and rt:,i, the exergetic substitution efficiency defined by r/_i . -

e x i Ziu

(5.19)

ex u

If rtz,;, < rti, the value r/, > 1 results from Eq. (5.18). Some typical values of the cumulative degree of thermodynamic perfection rt are given in Table 5.1. For small values of r, that result from the substitution ratio z;,, the cumulative exergy consumption will be large for the major productj of the specified process technology. Equation (5.14) can be transformed as follows:

tl

where a,m - wital/) and f,m -- f~!/) For technology l, every subscriptj corresponds to some subscript m while for technology t, every subscript i corresponds to some subscript n. Equation (5.20) can also be formulated for semifinished products, which are consumed in other links of the technological network. In matrix form, Eq. (5.20) becomes •

•

tl

•

r ( l - A + F) = (I)

(5.21)

282

5.

Thermoeconomics

Table 5.1 Cumulative degrees of thermodynamic perfection for some production technologies Product Aluminum

Specific exergy (MJ/kg)

r/(%)

32.9

9.6 13.2 44.0 10.3 6.2 3.2 2.6 1.5 0.8 0.5 45.4 64.8 41.5 18.7 27.5 74.3 7.6 6.7 2.7 4.2 18.3

Iron Cement

8.2 0.635

Copper

2.11

Glass

0.174

Ammonia gas

20.03

Paper

16.5

Zinc

5.19

Sulfuric acid

1.66

Source:

Production technology Bayer process and Hall cell 50% bauxite ore Electrolytic method from A1203 From hematite ore in the Earth From raw materials with dry method Medium rotary kiln with wet method From ore containing Cu2S, smelting and refining Hydrometallurgical method Electrolytic From raw material From panels Steam reforming of naphtha Steam reforming of natural gas Semicombustion of natural gas From timber Integrated plant with fuels from waste products From waste paper From ore with vertical retort Electrothermic method From ZnS, metallurgical method From ZnS, electrolytic method Frasch process with sulfur combustion

Szargut (1990).

where I is the diagonal unit matrix, A and F the square matrices with elements o f M × M, M = mmax, and r and B the rectangular matrices with elements of K × M, K = kmax- From Eq. (5.21), we obtain r = @ ( I - A + F)-1

_

_

(I~S*

(5.22)

The elements of the inverse matrix S* represent the cumulative net consumption of intermediate products per unit of the major product leaving the system =

+""

(5.23)

Here, S ~ shows the net amount of product 1 per unit of product n.

5.2.6

Cumulative Exergy Loss

The difference between cumulative exergy consumption r and exergy consumption of a natural resource represents the cumulative exergy loss (&h) involved in all parts of a manufacturing technological network (5.24)

64) = r - ex

The components 64~ provide information for improving the technological network. The difference ( r - ex)~ defines the constituent exergy loss or a particular semifinished product n, and results from the thermodynamic imperfection of the constituent technological network. In complex processes, raw materials and semifinished products are partially used for the manufacturing of byproducts. Hence, the coefficient of net consumption A~m of semifinished products and raw materials per unit of the major product should be determined by A~m = anm -~_.f.mA~. = a~m -~_.f.mzp.A~p u

(5.25)

p

where p is the index of the major product substituted by the byproduct u and Zp, the substitution ratio of the product p by the product u. In a process substituted by the utilization of the byproduct the coefficient Anm can be negative if the consumption of the semifinished product n is greater, than in the principal process considered.

5.2

283

Thermodynamiccost

The constituent exergy losses are calculated from the coefficients Anm 6O,,m - A~m(r, - e x n ) '

n=/=m

(5.26)

Some losses can be negative due to the elimination of the constituent exergy losses in the substituted process.

5.2.7

Local Gross Exergy Loss

Local gross exergy loss 6 ~ m represents the sum of internal and external exergy losses in the used technology for the major product and byproduct, and it can be calculated from the following steady-state exergy balance" Z

amneXn =

eXm

-1" Z

tl

(5.27)

L m e X u q- dd/~m

ll

Here, the energy exchanged with surroundings should be treated as one of the useful product exergies represented by eXm, exn, or ex,. The local net exergy loss refers to the complex of useful products containing a unit of the major product, and results from the following difference:

6&,,,,, = 6&,, - ~__~.f,,maO,,

(5.28)

ll

where 6q~,, is the local exergy loss due to the byproduct u. The local exergy loss due to the byproduct results not only from the local net exergy loss in the substituted process, but also from the difference between the exergy of the byproduct and the substituted major product 6&,, = ,.~",.6Om, - (ex,, - zp.xp) = zp,,adppp - ( 1 - ~.,p. )ex.

(5.29)

where 6Chpp is the local net exergy loss due to the major product in the substituted process and rL-,p, the exergetic substitution efficiency (Eq. 5.19). After combining Eqs. (5.28) and (5.29), we obtain 6Omm = 8~,, -- ~ ~,m'->,,aOpp + ~ f,~ ( 1 - ~7-,pu)eX, l~

(5.30)

u

The above equation (5.30) determines the local net exergy losses for all major products.

5.2.8

Partial Exergy Loss

Partial exergy loss 6chx~,, expresses the local net exergy loss in link k of the technological network for the product m km6&kk

(5.31)

where S*km is the coefficient of cumulative net consumption in Eq. (5.23). If the product of link m is consumed in the preceding links of the technological network,• then S,~,m > 1 and the partial exergy loss 6Zhmm is greater than the local net exergy loss 64'ram. A partial exergy loss in a complex process can be negative, if the production of the intermediate product k is smaller in the main network than in the network substituted by the utilization of byproducts. The sum of partial exergy losses equals the cumulative exergy loss

6&,,, - ~ 6~# m

(5.32)

k

Relative constituents &c, local exergy loss 4'i, and partial exergy loss 6Zbkm can be defined by

~ c , nm

.

. . (~dpnm

.

6 l ,m m

1;,,

8 d/~mm ", rm

-

~ba,,--

~4~

(5.33)

rm

From Eqs. (5.31) and (5.33), we obtain

,...+ n~ m

(5.34)

284

5.

Thermoeconomics

't~m + E ~km ~= 1 k

(5.3 5)

The analysis of constituent and partial exergy losses may improve the processes of a technological network. If some constituent exergy loss is very large, the possibility of changing the production technology or substituting the semifinished product by another more convenient one should be considered. If a partial exergy loss is considerably large, reducing the thermodynamic imperfections of the technological network should be investigated. Decreasing the internal irreversibilities, better utilizing the waste products or changing the technology may be considered. Designing more suitable operational conditions and reducing the consumption of some expensive intermediate products are additional possibilities.

5.2.9

Exhaustion of Nonrenewable Resources

The utilization of domestic natural nonrenewable resources is inevitable, and analyzing these resources helps to assess the profitability of importing raw materials, fuels, and semifinished products as well as utilizing secondary raw materials. In analyzing the exhaustion of nonrenewable natural resources, the balance equations of Eq. (5.20) should be modified if domestic nonrenewable resources are of interest. In this case, imported raw materials, fuels, and semifinished products should be taken into account separately

ekm = ~_. (a.. - f~m)e~ + ~_. armekr -k-dPkm n r

(5.36)

where ekm, e~, and egr are the exhaustion of the domestic nonrenewable natural resources k per unit of the products m, n, and r, respectively, arm the coefficient of gross consumption of imported raw material, fuel, or semifinished product r, and ~km the gross consumption of the domestic nonrenewable natural resources k within the link m. The index e~ should be determined by assuming that the import is economic, and the unit value of the exported and imported products is considered with the same exhaustion of nonrenewable natural resources

E Enekn % = e(S )D r = ~ Dr EnD, ,

(5.37)

n

where e(kd) is the exhaustion of nonrenewable natural resources per unit of the monetary values of exported products, Dr and D~ the specified monetary value of the imported product r and exported product n, respectively, and E~ the export of the product n. Introducing Eq. (5.37) into Eq. (5.36), the balance equations become e ~ = ~ [(a,,n - rum)+ d~m]%, +dp~

(5.38)

n

where

Drarm dnm = E n ~ EnD n n

In matrix form, Eq. (5.38) becomes e ( l - A + F - D) = @

(5.39)

e = • ( I - A + F - D)-1

(5.40)

and

where A, F, and D are the square matrices with M × M elements and e the rectangular matrix with K × M elements. Set of Eq. (5.39) comprises all the domestic semifinished products and the exported products. An approximate solution is

e~d) - ~-~*k DN

(5.41)

5.3

Ecological cost

285

where E~k is the total exhaustion of domestic natural resources k and DN the economic value of all domestic products and exported products. Equation (5.41) enables one to formulate the balance equations of Eq. (5.39) for semifinished products. Equation (5.37) defines the exhaustion of nonrenewable natural resources for imported products, so Eq. (5.36) is fully known. In the analysis of material production, the utilization efficiency of nonrenewable domestic natural resources can be defined as rl' -

CX

(5.42)

g

Usually, e > ex and rl' < 1, but for secondary raw materials rI' > > 1, and for imported raw materials and fuels, we usually have rl > 1. For secondary raw materials, the exhaustion of nonrenewable natural resources is related to the consumption ofexergy for processing and transportation, and usually it is much smaller than the exergy of the materials under consideration. The inequality rI' > > 1 suggests that the utilization of secondary raw materials may be beneficial, since they substitute the semifinished products requiring a large amount of exergy for production. We may have ex > e for imported raw materials, fuels, and semifinished products if the exported goods are more advanced than the imported ones.

5.2.10

Exergy Destruction Number

The use of an augmentation device results in an improved heat transfer coefficient, thus reducing exergy destruction due to convective heat transfer; however, exergy destruction due to frictional effects may increase. The exergy destruction number N~ is the ratio of the nondimensional exergy destruction number of the augmented system to that of the unaugmented one

EXa

N,- - _-7-7

(5.43)

Ex s

where subscripts "a" and "s" denote the augmented and unnaugmented cases, respectively, and Ex* the nondimensional exergy destruction number, which is defined by E x * --

CXfd

mToCp

(5.44)

Here, exfd is the flow-exergy destruction, or irreversibility, and To the reference temperature. The system will be thermodynamically advantageous only if the N~ is less than unity. The exergy destruction number is widely used in secondlaw-based thermoeconomic analysis of thermal processes such as heat exchangers.

5.3

ECOLOGICAL COST

The production, conversion, and utilization of energy may lead to environmental problems, such as air and water pollution, impact on the use of land and rivers, thermal pollution due to mismanagement of waste heat, and global climate change. As an energy conservation equation, the first law of thermodynamics is directly related to the energy management impact on the environment. One of the links between the principles of thermodynamics and the environment is exergy, because it is mainly a measure of the departure of the state of a system from that of the equilibrium state of the environment. Performing an exergy analysis on the Earth's natural processes may reveal disturbances due to large-scale changes, and could form a sound base for ecological planning for sustainable development. Some of the major disturbances are: 1. Chaos due to the destruction of order is a form of environmental damage. Entropy is fundamentally a measure of chaos, while exergy is a measure of order; the exergy of an ordered system is greater than that of a chaotic system. 2. Resource degradation leads to exergy loss. A natural resource with exergy is in nonequilibrium state compared with the environment. 3. Uncontrollable waste exergy emission can cause a change in the environment. Exergy analysis may be an important tool to interrelate energy management, the environment, and sustainable development in order to improve economic and environmental assessments. Ecological cost analysis may minimize

286

5.

Thermoeconomics

the depletion of nonrenewable natural resources. Determining the exhaustion of nonrenewable natural resources connected with the extraction of raw materials and fuels from natural resources is not sufficient in fully understanding the ecological impact of production processes. The influence of waste product discharge into the environment should also be considered. Waste products may be harmful to agriculture, plant life, human health, and industrial activity.

5.3.1

Index of Ecological Cost

The exhaustion of nonrenewable natural resources is called the index of ecological cost. To determine the domestic ecological cost q~, the impact of imported materials and fuels is taken into account

where s is the index of harmful waste product, dnmis defined in Eq. (5.38), ~ the immediate gross consumption of the nonrenewable domestic natural resource k per unit of complex useful products containing a unit of the major product m, ~ m the exergy of harmful waste product s, x~ the destruction coefficient of the product n per unit of the exergy of waste product s, Yks the destruction coefficient of the nonrenewable natural resources k per unit of the exergy of waste product s, and z~ the multiplier of exergy consumption to eliminate the results of human health deterioration per unit of exergy of the waste product s. The destruction coefficients x and y are

x,s

~s'

Y~

~

(5.46)

where 6 ~ , is the exergy of the destroyed useful product n and 6~ k is the exergy decrease of the damaged natural resources. The coefficient xns should also take into account the reduction of agricultural and forest production. The global ecological cost can be calculated. The degree of the negative impact of the process on natural resources can be characterized by means of the ecological efficiency tie, and from Eqs. (5.18) and (5.45) we have

1% - -

q~

(5.47)

Usually, r/e < l, but sometimes values of r/e > 1 can appear if the restorable natural resources are used for the process. The transition from one form of exergy to another, for example, from chemical to structural, may create economic value. Self-organization is a production process, and exergy is necessary to build a structure with a value, that may not be measured and described by exergy.

5.3.2

Global Warming Potential

The global warming potential is a measure of how much a given mass of a chemical substance contributes to global warming over a given period of time. The global warming potential is the ratio of the warming caused by a substance to the warming caused by a similar mass of carbon dioxide. Therefore, the global warming potential of carbon dioxide is defined as 1.0, while water has a global warming potential of 0. Chlorofluorocarbon-12 has a global warming potential of 8500, while chlorofluorocarbon-ll has a global warming potential of 5000. Various hydrochlorofluorocarbons and hydrofluorocarbons have global warming potentials ranging from 93 to 12,100. These values are calculated over a 100-year period of time.

5.4

AVAILABILITY

One of the important definitions in finite-time thermodynamics is the definition of finite-time availability A given by

(5.48)

5.5 Thermodynamicoptimum

287

Here, ti and tf are the initial and final times of the irreversible process and To the environmental temperature. The maximization is carried out with the constraints imposed on the process. Equation (5.48) represents the second law of thermodynamics in equality form by subtracting the work equivalent of the entropy produced, which is the decrease in availability in the process. Availability depends on the variables of the system as well as the variables of the environment A = U + Po V - T o S - ~ txoiN i

(5.49)

Here, the temperature, pressure, and chemical potential are estimated at ambient conditions. For an optimal control problem, one must specify: (i) control variables, volume, rate, voltage, and limits on the variables, (ii) equations that show the time evolution of the system which are usually differential equations describing heat transfer and chemical reactions, (iii) constraints imposed on the system such as conservation equations, and (iv) objective function, which is usually in integral form for the required quantity to be optimized. The value of process time may be fixed or may be part of the optimization.

5.4.1

Essergy

The potential work of any system is given by E~ - E - roS + PoV- Y_..;oNi

(5.50)

i

where/x and N are the chemical potential and number of moles of substance i, E the total energy including all kinetic and potential energies in addition to internal energy, and the subscript "0" denotes the reference state representing the environment of the system. The term Ees is the essential energy in the form essential for work (power) production, so that Ees shows the essergy (essential energy). The corresponding flow of essergy ~0e~, excluding kinetic and potential energies for any uniform mixture of substances, is 4 ~ - H - ToS - ~ I~ioN i

(5.51)

i

5.5

THERMODYNAMIC OPTIMUM

Thermoeconomics formulates an economic balance through exergy cost and optimization. The minimization of entropy generation plays only a secondary role in thermoeconomics, mainly because economic performance is always expressed in economic values of money and price. Therefore, the thermodynamic optimization problem may not be expressed in terms of the problem of the minimization of irreversibility. For example, the problem of minimum overall exergy consumption may not be equivalent to the problem of minimum dissipation because of the disregarded exergy of the outgoing flows and changing prices of exergy unit. This problem mainly belongs to the areas of energy management and the cost of energy. Industrial systems consist of various resource consumption processes and supporting processes to supply and remove resources. The supporting processes may involve exergy loss and exergy transfer between resources, new resource upgrading, postconsumption recovery, and the dispersion and degradation of resources released to the environment. The contemporary theory of optimization can be used for analyzing these systems. The first approach is to optimize the system by adjusting the design and operating parameters through governing equations that describe internal changes, and by imposing control through system boundaries. The second approach aims to predict system behavior under a set of specified external conditions with governing equations derived from certain variational or extremum principles.

Example 5.4 Minimization of entropy production For a fixed design, the minimization of the rate of entropy production may yield optimal solutions in some economic sense. Such a minimization comes with certain set of constraints. For a single force-flow system, the local rate of volumetric entropy production is dp - 5v L X 2 d V

(5.52)

288

Thermoeconomics

5.

where L is the phenomenological coefficient and is not a function of the driving force X. The minimization problem is the optimization of the system with a finite size V, and the solution is the homogeneous distribution of the force over the system. Assuming a steady-state heat transfer operation with no momentum and mass transfer, the expression of total entropy production is (Tondeur, 1990)

: -I ~2 (qVT)dV : -k I v

dV

(5.53)

v

where the heat flux is obtained from the Fourier law (5.54)

q = -kVT

and k is the thermal conductivity assumed as a constant. The entropy production is a function of the temperature field. Then, the minimization problem is to obtain the temperature distribution T(x) corresponding to a minimum entropy production • using the following Euler-Lagrange equation:

Minimizing the entropy production fianction with the constraim expressed in Eq. (5.54), the above equation becomes

ZLr oxj - 7

=o

(5.56)

For a heat exchanger, a characteristic direction related to the temperature field is the direction Z(x) normal to the heat transfer area, and the above equation yields

~-~XIT 0-~-xT ]Z(x) =0

(5.57)

VT)

(5.58)

and we obtain

T

= constant Z(x)

The above equation shows that by keeping the driving force VT/T uniformly distributed along the space variables, entropy production will be minimum. For an optimum design, we may consider VT

AT

T

T

- constant

(5.59)

where AT= Th -- Tc or (p - 1)Tc, and the temperature gradient is a function of the temperatures Th and Tc of hot and cold streams, respectively Th = p T c

(5.60)

where p is a constant with value > 1. The following expression also produces a constant C: ~)Op

- Tc) 2 - U(ThVhVc = C = constant

(5.61)

5.6

Equipartitionand optimization in separation systems

289

where 6 is a small change, U the total heat transfer coefficient, and A the heat exchanger area

6~_6q(1

1)

Tc

6A-

and

Th '

6q g(Th

-Tc)

From Eqs. (5.60) and (5.61), we have (2 +

C/U) +[(2 +

P=

C/U)

2 -

4] 1/2

(5.62)

2

The energy balance is

(5.63)

whdTh = wcdTc

where 1/i2h and Wc are the products of heat capacity and hot and cold streams flow rates, respectively. From Eqs. (5.60) and (5.63), we have d~h dL

- - p --

Wc Wh

--

Th Tc

(5.64)

which are the matching conditions to minimize the entropy production in any heat exchanger. For example, for a specified heat exchanger area and hot stream input and output temperatures Ti and To, respectively, the minimum entropy production is obtained when We _ Ti _ constant % To

(5.65)

We can extend this approach for a network of heat exchangers.

5.6

EQUIPARTITION AND OPTIMIZATION IN SEPARATION SYSTEMS

Thermodynamic cost analysis relates the thermodynamic limits of separation systems to finite rate processes, and considers the environmental impact through the depletion of natural resources within the exergy loss concept. Still, economic analysis and thermodynamic analysis approaches may not be parallel. For example, it is estimated that a diabatic column has a lower exergy loss (39%) than that of adiabatic distillation; however, this may not lead to a gain in the economic sense, yet it is certainly a gain in the thermodynamic sense. The minimization of entropy production is not always an economic criterion; sometimes, existing separation equipment may be modified for an even distribution of forces or an even distribution of entropy production. Thermodynamic analysis requires careful interpretation and application. The results of thermodynamic analysis may be in line with those of economic analysis when the thermodynamic cost optimum, not the maximum thermodynamic efficiency, is considered with process specifications. Figure 5.3 shows pinch technology in terms of optimum hot and cold utilities by accounting for the investment costs and exergy cost. With an optimum approach temperature ATmin, the total cost may be optimized.

Example 5.5 Equipartition principle in separation processes: Extraction Since the minimization of entropy production is not always an economic criterion, it is necessary to relate the overall entropy production and its distribution to the economy of the process. To do this, we may consider various processes with different operating configurations. For example, by modifying an existing design, we may attain an even distribution of forces and hence an even distribution of entropy production. Consider a simple mixer for extraction. In minimal entropy production, size V, time t, and duty d are specified, and the average driving force is also fixed. We can also define the flow rate Q and the input concentration of the solute, and at steady state, output concentration is determined. The only unknown variables are the solvent flow rate and composition, and one of them is a decision variable; specifying the flow rate will determine the solvent composition. Cocurrent and countercurrent flow configurations of the extractor can now be compared with the

290

5.

Thermoeconomics

Hotutility O------

/

r~ 0

~

4

I-Iotutility

ATtain ~ - - - ' J / ~ Coldutility / J ~./f/~ Hotutility

\

.

r~

~~

O ~

.

.

.

C ~ l d ~ ~

~ ~"0---

Coldutility

Operatingcost Exergyloss Figure 5.3. Principleof pinch technology. same initial specifications (V, t, J, Q, c). Cocurrent operation will yield a larger entropy production P2 than the countercurrent operation P1, and investigating the implications of this on the decision variable is important. For a steady-state and adiabatic operation, we have the following relations from Tondeur and Kvaalen (1987). For processes 1 and 2 with the solvent flow rates of QI and Q2, we have

Q1As1 = - AS--bP1

(5.66)

Q 2 ~ s 2 -- - A s -[- P 2

(5.67)

where AS is the total entropy change, and AS 1 and AS 2 the changes in specific entropies of the solvent. Subtracting Eq. (5.67) from Eq. (5.66), we have

Qlz~,s1-Q2As2 = P1-P2 <0

(5.68)

The load is defined as

J

= Q1Ac1 = Q 2 A c 2

(5.69)

where Ac is the concentration change of the solute in the solvent throughout the process. Combining Eqs. (5.68) and (5.69), we obtain

As2 ~C1 mc2 ASa<

(5.70)

The specific entropy of a solvent increases with the solute concentration, and if the input solvent is the same, inequality (5.70) yields Aca > Ac2, and hence Eq. (5.69) shows that Q1 < Q2- This means that the solvent flow rate is smaller in the less dissipative operation, and the solvent at the outlet is more concentrated. That is, the operating conditions of solvent determine the less dissipative operation. Whether this optimum is an overall economic optimum will depend mainly on the cost of the technology. We can also compare the two processes with the same total entropy productions, the same size and duration, and the same phenomenological coefficients. Process 1 has only equipartitioned forces; therefore, the duties of these processes will be different. The total entropy productions for the processes are expressed as

el = earl -- Z(Xavl)2( V t )

P2 = I I LX2dVdt >

Pav2 = L(Xav2) 2(Vt)

(5.71) (5.72)

5.6

Equipartition and optimization in separation systems

291

Since P1 -- P2, combining Eqs. (5.71) and (5.72) yields Xa2vl ~> Xa2v2, and hence J1] > IJ2 • That is, the flow rate for equipartitioned process 1 is larger than that of process 2 at a given size, duration, and entropy production. In another operating configuration, we can compare the respective size and durations for specified duty and entropy production. Equations (5.71) and (5.72) are still valid, and we have P1 > Pav2 and J1 = J2, which yield P1 > Pav2

J1

(5.73)

J2

and thus Xav 1 > Xav 2

(5.74)

(Vt)l < (Vt)2

(5.75)

This result indicates that for a given flow and entropy production, the equipartitioned configuration is smaller in size for a specified operational time. Alternatively, it requires less contact time for a given size, and thus a higher throughput. To determine an economic optimum, we assume the operating costs are a linear function of the solvent entropy change and entropy production, and the investment costs are a linear function of the space and time of the process. The total cost is

C T : aP + b + czVt - I ~ (aLX2 + c'r)dVdt + b

(5.76)

where ~"is the amortization rate and a, b, and c the constants related to the costs. The integral in Eq. (5.76) is subject to the constraint of a specified flow

J = I ~ LXdVdt

(5.77)

The variational technique minimizes the total cost, and the Euler equation for variable X is given by 0

OX

(aLX 2 + cz + ALX) = 0

(5.78)

where A is a Lagrange multiplier. The above equation yields

2aLX + AL = 0 X =-

A 2a

- constant

(5.79) (5.80)

The obtained value of X that minimizes the total cost subject to J is a uniform distribution. This illustrates the economic impact of the uniform distribution of driving forces in a transport process.

Example 5.6 Thermoeconomics of extraction Consider a steady-state operation in which the forces are uniformly distributed; the investment cost Ci of a transfer unit is assumed to be linearly related to size V, and operating costs Co are linearly related to exergy consumption C v -- C i - Cif - a V

(5.81)

C o = Cof Jr- bz~kEx

(5.82)

where Cv is the variable part of the investment cost, Cif a fixed investment cost, Cof a fixed operating cost, and a and b the cost parameters. Exergy loss AExc is expressed as

292

5.

Thermoeconomics

AEx~ = AEx m + TocI3av

(5.83)

Here, To is a reference temperature (dead state) and Z ~ X m a thermodynamic minimum value. The total flow J = LVXav can be written by using Eq. (5.81)

J = LXavCv

(5.84)

a

Eliminating the constant (average) force Xav between Eq. (5.83) and the total entropy production (I)av = JAXav, we obtain

aJ 2 ~av -

(5.85)

LCv

Substituting the above equation into Eq. (5.81) and the latter into Eq. (5.82), a relationship between the operating and investment costs is obtained Co

ab ToJ2

: ~

LCv

at- Cof at-

bAExm

(5.86)

The optimal size is obtained by minimizing the total operating and investments costs, which are linearly amortized with the amortization rate z. CT(Ci) = zCi = Co. The minimum of CT is obtained as dCT/dCi = 0, and we have b,I, av

(Cv)op t

_

b(I'av

_

8Vop t

To

(5.87)

According to Eq. (5.87), the quantities bT0~av, which are related to irreversible dissipation and q'Vopt, should be equal in any transfer unit. Generally, operating costs are linearly related to dissipation, while investment costs are linearly related to the size of equipment. The optimum size distribution of the transfer units is obtained when amortization cost is equal to the cost of lost energy due to irreversibility. The cost parameters a and b may be different from one transfer unit to another; when a = b, then (I)av/Vop t is a constant, and the optimal size distribution leads to equipartition of the local rate of entropy production. The optimal size of a transfer unit can be obtained from Eq. (5.78)

(Cv)°pt:Ci'°pt-Cif :j(abT°] 1/2Ln" ( bT° 11/2

V°pt = J k ,

(5.88)

(5.89)

Distributing the entropy production as evenly as possible along space and time would allow for the design and operation of an economic separation process. Dissipation equations show that both the driving forces and flows play the same role in quantifying the rate of entropy production. Therefore, the equipartition of entropy production principle may point out that the uniform distribution of driving forces is identical to the uniform distribution of flows.

Example 5.7 Equipartition principle: Heat exchanger For a heat exchanger operating at steady state, the total entropy generation P is obtained by integrating over the surface area P =

L~y2dA A

(5.90)

5.6

Equipartition and optimization in separation systems

293

We consider that the duty of the exchanger is specified as qs qs - L I XdA

(5.91)

A

An average driving force over the surface area is obtained as 1 IXdA Xav = -A A

(5.92)

Thus, Eqs. (5.91) and (5.92) yield the specified duty qs qs = LAXav

(5.93)

The above equation shows that for a given surface area A and constant L, specification of the duty leads to the average driving force. Minimizing the integral in Eq. (5.90) subject to a constraint given in Eq. (5.91) is a variational problem, and the solution by the Euler equation in terms of the force is given by (Tondeur, 1990) 0 (X 2 + A X ) = 0 OX

(5.94)

where A is a Lagrange multiplier (a constant). The above equation is satisfied by X = -A/2, i.e., by a constant value of X. The second derivative yields 02

0)(22

(X 2 + AX)>0

(5.95)

The above equation implies that the extremum is a minimum. Thus, with a constant transfer coefficient, the distribution of the driving force that minimizes the entropy generation under the constraint of a specified duty is a uniform distribution. The minimal dissipation for a specified duty implies the equipartition of the driving force and entropy generation along the time and space variables of the process. When the linear phenomenological equations do not hold (Tondeur, 1990), we have P = I JXdV;

J-

Jo--

VJav with X = X ( j )

(5.96)

where j is the specific flow per unit volume. The constraints on these relations are JX>0;

X'-

dX dJ

>0;

J(X=O)=O

(5 97)

The Lagrangian expression is given by F ( J ) = P + A(J - J0 ) = I ( J X + AJ - AJav )dV

(5.98)

The Euler equation corresponding to an extremum of P is given by OF oJ

t'

(5.99)

- J ( J X ' + x + A)dV = 0

which yields JX' + X + A - ~0 Oj[(X+A)J ]

0

(5.100)

294

5.

Thermoeconomics

y

v

X

X

(a)

Figure 5.4.

(b)

Concave and convex relationships between the flow and the force.

The above equation shows that (X = h ) J = constant, and hence the solution yields J = constant and X = constant. Therefore, P is stationary when the flow, force, and entropy generation are uniformly distributed. The sign of the second derivative reveals whether this stationary value is a minimum or not 02

OJ 2 [(X + h)J] = X " J + X '

(5.101)

Since J and X' are positive, the quantity in the above equation (5.101) is always positive when X " > 0 which means when Xis a convex function of J. When the flow J is a linear or a concave function of the driving force X (02J/OX2 < 0) (Figure 5.4a), then equipartition of entropy generation corresponds to minimal dissipation. On the other hand, when X" is negative in the above equation, the sign of 02[(X+ J)J]/OJ 2 may be positive or negative, and may change along the process. When 02F/OJ 2 < O, the value of entropy generation is maximum. When the flow versus force curve is sufficiently convex (Figure 5.4b), nonuniformity may lead to an economic configuration. Such a situation may arise in an electrochemical cell that does not obey the Ohm's law. A strongly convex flow-force curve corresponds to ordered structures, which are dissipative and constantly require a supply of matter and energy from the outside. For example, B6nards cells occur during a natural convection in a fluid system heated from the bottom; after the difference between the surface temperature and the fluid bulk temperature exceeds a certain limit the system moves into the nonlinear region in the thermodynamic branch, and the fluid shows a structured state as long as the temperature difference is maintained. The generality of the equipartition principle should be investigated within flow-force relationships. In near equilibrium phenomena, linear flow-force relations are valid, and optimization criteria (e.g., for coupled heat and mass transfer)generally lead to a constant level of entropy generation along an optimal path provided that there is no constraint imposed on the parameters controlling the system. For systems far from equilibrium, the most stable configurations may correspond to unsteady, dissipative structures. Therefore, equipartition or stability should be considered in the economics of industrial systems or the evolution of natural systems. Equipartition may also help to improve existing designs and avoid flaws in the new design of processes.

Example 5.8 Characterization of the deviation from equipartition Consider a heat exchanger with surface area A and specified heat duty qs. The driving force is constant and equal to its average Xav _ qs

(5.102)

LA

The overall entropy production is

Po -- LfA X2dA

= L(Xav) 2 =

(5.103)

qsXav

Consider a real heat exchanger with the same area A, coefficient L, and qs. The average driving force is Xav. The difference in the entropy production between the real heat exchanger and the one with equipartition configuration is

-

]

(5.104)

5.6

Equipartition and optimization in separation systems

295

where the average driving force is defined in Eq. (5.92). The term (Y2)av is the mean of X 2 over the surface A. Therefore, the contents of square brackets represent the mean quadratic deviation from the mean and hence the variance s 2 of distribution X. The variance is a positive quantity, and we have

P - P o = LAs2(X) = LA[(X-Xav)2]av >0

(5.1o5)

The above equation suggests that the entropy production would be higher if the system deviates from the uniform distribution of the driving force. We can find an equivalent form to the above equation in terms of heat flow q P - Do _ L s 2 ( X )

A

--

1 $2

-L

(q)

(5.106)

where s2(q) is the variance of the distribution of the heat flow. Configurations that minimize s2(X) and s2(q) also minimize entropy production and lead to thermodynamically optimum configurations. Such thermodynamic analysis will contribute to the study of feasibility and economic analysis after relating the level of entropy production to engineering economics.

Example 5.9 Distribution of driving forces Consider two identical heat exchangers 1 and 2 operating at steady state with the same total entropy production P. Assume that the distributions of the driving forces are different and are characterized by s ( < s2

(5.107)

From Eqs. (5.105) and (5.107), we find

-Pol LA

<

P2 -Po2 LA

(5.108)

From the assumption P1 = P2, we have Pol > Po2, since Pol and Po2 are the exchangers operating with an equipartition configuration (s 2 = 0). The heat exchangers may use different temperatures and flow rates for hot and cold streams. Equation (5.108) yields

(Xl,av) 2 > (X2,av) 2

(5.109)

IX,,avl > IXz,av[ and ]qll > q21

(5.110)

The above equation suggests that in heat exchanger 1, for example, the cold fluid would be heated more or the use of a larger cold flow rate is possible. Therefore, the heat exchanger with the smallest s 2 would achieve the largest duty and be more economic in practice. This simple analysis suggests that the distribution of entropy production may play a more important role than total entropy production.

Example 5.10 Variance and heat exchangers Consider two heat exchangers with the same heat duty and total entropy production. They have different heat transfer areas and different variances (Sl) 2 < ($2)2, and hence

Pol > ~Po2

A1

A2

(5.111)

Show that the heat exchanger with the smallest variance is more economic. Equation (5.111) leads to

(Xl,av) 2

(X2,av) 2

4

A2

(5.112)

296

5.

Thermoeconomics

Since the heat duties are the same, we have

q = LAl ( Xl,av ) = LA2 ( X2,av )

(5.113)

By multiplying Eqs. (5.112) and (5.113) side by side, we get (Xl,av) 3 > (X2,av) 3 :=~ Xl,av > X2,av

(5.114)

From Eq. (5.113), we obtain A 1< A2. Therefore, the heat exchanger with the smallest variance requires the smallest heat exchanger area and would be more economic.

Example 5.11 Hot fluid flow rate effect Consider two heat exchangers 1 and 2 operating at steady state and constant pressure with the same heat duty. The total entropy change of the cold fluid is the same for both heat exchangers and determined by the specified heat duty qs. There is no heat loss to the environment. The overall entropy balances for the heat exchangers are

rhlAs1 - P1 = AS

(5.115)

rh22~s2 - P2 = AS

(5.116)

where rh is the mass flow rate, As the specific entropy change of the fluid between output and input, P the total entropy production, and AS the total entropy change of the cold fluid. The heat duty is based on the enthalpy changes of the hot fluid Ah q = &lZ~hl = th2~h 2

(5.117)

If we assume that P1 < P2, and subtracting Eq. (5.117) from Eq. (5.116), we have /'hi ~s1 -/'h2 As2 =-- P1 - P2 < 0

(5.118)

Since the hot fluid becomes colder, As < 0, and we have

thl IZXSl[ >

rh2 IZ~S21

(5.119)

Combining Eqs. (5.117) to (5.119), we find

Ahl < Xh2 2~Sl

As2

(5.120)

On an enthalpy versus entropy diagram (Mollier diagram), the above equation shows the slopes of chords to the constant pressure curve between input and output conditions. The constant pressure curves are convex (02h/0s2). If the input conditions are the same for both exchangers, inequality (5.120) and Figure 5.5 show that ]Ahl] > ]2~h2]

(5.121)

and because of Eq. (5.118), we have rh1 < rh2 . Therefore, exchanger 1, having the smallest entropy production, requires a smaller flow rate of hot fluid. The condition (02h/Os 2) is always satisfied for pure fluids. For mixtures, however, this condition may not always be satisfied and should be verified.

5.6

Equipartition and optimization in separation systems

297

Constant pressure line

.......i/ inlet

Ah t

Ah 2 ............... ... /!

~ /:

y"

1......./

.... ~2........

()utlct 1 Figure 5.5. Mollier diagram for fluids 1 and 2 at constant pressure curve.

Example 5.12 Equipartition principle in an electrochemical cell with a specified duty We desire the electrode to transfer a specified amount of electricity Q over a finite time to

f'"Zdt= Iavt0

(5.122)

~O=-d0

where I is the instantaneous electric current and I~v its average over the time interval to. If Ohm's law holds within the cell, we have AE - R1

(5.123)

where AE is the electric potential difference between the electrodes and R the resistance. The power dissipated within the cell is P =

f tl) AEldt o

= R IO0 I2dt

(5.124)

As before, minimizing the power dissipated for a specified Q implies that I - constant. The deviation from this optimal configuration is similar to Eq. (5.106) P-Po

q

~

Rs2(I)--R

1 $2

(~E)

(5.125)

l0 The above equation shows that the steady state is less dissipative for a specified duty in a finite time. This conclusion is in line with the minimum entropy production principle of Prigogine. If we have multiple working electrodes or the electrode is divided into N different zones, which may work at different potentials, the overall specified duty becomes [o

Q--~-~ fo Idt - iavNt o i=l

(5.126)

Here, the discrete summation is over the N zones. The average Iav is defined over time and number of zones. Repeating the previous procedure and using Po as the entropy production corresponding to Iav at all times and in all zones (equipartition configuration), we have P - Po _ Rs2(i) Nt o

(5.127)

298

5.

Thermoeconomics

The above equation shows the minimal dissipation of power for a specified duty corresponding to an equipartition configuration of flow, driving force, and entropy production along the time and space variables of the process.

5.6.1

Thermoeconomics and Distillation Columns

Tables 4.17 and 4.18 show the performances of distillation columns with diabatic, adiabatic, and isoforce configurations. Diabatic and isoforce column operations reduce exergy losses as well as the amount of hot utility required considerably. These modifications may lead to a mass transfer unit more reversible but may require more transfer units and hence more column height and heat transfer area. For example, in a diathermal distillation column with heat exchangers at every stage, it is possible to adjust the flow ratio of the phases and thus the slopes of operating lines, and the driving forces along the column. This also affects the driving force distribution and entropy production. The task of a process engineer is to decide the target cost or costs to be optimized in a new design or an existing operation. Energy saving in distillation systems has attracted considerable innovative approaches. Such approaches incorporate the principles of thermodynamics and have reached an advanced stage through pinch analysis, exergy analysis, and the equipartition principle. Thermodynamic analysis considers the critical interrelations among energy cost, thermodynamic cost, and ecological cost. Thermodynamic analysis is becoming popular for other separation systems, such as supercritical extraction, desalination processes, hybrid vapor permeation-distillation, and cryogenic air separation. For example, the energy requirement analysis of common cycles used in supercritical extraction has utilized exergy losses and an optimum extraction pressure, which produces a minimum in exergy loss for specified temperature and separation pressure. Distillation columns should be optimized considering both capital cost and operating (energy) cost. The heuristics of using a reflux ratio of 1.03-1.3 times the minimum reflux ratio is in line with both the capital cost and the operating cost for binary distillation systems.

Example 5.13 Optimal distillation column: Diabatic configuration Consider a distillation column made of N distinct elements. The heat and mass transfer area can be defined separately. Defining an investment cost Ci for element i as a linear function of the size Ai, we have (5.128)

Ci -- ceiAi + fli

where a is proportionality cost factor and/3 a fixed cost (including the heat transfer area). We assume that the average operating cost Ci is a linear function of the exergy loss in the element i per year

Ci,av = "Yi [Exi + ToPi ] + 6i

(5.129)

where Yi is a proportionality cost factor and 6 i a fixed cost, while To is the reference temperature (environmental temperature). Here, the exergy loss is split into a thermodynamic minimum Ex i (e.g., minimum separation work), and an irreversible contribution ToPi, where Pi is the entropy produced in element i. In the diabatic distillation column, each element is small enough that equipartition of entropy production may be approximately achieved by adjusting the heat flows and thus the liquid and vapor flow rates. We assume that each element performs a specified duty of J/0. The total cost function Ct for all N elements is N

C t =Zfi,av-+-7"Ci

(5.130)

i

where z is the yearly amortization rate. Using the variational approach, we minimize a Lagrangian as N

N

~'~ = Z Ci,av + TCi + Z ~i (di - dio ) = O i i

(5.131)

This should satisfy the following conditions: -0 OA i

0212 and ~ > 0

(for alli)

(5.132)

5.6

Equipartition and optimization in separation systems

299

Using Xav = Jo/LA, Pi becomes

j2 (5.133)

L,A,

So, f~ becomes

f~ = ~Z~ rczi~ + r~i + yiExi + 6i + •

L,4

+ Ai (Ji - Jio )

(5.134)

The derivative with respect to A/yields a system of independent equations 0~ -

OAj

J J-r°2 =0 TOLj - -

2

LiAj

(5.135)

or

r =

/ToP:

~jAj

(5.136)

and we have

02D.

2r

A,

>0

(5.137)

Since r is a constant depending only on economic conditions, the right of Eq. (5.136) must be independent of the element j, and thus an invariant throughout the process. The product rjaAj is the annual cost related to irreversible energy waste in terms of exergy loss. These two quantities should be equal in any element. We may conclude that under these assumptions, the optimal size distribution of the elements requires the equipartition of the ratio TToP/aA on all the elements, and the cost of exergy loss is equal to the amortized proportional investment cost in that element. The equipartition principle is mainly used to investigate binary distillation columns, and should be extended to multicomponent and nonideal mixtures. One should also account for the coupling between driving forces since heat and mass transfer coupling may be considerable and should not be neglected especially in diabatic columns.

Example 5.14 Optimal feed state for a binary distillation Consider a binary distillation column with specified distillate and bottom compositions. The feed composition is 30 mol% of the more volatile component. Investigate the problem of conditioning the feed. Should the feed be in saturated liquid or saturated vapor state? Figure 5.6 shows the feed lines and operating lines for a saturated liquid feed and a saturated vapor feed. The feed line for liquid comes with an infinite slope as a vertical line starting from the feed location on the diagonal line on a McCabe-Thiele graph. On the other hand, the slope of the feed line for a saturated vapor is zero and represented by a horizontal line starting from the feed location. As shown, saturated liquid feed yields a better distribution of area between the equilibrium curve and the top and bottom operating lines compared with the area of a saturated vapor feed. The departure from equilibrium is more evenly distributed for the liquid feed and hence yields a better performance. For example, the trays are evenly distributed above and below the feed stage and the reflux ratio is smaller in the liquid feed. This configuration may be different for a different feed composition. 5.6.2

Retrofit of a Distillation Column

Retrofits are modifications of existing distillation columns to reduce the cost of operations by increasing the efficiency in energy utilization. Thermodynamic analysis is one method for determining the appropriate retrofits. Thermodynamic analysis mainly seeks modification targets for reducing thermodynamic losses due to heat and mass transfer, pressure drop, and mixing. For example, in a binary distillation, operating curves come closer to the equilibrium curve, and the reflux ratio approaches its minimum value. However, multicomponent distillation may be a more difficult

300

5.

Thermoeconomics

Y

Feed

x

Figure 5.6. State of the feed for a distillation column. The bold line is the operating line for a saturated liquid feed.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

t °"EF, t.............................................. t sect,o.5

-~ ,

,[ OFFGAS .... 1

HIERARCHY

I

SECTION1 HIERARCHY

---4 Asl

SECTION2

-t

I-->

I

SECTION3

407 ] - - ~

HIERARCHY

SECTION4 HIERARCHY

HIERARCHY

A ..................................................

A

I QTOP ~..................................................

..............................................................[QREB t..............................................................

Figure 5.7. Connection of the sections of the methanol plant with material and heat streams. problem; the sharpness of multicomponent separation is limited, and near reversible operating conditions may be difficult to achieve. To analyze the performance of an existing column quantitatively for exploring the energy saving potential, it is customary to construct the temperature-enthalpy and stage-enthalpy curves (called the column grand composite curves), and the stage exergy loss profiles. The column grand composite curves display the net enthalpies for the actual and ideal operations at each stage, and the cold and hot heat utility requirements, while the exergy loss profiles indicate the level of irreversibility at each stage, including the condenser and reboiler. Therefore, the area between the actual and the ideal operations in a column grand composite curve should be small, and exergy losses should be lower for a thermodynamically efficient operation. The column grand composite curves are constructed by solving the mass and energy balances for a reversible column operation. The stage exergy loss profiles are generated by the stage exergy balance calculations with a reference temperature.

Example 5.15 Retrofits of distillation columns by thermodynamic analysis The synthesis of methanol takes place in a tube reactor in section 3 in the methanol plant shown in Figure 5.7. The reactor outlet is flashed at 45°C and 75 bar, and the liquid product (stream 407) containing 73.45 mol% of methanol is fed into the separation section (see Figure 5.8), where the methanol is purified. Stream 407 and the makeup water are the feed streams to the section. Table 5.2 shows the properties and compositions of the streams in section 3. The converged simulations are obtained from the Redlich-Kwong-Soave method to estimate the vapor properties, while the activity coefficient

301

5.6 Equipartition and optimization m separation systems

,LA

H

--

i40N,EED, 1

,'

lii

il

Figure 5.8. Separation section of the methanol plant with subsystems used in thermodynamic efficiency estimations: ($1) column 1; ($2) column 2; ($3) columns 1 and 2. Table 5.2 Material and heat streams for the separation section of the methanol plant

Stream FEED4 MKWATER BTMS SIDE1 LIQ2 VAP1 OFFGAS METHANOL LIQ1 FEED2 FEED1 ql q2

n

ii7

T

H

I:t

S

Ex*

(kmol/h)

(kg/h)

(K)

(kJ/mol)

(MW)

(J/(mol K))

(kJ/mol)

-186.5 -355.0 -81.29 -1.254 -0.01011 -3.260 -5.243 - 124.7 -0.02267 -207.7 -216.2 15.299 18.900

-215.49 - 166.95 -142.06 -207.77 -224.70 -21.15 -32.75 -224.60 -237.28 - 192.77 -208.38 -

2655.32 444.21 1050.96 18.43 0.1560 33.80 70.25 1925.59 0.3414 2995.14 3029.28 -

76938.66 8002.62 18955.20 550.00 5.00 1388.90 2330.66 61700.40 11.10 81210.60 82610.60 -

318.15 313.15 393.17 359.77 348.00 305.91 318.86 348.29 305.91 359.00 323.15 377.00 409.00

-252.82 -287.73 -278.46 -244.99 -233.10 -347.23 --268.66 -233.07 -239.07 -249.61 --256.89 -

-188.44 -237.77 -235.93 -182.88 -165.97 -340.66 -258.69 - 165.97 -168.19 -- 192.00 - 194.68 3.199 5.122

*To= 298.15 K.

model NRTL and the Henry components method are used for predicting the equilibrium and liquid properties. Assessments of the performances of the existing columns, suggested retrofits, and the effectiveness of the retrofits with minimum or no change in column pressure and stage numbers are discussed below: C o l u m n 1: As the base case design in Table 5.3 shows, column 1 has 51 stages, and operates with a partial condenser with a duty of 1.3 71 M W at the top, and a side condenser with a duty of 8.144 M W at stage 2. It has no reboiler; however, it receives a side heat stream with a duty of 15.299 M W to the last stage from section 2 of the plant. The temperature profiles of both columns are shown in Figure 5.9. The temperature-enthalpy and the temperature and composition profiles may help in assessing the operation and determining the extent and position of side heating or condensing for the column. Figure 5.10a shows the temperature-enthalpy curve. There exists a significant area difference between the ideal and actual enthalpy profiles, which identifies the scope for side condensing. As the temperature change after stage 3 is very small, and a side condenser at stage 2 already exists, a second side condenser at stage 4 with a duty of 2.1 M W has been installed. As Figure 5.10b shows, the side condenser has reduced the area between the ideal and actual enthalpy profiles to some degree without increasing the number of stages. Figure 5.11 also shows that the actual vapor flow closely follows the thermodynamic ideal minimum vapor flow at stages 2 - 4 after the retrofitting. The duty of 2.1 M W is in the range of enthalpy difference between the hot duty of 15.299 M W and the total cold duty of 9.51 M W

302

5.

400 -

Thermoeconomics

J

I

I

390 - ] -o-- Column 1 _- Column2 380 -

J

J

370 360-

~ 340 -c 330 320 310 300 0

10 20 30 40 50 60 70 80 90 100 Stage

Figure 5.9. Temperature profiles for the columns in Example 5.15. Table 5.3 Operating parameters of designs 1 and 2 for column 1 Parameter No. of stages Feed stage Feed temperature (°C) Reflux ratio Condenser duty (MW) Distillate rate (kmol/h) Condenser temperature(°C) Side condenser 1 stage Side condenser 1 duty (MW) Stage 2 temperature(°C) Side condenser 2 stage Side condenser 2 duty (MW) Stage 4 temperature(°C) Heat stream (ql) duty (MW) Heat stream (ql) stage Heat stream (ql) temperature(°C) Boilup rate (kmol/h) Bottom rate (kmol/h) Bottom temperature(°C)

Design 1 (base case) 51 14 43.74 3.7 - 1.3717 34.141 32.75 2 -8.144 69.4 74.37 15.299 51 104 1551.284 2995.144 85.85

Design 2 (retrofitted) 51 14 65 4.56 - 1.691 34.140 32.75 2 -7.7 70.26 4 -2.1 74.36 15.299 51 104 1551.560 2995.144 85.85

(side condenser + partial condenser shown in Table 5.3). The existing side condenser duty is reduced to 7.7 from 8.144 MW, so that the new total duty of 11.49 MW is close to the previous total of 9.51 MW. After the retrofitting, therefore, the total cost would not change much, and the need for extra stages would be negligible as the heat changes sharply below the first side condenser. Since side heat exchangers are more effective at convenient temperature levels or stages for exchanging heat using cheaper utilities, care should be exercised in positioning them. Another approach may be based on the uniform distribution of the driving forces that cause the separation, leading to less entropy production and hence less exergy loss in the column where the coupling of heat and mass transfer may not be negligible. Figure 5.10 also displays a sharp change of the enthalpy on the reboiler side. The extent of the change determines the approximate feed preheating duty required, as the feed at 43.74°C is highly subcooled (Table 5.3). Therefore, a new heat exchanger (HEX, in Figure 5.8) with a duty of 1.987 M W is used as the second retrofit for the column, and the feed temperature has increased to 65 from 43.74°C. Figure 5.12 compares the enthalpies for the base case and retrofitted designs. The difference between the hot and cold duties is lower, and the actual and ideal profiles are closer to each other after the retrofits.

5.6

Equipartitionand optimization in separation systems

303

360

350-

340 -

330

320

310]

,

300 -2.5

i

0

2.5

5

7.5 10 12.5 Deficit, MW (a)

Enthalpy

,

15

,,

17.5

360

350

34()

~D

330

~D

32O

310 -.o-- Ideal Profile ] 300 -2.5

,

i

i

()

2.5

5

Enthalpy

i

i

7.5 10 i2.5 Deficit, MW (b)

,

15

17.5

Figure 5.10. Temperature-enthalpy curves for column l'(a) design 1 and (b) design 2.

The suggested retrofits also aim at reducing the irreversibility due to mixing of streams at different temperatures at the feed stage, which is at 80.18°C, and throughout the column. The exergy loss profiles of Figure 5.13 show that the reduction in exergy loss at the feed stage is - 6 0 % with the values of 0.3865 MW in design 1 and 0.1516 MW in design 2. However, the exergy loss at the partial condenser increases by 28%, and becomes 0.150 MW in design 2 instead of 0.117 MW in design 1. As Table 5.5 shows, the reduction in the total exergy loss or the recovered available energy is 21.5% with total column exergy losses of 0.837 and 0.656 MW in designs 1 and 2, respectively. Table 5.3 compares the base case design and the retrofitted design. Column 2: As the base case design in Table 5.4 shows, column 2 has 95 stages, and a total condenser with a duty of 281.832 MW. It operates with a high reflux ratio, and receives a side heat stream of 18.9 MW to the last stage from section 2 of the plant. One of the side products is the methanol stream described in Table 5.2, and drawn at stage 4 at 348.3 K. The second side product is drawn at stage 86 at 361.2 K. The temperature profile in Figure 5.9 shows that the temperature increases sharply after stage 84, and the separation system resembles a stripping column with the feed close to the bottom at stage 60. Figure 5.14a shows a significant area difference between the ideal and the actual enthalpy profiles above the feed stage representing the pinch, and hence suggests side reboiling at appropriate temperature levels to decrease the difference. The existing reboiler duty is 282.28 MW (Table 5.4). Besides that, there is a side product at stage 86 and a side heat input of 18.9 MW at stage 95. Therefore, the decision has been made to install two side reboilers

304

5.

Thermoeconomics

0 .

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

,

,,,

,,,,

~

, .....

,,,

ag ID

gq

og > 1

6

L

11

16

21

26

31

36

41

46

51

31

36

41

46

51

Stage (a)

~g o

oo° > 6

11

16

21

26 Stage

(b) Figure 5.11. Vapor flow profiles of column 1" (a) design 1 and (b) design 2 in Example 5.15 (Table 5.3 describes designs 1 and 2).

Table 5.4 Operating parameters of designs 1 and 2 for column 2 Parameter No. of stages Feed stage Feed temperature (°C) Reflux ratio Condenser duty (MW) Distillate rate (kmol/h) Condenser temperature (°C) Reboiler duty (MW) Boilup rate (kmol/h) Bottom rate (kmol/h) Reboiler temperature (°C) Side reboiler 1 stage Side reboiler 1 duty (MW) Stage 87 temperature (°C) Side reboiler 2 stage Side reboiler 2 duty (MW) Stage 92 temperature (°C) Heat stream (q2) duty (MW) Heat stream (q2) stage Heat stream (q2) temperature (°C)

Design 1 (base case) 95 60 85.84 188765 -281.832 0.156 74.85 282.283 24890.68 1050.959 119.71 90.97

110.91 18.90 95 136

Design 2 (retrofitted) 95 60 85.84 188765 -281.833 0.156 74.85 52.292 4633.93 1049.66 120.02 87 180 93.35 92 50 110.94 18.90 95 136

5.6

Equipartition and optimization in separation systems

305

360 O

j

i

12

14

35O

:~ 34O

~D

sso

~" 3 2 0 --o2- D e s i g n 1 310

~Design

2

'

300 0

2

4

6

8

10

16

Enthalpy deficit, MW (a)

50

40 ---o-- D e s i g n 1 •

3O

~

+

Design 2 i

20

~

,

•

10

12

14

() 0

~

4

(5

8

Enthalpy deficit, MW (b)

Figure 5.12. Column grand composite curves for column 1 in Example 5.15: (a) temperature-enthalpy deficit curves and (b) stage-enthalpy deficit curves (Table 5.3 describes designs 1 and 2).

at stages 87 and 92 with the duties of 180 and 50 MW, respectively. Obviously, these two side reboilers are more economic as they operate at lower temperatures and require less expensive steams compared with the steam used in the existing reboiler. With the two side reboilers, the duty of the reboiler decreases to 52.3 from 282.3 MW. Extra stages due to the side reboilers would be minimal since the enthalpy rises sharply at each stage after stage 84. Figure 5.14b shows a considerable reduction in the area between the ideal and actual enthalpy profiles after the retrofits. Moreover, Figure 5.15 shows that the side reboilers have reduced the gap between the ideal and actual vapor flows between stages 84 and 95, where the stage temperatures change sharply. The enthalpy curves in Figure 5.16 also show that the retrofitted design is closer to ideal operation than design 1. Figure 5.17 compares the exergy loss profiles in designs 1 and 2. The base case design operates with rather large exergy losses at the feed stage and around the reboiler. The rest of the column has negligible exergy losses mainly due to the flat methanol concentration profile. The retrofits reduce the total exergy losses by --~41.3%, and hence save a considerable amount of the available energy. Table 5.4 compares the two designs of the colunm. Using data from Table 5.2, the minimum values of exergy for the required separation and thermodynamic efficiencies for designs 1 and 2 are estimated, respectively, and compared in Table 5.5. The estimations are based on the value T0 = 298.15. Figure 5.8 identifies the subsystems considered in Table 5.5. The reductions in the exergy losses range from 21.5% to 41.35%. The thermodynamic efficiencies have increased considerably in the retrofitted designs, although the efficiencies remain low, which is common for industrial column operations. For column 1, the efficiency increases to 55.4% from 50.6%, while the efficiency increases to 6.7% from 4.0% in column 2.

306

5.

Thermoeconomics

360 ¢

~_________._____._____._______.~

350 ~ [ ~N~,~[

1---O--Design 1[ [ ~ D e s i g n 2[

340 330 e-

320

310 300 0

, 0.1

, , 0.2 0.3 Exergy loss, MW

0.4

(a) 50 ,~ I

40

---O-- Design 1 Design 2

~30

I

20

0 0

0.1

0.2 0.3 Exergy loss, MW

0.4

(b)

Figure 5.13. Exergy loss profiles for column 1 in Example 5.15: (a) temperature-exergy loss profiles and (b) stage-exergy loss profiles (Table 5.3 describes designs 1 and 2).

Approximate economic analyses shown in Tables 5.5 and 5.6 compare the fixed capital costs of the retrofits with the savings in electricity due to the reduced exergy losses. The fixed capital cost consists of equipment, materials, construction, and labor. Table 5.6 shows the approximate values of fixed capital costs for the heat exchangers needed in the retrofits. The costs are estimated by using the current chemical engineering plant cost index of 420, and the approximate areas obtained from the individual duties. The energy saving estimations are based on the unit cost of electricity of US$ 0.060/(kWh) and a total 8322 h/year of plant operation. The costs of related retrofits and the yearly saved exergy equivalent of electricity for each subsystem are compared in Table 5.5, which shows that the retrofits are effective and save a considerable amount of energy in electricity per year. Column grand composite curves and exergy loss profiles enable process engineers to assess an existing operation, suggest retrofits if necessary, and determine the effectiveness of the retrofits. The suggested retrofits consist of an additional side condenser at stage 4 and feed preheating for column 1, and two side reboilers at stages 87 and 92, respectively, for column 2. The effectiveness of the retrofits has been assessed by the improved column grand composite curves and exergy loss profiles as well as by an approximate economic analysis. After the retrofits, actual and minimum vapor flow profiles have become closer. Also, the difference between the ideal and actual profiles of the enthalpies in the column grand composite curves has become smaller. The range of reductions in the total exergy losses is 21.5-41.3%, which leads to a considerable reduction in the available energy losses. The

5. 7

307

Thermoeconomics of latent heat storage

t11 ~ 0

tb i

t

.,

I /

1

I

=

7'-

Actual Profil© f

t

i,i

IL

............

-50

0

50

100

150

FalfllalpyDeficit (a)

200 MW

250

300

o

o

1 i--:-c~° __~ ÷

I 1 I Ideal Profile Actual Profile

.J"1

1

o

o

.J ee5 ,,

~;

~.,

:->-.~

T 1

0

0

50

100

.......

150

200

EnthalpyDeficit M W (b)

250

300

Figure 5.14. Temperature-enthalpy deficit curves for column 2 in Example 5.15: (a) design 1 and (b) design 2 (Table 5.4 describes designs 1 and 2).

thermodynamic efficiencies also increased considerably as thermodynamic imperfections decrease. The savings in electricity can recoup the initial cost of the retrofits in a short time. Government incentives for environmentfriendly designs may reduce the cost of the retrofits further. 5.7

THERMOECONOMICS

OF LATENT HEAT STORAGE

Latent heat storage is a popular research area with industrial and domestic applications, such as energy recovery of air conditioning, and under-floor electric heating by using a phase changing material. Figure 5.18 shows the charging and discharging operations with appropriate valves, and temperature profiles for countercurrent latent heat storage with subcooling and sensible heating. An optimum latent heat storage system performs exergy storage and recovery operations by destroying as little as possible of the supplied exergy. A charging fluid heats the phase changing material, which may initially be at a subcooled temperature Tso and may eventually reach a temperature Tsh after sensible heating. Therefore, the latent heat storage system undergoes a temperature difference of Tsh - Tsc, as shown in Figure 5.19. Heat available for storage would be qc - UA(ATu~)~ - thcCpc (Tci-Tco)

(5.138)

where U is the overall heat transfer coefficient, A is the heat transfer area, rnc is the charging fluid flow rate, Tci and Tco are the inlet and outlet temperatures of the charging fluid flow and z~Tlm the logarithmic mean temperature difference expressed by (Tci- Tsc)- (Tco - Tsh)

Tci - Tco

( Tci - T~c ) In T~o _ Tsh

NTU c

(5.139)

308

5.

Thermoeconomics

CD O g

.

.

~r

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

---c--- ThermodynamicIdeal MinimumFlow * Actual Flow

0

> 2 _/

.............................................. .

1

21

.

.

41

.

.

.

.

.

.

61

81

101

Stage (a) O . . . . . . . . . . . . . . . . .

C O

~..,

•

............................ 1

•

ThermodynamicIdeal MitKrnumFlow ActualFlow

t

,,

l

,

g--

{

O

tD

,aoo o~ o

! i

.2° Ph

I .......l............

. . . . . . . . . . . . . . . . . . . . . . . . . .

1

21

zl

, ..........

,,

,,

61

81

101

Stage (b)

Figure 5.15. Vapor flow profiles for column 2: (a) design 1 and (b) design 2 in Example 5.15 (Table 5.4 describes designs 1 and 2). where N T U = UA/(rhcCpc ) = (Tci- Tco)/Arlm is the number of transfer units. Equation (5.139) relates the value of NTU with temperature. Heat lost by the charging fluid qc will be gained by the phase changing material qs qc = qs = ms [Cps ( Tl - l'si ) + A H m q'- Cpl ( l"sh -- Th ) ]

(5.140)

where AHrn is the heat of melting, T1 and Th the lowest and highest melting points of the phase changing material, and @s and Cpl the specific heats of solid and liquid states of the phase changing material, respectively. The net rate of exergy E x of the charging fluid is

A/~x c - (/~Xco-/~Xci ) - th c

Cpc

( T c i - Tco ) - T O In

(5.141)

Exergy stored by the phase changing material is

Os where Ts is an average temperature of storage, which may be approximated by (Tsc + Tsh)/2. The first and second law efficiencies are r/=

actual heat stored maximum energy gain

_

Tci-

- ~

Tco

Tci-T s

(5.143)

5. 7

309

Thermoeconomics of latent heat storage

100 90 80 70 6O 50 40

i

--<>--- Design 1

30

~

20

............

Design 2

,!

-~

10 0 0

50

100 150 200 Enthalpy deficit, MW (a)

100

I

!

250

300

i

90 80 70 60 ~D

50 40

Design 1 .........

3O

Design 2 1

i' 20

.

+

_

!

i . . . . . . . . . . . . .

I

{

........

!

10

0

50

100

150

200

250

300

Enthalpy deficit, MW (b)

Figure 5.16. Column grand composite curves for column 2 in Example 5.15: (a) temperature-enthalpy deficit curves and (b) stage-enthalpy deficit curves (Table 5.4 describes designs 1 and 2).

exergy of PCM

(T~i - Tc° ) 1 - T~h [ (Tci)j rbh - exergy of charge fluid = (Tci _ Tco ) - To In Tc°

If it is assumed that the phase changing material is totally melted and heated to a temperature the discharge flow qd is estimated by

qd -- U A A ( Tim )d --

thdCpd(Tdi -- Tdo )

(5.144)

Tsh ,

recovered heat by

(5.145)

310

5.

Thermoeconomics

400 390 380 370 [.., 360 Design ~1__ Design

350 340 0.00

t

1.00

2.00 3.00 Exergy loss, MW

4.00

(a) 10090- ~'L-~ - 807060

---o- Design 1 ~ Design 2

1

5040302010

oi

0.00

1.00

2.00 3.00 Exergy loss, MW (b)

4.00

Figure 5.17. Exergy loss profiles for column 2 in Example 5 15: (a) temperature-exergy loss profiles and (b) stage-exergy loss profiles (Table 5.4 describes designs 1 and 2).

Table 5.5 Assessments of the effectiveness of the retrofits System

Design 1 (base case) Exmin (MW)

(S1) Column 1 ($2) Column 2 (S1 + $2) Columns 1 + 2

Exloss (MW)

Design 2 (retrofitted)

llth (%)

Exmin (MW)

0.815

0.856

0.837

50.6

1.136

26.979

4.0

1.992

27.814

6.7

EXloss (MW)

'Tith (%)

Saved EXloss (MW)

Change Exloss (%)

FCC a of retrofits (US$)

Electricity savingb (US$/year)

0.656

55.4

0.179

21.5

183500

89578

1.135

15.847

6.7

11.133

41.3

409000

5558829

1.950

16.502

10.6

11.312

40.7

592500

5648407

(Exloss) total column exergy loss from the converged simulation by Aspen Plus with Soave-Redlich-Kwong, NRTL, and Henry component methods. aFCC" fixed capital cost. bElectricity equivalent of energy savings is based on a unit cost of electricity of US$ 0.060/(kWh).

5. 7

311

Thermoeconomics of latent heat storage

Table 5.6 Approximate fixed capital cost calculations for the retrofits Heat exchanger

Type

Duty (MW)

Preheater (HEX), column 1 Side condenser, column 1 Total for column 1

S/T b, fixed tube sheet S/T b, fixed tube sheet

Side reboiler 1, column 2 Side reboiler 2, column 2 Total for column 2

Floating head Floating head

Material

P (bar)

Area (m 2)

FCC ~

(us$)

1.9 2.1

5.0 1.5

Carbon steel Carbon steel

130 130

90500 93000 183500

180.0 50.0

2.0 2.0

Carbon steel Carbon steel

600 170

294000 115000 409000

~'Approximate fixed capital cost with the: chemical ~::ngineering plant cost index = 420. bS/T: shell and tube.

Solar Energy

Solar Energy Exp2c, cp2c

Exse, cse

1_ 4 ~a~ ca ]

Solar Air Heaters

<mfl "-I I ~

1

,

Latent Heat Storage 2

Ex3p, c3p

T

_.. Ex3a, c3a

Greenhouse

Exp2d, cp2d

Figure 5.18. Units of the latent heat storage system.

T

d '

/~xp 1, cpl

~

Tco

/~xp2c, cp2c Ts p¢,~,

,,

Tsc

Exp2d, cp2d

Ex3a, c3a

i

Figure 5.19. Approximate temperature profiles for a latent heat storage unit.

The net exergy change of the charging fluid would be

~L'X d -- (L'Xdi --/~Vdo ) -- mdCpd [(Tdi -- Tdo )-- TO In

rdo

(5.146)

The first and second law efficiencies are Tdo -- Tdi

(5.147)

Tdi -- Tsl

exergy given to discharge fluid 7Ith =

exergy of PCM

rd o -- Td ~ -- To I n

i r0 ~--

(Tdo-Tdi)(1--~sl)

(5.148)

312

5.

Thermoeconomics

All the temperatures are time dependent, and the charging and discharging cycles need to be monitored over the time of operation. Structural theory facilitates the evaluation of exergy cost and the incorporation of thermoeconomic functional analysis. Structural theory is a common formulation for the various thermoeconomic methods. It provides costing equations from a set of modeling equations for the components or units of a system (Figures 5.18 and 5.19). Structural theory needs a productive structure displaying how the resource consumption is distributed among the components of a system. The flows entering a component in the productive structure are considered fuels F and flows leaving a component are products P. The components are subsystems with control volumes as well as mixers and splitters. Therefore, the productive structure is a graphical representation of the fuel and product distribution. For any component j, or a subsystem, the unit exergy consumption eXc is expressed on a fuel/product basis by

eXcj -

Fj_Fj Ps Exs

(5.149)

For linear modeling, the average costs of fuels and products are defined by C*

-

OF° •

sin-Ore----f,

*

(5.150)

OF°

C jp - op j

where F o is the fuel to the overall system expressed as a function of the flow other related parameters. The total annual production cost Cr in $/(kW h) is N

N

j=l

j=l

mj or product pj, respectively, and the

(5.151)

where ci is the specific cost of product i in $/(kW h), CjF is the cost of fuel, and of componentj in kW and is expressed in terms of NTU using Eq. (5.139)

Exj : rhjCp[NTUjA~mj-T o ln(TJl]]

r0)j

Exj is the rate of exergy as a product

(5.152)

The optimum total production cost rate with respect to NTU is obtained from dC¢ - 0 dNTU

(5.153)

Including an ecovector to account for the exergoeconomic costs or environmental impact can extend the thermoeconomic approach. An ecovector is a set of environmental burdens of an operation, and can be associated with input flows; it includes information about natural resources, the exergy of these resources, and monetary costs. The external environmental costs associated with the environmental burdens may also be added into the ecovector. Extended exergy accounting includes the exergetics flowcharts for nonenergetic costs of labor and environmental remediation expenditures.

Example 5.16 Cash flow diagram for seasonal latent heat storage Figure 5.18 shows the three basic components of a latent heat storage system: solar air heaters, latent heat storage unit, and the greenhouse. Whenever the temperature in the greenhouse drops below a set point, a fan circulates the air from the greenhouse through the latent heat storage unit until the temperature reaches the required level. Costs are the amount of resources consumed to produce a flow or a product. When exergy is added into a flow, the cost of the flow leaving a component is equal to the cost of the flow entering plus the fuel value of added exergy. When exergy is removed, the fuel value of exergy is subtracted. The products and their average costs in the productive structure shown in Figure 5.18 are summarized below.

5. 7

313

Thermoeconomics of latent heat storage

Unit 1: Solar air heater system- Added exergy provided by the solar air heater system can be expressed in terms of NTU and z~Tlm using Eq. (5.152) as (5.154)

The airflow leaving the solar air heaters adds exergy, and therefore the cost is

Clp = Cli -+-C1v

(5.155)

where Cli is the cost of the flow entering unit 1 and CIF the fuel value of added exergy. The specific costs of warm air Ca and exergy Cex are

Clp

ClpF

Ca - /~Xlp- EXlp'

-- C1F %

(5.156)

A/~xl

where Clpv is the fuel value of the product leaving the solar air heating system.

Unit 2: Latent heat storage system- Figure 5.19 shows an approximate temperature profile within the storage unit. The removed exergy by the latent heat storage system during charging is

[NTUclmc In

(5.157)

The cost of the product after charging is

Ccp = Cci - Ccv

(5.158)

The specific costs of the product leaving the latent heat storage unit Cc and the removed exergy Cexc are

CC

m

Ccp _ CcpF /~Xcp ]~Xcp

CcF

,

2cExc

Cexc

°

(5 159)

Discharging flow adds exergy form the latent heat storage, and is given by

A/~x d = (/~Xdp --/~Xdi ) -- th3 Cp [ NTU d ATlmd-Toln(

TdO]

~,Tdi )

J

(5.160)

The cost of discharging flow is

Cdp = Cdi -t- CdF

(5.161)

The specific costs of discharging flow leaving the latent heat storage c d and the added exergy Cexd are

cd -

Cdp _ CdpF . --, EXdp EXdp

¢exd

-

CdF A]~Xd

(5.162)

Unit 3: Greenhouse- The exergy change within the greenhouse is A]~x3 - L'x3p -/~x3i - lCvl3CpINTU3z~Tlm3

t, T3i )

(5.163)

314

5.

Thermoeconomics

The exergy from the discharge flow is removed in the greenhouse, and the cost of the flow leaving the greenhouse becomes C3p = C3i - C3F

(5.164)

The specific costs of flow leaving the greenhouse Cg and the exergy removed Cexgare C3p __ C3pF Cg -- /~X3p /~X3p '

C3F ¢exg -- ~ X

(5.165)

3

The total cost of products of the three components would be CpT = Clp + Ccp + Cdp + C3p -- Ca/~Xlp Jr Cc/~Xcp Jr Cd/~Xdp Jr Cg/~X3p

(5.166)

The cost of a product for componentj is based on a fuel/product basis Cjp = Cjpv, SOthat the total cost of products is

= C,p + % + Cap + C3p = clp +

+

+ C3p

(5.167)

Cost optimization basically depends on the trade-offs between the cost of energy (fuel) and capital investment as seen in Figure 5.20. Fines for pollution and incentives for environment-friendly technologies may reduce the cost associated with exergy loss. The thermoeconomics of the latent heat storage system involves fixed capital investment, operational and maintenance cost, and exergy costs. The total fixed capital investment consists of (i) direct expenses, which are equipment cost, materials, and labor, (ii) indirect project expenses, which are freight, insurance, taxes, construction, and overhead, (iii) contingency and contractor fees, and (iv) auxiliary facilities, such as site development and auxiliary buildings. Table 5.7a shows the data used in the thermoeconomic analysis. Discounted cash flow diagram can determine the profitability criteria in terms of the payback period, net present value, and rate of return from. In the discounted cash flow diagram each of the annual cash flow is discounted to time zero for the latent heat storage system. The payback period is the time required, after construction, to recover the fixed capital investment. The net present value shows the cumulative discounted cash value at the end of useful life. Positive values of net present value and a shorter payback period are preferred. The rate of return is the interest rate at which all the cash flows must be discounted to obtain zero net present value. If rate of return is greater than the internal discount rate, then the latent heat storage system is considered feasible. Figure 5.20 shows the discounted cash flow diagram obtained from Table 5.7b using the data in Table 5.7a. A net present value ofUS$102,462.21 is obtained at the end of 15 years of useful life, which shows a profitable investment. The approximate discounted payback period is about eight years. The discounted rate of return is --~10.485%, which is greater than the internal interest rate of 8%. By changing the values of exergy costs, or the tax rate the cash flow diagram can be modified easily (Demirel and Ozturk, 2006).

Table 5.7a Economic data used for the thermoeconomic analysis of the seasonal heat storage system

Fixed capital investments for the components FCI1 + FCI 2 + FCI3 = US$ 200000 + US$ 200000 + US$100000 = US$ 500000 Cost of land: L = US$ 50000 Working capital: WC = 0.2(US$ 500000) = US$100000 Yearly revenues or savings: R = US$160000 Total cost of production (Cop) from Eqs. (5.166) and (5.167) C o p - - CpT = Clp -q- Ccp -+- Cdp -Jr-C3p = Clp F q- Ccp F + Cdp F Jr- C3p F -- US$ 55000 Clp F -- U S $

20000, Ccp F -- U S $ 1 5 0 0 0 ,

Cdp F = U S $ 1 0 0 0 0 , C3p F -- U S $ 1 0 0 0 0

Taxation rate: t = 35% Salvage value of the whole seasonal storage: S = US$ 50000 Useful life of the system: n = 15 years; depreciation over 10 years Discount rate: i = 8%

315

Problems

Table 5.7b Discounted cash flow estimations for the seasonal latent heat system n

FCI

D=

A t`'

Cop

R

Bc

DCF

CCF -50000

0

-50000

0

500000

0

-500000

-50000

1

-500000

0

500000

0

-600000

-555556

-605555.56

2

0

50000

450000

160000

55000

85750

73516.8

-532038.75

3

0

50000

400000

160000

55000

85750

68071.11

-463967.64

4

0

50000

350000

160000

55000

85750

63028.81

-400938.83

5

0

50000

300000

160000

55000

85750

58360.01

-342578.82

6

0

50000

250000

160000

55000

85750

54037.05

-288541.77

7

0

50000

200000

160000

55000

85750

50034.3

-238507.47

8

0

50000

150000

160000

55000

85750

46328.06

-192179.41

9

0

50000

100000

160000

55000

85750

42896.35

-149283.07

10

0

50000

50000

160000

55000

85750

39718.84

-109564.22

11

0

50000

50000

160000

55000

85750

36776.71

-72787.51

12

0

0

50000

160000

55000

68250

27103.01

-45684.50

13

0

0

50000

160000

55000

68250

25095.38

-20589.12

14

0

0

50000

160000

55000

68250

23236.47

2647.34

15

0

0

50000

160000

55000

68250

21515.25

24162.591

16

200000

0

50000

160000

55000

268250

78299.62

102462.20

~Depreciation- straight line method: D = (FCI - S)/n. bBook value: A = FCI - ED/,. CAfter tax cash flow - net profit + depreciation: B = (R - Cop - Dl<)(1 - t) + D~.

2ooooo I 1ooooo -1

2

4

6

- 100000 i

-200000 -300000

/

S

f

-400000 -500000 -600000 -700000 year

Figure 5.20. Discounted cash flow diagram based on the economic data in Table 5.7a and cumulative cash flows in Table 5.7b for the latent heat storage system.

PROBLEMS 5.1

In a steam power generation plant (see the following schematic), the boiler uses natural gas as fuel, which enters the boiler with an exergy rate of 110 MW. The steam exits the boiler at 6000 kPa and 673.15 K, and exhausts from the turbine at 700 kPa and 433.15 K. The mass flow rate of steam is 20.2 kg/s. The unit cost of the fuel is US$ 0.016/(kW h) of exergy, and the specific cost of electricity is US$ 0.05/(kW h). The cost of fixed capital and operating costs of the boiler and turbine are U S $ 1 2 0 0 and 90 h -1, respectively. The exhaust gases from the boiler are discharged into the surroundings with negligible cost. Heat transfer with the surroundings and kinetic and potential energy effects are negligible. The environmental temperature is 300 K. Determine the cost rate of process steam discharged from the turbine.

316

5

Thermoeconomics Exhaust gases

Fuel •

t ......

I

Boiler

Air

I .......

Work I

Feedwater Exhausted steam 5.2

In a steam power generation plant, the boiler uses a fuel, which enters the boiler with an exergy rate of 85 MW. The steam exits the boiler at 6000 kPa and 673.15 K, and exhausts from the turbine at 700 kPa and 433.15 K. The mass flow rate of steam is 19.5 kg/s. The unit cost of the fuel is US$ 0.017/(kW h) of exergy, and the specific cost of electricity is US$ 0.06/(kWh). The cost of fixed capital and operating costs of the boiler and turbine are US$1150 h-1 and 75 h-1, respectively. The exhaust gases from the boiler are discharged into the surroundings with negligible cost. Heat transfer with the surroundings and kinetic and potential energy effects are negligible. The environmental temperature is 298.15 K. Determine the cost rate of process steam discharged from the turbine.

5.3

A turbine produces 55 MWh of electricity per year. The annual average cost of the steam is US$ 0.017/(kWh) of exergy (fuel). The total cost of the unit (fixed capital investment and operating costs) is US$ 2.6 × 105. If the turbine exergetic efficiency increases from 80% to 88%, after an increase of 3% in the total cost of the unit, evaluate the change of the unit cost of electricity.

5.4

A turbine produces 60 MWh of electricity per year. The annual average cost of the steam is US$ 0.0175/ (kWh) of exergy (fuel). The total cost of the unit (fixed capital investment and operating costs) is US$ 2.5 × 105. If the turbine exergetic efficiency decreases from 90% to 80% after a deterioration of the turbine with use, evaluate the change of the unit cost of electricity.

5.5

Thermal analysis of the Aspen Plus simulator produces column grand composite curves of temperatureenthalpy and stage-enthalpy curves for rigorous distillation column simulations. These types of calculations are performed for RADFRAC columns. Using the following input summary for a RADFRAC column, construct the temperature-enthalpy, stage-enthalpy curves, and the stage exergy loss profiles. (a) Assess the thermodynamic performance of the column. (b) Suggest retrofits. Input summary: General Simulation with English Units: F, psi, lb/hr, lbmol/hr, Btu/hr, cuft/hr. Flow basis for input: Mole Stream report composition: Mole flow COMPONENTS C3 C3H8 / IC4 C4H10-2 / NC4 C4H 10-1 / IC5 C5H12-2 / NC5 C5H12-1 / NC6 C6H 14-1 FLOWsHEET BLOCK RADFRAC IN - FEED OUT - DIST BOTTOM

Problems

317

PROPERTIES PENG-ROB PROPERTIES NRTL-2 STREAM FEED S U B S T R E A M MIXED P R E S - 4 . 4 < a t m > V F R A C = 0 . M O L E - F L O W = 100. MOLE-FLOW C3 5. / IC4 10. / NC4 30. / IC5 20. / NC5 & 15. / NC6 20. B L O C K RADFRAC tL&DFRAC PARAM NSTAGE = 28 COL-CONFIG C O N D E N S E R = TOTAL FEEDS FEED 14 PRODUCTS DIST 1 L ! BOTTOM 28 L P-SPEC 1 4.4 < a t m > / 2 4 4.4 < a t m > COL-SPECS D : F - 0 . 4 4 M O L E - R R - 1 . 8 T-EST 1 308. < K > / 2 8 367. < K > EO-CONV-OPTI

5.6

Hydraulic analysis of the Aspen Plus simulator produces "thermodynamic ideal minimum flow" and actual flow curves for rigorous distillation column simulations. These types of calculations are performed for RADFRAC columns. Using the input summary given m Problem 5.5 construct the stage-flow curves. Assess the thermodynamic pertbrmance of the column.

5.7

Using the following input summary for tLA.DFRAC columns, construct the column grand composite curves and stage exergy profiles with the property methods of Peng-Robinson. Discuss the results. TITLE 'RADFRAC SIMULATION' IN-UNITS ENG COMPONENTS ETItAN-01 C 2 H 6 0 - 2 / 1-PRO-01 C 3 H 8 0 - 1 / ISOBU-01 C4H 100-3 / N-BUT-01 C4H 100-1 FLOWSHEET B L O C K D 1 IN=FEED OUT = DIS 1 BOT 1 B L O C K D2 IN=DIS 1 OUT = DIS2 BOT2 IN-UNITS ENG STREAM FEED S U B S T R E A M MIXED P R E S - 2 0 . VFRAC =0. M O L E - F L O W = 100. MOLE-FRAC ETHAN-01 0.25 / 1-PRO-01 0.5 / ISOBU-01 0.1 ! & N-BUT-01 0.15 B L O C K D 1 RADFtLA~C PARAM NSTAGE =41 HYDRAULIC = YES COL-CONFIG C O N D E N S E R = TOTAL FEEDS FEED 19 PRODUCTS DISI 1 L / BOT1 41 k P-SPEC 1 20. COL-SPECS D P - C O L = 0 . M O L E - D - 7 4 . 7 M O L E - R R = 3.65 SC-REFLUX DEGSUB =0. REPORT STDVPROF TARGET HYDANAL B L O C K D2 RADFRAC PARAM NSTAGE = 23 H Y D R A U L I C = YES COL-CONFIG C O N D E N S E R = TOTAL FEEDS DIS 1 12

318

5.

Thermoeconomics

PRODUCTS DIS2 1 L / BOT2 23 L P-SPEC 1 20. COL-SPECS DP-COL=0. MOLE-D=25. MOLE-RR=3.64 SC-REFLUX DEGSUB=0.

5.8

Use the following economic data and prepare a discounted cash flow diagram. Asses the feasibility of the investment on the latent heat storage system: Economic data used for the seasonal heat storage system Fixed capital investments for the components: FCI 1 + FCI2 + FCI 3 - US$200000 + US$200000 + US$200000 = US$600000 Cost of land: L = US$50000 Working capital: WC = 0.1 (US$600000)= US$120000 Yearly revenues or savings: R = US$150000 COp = CpT = Clp + Ccp nt- Cdp nt- C3p = Clp f nt- Ccp f nt- Cdp f + C3p f = U 8 5 5 0 0 0 0

Taxation rate: t = 30% Salvage value of the whole seasonal storage system: S = US$50000 Useful life of the system: n = 15 years; Depreciation over 10 years Discount rate i = 5.5%

5.9

Use the following economic data and prepare a discounted cash flow diagram. Asses the feasibility of the investment on the latent heat storage system: Economic data used for the seasonal heat storage system Fixed capital investments for the components: FCI~ + FCI2 + FCI3 = US$250000 + US$250000 + US$250000 = US$750000 Cost of land: L = US$50000 Working capital: WC = 0.2 (US$750000)= US$150000 Yearly revenues or savings: R = US$140000 COp = CpT = Clp nt- Ccp -Jr-Cdp nt- C3p = Clp f + Ccp f + Cdp f + C3pf = U 8 5 5 0 0 0 0

Taxation rate: t = 25% Salvage value of the whole seasonal storage system: S = US$50000 Useful life of the system: n = 12 years; Depreciation over 8 years Discount rate i = 6.5%

REFERENCES R.U. Ayres, Ecol. Econ., 26 (1998) 189. Y. Demirel, Sep. Sci. Technol., 39 (2004) 3897. Y. Demirel, Sep. Sci. Technol., 41 (2006) 791. Y. Demirel, Int. J. Exergy, 3 (2006) 345-361. Y. Demirel and H.H. Ozturk, Int. J. Energy Res., 30 (2006) 1001. R. Domanski and G. Fellah, Appl. Therm. Eng., 18 (1998) 693. B. Erlach, L. Serra and A. Valero, Energy Conversion Manage., 40 (1999) 1627. M.J. Moran and H.N. Shapiro, Fundamentals of Engineering Thermodynamics, 4th ed., Wiley, New York (2000). J. Szargut, Finite-Time Thermodynamics and Thermoeconomics, Eds. S. Sieniutcyz and E Salamon, Taylor & Francis, New York, 1990. D. Tondeur, Finite-Time Thermodynamics and Thermoeconomics, Eds. S. Sieniutcyz and P. Salamon, Taylor & Francis, New York, 1990. D. Tondeur and E. Kvaalen, Ind. Eng. Chem. Res., 26 (1987) 50. G. Tsataronis, Prog. Energy Combust. Syst., 19 (1993) 227. R. Turton, R.C. Bailie, W.B. Whiting and J.A. Shaeiwitz, Analysis, Synthesis, and Design of Chemical Processes, 2nd ed., Prentice Hall, Upper Saddle River (2003). A. Valero, C. Torres and M.A. Lozano, AES Vol. 9/HTD, Vol. 124, ASME, New York (1989).

REFERENCES FOR FURTHER READING J. Beusa and G. Tsatsaronis, Comp. Chem. Eng., 25 (2001) 359. M.A. Rosen and I. Dincer, Int. J. Energy Res., 27 (2003) 415. A. Valero, L. Serra and J. Uche, J. Energy Sources Technol., 128 (2006) 1, 9.

6 DIFFUSION 6.1

INTRODUCTION

We may describe multicomponent diffusion by (1) the Maxwell-Stefan equation where flows and forces are mixed, (2) the Chapman-Cowling and Hirschfelder-Curtiss-Bird approaches where the diffusion of all the components are treated in a similar way, and (3) a reference to a particular component, for example, the solvent or mass average (barycentric) definition. Frames of reference in multicomponent system must be clearly defined. Binary diffusion coefficients are often composition dependent in liquids, while they are assumed independent of composition for gases. Under mechanical equilibrium on a molecular scale, the exchange of momentum proceeds faster than the exchange of mass and heat for liquids. On the other hand, the molecular exchange of momentum, matter, and heat is on the same order as gases. The rate of exchange of transport processes is measured by the Schmidt number Sc and the Prandtl number Pr. Usually, the assumption of mechanical equilibrium in gases for heat and mass transfer is not reliable.

6.2

MAXWELL-STEFAN EQUATION

Maxwell-Stefan equations describe steady diffusion flows, assuming that sheafing forces for each species are negligible. As there are no velocity gradients assumed, the Maxwell-Stefan equations can be written in the forms of fluxes. For a ternary mixture of components 1, 2, and 3, the flow of component 1 in the z direction is dq dc 2 JI -- Dl l - 7 - + D 1 2 clz dz

(6.1)

Similar equations can be written for components J2 and J3. The coefficients Dll and D22 are the main coefficients; they are not self-diffusion coefficients. D~2 and D2~ are the cross-coefficients and assumed to be equal to each other for binary gas mixtures. Multicomponent diffusion by the Maxwell-Stefan equation is dx e a~

j=l c 2 D(j,

/ '/ CJ

ci

(6.2)

where c; is the concentration of species i, c is the total concentration, J; is the flow of species i, and D£- is the Maxwell-Stefan diffusivity. Maxwell and Stefan proposed a method to describe the diffusion in multicomponent gas and liquid mixtures of isotropic systems J -

LklPvt

(6.3)

l=1

where vl is the velocity of component l, P is the pressure, and Lkt is the Onsager reciprocal relations. Equation (6.3) shows the phenomenological equations used to describe diffusion where the Onsager reciprocal relations can be expressed in terms of the diffusion coefficients. Maxwell suggested an equation for dilute gases and Stefan suggested one for liquids, and hence Maxwell-Stefan equations cover both vapor and liquid phases. It is assumed that the diffusion results from equal and opposite forces that are proportional to the velocity differences of the components. The Maxwell-Stefan equations do not depend on choice of the reference velocity, and therefore they are a proper starting point for other descriptions of multicomponent diffusion. For ideal gas mixtures, diffusivities i, and D~l are

320

6.

Diffusion

independent of the composition, and equal to diffusivity D[t of the binary pair kl. In an n-component system, only n(n - 1)/2 different Maxwell-Stefan diffusivities are required as a result of the simple symmetry relations. Some advantages of the Maxwell-Stefan description of diffusion are as follows: 1. 2. 3. 4.

Diffusion is independent of choice of the reference velocity. The diffusion of all the components is treated equally. Diffusion is in agreement with the results of the kinetic theory of dilute monatomic gases. The Maxwell-Stefan diffusion coefficients represent binary diffusivities; for ideal and many nonideal mixtures, they are independent of the concentration of the species in the multicomponent mixtures.

Mass and molar diffusion are important in practice, and can be derived from the Maxwell-Stefan description of diffusion. The Maxwell-Stefan multicomponent diffusivities are obtained from the binary diffusivities, which are easy to measure.

6.2.1

Isothermal Systems

For an isothermal mixture, the dissipation function is l-1 £ air = _ p 2 Z

)2 Zkl (V l -- v k

>-- 0

(6.4)

k=l /=2

where v t - vk is the velocity difference between the species I and k. For a three-component system, Eq. (6.4) yields 2 XI'r=-p2(L12v21 nt-L13V~l-[-L23v~2) ~ 0

(6.5)

where vkt2 _ (Vk--Vt) 2. Eliminating v32 with v31 - v21, and dividing the result by v321leads to the following quadratic constraint

r/2

-(L12 -Jr-L23) v21 v31

r

+ 2L23 v21 -(/-q3 + L23) ~ 0 ~ v31

(6.6)

Necessary conditions for satisfying this inequality are - L 2 1 - L 2 3 - L22 -> O,

- L 3 1 - L 3 2 = L33 -> O,

-L12 -L13 - Lll -> 0

(6.7)

If Eq. (6.7) is satisfied, the constraint for the three-component system becomes L21L13 -nt- L23 (L12 nt- t13 ) ~ 0

(6.8)

Equation (6.8) is a sufficient constraint irrespective of the choice of independent velocity difference with --Lkt>--O for k 4: l, and shows that it is possible to have negative phenomenological coefficients without violating the condition of total positive dissipation. The value of-L23 does not need to be positive and is subject to the following constraint L12L13 -L23 z> ~ < 0 L12 -+-L13

(6.9)

by assuming that the other coefficients satisfy -L12 > 0 and -L13 > 0. The constraints for n-component system is generalized as --Lkk >-- 0 and LkkLtt - LZt >- 0

for n -> 3

(6.10)

Maxwell described diffusion by velocity differences, which yield forces from the friction between the molecules of different species. He considered a chemical potential gradient caused by friction, which is proportional to the concentration. The diffusion coefficient of Maxwell-Stefan can be defined as t

Dkt =--

XkXl PLkl

where xk - ck/c are the mole fractions. The following relations hold for the diffusion coefficients

(6.11)

6.2

321

MaxwelI-Stefan equation

D'kl >-- O, D~.+ - Dl'k,

and D£k --> 0

(6.12)

Thus, the Maxwell-Stefan diffusion coefficients satisfy simple symmetry relations. Onsager's reciprocal relations reduce the number of coefficients to be determined in a phenomenological approach. Satisfying all the inequalities in Eq. (6.12) leads to the dissipation function to be positive definite. For binary mixtures, the Maxwell-Stefan diffusivity has to be positive, but for multicomponent system, negative diffusivities are possible (for example, in electrolyte solutions). From Eq. (6.12), the Maxwell-Stefan diffusivities in an n-component system satisfy the following inequality xl ->0

(6.13)

l-I D~.l kvl

Negative Maxwell-Stefan diffusivities are allowed if they satisfy !

!

D[~ ~-- DkkDl¢

(6.14)

It is useful to express Eq. (6.14) in terms of D'a-I as follows Xk ~ --

Xl

XkX 1

k:, D~., k:, D'kl 6.2.2

(6.15)

>--0

D'k~

Nonisothermal Systems

For nonisothermal systems, diffusion flow becomes H

Jk

~" XkXl -- z__, (V l -- Vl,. ) -

Lkq T

- 1

grad T

(6.16)

+--t D[.t

If grad T 4=0, thermal diffusion flow appears, and is given by

Oh.vk - - p D [ grad T

(6.17)

T T

Here, the coefficient Da. is defined as the barycentric coefficients of thermal diffusion. Due to ~pkVk - 0

(6.18)

thermal diffusion coefficients satisfy the following constraint ~DT-0

(6.19)

k

From Eq. (6.16), the phenomenological coefficients are

_Lkq __ Lqk _ ~ xkx___L[ D[ D'kt

wl

(6.20) Wk

Using Eq. (6.20), the Maxwell-Stefan diffusion becomes

aa. = ~_, xkxl (v~ - v~ )

(6.21)

where vko is defined as

D[ grad T

~. - vk + - wk

V°

T

(6.22)

Equation (6.21) is considered a generalization of the Maxwell diffusion equation. Curtiss and Hirschfelder derived similar equations for dilute gases by using the kinetic theory of gases.

322

6.

Diffusion

Example 6.1 Maxwell-Stefan equation for binary mixtures For an application of the Maxwell-Stefan description of diffusion, we consider a binary isotropic mixture with components 1 and 2. To solve the mass balance equations, the diffusion flow has to be known. The binary diffusion flow without the electromagnetic field and external forces is given by J1 = - J 2

=

Pl

-~- [(grad ~1 )T,P q- (Vl -- V) grad P]

(6.23)

where Vl-V = (Wl + W2)V1--W1V1--W2V2 -" W2(V1--V2). Thermodynamic correction factor F is defined using the Gibbs-Duhem relation (6.24)

P ~. OXl )T,P

-- ~

~. OX2 ) T,P

The thermodynamic factor is a measure of deviation from ideal behavior, and equal to unity for ideal systems. Equation (6.24) can be rearranged as

_.•

t91

(grad ~I)T,P = --p

(.0&]

OXl ) T,P

grad x 1 = F grad X 1

(6.25)

Diffusion flow in Eq. (6.23) can be expressed in terms of F al = -a2

-- F

(grad x 1) +

pW 1W2 (V 1 - -

V2 )

grad P P

(6.26)

For a binary system, grad w 1 is related to gradxl by using the summation relationxl + x2 = 1 and M = M i x 1 + m2x2, and we obtain dw 1 =

d

X1 = ---~ dx 1 - x 1

;

and dw 1 - M1MM2 2 dXl

where M1 and M2 are the molar masses of components 1 and 2, respectively. From Eq. (6.27), we have

J1

M1M2 F grad w1 + PWlW 2 (V 1 -- V2)

( / gradp

P

(6.28)

From Eq. (6.21), we have V 2o -- V o1 _

D[2 J1

(6.29)

XlX2

Finally, the diffusion flow J1 becomes J1 = PlVl -- -PD12

/~~rd grad T grad W1 -- pO~' ,:,_a______~_pD( P T

(6.30)

where the binary diffusivity D12 is called the Fick diffusivity, which can be used in the barycentric description (mass average) as well as in the molar description. The binary pressure diffusivity D1P is defined by D1P -. PO{2 MM1 M e 2 w 1We (V1 - V2)

(6.31)

where D~2 = D12/1-'. Equation (6.30) shows that diffusion occurs due to concentration difference (ordinary diffusion), pressure difference (pressure diffusion), and temperature difference (thermal diffusion) without other external forces and electromagnetic field effects.

6.2

323

M a x w e l I - S t e f a n equation

The behavior of the Fick diffusion coefficient in nonideal systems may be complicated, while the MaxwellStefan diffusion coefficients behave quite well, and are always positive for binary systems. In nonideal binary systems, the Fick diffusivity varies with concentration. As seen in Figure 6.1, water-acetone and water-ethanol systems exhibit a minimum diffusivity at intermediate concentrations. Table 6.1 displays the dependency of binary diffusivity coefficients on concentration for selected alkenes in chloroform at 30°C and 1 atm. As the nonideality increases, mixture may split into two liquid phases at certain composition and temperature. A solution is a condensed phase of several components, which may be subject to strong intermolecular forces. Despite the fundamental differences between solutions and gases, some laws for solutions are analogous to those for gases. If the solution is sufficiently dilute, the osmotic pressure is described by an equation similar to that for an ideal gas, and ideal solutions are treated as a special case of ideal gas.

6.2.3

Ideal Solutions

The chemical potential of an ideal solution may be expressed by Izk

-

O

I~-

(6.32)

(T,P)+ RkTlnxk

where R k = R/Mk; R is the universal gas constant and Mk the molar mass. The behavior of dilute solutions may be close to that of ideal solutions depending on the nature of the solvent and dissolved substances. On the other hand, in electrolytes, deviations from ideal behavior may occur even in very dilute solutions due to the spatial range of the electromagnetic forces. Solutions of molecules with normal size and similar chemical structures can behave ideally over a large range of concentrations. The principal difference between ideal gases and ideal solutions is caused by molecular interactions

-.,.,.

2

O

0.5

,

i

,

0.2

I

I

I

0.4

J

I

0.6

i

0.8

1

XA (a)

"-"

4

E

% 3 g3

2

1

I

0.2

0.4

,

I

0.6

,

I

0.8

b

1

XA

(b) Figure 6.1. Concentration dependence of the Fick diffusivity for the binary mixtures: (a) water (A)--acetone (B) at 45°C, (b) water (A)methanol (B) at 40°C in terms of water mole fraction (Tyn and Caius, 1975).

324

6.

Diffusion

Table 6.1 Mutual diffusion coefficients as a function composition for binary mixtures of some alkanes in chloroform at 30°C and 1 atm Solute (1)

xl

D × 109 m2/s

n-Hexane

0.00413 0.2034 0.4028 0.6000 0.8009 0.9894

2.43 2.42 2.59 3.18 3.75 4.44

n-Heptane

0.00543 0.2007 0.4003 0.6006 0.7995 0.9895

2.23 2.21 2.38 2.85 3.27 3.52

n-Octane

0.00811 0.2016 0.4000 0.6024 0.7847 0.9870

2.03 1.94 2.13 2.44 2.83 3.08

3-Methylpentane

0.00814 0.2029 0.3998 0.6000 0.8009 0.9906

2.28 2.35 2.53 3.00 3.60 4.53

2,3-Dimethylpentane

0.00733 0.1507 0.3024 0.5015 0.7003 0.9883

2.08 2.09 2.20 2.54 2.80 3.55

2,2,4-Trimethylpentane

0.00641 0.2009 0.4005 0.6000 0.8012 0.9866

1.95 1.92 2.03 2.29 2.67 3.06

Source: R.L. Rowley, S.-C. Yi, D.V Gubler and J.M. Stoker, J Chem. Eng. Data, 33 (1988) 33.

in short time for gases and longer time for liquids. Therefore, the pressure effect on/x] (T, P) for ideal gases is different from the corresponding quantities for ideal solutions. This can be shown through the isothermal compressibility coefficient Kk

Kk

1 ( Ovk )

(6.33)

The compressibility of a liquid is practically independent of the pressure, and we have

vk (T,P)= vk (T, 0)(1-KkP ) The specific volume

vk does not depend on the concentration in an ideal system.

(6.34)

6.2

325

Maxwell-Stefan equation

For all ideal systems of gas and liquid mixtures, we have -~

/

'

7~

OP

= vk

(6.35)

T

From Eqs. (6.34) and (6.35), we get tx k ( T ) -

tx k ( T , P ) -

P

1---~KkP

vk

,

Isothermal compressibility for an ideal gas mixture K is given by 1/P, whereas for liquids the compressibility is negligible. The activities can be introduced to describe the deviations from the ideal behavior of solutions; the activities are expressed in terms of the activity coefficients.

6.2.4

Nonideal Solutions

For nonideal systems, intermolecular interactions may be simplified by introducing the activities into the diffusion potentials. Deviations from ideal behavior can be estimated by the Fick and Maxwell-Stefan diffusivities and the thermodynamic factor. The chemical potential/xx, can be expressed in terms of temperature, pressure, and activity a~ /x~: - ~[I (T, P) + RkT In a k

(6.37)

and the total differential is given by

d # k .... . O a k T.t'

OP ),l,,r

~, OT )p,,:

The partial differentiation of chemical potential with respect to the activity is obtained from Eq. (6.37), and given by ( 0#/;. ]

_ RkT

Oak )T,P

(6.39)

ak

For nonideal systems, Eq. (6.39) may be rearranged as __ P~ (grad/xx. )r.P ._. . .Pk . R~.T (grad a k )r,P = -xk - (grad a k )r,P P Pa~. ak

(6.40)

where Px_ = PkRkK and xk = Pk/P. Diffusion flow in terms of activity for nonideal systems is Jk - xk (grad a k )r,P + ( p x . v x . - w k ) grad P

a~.

(6.41)

P

The description of diffusion may be complex in mixtures with more than two components. Diffusion coefficients in multicomponent mixtures are usually unknown, although sufficient experimental and theoretical information on binary systems is available. The Maxwell-Stefan diffusivities can be estimated for dilute monatomic gases from Dik-~Dx-t when the Fick diffusivities are available. The Maxwell diffusivity is independent of the concentration for ideal gases, and almost independent of the concentration for ideal liquid mixtures. The Maxwell-Stefan diffusivities can be calculated from D~.l - Dkt F

(6.42)

The thermodynamic correction factor F can be expressed in terms of the activity using Eq. (6.39) F -x k

~ In a k Ox k

-

0 In a k Olnx k

(6.43)

326

6.

Diffusion

Using the definition of activity ak = xkYk, Eq. (6.43) becomes F=I+~

Oln yk Olnx k

(6.44)

The activity coefficients of nonideal mixtures can be calculated using the molecular models of NRTL, UNIQUAC, or the group contribution method of UNIFAC with temperature-dependent parameters, since nonideality may be a strong function of temperature and composition. The Maxwell-Stefan diffusivity for a binary mixture of water-ethanol can be considered independent of the concentration of the mixture at around 40°C. However, for temperatures above 60°C, deviation from the ideal behavior increases, and the Maxwell-Stefan diffusivity can no longer be approximated as concentration independent. For highly nonideal mixtures, one should consider the concentration dependence of the diffusivities.

6.2.5 Binary Systems For a binary mixture, if experimental diffusivities do not exist over the whole range of concentration, an interpolation of the diffusivities at infinite dilution DkzfOC is used. In calculating the diffusivities at infinite dilution by the StokesEinstein relation, we consider small isolated hard spheres, submerged in a liquid, that are subjected to Brownian motion. The friction of the spheres in the liquid is given by the Stokes' law; Einstein used the Stokes' law to calculate the mean-square displacement of a particle. The displacement increases linearly with time, and the proportionality constant is the Stokes-Einstein diffusivity kBT DO -

6rrr/j~

(6.45)

where r i is the radius of the particle i, kB is the Boltzmann constant, and fly is the viscosity of the solvent j. The generalized Stokes-Einstein diffusivity is modified to account for the particle sizes of solute and solvent, and is given by

where nc is around 3.5. For nonassociating and associating binary organic mixtures, the values of nc are 3.54, and 3.53, respectively; nc is 3.65 for self-diffusivities of organic species and 3.47 for polar organic species in water, where hydrogen bonding occurs. The radius is calculated from the van der Waals volumes. The estimates of the diffusivities with the modified Stokes-Einstein relation are comparable to or more accurate than those estimated with the Wilke-Chang relation. A modified version of the Vignes interpolation for concentrated binary solutions is adequate for nearly ideal systems, and is given by 1 ,t,DPk~ D 'kz = -~ kZ 1rt ~ ) x ~ ( D ,kll-< rl z ) x ,

(6.47a)

where rl is the viscosity of the mixture. The accuracy in estimating the diffusivity of moderately ideal mixtures is quite high. The simple linear interpolation with the viscosity correction correlates the diffusivities to within the experimental error 1 (DPk_~l r~d~ 1~qtxt) D'kl=-~ kt nkxk+~kt

(6.47b)

The interpolations of Eqs. (6.47a) and (6.47b) are satisfactory for nonideal nonassociating systems, however, for associating mixtures, they yield relatively large errors.

6.2.6 Ternary Systems Estimations for ternary diffusion are more complicated than for binary diffusion, and there is a lack of experimental data. To obtain estimates for ternary mixtures, the interpolation relations, given in Eqs. (6.47a) and (6.47b), are extended as Dkt,

--

)xi -1~ (D,k~l kZ ~k)xk (D,kZz-~lrlz)x, ~(D,i~l kl Tli

(6.48)

6.2

327

Maxwell-Stefan equation

,~ 1 /-~,I~1 r-~,i~l Dkl - ~ ( D~-~-~lrl~xk + "-'k/ rl/x / -Jr-l_)kl TliX i )

(6.49)

Estimations for nonideal mixtures need six diffusivities at infinite dilution and three diffusivities of the type Fltk--*l .L~kl • Negative diffusion coefficients can exist in ternary systems and are consistent with the nonequilibrium thermodynamics approach. Some of the molecular theories of multicomponent diffusion in mixtures led to expressions for mass flow of the Maxwell-Stefan fonn, and predicted mass flow dependent on the velocity gradients in the system. Such dependencies are not allowed in linear nonequilibrium thermodynamics. Mass flow contains concentration rather than activity as driving forces. In order to overcome this inconsistency, we must start with Jaumann's entropy balance equation DS p~ = -(V.s) + • Dt

(6.50)

where p is the density of the fluid mixture, S the entropy per unit mass, entropy production per unit volume. The operator D Dt

s

the entropy flow vector, and • the rate of

0 ---+(v.V)

(6.51)

Ot

is the substantial derivative. From the balance equations of mass, momentum, energy, and the Gibbs relation, one obtains explicit expressions for s and O. For multicomponent diffusion, and the mass flow expressions, we mainly use Fick's law and the Maxwell-Stefan forms • Using the symmetric difthsivity, in 1ength"/tlme, ' " we have cRTLij

D!) -

(6.52)

PiPi where p; is the density of species i, L!j are the phenomenological coefficients, and c is the total molar concentration P" c=Z; =ZM,

(6.53)

where M,. shows the molecular weight of component i. The diffusivity coefficients have the following properties D;) - D);

( i - 1,2 .... , n)

~ w;D!} = 0

(j = 1,2,..., n)

(6.54)

(6.55)

i=-I

There are n(n - 1)/2 independent diffusivities DIj, which are also the coefficients in a positive definite quadratic form, since according to the second law of thermodynamics, the internal entropy of a single process never decreases. In terms of these symmetric diffusivities, the mass flow becomes

J; - -D,rV In f - - p L D$@

(i - 1,2,..., n)

(6.56)

j--I

where D r is the generalized thermal diffusion coefficient in mass/(length)(time). The generalized driving force X, is given by

cRTX, - VP i - w i V P - piF~ + w i ~ p/Fi j=-I

(6.57)

328

6.

Diffusion

where F i is the force per unit mass acting on the ith species. Using Eqs. (6.56) and (6.57), we can express the mass flow in terms of general driving force, which is the Fick form

Ji = - D i T V l n T -

cRr

Dij V P j - w j V P - p j F

i +wj

j=l

(6.58)

PkFk k=l

It may be convenient to express Xi as a linear function of Ji, which is in the Maxwell-Stefan form

c R T Z C;,k Jk k4:i Pk

Ji : v e i _ w i V P _ ~ F i Pi .

_+_wi

pjFi-cRT~.~

j=l

k~i

Cik D~

D~i

Pk

Pi

(VlnT)

(6.59)

! t where Ci~ is the inverse diffusivity, and sometimes is expressed as C 0.! = xixk/Di~, and C/kare the Maxwell-Stefan diffusivities. Equations (6.58) and (6.59) contain the same information and are related through multicomponent diffusivities D~. and multicomponent inverse diffusivities C~.. For low-density gases, we have VPi = V(ciR7). For polymeric liquids, a similar form to Eq. (6.59) can be found from a molecular theory by replacing the pressure P and the partial pressure Pi with the total stress tensor and the partial stress tensor. The mass flow is related to the velocity gradient via the stress tensor, temperature, and concentration gradients.

6.2.7

Generalized Matrix Method

Fast and satisfactory mass transfer calculations are necessary since we may have to repeat such calculations many times for a rate-based distillation column model or two-phase flow with mass transfer between the phases in the design and simulation process. The generalized matrix method may be used for multicomponent mass transfer calculations. The generalized matrix method utilizes the Maxwell-Stefan model with the linearized film model for diffusion flux, assuming a constant diffusion coefficient matrix and total concentration in the diffusion region. In an isotropic medium, Fick's law may describe the multicomponent molecular mass transfer at a specified temperature and pressure, assuming independent diffusion of the species in a fluid mixture. Such independent diffusion, however, is only an approximation in the following cases: (i) diffusion of a dilute component in a solvent, (ii) diffusion of various components with identical diffusion properties, and (iii) diffusion in a binary mixture. For binary diffusion, there is only one independent flow, force or concentration gradient, and diffusion coefficient. On the other hand, multicomponent diffusion differs from binary diffusion because of the possibility of interactions among the species in mixtures of three or more species. Some of the possible interactions are (1) a flow may be zero although its zero driving force vanishes, which is known as the diffusion barrier; (2) the flow of a species may be in a direction opposite to that indicated by its driving force, which is called reverse flow; and (3) the flow of a species may occur in the absence of a driving force, which may be called osmotic flow. The theory ofnonequilibrium thermodynamics indicates that the chemical potential arises as the proper driving force for diffusion. This is also consistent with the condition of fluid phase equilibrium, which is satisfied when the chemical potentials of a species are equal in each phase. In the generalized Maxwell-Stefan equations, chemical potential gradients, which are the thermodynamic forces, are linear functions of the diffusion flows Xi = Z XiJm - xmJi - Z XiNm - xmNi R T vTt'/ci , , m cDim m cDim

(i = 1,2, ..., n - 1 ) , i 4: m

(6.60)

where x is the mole fraction, R is the universal gas constant, T is the temperature, and c is the total molar density. The D/m denotes the generalized Maxwell-Stefan diffusion coefficients, and according to the Onsager reciprocal rules, there are n(n - 1) of such coefficients, which exhibit the symmetry property Din , = Dmi. The first equality of Eq. (6.60) represents the diffusion irrespective of the reference velocity frame chosen, while the second requires one to define the reference velocity. In matrix notation, Eq. (6.60) becomes -c[FJ[B] -l(Vx) = (J)

(6.61)

where the elements of the matrix of thermodynamic factor [F] is defined by

I'ij - ~ij -~- xi 0 In 7i xj Olnx i

(i,j = 1,2, ..., n - l )

(6.62)

6.2

329

Maxwell-Stefan equation

where 6 Uis the Kronecker delta and equal to 1 for i = j, and 0 for i 4=j, and 3'; is the activity coefficient of species i. [B] is the matrix of diffusion coefficients, and defined by Bii = xi ..... .-{-n,~L ''Xm' O~'n m=l Dim

/' 't

ii ---taXi

t

Dij

(i = 1,2,..., n - 1), i # m

(i,j = 1,2,...,

n-l),

i 4= j

Di',,

(6.63a)

(6.63b)

Comparing the matrix notation of Fick's law (J) = -c[D](Vx) with Eq. (6.61) yields [D] = [B]-I[F]

(6.64)

Equation (6.64) provides a general method of predicting the elements of Fick's law of diffusion coefficient matrix.

6.2.8

Diffusion in Mixtures of Ideal Gases

Consider the problem of steady-state one-dimensional diffusion in a mixture of ideal gases. At constant T and P, the total molar density, c = P/RT is constant. Also, the Maxwell-Stefan diffusion coefficients Dim reduce to binary molecular diffusion Dim , which can be estimated from the kinetic theory of gases. Since Dim is composition independent for ideal gas systems, Eq. (6.61) becomes - c [ B ] - I (~Ty) -- ( J )

(6.65)

There are n - 1 independent relations in the Maxwell-Stefan formulation. For a one-dimensional diffusion in direction z, Eq. (6.65) becomes

dyi - ~

drl

m=l

YiJm -- YmJi kim

(i = 1,2, ..., n - 1), i 4: m

(6.66)

where r / = z/6 and kim = cDim/~. Here, r/is the dimensionless distance along the diffusion path, and kim is the transfer coefficient of the binary pairs of species i and m. Krishna and Standard (1976) suggested a general solution to the n-species system and defined the following parameters: (1) Matrix of dimensionless mass transfer rate factors [~] with the elements defined as

f~ii = ~Ni +

~ Nm

(i=1,2,

- ,

. n-l),

i4=m

(6.67)

m=l

~J=-Ni[

lkO.

k~l) (i,j

= 1,2, ..., n - l ) ,

i :/= j

(6.68)

(2) Column vector to with elements defined as

N~

(6.69)

(3) Square matrix [B0] with the following elements

Yoi ~ YOm

B°ii - kin -+- ,,=1 ~

(i = 1,2, ..., n - 1), i 4= m

(6.70)

330

6.

Diffusion

(1 1/ B°ij = - Y ° i

ko.

(i, j = 1,2, ..., n - 1), i 4: j

ki n

(6.71)

Using the parameters above, Eq. (6.66) may be converted into n - 1 dimensional matrix notation

d(y) dr/

- [~] (y) + (to)

(6.72)

l(d(Y) I

(6.73)

The diffusion flows are estimated from

(Jo) = -[Bo 1- [ , - ~ / , : o

The composition profiles can be obtained from Eq. (6.72), which also facilitates the estimation of the gradient (d(y)/drt)n=o; using the gradient in Eq. (6.73), we obtain

(J0) = -[k~] (y0 - y 6 )

(6.74)

Here, [1~] is called the matrix of high flux mass transfer coefficients, and is defined by [k~] = [B0]-I[~I t]

(6.75)

[ ~ ] : [ ~ ] { e x p [ ~ ] - [I]} -1

(6.76)

where [ ~ ] is the matrix of correction factors

The [I] is the identity matrix. When N i --~ 0, the matrix of correction factors [~] reduces to the identity matrix, and the matrix [1~] becomes [k0], which is defined as the matrix of zero flux mass transfer coefficients [k ° ] = [B ° ]-1

(6.77)

Using the diffusion fluxes J0i, the total molar fluxes are defined by N = J + N t (Y0)

(6.78)

where Nt is the total flux --1

Nt = _

znm : l ( P k - - P n ) J m m m=/=k Z" l)mYOm

(6.79)

m=l

Using the determinacy condition below, the total fluxes are estimated by a trial and error procedure.

~voN i = 0

(6.80)

i=1

where vi = 1 for equimolar countercurrent diffusion, and vi = 0 for diffusion of n - 1 components through a stagnant nth component.

Example 6.2 Diffusion in a ternary ideal gas mixture Methane is being absorbed from a mixture of argon and helium by a nonvolatile liquid in a wetted wall column operated at 25°C and 1 atm. The following boundary conditions and data may be used: System: methane (1)-argon (2)-helium (3)

6.2

MaxwelI-Stefan equation

331

Boundary conditions: z = 0(bulk gas), Y01 = Y02 = 0.4, Y03 = 0.2

(6.81)

z = 6 = 0.1 (interface), Y61 = 0.1, Y62 and Y63 unknown

(6.82)

Ordinary diffusion coefficients: Dye2 = 0.202, Dy,3 - 0.675, Dy23 = 0.729cm2/s Assume that a simple film model exists for the mass transfer, equilibrium is established at the gas-liquid interface, and the diffusion occurs at isobaric and isothermal conditions. Also assume that neither helium nor argon is absorbed, so that N2 = N3 = 0. Then, the Maxwell-Stefan equations for the diffusion of argon and helium are dy2 _ y2N1

dr/

(6.83)

k21

(6.84)

dy3 _ y3N1 dr/

k13

where the terms r/and k/j are defined in Eq. (6.66). By integrating Eqs. (6.83) and (6.84), and using the boundary conditions described in Eqs. (6.81) and (6.82) with the summation equation ~Y6i 1, we obtain =

(6.85) The solution to Eq. (6.85) yields the interfacial composition of argon and helium: Y~2 = 0.675 and Y,53 = 0.225

The rate of absorption of methane is N 1 = 4.29 × 10 .5 gmol/(cm 2 s) For the system, the flux of methane may also be estimated by defining an effective diffusivity Dl,eff J1

=

( gYl)

-CDl,eff ~. dz

(6.86)

where 1--y 1 Dl'eff - (Y2/D12 ) -+- (Y3/D13 )

We take the arithmetic average of Dl,eff as

Dl'eff --

Dl,eff (z = 0) + Ol,eff (z = ~) 2

(6.87)

So, Eqs. (6.83) and (6.84) can be integrated, and we have N 1 = 4.13 × 10-5 gmol/(cm 2 s). The newly obtained value of N1 is close to the previous value of N1. For a species absorbed from a mixture of inert gases by a nonvolatile liquid, the Dl,eff depends on the composition and diffusion coefficients, and may be generalized by 1-y 1

Dl,efe

~ ~=2(Y~/Dli)

(6.88)

332

6.

Diffusion

Example 6.3 Diffusion of species from a gas mixture to a falling liquid film In a wetted wall column, the falling liquid film consists of a mixture of acetone (1) and benzene (2). This falling film is in contact with a downward flowing gas mixture of acetone, benzene, and helium (3). The pressure, temperature, and thickness of the film are P - 1 atm, T - 28.1 °C, and 6 = 0.044 cm. The gas entering at the top of the column contains 10 mol% acetone and no benzene, and the composition of the vapor flow at the interface is Yll = 0.02, Y12 = 0.28, Y13 = 0.70 Y01 = 0.10, Y02 = 0.0, Y03 = 0.90

The diffusion coefficients are D12 -- 0.04, D13 -- 0.41, D23 -- 0.39(cm 2 s)

We assume that helium is insoluble in the liquid phase. The interfacial rates of diffusion of acetone and benzene can be calculated at the top of the column in an iterative manner using the following steps: Step 1: Calculate the matrix of zero flux mass transfer coefficients from the inversion of the matrix [B0], which is obtained from Eqs. (6.70), (6.71), and (6.77) °~Y/Col -" k13[k12 nt- f°l(k23 - k12)] ' S

(6.89)

,~yko, = k23Y01(k13 - k]2) 2 S

(6.90)

'~Y/g021 -'- k13Y°2 (k23 - k12 )

(6.9 1)

S k23 [k12 + )/'°2 (k23 - k12)]

ky°22 =

(6.92)

S

w h e r e S = Yolk23 Jr- Yo2k13 -k- Yo3k12

Step 2: For the first iteration, assume that the matrix of correction factors is the same as the identity matrix, and therefore the finite flux mass transfer matrix, given by Eq. (6.75), becomes [k;] = [Bo]-l[ko]

(6.93)

Step 3: Calculate diffusion fluxes Joi from Eq. (6.74) using the condition ~Joi = O. Step 4: Estimate the total molar fluxes Ni from Eq. (6.78) (6.94)

N 3 = 0 = J03 + Yo3Nt /

"x

Ni = J o i - [ Yoi ] J o 3

(i--1,2)

(6.95)

k, Y03 }

Step 5: Calculate the elements of the matrix of mass transfer rate factors [~] from Eqs. (6.67) and (6.68). Step 6: Calculate the elements of the matrix of correction factors [~] using Eq. (6.76), which may be obtained from the Sylvester expansion [~I~]-

A1

eAl-1

[(I~] -- A2[I ] +

A1-A 2

A2

e'~2-1

[CI~]-- AI[I] A2-A 1

(6.96)

where A1 and A2 are the eigen values of the matrix [~]. After completing the first iteration, the calculations are repeated starting from step 2 by updating the matrix of correction factors in a new finite flux mass transfer coefficient evaluation. When the two successively estimated values of Ni are close enough to each other, the iterations are

6.2

MaxwelI-Stefan equation

333

Table 6.2 Interfacial diffusion rates of acetone (1) and benzene (2) at the top of the column in Example 6.3

Nl•/C

N2t%/c

:¥,6/c

k,~, 6/c

k,._ 6/c

(mm:~-/s)

(mm<'s)

k,, 6/c (mm2/s)

k,. 6/c

(mm2/s)

(mm2/s)

(mm2/s)

(mm2/s)

1 2

--2.98 1.07

--5.82 -10.68

-8.81 ....9.61

41.00 65.26

19.24 11.37

0 ....l 8.68

20.80 32.83

5

-0.94

-8.91

9.84

81.33

23.07

-33.42

22.26

10 15

-0.28 -0.41

--9.52 -9.41

-9.81 -9.82

78.38 78.93

19.91 20.50

-30.60 --31.13

25.28

20 25 26

-0.38 --0.39 -0.39

-9.43 -9.43 -9.43

-9.82 0.82 -9.82

78.83 78.84 78.85

20.39 20.40 20.41

-31.03 -31.04 -31.05

24.82 24.82 24.81

Iteration

24.72

terminated. Some selected estimates of the total molar fluxes and finite flux mass transfe; coefficients are given in Table 6.2, which shows that convergence at the beginning is very fast but later becomes very slow. An oscillation is possible, especially at the early stage of iterations, which may be due to the over-correction caused by the matrix of correction factors. If nitrogen is used instead of helium, then the values of D13 and 023 approach D12. Therefore, matrix [1%] becomes a diagonal matrix as the c r o s s - c o e f f i c i e n t s kyl2 and ky21 vanish. This is called the pseudo-binary case, and we have N1 = - N2.

E x a m p l e 6.4 Wetted wall column with a ternary liquid mixture We have the experimental data on distillation of

ethanol (1), tert-butanol (2), and water (3) in a wetted wall column reported by Krishna and Standard (1976). As Figure 6.2 shows, the column operates at total reflux with countercurrent flow. Therefore, at steady state, the compositions of the liquid and vapor phases at any point in the column are equal to each other. The measured compositions of the phases at the bottom are Xbl =Ybl -- 0.1862,

Xb2 --Yb2 = 0.1405,

Xb3 =Yb3 = 0 . 6 7 3 3

The zero flux mass transfer coefficients of the binary pairs are ky,2 = 1.0 × 10 - s , ky,~ - 2.25 × 10 -s

kv, = 1 86 × 10 -s gmol/(cm 2 s) '

k~, = 6.58 × 10 -5 "

2

~

kx, = 1.45 × 10 - 4 '

3

3

"

k v = 1 16 × 10 - 4 gmol/(cm 2 s) '

.!3

"

The phase equilibrium is described by the following approximate relations Yll = 0-4862Xll - 0.1587x12 + 0.2256 Y12 -- 0-4338xl 1 - 0.5728x12 + 0.2773 where Xl2 = 1.125xll - 0.069. The diffusion process in each phase may be described by a film model. By applying the Maxwell-Stefan equations for each phase, the interfacial compositions and the rates of interface transport at the bottom of the column can be estimated using the following steps: Step 1: From Eqs. (6.89) to (6.92), estimate the matrix of zero flux mass transfer coefficients for both the liquid and vapor phases at the interface [kx,] and [ky,]. Since the interface compositions are needed, first assume a value for xil. Step 2: For the first iteration, assume the matrix of correction factors [ ~ j and [~,,] equal to the identity matrix [I]. Step 3: Calculate the diffusion fluxes in each phase using Eq. (6.74). Step 4: Calculate the total molar fluxes. Due to interface invariance N~ = Nv, we have

Nti =

Jxli - Jyli

(y. - x . )

(6.97)

334

6.

Diffusion

condenser

,v

reboiler

Figure 6.2. Schematic of the wetted wall column at total reflux.

Table 6.3 Interfacial compositions and rates of interface diffusion at the bottom of the column in Example 6.4 Xll

0.1754 0.1755 0.1756 0.1757 0.1758

Nl × 10 -5 mol/(cm 2 s)

N2 × 10 -5 mol/(cm 2 s)

N3 X 10 -5 mol/(cm 2 s)

Ntl × 10 -5 mol/(cm 2 s)

Nt2 × 10 -5 mol/(cm 2 s)

0.1330 0.1409 0.1489 0.1569 0.1650

0.0783 0.0849 0.0915 0.0981 0.1048

1.4132 14~39 1.4547 1.4755 1.4968

1.6246 1.6598 1.6952 1.7306 1.7667

1.6320 1.6636 1.6954 1.7273 1.7597

Ntl ~ Nt2 N,1 > N,2

Step 5" Calculate the matrix of correction factors. At r/= 1, the [~x] is defined by [~I~l ] = [ ~ ] { e x p [ ~ ] - [I]} -1

(6.98)

Step 6: Using the Sylvester expansion, estimate Eq. (6.98) by [xI~l] = A1expA1 [~]-AI[I] expAl-1 Aa-A 2

expA2 [~]-A2[I] expA2-1 A2-A 1

+/~2

(6.99)

Calculations are repeated starting from step 2 until the values NI in two successive iterations are sufficiently close to each other for every assumed value of the Xn. A computer code prepared to perform the steps above yields Xn = 0.1757, at which the inequality Nt2 > Ntl changes to Nil < Nt2. The results are summarized in Table 6.3. This method provides the exact solutions for ideal systems at constant temperature and pressure. It is successful in describing diffusion flow in (i) nearly ideal mixtures, (ii) equimolar counter diffusion where the total flux is zero (Nt - 0), (iii) diffusion of one component through a mixture of n - 1 inert components, and (iv) pseudo-binary case and the diffusion of two very similar components in a third.

6.3

6.2.9

335

Diffusion in nonelectrolyte systems

Generalized Matrix Method for Diffusion in Nonideal Mixtures

In nonideal mixtures, the thermodynamic nonldeality of the mixture has to be considered. We still need to predict the concentration dependence of the mutual diffusion coefficient D!/of a binary pair of nonelectrolytes. The concentration dependency of D!i in liquid mixtures may be calculated by using the Vignes equation or the Leffler and Cullinan equation. Besides these, we may also use a correlation suggested by Dullien and Asfour {1985), given by tg i

Y#

.......

~7.~

X..i

.

,

.

.

.

rl

(6.100)

Xi

where D Oand D ~ arc the mutuai (l/ffus~oi~ cocif~,:~cnts at x/--, 1 and x,--, 1, respectively. The rl~ and ~; are the absolute viscosities of pure components i and j, respectively. In Eq. (6.100), In (D!//rl') is assumed as a linear function of x/. The authors suggested that Eq. (6.100) does not require the activity correction for the nonideality in regular solutions. However, the equation is not recommended lbr mixtures containing n-alkanes and polar species. The generalized corresponding states principle also may be used to predict the D!/in nonideal liquid mixtures. The extension of ideal phase analysis of the Maxwell-Stefan equations to nonideal liquid mixtures requires the sufficiently accurate estimation of compositiot>dependent mutual diffusion coefficients and the matrix of thermodynamic factors. However, experimental data on r;mtual diffusion coefficients are rare, and prediction methods are satisfactory only for certain types of liquid mixtures. The thermodynamic factor may be calculated from activity coefficient models such as NRTL or UNIQUAC, which have adjustable parameters estimated from experimental phase equilibrium data. The group contribution method of UNIFAC may also be helpful, as it has a readily available parameter table consisting of many species. It, ihowever, reliable data are not available, then the averaged values of the generalized MaxwellStefan diffusion coefficients and the matrix of thermodynamic factors are calculated at some mean composition between x0; and x_~. Hence, the matrix of zero flux mass transfer coefficients [k0] is estimated by [ko] = [Bo]

....l[r]

(6.101)

The modified matrix of dimensionless mass transfer rate factors is defined as [@'] = [F]- ~[{I}]

(6.102)

The total fluxes N; are calculated as for ideai gases using the: fimtc timexmass trmxsfer coefficients defined by {6.103)

[kl} ]-- Ik o ] [{I}'] {exp [{I}']- [I] }--1

For nonideaI liquid mixtures, the generalized matrix method leads to only approximate solutions. The method is sensitive to the accuracy of the thermodynami~ factor.

6.3

DIFFUSION IN NONELECTROLYTE SYSTEMS

The linear phenomenological law' of diffusion tbr a binary system is given by

Jl - o~{ v l - v2 ) - 7

F1 -

/

(6.104)

ox I )

For a perfect gas or an ideal solution, we have {}

p,/ - Ac/{T,P)+RTlnq

(6.105)

Inserting Eq. (6.105) into Eq. (6.104) yields

L{ Opq) LRT{Oq cl(v 1 - v. ) -:/: V, J4, -~ - : ' i}x~ T c I Ox1

~Mlcl } RT

(6.106)

336

6.

Diffusion

Here, we can distinguish the following two separate systems: (i) For a uniform system, where Ocl/Oxl = 0, we have L

V1 --V 2 = ~ F 1 M

Tq

1

(6.107)

The coefficient of proportionality between the relative velocity (vl-v2) and the force F1M1 is called the mobility of component 1 B* and is defined by B*-

(6.108)

L Tc 1

(ii) For a system without external forces F 1 -- 0, we have q (v 1 _ v2)

L R T Oq T q Ox1

=

(6.109)

The coefficient of proportionality between the flow of diffusion cl(vl -v2) and the concentration gradient is the diffusion coefficient D-

LRT

(6.110)

T q

Comparing Eqs. (6.108) and (6.110) yields the Einstein relation between the mobility and diffusion coefficient

(6.111)

D = RTB*

For a system without an external force, Eq. (6.106) can be written as CI(V 1 _V2)__

L 0~1 O N 1 T ON1 0 x 1

(6.t12)

The phenomenological law defines the diffusion coefficient, D as Cl(V1 - v 2 ) = _DcON1

Ox1

(6.113 )

so that D = ~1 L O~I Tc ON1

(6.114)

This definition is equivalent to Eq. (6.110) for a perfect gas or for an ideal solution. Equation (6.114) shows that the diffusion coefficient is the product of the phenomenological coefficient L and the thermodynamic quantity (Otzi/ON1)/cT. The coefficient L is positive, and so is the quantity Otxl/ON1 for all ideal systems. This means that the diffusion coefficient is positive, and according to Eq. (6.113), the diffusion flow has the direction imposed by the existing concentration gradient. In some highly nonideal systems of partially or completely immiscible mixtures, such as water-butane and water-benzene, the quantity Oi~l/ON1 may be negative, corresponding to thermodynamic instability. Such systems may split into two liquid phases, and may have negative diffusion coefficients in the immiscible region. In contrast, the thermal conductivity is always positive. The diffusion coefficient is a product of two quantities, only one of which, the L, has a definite sign. 6.4

DIFFUSION IN ELECTROLYTE SYSTEMS

For electrostatic potentials and electric current of charged ionic species, we start with the fundamental Gibbs equation d U = TdS - 6 W

(6.115)

6.4

Diffusion in electrolyte systems

337

and reconsider the work term 6 W. Usually, the 6 W means the work of compression P d V and the work involved in changing the number of moles of the components (chemical work: - ~ t z i d N i ) . However, when we have a region with an electrostatic potential 48, a change in the charge de results in the electrostatic work, and hence Eq. (6.115) may be extended as follows d U = T d S - P d V + E t'tidNi + Ode

(6.116)

When the de is due to changes in the concentration of ionic species, we get (6.117)

de - y_, z i F d N i

and then Eq. (6.116) becomes d U - TdS - P d V + ~_~ t.tidN i + ~ , z i F O d N i i

(6.118)

i

Combining the last two summations, we have d U - T d S - P d V + E (tx~ + z~F6)dNi

(6.119)

i

Equation (6.119) indicates that the chemical work in electrolytes contains a chemical term txdN~ and an electrical term ziFg, dNi and the sum is called the electrochemical potential pci of the ionic species i [-*i - Izi + ziFO

(6.120)

If we have a phase in which the composition is identical at points 1 and 2 but ~1 =/=~2, then we have 1

#i-/x; -1

2

-2

#, - #i - ziF(Ol - ~2 )

(6.121)

(6.122)

Substituting Eq. (6.120) in Eq. (6.119), we get dU - T d S - P d V + E ficidNi

(6.123)

We may also use the change in the Gibbs free energy in terms of the chemical potential dG - - S d T + VdP + E fzidNi

(6.124)

or the Gibbs-Duhem relation, SdT-

VdP + E N i d ~ i - 0

(6.125)

Under isothermal and isobaric conditions, Eq. (6.125) reduces to ~__ Nidfic ~ - 0

(6.126)

i

It is often useful to express the electrochemical potential as a sum of explicit terms of activity and electrostatic potential as follows P~i - #~' ( T) + VP + R T In a i + z ; F ~

(6.127)

Thus, in the case of an ion distributed between two phases c~ and/3, the condition of thermodynamic equilibrium is /2~~-/2,./3

(6.128)

338

6.

Diffusion

Introducing Eq. (6.127) into Eq. (6.128) yields Atx° +(Vi~P '~ -Vi¢3P¢3)+ R T l n A a i + ziFAqt = 0

(6.129)

If, for example, c~ and/3 are aqueous phases separated by a membrane, then (/x°) ~ = ( o)/3 and V/~ = V/t~. So, Eq. (6.129) becomes V~AP + R T l n A a i + ziFA¢/ = 0

(6.130)

In most cases of interest, V;AP is negligible in comparison with the other terms, so that the condition of phase equilibrium across a membrane becomes (6.131)

a~__[_.= _ z i F A ~ R T In a/t3

For an ideal solution, the condition of phase equilibrium is (6.132)

c~__~_= _ z i F A O RT In c/~

Equation (6.132) may also written in base 10 logarithms

- M p - 2.3RT log

ziF

ci/3

: 58 log Zi

(mV at 20°C)

(6.133)

C~

Example 6.5 Diffusion in aqueous solutions Consider an aqueous solution with N1 moles of sodium chloride and N2 moles of calcium chloride. An increase in the concentrations of both salts by amounts dN1 and dN2 causes the following changes in the ionic concentrations dNNa = dN1, dNca = dN2, dNc1 = d N 1 + 2 d N 2

(6.134)

Introducing these relations into Eq. (6.123), we obtain d U = T d S - P d V +/.~NadNNa nt- ~ c a d N c a

+ ~CcldNcl

(6.135)

or

d U = TdS - P d V + (/'~Na at- ~C1 ) dN1 + (~Ca at- 2/2Cl) dN2

(6.136)

The corresponding chemical potentials of the electroneutral combinations are /'~Na at- ]~C1 --/'LNa + FO +/ZC1 -- F~p = ~Na at- ~C1

(6.137)

/'~Ca at- 2/2Cl = ~Ca at- 2F~b + 2/xCl - 2F~, = ~Ca + 2/xCl

(6.138)

and

The physical significance of these combinations stems from the dissociation equilibria NaC1 ¢:~ Na + + C1- and CaC12 ¢,. Ca 2+ + 2C1which are characterized thermodynamically by the following definitions NaCI -- ]~Na +/~C1 =/ZNaCl o + PVNaCl + R T In aNaac1

(6.139)

6.4

339

Diffusion in electrolyte systems

/'~CaCI~_ = ~Ca + 2/~C1 -- ~CaCI2 o + P Vcac12 Jr- R T In aca acl2

(6.140)

Therefore, for electrically neutral species, Eq. (6.136) becomes d U = TdS - P d V + (/&NaC1) dN1 + (~CaC12 ) dN2

(6.141)

Example 6.6 D i f f u s i o n a c r o s s a membrane The conditions of phase equilibria across a membrane separating two salt solutions a and 13 are /3 /U'NaC1 = [/'NaC1

(6.142)

/.L~aC12 -- /.Z/3CaC12

(6.143)

Let us assume that one side of the membrane contains a chloride salt of a macromolecule to which the membrane is impermeable. The other side contains a solution of CaCI2 alone. The concentration of the macromolecule is Cm, and the number of charged groups per molecule is v. The concentration of CaC12 in the solution containing the macromolecule is Cs, and the concentration in the other phase is Cs;3. For the equilibrium between phases a and/3, we have ~aCl2 if- VCac12P °~ Jr- R T l n c ~ a (C~I) 2 -- ~aC12 if- VCaC12P]3 q- RT lnccPa (Cc~I) 2

(6.144)

Since the pressure terms are negligible, this expression reduces to C~ a (C~I)2 _ CC~a(Cc~1)2

(6.145)

~ = l?Cm -+-2c s CCa - Cs~ and CC1

(6.146)

CC~a -- Cs/3 and Cc~a = 2Cs~

(6.147)

We know that

and hence Cs~(1}Cm if_ 2C 2 )2 =_ Cs]3(2Cs~ )2 = 4(Cs~ )3

(6.148)

Equation (6.148) is the well-known Donnan equilibrium of salt across a membrane in the presence of a polyelectrolyte, to which the membrane is permeable. It demonstrates the characteristic properties of the chemical potentials of neutral salts.

6.4.1 PhenomenologicalApproach in Electrolyte Systems The local dissipation function for a system with charged species is xp - - j , . V T - ~ JiV/2 + J,.A i

(6.149)

where A is the electrochemical affinity, and is given by A - -Z

vi#ci - - Z

vi(#i + ziFd/) - - Z

vi~i - F # ; Z viz i

(6.150)

Since the charge is conserved in the reaction, ~v;z; = 0, so that d -- - - Z

l?i[&i -- d i

(6.151)

340

6.

Diffusion

For an isothermal system excluding chemical reactions, Eq. (6.149) reduces to (6.152)

= -~_.JiV[zi

i Electrochemical potentials also obey the Gibbs-Duhem equation (6.153)

Z Cirri ~-"0 i

In an n-component system, there are n - 1 independent forces (-7/2/). Equation (6.153) is used to eliminate the force for the solvent, and we have ) n-1 n-1 Ci xI'r -- E Ji - - - Jw V(-]~i ) --- £ J d v (--]'~i) Cw i -7•

(6.154)

where j/a is the flow of solute relative to that of the solvent. For a solution of a single electrolyte dissociating into two ions, the dissipation function is (6.155)

xI* = - - J d V ~ l - JdV/22 The phenomenological equations relating the flows and forces defined by Eq. (6.155) are j f -- - L l l V ~ I - L12V~ 2

(6.156)

jd = _L21V/~I _ L22V/~2

(6.157)

If V ~ 2 -- 0 , then J1a = - E l 1V]~I, which indicates that Lll is the generalized mobility of the cation, since it is the proportionality coefficient relating the flow to its conjugate force. Then, J2a is not zero, but is given by J2a = -LalV/21, indicating that the diffusion of the cation causes a drag effect on the anions, and such interactions are determined by the coefficient L12 or L21. Equations (6.156) and (6.157) can be used in the special case of an electrical conductance measurement. This analysis is usually carried out under isothermal, isobaric, and uniform concentration (Vtx;= 0) for all species in the cell. The electric current I is driven by a potential difference between two nonpolarizable electrodes, and the local field intensity e is defined by = -Vqt

(6.158)

Then, the forces acting on a Zl-valence cation and a Zz-valence anion become

-zlFe

(6.159)

= ~/X2 .at-z 2 F V ~ =-z2Fe"

(6.160)

V~I = V/X 1 -t- ZlFV ~ = ~2

So, Eqs. (6.156) and (6.157) become jd = (Z1Lll + 22L12)Fg

(6.161)

jd = (ZlL12 + z2L22)Fe

(6.162)

For a monomonovalent salt such as NaC1 or KC1, for which zl

= -z2 =

1, we have

J f =(i[,11-Zl2)Fg

(6.163)

jd __ (L12 _ L22)F~

(6.164)

6.4

Diffusion in electrolyte systems

341

The electric current due to the transport of all ionic species is given by the sum over all the charges carried by the ionic flows I

- ~ ziFjdi

(6.165)

i=1

For a single electrolyte, we obtain I - z, FJ'l + + z?FJd2 - ( z ( L 1 ,

+ 2z, z2/_q2 + z 2 L 2 2 ) F 2 g ,

(6.166)

Due to the condition of electroneutrality, diffusion flows can be used in Eq. (6.166). The flows relative to the water velocity are jja _ c, (v, - v w) and JJ We also have cl = vlcs and c2 =

l?2Cs,

:

C2 (V 2 - - V w )

(6.167)

SO Eq. (6.166) becomes

I = z l F v l c s v 1 + z 2 F v 2 c s v 2 - v w F c s ( V i Z 1 + v2z2)

(6.168)

However, electroneutrality implies that vjz~ + v2z2 = 0, so I = ZlFJ 1 + z2FJ2, where J1 and J2 are the absolute flows, J1 =ClVl and J 2 =C2V2, respectively. Ohm's law holds for homogeneous, isothermal salt solutions, therefore, the relation between the current and the electric field intensity may be reduced to l=Kt

(6.169)

where K is the electrical c o n d u c t a n c e of the solution. Comparing Eqs. (6.165) and (6.169) indicates that (6.170)

K = L'F 2

where L' is given by L'-

Z2Lll q- 2zlz2L12 q- z2L22

(6.171)

Equation (6.171) shows the direct relation between the electrical conductance of the solution and the phenomenological coefficient. Similar relations are obtained by measuring the fraction of the total current that is carried by each ion, also under the conditions V/xi = 0. This fraction is called the H i t t o r f t r a n s f e r e n c e n u m b e r (ti) and is expressed by

, z(z

(6.172) I

V/~;=O

For the case of a single electrolyte, tl and t2 may be evaluated by introducing Eqs. (6.161) and (6.162) into Eq. (6.172) t 1 - zlFJ/~- -- zI(Z1Lll if-22/-"12) I L'

(6.173)

t2 - z2FJ2'J -- z 2 ( z l L I 2 + z2L22)

(6.174)

I

L'

It is apparent that the two transference numbers are not independent since tl + t2 = 1. Therefore, an additional expression is required to evaluate the three coefficients, Lll, L12, and L22. Such a relation may be obtained from the diffusion of the electrolyte. In this case, there is no electric current in the system, and the total transport of charge must vanish z, ji , + z2j~ _ I _ 0 F

(6.175)

342

6.

Diffusion

By introducing Eqs. (6.156) and (6.157) for ,I d and a2d, we obtain a relation between the forces acting on the two ions --Z1 (L1 lV/~ 1 -Jr-L12V~2 ) - z2 (L12V].~ 1 -+- L22V~2 ) -- 0

or using the general form of Eq. (6.139) have

191V/~1 nt- 192V/.~2 -- V ~ s ,

V~I=

and from the neutrality condition

(6.176) V1Z 1 -+- V2Z2 =0,

we

Z--~2( Z1L12L' 'k-Z2L22 VI'Zs

(6.177)

Z1 ( Z1Lll -k-z2L12)

(6.178)

192

Introducing Eqs. (6.177) and (6.178) into Eqs. (6.156) and (6.157), we obtain

,1zlz2(lL22 I 2=zlze( , lL22 I 192

L'

gl.Zs

191

L'

V/Zs

(6.179)

(6.180)

The flows of Jsd of the neutral salt is

jd _ Jd _ Jd z1z2 ( ZllZ22-I~2 ) 19--71 192 VlV2 L' Vies

(6.181)

Fick's law describes the diffusion of a neutral salt in a binary solution as (6.182)

Jd1 = - D V c s

Comparing the resulting Eqs. (6.181) and (6.182), we may express the diffusion coefficient in terms of the L/j as

ZlZ2(1L22 /

1211)2

L'

(6.183)

The electrical conductance, transference number, and diffusion coefficient provide the three relations from which the phenomenological coefficients can be determined, and for a monomonovalent salt we have

D

Kt(

/Xss

F 2

LI1 = ~ +

D

L22 -

Kt~

+ __F---7

(6.184)

(6.185)

/Zss D

LI2 = L21 -

Ktlt 2 F2

(6.186)

/Zss

Straight coefficients L ll and L22 are nearly linear functions of concentrations, while the cross-coefficient L 12 is highly dependent on concentration and becomes quite small at high dilution, where the interactions between the ions are minimal. For determining the properties of the phenomenological coefficients, it may be advantageous to consider the mobilities, which express the behavior of ions. This description of the behavior is similar to the one gained by considering the frictional coefficients in the case of membrane permeability. The mobility may be defined by using the explicit expressions for the flows under uniform chemical potentials

6.4

343

Diffusion in electrolyte systems

jd

__ c1 ( v 1 _ V w ) = VlCsOOlZ1F E

(6.187)

where VlCs=Cl is the concentration of ion 1, and ~ol is the ionic mobility. Equation (6.143) shows that tO 1 is the relative velocity of the ion per unit electrical force. It is the velocity acquired by a force of 1 dyne. In practice, the mobilities ui are defined as the velocity of the ions acquired in a field of e = 1 V/cm, and mobilities Ua and u2 of the cation and anion, respectively, are U1 = Z 10A 1F

and

-- z2 (.0 2 F

(6.188)

=--V2Csb/2 E

(6.189)

-//2

Therefore, from Eq. (6.187) we may write

Jd1 -- VlCsUle and J2d and the total electric current becomes

I = zaFjdl + z2Fa J = VlZlCsF(u 1 +U2)E

(6.190)

Therefore, in terms of the mobilities, the electrical conductance is given by (6.191)

K -- VlalCsF(Ul -+- u 2 )

It is often convenient to consider the equivalent conductance

-

~eq --

heq

F(u 1

instead of K

+ lg2 )

(6.192)

VlZ1Cs

Similarly, the conductance of a single ion can be defined as (6.193)

hi = Fum and A2 = Fu 2 So, heq = h 1 + h2, which is the well-known expression of Kohlrausch. For a cation, we can express the diffusion in terms of the mobility af

-- PlCsUl E -- ( Z 1 L l l q- z 2 L 2 2 ) F E

(6.194)

Therefore, we have __ (Z1Lll J r - z 2 L I 2 ) F U1 -C1

_ zZL11F _ ~ n t C1Z1

z1z2L12 F ~ C1Z 1

(6.195)

-- u12 -- ~ZlZ2L12F PlZlCs

(6 196)

-

We now define the reducedphenomenological mobility (uij)

Igl 1 --

z2

1F

and

Pl Z1Cs

where Ull is the reduced phenomenological mobility of ion 1, and/A12 is a measure of the interactions between ions 1 and 2, so that Ul -- Ul 1 --U12 a n d u 2 -- u22 - u 1 2

(6.197)

Introducing Eq. (6.197) into Eq. (6.191), we have K -- C l 2 1 F ( U l l - 2 u 1 2

-+- u22 )

(6.198)

The equivalent conductances become Aeq -- F ( U l l

- 2u12 -+- u 2 2 )

(6.199)

344

6.

Diffusion

AS - F(H11 -+-U12 )

(6.200)

A2 = F(u22 + u12)

(6.201)

The transference numbers are defined by Hll--H12

t1 =

(6.202)

Ull -- 2u12 + u22

t2 =

U22--H12

(6.203)

u]] -- 2u12 + b/22 Finally, the diffusion coefficient of the salt can be expressed in terms of the mobilities

ss/",,"22 cs

(6.204)

O =

From Eqs. (6.199) to (6.204), the reduced phenomenological mobilities are obtained as ull = F v~zlD + A2

u22

Cs/Zss

AeqF

F VlZlD ~

- A2 ~

Cs~ss

AeqF

H12 = F PlZlD

(6.205)

(6.206)

(6.207)

AIA2

CsP'ss

heqF

These expressions can be used to calculate u O. from known values of the other parameters. Calculations for NaC1 show that uli and u22 remain approximately constant over a relatively wide range of concentrations, while u~2 changes considerably.

6.5

DIFFUSION WITHOUT SHEAR FORCES

Following Kerkhof and Geboers (2005), an approximation of the Boltzmann equation for a multicomponent monatomic gas system is OVi _ /9/-~ --fli {Vi "VVi }-- r e / +

f n ~ [ DiE - P 2 XiXj ~ DT'i j=l D/~ Pi

" (vj

Dr'J V l n T + V • [2TIiSi]+P~_~ ~xixj

Pj

.-vi)

(6.208)

j:l

or

Acceleration force =-convected momentum change- partial pressure gradient + external forces - thermal diffusion force + shearing force + intermolecular friction force In Eq. (6.208), D~. are the Maxwell-Stefan diffusivities and D~ = D~, ~/is the Newtonian viscosity, and S is the rate of the deformation tensor defined by

1{

S i = ~ VV i "[-(VVi

)v- - -2~ ( V ' v i ) I }

(6.209)

6.5

345

Diffusion without shear forces

where I is the diagonal unit tensor. By neglecting shear forces and bulk viscosity effects at isothermal conditions, Eq. (6.208) reduces to OV i __ Pi--~----Pi{Vi'~Vi}--~Pi

n +PiFi +P~_. Xixj (Vj--Vi)

./=~

D:/

(6.210)

For a fixed coordinate system, we consider steady transport and have n

0 . - P i { V. i Vvi}.

VP,. +piFi + P ~ x i x j (Vj --Vi) j=l D~j

(6.211)

Here, the first term represents the change of the convection flow, which is small compared with the other forces, so that Eq. (6.211) becomes n

xix--j-j (Vj -- Vi)

0 -- - - V ~ . -F PiFi "Jr-P E

j=l D!~

(6.212)

When external force is absent, and using the flows defined by N/= vici, we find _ xiNi -xjNi

_

1 Vp i = Vc i

D!/!

,i=~

(6.213)

RT

Here, the system is isobaric and the total concentration is constant. Also, the total flux is constant in direction z, and we have ON_ " -

0

0z

(6.214)

The definition of molar average velocity 1

N

c .i=1

c

(6.215)

V u - - )[_civ / - -

leads all the other spatial derivatives to be

OVM,=

0VM.= --

0,

Oz

-

0,

Ox

OVu,:

-

0

(6.216)

Oy

So, we have ( n - 1) independent fluxes and concentrations n-1

H-l

N. - - E

N/,

and V c . = - ~ _ . VC/

/ 1

(6.217)

,/=1

The matrix from Eq. (6.213) becomes [B](N)

(6.218)

= -(Vc)

with the coefficients

Bii - xi

1 D' m

Di~

Din

k=l D:k k~i

(6.219)

Equation (6.218) becomes (N) - [B]- J(Vc) - [D] (Vc)

(6.220)

346

6.

Diffusion

where [D] is the matrix of Fickian diffusion coefficients, which are not symmetric. For species i, we have n-1

n-1

Ni = - - E D/j ( g c j ) = --c E D O.( g x j ) j=l j=l

(6.221)

Equation (6.221) shows that the flux of a component is dependent on the concentration gradients of all components in the mixture.

Example 6.7 Binary and ternary isothermal gas mixtures For a binary mixture of gases under isothermal and isobaric conditions and without shear forces, from Eq. (6.213) we have

By substituting

N 2 + N 1 = N, we

XlN 2 - x2N 1 = D12VCl

(6.222)

N 1 -- X l N - D12~rCl

(6.223)

find

For an equimolar diffusion (no net flow), Eq. (6.223) becomes

(6.224)

N 1 = -D12~7c1

For a ternary mixture under the same conditions, we have XlN2 - x2N1 + XlN3 - x3N1 = Vc1 D{2

(6.225)

D~3

x2N1 - XlN2 + x2N3 - x3N2 = ~7c2

D;2

(6.226)

D5

On the other hand, the Fickian-type relations are N 1 = -Dll~TC 1 - D12~7c 2

(6.227)

N2 -- -D21Vc1 - D22Vc2

(6.228)

From the inversion, we then have the diffusion coefficients defined by Dll =

O12 =

D13[XlO23 + ( l - x l ) O l 2 ]

XlO23 (D13 - D12)]

021 -- x2D13(D23 - D12)]

S 022 =

D23[x2D13+(1-

x2)D12 ]

with S -- XlD23 -Jr-x2D13 -Jr-x3D12

(6.229)

6.5

347

Diffusion without shear forces

Example 6.8 Diffusion in a dilute isothermal gas mixture Suppose that in a gas mixture, component n is in abundance and other components are in trace amounts. Then, we have Pn -~ P,C,,Vv,etZ,, ~O,c,,V,, ~ 1, and for a steady transport, Eq. (6.208) becomes the Navier-Stokes equation for the single component, xi ~ 0, i 4: n. m

0--p{v,.Vv,}i-VP+p,F,

+V'[2niS i + ¢ ( V ' v i ) I ]

(6.230)

This equation can be solved with appropriate boundary conditions. For a trace component i, the i-i momentum exchange will be negligible compared with the i - n exchange. This leads to a smaller shear effect than the diffusive friction effect. Therefore, for component i, we have 0 ~ V P i + PiE + P xix" (v,

- v i)

(6.231)

D;', When there is no external force, we get 0 .~. VP;. + P xix" (v, - v i)

(6.232)

D;.', or

1

0 -~ Vc; +

( x i N . - N i)

(6.233)

Di'Vc i

(6.234)

D/', N i ~. N n x i

6.5.1

-

-

Gas Diffusion in Meso-and Macroporous Media

Modeling of diffusion of gases in porous media involves averaging mass and momentum balances and considering the three-dimensional nature of the medium. In a practical engineering approach, we consider the counterdiffusion of gases through a porous medium, and assume that we can describe the geometry by means of a single effective pore radius, the porosity e, and a tortuoisity factor ~-. Following Kerkhof and Geboers (2005), the flux of a species with respect to a unit area of the medium is 8 t

- - -

N,,i,,v

2 Ni,av

m

(6.235)

T

where Nx.;,av = vx,;,avC; is the cross-section averaged molar fluxes based on cross-section averaged velocities, which depend on driving forces defined by

B,-

(6.236)

dP~

dx The averaged molar flux of a gas is ,

r~ Pay + DK

N;~'v - -~2 ~

AP RTL

(6.237)

where L is the length of the tube, 31is the dynamic viscosity, rp is the channel radius, and DK is the Knudsen coefficient, and is approximated by

[2 8TJ2]

DK ~ 0.89 ~r;)

where M is the molar mass.

(6.238)

348

6.

The driving forces may be defined by

B1

dP1

-RT

dx

B2

dP2 dx

Diffusion

[, [, gD

(x2Ntx,l,av - XlNrx,2,av ) + flmNx,l,av

-e,

(6.239)

(XlNtx,2,av -- x2Nx,l,av ) -+-f2mN.tr,2,av

--

(6.240)

D( 2

-RT

gD

D~ 2

8

where gD is the diffusion averaging factor, and fro is the wall friction factors. These equations show that the force on a component per unit volume is due to friction with the other components and due to shearing friction with the tube wall. These equations may be solved in an iterative manner. These driving forces can be extended to multicomponent mixtures.

,i

]2

dx

(6.241)

~ (xjN'x'i'av - xiN'x'j'av ) + fiimN'x'i'av m j=l D~ e

Here, fire is obtained from the binary friction model. For isobaric counterdiffusion, from the above equations, we have N tx,l,av

Am

~

N'x,2,av

(6.242)

Am

For the case of equimolar diffusion through a porous medium, we have a net total pressure gradient defined by dP

7.2

(6.243)

- -RTN'x 1 av(Am -- Am) - -

dx

''

8

For counter diffusion in large pores, the friction term dominates the wall-friction term.

6.5.2

Diffusion in Liquid Mixtures

A generalization of the Boltzmann equation for liquids and dense gases is / OV i __ __ n ~ [ [3i - - ~ --Pi {Vi "VV i } CiVT, PI,Zi -- ciViVP i @ p i g - c R r E XiXj L DT'i j=l D~. Pi

+V.[2r/iS i + ¢ ( V . v i ) l ] + c R T

~

DT'j ] V In T Pj

)

(6.244)

xixj

~ (vj-vi) j-1 D/~

where ¢ is the bulk viscosity and p, is the chemical potential. Equation (6.208) for monatomic gases differs from Eq. (6.244) for liquids. In Eq. (6.244), the total pressure is replaced by cRT; the partial pressure gradient has been replaced by the chemical potential gradient ci7/, i = ciVr,pt, i + ciV~VP, and the bulk viscosity is introduced. By disregarding convection and shear forces at steady state, we have

n .

.

.

.

.

.

(v;- vi)

(6.245)

We may remove the concentration and chemical potential of component n using the Gibbs-Duhem equation

• i=1

Ci V T,p ld6i --

0

(6.246)

6.5

349

Diffusion without shear forces

Equation (6.245) can be inverted to find the generalized Fickian formulation 1

1

(N) = N ( x ) - ~-f[B]- [ c R T [ F ] ( V x ) + ( c V ) V P - ( p F ) + ( c ~ ) R T V l n T ]

(6.247)

For many liquid mixtures, it is assumed that there is equivolumetric transport, and hence molecular volume contraction is negligible. For isothermal conditions and without external forces, the pressure gradient vanishes, and we have (N) = N(x)-[B]-Ic[F](Vx)

(6.248)

(N) = N(x)-[D]c[F](Vx)

(6.249)

Then, the Fickian formulation is

where [D] = - [B]-l[F].

6.5.3

Diffusion in Mixture of Electrolytes

Considering the force by an electric field on ion transport at isothermal conditions, we find from Eq. (6.245) ~ XiXj

0 --- --CiVT, pI,.Li -- ci~.VP i + c i z i F V ~ + cRT

~

(vj - vi)

j=~ D[/

(6.250)

where z i is the charge and F is Faraday's constant. Many electrolytes are in electroneutrality, given by 0 - Y__dn__lCiZi . The addition of Eq. (6.250) over all components yields VP = 0, and hence the system is isobaric. Then, Eq. (6.250) becomes n

0

--

--Cil~T,pl.tii + ciziFVO + cRT~_~ ~xixi j=l

(Vj - - V i )

D!}

(6.251)

Equation (6.251) can also be written as n

F,iVXi +ciziFVO+RT~_~ xiNj - x j S i

O=-ciRT i=l

j=l

!

(6.252)

D!/

The activity coefficients to be used in the thermodynamic factor F of ions are generally concentration dependent. For a very dilute mixture and taking the solvent as component n, and xn ~- 1, Eq. (6.252) becomes

0 = -cRTF~/Vx i + ciziFV~t- RT xnNi D~',

(6.253)

When the activity coefficients are equal to unity, we find the Nernst-Planck equation

N: -=-D,,

6.5.4

Vc i +ciz i --~V4~

(6.254)

Liquid Diffusion in Meso- and Macroporous Media

Modeling of diffusion of liquids becomes more complex when the steric effect of molecular exclusion inside the pores is accounted for. Following Kerkhof and Geboers (2005), a distribution coefficient between the pore and free liquid may be defined by K ; - cp f

(6.255)

Ci

Here, we assume a liquid mixture inside a pore, which is in equilibrium with the free liquid outside, leading to

350

6.

Diffusion

/~P =/,/f

(6.256)

In modeling diffusion, we assume that the local thermodynamic equilibrium holds and variables can be estimated from equilibrium relations. The concentration-dependent part of the gradient of chemical potential is

cp

RT

'

nl

(P/

j=l

-~j

= Ki Z L, jV

(6.257)

Here, we estimate the thermodynamic factors F at the hypothetical free liquid values, which would be in equilibrium with the actual pore liquid concentration. If Ki is not dependent on concentration, Eq. (6.257) becomes

cp ,,-1 Fc O VT PtAP = Ki £ ' V(c p) RT ' j=l -~j

(6.258)

For a binary liquid mixture, we have

cp dl'zp

P ? _RT[gD (xPNx,l,av XlNrx,2,av) flmNrx,l,av] 7"2

_

?

_

_

?

D~2

dx

_

_

e

cP dtZPdx - - R T [ gDD; 2 (xPNx'2'av' - x2Nx'la' v)P'

- f2mNx'2'av 1 ' "r2e,

(6.259)

(6.260)

From the definition

Bi

-- Ci ~ dx )r = Ci ~, dx

T,P + ciVi

~

(6.261)

we add Eqs. (6.259) and (6.260) to find

q T2 F dP --RTlflmN'x,l,av + f2mN'x,2,av ] dx t_ _~ e

(6.262)

For isobaric counter diffusion of liquids, we have the same relation as Eq. (6.242) NPx,l,av __ Nx,l,av __

flm

Nrx,2,av

f2m

N x,2,av

(6.263)

For equivolumetric liquid diffusion through a porous medium, we have 0 = V1Nrx,l,av +

(6.264)

V2Nrx,2,av

The total pressure gradient becomes

dP - - R

Vl) ,2

f lm -- f 2m -~2

8_.

(6.265)

and

dx

--RTN'x'l'avCPVl ( flm - f2m ~2 ) 'r28

(6.266)

6.6

351

Statistical rate theory

Using Eq. (6.259), we find equivolumetric diffusion

: -

N:,,,av

~m;i~) + C(.fim~2V2 +

(6.267) gD/D(21

7

where 05 is the volume fraction. For larger pores, the values o f fire become small and gD approaches unity, and we obtain a Fickian diffusion equation, with the thermodynamic correction estimated by

F c ij - 6ii + ci

'

6.6

.

.

.

.

OCj )ck P T

(6.268)

1-

C

-~n

STATISTICAL RATE THEORY

Onsager's reciprocal rules are valid for systems that are sufficiently close to global equilibrium. It is crucial to determine under what conditions the assumption of linearity will hold. Onsager's reciprocal rules hold if the flows and forces are independent of one another and are identified from the rate entropy production or dissipation function. Statistical rate theory may help in verifying Onsager's reciprocal rules and understanding the linearity criteria. Statistical rate theory is not based on the assumption of near equilibrium, and leads to rate equations consisting of experimental and thermodynamic variables that may be measured or controlled. Statistical rate theory is based on the local thermodynamic equilibrium. It is derived from the quantum mechanical probability that a single molecule will be transferred between phases or across an interface or that a forward chemical reaction will occur in a single reaction step. Therefore, it should be modified to apply to systems in which simultaneous multiple molecular phenomena would be significant. For a transport process or a chemical reaction process involving single molecular phenomena at some time scale, the statistical rate theory equation for the net rate of the flow J is

q[exp(/exp( 'b ASr

(6.269)

where Jeq is the equilibrium exchange rate of molecules between the phases, ASf and kSb are the entropy changes in the isolated systems as a result of a single molecule been transferred forward and backward, respectively, and k is the Boltzmann constant. In statistical rate theory, the microscopic transition rates between any two quantum mechanical states of molecular configurations that differ by a single molecule having been transferred between phases (or having undergone a chemical reaction) are equal. That means that the average of these rates does not change, and Jeq is a constant throughout the process and equal to the equilibrium exchange rate. As long as the entropy changes are large, Eq. (6.269) cannot be linearized. For example, chemical reactions and interfacial transport between two phases yield large entropy changes. Statistical rate theory leads to well-defined coefficients that can be measured or controlled, and hence the criteria for linearization may be explicitly expressed.

Example 6.9 Transport in biological cells: osmotic and pressure driven mass transport across a biological cell membrane After Elliott et al. (2000) consider a compartmental system shown in Figure 6.3. Here, a biological cell containing a dilute solute and water solution is immersed in the same solute and water solution. The cell is placed in a thermal reservoir with temperature TR. The cell exchanges the solute and water across the wall, and therefore, undergoes osmotic shrinkage or swelling. We assume that both the water and the solute are incompressible and the saturation concentration of the solute in water does not depend on pressure. The cell is in mechanical equilibrium, although the water concentration or pressure inside and outside the cell is different. The pressure difference inside and outside the cell causes and is balanced by a tension in the cell membrane. The cell and its surroundings are at constant temperature. The derivation of the transport equations starts with the formulation of the entropy production rate. A differential change of the entropy of the isolated system dSsy s is dSsy s -- dS ° + dS i nt- d S m + d S R

(6.270)

352

6.

Figure 6.3.

Diffusion

Schematic mass transport in a biological cell in a thermal reservoir.

where So, Si, Sm, and SR are the entropies of the fluid outside the cell, the fluid inside the cell, the cell membrane, and the reservoir, respectively. The differential entropy of the fluid outside the cell is 1 Po dVo dSo :FdUo +T

/Xw,od N w o _

r

~/Xs'° d N s o

T

(6.271)

'

where Uo, Po, and Vo are the internal energy, the pressure, and the volume, respectively, of the fluid outside the cell, /Xw,o and Nw,o are the chemical potential of the water and the number of moles of water outside the cell, and/Xs,o and Ns,o are the chemical potential of the solute and the number of moles of solute outside the cell. For the fluid inside the cell, we have 1 d U i + Pi d V i

dSi :--T

--T

dN w i -

~w,i

-

T

'

~s,i

T

(6.272)

dN s i

'

The subscript i indicates the properties for the fluid inside the cell. For the membrane, we have dSm : 7 d g m -

(6.273)

dam - Z tZm'k dNm,k k

T

where Urn, 7m, and Am are the internal energy, the tension, and the surface area, respectively, of the cell membrane. Here, the cell membrane is treated as a two-dimensional phase,/Xm,k is the chemical potential of the kth molecular species in the membrane, and Nm,k is the number of molecules of the kth species in the membrane. For a quasi-static heat transfer in the reservoir, we have dS R -- ---'~1 d U o _ F1d U i - - ~1d U m

(6.274)

After substituting Eqs. (6.271)-(6.274) into Eq. (6.270) and applying the following constraints d V o = - d V i, d N w ,o = - d N w , i, d N s ,o

=-dNs,i, d N m, k = 0

(6.275)

we have dS~y s _ P i - e o T

dVi

"Ym dA m +

~w,o --/£w,i

~w,o --/Zs, i dNw i + '

T

T

dNs T

'

i

(6.276)

By assuming that mechanical equilibrium holds for the membrane and that the cell is spherical with the radius r, we have Pi -

Po

-

2ym F

(6.277)

6.6

353

Statistical rate theory

Substituting Eq. (6.277) into Eq. (6.276), we obtain the rate of entropy production dSsys - / Z w ' ° -/'l'w'i '-'~"AA'lwi nt-/-/~w,o -- ~s,i ,.,l,A£fs dt

T

'

T

'

i

(6.278)

where Nw,i and Ns,i are the rates of change of the numbers of water and solute molecules inside the cell, respectively. The forces in Eq. (6.278) are related by the Gibbs-Duhem relation and are not independent. For a dilute solution, the difference in the chemical potentials of an incompressible solvent across the membrane is (6.279)

~ w , o - ]~w,i = V w ( P o - P i ) - k T ( x s , o - X s , i )

where Vwis the partial molecular volume of water, k is the Boltzmann constant, and Xs is the mole fraction of solute, which is approximately defined by x s = Cs/C w, where Cs is the concentration of the solute and cw is the concentration of pure water. For an incompressible solute with a pressure-independent saturation concentration, the difference in the chemical potentials of the solute across the membrane is ~s,o -- ~s,i -- Vs (/Do -- P i ) -

kT[ln(xs,o)-ln(xs,i)]

(6.280)

where Vs is the partial molecular volume of solute, and In (Xs, o ) - In (Xs, i ) = cs'°

_--Cs'i

(6.281)

Cs

where Cs,o m Cs,i CS z

In (Xs,o) - In (Xs,i) Substituting Eqs. (6.279)-(6.281) into Eq. (6.278), and rearranging, yields dSsys -[Vs]Vs,i--1- Vw~/w i](Po - Pi)--tdt

'

[ ]~rs,i

-

F

]Vwi] , kT(cs,o-Cs,i )

(6.282)

cw

We can identify the flows and forces from Eq. (6.282) and establish the following phenomenological equations Vs]Vs, i -4- Vw]Vw, i = L I I ( P o - Pi ) q- Ll2kT(cs,o - Cs,i)

[

/NTs i -' c

iN~w i ] ,' -- L21 (/Do -- Pi ) + L22 k T cw

(c s o

- Cs,i )

(6.283)

(6.284)

On the other hand, from statistical rate theory, we have ~rs~' - Js'eq [exp ( ASf-T]- exp ( ]]--~ASb

(6.285)

where Js,eq is the equilibrium exchange rate of solute molecules across the membrane. The forward entropy change is ASf - AS o + AS i --I-AS m -4- AS R

(6.286)

Each phase is a simple system, and we may write the appropriate Euler relations 1

PoA Vo -/Xw'---~°AN w o -/zs'° ZL/V T ' T s,o

ASo = rAC/o + T

(6.287)

354

6.

Diffusion

~si=F1 AUi + ?-Pi AVi

/'/'w,i _ i z~TVsi r ANwi, _ _/'Ls, T ' AAm - Z/zm'k ~Nm k k T

(6.289)

1AU i 1 -T-T ~um

(6.290)

AS m = -~- A Um -

ASR=

-T

1AU °

(6.288)

We formulate the ASb in a similar manner. Using Eqs. (6.286)-(6.290) and the following constraints A Vo = - A V i, ANw,o = ~/Ww,i - 0 ,

(6.291)

ANs, o - - - 1, ~Ws, i --1, L~/Wm,k - 0

in Eq. (6.285), we obtain

[ (so si/exp/ Si [/ wo wi) exp/Wiwo)]

]Vs,i - Js,eq exp

kT

]~rw,i = Jw,eq exp

kT

(6.292)

kT

(6.293)

kT

Equations (6.292) and (6.293) are the formulations of nonequilibrium thermodynamics and describe the osmotic transport of solute and water across the membrane. These equations can be linearized for small chemical potential differences, and we obtain ]~rsi

2Js'eq

]Vw,i

2Jw'eq

'-

kT

-

kT

(6.294)

(~s,o--~s,i)

(6.295)

(~w,o-~w,i)

Combining Eqs. (6.294) and (6.295) with Eqs. (6.283)-(6.286), we have 2 (Js eqVs Jw'eqVwcw) kT(cs'° -- Cs,i) Vsl~rs,i "k-VwJ~rw,i - K2 (Js,eqV? . k _ J w , e q V 2 ) ( p o _ P i ) + K Cs

Nsi c

cw

2(,seqs

kT

cs

i,+2('seq "weq)

cw

--k-f

~2Cs -+- Cw2

kT(cs,o

-- Cs,i)

(6.296)

(6.297)

Comparing these statistical rate theory equations, with Eqs. (6.283) and (6.284), we obtain the following phenomenological coefficients z~i = - f2f

(Js,eqW?+Jw,eqW2)

(6.298)

(

(6.299)

2 Vs /-'12 = k-T Js,eq c7

/

--Jw,eq-~ Cw, }

6.6

355

Statistical rate theory

L21 -- ~

Js,eq ~ -- Jw, eq Cs Cw )

2 (Js,eq Jw, eq) L22 = ~ k C2 -Jr- C; 2

(6.300)

(6.301)

Equations (6.299) and (6.300) show that Onsager's reciprocal rules hold. The Js,eq and Jw, eq have a microscopic definition represented by perturbation matrix elements and a macroscopic definition represented by the equilibrium exchange rate. As long as the criteria of linearization are satisfied, the statistical rate theory may be used to describe systems with temperature differences at an interface besides the driving forces of pressure and concentration differences.

6.6.1

Diffusion in Inhomogeneous and Anisotropic Media

Macroscopic diffusion model is based on underlying microscopic dynamics and should reflect the microscopic properties of the diffusion process. A single diffusion equation with a constant diffusion coefficient may not represent inhomogeneous and anisotropic diffusion in macro and micro scales. The diffusion equation from the continuity equation yields OP

Ot

- -v.a

(6.302)

where P and J are the density (probability or number) and diffusion flow of the particles. Following Christensen and Pedersen (2003), a definition for the diffusion flow J is a - -(P~2v v + fivP)

(6.303)

where V is an external potential,/2 is the mobility, and I) is the diffusion tensor given by the Einstein relation [zkT = l}

(6.304)

In Eq. (6.303), the first term represents the drift in the potential force field Vand the second is the diffusional drift given by Fick's law. Combining Eq. (6.302) with Eq. (6.303), we have OP

- V.{P/2V V + I}VP)

(6.305)

Since Eq. (6.305) cannot represent systems with inhomogeneous temperatures, we may have the following alternative equation OP Ot

- v . {Pt2v~ ~ + v. tiP) = v . ( P ( ~ 2 v v + v . f i ) + ISVP)

(6.306)

Equations (6.305) and (6.306) are different because of the drift term V.(PV. 1}), which is sometimes called a "spurious" drift term. These diffusion equations have different equilibrium distributions and are two special cases of a more general diffusion equation.

6.6.2

van Kampen's Hopping Model for Diffusion

The hopping model was originally introduced to discuss electron transport in solid materials, but it may be useful as a general model for diffusive motion~ In a one-dimensional diffusion equation based on hopping model, the diffusion medium is modeled by a large number of wells/traps in which the particles can get temporarily caught. The density of traps (o-) is the density times the cross-section of traps and may change throughout the media. In solvents, for example, the density represents the capability of the solvent molecules to form a cage around the suspended particle. The rate of escape of particles (a) is controiled by the local energy barrier • of the trap and the local temperature T

356

6.

Diffusion

a = a exp -~-T

(6.307)

Here, a defines the global time scale for escape out of the traps, and incorporates the spatial variation of the escape a into the potential barrier ~. Large values of o~ signifies shallow wells and hence fast diffusion, while large values of o- signifies small mean free paths and hence slow diffusion. Inhomogeneities in the medium may cause spatial dependencies of c~ and o-, such as in micelles, or by the interaction of two diffusing molecules. The isotropic diffusion equation based on van Kampen's one-dimensional hopping model may be extended to three dimensions using Cartesian coordinates in flat Euclidean space

OP_av.[exp(-~/kT ) (Vo- VV) (exp(-~/kT) 0.2 P o- + kT +V Ot o-2

P

)]

(6.308)

Equation (6.308) implies that the isotropic diffusive motion along the coordinate axes is independent. Here, 7V/KT is the drift due to an external potential force field V, while 7o-/o-represents an internal drift caused by a concentration gradient of the traps. The term PV(e-*/XT/O-2) is the "spurious" drift term. Equation (6.308) allows spatial variations of all parameters T, V, ~, and o- with inhomogeneous temperature. From Eq. (6.308), the diffusion coefficient becomes D= a

exp ( - • / kT)

(6.309)

o -2

The stationary solution of Eq. (6.308) for systems with a uniform temperature is

Ps = C o-exp(-V/kT) exp(-~/kT)

(6.310)

where C is the normalization constant. The stationary distribution depends on the local value of the macroscopic diffusion coefficient D and on the local value of one of the microscopic trap parameters o- or ~. Consider three special cases based on a simplification of Eq. (6.308): 1. o- oc exp (-~/KT). For this case, Eq. (6.308) becomes

OP Ot

(6.311)

This is the traditional diffusion model given in Eq. (6.305) with the diffusion coefficient D proportional to l/o-. For this case, the so-called "spurious drift" term vanishes because the effects of c~ and o- cancel each other out in the stationary state. The stationary distribution is proportional to the Boltzmann distribution exp(-V/kT) and independent of D. 2. o- = constant. Then, Eq. (6.308) becomes

OP Ot

+ ~

D

+ DVP

which is similar to the relation given in Eq. (6.306). The stationary solution is proportional to example, the particles would experience very slow diffusion in regions of low mobility. 3. • = constant. Internal drift does not vanish, and Eq. (6.308) becomes

OP - v'[DP ( VV kT + -~2FD ) +

(6.312)

exp(-V/kT)/D, for

(6.313)

which is different from both Eqs. (6.305) and (6.306). The stationary solution is V]x/-D Ps =exp -~--~

(6.314)

For isotropic systems, the diffusion equations for these three cases are mathematically equivalent since they can be transformed into each other by introducing effective potentials.

6.6

Statisttcal rate theory

357

Equation (6.305) has been used widely to model diffusion in liquids, but the above discussion shows that it is valid only where c~ ,x o-. Equation (6~306) is valid when the concentration of traps is constant, a situation that is more realistic. In all other cases, the diffusion equation is a combination of Eqs. (6.305) and (6.306).

6.6.3

Anisotropic Diffusion

The general diffusion equation, based on the hopping model, is

OP--g('cP)+av'!ex--p-P(-OP/kT)p{gCrat ,_ (re ---~- ~kTF}+ g . exp(-dP/kT)o 2 P~

(6.315)

where F is an external fbrce and ~ is the velocity held of the medium. If we assume that the parameter cr is isotropic while the trap potential is anisotropic and represented by the tensor D -- o

exp(-~/kT)

(6.316)

o"

The tensor + is required to be symmetric because of its relation with the diffusion tensor. Of course, the o- can also be anisotropic. The above equation may cover most phy~;ical systems, and it can even be used on curved manifolds. The anisotropy introduces two new f;eamres: (i) equations (6.305) and (6.306) cannot in general be transformed into each other, as the drift term g. 1) may not be a gradient field. Equation (6.306) can describe systems where the directions of the principal axes depend on the spatial position. (ii) Detailed balance implies that the diffusion flow a vanishes everywhere in the stationary state. However. this is not automatically satisfied for anisotropic systems and one needs to exercise extra care in the modeling of such systems. Inhomogeneity does not affect the detailed balance. (ii!) The diffusive part of the diffusion flow must be represented by a - - V l i ) P , while the drift is represented by (PV. D). In general, the difRlsion equation depends on all the microscopic parameters. The microscopic parameters of van Kampen's model are the local values of the effect!re trap density or, which is density times cross-section and work function • . The traditional dift~asion relation of Eqo (6.305) is valid only for isotropic diffusion and under the restrictive conditions that o-~cexp(-~/kT)~ It may be unsatisfactory even in a homogeneous system with nontrivial geometry. Equation (.6.306) is valid when the effective trap concentration is constant, which is more realistic fbr liquids.

6.6.4

Diffusion in Biological Solutes in Liquids

The diffusion of small molecules and macromolecules (e.g., proteins) in aqueous solutions plays an important role in microorganisms, plants, and animals. Diffusion is also a major part in food processing and in the drying of liquid mixtures and solutions, such as diflk~sing aroma constituents in fruit juice, coffee, and tea from solutions during evaporations. In fermentation, nutrients, oxygen, and sugar diffuse to the microorganisms, and products, waste, and sometimes enzymes diffuse away. The kidneys ~emove waste products like urea, creatinine, and excess fluid from the blood. Kidney dialysis removes ,,~aste products from the blood of patients with improperly working kidneys. During the hemodialysis process, the patient ~sblood is pumped through a dialyzer, and waste diffuses through a semipermeable membrane to the aqueous SOJtltlOl~l cleaning i!uid Macromolecules have large molecular weights and various random shapes that may be coil-like, rod-like, or globular (spheres or cllipsoids~ They l:b,~m t~t~e solutio>;. ]-~heir sizes and shapes affect their diffusion in solutions. Besides that, interactions of large molecules w|th the smal! solvent and/or solute molecules affect the diffusion of macromolecules and smaller molecules. Sometimes, reacts.on-diffusion systems may lead to facilitated and active transport of solutes and ions in biological systems. These type> ot'transport will be discussed in Chapter 9. Macromolecules often have a number ot sites i'or interactions and binding of the solute or ligand molecules. For example, hemoglobin in the blood binds oxygen at certain sites. Surface charges on the molecules also affect the diffusion. Therefore, the presence of macromolecules and small solute molecules in solutions may affect Fickian-type diffusion. Most of the experimental data on protein diffusivities have been extrapolated to very dilute or zero concentration since the diffusivity is often a function of concentration. Table 6.4 shows diffusivities of some proteins and small solutes in aqueous solutions. The diffusion coefficients for the macromolecules of proteins are on the order of magnitude of 5 × 10 l i me/s. For small solute molecuies, the diffusivities are around 1 × 10-9 m2/s. Thus, macromolecules difi\ise about 20 times slower then small moiec~ies. Small soi~f:cs such as urea and sodium caprytate ofien coexist with protein lnacromolecules in solutions. When these small molecules d~ffuse the i~rotein soitmc,n~ the diffusivity of the molecules decreases with increasing protein

358

6.

Diffusion

Table 6.4 Diffusivities of dilute biological solutes in aqueous solutions Solute

Molecular weight

Sucrose

Temperature (K)

342.3

Urea

310 293 293 298 293 298 310 298 298 298 293 293 293 293 293 293

60.1

Glycerol Glycine Creatinine Sodium caprylate Bovine serum albumin Urease

92.1 75.1 113.1 166.2 67500 482700

Soybean protein Lipoxidase Fibrinogen, human Human serum albumin y-Globulin, human

361800 97440 339700 72300 153100

Diffusivity (m2/s) 0.697 x 0.460 x 1.20 x 1.378 x 0.825 x 1.055 x 1.08 x 8.78 × 6.81 x 4.01 x 3.46 x 2.91 x 5.59 x 1.98 X 5.93 × 4.00 x

10- 9 10-9 10.-9 10 9 10-9 10-9 10-9 10-l° 10-11 10-ll 10-ll 10 ~1 10--11 10 1~ 10-al 10-11

Source: C.J. Geankoplis, Transport Processes and Separation Process Principles, 4th ed., Prentice Hall, Upper Saddle River (2003).

concentrations. This reduction is partly because of the binding of small molecules to proteins and is partly due to blockage by the large molecules.

6.6.5

Prediction of Diffusivities of Biological Solutes

For predicting the diffusivity of small molecules (with molecular weights less than about 1000 or molar volumes less than about 0.500 m3/kg) in aqueous solution, we may use the Wilke-Chang correlation to estimate the diffusivity in mZ/s 1.173 x 10-16 (~tMB)1/2 T DAB = ~B VO'6

(6.317)

where ~ is an association parameter of the solvent (~ is 2.6 for water, 1.9 for methanol, 1.5 for ethanol, 1.0 for benzene, ether, and n-heptane, and 1.0 for other unassociated solvents). MB and/ZB are the molecular weight and viscosity of solvent B in Pa s or kg/m s. VA is the molar volume of the solute at the boiling point, which may be obtained from Table 2.10 in Chapter 2. This relation should be used with caution outside temperature ranges of 278-313 K. For larger molecules, the equations for diffusivity estimations may not be too accurate. As an approximation, the Stokes-Einstein equation can be used 9.96 X 10-16T DAB =

~ V 1/3

(6.318)

where/x is the viscosity of the solution, and VA is the molar volume of the molecule. For a molecular weight above 1000, the following equation may be used 9.40 x 10-15T DAB----

~(MA)I/3

(6.319)

where M A is the molecular weight of the large molecule A. When the shape of the molecule deviates greatly from a regular spherical shape, this equation should be used with caution. During the diffusion of small molecules (with molecular weights less than about 1000 g/mol or molar volumes less than about 0.500 m3/kg) in protein solution, the diffusion may be blocked by the large molecules. In order to account for this effect, we need the diffusivity DAB of solute A in water alone, the water of hydration on the protein, and an obstruction factor. A semitheoretical relation to approximate the diffusivity of solute in a globular-type protein solution is DAp -- DAB (I -- I .8 I X l 0 -3 Cp)

(6.320)

6.6

359

Statistical rate theory

where Cp is concentration of P in kg/m 3. The approximate diffusion equation is

Nx=

OAp (CA1--CA2) z2

-

-

(6.321)

z1

When, however, a solute molecule A is bound to a protein, the diffusion flux of A is equal to the flux of unbound solute A and the flux of the bound protein-solute complex. This type of flux estimation requires data on binding. The equation used is

D A p - [DA~ (1 - 181 x l 0-3 CP)( free A%/100 + Dp (bound A%)]100

(6.322)

where Dp is the diffusivity of protein alone in the solution m m2/s. The percentage of free A can be determined from the experimental binding coefficient.

Example 6.10 Prediction of diffusion coefficients of macromolecules Predict the diffusivity of human serum albumin at 293 K in water as a dilute solution and compare with the experimental data in Table 6.4. Table 6.4 shows the molecular weight of human serum albumin A as MA = 72300 kg/kg tool. The viscosity of water at 298 K is 0.897 X 10.3 Pa s. Using Eq. (6.319)

DAB =

9.40 × 10-15 T /~(MA ) 1/3

=

9.40 × 10-15 (293) (0.897 × 10-3)(72300) 1/3

= 7.37×10 -11 m2/s

This value is 24% higher than the experimental value of 5.93 × 10-11 m2/s.

6.6.6

Diffusion in Biological Gels

Gels are semisolid and porous materials. Some typical gels are agarose, agar, and gelatin. Also, a number of organic polymers exist as gels in various types of solutions. They are composed ofmacromolecules. For example, the gel structure of agarose is loosely interwoven, and is composed extensively of hydrogen-bonded polysaccharide macromolecules. The pores that are open spaces in the gel structure are filled with water. The rates of diffusion of small molecules in the gels are usually less than in aqueous solutions. When there are no electrical effects, the gel structure mainly increases the path length for diffusion. Table 6.5 shows a few typical values of the diffusivity of some solutes in various gels. In some cases, the diffusivity of the solute molecule in pure water (wt% = 0) is given in Table 6.4. This shows how much the diffusivity decreases due to the gel structure. For example, at 293 K, Table 6.4 shows that the diffusivity of sucrose in water is 0.460 ;4 10.9 m2/s, while it is 0.107 × 10-9 m2/s in 5.1 wt% gelatin. This indicates a considerable decrease of 77%.

Example 6.11 Diffusion of solutes in biological gels A 0.02-m-long tube of a gel solution connects two chambers of agitated solutions of dextrose in water. The gel solution is 0.79 wt% agar in water and is at 278 K. The dextrose concentration in the first chamber is 0.4 g mol dextrose per liter solution and the other chamber concentration is 0.01 g mol dextrose per liter solution. Estimate the flow of dextrose in kg mol/s m 2 at steady state. Solution" Assume that the system undergoes steady state and one-dimensional diffusion. From Table 6.5, we read the diffusivity coefficient for solute dextrose at 278 K as DAB = 0.327 × 10.9 mZ/cm. The concentrations are

Cal--

0.4 1000

- 0.0004 g mol/cm 3 = 0.4 kg mol/m 3

360

6.

Diffusion

Table 6.5 Typical diffusivities of solutes in dilute gels in aqueous solutions Solute Sucrose

Urea

Methanol Urea

Gel

Wt% gel in solution

Gelatin Gelatin Gelatin Gelatin Gelatin Gelatin Gelatin Gelatin Gelatin Gelatin

Temperature (K)

Diffusivity (m2/s)

0

278

0.285 x 10 9

3.8

278

0.209 × 10 -9

5.1

293

0.107 x 10 -9

10.35

278

0.252 × 10 9

0 2.9 5.1 5.1 10.0

0.880 0.644 0.609 0.542 0.859

3.8

278

X X X X x

10 9 10 9 10 9 10-9 10 9

0.626 x I0 9

Agar

278

0.727 × 10 9

Agar Agar

278 278

0.591 × 10 -'~ 0.472 × 10 9

Agar

278

0.297 x 10-9

Agar

278

0.199 × 10

Agar

278

0.297 × 10

Dextrose Sucrose Ethanol

Agar

278

0.327 × 10

Agar

NaC1 (0.05 M)

Agarose Agarose

278 278 278 298

0.247 0.393 1.511 1.398

Glycerin

Agar

× x x x

10 10 10 -9 10 ...9

298

Source: C.J. Geankoplis, Transport Processes a n d Separation Process Principles, 4th ed., Prentice Hall, Upper Saddle River (2003).

(:12-

0.01

- 0.00001 g mol/cm 3 = 0.01 kg mol/m 3

1000

Since the urea concentration is very low, for dextrose diffusing through stagnant water in the gel, we may use NA~=--DAB

d f Az ..... _ D A B A C

dz

--

~

_

Az

_

DAB(CA1 --CA2) 2 2 -- Z 1

0.327×10-9(0.4-0.01) 0.02-0 NAz = 6.377 × 10-9 kg mol/s.

PROBLEMS 6.1

Derive modeling equations for diffusion through a stagnant phase.

6.2

Derive modeling equations for diffusion into a falling liquid film.

6.3

Two large vessels containing binary mixtures of gases A and B are connected by a truncated conical duct, which is 2 ft in length and has internal diameters of larger and smaller ends of 8 and 4 in, respectively. One vessel contains 80 mol% A, and the other 30 mol% A. The pressure is 1 atm and the temperature is 32°E The diffusivity under these conditions is 0.702 ft2/h. By disregarding the convection effects: (a) Calculate the rate of transfer of A. (b) Compare the results with those that would be obtained if the conical duct was replaced with a circular duct with a diameter of 6 in.

361

Problems

6.4

Oxygen in muscles is used for the oxidative removal of lactic acid. One theory suggests that a slab of muscle in contact with oxygen will possess a recovered oxygen zone at the muscle-oxygen interface, and an unrecovered lactic acid zone. The interfacial boundary between the two zones will advance with time into the lactic acid zone. Consider semi-infinite muscle tissue region x > 0 through which oxygen diffuses. At the external boundary x = 0, the oxygen concentration is Co - constant. The oxygen zone is 0 < x < L, where the boundary L between the oxygen zone and the lactic acid zone depends on time t, and L(0) = 0. The boundary conditions are C = C(O,t) = Co and C(L,t) = 0. The velocity of the advancing front dL/dt is assumed to be proportional to the oxygen flux at L

Ox

=L

dt

where A is a constant of proportionality. (a) When the front advances slowly, and in the oxygen zone, the concentration C satisfies the steady-state diffusion, find the velocity of the advancing f~ont. (b) Find the velocity if the oxygen diffusion is not assumed to be in a quasi-steady state. 6.5

Predict the diffusivity of lipoxidase at 293 K in water as a dilute solution and compare with the experimental data in Table 6.4.

6.6

Predict the diffusivity of soybean protein at 293 K in water as a dilute solution and compare with the experimental data in Table 6.4.

6.7

Estimate the diffusion coefficient of hemoglobin in water at 293 K. The globular hemoglobin molecule has a radius r ~ 30 A and the Boltzmann constant k = 1.38 × 10-16 erg/K. The viscosity is 0.01 E

6.8

A 0.02-m-long tube of a gel solution connects two chambers of agitated solutions of dextrose in water. The gel solution is 0.79 wt% agar in water and is at 278 K. The dextrose concentration in the first chamber is 0.4 g mol dextrose per liter solution and the other chamber concentration is 0.01 g tool dextrose per liter solution. Estimate the flow of dextrose in kg mol/s m 2 at steady state.

6.9

A 0.05-m-long tube of a gel solution connects two chambers of agitated solutions of urea in water. The gel solution is 1.05 wt% agar in water and is at 278 K. If the urea concentration in the first chamber is 0.25 g mol urea per liter solution and the other chamber concentration is zero. Estimate the flow of urea in kg mol/s m 2 at steady state.

6.10

Consider a thin rectangular region in a catalyst particle shown below. Component A diffuses across the top surface. After reaching one of the other three surfaces, component A undergoes an instantaneous reaction. Therefore, the concentration of A at the three surfaces of the region will be zero as shown in the figure below. There will be no net bulk motion within the region of catalyst. Derive the Laplace equation for a two-dimensional model of a catalyst passage. Use the separation of variables technique to obtain the exact solution.

CA= C(x)

CA =0

CA= 0

0

6.11

CA=O W

X

Consider the drying of a large sheet of wood with a uniform thickness of z. For a one-dimensional diffusion problem, the initial concentration profile through the wood will be a function of z. Develop an analytical method to describe the concentration profile of water moisture within the wood.

362 6.12

6.

Diffusion

Consider the absorption of oxygen from air in the aeration of a lake or the solid surface diffusion in the hardening of mild steel in a carburizing atmosphere. Both these processes involve diffusion in a semi-infinite medium. Assume that a semi-infinite medium has a uniform initial concentration of CAo and is subjected to a constant surface concentration of CAs. Derive the equation for the concentration profiles for a preheated piece of mild steel with an initial concentration of 0.02 wt% carbon. This mild steel is subjected to a carburizing atmosphere for 2 h, and the surface concentration of carbon is 0.7%. If the diffusivity of carbon through the steel is 1 × 10-11 m2/s at the process temperature and pressure, estimate the carbon composition at 0.05 cm below the surface.

REFERENCES R.B. Bird, W.E. Stewart and E.N. Lightfoot, Transport Phenomena, 2nd ed., Wiley, New York (2002). M. Christensen and J.B. Pedersen, J. Chem. Phys., 119 (2003) 5171. C.E Curtiss and R.B. Bird, Ind. Eng. Chem. Res., 38 (1999) 2515. E.L. Cussler, R. Aris andA. Bhown, J. Memb. Sci., 43 (1989) 149. J.A. Daoud, S.A. E1-Reefy and H.E Aly, Sep. Sci. Tech., 33 (1998) 537. Y. Demirel, Chim. Acta Turcica, 14 (1986) 114. Y. Demirel and S.I. Sandler, Int. J. Heat Mass Transfer, 44 (2001) 2439. EA.L. Dullien and A.EA. Asfour, Ind. Eng. Chem. Fundam., 24 (1985) 1. J.B. Duncan and H.L. Toor, AIChE J., 8 (1962) 38. J.A.W. Elliott, H.Y. Elmoazzen and L.E. McGann, J. Chem. Phys., 113 (2000) 6573. A. Katchalsky and EE Curran, Nonequilibrium Thermodynamics in Biophysics, Harvard University Press, Cambridge (1967). B.C. Eu, Kinetic Theory and Irreversible Thermodynamics, John Wiley, New York (1992). P.J.A.M. Kerkhof and M.A.M. Geboers, Chem. Eng. Sci., 60 (2005) 3129. D. Kondepudi and I. Prigogine, Modern Thermodynamics, From Heat Engines to Dissipative Structures, Wiley, New York, 1999. R. Krishna and G.L. Standard, AIChE J., 22 (1976) 383. J. Leffier and H.T. Cullinan, Ind. Eng. Chem. Fundam., 9 (1970) 84. R. Taylor, Ind. Eng. Chem. Fundam., 21 (1982) 407. R. Taylor and R. Krishna, Multicomponent Mass Transfer, Wiley, New York (1993). M.T. Tyn and W.E Calus, J. Chem. Eng. Data, 20 (1975) 310. A. Vignes, Ind. Eng. Chem. Fundam., 5 (1966) 189.

REFERENCES FOR FURTHER READING V. Alopaeus, Comp. Chem. Eng., 26 (2002) 461. A. Eftekhari, Chem. Phys. Lett., 374 (2003) 164. R. Krishna and J.A. Wesselingh, Chem. Eng. Sci., 52 (1997) 861. O.O. Medvedev and A.A. Shapiro, Fluid Phase Equilib., 208 (2003) 291. L. Minkin, Radiat. Prot. Dosimetry, 106 (2003) 267. G.L.J.A. Rikken and P. Wyder, Phys. Rev. Lett., 94 (2005) 016601. M.O. Vlad, J. Ross and EW. Scheneider, Phys. Rev. E, 62 (2000) 1743.

7 HEAT AND MASS TRANSFER 7.1

INTRODUCTION

Simultaneous heat and mass transfer plays an important role in various physical, chemical, and biological processes; hence, a vast amount of published research is available in the literature. Heat and mass transfer occurs in absorption, distillation extraction, drying, melting and crystallization, evaporation, and condensation. Mass flow due to the temperature gradient is known as the thermal diffusion or Soret effect. Heat flow due to the isothermal chemical potential gradient is known as the diffusion thermoeffect or the Dufour effect. The Dufour effect is characterized by the heat of transport, which represents the heat flow due to the diffusion of component i under isothermal conditions. Soret effect and Dufour effect represent the coupled phenomena between the vectorial flows of heat and mass. Since many chemical reactions within a biological cell produce or consume heat, local temperature gradients may contribute in the transport of materials across biomembranes. Various formulations and methodologies have been suggested for describing combined heat and mass transfer problems, such as the integral transform technique, in the development of general solutions. In this chapter, cross phenomena or coupled heat and mass transfer are discussed using the linear nonequilibrium thermodynamics theory. 7.2

COUPLED HEAT AND MASS T R A N S F E R

Using a dissipation function or entropy production equation, the conjugate flows and forces are identified and used in the phenomenological equations for simultaneous heat and mass transfer. Consider the heat and diffusion flows in a fluid at mechanical equilibrium not undergoing a chemical reaction. The dissipation function for such a system is

- -Jq' V In T -

~_~ Ji "aik

i.k=l

/=1

_>0

I Ow/

(7.1)

T./'.wi~ I

where aik = gik + w~/w,, 8~k is the unit tensor, and j; is the diffusion flow of component i. The heat flows are related through the internal energy flow J,,

Jr,-- J// q- £ hiJi - J; + £HiJi i=1

Jq and Jq !

tt

(7.2)

i--1

Similarly, the entropy flow is expressed by

J; £

J,,---T-+

sij i

(7.3)

i=1

where s; is the partial specific entropy and h; and u; are the partial specific enthalpy and partial specific internal energy, respectively. 7.2.1

Binary Systems

The independent forces and flows are identified by the dissipation function of Eq. (7.1). Therefore, the forces and heat and mass flows for a binary system are

364

7.

Heat and mass transfer

Xq = - V l n T

(7.4)

Xl = - 1--~ ~-------~1 0

VW1

(7.5)

w2 OWl T,P The linear phenomenological equations describe the flows (7.6)

-Jl = LlqVlnT +Lll w2 ~,Ow1)r,P

1(

0~1 ]

-Jq = LqqV lnT + Lql W2 OWl)T,P

VW1

(7.7)

By the Onsager reciprocal relations, the matrix ofphenomenological coefficients is symmetric, Llq -- Lql. Since the dissipation function is positive, the phenomenological coefficients must satisfy the inequalities

Lqq > O, Lll > 0,

LqqL11-L2ql > 0

(7.8)

Fourier's law describes heat conduction caused only by the temperature gradient

Jq

-kVT

(7.9)

where k is the thermal conductivity in the absence of a concentration gradient. Comparison of Eqs. (7.7) and (7.9) yields the relationship between the phenomenological coefficient Lqq and the thermal conductivity coefficient

Lqq = kT

(7.10)

Fick's law describes the diffusion flow caused only by the concentration gradient for an isothermal fluid Jl -- -pDVwl

(7.11)

which contains the diffusion coefficient D given by

D=Dll--Lll

1 ( 0/xl )

(7.12)

Pw2 ~,OWl T,P

The diffusion caused only by the temperature gradient is called the thermal diffusion (Soret effect). When the concentration gradient vanishes, Eq. (7.6) reduces to

Jl -- -Llq V In T

=

- p DT1VT

(7.13)

where DT1 is the thermal diffusion coefficient of component 1, and is related to the cross coefficient Llq by

Llq

DT1 --

(7.14)

P The Soret effect D~ may be defined by

D~ - DT1

-

-

Llq

(7.15)

WlW2T PWlW2T The thermal diffusion coefficient is usually smaller by a factor of 102-10 3 than the ordinary diffusion coefficient for nonelectrolytes and gases.

Z2

365

Coupled heat and mass transfer

The heat flow due to the Dufour effect arises only from a concentration gradient, and is expressed by

_, Jq - - L q l

( Otxl) TD('Vw 1

w,

Ow1 r,P VWl -- --11 ~,OW1 T,P

(7.16)

where D~ is the Dufour effect for component 1, and is related to the phenomenological coefficient Lql by Lql

D(, =

(7.17)

pwlw2 T

and Onsager's reciprocal relations yield (7.18)

D('-D(

So, Eqs. (7.6) and (7.7) can be expressed in terms of the transport coefficients of thermal conductivity k and diffusivity D -Jl

- J2 -

P(WlW2D(VT + D V W l )

(7.19)

p(W2WlD2 V T - DVwl)

(7.20)

or, as the sum of mass fractions is unity --J2 --

Jl

-

- J q - k V T + Pl ~, Ow 1 )

T,P

TD(' Vw I

(7.21)

Equations (7.19) and (7.20) suggest that D( - - D ;

(7.22)

The thermal diffusion ratio KT1 of component 1 is defined by

KT1-- D where ST1 is called the Soret coefficient for component 1, and is given by

D(]_ STI-

7

KT1 wlw2 T

L,, )

k aw, )j

(7.24)

Table 7.1 shows some experimental values of thermal diffusion ratios for liquids and gases at low density and pressure. IfKT1 is positive, component 1 diffuses to a cooler region; otherwise, it diffuses to a hotter region. The thermal diffusion .factor c~1 for component 1 is mainly independent of concentration for gases, and is given by

~l - T --5 - Ts T

(7.25)

The inequalities in Eq. (7.8) can now be written in terms of the transport coefficients of the thermal conductivity and mass diffusivity by using the thermodynamic stability condition (Otz~/Owl)r,p >- 0 k > 0 , D > O , (D{)2 <

kD p w ( w z T (at*l / awl )r,e

(7.26)

366

7.

Heat and mass transfer

Table 7.1. Experimental values of thermal diffusion ratios for liquids and gases at low density and pressure a

Components (liquids) A-B

T (K)

C2H2C14-n-C6H14 C2H4Br2-n-C2H4C12 C2H2C14-CC14 CBr4-CC14 CC14-CH3OH HzO-CH3OH

298 298 298 298 313 313

0.50 0.50 0.50 0.09 0.50 0.50

330 330 264 264 327

0.20 0.60 0.294 0.775 0.10 0.50 0.90

KT Ma2v (1/K)

XA

M1M 2

Components (gases) H2-Ne Hz-N 2 Hz-D2

1.080 0.225 0.060 0.129 1.230 -0.137 0.0531 0.1004 0.0548 0.0633 0.0145 0.0432 0.0166

aR.L. Saxton, E.L. Dougherty and H.G. Drickamer, J. Chem. Phys., 22 (1954) 1166. R i . Saxton and H.G. Drickamer, J. Chem. Phys., 22 (1954) 1287. L.J. Tichacek, W.S. Kmak and H.G. Drickamer, J. Chem. Phys., 60 (1956) 660. T.L. Ibbs, K.E. Grew andA.A. Hirst, Proc. R. Soc. (Lond.),A173 (1939) 543. H.R. Heath, T i . Ibbs and N.E. Wild, Proc. R. Soc. (Lond.),A178 (1941) 380.

For a binary mixture of ideal gases, the molar fraction Xl and the chemical potential/Xl of the first component are given by x1 =

(7.27)

Mzwl

M2 Wl + M lw2

(7.28)

~1 : tz~ ( T , P ) + R T l n P x l

The partial derivative of chemical potential of the mixture at constant temperature and pressure is 0/Zl)

RT

=

OWl T,P

(7.29)

Wl ( M2 W1 -t- M l w 2 )

So, the diffusion coefficient, given in Eq. (7.12), becomes RT

1

P

wlWz(M2w1 + MlW2)

D = Lll-

(7.30)

and the phenomenological equation of the heat flow is given by -J"q - k V T Jr- O t

RT2 p M2 Wl +

Vw 1

(7.31)

M lw2

With the thermal diffusion ratio KT, Eq. (7.19) becomes -Jl -- pD (KT1 V In T + Vw 1)

(7.32)

Table 7.2 shows the viscosity, mutual diffusion coefficient, and thermodynamic factor for aqueous solutions of ethylene glycol and polyethylene glycol (PEG) at 25°C; the diffusivity decreases considerably with increasing molecular weight, while the viscosity increases. Table 7.2 shows the thermal diffusion ratios for liquids and gases at low density and pressure; the thermal diffusion ratios are relatively larger in liquids.

7.2

367

Coupled heat and mass transfer

Table 7.2a. Viscosities, mutual diffusion coefficients, and thermodynamic factors for aqueous solutions of ethylene glycol and PEG at 25°C a Molecular

q (cP)

Olnyl

D (10 5 cm2/s)

weight of PEG, M

1+

01nx~ r

62.1

0.920

1.138

1.00

106.1 150.2 194.2 400 600 1000 1500

0.920 0.921 0.922 0.924 0.927 0.938 0.946

0.886 0.753 0.663 0.440 0.349 0.269 0.223

1.01 1.01 1.01 1.03 1.06 1.14 1.24

2000 3400 4600 8000

0.953 0.985 1.005 1.065

0.187 0.145 0.124 0.093

1.33 1.55 1.69 1.95

10000 20000

1.103 1.291

0.082 0.058

2.06 2.32

"Chan et al. ( 1993 ). bThe thermodynamic factor is estimated from the van Laar equation.

Table 7.2b. Thermal diffusion ratio, KT, thermal diffusion coefficients DT, and heats of transport Q* for aqueous ethylene glycol and polyethylene glycol (PEG) solutions at 25°C a Molecular weight ofPEG, M 62.1 106.1 150.2 194.2 400 600 1000 1500 2000 3400 4600 8000 I0000 20000

n':

(r (IK)

Kr (l/K)

D-r (10 5 cm2/(sK))

Heat of transport, Q* (kJ/mol)

1 2 3 4 8.7 13.2 22.3 33.6 45.0 76.8 104 181 227 453

0.00380 0.00566 0.00736 0.00842 0.01350 0.(1170 0.0209 0.0261 0.0313 0.0383 0.(1460 0.0595 0.0639 0.0899

0.00351 0.00535 0.00704 0.00809 0.0131 0.0166 0.0205 0.0256 0.0308 0.0377 0.0453 0.0587 0.0631 0.0889

0.00399 0.00474 0.00530 0.00536 0.00578 0.00580 0.00552 0.00571 0.00575 0.00546 0.00560 0.00546 0.00515 0.00516

2.82 4.23 5.49 6.28 10.2 13.4 17.7 24.0 30.9 43.8 57.4 85.9 97.1 154

~Chan et al. (2003). n ~:is the average number of ethylene glycol segments per molecule: n = ( M - 18.02)/44.05. In Table 7.2b, the values of heat of transport is obtained from

~) = Rr~-,,- 1 ~ ~n~, ) T

1

where ~/i and xl are the activity coefficient and mole fraction of ethylene glycol or PEG. (r is the molality based thermal diffusion ratio (Soret coefficient)

7.2.2 MulticomponentSystems The dissipation function resulting from heat and mass transfer is expressed by T--J,,.

VlnT-~

JiLTv[-~I-Fi]

(7.33)

i=1

A transformation of Eq. (7.33) may be useful by using the total potential tt', which is the summation of the chemical potential and the potential energy

368

7.

Heat and mass transfer

t.L' = I~ + ep

(7.34)

so that we have

hiVT -

VT~;

hiVT T

(7.35)

where VT/X~is the isothermal gradient of the total potential, and given by YT[.L ~ --" Y/.Li @ siVT -k-Vepi = Vt.t,~ -4-siVT

(7.36)

Using the total potential together with the internal energy flow J, J u = J ; r -+-~

hiji

(7.37)

i=1

The dissipation function, given by Eq. (7.33), becomes -- - J q . V l n T -

(7.38)

Ji "YT~i i=1

Since only n - 1 diffusion flows Ji are independent, we have n--1

Ji "VT/X; : E Ji "VT (/x;-/X'n) i=1

(7.39)

i=1

So, we transform Eq. (7.38) into the form that shows each of the independent forces n-1

= - J q . V In T - ~ Ji "YT (~b[~;--

~LL;)

(7.40)

i=1

We may consider the phenomenological equations for the n + 1 vector flows Of Jq and ji, and n + 1 forces of V(1/T) and VT/X'.Assuming linear relations between the forces and the flows, we have the following phenomenological equations

- J q - LqqVlnT +

LqjVTtX j

(7.41)

j=l

--Ji = LiqVlnT + ~ , L/jVT/Z)

(i = 1 , 2 , . . . , n )

(7.42)

j=l

For a system in mechanical equilibrium in which the pressure gradient is balanced by the mass forces, the GibbsDuhem relation becomes

~ pjVT/x'j

-0

(7.43)

j=l

Summation of the diffusion flows ji yield

~ Ji -- ~ Pi i=1

(Vi -- V) -- 0

(7.44)

i=1

where v refers to the mass-average velocity. Therefore, the forces -VT/Xj' and the flows are not all independent. The coefficient Lqq is related to the thermal conductivity, while Liq and Lqi define the thermal diffusion and heat transferred by mass diffussion (Dufour effect) of component i, respectively. The coefficient Lii determines that part of the diffusion current j; arising from its own chemical potential gradient of component i, while the codiffusion coefficient LO.defines

7.3

369

Heat of transport

that part o f j i arising from the chemical potential gradients of componentj. The codiffusion coefficients L~j are affected by the forces acting between the dissimilar molecules. If the average intermolecular force between i andj is repulsive, the diffusion ofj induces a diffusion current of i in the opposite direction, and L O.is negative. Otherwise, L!j is positive and the diffusion of componentj induces a diffusion current of component i in the same direction. We may determine each phenomenological coefficient experimentally. The Onsager reciprocal relations reduce the number of coefficients to be determined. If we substitute Eq. (7.42) into Eq. (7.44), we find that the coefficients Liq and L!/obey the following relations:

After deriving

--~TT~ ~ from

• Liq - O.

~ L(i - 0

i=1

i 1

(j - 1,2,...,n)

(7.45)

Eq. (7.41), and substituting into Eq. (7.40), we obtain

-j,-L,qVlnT+~(L!i---

piLi'p, ] VT~ )

( i - 1,2,...,n)

(7.46)

/=1

Since the (n - 1) chemical potential gradients are independent, the coefficients (L O.- p/L~n/Pn) are determined. In order to find the independent forces for the heat and diffusion flows in a system at mechanical equilibrium, we express the dissipation function due to heat and diffusion in the form given in Eq. (7.40). Later, we establish the linear relations for the flows and the forces, in which all the forces are independent /7--1

- J q - LqqVIn T + ~

LqjVt(tz) -/z',)

(7.47)

j--1 tl- ]

--Ji

L,qVlnT + ~ LuXTt(#; -/z,',)

--

(i - 1,2,...,n-I)

(7.48)

I=1

From these linear relations, we can define the following relations between the phenomenological coefficients"

n-1

n-I .j= n-I

l

Lni - - E Lii

n-1 n-1

n-1

j=l

j=l

z,,,,,-Z EL,;, z,,,--EL,;,

j= 1

i =1

(7.49)

( i -- 1.2..... n - 1)

i=1

7.3

HEAT OF TRANSPORT

The transportation quantities are useful in describing the transport phenomena in a multicomponent fluid. Consider rewriting Eq. (7.48) as n-1

~,L!jVT(/.~;

-(j; + L ; q V l n T ) -

-#')

(i-1,2,...,n-I)

(7.50)

+LiqVlnT ) (j-1,2,...,n-I]

(7.51)

/--: ]

Then, the thermodynamic forces can be determined by /l- 1

Vt(~} - # , ' ~ ) - - ~ K j ; ( ] ; i=l

Since the matrix of resistance coefficients Ki; is the inverse of the matrix of conductance coefficients L!j, we have ~ i - L~ 1. Combining Eqs. (7.51) and (7.47) yields

~ -J~ -

)

17--1

LiqLqjK/i V In T - ~,~ Lq./K/iJi

Lqq ~,/-I

i,/=-1

(7.52)

370

Z

Heat and mass transfer

From Eq. (7.52), the heat of transport Q~ of species i is defined by n-1

n-1

Q; -- Z Lqj K ji -- E LqjL~1 (i = 1,2,...,n-I) j=l

(7.53)

j=l

Using the Onsager reciprocal relations, Eq. (7.53) can be rewritten as n-1

n-1

Q; -- E Ljqgij : Z LjqLj il j=l

(i = 1,2,...,n- 1)

(7.54)

j=l

The heat of transport can be used in the phenomenological equations to eliminate the coefficients Lqj or Ljq. After introducing Eq. (7.54) into Eq. (7.52), we obtain the expression for heat flow in terms of the heat of transport n-1

-Jq

Zqq

/

Z LiqQi i,j=l

n-1

Z

VlnT-

Qi Ji

(7.55)

i,j=l

For an isothermal system where Vln T = 0, we have

nl

f':/

tt

*

aq : Zi=I

*

Qi Ji and Qi

(7.56)

: ~-~/J T

Equation (7.56) shows that the heat of transport Qi* is the heat carried by a unit diffusion flow of component i when there is no temperature gradient and no diffusion of other components. For a binary fluid, the heat of transport is expressed by

Q1

• -

Lql Zll

-

Llq /~l

*

- U 1 - ( h 1 - h2)

(7.57)

where U~ is called the energy of transport. U~ is the internal energy carried by a unit diffusion of component 1 when there is no temperature gradient and no diffusion of the other components. The heat of transport is the flow of heat entering through the surface of contact to maintain isothermal conditions if a unit of mass leaves the region in equilibrium. The value of Q~ can be calculated analytically when the energy field of molecules crossing the surface is known. n-1

J. : ~ U;j i

(7.58)

i=1

Similarly, we may also define the entropy of transport S[ by n-1

Js -- Z S;ji

(7.59)

i=1

We can relate the heat of transport and energy of transport by using the relation n-1

(7.60)

J'q = Ju - 2 (hi - hn) Ji i--1

and obtain

Qi = U i - ( h i - h n )

(VT=0)

(i=l,2,...,n-1)

(7.61)

Likewise, we can relate the heat of transport and entropy of transport by using the relation

[ n,

Jq = T Js -~_,(si - s , ) Ji i=1

1

(7.62)

Z4

371

Degreeof coupling

and find

Si _ Qi + (s i _ s,, ) - 1 (U~ - tx i + ) T T tx~

(7.63)

Using the heat of transport given in Eq. (7.57), we can eliminate the cross coefficients Lql and L|q in Eqs. (7.6) and (7.7), and obtain ~ J~

- LqqVlnT + L,,Q7 ~ ( Otxl ] Vw 1 w2 t, Owl ) r,P

(7.64)

(7.65)

T,P Two stationary states in the coupled processes can be identified as the level flow where X i = 0, and the static head where Ji = 0. Examples of the static head are open-circuited fuel cells and active transport in a cell membrane, while examples of the level flow are short-circuited fuel cells and salt and water transport in kidneys.

7.4

DEGREE OF COUPLING

From Eqs. (7.47) and (7.48), we can define the ratio of forces A - V In T/Vr(/x(-/x~), and the ratio of flows r / = Jq/jl for a binary system. Dividing Eq. (7.47) with Eq. (7.48), and further dividing both the numerator and denominator by (LqqLll)l/2[Vr(kt{ - kt~)], we obtain

)1/2 A + Llq/(LqqLil 1 Llq/(LqqLll )1,/2 A + (L 11/Lqq )1/2

Yl -- ( Lqq/Ll l

)1/2

(7.66)

Equation (7.66) shows that the ratio of flows rl varies with the ratio of forces A. As the quantity Llq/(LqqLll) 1/2 approaches zero, each flow becomes independent, and we have the ratio of flows approaching rl ~ (Lqq/Lll)A. If Llq/(LqqLll) 1/2 approaches to + 1, then the two flows are not associated with the forces, and the ratio of flows approaches a fixed value when rl ~ __+ (Lqq/Lll ) 1/'~"-. This is the case where the matrix of the phenomenological coefficients becomes singular. The ratio

Llq q-

(LqqLll

)1/2

(7.67)

is called the degree of coupling, for a two-flow system. The degree of coupling is a dimensionless parameter, and quantifies the coupling of the energy conversion in a process. As the heat and mass flows are both vectors, the sign of q indicates the direction of forces on the substances. If Liq > 0 and hence q > 0, a substance may drag another substance in the same direction; however, the substance diffuses in the opposite direction if Liq < 0 and q < 0. For heat and mass flows, the dissipation function, given in Eq. (7.40), defines the two limiting values of q between + 1 and - 1. An incomplete coupling takes a value between these two limits. Using the phenomenological stoichiometric coefficient z 2 _ Lqq

LI~

(7.68)

Equation (7.66) reduces to r/=

q+Az Aq+(1/z)

(7.69)

Equation (7.69) indicates that for the known values ofz and A, the ratio of flows is determined by the degree of coupling q. With complete coupling, A is equal to z and q becomes + 1 or - 1. The value of A may be negative when the chemical potential may a have negative value due to the nonideality of the mixture. In nonideal systems, the change

372

7.

Heat and mass transfer

of the Fick diffusivity with respect to the concentration may be zero, and sometimes negative, and phase splitting may occur in the liquid flows. This complex behavior needs to be determined from thermodynamic models. The reduced force (zk), and the reduced flow (B/z) can be related by

7q _ q + zA _ l + zA/q z qzA + 1 zA + 1/q

(7.70)

The term zk/q changes between 0 and - 1 .

7.5 7.5.1

COUPLING IN LIQUID MIXTURES Coupling in Binary Liquid Mixtures

For a binary fluid at mechanical equilibrium and for diffusion based on the mass-average velocity, we can now establish a set of phenomenological equations (Eqs. 7.6 and 7.7) with nonvanishing cross coefficients, and hence represent the coupled heat and mass flows ,

1._~ ( 0 ~ 1 ) V W

1

-Jq = LqqVlnT +Llq w2 ~,Owl )v,e

--Jl = LqlV In T + Lll

1 (0.1/ w2 ~,OWl )T,P

These equations obey the Onsager reciprocal relations, which state that the phenomenological coefficient matrix is symmetric. The coefficients Lqq and L ll a r e associated with the thermal conductivity k and the mutual diffusivity D, respectively. In contrast, the cross coefficients Llq and Lql define the coupling phenomena, namely the thermal diffusion (Soret effect) and the heat flow due to the diffusion of substance i (Dufour effect).

Example 7.1 Mass diffusion flow in term of mole fractions Derive the coupled mass and heat flows in terms of mole fractions. In terms of mass fractions, the diffusion flow and heat flow for a binary mixture are --Jl = pD (KrlV In T + Vw1) tl

*

-Jq = kVT + pQ1 D11VWl Mass fractions and the concentration of the solution are obtained from Wl:

XlM1

_ XlM1 --~, XlM1 + x 2M2 Mav

C--

P Mav

Therefore, the gradients of mass and molar fractions are related by ~Tw1 =

M1M2 _ M1M2 (xlm 1 + x2M2) 2 Vx1 (Mav)2 Vx1

Using the gradient above and the concentration of the solution, we get -Jl = cMavD K n V In T +

(Mav) 2

Vxa

M1M2 - J ; : kVT + cMavQ1Dll (Mav)2 Vx1

7.5

7.5.2

Coupling in liquid mixtures

373

Degree of Coupling and Heats of Transport

The heat of transport of component 1, Q) - L Iq/L~l, can be used in the phenomenological equations to eliminate the coefficients Lql or Llq. If we express Llq in terms of the heat of transport (Llq - Q~Lll), Eq. (7.66) becomes

(Lqq/Lll)1./'2 A+Q1 * (Lll/Lqq )1/2 rl - Ql (Lll/Lqq) 1/2A +(Lll/Lqq) 1/2

(7.71)

We can define the degree of coupling in terms of the heat of transport

q-Q1

L1---!-I

Lqq

(7.72)

The degree of coupling may be a basis for comparison of systems with various coupled forces. The phenomenological coefficients are expressed in terms of k, D, and Q1 • pDM 1M2 Wl w2 Llq - Q1 [MRT(1 + Fll)]

(7.73)

Lqq = kT

(7.74)

L ~ - p w 2 D ( ' t x )j o w l

(7.75)

where Fll = (0 In yl/0 lnxj)r,p is the thermodynamic factor, and can be determined from an activity coefficient model such as NRTL. Table 7.2a shows the thermodynamic factors, while Table 7.2b displays the Soret coefficients, thermal diffusion coefficients, and heats of transport for aqueous solutions of ethylene glycol and PEG at 25°C. Table 7.3 shows the thermal conductivity of various alkanes in chloroform at 30°C and 1 atm. We observe a considerable decrease in the values of k with decreasing concentration. For the branched alkanes, the values of k reaches a minimum at around x l --~ 0.5. We can describe the degree of coupling q and the thermal diffusion ratio of component 1 KT~ in terms of the transport coefficients and thermodynamic factor (F)

Qj

pDM~ M 2w l w 2

kM1M2WlW2 j/2 KT1 -- q pDMR(1 + Fll)

(7.76)

(7.77)

Equation (7.76) shows that the degree of coupling is a function of the heat of transport, the thermal conductivity, * 1/'~. ~ The value Llq is independent of and the diffusion coefficient, and is directly proportional to the product Q1(D/k) the thermal conductivity. As the heat and diffusion flows are both vectors, the sign of q is related to the direction of flow of species. If q > 0, the flow of a species may drag another species in the same direction; however, it may push the other species in the opposite direction if q < 0. For heat and mass flows, for example, the two limiting values of q are + 1 and - 1 . Since the degree of coupling is directly proportional to the product Q~(D/k) 1/2, the error level of the predictions of q is mainly related to the reported error levels of Q~ values. The polynomial fits to the thermal conductivity, mass diffusivity, and heat of transport for the alkanes in chloroform and in carbon tetrachloride are given in Tables C 1-C6 in Appendix C. The thermal conductivity for the hexane-carbon tetrachloride mixture has been predicted by the local composition model NRTL. The various activity coefficient models with the data given in DECHEMA series may be used to estimate the thermodynamic factors. However, it should be noted that the thermodynamic factors obtained from various molecular models as well as from two sets of parameters of the same model might be different.

374

7.

Heat and mass transfer

Table 7.3. Thermal conductivities of binary mixtures as a function of mole fraction of selected alkanes in chloroform at 30°C and 1 atm a Solute(I)

xl

k (mW/(m K))

n-Hexane

0.0 0.2572 0.4801 0.5808 0.6751 0.8471 1.0

110.79 104.28 104.90 105.51 107.25 112.06 115.19

n-Heptane

0.0 0.2295 0.4427 0.5437 0.6412 0.8266 1.0

110.79 105.50 106.68 108.75 110.62 115.55 120.41

n-Octane

0.0 0.2071 0.4106 0.5110 0.6105 0.8070 1.0

110.79 106.79 108.74 110.60 113.71 117.92 123.81

3-Methylpentane

0.0 0.2572 0.4801 0.5808 0.6751 0.8471 1.0

110.79 99.66 97.83 99.40 100.36 104.01 106.94

2,3-Dimethylpentane

0.0 0.2295 0.4427 0.5437 0.6412 0.8266 1.0

110.79 103.99 101.32 100.93 102.11 103.04 105.26

2,2,4-Trimethylpentane

0.0 0.2071 0.4106 0.5110 0.6105 0.8070 1.0

110.79 102.77 97.27 96.21 95.67 95.69 96.02

aRowley et al. (1988).

The degree of coupling and the thermal diffusion ratio KT1for the liquid mixtures are calculated from Eqs. (7.76) and (7.77), and shown in Figures 7.1 and 7.2. The liquid mixtures consist of six to eight carbon alkanes of n-hexane, n-heptane, n-octane, 3-methylpentane, 2,3-dimethylpentane, and 2,2,4-trimethylpentane in chloroform and in carbon tetrachloride. These systems represent straight and branched chains of the alkanes in two solvents. As the degree of coupling and the thermal diffusion ratio depend on the heat and mass transfer coefficients, the plots of q and KT1 versus the alkane compositions Xl show the combined effects of the transport coefficients on q and KT1 in various solvents. The figures reveal some important properties of coupling. The first is that the absolute values of q and KT1 reach peak values at a certain concentration of the alkane, and these peak values decrease as the molecular weight increases. Second, the solute concentrations at the peak values of coupling decrease gradually as the molecular weights increase. The third is that the behavior of alkanes is similar up to a certain concentration of solute depending

7.5

375

Coupling in liquid mixtures

in chloroform . . . . . . . . . . . .

0

= . . . .

.

. -

.

.

.

.

.

.

.

.

. . - _

0

-0.01

-0.01

-0.02

-0.02

q -0.03

-0.03

-0.04

-0.04

-0.05

\,..

-- , , " /

-0.05 0

0,2

0.4

0,6

(a)

0.8

!

0

0.2

0.4

0.6

0.8

1

(b)

XI

in carbon tetrlchloride 0

0 /

-0.01

-0.0I

k,

-0,02 q -0.03

/

-0.02

,:,: //

q -0.03

"- -_

~ S / / /

-0.04 -0.04

-0.05 0

0.2

0,4

(c)

0,6

0,8

i

0

0.2

0.4

0.6

(d)

X!

0.8

t

x,

Figure 7.1. Change of the degree of coupling q with the alkane concentrations xl at 30°C and ambient pressure: (a) and (c) straight chain alkanes, (--) n-hexane, (---) n-heptane, (---) n-octane; (b) and (d) branched-chain alkanes, (--) 3-methylpentane, (---) 2,2dimethylpentane, (---) 2,2,4-trimethylpentane. Reprinted with the permission from Elsevier, Y. Demirel and S.I. Sandier, Int. J. Heat Mass Transfer, 43 (2002) 75. in chloroform 0

0

-0,I

-0,1

-0.2

-0.2

-'--0,3

-0.4

-0.4

\%.

-0,5

,,'/

",,,L

-0,3

-'0.5

-0.6

-O.6 0

0.2

0,4

(a)

0,6

0.8

1

0

0,2

0.4

0.6

(b)

Xl

0.8

1

xj

in carbon t e t r a c h l o r i d e 0

0

-0.1

-0,I

-0,2

.

.

.

.

.

.

.

.

, .

.

.

.

.

: .

.

.

.

.

.

.

.

.

.

..-......._

_

_

.

.

.

_

-0,2

'\

~" -0,3

/2

i" -0,3

-0.4

-0.4

-0.5

-0.5 0

(a)

.

0.2

0.4

0.6 Xt

0.8

1

0

(b)

0.2

=

0.4

0.6

0.8

1

x,

Figure 7.2. Change of the thermal diffusion ratio KT1 with the alkane concentrations xl at 30°C and ambient pressure: (a) straight chain alkanes, (--) n-hexane, (---), n-heptane, (---) n-octane (b) branched-chain alkanes, (--) 3-methylpentane, (---) 2,2dimethylpentane, (---) 2,2,4-trimethylpentane. Reprinted with the permission from Elsevier, Y. Demirel and S.I. Sandier, Int. J. Heat Mass Transfer, 43 (2002) 75.

upon the combined effects of branching and the solvent on q and KTI (through approximately xl = 0.2), but at higher concentrations they behave differently. The fourth is that the absolute maximum extent of coupling is small, as expected, and the branching of alkanes has only marginal effects on the coupling phenomena. Tables 7.4 and 7.5 show the peak values of q and KT1 for both the straight and branched alkanes separately in both the solvents chloroform and carbon tetrachloride. The general trend is that the branching of the solute molecule has a minimal effect on the coupling for the considered alkanes. Tables 7.4 and 7.5 also show the effect of the solvent on q and KT1. The alkane concentrations where the peaks of q and KT1 occur are lower in chloroform than in carbon

376

7

Heat and mass transfer

Table 7.4. Degree of coupling q, and maximum ratio of dissipation ('r/,~)maxa

Straight-chain alkanes Solute

- qmax

Branched-chain alkanes

x~

(~TX)max

Solute

- qmax

x~

(XlO 4) (a) For alkanes in chloroform n-Hexane 0.056 n-Heptane 0.046 n-Octane 0.046

( 'r/)k)max

(XlO 4)

0.534 0.408 0.337

7.993 5.365 5.295

3-Methylpentane 2,3-Dimethylpentane 2,2,4- Trimethylpentane

0.053 0.048 0.046

0.475 0.392 0.340

7.085 5.815 5.226

(b) For alkanes in carbon tetrachloride n-Hexane 0.048 0.588 n-Heptane 0.045 0.527 n-Octane 0.041 0.341

5.718 5.203 4.185

3-Methylpentane 2,3-Dimethylpentane 2,2,4-Trimethylpentane

0.055 0.040 0.038

0.570 0.467 0.513

7.574 4.083 3.574

aDemirel and Sandler (2002).

Table 7.5. Thermal diffusion ratio for solute KT1a

Straight-chain alkanes Solute

Branched-chain alkanes

--KTj (l/K)

x~

(a) For alkanes in chloroform n-Hexane n-Heptane n-Octane

0.679 0.565 0.675

0.569 0.488 0.484

(b) For alkanes in carbon tetrachloride n-Hexane n-Heptane n-Octane

0.564 0.572 0.503

0.629 0.628 0.441

Solute

--KT1 (l/K)

x~

3-Methylpentane 2,3-Dimethylpentane 2,2,4-Trimethylpentane

0.596 0.591 0.622

0.527 0.449 0.402

3-Methylpentane 2,3-Dimethylpentane 2,2,4-Trimethylpentane

0.579 0.476 0.503

0.569 0.550 0.557

aDemirel and Sandler (2002).

tetrachloride. Generally, the peak values qmaxare also smaller in carbon tetrachloride than in chloroform. Consequently, it appears that concentration affects the degree of coupling in fluid mixtures. Concentration effects on the heats of transport and the thermal diffusion ratio of chloroform with various alkanes at 30°C and 1 atm are seen in Table 7.6. Table 7.7 shows the experimental heats of transport at various concentrations and at temperatures 298 and 308 K for binary mixtures of toluene (1), chlorobenzene (2), and bromobenzene (3) at 1 atm. The absolute values of heats of transport decrease gradually as the concentrations of the alkane increase. Table 7.7 also contains values of cross coefficients obtained from easily measurable quantities and the thermodynamic factor. Table 7.8 shows the thermal diffusion ratios and thermal diffusion coefficients obtained from Onsager's reciprocal rules for toluene, chlorobenzene, and bromobenzene at 1 atm and at 298 and 308 K. Thermal diffusion or heats of transport may be extremely sensitive to the molecular interactions in solutions (Rowley et al., 1988).

7'.5.3 Efficiency of Coupling Phenomenological stoichiometry z is defined by

ILqq

z--~

/ 1/2

(7.78)

With the definitions of the degree of coupling q and the phenomenological stoichiometry z, Eq. (7.66) can be written as

~7 -

zA + Q~/z 1/z

Q1A / z +

(7.79)

7.5

377

Coupling in fiquid mixtures

Table 7.6. Heats of transports and thermal diffusion ratio ET a of chloroform in binary mixtures with selected alkanes at 30°C and 1 atm b

Solute(l)

x,

- Q ~ (kJ/kg)

KT1 (l/K)

n-Hexane

0.1334 0.3725

60.0 49.5

2.10 2.29

0.5808

46.3

2.40

0.7637

40.0

2.26

0.9257

34.1

2.13

0.1169

56.5

2.30

0.3380 0.5437

45.8 41.1

2.33 2.27

0.7354

35.2

1.99

0.9147

31.9

1.80

0.1040

54.3

2.92

0.3093 0.5110 0.7092 0.9039

45.0 40.3 34.2 25.7

2.82 2.46 1.91 1.32

n-Heptane

n-Octane

3-Methylpentane

2,3-Dimethylpentane

2,2,4-Trimethylpentane

~'K~ -

D~ _ Dxtx2

0.1334

58.3

1.91

0.3725

48.5

2.26

0.5808 0.7637

44.3 40.1

2.43 2.40

0.9257

36.3

2.31

0.1169

55.0

2.34

0.3380

45.0

2.44

0.5437

39.9

2.31

0.7354 0.9147 0.1040

33.5 27.4 53.1

1.92 1.55 2.87

0.3093

42.6

2.69

0.5110

39.7

2.43

0.7092

31.0

1.73

0.9039

23.7

1.22

O('M,:MI RTM3 ( | + |~J i )

~Rowley et al. (1988).

Equation (7.79) shows that as the degree of coupling approaches zero, each flow becomes independent, and we have r / ~ z2A. If q approaches _+ 1, then the two flows are no longer associated with the forces, and the ratio of flows approaches a fixed value: A --. _+z. This case represents a complete coupling. Negative values of rt occur when the differentiation of chemical potential with respect to concentration is negative due to the nonideality of the mixture. The degree of coupling is not a unique characteristic of the system since there may be various ways of describing flows and forces consistent with a given equation for entropy production. For a complete coupling, q = _+ 1 for any choice of flows and forces, and z reaches a unique value. We can define the ratio of dissipations due to heat and mass flows in terms of the reduced force ratio and flow ratio

JqXq ~/A - - ~

(7.80)

jlY,

Equation (7.80) may be called the e f f i c i e n c y o f e n e r g y conversion. When jlX1 shows the input and J'q'Xqthe output power, then diffusion drives the heat flow. Since ~X is zero when either JqIt or Xq is zero, then it must pass through a maximum at an intermediate value. The values of ~/X are often small in regions of physical interest, and the maximum value depends on the degree of coupling only 2 ( T/Jt)ma x =

q

(1 + x/l_ q2 )2

(7.81)

378

7.

Heat and mass transfer

Table 7.7. Experimental heats of transports in binary mixtures of toluene (1), chlorobenzene (2), and bromobenzene (3) at 1 atm a i

j

Wi

T (K)

- Q ~ (kJ/kg)

1 + I'ii

-Loi (X 107) (kg/(m S)) b

1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2

2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3

0.3000 0.3996 0.5000 0.6025 0.3000 0.3000 0.4000 0.5500 0.6000 0.2010 0.7000 0.2000 0.3000 0.5000 0.6000 0.2000 0.3000 0.5000 0.6000 0.7000

298 298 298 298 308 308 308 308 308 308 308 298 298 298 298 308 308 308 308 308

6.21 12.08 10.50 14.98 16.19 15.38 16.59 20.81 24.82 23.35 19.33 14.20 9.41 8.55 9.31 23.71 24.84 12.30 15.08 16.57

1.020 1.020 1.020 1.026 1.025 1.020 1.024 1.026 1.025 1.019 1.018 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0

1.07 2.29 1.99 2.63 3.07 2.92 3.45 4.22 4.80 3.90 3.30 2.34 2.07 2.14 2.24 4.20 5.74 3.37 3.98 3.84

aRowley and Hall (1986).

bLoi = Pvo;gigjwiwj MavRT (1 - F ii)

Table 7.8. Thermal diffusion ratios and thermal diffusion coefficients from Onsager's reciprocal rules for toluene (1), chlorobenzene (2), and bromobenzene (3) at 1 atm a i

j

wi

T(K)

-KTi(1/K)

--DTi(×lolO)(m2/s)

1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2

2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3

0.3000 0.3996 0.5000 0.6025 0.3000 0.3000 0.4000 0.5500 0.6000 0.2010 0.7000 0.2000 0.3000 0.5000 0.6000 0.2000 0.3000 0.5000 0.6000 0.7000

298 298 298 298 308 308 308 308 308 308 308 298 298 298 298 308 308 308 308 308

0.0565 0.120 0.104 0.138 0.141 0.134 0.158 0.193 0.220 0.167 0.149 0.124 0.109 0.114 0.120 0.192 0.263 0.155 0.185 0.180

1.05 2.31 2.06 2.79 3.05 2.90 3.52 4.47 5.14 3.04 3.38 1.68 1.48 1.65 1.74 3.05 4.29 2.62 3.12 3.04

aRowley and Hall (1986).

7.5

7.5.4

379

Coupling in fiquid mixtures

Dissipation Function and Degree of Coupling

By substituting Eqs. (7.6) and (7.7) into the dissipation function of Eq. (7.1), we obtain the three contributions due to heat flow, mass flow, and the coupled transport, respectively

aJlt--Lqq(7 V1 T )

2

+Lll

~

l

0/x1

~

14,'. OW1

Vw 1

1

+2Llq

T

VT

~

1

0~1

~

w2 Ow1

Vw 1

(7.82)

We can also use the transport coefficients, given in Eqs. (7.73) to (7.77), and the degree of coupling in the dissipation function, and obtain 1

- kT T V T

1 [/ ) ] 1

)2

0/£1

+ pD --w, Owl~ (Vw 1

+

/ )E () 1

2pDQ 1 1

1

Z

~VT

0/x1

OWl~ Vw 1

(7.83)

By differentiating the chemical potential in terms F

(OIX---!-l) =MRT(I+Fll)OW1 WlM1M2

(7.84)

and using the definition ofq given in Eq. (7.78), we can express the dissipation function in terms of the degree of coupling q, the thermodynamic factor F, and the transport coefficients of thermal conductivity k and diffusion D

(1)2

* - kT ~ V T

+

p D M R T ( I + F l (I~7W l ) ) 21 WlW2MIM2

+2q

(pDkMR(I+F1)11/2 1 wlWzM1M2

(VT)(Vw 1)

(7.85)

The terms on the right of Eq. (7.85) show the dissipation due to heat flow and mass flow and coupling between the heat and mass flows, respectively.

7.5.5

Coupling in Multicomponent Mixtures

The heat of transport of component i is a measure of the local heat addition or removal required to maintain isothermal conditions as molecular diffusion of component i takes place from a higher chemical potential to a lower one. Since there are only n - 1 independent diffusion flows in an n-component mixture, there are only n - 1 independent heats of transport. Using the forces and flows identified in Eq. (7.1), and the Gibbs-Duhem equation for an n-component system at constant temperature and pressure, we obtain

VT]&n----Z Wk VTtZk

(7.86)

k=l

For isotropic, n-component, nonelectrolyte mixtures without external fields and pressure gradients, the phenomenological Eqs. (7.47) and (7.48) are expressed by

- k=l 1=1 -j; - L; v in r + Z Z Z L;j j=l k=l /=1

w,

+--

Wn

OWl T,P

Owl )r,e Vwt

(i = 1,2,...,n-I)

(7.87)

(7.88)

Coefficients Lqq and L~ are associated with the thermal conductivity k and the mutual diffusivity D, respectively, while the cross coefficients Liq and Lqi define the coupling. Thermal conductivity (k) is related t o Lqqby k = Lqq/T,while the thermal diffusion coefficient is related to Liq by L;q = pDri. Tables 7.9 and 7.10 show the values of the phenomenological cofficients Lij for the ternary mixture of toluene (1)-chlorobenzene (2)-bromobenzene (3) at 298.15 and 308.15 K. In ternary mixtures, there are two independent heats of transport related to the two independent crossphenomenological coefficients Lql and Lq2

380

7.

Heat and mass transfer

Table 7.9. Phenomenological coefficients in Eqs. (7.87) and (7.88) for ternary mixture of toluene (1)-chlorobenzene (2)-bromobenzene (3) at 1 atm a T(K)

wl

w2

Lll (X 109) (kgZ/(m s kJ))

298.15 298.15 298.15 298.15 298.15 298.15 298 15 298 15 308 15 308 15 308 15 308 15 308 15 308 15 308.15 308.15 308.15

0.100 0.200 0.200 0.200 0.362 0.397 0.450 0.600 0.200 0.200 0.200 0.325 0.326 0.397 0.400 0.450 0.600

0.500 0.200 0.400 0.600 0.317 0.303 0.174 0.200 0.200 0.400 0.600 0.175 0.274 0.303 0.100 0.450 0.250

6.85 13.4 13.0 12.5 19.9 20.8 22.4 21.9 14.9 14.4 13.9 21.5 21.1 24.2 23.1 22.5 23.6

L12 (Xl09) (kgZ/(m s kJ))

L21 (Xl09) (kgZ/(m s kJ))

-3.70 -3.23 -6.01 -8.78 -8.82 -9.28 -6.30 -9.51 -3.58 -6.68 -9.77 -5.15 -7.72 -3.76 - 10.3 - 17.2 -13.1

-3.44 -3.05 -5.79 -8.53 -8.76 -9.26 -6.29 9.63 -3.39 -6.43 -9.48 -4.99 -7.60 -3.65 - 10.3 - 17.0 -13.2

L22 (Xl09) (kgZ/(m s kJ)) 23.7 14.1 21.7 21.1 18.8 18.2 12.4 13.5 15.8 24.2 23.5 14.0 19.4 8.61 20.3 22.3 17.4

aplatt et al. (1983).

Table 7.10. Phenomenological coefficients in Eqs. (7.87) and (7.88) for ternary mixture of toluene (1)-chlorobenzene (2)-bromobenzene (3) at 1 atm a T (K)

wl

W2

Lqq (W/m)

Lql (Xl07) (kg/(m s))

Lq2 (X 10 7) (kg/(m s))

298.15 298.15 298.15 298.15 298.15 298.15 298.15 298.15 308.15 308.15 308.15 308.15 308.15 308.15 308.15 308.15 308.15

0.100 0.200 0.200 0.200 0.362 0.397 0.450 0.600 0.200 0.200 0.200 0.325 0.326 0.397 0.400 0.450 0.600

0.500 0.200 0.400 0.600 0.317 0.303 0.174 0.200 0.200 0.400 0.600 0.175 0.274 0.303 0.100 0.450 0.250

74.98 72.39 75.55 79.18 77.94 78.57 77.43 81.96 75.50 78.80 82.58 77.76 79.56 78.16 81.96 86.65 86.65

--0.937 --2.493 -- 1.543 --1.531 --2.620 --2.869 -3.154 -4.606 -3.430 -2.246 - 1.699 -3.037 -3.329 -4.553 -3.660 --4.025 --4.022

--1.510 --0.190 - 1.252 -0.647 -0.253 -0.081 -0.155 2.390 0.861 -- 1.013 -- 1.001 --0.510 -- 1.157 -0.627 -1.910 1.115 --0.024

aplatt et al. (1983).

Lql = l_qlQ1 + LzlQ2

(7.89)

Lq2 = I-q2Q~ + L22Q2

(7.90)

W e relate the t h e r m a l c o n d u c t i v i t y k a n d the t h e r m a l d i f f u s i o n c o e f f i c i e n t as

Lqq = kT

and

Liq = pOTi.

Dyi to

the p h e n o m e n o l o g i c a l c o e f f i c i e n t s

T h e r e f o r e , E q s . ( 7 . 8 7 ) a n d ( 7 . 8 8 ) c a n b e e x p r e s s e d in t e r m s o f t h e t r a n s p o r t c o e f f i c i e n t s

n-1 n-1 - J q = k V r + Z Z P Q ; D k ' Vw' k=l l=l

(7.91)

7.5

381

Coupling in liquid mixtures

Table 7.11. Thermal diffusion ratios from Onsager's reciprocal rules for toluene (1), chlorobenzene (2), and bromobenzene (3) at 1 atm and 35 °a u'l

u'2

-Kll

(I/K)

(I/K)

-KI2

0.300

0.700

0.141

0.300

0.700

0.134

-

0.400

0.600

0.158

-

0.550

0.450

0.193

-

0.600

0.400

0.220

-

0.201

0.799

0.167

-

0.700

0.300

0.149

-

0.0

0.200

-

0.192

0.0

0.300

-

0.263

0.0

0.500

-

0.155

0.0

0.600

-

0.185

0.0

0. 700

-

0.200

0.200

0.145

-0. 040

0.180

0.200

0.400

0.095

0.048

0.200

0.600

0.074

0.049

0.325

0.175

0.128

0.024

0.326

0.274

0.141

0.055

0.400

0.100

0.194

0.029

0.400

0.300

0.156

0.091

0.450

0.450

0.175

- 0.055

0.600

0.250

0.177

0.001

~Rowley and ttall ( 1 9 8 6 ) .

n-1

-Ji - pDTiV l n T + ~ , p D i t V w l

(7.92)

1=1

Equations (7.91) and (7.92) are valid for mixtures at mechanical equilibrium, containing no external body forces, and with negligible surface effects. Also, mass-average velocity is small even under an initially large concentration gradient. For a ternary mixture, Eqs. (17.91 ) and (7.92) become -J

- k V T + p(Q1Dl l + Q2D21 )VWl + p(Q1D12 + QzDzz)VW2 -Jl

-J2

(7.93)

--

PDTI V

T + pDl l VW 1 -t- PD12 g w 2

(7.94)

-

P D T 2 V l n T + pD21Vwl + pD22Vw2

(7.95)

In

where Dit is the diffusion coefficient, and related to the phenomenological coefficients as 1 Dil -- -- Z /9 i =1 k=l

i

wk Lq.i (~i/, + ~ Wn

/{ ) O~k -OWl

(7.96)

T,P

Table 7.11 shows the thermal diffusion ratios obtained from Onsager's reciprocal rules for toluene (1), chlorobenzene (2), and bromobenzene (3) at 1 atm and 35°C. The heats of transport for the ternary mixtures are shown in Tables 7.12 and 7.13. For the ternary mixture of toluene ( 1 )-chlorobenzene (2)-bromobenzene (3), the heats of transport are tabulated at 298 and 308 K. The temperature- and composition-dependent heats of transport values are fitted by the following equations by Platt et al. (1982) with a deviation below 25%: Q1 - (MavWlW2)[-29.687 + 0 . 0 7 0 6 T + 0 . 0 8 7 1 5 T w l - ( 9 8 . 7 0 - 0 . 4 1 8 T ) w 2 + ( 7 8 7 . 3 - 2.765) % w 2 - 21.59w~ - 0.1071Tw~ ]

Q2 - (Mavwl w2 ) [ - 39.370 + 0 . 1 3 0 2 T + 0.01679Tw~ - (120.6 - 0 . 4 1 9 9 T ) w 2 +(993.7-

3 . 3 0 8 T ) w , u' 2 + 7.006Wl 2 - 0.04601Tw22 ]

382

7.

Heat and mass transfer

Tables 7.14 shows the fitted parameters ai for the phenomenological coefficients in Eqs. (7.87) and (7.88) to the following equation: L i k = a o + a 1w 1 nt- a 2 w 2 -k- a 3 w 2 nt- a 4 w 1w 2 nt- a 5 w 2

for the ternary mixture of toluene (1)-chlorobenzene (2)-bromobenzene (3) at 1 atm, and at 298.15 and 308.15 K.

Table 7.12. Heats of transport of ternary mixture of toluene (1)-chlorobenzene (2)-bromobenzene (3) at 1 atm a

T (K) 298 298 298 298 298 298 298 298 308 308 308 308 308 308 308 308 308

W1A

W2A

-Q~ (kJ/kg)

-Q2 (kJ/kg)

0.100 0.200 0.200 0.200 0.362 0.397 0.450 0.600 0.200 0.200 0.200 0.325 0.326 0.397 0.400 0.450 0.600

0.500 0.200 0.400 0.600 0.317 0.303 0.174 0.200 0.200 0.400 0.600 0.175 0.274 0.303 0.100 0.450 0.250

18.2 20.0 16.5 20.1 15.9 18.1 16.8 19.3 23.0 19.8 21.1 16.4 20.9 25.9 21.3 33.9 29.6

9.3 5.9 10.3 11.4 6.1 9.7 9.8 -4.1 -0.2 9.7 13.0 9.7 14.3 22.6 16.6 21.2 22.4

aG. Platt, T. Vongvanich, G. Fowler and R.L. Rowley, J. Chem. Phys., 77 (1982) 2121.

Table 7.13. Heats of transport of ternary mixture of carbon tetrachloride (1)-benzene (2)-acetone (3) at 1 atm and 298.15 Ka W1

W2

Q] (kJ/kg)

Q2 (kJ/kg)

0.1 0.4 0.6 0.8

0.3 0.1 0.1 0.1

17.3 23.9 18.3 28.0

3.6 13.4 18.7 30.2

aS.C. Yi and R.L. Rowley, J. N o n - E q u i l i b . T h e r m o d y n . , 14 (1989).

Table 7.14. Parameters in the fitted equation for phenomenological coefficients for the ternary mixture of toluene (1)-chlorobenzene (2)-bromobenzene (3) a Ljk = ao + al wl + a2 w 2 + a3 w2 4- a4 w~ w2 + a5 w2, where w; is the weight fraction Lik

Lqq

Lll L22

T

(K)

298.15 308.15 298.15 308.15 298.15 308.15

a0

a1

a2

a3

a4

67.0426 69.9331 -3.2126 -3.2166 -3.37541 -3.19829

10.8125 11.2839 101.41 111.835 7.13475 7.1013

11.9207 12.451 93.8366 -104.64 -1.68174 -1.69981

8.81046 9.11118 6.23984 5.10796 108.765 118.181

17.9015 18.7197 -25.0849 -24.6334 -36.1574 -36.1694

aDemirel and Sandler (2002).

a5

5.73382 6.05464 -4.27267 -3.37569 -104.986 -114.52

7.5

7.5.6

383

Coupling in liquid mixtures

Degrees of Coupling in Multicomponent Mixtures

For a ternary mixture, there are two independent degrees of coupling between the heat and mass flows, and are given by Lql qql

Lq2

( LqqLl l

)1/2 '

qq2

(LqqL22

)1/2

(7.97)

These equations show the relationships between the degrees of coupling and the cross coefficients Lqi. The sign of heat of transport is an artifact of numbering the substances since -Q~ - Q2 in a binary mixture of substances 1 and 2. The negative sign in the numbering system used here indicates that the heat flows toward the more concentrated substance. The absolute values of the degree of coupling decrease with increasing temperature for binary mixtures, while the effect of the composition on the degree of coupling is more complex. The degree of coupling decreases gradually with increasing concentration of toluene for toulene (1) and chlorobenzene (2), while it increases with increasing chlorobenzene concentration at 35°C, and it remains almost the same at 25°C for chlorobenzene (1) and bromobenzene (2). The heats of transport have a complex composition dependence, and are sensitive to the composition of the heavy component bromobenzene. For the ternary mixture, the parameters of the fitted equations for the phenomenological coefficients computed from the diffusion coefficients are given in Table 7.14. The fitted values of Lql and Lq2 in kg/(m s) are estimated from Eqs. (7.89) and (7.90) as functions of composition and temperature as follows Lql - 10-7 ( 1 5 . 6 1 - O . 0 5 9 T - O . O 5 0 1 T w l

+ 2.687Tw2)

(7.98)

Lq2 - 10-7wlw3 [ - 5 3 3 . 0 + 2.185T-441.5w 1 + ( 7 1 8 . 1 - 4.025T)w 2 + 1011 w,w 2 + (4096-12.53T) w 2 + 300.3w~ ]

(7.99)

Besides the cross coefficients, the straight coefficients Lqq, L l l, and L22 should also be calculated. The values of Lqi, Lqq, and Lii, are used in Eqs. (7.125) and (7.126) to calculate the degrees of coupling in the ternary mixture. Figure 7.3 shows the degrees of coupling for 0.1 < wl < 0.6 (i = l, 2) and w3 > 0.1 at 25°C and 35°C and ambient pressure; the degree of coupling qq2 changes direction with changing bromobenzene composition. At high concentration of bromobenzene, qq2 is positive and the flows of the components are in the same direction; at lower concentration, however, qq2 becomes negative and hence the components flow in the opposite directions. The cross-phenomenological coefficient Lq2 changes its sign as a function of the mass fraction of the heavy component bromobenzene. This means that the direction of coupling due to the heat transported by the flow of chlorobenzene relative to the mass-average velocity in the toluene-chlorobenzene mixture can be reversed by controlling the mass fraction ofbromobenzene in the mixture. From the standpoint of thermal diffusion, the addition of bromobenzene to the toluene-chlorobenzene mixture can change the magnitude and direction of the separation. The effect of temperature on qql and qq2 is mostly marginal. As the number of components increase, the relative compositions of each component may play an important role in the coupling between two-flow systems.

o,3/ o,ooo°. 0 . 2 /

r

A 0,000

0.0002 qq2

qq2

V

O.l

•

.... j /

?

_o_

0.3

Figure7.3. Change of degree of coupling %1 and qq2 with weight fraction of toluene wl and chlorobenzene w2 at (a) 25°C, (b) 35°C. Reprinted with the permission from Elsevier, Y Demirel and S.i. Sandier, Int. J. Heat Mass Transfer, 43 (2002) 75.

7.

384 7.6

Heat and mass transfer

COUPLED MASS AND ENERGY BALANCES

We consider the general balance equations of mass and energy in the absence of chemical reactions, and electrical, magnetic and viscous effects. The partial differential equations of these general balance equations represent the mathematically and thermodynamically coupled phenomena, which may describe some complex behavior due to interactions among various forces and flows within a system. 7.6.1

Binary Mixtures

For a binary mixture under mechanical equilibrium and without chemical reaction, the general balance equations are P ~ at

= - V'nl

(7.100)

OT --V.q Ot

pCp

(7.101)

where nl and q denote the total flow of species 1 and total heat flow defined by nl = Jl +Pl v

(7.102)

Ju = q = Jq + EJihi

(7.103)

i where v is the mass-average velocity and h i is the partial molar enthalpy of species i, and Ji and Jq' are the diffusion flow and conduction heat flow, respectively. These equations show that the changes in concentrations and temperature are due to diffusion, conduction, and bulk motion. Using the molecular transport only in these differential equations, we have

Owl 3= - V.jl p .--~pep

OT = - V . J q

(7.104)

"

(7.105)

• hi). By using the Fick and Fourier laws in where Jq,, is also called the vector of reduced heat flow (Jq,, = q _ ~ni=1]i one-dimensional uncoupled molecular transport, Eqs. (7.104) and (7.105) become

Owl)

P ~-~)

--

02 Wl

pD11 0),2

OT)

02T pCp --~ = k~oy 2

(7.106)

(7107)

Without the external mass and heat transfer resistances, the initial and boundary conditions with the y-coordinate oriented from the centerline (y = 0) to the surface (y = L) are w(O,+_L) = Wls,

dw 1(t, 0) = 0 t > 0, dy

and

dT(t O) T(0,_L) = Ts, ~ = 0 t> 0 dy

(7.108)

By substituting Eqs. (7.64) and (7.65) into Eqs. (7.104) and (7.105), we have

0Wl] _ V. /-qlQl*VlnT + / - 1 1 -

P Ot )

W2 ~OW1)T,P

(7.109)

Z6

385

Coupled mass and energy balances

•

pCp ~ t - V. LqqVlnT+Lq,Q, 1_~(Ol~l

VW 1

/

(7.110)

W2 ~ OWl T,P Here, the cross coefficients are eliminated by using the heats of transport. These equations may be solved by using appropriate initial and boundary conditions in Eq. (7.105). If we use Eq. (7.101) instead of Eq. (7.105) and a thermal diffusion coefficient for one-dimensional heat and diffusion flows for a binary mixture, we have the following coupled balance equations:

(o.:,)_o(

or +pD-~yow, ])

(7.111)

pC p - ~ - --~y ,k -~v. + --~v. PD Q; oy ) (M1 M2 )2

Ox2

03y )

(7.112)

where Mav is the average molecular weight of the mixture, M~ is the molecular weight of component i, x 1 is the mole fraction of component 1, and hE is the molar excess enthalpy. At steady state and considering molecular transport, coupled mass and energy balances in Eqs. (7.109) and (7.110) become (7.113)

O - V . LllQl VlnT+Lll-w2 _Owl

(7.114)

7.6.2

Multicomponent Mixtures

By substituting Eqs. (7.87) and (7.88) into the following balance equations:

/

p --07

--V'ji , pCp .--07 - - V ' J q

"

we have the thermodynamically and mathematically coupled mass and heat balance equations

P k,--07-} V" PDTiVIn T +

E

Li/ 6/k + -

Wn

j=l k=l l=1 '

OWl ) T,P

0% -;7 -v. krVlnr+Z Z Z L. a:+-/=I k=l

/=I

Wn

Vw/

(i = 1,2,...,n-I)

~ OWl ) T,P

w,

(7.115)

(7.116)

Here, the phenomenological coefficients Lqq and L;,/are related to thermal conductivity k and thermal diffusion coefficient DT by Lqq = kT and Liq = pDvi. For a ternary mixture under mechanical equilibrium and without chemical reaction, mass and heat balance equations are

WI )--

p --~

-V'n 1

(7.117)

Ow2 ) Pk, Ot - - V'n2

(7.118)

386

7.

Heat and mass transfer

OT pCp at - - V . q

(7.119)

where nl and n 2 denote the total flows of species 1 and 2 defined by n i = Ji +Piv. Using the diffusion flows in Eqs. (7.117) and (7.118), we have

Owl ) P ~.-~-- = - V ' J l

w2 )=

P -~

(7.120)

-V'j2

(7.121)

OT pCp Ot - - V . q

(7 122)

By substituting Eqs. (7.87) and (7.88) into Eqs. (7.120) to (7.122), we have

r )

2

P ~. O~ = V" ~-~(PDTiVlnT +

(7.123)

(i=1,2)

pCp OT = 17. kVT + p ~ Z QkDkiVwi + p Z Z hkkDkiVwiVWk E * i=1 k=l

(7.124)

i=1 k=l

where hE is the excess specific enthalpy defined in terms of excess enthalpy he

(2h)

OW2 T,P

If we consider Jq' instead of q in Eq. (7.122) then Eqs. (7.123) and (7.124) become

P ~. Ot = V'(pDT1VlnT + PDllVW1 + PD12Vw2)

OW2 )

p --~

(7.125)

= V'(pDT2V In T + pD21Vw1+ pD22Vw2 )

pCp OT = V.[kVT+p(Q~D11+Q2D21)Vw1+P(Q1D12 +Q2D22)Vw2 ]

-05-)

*

*

Under steady-state conditions, equations above would be

0 = V'(pDT1VlnT + POllVW1+ POl2VW2) 0 = V •(pDy2 V In T + pD21VW1 + PD22Vw2 )

(7.126)

0 = V'[kVT + p(Q1D11 + Q£D21)Vw1 + p(Q~D12 + Q2D22)Vw21 For a ternary mixture, equations above can describe thermodynamically and mathematically coupled mass and energy conservation equations without chemical reaction, and electrical, magnetic and viscous effects. To solve these equations, we need the data on heats of transport, thermal diffusion coefficient, diffusion coefficients and thermal conductivity, and the accuracy of solutions depend on the accuracy of the data.

7. 7

7.7 7.7.1

Separation by thermal diffusion

387

SEPARATION BY THERMAL DIFFUSION Thermal Field-Flow Fractionation

Thermal diffusion plays an important role in achieving the purification of macromolecules, isomeric substances, and isotopic elements. It is also important in models for predicting the composition profiles of oil fields. Thermal field-flow fractionation separates macromolecules and colloids. The separation is based on the Soret effect. A relatively large temperature gradient is applied across a solution flowing through the narrow gap of a concentric tube. The thermal gradient drives solute molecules with larger thermal diffusion ratio from the solutes with smaller thermal diffusion ratio to the concentric tube walls where friction slows down the flow. Therefore, solutes with large Soret coefficients are retained and separated from the solutes with smaller Soret coefficients.

7.7.2

Soret Coefficients for Aqueous Polyethylene Glycol Solutions

The flow of a solute in a nonisothermal solution is caused by the solute concentration gradient and by the temperature gradient J - -DVc--cDrVT

(7.127)

where ,I is the molar flow of the solute in the volume-fixed frame of reference and DT is the thermal diffusion coefficient. When the ordinary diffusion (-DVc) counterbalances the thermal diffusion flow (-CDTVT) at steady state, we have DT _ _ I ( V @ , ) _ KT-- D c

(7.128)

where KT is the thermal diffusion ratio and represents the relative change in solute molarity per degree of temperature. Equations (7.127) and (7.128) are applied to molecular transport in a binary and dilute multicomponent solution of noninteracting solutes. When the values of KT and DT are positive, the solute diffuses from a warmer to a cooler region of a nonisothermal solution, and the solvent simultaneously diffuses toward the warmer region. The separation of polymers due to thermal diffusion may be quite large. For example, the thermal diffusion ratio for dilute solutions of polystyrene in tetrahydrofuran is around 0.6 K-1. This indicates that the change of polystyrene concentration per degree is 60%. The type of solvent and polymer pair may have a considerable effect on both the thermal diffusion ratio and the thermal diffusion coefficient. The segmental model predicts the DT independent of the polymer molecular weight, and is given by

UsDseg

DT = - -

RT 2

(7.129)

where Dsegis the polymer segmental diffusion coefficient, Us is the activation energy for solvent viscous flow, and R is the gas constant. Schimpf and Semenov (2004) developed the following model for the thermal diffusion coefficient for dilute solutions of flexible polymers D~ =

16c~rZAps 27r/Vs

(7.130)

where r/, Vs, and c~ are the viscosity, molar volume, and thermal expansivity of the solvent, respectively, rp is the effective radius of the polymer segment, and Aps is the Hamaker constant for polymer-solvent interactions, which can be estimated from the Hamaker constants of the pure polymer App and pure solvent Ass using Aps = (AppAss) °5. Schimpf and Giddings's (2003) correlation is DT

_

1.19X10

-4

k p _ k s 0.374

U0.623

(7.131)

where kp - ks is the difference of thermal conductivities of the polymer and solvent, in W/(m s), Us is in J/mol, and DT is estimated in cm2/(s K).

388

7.

Heat and mass transfer

Another model based on Emery and Drickamer's theory (Schimpf and Semenov, 2004) yields

D T = ~ s -s

14---

Us -

tiP

~PP

Up

2RT 2

(7.132)

where Us and Up are the activation energies for the liquid polymer and solvent, Mp and Ms are the molar weights for the polymer and solvent, pp and Ps are the densities of the polymer and solvent, and Vp and Vs are the molar volumes of the polymer and solvent. Equation (7.132) is capable of predicting negative DT values, and shows that DT is proportional to the product MpD for a given polymer solution. The molality-based thermal diffusion ratio tr is related to molarity-based thermal diffusion ratio K x = CoVotr+ c~

(7.133)

where Co and V0 are the solvent molarity and molar volume, respectively, and a is the thermal expansivity. Table 7.2a shows the mutual diffusion coefficients and thermodynamic factors and Table 7.2b shows the experimental values of thermal diffusion ratios, thermal diffusivities, and heats of transport for aqueous ethylene glycol and PEG at 25°C.

Example 7.2 Separation by thermal diffusion Consider two vessels connected by a thermally insulated conduit. The system is filled with a solution of hydrogen and nitrogen. The hydrogen mass fraction is Wn2 = 0.2. Estimate the difference between the mass fraction of the components in the two vessels at stationary state when one of the vessels is at 200 K and the other at 370 K. At stationary state JH2 -"--JN2 = 0, and from Eq. (7.19), we find K~

VWH2 = -- --'H~2 VT

(7.134)

T

The integration of this equation for a one-dimensional system between the temperature limits T1 and T2 yields

WH2,II

-- WH 2

,I

=--f TIIKT JT~

T

dr

Since the thermal diffusion ratio is temperature dependent, the integral above can be integrated by assuming a constant thermal diffusion ratio for a reference temperature obtained from Tr =

Tn-----~TIIn Tn - ( 3 7 0 ) ( 2 0 0 ) In 370 = 268 K TI 370- 200 200

TII - TI

TII WH2,II -- WH2,I -- - K T l n ~ ri

Table 7.1 shows the values of the thermal diffusion ratio in the form: which is very close to reference temperature 268 K. Therefore, we have K T = 0.0548 MH:MN2 = 0.0548

KT(MZv/M1M2)=0.0548

at T = 264 K,

(MN:WH2+Mn2WN:)2

Ma2v

MN 2 MH 2

K T = 0.0548 [28(0.2) + 2(0.8)] 2 = 0.0507 28(2) As the thermal diffusion ratio is positive, the hydrogen diffuses into the vessel at a lower temperature. We estimate the difference between the hydrogen mass fractions in two vessels by WH2,II --WH2,I - - - K T In TI-LI= - 0 . 0 5 0 7 1 n

TI

370 = -0.0312

200

7. 7

389

Separation by thermal diffusion

Example 7.3 Total energy flow and phenomenological equations For mixtures, the energy flow contains the conductive flow qc, and the contributions resulting from the interdiffusion qd of various substances and the Dufour effect qD; we therefore express the total energy flow relative to the mass-average velocity q = qc + qd + qD

(7.13 5)

where qc = -kVT, and qd = £ h i J i , and h i is the partial molal enthalpy. When we express the energy flow e with respect to fixed stationary coordinates by disregarding the Dufour effect, the viscous effect, and kinetic energy, we have e = -kgT

+ Z

hiNi

(7.136)

Equation (7.136) is the usual starting point for simultaneous heat and mass transfer. Mass flow is associated with the mechanical driving forces and thermal driving force (7.137)

Ji = Ji, x + Ji, P + Ji, g q- Ji, r

where Ji,x is the ordinary diffusion, c 2 ~_, ai, x

-

pRT

MiMiDi/ i

"

"

,, OGj xj Z k = 1 " k~j

Vx k

(1 4= j , k )

(7.138)

C)Xk T,P,x:

Ji, P is the pressure diffusion,

Ji, e -

c pRT

i

M i M jD!/ x j V j

Jig = - c - - - pRT

i

V/ Mj

1 VP p

(7.139)

--gk

(7.140)

Ji, g is the forced diffusion

'

MiMiDii . .

x j Mj gj . .

k=l

1)

where g is the body force, and Ji, r is the thermal diffusion, J i,r - -DTi V ln T

(7.141)

where Gj and ~ are the partial Gibbs free energy and partial volume, respectively, D# are multicomponent diffusion coefficients, and Dr/- are multicomponent thermal diffusion coefficients; these coefficients show the following properties: Dii - O

and

~ DTi -- O

(7.142)

i

and

•

(MiM/Dil - MiMkDik ) - 0

(7.143)

i=1

Ordinary diffusion depends on the partial Gibbs free energy and the concentration gradient. The pressure diffusion is considerable only for a high-pressure gradient, such as centrifuge separation. The forced diffusion is mainly important in electrolytes and the local electric field strength. Each ionic substance may be under the influence of

390

7.

Heat and mass transfer

a different force. If the external force is gravity, then 'Ji,g vanishes since all gi are the same. Thermal diffusion leads to the separation of mixtures under very steep temperature gradients. For a binary system, ordinary, pressure, forced, and thermal mass flow is expressed as follows:

OlnxA

RT

1) VP - MApSxA ] p pR----------~(gA-g~)+K~'VlnT

MA

(7.144)

where (dGA)T,p= RTd In aA, and KT is the thermal diffusion ratio defined by P DTA K y = C2MAMB DAB

(7.145)

When KT is positive, component A moves to the colder region; otherwise, it moves to the warmer region. Some typical values of thermal diffusion ratios for binary fluid systems are given in Table 7.1.

Example 7.4 Modified Graetz problem with coupled heat and mass flows The Graetz problem originally addressed heat transfer to a pure fluid without the axial conduction with various boundary conditions. However, later the Graetz problem was transformed to describe various heat and mass transfer problems, where mostly heat and mass flows are uncoupled. In drying processes, however, some researchers have considered the thermal diffusion flow of moisture caused by a temperature gradient. Consider a fully developed flow of a Newtonian fluid between parallel plates with a parabolic velocity distribution (Coelho and Telles, 2002) v = Vmax(1-- ~'/2)

(7.146)

where ~ = y/H, and H is the distance from the wall to the center line. The fluid consists of a solvent and n number of solutes. The flow enters the channel with uniform concentrations Cio and uniform temperature To. At the inlet, the confining walls and the fluid are in thermodynamic equilibrium. The wall temperatures vary. Steady-state mass and energy balances are

pv

OCi =-V.ji Ox

(i=l,2,...,n-1)

OT = Ox

pCpv ~

-V'Jq

(7.147)

(7.148)

where Ji and Jq are the mass and heat flows, respectively. Assuming that the local equilibrium holds, we have the following linear phenomenological relations for n - 1 independent concentrations: n-1

Ji = -DsiqVT- ~ DikVCk

(7.149)

k=l n-1 Jq =

-kVT- ~

DDqkVC k

(7.150)

k=l

Here, Ds and Do are the coefficients representing the Soret and Dufour effects, respectively, Dii is the self-diffusion coefficient, and Dik is the diffusion coefficient between components i and k. Equations (7.149) and (7.150) may be nonlinear because of, for example, reference frame differences, an anisotropic medium for heat and mass transfer, and temperature- and concentration-dependent thermal conductivity and diffusion coefficients.

7. 7

391

Separation by thermal diffusion

Substitution of Eqs. (7.149) and (7.150) into mass and heat balance equations yield n-1

E ~ikZ~2@k + ~iq A20

=

(1-- r/2) O@i

(7.151)

OZ

k=l

n-1

Z ~qkA2q)l< nt- A 2 0

:

(l--

00 T/2) --

(7.152)

OZ

k=l

where

T-T o Ci =--, O - AT ' qPi Ci 0

Z=

x DD@Cko , ~)qk = ~ , HPe kAT

_ DsiqATCio , P e = ~H, v m a x DikCpCk° d/~iq~ik : kCio , kCio ol

a--

k

(7.153)

oG

and the dimensionless Laplacian operator in Cartesian coordinates is

A2

02

1

02

= ~ + ~ -3T/2 pe 2 Oz2

(7.154)

The boundary conditions for temperature of the upper (U) and lower (L) plates are

O(Z,--1)=OL(Z), O(Z,+I)--Ou(Z ) lim:_,~ 0L(z) = Ou(z ) = 0

(7.155)

Both plates are held at specified, variable temperatures. For asymmetric wall temperature boundary conditions, the lower plate may be held at the input temperature, while the top wall temperature would be at a specified variable value. The boundary conditions for concentration reflect the asymptotic approach to the fluid composition at the inlet and the permeability properties of both walls lim:_+~ ~ i ( Z ,

1'1) --- 1

ji(z,--

1).n

ji(z,+

1).n - Kui (Ci(z,+ 1)- Cui ) = 0

-

KLi(Ci(z,-1)-CLi ) = 0 (7.156)

Here, Ki are the mass transfer coefficients (permeabilities) for each wall, and CLi and Cui a r e the ambient concentrations of each component i outside the lower and upper walls, respectively. Sometimes, selective membranes may be used as the walls. These membranes may be permeable to selected components only. For example, in a purification process, the membrane would be permeable to one of the solutes only. In a concentration process, both walls can be impermeable to the selected solute. Equations (7.151) and (7.152) describe the thermodynamically and mathematically coupled heat and mass flows at stationary conditions and may be solved with boundary conditions and with some simplifications (Coelho and Telles, 2002).

Example 7.5 Cooling nuclear pellets Spherical nuclear fuel pellets generate heat at a rate per unit volume, q, and being cooled at the boundary by convection heat transfer. For a single pellet on start up, we have

at

-ol

~ -~ Or

-I-~ pCp

392

7.

Heat and mass transfer

subject to initial and boundary conditions

~ = R , - k av~= h(v- vf) Or ~0T = 0 Or T(0,r) = To

r=0, t=0,

where a = k/(pCp), h is the heat transfer coefficient, and Tf is the flowing coolant fluid temperature. Derive an expression to predict transient response T(r, t). Solution: Assume constant physical properties, constant heat source. Temperature with steady T(r) (where t approaches infinity) and unsteady T'(r,t) parts as follows T(r,t) = T(r) + T'(r,t)

(7.157)

For steady state case 0 = a~-~r

q +0%

r2

(7.158)

m

Integrating directly with the symmetry condition

OT

= 0 at r = 0, we obtain

Or

T(r) . . . . q r2 + I k6

(7.159)

From the first boundary condition describing the cooling at the surface we obtain the constant/, and temperature becomes T(r) = Tf + q_fi_R+ qR 2 1 3h 6k

(7.160)

For unsteady state heat transfer, we have

OT' Ot

1 0 (r2 OT' ~ -~-&r Or )

-a

OT'

r=R,

-k

r=0,

OT' ~=0 Or

t = o,

T ' ( o , r) - ro - ~'(r)

Or

(7.161)

= h ( T ' - rf)

m

Equation (7.161) becomes OT' Oz

E2 82 OS

08 J

(7.162)

with r R'

at R2

s=0,

~=0T 0 0e

e = l,

~OT = BiT 0s

Bi-

hR

k

(7.163)

7. 7

393

Separation by thermal diffusion

Apply the separation of variables method:

T'(/3,r) = X(/3)Y(r)

(7.164)

Substituting Eq. (7.164) into Eq. (7.163), we get

l d(/3 2dx ) A2 e2d/3 ~ + X = 0 dY+A2y dr With X =

u(/3)//3,Eq.

(7.165)

0

(7.166)

(7.165) becomes d2bl ~ - + - A2u = 0 d/32

(7.167)

The solutions for u, X, and Y are u(/3 } - ,4 sin(A/3) + B cos(A/3}

X(/3)- A sin(hf_)_) + B cos(h/3_________~)) /3

(7.168)

/3

y - C exp(-A2r) The combined solution becomes"

T ' - ( A sin(A~3)~3+BC°S(A/3~))exp(-A2r)+C+D/3 --/3 T' is finite at the center

dT'/dr- O, and B -

(7.169)

C - D - 0 (r --+ ~c)

T'( /3,r) - A sin(a/3) exp(-A2r) /3

at/3 = 1 - dT'/d/3 = BiT and sin(A) - acos(A ) - BiSin(A). There are infinite numbers of values of A, which satisfy this solution, and the general solution is a superposition of all possible solutions 7~

T'(/3,r)- E A,, sin(An/3~)e x p ( - a , 2 r ) /3

tl-- 1

Finally, from the orthogonality condition and the initial condition, we have Y'(/3, r ) = 2Bi~[~ ~} - rr

{N / AT, - 1)

,,=, ( B i - 1 + cos2 An)

cos a, sin(an/3 ) ~ exp(-A2z) A,

e

where

qR2 N

With specified values of Bi and

N, the temperature

__

profiles may be constructed.

(7.170)

7.

394

7.8

Heat and mass transfer

NONLINEAR APPROACH

Nonlinear systems of transport and chemical kinetics analyzed by the generalized Marcelin-de Donder equations consider two competing forward and backward directions of an elementary process. These equations characterize the flow of matter and energy through the energy barrier and contain potentials F = (-ix~T, 1/T) in exponential forms

Jr = Jrf - J r b = Jr0

exp -

l)fi

-exp

(

bi)]

Pbi~

-~ i

where l)ik is the stoichiometric coefficients that are positive for products and negative for a chemical reaction and satisfies the mass balance EviMi = 0. In Eq. (7.171),/x = 0 for i = 0, which corresponds to the energy transfer, while i = 1, 2, 3,..., n refers to species transfer. For elementary transport processes of heat and mass, stoichiometric coefficients in both directions are equal vfi = Pbi l)i" The term Jr0 denotes the exchange current. The Jr0 and Pik are common for both directions. The ratio of absolute flows is -

-

Jrf Jrb

exp

(

/

R

(7.172)

Based on Eq. (7.171), generalization of the isothermal kinetics of the Marcelin-de Donder yields Jr = Jr0 exp ~ . v fi - ~ f

exp

(7.173)

-I)bi ~ b

The generalized form can represent slow transport processes and nonisothermal effects, and satisfies the detailed balance at thermodynamic equilibrium. The exchange current Jr0 is

Jr0=kfexp

/~i -l"fi~Tf)=kbexp

.i0/

--Pbi--~b

and assures vanishing affinities at equilibrium. For the isothermal chemical kinetic system of S = P, using Eq. (7.173), we have

r: fexp ) (~so

(exp

~s

exp /xe (7.175)

For the cross symmetry property, the partial derivatives of flows with respect to potentials are

OFk

--Lik = - J i o ~

exp -

R (7.176)

OFi

-L~i - --Jko --~ exp -

(7.177)

R

Lik=

L~. Not being confined to linear rate relations, the general symmetry yields Equation (7.171) in terms of the generalized forward and backward potentials of 1-Ifand li b is

= (exp( / exp/ )/ (7.178) where

1--If-

and l i b

--

i=1

f

f

=

-

i=1

b

b

7.8

395

Nonlinear approach

Both the generalized potentials are state functions. If the chemical kinetics represented by the chemical potentials is ignored in Eq. (7.178), heat effects are described by the generalized potentials as follows (Hfx. - l-Ibk )heat -- P0,bk

P0,fk _ PO,bk -- P0,fk _+_ PO,fk (Tf - T b )

Tb

Tf

Tb

(7.179)

TfT b

Based on Eq. (7.171), kinetic equations for mass and heat flows are

Js-Js0

r +Vqs

exp Vss ~

-U

r - e x p Vss ~--~

+Vqs b

Jq-Jqo

exp

VSq - ~

+vqq - - ~ f

-exp

VSq - ~

f

(7.180) b

+Vqq --R-T b

(7.181) b

The Onsager coefficients of Eqs. (7.180) and (7.181) are Lik, eq -- Jio

~

exp --Vki Fk'eqR ) = Ji, eq PkiR

(7.182)

For a symmetric matrix uki,both absolute equilibrium flows Js,eq and Jq, eq m u s t be identical and replaced by a universal constant Jeq. However, if the matrix ukiis not symmetric, which is usual, the equilibrium flows are related to each other so that the Onsager symmetry is achieved PSq Lsq, eq - LqS, eq - J s , eq R

PqS - Jq, eq R

(7.183)

Therefore, the generalized kinetic equations for exchange (transport) processes and chemical reactions are of similar structure. During a diffusion-controlled reaction, matter is transported around an interface, which separates the reactants and the product. The progress of the reaction is strongly determined by the morphology of the interface with a complicated structure, which controls the boundary conditions for the transport problem. The morphological stability of interfaces with nonequilibrium systems may undergo self-organization or pattern formation arising in biology, physics, chemistry, and geology.

Example 7.6 Fokker-Planck equation for Brownian motion in a temperature gradient: short-term behavior of the Brownian particles The following is from Perez-Madrid et al. (1994). By applying the nonequilibrium thermodynamics of internal degrees of freedom for the Brownian motion in a temperature gradient, the FokkerPlanck equation may be obtained. The Brownian gas has an integral degree of freedom, which is the velocity v of a Brownian particle. The probability density for the Brownian particles in velocity-coordinate space is

f ( v , r , t ) = p(v,r,t) m

(7.184)

where r is the position, t is the time, and p is the mass density of the Brownian particles. The mass density of a system consisting of Brownian gas and a heat bath is P -- PH + PB -- PH +

mlf(v,r,t)dv

(7.185)

For a constant PH, the Gibbs equation is

6ps - ~1S p e - - ~m S /x(v,r,t) 6f (v,r,t)dv (7.186)

396

7.

Heat and mass transfer

where 6 represents the total differential of a quantity,/z(v, r, t) is the chemical potential gradient of the Brownian gas component with internal coordinate v, and T(r) is the temperature of the bath at position r. The chemical potential is related to energy (e) and entropy (s) per unit mass

pe- Tps + P = f/.L(v,r,t) p(v,r,t)dV + PH~H

(7.187)

Here,/~H is the chemical potential of the heat bath and P is the hydrostatic pressure. The mass energy and entropy balance equations are needed. The rate of change of probability density with time is

O f = - r . Of-O-~-.J v = - v . O f - O-~.J v Ot Or Ov Or Ov

(7.188)

The conservation of mass for the Brownian particles (B) is obtained from integrating Eq. (7.188) OPB

at

--

--

V"

(7.189)

PBPB

where VB is the average velocity of the Brownian particles obtained from 1, VB(r,t ) = --:-J p(v,r,t)vdv PB

(7.190)

Ope _ -- - V ' J q Ot

(7.191)

The energy conservation is

where Jq is a heat flow in the reference frame in which the heat bath is at rest. The entropy balance equation is derived assuming that the gas is at local equilibrium. We also assume that the suspension of Brownian particles in the heat bath may be a multicomponent ideal solution, and the thermodynamic potential is expressed by

Iz(v,r,t) = kT ln f(v,r,t)+C(v,r,t)

(7.192)

m

where the potential function C(v, r, t) can be a function of the local thermodynamic state variables T(r) and pB(r, t). Then, the chemical potential must satisfy the following conditions: 1. Entropy at constant energy and density of Brownian particles has a maximum, is uniform in velocity space, and equal to the thermodynamic potential of a Brownian ideal gas, so we have

tZB=tZl'eq(V'r't)=kT( -m - lnpB -m - - - 23In (2~rkT)/ m

(7193)

2. The distribution functionf(v) is Maxwellian at local equilibrium, and is defined by

fl, eq(V,r,t) -- exp ( l/zvB2-)~m

kT

(7.194)

Using Eqs. (7.193) and (7.194) in Eq. (7.192), we find C(v) =(1/2)v 2, and Eq. (7.192) becomes /z(v,r,t) = k--T-Tlnf(v,r,t)+ lu2 m 2

(7.195)

From Eq. (7.186), conservation of mass and energy, and the chemical potential, the rate of change of entropy per unit volume is obtained as

Ops_ - - V . j s +~ Ot

(7.196)

7.8

397

Nonlinear approach

where the entropy flow as, the entropy source strength ~, and the modified heat flow (J'q) are obtained from the second law of thermodynamics J~ - a'q - k I f(v, r, t)(ln f ( v , r , t ) T

~=

T2 V T -

0

1)vdv

(7.197)

(7.198)

f

1

(7.199)

J'q - Jq - mI-~v2vf (v,r,t)dv

One of the contributions to the modified heat flow is the motion of the Brownian particles. The entropy source strength is due to heat flow and due to diffusion in velocity space (internal degree of freedom), which is the contribution of the motion of the Brownian particles in the heat bath. Equation (7.197) is based on the identity 0

df k l V.-d~r l n f d v = k -~r . ~ v f ln f dv -

kJ v . -df t'

(7.200)

while Eq. (7.197) results from a partial integration over velocity space by assuming Jv ~ 0 as v --, _+oc.

7.8.1 Phenomenological Equations Since the system is isotropic and assuming locality in velocity space, and using the linear nonequilibrium formulations based on the entropy production relation in Eq. (7.198), we have the linear phenomenological equations

j:,

-j'kL v 7vv

(7.201)

dv

(7.202)

Jv--L 7 r -

The Onsager relations yield Lvq - -Lqv. The heat conduction is expressed by A Lqq/T 2 and the friction coefficients are y - Lvq/fT, ~ - mLv~/fT. With these relations, Eqs. (7.201) and (7.202) become =

(7.203)

VT 1

J, - -~f T -

~

(

k T O- f- )

fv +--

m

(7.204)

0v

Using Eq. (7.204) in Eq. (7.188), the Fokker-Planck equation for the Brownian motion in a heat bath with a temperature gradient is obtained Oj O [ kT Of ) 3' 0 0 T Of _ _ v . _ _ + ~ • fv + ~ B + .f __ at Or -~v ~ m Ov T -~v Or

(7.205)

Using Eq. (7.203) in the energy conservation. Equation (7.191), a differential equation is obtained

02T Ope _ A ~ T _ Ot c)r~

0~(kTOf) 02T my--" fv+~ -- dv- A--T-y Or , m Ov Or

c)

-~r

"PBVB

(7.206)

398

7.

Heat and mass transfer

Equation (7.206) disregards the small contribution to the heat flow arising from the kinetic energy of the Brownian particles. Equation (7.206) is mathematically and thermodynamically coupled and describes specifically the coupled evolutions of the temperature field and the velocity-coordinate probability distribution of the Brownian particles. However, for larger times than the characteristic time/3-1, the system is in the diffusion and thermal diffusion regime.

7.8.2

The Thermal Diffusion Regime

Conservation of momentum may be used to simplify the equation of motion for the Brownian gas for long time behavior: t >>,8 -~. At this regime, the Brownian gas will reach an internal equilibrium with the heat bath. From Eq. (7.188) and the mean velocity in Eq. (7.190), the equation of motion for the mean velocity becomes dVB _ _ V.PB (r, t) + mf Jvdv

(7.207)

PB ( r , t ) = m f f ( v - v a ) ( V - V a ) d v

(7.208)

PB dt where PB is the pressure tensor given by

And the substantial derivation is d/dt = O/Ot+ Va" (O/Or). By substituting Eq. (7.205) into Eq. (7.207), the equation of motion becomes VT dva+ 1 V.Pa(r,t)+y__f__=_[~v a dt PB /

(7.209)

For the Brownian gas at internal equilibrium, the distribution function is approximated by

{ II~B--1/2(V--VB)2]} kT

f ( v , r , t ) ~ f, eq(V,r,t) = exp m

(7.210)

and the pressure tensor is reduced to gas pressure PB: PB = PBU, PB = pBkT/m, where U is the unit tensor. The inertia term on the left side of Eq. (7.209) can be neglected, and we have

JD = PBVB = --DVPB -- DT

VT

(7.211)

where the diffusion coefficient D and the thermal diffusion coefficient (DT) are defined by

kT (Tm) D - m[~' D T = pBD l+-k-~-

(7.212)

With Eqs. (7.194) and (7.210), and mf Jvdv = VPB, the entropy production equation becomes VT VPB (I) -- - J q " - ~ - J D " ~

pT

(7.213)

Using the relation PB = pBkT/m, Eq. (7.213) becomes ~ : _j,q VT (k/m)VpB " U - - JD" PB

(7.214)

k VPB j'q = Jq + P B J D =-A'VT-DTT--~ m PB PB

(7.215)

where the modified heat flux is

399

7.8 Nonlinearapproach

where the heat conduction coefficient A' is defined by A' = A + (k/m)(D2T/DpB). Both Eqs. (7.213) and (7.214) can identify the conjugate forces and flows in ordinary space for which the Onsager relations will hold, and the linear phenomenological equations become J) - - A ' V T -

DTT

k VPB

-- - L q q V T - LqDVPB

m PB

(7.216)

JD -- --DT V__TT_ DIFp B = _LDqlF T T

_

(7.217)

LDDIFpB

Example 7.7 Absorption of ammonia vapor by lithium nitrate--ammonia solution The following modeling is from Venegas et al. (2004). For simultaneous heat and mass transfer during the absorption of ammonia vapor by lithium nitrate-ammonia (A) solution droplets, the ammonia concentration profile in the liquid phase can be estimated from the continuity equation without a source term OPA ~ Ot

(7.218)

V ' ( p d D V x ) + V" (PAV) = 0

where Pd is the density of the dispersed phase and PA is the density of ammonia. As there is no source term, the ammonia production and sink do not exist. In an adiabatic chamber, when the solution is dispersed in droplets, we may assume that the density and diffusion coefficient are constant. Therefore, Eq. (7.218) becomes Ox _ DV 2x + v" Vx = 0 Ot

(7.219)

In spherical coordinates and considering symmetry in & direction, Eq. (7.219) reduces to OZx 2 0 x cot00x Ox Ox + V r ' m + v o Ox ~ + ~ 2~ +O0 Ot Or r O0 - D ~ o r 2 + r- Or

02X)

1

m

)

-7 j

(7.220)

Energy conservation equation for performing the same analysis in spherical coordinates yields

~+v o,

r'~+

,0

cot0

- D / ~ + - ~ + ~ 0o t, or r Or

~+ 00

1

-

- |

(7.221)

77 00

The second and third terms on the left of Eq. (7.221) represent the temperature variation in the radial and angular directions, respectively, due to the convection effect. The terms on the fight correspond to the temperature variation in the radial and angular directions due to conduction. Initial and boundary conditions for the absorption of refrigerant vapor by solution droplets are as follows: Att=0x=x

0T=T o forallr

OT - 0 for t > 0 Or Or Atr=Rx=xeqT=Teq fort>0

Atr=0

At0=0

Ox

and

0=Tr

Ox

00

-

OT

00

-0

(7.222) fort>0

For a low-pressure absorber at constant pressure, it is common to relate the temperature and concentration at saturated equilibrium by a linear function Xeq = -0.00372Teq + 1.58226. Another boundary condition suggests that the heat flux at the droplet surface is proportional to the absorbed mass of refrigerant vapor, and we have (7.223)

400

Z

Heat and mass transfer

where M-Iv is the latent heat. Equation (7.223) is based on the assumption that absorption heat release occurs only at the droplet surface and that no heat is exchanged between the liquid and vapor phases. Relative movement between the droplet and the surrounding fluid can induce circulatory motion inside the droplet and affect the mass and heat transfer. These circulatory velocities depend on the Reynolds number, and start to occur for Reynolds numbers higher than 20. For Re < 1, the stream function q~ in spherical coordinates with the origin on the sphere center is

v ( R2r2-r4 ) q~= ~-2

4(1+3,)

sin 2 0

(7.224)

where v is the droplet velocity and 3' is the ratio of liquid and vapor viscosity. The velocity components are

1 ]2 r

---

Pe

0q~

r 2 sin 0 00

1 VO - -

/ /2/cos0

(7.225)

Pe

0q~

r sin 0 Or

(7.226)

where Pe' is the modified Peclet number defined by Pe

Pe ' =

(7.227)

4(1+~,)

and used to correct the effects of the liquid and vapor viscosity ratio. To simplify the modeling, the following dimensionless variables are introduced _ X--

x

T -

,

T ,

xo

_ r r = ~

TO

Dt ,

(7.228)

7"-

R

R2

Using Eqs. (7.225) and (7.226) in Eqs. (7.220) and (7.221), we have

2 0 Y cot 0 0 g OE - -02y _ _ + _ ~ + ~ ~ OF 2 F OF F 2 O0 Or

~

1~ 02y F 2 002

(7.229)

The energy conservation equation for performing a similar analysis in spherical coordinates yields _--z+---+ F OF

cotoo 102 / ( t

F2

00

+Pe'

(1-F2)cosOOf

F 2 002

0F

+/

/ sn°OT/

272-1 F

(7.230)

O0

The initial and boundary conditions are Forr-0x=T=l

for allr m

Atr=0

OY

aT

- 0 for r > 0 OF 0g Atr=lY=xeqT=Teq forz>0

At0=0

-

and

0=rr

OY 00

-

OT 00

-0

(7.231)

fort>0

7.9

401

Heat and mass transfer in discontinuous system

Another boundary condition suggests that the heat flux at the droplet surface is proportional to the absorbed mass of refrigerant vapor, and we have

Xeq

0"00372T° )Teq -~- 1.58226 -

-

x0

AHvD

Cpc~

{

(7.232)

x0

/

xo

02 _

0.00372To O~=1

aT

(7.233)

O~=1

m

At:0=0

and

0-rr

02

O0

-

aT

-0

fort

(7.234)

O0

The simultaneous solutions of Eqs. (7.229) to (7.234) describe the mass and heat transport processes between refrigerant vapor and solution droplets at constant pressure and Reynolds numbers smaller than 1. Typical lithium nitrateammonia solution properties and ammonia vapor properties are available in Venegas et al. (2004).

7.9

HEAT AND MASS TRANSFER IN DISCONTINUOUS SYSTEM

Transport problems in discontinuous (heterogeneous) system discuss the flows of the substance, heat, and electrical energy between two parts of the same system. These parts or phases are uniform and homogeneous. The two parts make up a closed system, although each individual part is an open system, and a substance can be transported from one part to another. There is no chemical reaction taking place in any part. Each part may contain n number of substances. For example, thermal diffusion in a discontinuous system is usually called thermal osmosis. If the parts are in different states of matter, there will be a natural interface. However, if both parts are in liquid or gas phases, then the parts are separated by a porous wall or a semi-permeable membrane. The postulate of local thermodynamic equilibrium in a discontinuous system is replaced by the requirement that the intensive properties change very slowly in each part, so that the parts are in thermodynamic equilibrium at every instant. The intensive properties are a function of time only, and they are discontinuous at the interface and may change by jumps. In the following sections, thermomechanical effects and thermoelectricity are summarized. Considering the dissipation function below B , - Jil -AT ~ + £ JiArl.ti i 1

(7.235)

where

Artz i -- V;.~P + E

/--1

AW/

Ow i T , P , w / :¢=14'i

We can choose the following thermodynamic forces Xq -

AT

(7.236)

T

X i - Ar/Xi

(7.237)

Based on these thermodynamic forces, the linear phenomenological equations become

Aw/ /=1

/=!

i=1

Ow i )

T,P,w/~w i

(7.238)

z. Heatandmasstransfer

402

Jm=LmqXq+£LmiXi=Lmq--~-+£Lmi i--1

ViAP+ E

i=1

Otzi

j--1

~OWj ) T,P,wj =/=Wi

/

(7.239)

where m = 1,2,...,n. By Onsager's reciprocal rules of Lqi-- Liq and Lmi-- Lira, and the dissipation function is positive

xI~--ZqqX2 +£(Ziq-+-Zqi)XiXq+ £ LmiXmXi >0 i=1

(7.240)

i,m=l

Therefore, the phenomenological coefficients satisfy the following conditions"

Lqq > O, Lii > O, LmmLii-LZi > 0

(i,m = 1,2,...,n)

(7.241)

By introducing heats of transport of the components (when AT = 0)

Zmq = £ LmiQ;

(7.242)

i=1

into the phenomenological equations, we have

Jm = £ Lmi(Xi + Q~.Xq)

(7.243)

i=1

Jq = LqqXq +

Lim)(iQm= Lqq i,m=1

LmqQm Xq + m=l

QmJm

(7.244)

m=l

For an isothermal fluid, the thermodynamic force Xq vanishes, and the heat of transport for component m becomes

(7.245)

Qm = Ji~m =0,AT=0

This represents the heat transported per unit flow of component m without the flow of other components and without the temperature difference. With the explicit thermodynamic forces identified in Eqs. (7.236) and (7.237), the mass flow in Eq. (7.243) becomes

i=1

k=l ~, OWk)

T,P,wj~k

+ aT]

(7.246)

For a binary fluid, Eq. (7.246) yields the flows

(7.247)

(7.248)

where the heats of transport are related to the phenomenological coefficients by taking into account the Onsager rules

Llq = LllQ1 +/-qzQ2

(7.249)

7.9

Heat and mass transfer in discontinuous system

403

L2q - L21Q1 + L22Q2 - LI2Q1 + L22Q2

(7.250)

The heats of transport may be estimated from Eqs. (7.247) and (7.248)

Q1-

LIv

LI2

L2q

L22

, ]I~21Llq Q2 =

(7.251)

L2q

D

where D is the determinant D = L 1 IL22 - LI2L21. Assume that, at the beginning, the two parts are mixed homogeneously, and zXP = 0 and ture difference is the only remaining thermodynamic force, then Eq. (7.240) becomes j"

q-Lqq

Awj =

0. If the tempera-

AT = k A AT

T

(7.252)

(~

The thermal conductivity in this case becomes

k-

• Lqq

(7.253)

A T At stationary state, when the flows of substances vanish (Ji = 0), Eq. (7.244) yields

4 ' - - . Lqq - i=1 LiqQi

T -- ks --6AT

(7.254)

where ks is the thermal conductivity at stationary state, and related to the phenomenological coefficients by

(~

ks ~ ~-~

i=1

Liq Qi

- -~

i,m=l

LimQi m

(7.255)

Since the thermal conductivity is always positive, the heat flow is directed from the warmer part to the colder part. At stationary state, Eq. (7.246) becomes

0 - rAP+ k=l

OWk T,P,w/~x

Awk + Q__LAT T

(7.256)

Using the relations 2~=1Awk = 0 and E"k--t ~Xwk - 1, we can determine the pressure difference and (n - 1) mass fraction differences. 7.9.1

Thermal Effusion

For a binary fluid, we have

O-V1AP+(OtIx---L

Awt + Q__LAT T

(7.257)

O-V~AP+(°tx---L)

Aw~ + Q__LAT T

(7.258)

OWl T,P

"-

OWl T,P

Using the Gibbs-Duhem equation

OWl

T,P

-I-W2( 0~---~2 )OW 1 T,P

404

z.

Heat and mass transfer

Equation (7.258) becomes

O -- V2Ap _ W_____(~l°312'2] Aw 1 + ~ AT T w2 t OW1) r,p

(7.259)

After eliminating the pressure difference from Eqs. (7.259) and (7.257) and using the specific volume of the binary solution V = w l V 1 - w z V 2, we have

VT

(7.260)

t OWl )T,P

This equation describes the change of the difference of the mass fraction of component 1 with respect to a change in temperature at stationary states. This effect is called thermal effusion.

7.9.2

Thermomolecular Pressure

If we eliminate the mass fraction difference AWl from Eqs. (7.259) and (7.257), we obtain 'Sat' - - w2Q~ + w2Q2 AT VT

(7.261)

This equation describes the change of pressure difference with the temperature difference. This pressure difference is called the thermomolecular pressure.

7.9.3

Thermoosmosis

In a special membrane system, we may use a membrane, that is permeable only to component 1 so that J2 = 0. At a stationary state, we also have J1 = 0. If the membrane is movable and the pressure difference is zero, then from Eq. (7.257), we estimate the mass-fraction difference Aw 1

=

_

Ql*

AT

(7.262)

(0[D1/0W1)T,P T This equation describes a difference in the mass fraction arising because of a temperature difference. This phenomenon is called thermoosmosis, which is thermal diffusion in a discontinuous system.

7.9.4

Osmotic Pressure and Temperature

If the membrane is motionless and if there is no temperature difference, then from Eq. (7.257) we obtain the pressure difference kip = - (0~1 / Owl)v'e Aw 1

(7.263)

vl This equation describes the pressure difference because of the mass fraction difference when there is no temperature difference. This is called the osmotic pressure. This effect is reversible because AT = 0, J2 = 0, and at stationary state J1 = 0. Therefore, Eq. (7.244) yields Jq' = 0, and the rate of entropy production is zero. The stationary state under these conditions represents an equilibrium state. Equation (7.263) does not contain heats of transport, which is a characteristic quantity for describing nonequilibrium phenomena. The temperature difference arising from a mass fraction difference is called the osmotic temperature (7.264)

tOWl

Z9

7.9.5

Heat and mass transfer in discontinuous system

405

Thermomechanical Effects: One-Component System

Consider a system with two parts. The parts are separated by a permeable membrane. The two parts may have different temperatures and pressures. Therefore, two generalized flows of substance and heat occur, while the temperature difference and pressure difference are the two thermodynamic forces. In terms of entropy flow Js, the dissipation function is

qt - _ a , . V T - ~ Ji • V/x; i=1

(7.265)

This local equation should be integrated across the membrane to find an expression for a discontinuous system. For a steady state system with a single component, the integrated form is xp = J A T + J1A~I

(7.266)

After identifying the conjugate forces and flows, for small forces of AT and A/x, the heat flow and mass flows may be represented by the following linear phenomenological equations -J1 = L11A/Xl +LI2 A T

(7.267)

- J s = L21Atxl + L22AT

(7.268)

Here, the reciprocal rules hold, and we have L12 - L2~. The introduction of the explicit form of chemical potential for a single component Atx ~ = - S A T

(7.269)

+ VAP

into the phenomenological equations yields

-J1 = LI 1VAP1 + (L12 - L11S) AT

(7.270)

- J s = L21VAP1 +(L22 - L21S) AT

(7.271)

where S and V are the partial molar entropy and the partial molar volume of the component, respectively. One important case would be at steady state when a constant temperature difference is applied to the discontinuous system. Under these conditions, a pressure difference develops across the membrane that leads to dl = 0. The magnitude of this stationary pressure is obtained from Eq. (7.269)

z2LR) -~

-(LI2 - L11S)

-L12

S

VL I1

VL I1

V

.I,-:o

(7.272)

Equation (7.271) shows that the thermoosmotic effect is dependent on two factors. One is proportional to the ratio L12/Lll, and represents a coupling between the flow of the substance and the flow of entropy (heat). The other is proportional to the partial molar entropy S, since the difference in temperature causes a difference in chemical potential, as Eq. (7.269) shows. The coupling phenomenon may be described in terms of the entropy of transfer S*, which is the entropy transferred by a unit flow of substance under conditions of uniform temperature, and defined by

(7.273) Lll

~r=o

Combining of Eqs. (7.272) and (7.273) yields AP j •~

=-(S*-S) I~=0

V

(7.274)

406

7.

Heat and mass transfer

This equation describes the steady state pressure difference induced by a temperature difference. For a system containing a single component, the total heat flow Jq and the reduced heat flow Jq' are TJ~ = Jq - lXlJ 1

(7.275)

Jq' = Jq - H 1 J 1

(7.276)

where H1 is the partial molar enthalpy. Equations (7.275) and (7.276) help to derive the heat of transfer and energy of transfer for the discontinuous system, and relate them to S*. Introducing Eq. (7.275) into Eq. (7.276) yields

J~

1

Jq

(7.277) where (Jq/J1)Ar= o may be called the energy of transfer U* so that we have TS* = U* - tx

(7.278)

Using Eq. (7.276), we find

0-.1)-u--1

(7.279)

and the heat of transfer is (Jq'/J1)~V= 0 =Q*. Therefore, Q* = U* _ H. From the relation H =/z TS and the heat of transfer, we have Q* = T(S* - S). Using the heat of transfer, we may describe the steady state thermoosmotic effect by AP/J - -Q* ST- l=0 VT

(7.280)

A general example by Denbigh may clarify the concept of quantities of transfer. Consider two compartments separated by a barrier that is permeable only to molecules of a relatively high energy. If the molecules with higher energy penetrate from compartment 1 to compartment 2, then the energy of the transported molecules in compartment 2 will be greater than the average energy of the molecules in compartment 1. Compartment 1 will lose energy because of transfer, while compartment 2 will gain energy. In order to maintain a uniform temperature, heat equal to the transferred amount should be added to compartment 1 and removed from compartment 2. For the significance of thermal osmosis, consider the transfer of water in biological systems. An estimated heat of transfer of water across plant cell membranes is approximately 17,000 cal/mol. If this value is used in Eq. (7.280) along with appropriate values of V and T, we estimate that a temperature difference of 0.01 °C would cause a stationary pressure difference of 1.32 atm. However, the maintenance of a temperature difference of 0.01°C leads to a rather large temperature gradient of 10,000°C/cm as the membranes are about 100 A thick. Unless the membrane has a very low thermal conductivity, such a high temperature gradient may be difficult to maintain. In contrast, many chemical reactions taking place within a cell produce or consume heat, and therefore some local temperature gradients may exist and contribute in the transport of substances across biological membranes.

7.10

THERMOELECTRIC EFFECTS

Thermoelectric effects demonstrate the existence of coupling between electrical and thermal phenomena.

7.10.1

Seebeck Effect

In a thermocouple, heating one junction of a bimetallic couple and cooling the other produces electromotive force in the circuit. This observation was originally was made by Seebeck in 1821. Besides the use of thermocouples, transistor electronics and semiconductors are important areas of interest for thermoelectric phenomena. Thermocouples made of semiconductors can develop relatively large electromotive potentials and are used to convert heat into electricity.

7.10

7.10.2

407

Thermoelectric effects

Peltier Heat

In 1834, Peltier observed that the passage of electric current I through a bimetallic circuit caused the absorption of heat at one junction and rejection of heat at the other junction. The heat flow per unit current at constant temperature was called the Peltier heat qpe, and defined by heat added or removed

(7.281)

qPe =

I

With the use of semiconductors, it is possible to achieve rapid heating or cooling by using the Peltier effect. Relatively large temperature differences, as high as 70°C, can be maintained between hot and cold junctions.

7.10.3

Thomson Heat

Figure 7.4 shows the Thomson heat system. Consider a homogeneous wire heated to 373 K with two endpoints cooled to 273 K. If there is no current passing through the wire, the temperatures at points a and b would be the same. However, after passing a current (I), the temperatures at points a and b are different. Therefore, the electric current disturbs the temperature gradient, and the original gradient can be maintained only by adding or removing heat. The heat necessary per unit current and per unit temperature gradient is called the Thomson heat qTh, which is dependent on the nature of the wire. The Thomson heat for a metal wire a is defined by

qTh,.

-

1 6q

(7.282)

I dT

Figure 7.5 shows a composed of a bimetallic couple metal wires "a" and "b" with one junction maintained at temperature T and the other maintained at T+ dT. An electromotive force E causes a current I to pass through the wires. A Peltier heat qpe(T+dT) per unit current will be absorbed at the warm junction and an amount of heat qpe(T) will be given off at the cool junction. To maintain a temperature gradient, Thomson heat (qrh,a)(dT) must be supplied to the metal a, and an amount of heat (qTh,b)(dT) must be removed from b, since the current is in the opposite direction in metal wire b. In a closed work cycle, the electric energy is fully converted to heat. Therefore, the energy balance per unit current by the first law of thermodynamics is (7.283)

dE - qpe(T + d T ) - qpe(T) + qTh,adT-- qXh,bdT If we expand qpe(T + d T ) in a Taylor series and retain the first two terms, we have

qPe ( T + d T ) - qPe (T) + qPe dT dT 273 K CD

Ta a

373 K ~

Tb b

(7.284)

273 K ~ I

Figure 7.4. System for the Thomson heat demonstration. The uniform wire is at 373 K at the middle point. At the end points, temperatures are held at 273 K. After passing a current (/), the temperature at points a and b are measured.

qTh,b

qPe(T) T

Figure 7.5. A bimetallic couple of metals a and b, the two junctions (points 2 and 3) are held at different temperatures Tand T + dT. qPe and qmh show the Peltier and Thomson heats respectively, while E is the electromotive force.

408

7.

Heat and mass transfer

Using this relation in Eq. (7.283), we obtain dE

dqPe + qTh,a -- qTh,b

dT

(7.285)

dT

This is the first equation of Thomson for thermoelectricity. 7.10.4

Flows and Forces in a Bimetallic Circuit

The metallic circuit we consider has only electrons flowing, and the dissipation function in terms of entropy flow as is -- --Js " V r - J e

"V/&e

(7.286)

With the identified conjugate forces and flows, the linear phenomenological equations are - J s = L l l V T + L12V/&e -ae -

L21VT+

(7.287) (7.288)

L22V/&e

Consider the Seebeck effect resulting from two junctions maintained at two different temperatures as shown in Figure 7.5. Assume that points 1 and 4 are at the same temperature To. These points are connected to a potentiometer so that the electromotive force E can be measured with zero current Je = 0. Under these conditions and using the reciprocal rules, Eq. (7.287) yields li7/&e - - L 2 1

VT---S*VT

(7.289)

L22

where S* is the entropy of transfer and represents the entropy transferred per unit flow of electrons at uniform temperature

V/Xe

(7.290)

/Je) VT=0

L22

To estimate the total electromotive force of the circuit in Figure 7.5, Eq. (7.289) must be integrated between points 1 and 4. Assuming one-dimensional gradients, the result is obtained by summing the following integrals /&e2 -- ~&el -- -/&e4 --/&e3 -- --

I;a

* S a dT,

I;

* +dT Sa d T ,

/&e3 -/&e2

__

f T+dT , --a r Sb dT

rT+dT /&e4 - / & e l z AlL e z .IT

(7.291) *

($2 - Sb )dT

Since points 1 and 4 are at the same temperature, and due to electroneutrality in the circuit, there is no concentration gradient for the electrons, and we have A/&e = - F E , where F is the Faraday constant. Therefore, Eq. (7.291) becomes E = - - - 1 FT+dT(s a _ S b ) d T FaT

(7.292)

After differentiating Eq. (7.292) with respect to T, we find d E = _ Sa* - S b dT F

(7.293)

This equation is called the relative thermoelectric power of the metal "a" against "b". Since the transfer of entropy depends on the cross coefficients L12 or L21, this derivation represents coupling between the electrical and thermal phenomena.

7.10

Thermoelectric effects

409

junction a

b 1

J ,/,%,.

Jq, b ,~

qPe Figure 7.6. Schematic of a junction between metals "a" and "b";/is the electric current, and Jq,a and and qPe is the Peltier heat absorbed at the junction.

Jq,b are

heat flows in two metals,

Consider the Peltier effect where a heat flow accompanies a current under isothermal conditions. Figure 7.6 shows the junction between metals "a" and "b" at which the Peltier heat is absorbed. After applying Eq. (7.288) to both metals "a" and "b," we have (7.294)

Jc -- FL22,aVE = FL22,bVE

After introducing the current I, defined by I

(7.295)

- -FJ~

Equation (7.294) may be expressed in the form of Ohm's law - I = F 2L22,.VE

and

- I = F2L22,bVE

(7.296)

The heat flows passing the two metals a r e TJs, a and TJs, b and are not equal, since the Peltier heat must be absorbed at the junction to maintain constant temperature. Therefore, (7.297)

TJs, b - TJs, a - qPe

Using Eq. (7.287) at uniform temperature, we find TJ,,, - - TL~2,, V~c,

TJ~,,b = -TL12,b Vtxc

(7.298)

After dividing Eq. (7.298) by Eq. (7.296) side by side and using Eq. (7.290), we get [ TJ,,, ) I

TL12,a _ L22,a

VT=0

TL12,b _ L22,b

VT=O

TS; F

(7.299)

TS;

F

(7.300)

By inserting these expressions into Eq. (7.297), we have

qPe n

r(Sb --Sa F

(7.301)

Comparing Eq. (7.297) with Eq. (7.293), we find qPe-

T dE dT

(7.302)

This is known as the second equation of Thomson. This relation is based on the Onsager reciprocal rules, and the experimental verification of Eq. (302) would be additional confirmation of Onsager's rules.

z

410

Heat and mass transfer

Differentiation of Eq. (7.302) with respect to temperature yields dE + T d2 E dT dT 2

d q p e __

dT

(7.303)

Using the first equation of Thomson and Eq. (7.293), we have

qTh,a--qTh,a - - - - - -

dT 2

F

dT

dT

(7.304)

This relation represents the Thomson heat with specific entropies of transfer of individual metals "a" and "b"

qTh,a =

T dS a F d T ' qTh,b

T dS b F dT

(7.305)

Therefore, using Eqs. (7.302), through (7.304), the Peltier heat, its variation with temperature, and the Thomson specific heats may be estimated.

PROBLEMS 7.1

Derive equations that describe the temperature profiles for a plane wall, long hollow cylinder, and hollow sphere. Assume constant thermal conductivity, and temperature at the walls as T1 and T2.

7.2

Derive modeling equations for (a) Heat conduction with an electrical heat source. (b) Heat conduction with a nuclear heat source.

7.3

Derive modeling equations for (a) Heat conduction with a viscous heat source. (b) Heat conduction with a chemical heat source.

7.4

(a) Consider a gas mixture of species A and B confined between walls, which are apart from each other with a distance L, and maintained at different temperatures T(L) >> T(0). The walls are impermeable. The gas mixture is free of chemical reactions and convection flows. Derive a general equation to estimate the mole fraction difference of species A, XA(L) -- XA(0), induced by the temperature gradient between the walls. (b) Consider the same gas mixture. The species A is consumed by a fast, irreversible reaction 2A --+ B occurring on the surface of one of the walls at L = 0. The solid walls are impermeable, and there is no heat flow through the walls. Temperature and composition at L = 0 are To and CA0. Estimate the heat flow within the walls because of the Dufour effect.

7.5

Assume that the temperature distribution in a circular rod with internal heat source q may be represented by the following ordinary differential equation (ODE) d2T + 1 dT dr 2

r

dr

+q=0

The nondimensional radius changes over the range 0 < r < 1. The boundary conditions are

r ( r = 1)= 1,

dT )r = 0 -a-; -o

For the heat source values of q = 100 and 1000 J/m 2, plot the temperature profiles.

Problems

411

7.6

Seawater is pumped into a well-mixed tank at a rate of 0.5 m3/h. At the same time, water is evaporating at a rate of 0.02 m3/h. The salty seawater flows out at a rate of 0.5 m3/h. Seawater concentration is 6 g/L. The tank initially contains 1.5 m 3 of the input seawater. Determine: (a) The change of volume as a function of the time. (b) The salt concentration as a function of the time.

7.7

A sphere of ice with a 10 cm radius is initially at 273.15 K. This ice sphere is placed on a sponge, which absorbs the melted water. Determine the diameter of the ice sphere as a function of time. Assume that the heat transfer coefficient is 5 W/(m 2 K). The latent heat of melting is 333 kJ/kg, and the density of ice is approximately 0.917 kg/m 3.

7.8

Consider a tapered conical cooling fin. Assume that the temperature distribution of the cone can be described by the nondimensional differential equation should be d2Y+ dx 2

-aT

0

dx

where a is a nondimensional parameter defined by

a=~

hL/ 1 + ~ 4 / k

2m 2

where h is the heat transfer coefficient, k is the thermal conductivity, L is the length of the cone, and m is the slope of the cone wall. The boundary conditions are x - 0 T = 0 and x = 1 T = 1. Plot the temperature profile for oL- 30. T(x=O)=O

X

T(x=

7.9

Consider a 10-cm long thin rod. Solve the heat conduction equation OT 3t

-k~

02T dx 2

using the boundary conditions ofx - 0 T = 100°C, and x = 10 T = 25°C, and the initial condition of at t = 0 T = 0. The thermal conductivity k - 2.5 W/(cm K).

7.10

Consider a 10-cm long thin rod. Solve the heat conduction equation o30

oq20

37"

dy 2

using the boundary conditions of at v - 0 0 - 100°C, and y - 10 0 = 25°C, and the initial condition of at ~- = 0 0 - 0 (0 = < y - <0). The thermal conductivity k = 2.5 W/(cm K). Use the following nondimensional parameters to nondimensionalize the conduction equation x V=--,'r " L

-

r ( p e p L 2 / k)

ro , 0 - - ~ TL -- To

412

7.11

z

Heat and mass transfer

Consider a circular rod. The radial heat conduction in nondimensional form is described

020

1 00

00

Or 2

r Or

0~"

using the boundary conditions of at r - 0 (00(0, T)/Or) = 0, and r = 1 0 = 1, and the initial condition of at ~"= 0 0 = 0 (0 = < r = <0). Plot the temperature profile at various time intervals. 7.12

Consider a 80-cm long horizontal rod. The fight boundary (x = L = 80cm) is insulated, so that OT/Or = 0, while the left boundary (x = 0) is subject to convection heat loss =0

where h ( = 30 W/(m 2 K)) is the convective heat transfer coefficient, Ta ( = 50°C) is the ambient temperature, and k = 10 W/(m K) is the thermal conductivity. Determine the temperature change along the rod in the following heat conduction equation 00

020

Or

dy 2

where a = 2 × 10 .5 m2/s is the thermal diffusion. 7.13

Consider an insulated composite rod, which is formed of two parts of equal length. The thermal conductivities of parts a and b are ka and kb, for 0 --<x --< 1/2. The nondimensional transient, one-dimensional heat conduction equations over the length x of the rod are expressed by the following equations

OT

02T

Ot

dX 2

OT _ b 02T Ot

dx 2 '

1/2_<x_
where b = ka/kb. The initial and boundary conditions are T(x,0) = 0,

0<x
T ( O , t ) : 1,

T(1,t): 1

x = 1/2 a

b

Determine the temperature distribution as a function of time. 7.14

Assume that the following equation describes the two-dimensional heat conduction in an insulated pipe

OT Ot

m

02T OX2

j[_

02T Oy2

The boundary and initial conditions are: T ( x , O , t ) = O,

T(x,l,t) = 1

413

References

T(O,y,t) - O, T(1,y,t) : T ( x , y , O) = O,

l>x>--0,

1

l>y>--0

Determine the temperature change in time and distance.

REFERENCES R.B. Bird, W.E. Stewart and E.N. Lightfoot, Transport Phenomena, 2nd ed., Wiley, New York (2002). J. Chan, J.J. Popov, S.K. Kehl and D.G. Leaist, J. Sol. Chem., 32 (2003) 197. R.M.L. Coelho and A.S. Telles, Int. J. Heat Mass Transfer, 45 (2002) 3101. Y. Demirel and S.I. Sandler, Int. J. Heat Mass Transjer, 44 (2001) 2439. Y. Demirel and S.I. Sandler, Int. J. Heat Mass Transjer, 45 (2002) 75. D. Kondepudi and I. Prigogine, Modern Thermodynamics, Wiley, New York (1999). A. Perez-Madrid, J.M. Rubi and R Mazur, Physica A, 211 (1994) 231. G. Platt, T. Vongvanich, G. Fowler and R.L. Rowley, J. Chem. Phys., 77 (1982) 2121. G. Platt, T. Vongvanich and R.L. Rowley, J. Non-Equilib. Thermodyn., 8 (1983) 1. R.L. Rowley and M.D. Hall, J Chem. Phys., 85 (1986) 3550. R.L. Rowley, S.C. Yi, V. Gubler and J.M. Stoker, Fluid Phase Equilib., 36 (1987) 219. R.L. Rowley, S.C. Yi, V. Gubler and J. M. Stoker, J. Chem. Eng. Data, 33 (1988) 362. M.E. Schimpf and J.C. Giddings, J. Polym. Sci. Part B, 27 (2003) 1317. M.E. Schimpf and S.N. Semenov, Phys. Rev. E, 70 (2004) 031202. D. Tondeur and E. Kvaalen, Ind. Eng. Chem. Res., 26 (1987) 50. M. Venegas, M. Izquierdo, R Rodriguez and A. Lecuona, lnt. d Heat Mass Transfer, 47 (2004) 2653. S. Wisniewski, B. Staniszewski and R. Szymanik, Thermodynamics of Nonequilibrium Processes, D. Reidel Publishing Company, Dordrecht (1976). S.C. Yi and R.L. Rowley, J. Chem. Phys., 87 (1987) 7214.

REFERENCES FOR FURTHER READING A. Barletta and E. Zanchini, Int J Heat Mass Transjer, 40 (1997) 1007. J.M.O. de Zarate, J.A. Fornes and J.V. Sengers, Phys. Rev. E, 74 (2006) 046305. R.V. Devireddy, D.J. Smith and J.C. Bischof, J Heat I'ransfer, 124 (2002) 365. E Guillemet, J.P. Bardon and C. Rauch, hit. J Heat Mass Transfer, 40 (1997) 4043. J. guettmer-Strathmann, J Chem. Phys., 119 (2003) 2892. J.M. Simon, D. Bedeaux, S. Kjelstrup, J. Xu and E. Johannessen, J. Phvs. Chem. B, 110 (2006) 18528. S.E. Wright, D.S. Scott, J.B. Haddow and M.A. Rosen, Int..1 Eng. Sci., 39 (2001) 1961. J. Xu, S. Kjelstrup, D. Bedeaux, A. Rosjorde and k. Rekvig, ,l Colloid Intelf Sci., 299 (2006) 452. M. Zhang and F. Muller-Plahte, J. Chem. Phys., 125 (2{}{}61 124903.

8 CHEMICAL REACTIONS 8.1

INTRODUCTION

A chemical reaction is an irreversible process that produces entropy. The general criterion of irreversibility is diS > O. Criteria applicable under particular conditions are readily obtained from the Gibbs equation. The changes in thermodynamic potentials for chemical reactions yield the affinity A. All four potentials U, H, A, and G decrease as a chemical reaction proceeds. The rate of reaction, which is the change of the extent of the reaction with time, has the same sign as the affinity. The reaction system is in equilibrium state when the affinity is zero. This chapter, after introducing the equilibrium constant, discusses briefly the rate of entropy production in chemical reactions and coupling aspects of multiple reactions. Enzyme kinetics is also summarized.

8.2

CHEMICAL REACTION EQUILIBRIUM CONSTANT

A single reaction can be represented by

(8.1)

piBi i--1

where B i is species i, n is the number of different species, and u/are the stoichiometric coefficients for species i, which are positive for products and negative for reactants. The amount of Ni of species i at any time is given by N i -

Nio

(8.2)

+ l,, i E

where N/0 is the initial amount of species i, and e is the extent of the reaction, which is an extensive property. Differentiation of this equation is dN

(8.3)

i -- pid¢ "

For a single reaction in a closed system, the differential of the Gibbs energy is

dG - - fdt + VdP + { £ vi txi l

(8.4)

where/x; is the chemical potential of species i. At constant T and P, we have the Gibbs energy of reaction AG r

--

T,P

Pi[Li

= AG r

(8.5)

i=1

At chemical equilibrium, the Gibbs energy of the system is minimum, and from Eq. (8.5) we have

/

Pi[d,i,eq-- 0

--

T,P

i=1

(8.6)

416

8.

Chemical reactions

The chemical potential in terms of activity a is o

Id'i "-" [£i + R T In a i = t x ° + R T In c i')/i

(8.7)

where/x ° is the standard chemical potential (/xi =/z ° at a = 1), and activity is defined by ai = ciYi, where Yi is the activity coefficient of species i, which is a strong function of concentration. Assuming that the activity coefficient is either unity or close to unity, the chemical potential becomes ~ i -- t"L° nt- R T

In c i

(8.8)

Substituting Eq. (8.8) into Eq. (8.6), we have 1/

//

1,itx ° = - R T ~ _ ~ 1,i i=1 i=1

In Ci,eq

= -RT

In I-I (Ci,eq)ui

(8.9)

i=1

Using Eqs. (8.5) and (8.9), we relate the Gibbs energy of reaction and equilibrium constant K zXG ° = - R T

(8.10)

ln K

and

(8.11)

K = r I (¢i,eq ) vi i=1

Example 8.1 Equffibrium constant of a reaction Consider a homogeneous reaction between species B and P: B = P. An ideal solution of 1 L containing 1 mol of B initially has the concentration 1 M. Define the equilibrium constant K. The Gibbs energy of the reaction is G = N B (tx~3 + R T

ln[B] + Np (/x~, + RT[P])

Using the extent of reaction E, we find NB = 1 - e and Np = E. With these amounts, the relation above becomes G = ( 1 - e)/x~3 + e/X~ + RT[(1- E)ln(1- E) + e In E] At equilibrium (OG/Oe)p = 0, and the equilibrium extent of reaction is 8eq then we find

o

o

tXp - tXB = - R T

In

{eq / ...... 1 - Eeq

RT

In K

Example 8.2 Equilibrium compositions Consider the following chemical reaction 1,1A 1 + 1,2A 2 ~ 1,3A 3 + 1,4A 4

Determine the equilibrium compositions in terms of the extent of the reaction. The reaction coordinate 8 is: dE = dni/vi dE=

d/'tl _ d//2 _ d//3 _ d//4 1"1

d n i = 1,id8

1'2

dl/l i = 1,i t

1,'3

de

1,'4

n i = nio + 1"iE

8.2

Chemical reaction equilibrium constant

417

n = E ni = E nio + e , E Pi = no + e,P i

i

i ni _ nio + Pie,

Yi-

n

n o + t.,e,

For a multireaction, we have nio + ~ , j l"i,e,.j Yi z

n o + ~.,,j P j e , j

For the following reaction, we have 2 mol

CH4, 1 mol H20, 1 mol CO, and 4 mol H2 CH 4 + H 2 0 ~ C O + 3H 2

The summations of stoichiometric coefficients and the initial compositions are" u= ~u~ =-1-1+1+3=

2

i

no = ~ n i 0 = 2 + 1 + 1 + 4 = 8 i

We have the compositions in terms of the extent of reaction

YCtt4

8.2.1

2--e" 8 + 2e

1--e, 8 + 2e" YH2

YH,O

4+3e, 8 + 2e" YCO

l+e, 8+2e,

Changes of Thermodynamic Properties in Chemical Reactions

From Eqs. (8.4) and (8.5), the Gibbs energy becomes d G - - S d T + VdP + ~ @de,

(8.12)

The reaction entropy and volume change by (OS)

Oe

=

,e

=_AS r

OAG r

/

]r,~

= -AVr

OP

OS

(8.13)

(8.14)

(8.15)

where ASr is the reaction entropy and A Vr is the reaction volume. The Gibbs energy of reaction in terms of the enthalpy of reaction M-/r is (8.16)

AG r - AH r - TAS r

The enthalpy of reaction AHr can be calculated from the Gibbs-Helmholtz equation

•

-T 2 (O(AGr/T) OT

)

= p

~I

r

(8.17)

418

8.

Chemical reactions

The standard enthalpy of reaction &/_/oand the standard Gibbs energy of reaction AG ° can also be obtained from the standard enthalpy of formation of species i ~/~i and the standard Gibbs energy of formation of species i AG'~i by the following equations

A H ° -- ~ P i A H ~ i

and

AG ° - ~

i=1

pima~i

(8.18)

i=1

From the definition H = G + TS, we have ~ H r -~ A G r -+"T A S r

(8.19)

From Eqs. (8.10) and (8.17), we can relate the equilibrium constant with the standard enthalpy of reaction 0 In K ) _ M-/° OT p RT 2

(8.20)

This is the van't Hoff equation. If M-/r° is assumed to be independent of temperature, then the integration of this equation between T1 and T2 yields

1!

(8.21)

The derivative of the enthalpy of reaction with respect to temperature yields the heat capacity of the reaction at constant pressure (8.22)

ACpr -- ~ Pifpi i=1

If Cpi does not change significantly in the experimental temperature range, the enthalpy of the reaction will change linearly with T, while entropy changes logarithmically AH o = M-/°(298.15K) + A C por ( T - 2 9 8 . 1 5 )

(8.23)

A S ° ( T ) - AS°(298.15 K) + ACpln ~ 298.15K

(8.24)

Substituting these relations in the following relation AG ° = M-/° - TAS ° = - R T In K yields lnK =-AH°(298"15K) RT

+ '5S°(298"15 K) R

o(

ACpr 1R

298"15K~- In ~ T T 298.15K

(8.25)

Using this equation, we can estimate the effect of temperature on K.

Example 8.3 Temperature effect on equilibrium conversion Consider the elementary reversible reaction B ~- P with no initial product P, while the initial concentration of B is B0. The standard Gibbs energy and standard enthalpy of the reaction are AG ° (298.15K) - - 1 4 . 1 kJ/mol and AH ° (298.15K) -83.6kJ/mol. Assume that the specific heats of solutions are equal to that of water. Estimate the equilibrium conversion of B between 25°C and 120°C. For an ideal solution K = Cp,eq/CB,eq, AGr = - R T in K, and we have 14100 / K(298.15K) = exp 8.314(298.15) = 296.2

8.3

1

'

-

0.8

!

~

~

i

~

-~ . . . . .

I

!

i

0.7

----/

OOo,

:5" W

419

The principle of detailed balance

......

.....

.................

±

.......... i

04

.

.

.

.

.

.

.

ii

.

i

0.2

3()0 310

320

Figure 8.1.

330

340

350 T,K

360

370

380

390

400

Change of conversion with temperature.

Since the specific heats of the reactant and product are the same, AC~, = 0, and

AH°(T)

= AH°(298.15 K ) +

From Eq. (8.21) with Ta = 298.15 K and

lnK =

2~H°(Tz - T l)

RT~T2 lnK

=

T2 =

AC°p(T -

298.15) = - 8 3 . 6 kJ/mol

393.15 K, we find the value of K in terms of T2

+ In K(298.15) -

AH°(T2-298"15) RT2 (298.15)

AH°r (T2 - 298.15)

RT2(298.15)

10055.3 + ln(298.15) = ~ T2

+ ln(298.15)

28.0

On the other hand, the equilibrium constant in terms of conversion becomes

Cp,eq _ CB0 -- CB,eq _

CBoXB,eq

K

CB,eq where the conversion is X B , e q

-- (CB0 --

CB,eq

CB0 (1 -- XB,eq )

CB,eq)/CBo.From

_m

XB,eq 1 - XB,eq

the above equation, we have

XB,eq K

_.._

1 - XB,eq

or K

XB,eq (T2) -

I+K

_

exp[(10055.3/T2)-28.0 ] 1 - exp[(10055.3/T2) - 28.0]

Figure 8.1 shows the change of conversion with the temperature. In order to secure 80% conversion, the temperature should be below 340 K.

8.3

THE

PRINCIPLE

OF

DETAILED

BALANCE

In chemical kinetics, the reaction rates are proportional to concentrations or to some power of the concentrations. Phenomenological equations, however, require that the reaction velocities are proportional to the thermodynamic force or affinity. Affinity, in turn, is proportional to the logarithms of concentrations. Consider a monomolecular

420

8.

Chemical reactions

reaction between a reactant R and a product P in the vicinity of global equilibrium R = R The rates of reaction by the kinetic theory are d C R -- - k f C R nt- kb C P

(8.26)

dt

d C p _ NfCR _ kbCp

(8.27)

dt

where kf and kb are the forward and backward reaction rate constants. Chemical kinetics distinguishes two reaction flows with two constants. However, there is only one macroscopic flow of reaction from the left to the right, and is given by Jr =

dCR - dCp

dt

(8.28)

dt

At equilibrium, the reaction flow vanishes and we have kfCR,eq = k bcP,eq or K-

kb - CR'eq kf Cp,eq

(8.29)

where CR,eq and Cp,eq are the equilibrium concentrations of R and P, while K is the equilibrium constant of the reaction. If we define the following deviations, xi, of the concentrations from their equilibrium values when the reaction system is close to equilibrium, we have ~R --" CR -- CR,eq, ~P -- Cp - Cp,eq

(8.30)

with

~R

~P

CR,eq

~P,eq

<<1

If the reaction proceeds in a closed system where there is no exchange of matter with the outside, we have C R nt- Cp = CR,eq nt- Cp,eq and 6R + 6p = 0

(8.31)

Introducing Eq. (8.30) into Eq. (8.26) yields (CR,eq -k-•R ) -- kb(Cp,eq nt-ap)

J r = Td C R --- k f

(8.32)

Using Eq. (8.29), we obtain Jr -- dCR -- 6R (kf + kb ) = kf6 R (1 + K) dt

(8.33)

The same chemical reaction system may be considered by the thermodynamic approach, in which the affinity 2 A = -~

i=1

Pil.Li -- iC~R -- I&p

(8.34)

8.3

421

The principle of detailed balance

is the thermodynamic driving force for the reaction. By differentiating thermodynamic potentials with respect to the extent of the reaction, we obtain the affinity

At chemical equilibrium, we have A -- 0 - ~R,eq --/'ZP,eq and ]J'R,eq -- ~P,eq

(8.35)

We also have the reaction flow proportional to the force (8.36)

J~ - L~A - L~ (/.zR - ~p ) For ideal solutions, at constant temperature and pressure, we have

(8.37)

#i - tx° + R'T In C i

Where R' is the gas constant. Introducing Eqs. (8.37) and (8.30) into Eq. (8.34) and rearranging terms, we get

A = Ix~ +

R'T In CR,eq nt- R'T In[ 1 + CR,eq

- - ]J,p - -

R'T In Cp,eq - R'T In 1 + Cp,eq

(8.38)

From Eq. (8.35), we find

A-R'T

In 1+

- I n 1+ CR,eq

(8.39) Cp,eq

If 6R/CR,eq << 1, ~p/Cp, eq << 1 the logarithms can be expanded. If we retain the first term only, we have

A-R'T(

6R • C R ~q

(~P

]-R'T

Cp,eq )

6R ( I + K )

(8.40)

CR,eq

By introducing Eq. (8.40) into Eq. (8.36), we find J,. -

R'TL 6R (1 + K)

(8.41)

CR,eq

Comparing the equation above with Eq. (8.33) yields the phenomenological coefficient kfCR,eq

L- ~

R'T

(8.42)

This equation shows that the coefficient L depends on equilibrium composition. However, the coefficient L should be independent of the flows and forces it relates.

8.3.1

Microscopic Reversibility

As pointed out by Prigogine, the linear phenomenological equations as given in Eq. (8.36) are valid only when A/RT << 1, which requires that the reaction be quite close to equilibrium. In most reactions, A is much greater than RT. Consider the following cyclic reaction R< ~ >P 2

P<

:,B

B< 3 :~R

422

8.

Chemical reactions

Using kinetic coefficients, the reaction scheme is characterized by Jrl = klfCR - klbCp

(8.43)

Jr2 = k2fCp - N2bCB Jr3 = N3fCB - N3bCR

There are only two independent reaction velocities. The affinities are A1 = ~ R - - / g ' P ,

A2 =]AP--/AB,

(8.44)

A3 =]AB--IAR

These are interrelated by A3 = - A 1 - A z

(8.45)

From the dissipation function, we have ~Ir = T ( I ) = JrlA1 + Jr2A2 + Jr3A3 = ( J r l -

Jr3)A1 + ( J r 2 -

Jr3)A2

(8.46)

Based on this equation, the phenomenological equations become Jrl - Jr3 = LllA1 + L12A2

(8.47)

Jr2 - Jr3 = L21A1 + L22A2

(8.48)

At equilibrium, the affinities vanish (A 1 = 0, A 2 -- 0). Therefore, Jr1 - Jr3 = 0 and Jr2 -- Jr3 = 0 and the thermodynamic

equilibrium does not require that all the reaction velocities vanish; they all become equal. Under equilibrium conditions, then, the reaction system may circulate indefinitely without producing entropy and without violating any of the thermodynamic laws. However, according to the principle of detailed balance, the individual reaction velocities for every reaction should also vanish, as well as the independent flows (velocities). This concept is closely related to the principle of microscopic reversibility, which states that under equilibrium, any molecular process and the reverse of that process take place, on average, at the same rate. For the reaction system, the principle of detailed balance requires that Jrl -- Jr2 = Jr3 = 0

(8.49)

This condition requires that the coefficients L12 and L21 in Eqs. (8.47) and (8.48) equal. After applying Eq. (8.49) to Eq. (8.43), we have at equilibrium kl f CR, eq = kl b Cp,eq k2fCp,eq = k2bCB,eq

(8.5o)

k3fCB,eq -- k3bCR,eq

Using the deviations

6iin Eq. (8.43) and with the aid of Eq. (8.50), we have Jrl = klf~R - klb~P Jr2 = k2f6P - k2b~B

(8.51)

Jr3 = k3f~B -- k3b~R

From Eq. (8.40), the affinity may be represented in terms of the deviations from equilibrium

AI=RT{ 6R

6p j CR,eq Cp,eq

(8.52)

Using the principle of detailed balance given in Eq. (8.50), Eq. (8.52) becomes

RT

N1 -- ~ ( k l f ~ R NlfCR,eq

-- klb~ P)

(8.53)

8.4

423

Dissipation for chemical reactions

Combining Eqs. (8.51) and (8.53) yields

Jrl --

klfCR,eq A1

(8.54)

RT

By applying a similar analysis to A2 and A3, we obtain

Jr2 = k2fCp'eq A2

(8.55)

RT

Jr3 = - k3fCB'e~q (AI -+- -42 )

(8.56)

RT

Equations (8.47) and (8.48) become

klfCR,eq -Jr-k3fCB,eq A1 + k3fCB,eq A2

Jrl - J r 3 -

RT

(8.57)

RT

J r 2 - - J r 3 = k3fCB'eq A1 + k)..fCp,eq -+-ka__fCB,eq A2 RT

(8.58)

RT

Therefore, the principle of detailed balance leads to the phenomenological coefficients as explicit functions of the constant rate coefficients, and the equilibrium concentrations of R, P, and B

L I 1 = klfCR'eq -+-k3fCB'eq

(8.59)

RT

LI2-L21 =

k3fCB'eq

(8.60)

~ RT

L 22 = k2fCp'eq + k3fCB'eq

(8.61)

RT

This shows that L12 -- L21- This simple analysis suggests that a matrix obeying the general law of symmetric crosscoefficients could be derived from the principle of microscopic reversibility by applying the methods of statistical mechanics. 8.4

DISSIPATION FOR CHEMICAL REACTIONS

For a balanced reaction, the balance equations without kinetic and mechanical potential energies yield

dN

- piJ,-,

dU

dt

dS

- it + W,

dt

dt

-

4

+ dO

(8.62)

T

where Jr is the molar rate of reaction, and u; is the stoichiometric coefficient, which is positive for the product and negative for the reactant. The Gibbs function is -

dt

-

dt

+ I;.,

dt

i

k, dt )

This equation is valid for a homogeneous state of equilibrium. For slow reactions, it is reasonable to assume that the chemical process is proceeding through states that are in thermomechanical equilibrium at all times, although chemical equilibrium is not established. This assumption is supported by the fact that, for example, for a gaseous mixture, collisions not causing reactions are far more frequent than collisions that cause reactions. At a specified

424

8.

Chemical reactions

composition, the molecular distribution functions describing temperature and pressure are effectively the same as those at equilibrium. By combining Eqs. (8.62) and (8.63), and assuming that work results from reversible volume change, we have

(;t r where A denotes the instantaneous nonequilibrium chemical affinity of the reaction

(8.65)

A= --E Pi],Zi i

From the concept of the extent of reaction e, the instantaneous reaction rate is Jr = de/dt, and we have

j and

T,P

Equation (8.67) shows that the Gibbs function decreases as the reaction proceeds. The affinity is related to the Gibbs function k

1-I c~; = exp[(A- AG~)/RT]

(8.68)

i

where c i is the nonequilibrium concentration. At chemical equilibrium, the affinity vanishes. On the other hand, the change in the Gibbs energy of the ith reaction at temperature T is given by

AGr,i = Z Pijl~j

=

AHr,i - TASr,i

(8.69)

The chemical potentials in this equation refer to reactants at their original state before reaction and products at their final state after completion of the reaction. Therefore, the Gibbs function compares the two equilibrium states of reactants and products; in between these states, the reaction may proceed at constant T and P in a steady or unsteady state. In a nonviscous fluid, where I chemical reactions take place, the dissipation functions are limited to the scalar chemical affinities A and chemical rates Jr l

• = T ¢ = Z A j J >-0

(8.70)

j=l

The phenomenological equations are t

t

Jrj = E LJk Ak = Z LJk k=l

k=l

vik Mi tzi

(8.71)

i=1

and the Onsager reciprocal relations link the coefficients Ljk- Lkj. The dissipation function is a quadratic function of the chemical affinities l

- T ~ = __E LjkAiAk~ >- 0 j,k=l

(8.72)

8.5

425

Reaction velocity (flow)

The positive values of • imply the following inequalities

Lkx. > O, Lx.kLi i -L~./> 0

8.5

(8.73)

REACTION VELOCITY (FLOW)

The velocity of a chemical reaction is proportional to the product of the concentration of the species. For a reaction kf

A< the reaction rates for the forward Jrf and backward

/%

>B

(8.74)

Jrb directions

are

P-f

J~f - kr l-I ci,',: i

(8.75)

Pi.b

Jrb -- kb U Ci,b i

(8.76)

For a reaction j, the rate of reaction is given by

,i]

U i Ci Jr/ -- Jrf - Jrb -- Jrf i 1- -----~~

(8.77)

where K c - kf/kb is a constant at equilibrium conditions. For an ideal gas, the chemical affinity is given by iv[ A~

-

- - ~ uii tz~ /=-[

H i Ci

-

-RT

I n -

(8.78)

KC

where

HiCi

i

( A/ )

- exp -

(8.79)

so that the reaction rate becomes

.4/

(8.80)

If we expand the expression in brackets, and consider the case of a near-equilibrium state, which may be specified by the inequality IA//RT << 1, then we have a linear relationship between the reaction rate and the chemical affinity

.it~ _ ']rf,eq A~ - L,./ A RT • T

(8.81)

Therefore, the phenomenological coefficient is equal to

Lri

Jrt;eq R

(8.82)

426

8.

Chemical reactions

Very large affinity values may cause instability, and lead to new states that are no larger homogeneous in space. This causes a discontinuous decrease of entropy, and has important consequences in oscillating chemical reactions. Such reactions are far from equilibrium, and present undamped fluctuation on a macroscopic scale. Oscillations around a stationary state are possible as long as the total entropy production is positive. Some structures can only originate in a dissipative (nonequilibrium) medium and be maintained by a continuous supply of energy and matter. Such dissipative structures exist only within narrow limits due to the delicate balance between reaction rates and diffusion. If one of these factors is changed, then the balance is affected and the whole organized structure collapses. In a system of two simultaneous reactions, thermodynamic coupling allows one of the reactions to progress in a direction contrary to that imposed by its own affinity, provided that the total dissipation is positive.

Example 8.4 Affinity and heat of reaction Describe the relationship between the affinity and heat of a chemical reaction. The affinity is defined as a partial derivative of the Gibbs free energy

OG

883,

Using the Gibbs free energy G = H - TS in Eq. (8.83), we find

OH

(8.84)

where the first term (OH/Oe)p,T is the heat of reaction at constant pressure and temperature. If the second term (OS/Oe)p,v, which is the change of entropy with the extent of the reaction, is neglected, then the entropy production becomes proportional to the heat of reaction

dt

r Jr ~ - -T

P,T

(AHr)p,r r Jr

(8.85)

provided that the following condition is satisfied orl(AHr)p, r I>> T ~

T,P

(8.86)

The condition in Eq. (8.86) may not be easily satisfied. For simultaneous reactions, we have

diS dt

~

(AH~)p'T '

T

(8.87) Jr

where (zSJ-Pr)p,T is the heat of reaction for reaction i. When the condition in Eq. (8.86) is satisfied, the rate of entropy production can be measured by the heat of reaction of a system having complex simultaneous and coupled reactions, such as biological systems.

8.6

MULTIPLE CHEMICAL REACTIONS

Consider a simple system 1 consisting of n components and subject to r chemical reaction mechanisms, and having specified values U of energy, V of volume, and values of N1, N2,..., Nr of the amounts of components that are obtained from given values Nla, N2a,..., Nat. Such a system permits a very large number of states. But the second law requires that among all these states, the chemical equilibrium state is the only stable equilibrium state. In this state, we have

r,2., Nio - Nia + ),, u~j)ejO ( i - 1,2,...,r) j=l

(8.88)

8.6

427

Multiple chemical reactions

The values of U, V, iV,, and u determine uniquely the values of the entropy S, the reaction coordinate e, and each Nio

S-S(U,V,N,,u)

(8.89)

e/o - eio(U,V,N,,u)

(8.90)

N~o = N i o ( U . V . N . . v )

(8.91)

In an isolated system 1 with n components and r chemical reaction mechanisms, the system passes through a sequence of nonequilibrium states, and entropy is produced until the system reaches chemical equilibrium at which point the rate of change of each reaction coordinate is zero. The evaluation of entropy production as a function of time from a state that is not stable to a stable equilibrium state may not be possible. This evaluation requires the solution of a general equation of motion. However, we can approximate the rate of entropy production in terms of the r affinities of system 2, a surrogate for 1, and the rates of change of the reaction coordinates of the r chemical reactions. The surrogate system 2 is a simple system consisting of the same r reactions as system 1, but will have all the chemical reaction mechanisms inhibited. The following relations are from Gyftopoulos and Beretta (1991). At time t, the amounts of n species satisfy

N , - N;. +

£~,(J)e/

(i-1,2,.

. . ,

n)

(8.92)

/=1

The entropy of system 1 is a function of time. If Seq is the entropy at stable equilibrium, the values of ~(t) are smaller than the value Seq. The rate of entropy production • in the isolated system 1 is given by

(J)/x) j=l i=1

(8.93)

Teq

/=1

Since each of the functions • and ~. depends solely on the reaction coordinate ej, we have

c)(I)eq--_~ l"i(k)/'Li,eq 0ek

i

(8.94)

Teq

and c]2(IOeq __

02¢I)eq

Oe/Oe k

OekOe /

(8.95)

or equivalently (8.96) 0~1

L~,I',N a ,v,f

U,/',N a ,v,e

This equation indicates that the r x r matrix with elements (OXk/Oe l)u,~,,Na,,,,~ is symmetric for the matrix e. From the Jacobian properties, the matrix with elements (Oe k/OXz)u,~.Na,~, x is symmetric also for both zero and nonzero values of X, and we have

(8.97)

Equations (8.96) and (8.97) are among the many Maxwell relations that can be established for the stable equilibrium states of a multicomponent system 2. Assuming that the reacting system 1 belongs at each instant of time to the family of states 1e, and the rate of change of each reaction coordinate is a function of the element of vector X, we have J-

J(X)

(8.98)

428

8.

Chemical reactions

and J(O) = 0

(8.99)

Therefore, at equilibrium Xi = 0. Equation (8.97) shows that during the time evolution, the surrogate system 2 proceeds through stable equilibrium states, and system 1 proceeds through states X~. This condition is stated without any reference to microscopic reversibility, and applies for all values of X, which represent both the chemical equilibrium and nonequilibrium states. We can expand each of the r reactions into a Taylor series around the chemical equilibrium state at which X = 0 J = L. X + higher order terms

(8.100)

where L is a matrix of r × r dimension, and each element of it is defined by the relation

Lkl =

OJk ] (k = 1,2,...,r; O~l )g,V,Na,P,X:O

1 = 1,2,...,r)

(8.101)

Substituting Eq. (8.67) into Eq. (8.101) indicates that the matrix L is symmetric and the matrix elements Lkl obey the Onsager reciprocal relations. For small values of X (near chemical equilibrium state), the linear term predominates in Eq. (8.40), and the entropy production becomes ~=XT.L.X

(8.102)

where X T is the row vector of X. Because • -> 0 in general, the right side of Eq. (8.102) is a quadratic form, and the matrix L is positive semi-definite. Each 2(,. can be regarded as a driving force, and each rate of change of a reaction coordinate Jj as a flow that depends on all the forces; this indicates a coupled phenomenon. The reciprocal relations are valid both for the states that are in chemical equilibrium and for those that are not.

Example 8.5 Conservation of mass in chemical reactions Describe the mass conservation in closed and open systems with n number of components and the number of moles changing because of chemical reactions. (a) Closed system: Consider the following single reaction

N 2 +3H 2 ~ 2 N H 3

(8.103)

The change of mass with time is d m i "-

uiMide, or in terms of the number of moles n i d n

i -- pide.

(8.104)

where vi is the stoichiometric coefficient of species i in the reaction. The values of Pi are positive for reactants (N2, H2) , and negative for products (NH3), and Mi is the molar mass of component i. The term e is the degree of advancement or extent of the reaction. For the reaction in Eq. (8.103), we have

dmN----~2= dmH-------=-~2dmNH------=--~3de or dtIN2

-- d/'/H2

-1

-3

--MN2

--3MH 2

2MNH3

-- d/'/NH3

2

--

de

(8.1o5)

The total mass is m

=

Zmi

(8.106)

i

From Eqs. (8.106) and (8.104), we have (8.107)

dm = (E ~'iMi) de" =

8.6

429

Multiple chemical reactions

where ~ i P i m i = 0 is called the stoichiometric equation. From Eq. (8.104), we define the chemical reaction rate or reaction flow as follows d~3

Jr -

(8.108)

dt

The time change of the number of moles is (8.109)

piYr - dfli dt

For r number of simultaneous chemical reactions, we have

= ~ Uikdek

dn i

(8.110)

k=l

Consider the following simultaneous reactions: 2C+O 2~2CO C +0

(k-- 1)

(8.111)

(k = 2)

2 ---~C O 2

(8.112)

The changes of the number of moles from Eq. (8.110) are dn c = - 2 d e

1 -de2,

dno2 = - d e

I

-de2, dnco = 2de1,

dF/co 2 = d e 2

(8.113)

(b) Open systems" The change of mass of component i has two contributions (8.114)

dm i = d e m i - d i m i

where d e m i represents the mass exchanged between the system and its surroundings, and dimi represents the change within the system. Using Eq. (8.110), we have

dmi :demi ff-Mi~PikdF.k ordn i :deni+~PikdF, k k=l

(8.115)

k=l

and with the stoichiometric equations Ev/kM/= 0, we obtain (8.116)

dm - dem

Equation (8.116) shows that the change of the total mass is equal to the mass exchanged with the surroundings.

Example 8.6 Calculation of entropy production for a reversible reaction Consider the following reaction between a substrate S and a product P: S<

kf kb

>P

(8.117)

At time t, we have the concentrations of the species ~3

~3

C s - Cso - - - and Cp = Cpo + V V

(8.118)

The reaction velocity is

1 de J r -- J r f - J r b

-

V dt

(8.119)

430

8.

Chemical reactions

where the forward and the backward rates of reaction are Jrf = kf (Cso

-e) and Jrb

= kb (Cpo + s) for unit volume of V = 1

(8.120)

Combining Eqs. (8.119) and (8.120), we have d~ dt

- -~(kf

+ kb)+ kfCs0 -

(8.121)

kbCpo

~(o)=o. The solution is e = ~b[1- exp(-bt)]

(8.122)

where a

b = kf + kb, a = k f C s o - k bCpo, and 05 = -

b

After using these relations in Eq. (8.120), we obtain the affinity of the reaction considered A = R T ln Jrf - k f ( C s ° - s ) =

Jrb

kb (Cp0 +e)

kf[Cs0- 4)(1- exp(-bt))] kb[Cp0 + ~b(1- exp(-bt))]

(8.123)

The entropy production becomes

(i) = A 1 ( R T l n Jrf] T jr = ~(Jrf--Jrb) Jrb = R{kf[Cso

8.7

-

= R[kf(Cso-8)-kb(Cpo

+e)Jln kf(Cs° -e) kb (Cp0 -4-8)

(8.124)

~b(1- exp(-bt))]- kb [Cp0 + ~b(1- exp(-bt))]} In kf [Cs° - ~b(1- exp(-bt))] kb [Cp0 -[-~b(1- exp(-bt))]

STATIONARY STATES

At stationary state, all the properties of a system are independent of temperature. Stationary states resemble equilibrium states in their invariance with time; however, they differ in that flows still continue to occur and entropy is produced in the system. If a property is conservative, then the divergence of the corresponding flow must vanish; for example, Op/Ot = - d i v J. Therefore, the steady flow of a conservative quantity must be source-free and in stationary states; the flows of conservative properties are constant. If we consider the change of local entropy of a system at steady state Os/Ot = 0, the local entropy density must remain constant because external and internal parameters do not change with time. However, the divergence of entropy flow does not vanish: div ,Is = ~. Therefore, the entropy produced at any point of a system must be removed or transferred by a flow of entropy taking place at that point. A steady state cannot be maintained in an adiabatic system, since the entropy produced by irreversible processes cannot be removed because no entropy flow is exchanged with the environment. For an adiabatic system, equilibrium state is the only time-invariant state. Consider a monomolecular elementary chemical reaction S( klf ) X ( k2f > P klb

(8.125)

k2b

occurring under nonequilibrium conditions. In an open system, the substrate S is constantly supplied and the product P is constantly removed. The concentration of X is maintained at a nonequilibrium value. The evolution of X is defined by dCx

dt

-

k l f C S - k l b C X - k 2 f C X nt- k 2 b C P

(8.126)

8.7

431

Stationary states

Entropy production per unit volume in terms of reaction affinities is q~- A1 J~l + Az

T

(8.127)

T J~e

Flows of S and P keep the chemical potentials of/Xs and/Xp fixed leading to a fixed total affinity "41 -+ A2 - ( ~ s -/'Zx ) q- (/'Lx - ~P ) - ~ s - ~P -- A

(8.128)

Substituting this equation in Eq. (8.127), we get 1 diS

__A1Jrl

A-A1

Jr2

(8.129)

Jr1 - L l l -A-1 and Jr2 = L22 A - A 1 T T

(8.130)

VdtV

P-(I)-

T

+

T

The reaction velocities in the vicinity of global equilibrium are

Combining these relations with Eq. (8.129), the entropy production as a function of A1 becomes

Al2 ~ ( A1 ) - L11 _TT + L22

(A- 4 ) 2

(8.131)

T2

When the differentiation of • with respect to A~ is zero, we have a minimum O ~ ( A 1 ) _ LI 2 A 1 2 ( A - A1) OA 1 -17 + L22 T2 = 0

(8.132)

This equation shows that

A1

A2

(8.133)

Lll T + L22 --T-- - Jrl - Jr2 -- 0

The entropy production is minimized at the stationary state. The entropy production can also be expressed in terms of the concentrations. The value of Cx that minimizes the entropy production is the concentration of X at stationary state. The entropy production in terms of the reaction rates is

~ - R [ ( J r l f - - ' ] r i b )In( JrlbJrl----f-f) + (Jr2f - Jr2b) ln( Jr2j72~)]

(8.134)

The forward and backward reaction rates can be expressed in terms of concentrations Jrlf = klfCS, Jrlb - klbCx, Jr2f = k 2 f C x ,

Jr2b = k2bCp

(8.135)

At equilibrium, forward and backward rates of each reaction become equal to each other. The equilibrium concentrations are obtained from the principle of detailed balance

Cx'eq

-

klf k2b CS,eq Cp,eq kl b k2f

(8.136)

We may define small deviations in concentrations from the equilibrium 6 S = C S - Cs,eq, 6p -- Cp - Cp,eq , 6 X = C X -- Cx,eq

(8.137)

432

8.

Chemical reactions

These small deviations occur because of small fluctuations of the flows of S and R and 6s and 6p are fixed by these flows, while 6x is determined by the chemical reaction. Combining and rearranging Eqs. (8.134), (8.135), and (8.137), we have the entropy production in terms of the deviations (I)(aX) = R

[

(klfaS - klbaX )2 -Jr-(k2fax _ k2ba P )2 klfCs,eq

k2fCx,eq

J

(8.138)

If we set O~/06x = 0, the value of deviation 6x that minimizes the entropy production becomes 6x =

klfa S -Jr-k2ba P

(8.139)

klb q- k2f

Using Eq. (8.137) in Eq. (8.126), we have the stationary state

dCx dt

-

(8.140)

klfa S - klba x -- k2fa x -Jr"k2baP -- 0

This equation yields the stationary value of 6x given in Eq. (8.139). The stationary value of X minimizes the entropy production. At stationary state, the total entropy of the system is constant

dS - - 4S - +

4S

dt

dt

dt

(8.141)

=0

Therefore, the entropy change with the environment becomes negative

deS -

diS <

- ~

dt

0

(8.142)

dt

The entropy change with the environment involves energy flow J, and the flows of S and P, and the entropy current is Ju Ji Jq + SsJs+spJp Js = --T-+ E ~Li -l•

T

(8.143)

7'

where si is the partial molar entropy ( S i = (Os/Oni)T) and Jq is the heat flow. The total entropy flow exchange is determined by integrating Js. The entropy exchanged with the surroundings becomes

deS - 1 6q dt T dt

txs deN s T dt

~p

T

deN P <0 dt

(8.144)

where 6q/dt is the total flow of energy, and dNs/dt and dNp/dt are the total flows of the substrate and product. If there is no heat flow, the exchanged entropy flow is

deS ( d e N s deNp ) - ss +Sp < 0 dt dt dt

(8.145)

This equation shows that the species flowing out of the system must have more entropy than the species entering the system. The stationary value of Cx is obtained from the kinetic equations d C x - J r 1 - J r 2 = ( J r l f - J r l b ) - - (Jr2f - J r 2 b ) = klfCs - k l b C x

dt

- k 2 f C x + k2bCp = 0

(8.146)

where Jrlf and Jrlb are the forward and backward reaction rates. This equation shows that Jr1 = Jr2, and the steady-state condition is Cx =

k l f C S nt- k2bC P klb -+- k2f

(8.147)

8. 7

433

Stationary states

Example 8.7 Entropy production for series of reactions at stationary state For the following series of reactions with an entering flow S and exiting flow P, determine the rate of entropy production 1

2

S<

) X 1<

n

)X2...Xn_ 1 <

(8.148)

)P

For steady-state conditions, the reaction velocities are J~l - J~2 . . . . .

Jrn

(8.149)

I f A i << R T for every elementary reaction, the system would be close to global equilibrium, and the linear reaction velocities become

Jrl-

Jrlf,eq A1 , Jr~,_ _ Jr2f,eq A 2 , R T R T

Jrn

Jrnf,eq A n R T

(8.150)

Entropy production for a series of reactions is 1 diS_~ V dt

- 1

(8.151)

T (JrlAl +Jr2A2 + ' " + J r n A n )

The flows of S and P lead to a fixed nonzero value

A - ~ .47 i-I

(8.152)

This fixed affinity A moves the system from equilibrium. For using a fixed value of A, a linear reaction velocity is expressed by (8.153)

J r - Je A RT

where Je is the effective reaction rate obtained from 1 1 + ~ 1 +... + 1 Je Jrlf,eq Jr2f,eq Jrnf,eq

(8.154)

The linear reaction velocity is valid for an overall chemical reaction when A / R T << 1 for the elementary steps in the reaction and when concentrations do not change with time. If the overall reaction is not an elementary reaction, the relation Jr - Je [ 1 - e x p ( - A / R 1 ) ] is not valid. For reaction n, we have n-1 A, = A - ~ A , i=l

(8.155)

With this equation, the entropy production becomes

1 ais _ . = ± V dt

J,A. T

+'"+Jr. A - 2 4 "

(8.156)

i-1

Using linear reaction velocity Jr - Lii(A/T) and assuming that the condition A / R T << 1 is satisfied, Eq. (8.156) may be expressed in terms of the chemical affinity

E

L,,£

+...+L,,. A-Z i=1

)2]

(8.157)

434

8.

Chemical reactions

At stationary states, the minimum entropy production O ~ / O A i will be equal to each other as in Eq. (8.157). Consider the following set of unimolecular reactions 1

S<

0 would lead to Jri -- Jrn, and the reaction velocities

=

>X

2

(8.158)

X~._2P 3

S<

>P

Here, the third reaction is the summation of the first two; hence, only two out of three reactions are independent, and we have 4 + A2 = A3

(8.159)

The entropy production per unit volume is

A1

A2

(I) = Jrl T + Jr2 T

4

(8.160)

+ Jr3 - -

T

Using the relationship in Eq. (8.159), the entropy production becomes

4 A2 4 4 (I) = (Jrl +Jr3)T-k-(Jr2 + J r 3 ) T = J ; 1 T -+'Jrr2 T

(8.161)

The linear coupled reaction velocities are (8 162)

Jrl, = Lll T4 -{-LI2 TA2

4

Jr2 = L21 ~ + L22

A:

(8.163)

eq+R r3feq)T -+-/ r Rfeq) T2

(8.164)

Using the linear reaction velocity, we have

J/1 = Jrl -4-~r3 = Jrlf,eq "~-'k Jr3f,eq T = Comparing this equation with Eqs. (8.162) and (8.163), we have

Lll =

/r,feq+R r3feq/

'L12 -

r3Req

'L22 =

/r2feq+R r3feq/

'L21 -

r3feq R

8165>

This equation shows that the phenomenological coefficients are related to the reaction rates at equilibrium. The principle of detailed balance or microscopic reversibility is incorporated into Jr3f = J r 3 b - - J r 3 f , eq, and hence the Onsager reciprocal relations are valid. The second law requires the total entropy production resulting from all the simultaneous reactions to be positive. This has been verified experimentally. Sometimes, a system has two simultaneous coupled reactions, such that 4Jr1 < 0, A2Jr2 > 0

(8.166)

although the sum of the entropy production is always positive A1Jrl + A2Jr2 > 0 VT

(8.167)

8.7

435

Stationary states

Example 8.8 Entropy production in a homogeneous chemical system The oxidation of sulfur dioxide to trioxide over a vanadium pentoxide catalyst is 1 02 ---+SO 3 SO 2 + ~"

(8 168)

Sulfur trioxide is used to produce sulfuric acid, one of the most common chemicals used in industry. The reaction is strongly exothermic. Here, a tubular reactor is considered. The following relations are from Kjelstrup et al. (1999). The entropy production rate per unit volume of a chemical reactor is given by

/ A)

d~ - J~ - ~ =Jr X where Jr is the reaction rate, energy of reaction

(8.169)

(-A/T) is the thermodynamic driving force X, and A is the affinity, which is the Gibbs A = -AG r

(8.170)

The minimization problem is formulated as a Lagrange optimization

6

(8.171)

aX f (-JrX + AJr)dz= O where A is the Lagrange multiplier for the constraint of constant output J

J = Ac~Jr dz

(8.172)

where Ac is the cross-sectional area of the reactor. Since X can be related to concentration c, we can differentiate with respect to c instead of X, and obtain from Eq. (8.171) 2~ t"

6c

J(-Jr x + aJr)

dz -

(8.173)

0

The solution of this equation is possible for a particular expression for Jr. Considering a nonlinear chemical reaction, a practical solution may be obtained by the integration of the Gibbs-Helmholtz equation from equilibrium to optimal state

AHrd

Xopt = - f

(8.174)

eq

At equilibrium, Aeq = 0. The equilibrium temperature for a given conversion is found by using the condition Jr = 0. When we know the enthalpy of the reaction as a function of temperature, we can carry out the integral in Eq. (8.174). Since Yopt is known from Eq. (8.173), we can then find Topt. For a reactor operating with constant output, the criterion for optimal performance is for the cooling medium to have the highest possible temperature in the heat removal system. For a working example of the nonadiabatic reactor, there are 4631 cylindrical tubes with inner diameters of 7 mm packed with a catalyst and surrounded by a constantly boiling liquid at 703 K. Sulfur dioxide and air are fed into the reactor at a total pressure PT, in volume fractions of Yso2 = 0.11 and Yo2 - 0.10. The empirical expression of Jr takes into account diffusion and reaction kinetics, and we have

Jr = - k

Pso,

]

PSO3 )2

[Po2-{

]

(8.175)

The rate constant k has the dimension of mol/g of the catalyst, while the partial pressures are given in bar. The equilibrium constant is based on partial pressures, Kp (bar-°5). The rate of the reaction is negligible below 670 K,

436

8.

Chemical reactions

and the temperature of the materials in the reactor should not exceed 880 K; therefore, the temperatures of 670 and 880 K are practical boundaries for real operation. An expression for k is given by k = exp I - ~-97800 8 T 110 ) +ln(l"T

913]

(8.176)

where T is in K. Similarly, Kp in terms of temperature is given by - e x p -11800 --F-

_ 11.2]

(8.177)

The volume flow into the reactor is 3590 kmol/h. By introducing the degree of conversion x and the pressure P as variables, we obtain Jso3 =

[

kp 1- x Pso2 0.91-- 0.5X x

P

_

1 - 0.0055x

)2]

x

(1-x)Kp

(8.178)

where p is the mass density of the gas mixture. The affinity of the reaction is 1

A =/Xso 3 -/Xso 2 - ~/x02

(8.179)

and the driving force is

A REin[/ / x -

T

Pso2 P

~

-

In

0.91-0.5x

]

Kp

(8.180)

Equations (8.178) and (8.180) show that Jr is a complicated function of P, T, and composition, and it cannot be expressed in the form of linear phenomenological equations with constant coefficients. Estimations indicate that 80% of the conversion is accomplished within the first 20% of the reactor length. We obtain the optimum force Yopt as a function of A by solving Eq. (8.173) with the help of Eqs. (8.179) and (8.180) R F[(1/x(1- x))+(O.225/(1-O.O55x)(O.91-O.5x))] Xopt = -A - R [(F/Zx(1 - x)) + [(0.45Pso 3/(100.055x) 2)(P/Pr )] + (2x/(1 - x) 3K 3)] where

( F = P SO2 1 - 0.055x

(8.181)

)2

(1-x)Kp

The results for Yop t can be combined with Eq. (8.171) for a simultaneous determination of the conversion of x and P. The optimal driving force is small, around 10 J/(molK), and is almost constant through the reactor. However, the actual force is large at the inlet (about 30 J/(mol K)), and it passes a minimum (about 10 J/(mol K)) in the central region before it flattens out (about 16 J/(mol K)). This minimum is due to high temperature produced by the exothermic reaction. The large deviation between the actual and the optimal forces indicates that the lost work can be reduced without changing the rate of production. For example, maintaining stepwise control of temperature by using heat exchangers at the right positions may bring the temperature profile or the actual driving force close to the theoretical optimum, and reduce entropy production. When the optimal driving force is found for a given production rate, one may consider further optimization and, for example, further trade-offs can be made between the production rate and reactor size. Generally, a chemical reaction has a nonlinear relation between the rate Jr and the driving force -A/T. Chemical reactors are often designed to operate at the maximum rate of reaction. An alternative is a reactor operated with

8. 7

Stationary states

437

minimum useful work lost. The lost work per unit time in a chemical reactor is given by the Gouy-Stodola theorem, and is obtained by integrating the entropy production rate over the reactor volume W~ost - To~%IC~dx

(8.182)

where • is the entropy production per unit volume, Ac is the cross-sectional area that is assumed to be constant, and x is the length. Equation (8.182) implies a plug-flow reactor, since the integration is over the volume; in a batch reactor, integration is over time. The reaction takes place at temperature T, while To shows the temperature of the surroundings. Any reduction in lost work can result in economic gains, but these are not uniquely dependent on the reduction in lost work, since the reduction varies with energy and labor cost. So, it may be useful to undertake energy optimization studies separately from economic analyses. The minimization of Wlost(=Elost) entails the minimization of the integral in Eq. (4.182) with the constraint of constant production rate J (mol/h) and use of the Euler-Lagrange method

J = AcIJr dx

(8.183)

where Jr is the reaction rate (flow). When a chemical reaction is the only process in a system, the entropy production is the products of flows Jr and conjugate forces X

dp -- Jr ( - - ~ ) = JrX

(8.184)

Here, A is the affinity (the Gibbs energy of reaction) and T is the temperature. The flow for reaction is given by

Jr = L ( X ) X

(8.185)

where L is a phenomenological coefficient, which may be dependent on the force. An Euler-Lagrange minimization problem is formulated as

a f(JrX+AJr)dx=O 6X

(8.186)

where A is the Lagrange multiplier to be obtained in terms of Jr. For a required production rate, minimum lost work can be obtained by creating operating conditions compatible with the design parameters. This may be achieved when dcI)/dJr is constant. Using Eq. (8.184), the minimum lost work is obtained when

dX dJr

X + Jr --

=

constant

(8.187)

The solution to Eq. (8.186) is obtained with Eq. (8.187) when A = constant. Equation (8.187) means that the operating temperature should be parallel to the equilibrium temperature, when the enthalpy of the reaction is constant.

Example 8.9 Chemical reactions far from global equilibrium Consider the following chemical reaction far from equilibrium: A~B

(8.188)

The reaction rate is

Jr = klcA = k2CB

(8.189)

438

8.

Chemical reactions

where kl and k2 are the constants, and c A and CB show the concentrations of A and B, respectively. The following relations are from Kjelstrup and Island (1999). The normalized concentration c, due to mass conservation, is c = CA = 1 -- CB. The output of B per unit time J is given as

J = Act~[klC- kz(1- c)]dx

(8.190)

The driving force of the reaction is

- T (txA

tXB)

- R In

- In Keq

(8.191)

where Keq = kl/k2 is the equilibrium constant and R the gas constant, and the mixture is assumed to be ideal. The rate Jr, given in Eq. (8.189), is related to the force, Eq. (8.191), in a nonlinear way, as is the case for chemical reactions far from equilibrium. The total entropy production is determined by using the local value of the reaction rate and conjugate driving force in Eq. (8.184). Since the force X can be expressed in terms of concentration c, Eq. (8.186) becomes 0

- - ( J r X + AJ r) = 0 Oc

(8.192)

The result is given by

lneq] '1+'2(1+ lc The first term on the left of this equation is the thermodynamic force. The force is not necessarily constant when we have minimum lost work. The optimum force that gives the minimum total entropy production rate is obtained from

X°pt

(1 k1 + k2 1---~

To find the value ofh as a function of Jr and the rate constants of the reaction, we solve c(x) from Eq. (8.193) and substitute into Eq. (8.190). This value can be used with Eq. (8.194) to determine the value of the optimal force along the reactor. The optimum value of the Lagrange multiplier h is obtained from Eq. (8.186) (8.195)

0 In Jr By substituting Eq. (8.165) into Eq. (8.195), we have

A 2

1 2

Xopt = - - + - ( X + A ) ~

OlnL OlnX

(8.196)

The solution for A is obtained as - 2 J + JrAc ~ L*( - X ) 2 dx A=

I (L + (-X)L*

where L*=

OL OX

(8.197)

8.7

439

Stationary states

L being independent o f - X means that L can depend only on the state variables. The coefficient may depend on the temperature, pressure, and some of the concentrations. We should be able to vary - X independent of L. The optimal force is constant when the phenomenological coefficients are independent of the force. For reactions close to equilibrium, L is assumed to be independent of the forces, although it may vary along the path because the composition, temperature, and pressure vary along the reactor. By assuming L* = 0, Eq. (8.165) yields - (O In - l aJrX ]

-- - --TA

(8 .198)

From Eq. (8.196), we obtain [X(x)]Eo F = - -

A

(8.199)

2

Here, the subscript EoF means the equipartition of the force, and the force is constant. From Eq. (8.197), the Lagrange multiplier A is obtained as

(8.2oo)

t4 The specified production rate in terms of A becomes A J , - - -2 tAc I L ( x ) d x

(8.201)

Z

and the constant force is obtained as

JA

isL l

1

(8.202)

The force distribution that gives minimum lost work is uniform over the length of the steady-state reactor or over time in a batch reactor. Since the affinity A < 0 for a spontaneous reaction, A is a negative constant. Equation (8.195) indicates that the force is close to constant when we have L L* < < - - X

(8.203)

Near equilibrium, the local entropy production rate per volume is given by

(I)min ( X ) -

l ) 2[I L(x) dx] -2

LX2oF - L -JA ~

(8.204)

If the phenomenological coefficients are constant, then • is constant through the reactor, and (I)mi n becomes the summation of local entropy productions 2

• - 4]¢;.mm

'-gc

(8.205)

This equation shows that • becomes smaller for larger values of L(x) for a given reactor, and it can be used to determine the minimum lost work for a process. Equation (8.194) reduces to a constant force when the reaction is close to equilibrium. Consider the perturbation 8c on concentration at equilibrium Ceq c - Gq + 8c

(8.206)

440

8.

Chemical reactions

The concentration at equilibrium is obtained for Jr = 0

k2 Ceq

kl + k2

(8.207)

Close to equilibrium, we have Jr = (kl + k2)6c

(8.208)

X = R¢~c(klff-k2)2

(8.209)

and

k, k2

x

-

r(1 +1 3

~ k1 + k2 1 - Ceq

~ >0 Ceq

(8.210)

Equation (8.209) shows a constant force, while Eq. (8.210) shows that even i f - X i s constant the Jr need not be so since Ceq varies along the path. Therefore, the path of minimum lost work may not be a path of constant entropy production rate per unit volume. We can expect minimum lost work by the equipartition of forces when the reaction enthalpy varies little with composition, and if the reaction rate is a linear function of the conjugate force. The construction of particular operating path for a reactor with a specific output requires knowledge of reaction kinetics. The path that gives the minimum lost work entails a trade-off between the energy efficiency and the entropy production rate for a given reactor. It may be profitable to increase reactor investments and lower the energy cost for the same product, or to lower the production rate for a given energy input. For example, a 30% reduction in the lost work can be achieved by lowering the production by only 5%.

Example 8.10 Time variation of affinity We wish to derive the time variation of affinity in an open system at constant temperature and pressure. The affinity for reaction i is

4

----~lYi['Zk

(8.211)

k=l

The time variation o f A i at constant temperature and pressure is

dt

k=l

(8.212)

T,P dt

We use the change of concentration represented by (8.213)

dc k = dec k + dic k

where deck is the concentration exchanged through the surroundings, and dick is the change within the system. Using the following relations

l

Jri

i:1

V

(8.214)

d i M k = Mdic k = ~,,va. ~ d t

we have

d4_ dt

04 k= 1

deck+ T,P dt

Zl,,#m k=l m

V

(8.215)

8. 7

441

Stationary states

This equation shows that there are two contributions to the time variation of affinity. The first represents the supply of matter from the surroundings. The second is due to a chemical reaction within the system. Therefore, the two terms may counterbalance each other at a certain supply of matter. This leads to maintaining at a constant value. Such a phenomenon is only possible in an open system. In time, the system approaches a stationary state where all the reaction flows are zero except Jrm.

Ai

Example 8.11 Time variation of entropy production in simultaneous chemical reactions Consider two simultaneous reactions and derive relations for the time variation of the entropy production. Assuming that linear laws hold for a two-reaction system, we have (8.216)

Jr2-

L2~/~-)+L22(--~)

(8.217)

If Onsager's reciprocal rules hold, the rate of entropy production for two simultaneous chemical reactions is

P - diSdt - L~l

+ 2L12

T2

+ L22

(8.218)

D0

We also assume that the phenomenological coefficients are constant. The time variation of the entropy production is ldP

1 d

2 dt

2dt

dt }

L11T-k-L12T 7

-k- L12T+L22 T 7

>0

(8.219)

Using the linear laws of Eqs. (8.216) and (8.217) in the equation above, we have

2 dt

d-t

,,220,

-T + Jr2 7

For a closed system, the affinities A1 and A2 are expressed in terms of P and T, which are kept constant, and the extent of reactions el and e2, and we have 1 dip 2 dt

Jrl --[( OA1] Jrl "Jr-(OA1] Jr2 ] +---~---~ Jr2 [(OA21L 081 Jrl + ~.~08 (om2] 7"--)p,T 2)p T Jr2 ] T

(8.221)

From the definition of affinity, we write

08"2) P,T

C]el P,T

OelO~32 P,T

(8.222)

Using this equation in Eq. (8.161), we have

ldP

l{aA1 2 0A1 aA2 ) 0e I Jrl +2 JrlJr2 +~Jr22

2 dt = T

Oe2

Oe2

P,T < 0

(8.223)

Because terms OAilcgei belong to a definite quadratic form, the time variation of entropy production can only decrease with time.

442

8.

Chemical reactions

For an open system, affinities are functions of the mole numbers. From

dm= dem + dim, we have

dA1 (OA1]d#'li ~lOA1]deni (OAlldini:E(OA1) d131..p. (OA1]de#'li d' ~i t~ni)--~-:iT"~tOyli.JT"sr-~i COni)--~ i toni) pil--'~" ~i toni) 'dr

(8.224)

nt-E( OA1] dE2 fOAl)Jr1 .qt_ (OA1)det'li (OA1)dlli i t Oni)Pi2 -'~ = t 081) ~i t Oni) "-~ + t ~2 ) ~ Substituting this equation in Eq. (8,220), we have 1 dP

1 (OA1j21+20A 1 JrlJr2 +OA2jr2)+

2dr

]" ~,081

082

0~:2

1 l~{ OA1 7. Jrl ~-n/-t- Jr2

OA2] deni

Oni ) dt

(8.225)

In open systems, the time variation of the rate of entropy production has two contributions. The first contribution is an internal term, which is always negative 1 diP_ 1

(OA

T 081 Jr1 -k-20e2 JrlJr2 -k-~Jr20/32 "~ 0

2dt

(8.226)

The second contribution is an external term representing the exchange with the surroundings with no definite sign

l deP- l ~i (Jrl OA1 OA2]deni 2 dt T . -~n/-+-Jr2Oni ) dt

(8.227)

For transport processes, the entropy production always decreases in time.

Example 8.12 Minimum entropy production Consider the following synthesis reaction H 2 + Br2 = 2HBr

(8.228)

This results from the following intermediate reactions Br2 < 1 >2Br

(8.229)

H 2 +Br< 2 " H B r + H

(8.230)

H + B r e.. 3 " H B r + B r

(8.231)

The affinity of the net reaction is maintained at a constant value by the flows H 2 and Br2. The affinity of the first reaction is unconstrained. We wish to show that the stationary state leads to minimal entropy production. Assume that the system is homogeneous with a volume V = 1. The net reaction is obtained from the summation of reactions (8.230) and (8.231), and the affinity of the net reaction is A = A2 + A3

(8.232)

The value of A is maintained at a nonzero value, and hence at nonequilibrium conditions. The entropy production per unit volume is d~-

1 diS -

A1Jrl -[- A2 Jr2 + A3 Jr3 > 0

v dt

r

7-

-/-

(8.233)

-

The linear phenomenological reaction flows with vanishing cross-coefficients are

Jri = Lii

T

(8.234)

8.8

Michaelis-Menten kinetics

443

From Eqs. (8.228) and (8.229), we have (I)-- A1 Jr1 if- & 4 2 q- A - A2 Jr3 ~ 0

T

(8.235)

T

Using the linear reaction flows given in Eq. (8.234), and minimizing the entropy production by setting O~/OAi = O, for the independent affinities of A1 and A2, we find Jrl - 0 and Jr2 = Jr3

(8.236)

These equations support the kinetic equations at stationary states d C H - Jr2 -- Jr3 = 0

(8.237)

dt

dCBr -

2J~l - Jr2 if- Jr3 = 0

(8.238)

dt

8.8

MICHAELIS-MENTEN KINETICS

All chemical reactions in a biological cell take place with the direct participation of enzymes as catalysts. Enzymes are proteins, which are macromolecules composed of a combination of the 20 amino acids. Enzymes, as catalysts, are highly efficient and selective in binding small molecular species called ligands. A ligand that is acted upon by an enzyme to cause a chemical reaction is called a substrate. Only a certain, small portion of the amino acids that comprise an enzyme is involved in the catalytic reaction. This region is called the active site, and is directly involved in the formation of product(s). For example, the amino acid residues of proteins are greatly influenced by their local pH values, and the activity of proton acceptors and donors occurs in the active site. In 1913, Michaelis and Menten assumed that the enzyme and substrate react reversibly to form a substrate-enzyme complex. Later, the complex dissociates to form the free enzyme and product(s). The reactions are as follows k1

Fo

(8.239) k2

F1

~ >Fo + p

(8.240)

where Fo and F1 are the concentrations of the enzyme and substrate-enzyme complex, respectively. The terms kl, k2, and k3 represent the rate constants in the forward and backward directions. Michaelis and Menten also assumed that the first reaction is rapid and equilibrium is established quickly. The kinetic equations in accordance with the law of mass action are dS dt

- -lqFoS + k2F1

dE1 - k l F o S - ( k

2 + k3)F 1

(8.241)

(8.242)

dt dP dt

- k3F~

(8.243)

As the total concentration of the enzyme is constant, E - F0 + F1 - constant, Eq. (8.242) becomes - k

dt

ES-(k2 +

+ k S)Y

(8.244)

444

8.

Chemical reactions

Jr

Zero order behavior when S is large

Jr, max = k3E First order behavior when S is small .... ~

Jr=

L

\

_~ -

~ / ' / -

~t " Rate-concentration curve for Michaelis-Menten kinetics

max

2

l

~P~'atKM=S

KM=S

Figure 8.2. Some properties of the Michaelis-Menten kinetics. If S >> E in the stationary state, we assume that dFo/dt = dF1/dt = 0. From Eq. (8.244), we express F1 by

klES k2 + k3 + klS

F1

(8.245)

Since P = - S coincides with the reaction rate Jr, we obtain Jr -

aP at

-

k3klES k2 +k3 + klS

k3F1 =

(8.246)

Dividing the numerator and denominator of Eq. (8.246) by kl, and defining the Michaelis constant KM by

+ k3 kl

KM -

(8.247)

leads to the Michaelis-Menten equation

k3ES Jr =

(8.248)

KM +S

This equation shows that the reaction rate is directly proportional to E, and the relation between the reaction rate and the substrate concentration is not linear. However, if we study the reaction in certain regions, it can be approximated by linear kinetics. A plot of Jr versus S yields a rectangular hyperbola, and approaches Jr,max asymptotically

k3E

Jr,max =

(8.249)

at large values of S leading to a concentration-independent rate of reaction as seen from Figure 8.2. So, Eq. (8.248) can also be written as J r , max S Jr = ~

(8.250)

k +S

When the substrate concentration is large, the reaction rate is dependent on the substrate concentration. This represents zero-order kinetic behavior. When the concentration is very low, then the kinetics may be represented by first-order kinetic behavior. At Jr = Jr,max/2, the value of the Michaelis constant KM is obtained as S. The MichaelisMenten equation can be linearized, and the Lineweaver-Burk plot (Figure 8.3) is obtained from the following form 1 _

Jr

1

Jr,max

t

K M

1

(8.251)

Jr,max S

From the Lineweaver-Burk plot, we can obtain -1/KM from the abscissa intercept, and intercept.

1/Jr, ma x

from the ordinate

8.8

445

Michaelis-Menten kinetics

1

Jr

f J

f

/ /

"/,.max

1 KM

1 S

Figure 8.3. Lineweaver-Burk plot of the Michaelis-Menten equation.

The Monod equation describing the initial rates for microbial growth using the steady-state concentrations

j_j

{ cs j ,max

Ks + Cs

(8.252)

or dual Monod equation

rmax! CS l/ CP / Ks + Cs

"

(8.253)

K p + Cp

are well-known relations, although neither of them is general. Here, the concentrations Cs and Cp are the substrate and product concentrations, respectively, and Jr,ma×is the maximal growth rate. The terms Ks and Kp are the half saturation constants for growth, which are phenomenological constants. The Monod equation considers one irreversible step, although more steps are involved in the growth rate. Under relatively stable conditions, some chemical species remain uniform in concentration, or the overall reaction remains close to or far from global equilibrium. Michaelis-Menten kinetics leads to another rate equation to be considered using the Michaelis-Menten constants Ks and Kp, which are the ratio of reaction rate constants exp(A)- 1

Jrs - JrS,max O[exp(A) + (Jrs,max/JrP,max)]

(8.254)

where

(b

/ KS / ~ + 1

C S + Cp

Using Cs + Cp = constant, this equation relates the reaction velocity to affinity only. A plot of JrS/Jrs,max v e r s u s affinity shows an inflection point, representing a maximum slope. Therefore, the quasi-linear approximation between the reaction velocity and the affinity will be valid with a deviation of less than 15%.

E x a m p l e 8.13 Growth of a pathogenic bacterium Brucella abortus When the B. abortus is grown on an

agar medium, the colony has a smooth outline, and the constituents are called S-cells. During growth, mutations occur and form colonies having a wrinkled appearance. These cells are called R-cells. A flask of nutrient broth that is initially inoculated with S-cells contains a majority of R-cells by the time a stationary population size is attained. When the broth is analyzed, we see a decline of S-cells coinciding with the appearance of alanine,

446

8.

Chemical reactions

produced by S-cells, which kills S-cells but does not affect the growth of R-cells. Therefore, the growth curve exhibits a decline starting about 4 days after inoculation. Describe the growth kinetics for the S- and R-cell interaction. Discuss the equilibrium states of the system. Assume that A is the amount of alanine present in the broth at time t, and S and R denote the number of smooth and rough colony-forming cells. With very little or no interactions, S and R cells obey logistic growth laws. Meanwhile, because of alanine production, S cells die at a rate of AS. Since R cells arise because of mutation of S cells, there is an additional production of R cells at a rate proportional to S, if we disregard the effect of backmutations on the growth of S cells. We also assume that alanine is produced at a rate that is proportional to S. Therefore, we would have the following growth relations (8.255)

- - = k l S - fll $2 - aSA dt dR dt

dA dt

- k z R - f12R2 - AS

(8.256)

- yS

(8.257)

where kl, k2, ill,/~2, '~, and y are constants. The initial condition is (S, R, A) = (So, 0, 0). We can estimate the concentration from Eq. (8.257) and use it in Eq. (8.255), and we find t

(8.258)

dS _ k l S _ fllS2 _ a y S ~ S ( t ) d t dt o

So Eqs. (8.258) and (8.256) represent the growth kinetics. At stationary states, they become

(

o = s k~ - I ~ S - ~

)

S(t) dt

(8.259)

0

(8.260)

0 = R ( k 2 - t2 R) - AS

To satisfy Eq. (8.259), either S = 0, or the contents of the bracket equals zero. As the amount of S in the integral in Eq. (8.259) will increase with time, there is no fixed value of S that makes the contents of the bracket zero. Therefore, the only solution for Eq. (8.259) would be S - 0. Now, Eq. (8.260) becomes 0 = R ( k 2 -/32R )

(8.261)

and has two solutions with the roots, R = 0 a n d R - k2/f12. Hence, there are two stationary points: which is unstable, and (S, R) = (0, k2/~2), which is stable (Rubinov, 1975).

8.8.1

(S, R ) -

(0, 0),

Flow Force Relations for Microbial Growth

Consider a microbe coupling catabolism, which consists of processes that liberate free energy, and anabolism, which is biosynthesis. In the vicinity of global equilibrium (and for some far-from equilibrium systems) linear relations exist between thermodynamic forces, which are the corresponding Gibbs free energy differences A G, and flows. Therefore, a microbial growth may be represented by J=-LAG=-L(AG°+RTln

S)

(8.262)

where AG ° is the standard free energy difference, L is a phenomenological constant, which may depend on the enzyme, and R is the gas constant. Equation (8.262) can be used for catabolism C and anabolism A, and we have Jc = - L c A G c

(8.263)

8.9

447

Coupled chemical reactions

(8.264)

JA = --LAAGA

The dissipation function • is always positive, as the dissipation of the Gibbs free energy should occur because of the irreversibility of the system ~{r = _ J c A G C _ J A A G A > 0

(8.265)

Overall, the thermodynamic efficiency of growth rt is the ratio of the Gibbs free energy output and the Gibbs free energy input rl = -

JAAGA

~ < 1 JcAGc

(8.266)

Here, conventionally, biosynthesis is represented by a negative flow of anabolism. Thermodynamic efficiency can be estimated when the yield and actual concentrations of the substrates and products are known. The ratio JA/Jc is determined from the yield of the carbon and energy substrate, while AGA/AG C is determined by the standard free energy differences and concentrations of substrates and products. The theoretical efficiency of energy conversion may be optimized by varying AG A and the degree of coupling between catabolism and anabolism. Catabolism is the conversion of the growth-supporting energy source with the concurrent generation of ATE This ATP is utilized in anabolism for converting low-molecular-weight substrates via monomers into biomass. On the other hand, the leakage reactions are involved in all those processes that consume ATP without coupling to anabolism.

8.9

COUPLED CHEMICAL REACTIONS

Besides the transport, the most important processes in biological systems are those related to chemical reactions of metabolism. One of the typical aspects of such reactions is the requirement regarding the apparent stoichiometry of two partially coupled reactions, and the study of the efficiency of such reactions as limited by the constraints of the second law of thermodynamics. When we have the number of moles of species i and reactions j, we can express the Gibbs equation in terms of the extent of reaction ej and the affinity A/ P

dS - dU7/, + --T d V - ~_~. ~

de /

(8.267)

.!

where dej = dNij/vi/, and Aj = - Xiv(j~j In an isolated system in which U and V do not vary, d U = dV = 0, the dissipation function is given by the second law of thermodynamics q ~ - r dS dt

.

Ai "

>--0 dt )

(8.268)

According to the law of mass action of chemical kinetics, the rate of reaction is proportional to the product of the concentrations. The reaction velocity is d~3j

Jr; -

dt

(8.269)

With the respective affinities as the thermodynamic forces, Eq. (8.268) becomes

~ - E A /Jri >__0 / Therefore, A and Jr must have the same sign. In the chemical equilibrium state, we have A = 0.

(8.270)

448

8.

Chemical reactions

8.9.1 Two-reaction Coupling Despite its limitations for chemical reactions, the linear net theory has a useful conceptual base. Consider the linear phenomenological equations for two chemical reactions with flows Of Jrl and Jr2 Jrl =/-q 1A1 + L12A2

(8.271)

Jr2 = L214 + Lz2A2

(8.272)

These relations are based on the dissipation function

'~ = A1Jrl + A2Jr2 -> 0

(8.273)

For this sum to be positive, either both contributions can be positive, or one of them (A 1Jrl) can be negative while the other is positive and large enough to compensate for the negative effect of the first term. When, for example, A1Jrl is negative, the first reaction is carried out in the direction opposite to the direction imposed by its affinity. This is only possible if a coupling occurs between these reactions, and a sufficiently large entropy production is attained by the other reaction. This effect may cause one process to drive another process. Coupled processes are of great interest in biological systems, since in many situations the synthesis of reactions or the transport of substrates takes place in the direction opposite to that predicted by its thermodynamic force. The degree of coupling is defined by the ratio

q12

L12 (L11L22)1/2

(8.274)

The second law imposes L llL22 ~ (L 12)2, and therefore the degree of coupling is limited between - 1 and + 1. When q = + 1, the system is completely coupled and the two processes become a single process. When q = 0, the two processes are completely uncoupled and do not undergo any energy-conversion interactions. Let us consider Jrl as output and Jr2 as input flows, and the ratio Jrl/Jr2 is the stoichiometric ratio, which indicates the number of moles reacting in reaction 2 in order to produce a certain rate of reaction 1. From Eqs. (8.271) and (8.272), we have

Jr1 = LI1A1 + LlzA2 Jr2 L21A1+L22A2

(8.275)

Dividing the numerator and the denominator of this equation by (LllL22)l/2A2, we obtain Jr1 _

Jr2

q + ZX

Z~ 1+ qzx

(8.276)

where z andx are z = (Lll/L22) 1/2 andx = AlIA 2. This relation shows that the ratio of flows (rate of reactions) depends on the ratio of forces and the degree of coupling. When the degree of coupling q goes to zero, J r l / J r 2 - - Z2X, SO that z is a stoichiometric parameter. In the case of two coupled reactions, one spontaneous and the other forced, it is customary to define efficiency as the ratio of the dissipation in reaction 1 to the dissipation in reaction 2

JrlA1 r/= - ~ Jr:A2

(8.277)

From Eq. (8.276), we can write

r/= -

zx(q + zx) 1 + qzx

(8.278)

449

Problems

For the maximum efficiency of the energy transfer from reaction 2 to 1, we differentiate ~ with respect to x and equate to zero, and then we have -l+x/1-q 2 Xmax =

(8.279)

qz

After rearranging terms, the maximum efficiency is obtained as

~ma×

z

q2 [1 + X/1- q2 ]2

(8.280)

The value of TImaxdepends only on the extent of coupling. When two reactions are completely uncoupled, the efficiency must be zero. When q = 1, we have the maximum efficiency of unity.

PROBLEMS 8.1

Calculate the entropy production for racemization reaction S = E and plot the change of entropy production with time. Use kr = 2.0, kb = 0.01, Cs0 = 2.0 tool/L, Cp0 = 0.01 tool/L, V = 1 L, T = 300 K, and R - 8.314 J/(mol K).

8.2

Estimate the change of concentrations with time for the following elementary reaction: X + Y = 2 Z Use k~ = 0.5, kb = 0.01, C~0 = 1.0, C,,0 = 2.0, and (7_-o= 0.

8.3

Estimate the change of entropy production • with time for the following elementary reaction: X + Y = 2 Z Use kr = 0.5, kb = 0.01, C~0 = 1.0, C,,0 = 2.0, and Go = 0.

8.4

In the gas phase, isopropyl alcohol is dehydrogenated to produce propionaldehyde in the following reaction (CH3)2CHOH(g) = CH3CH2CHO(g ) ÷ H2(g) The standard heat and the Gibbs free energy for isopropyl alcohol are AH ° - 55.48 kJ/mol, AG ° - 17.74kJ/mol,

ACp- 16.736 J/(molK)

Determine the equilibrium composition of isopropyl alcohol at 500 K and 1 atm if ACp = 16.736 J/mol. 8.5

The following reaction shows the production of ethylene dichloride using ethylene dibromide as a catalyst C2H4 + C12 - C2H4C12 The normal boiling point of ethylene dichloride is 83.47°C. Estimate the equilibrium conversion of ethylene if the reaction takes place at 50°C and Iatm.

8.6

We heat a component A. The pure component A, originally at 298.15 K and 1 atm, decomposes upon heating and the following reactions occur in the gas phase: A=B+C A=D+E (a) Estimate the equilibrium composition of component A at a pressure of 1 atm over a temperature range of 1000-1500 K. (b) Estimate the equilibrium composition of A over a temperature range of 1000-1500K if pure A at 298.15 K and 1 atm is heated in a constant-volume reactor.

450

8.

Chemical reactions

For this temperature range, the following K values may be used: T(K)

Kal Ka2 8.7

1000 2.907 534.3

1200 38.88 2581

1400 246.0 7754

1500 512.6 11950

Ethyl benzene is produced in a constant-volume reactor by the following reaction: C6H6(g) + C2H4(g ) = C6H5C2H5(g ) For this purpose, 1 mol of ethylene and 1 mol of benzene are heated to 600 K. The initial pressure within the reactor is 1 atm. At 600 K, Ka = 345.0 and AHr = 103940 kJ/mol for the reaction. Determine: (a) The equilibrium compositions of ethyl benzene, benzene, and ethylene. (b) The heat to be removed to maintain the isothermal reaction above at 600 K.

8.8 8.9

Determine the entropy and enthalpy change of an electrochemical cell reaction. The kidneys transport glucose from the urine to the blood against a concentration gradient, in which the transport occurs from a low concentration to a high concentration. This is called active transport, and can occur only if the transport is coupled with a spontaneous chemical reaction. Assume that the initial concentration of glucose in the urine is 5 × 10-5 mol/kg. The glucose concentration becomes 5 × 10-6 mol/kg after leaving the kidneys. The blood contains an almost constant concentration of glucose of 5 x 10-3 mol/kg. Estimate the minimum work or Gibbs free energy supplied per mole of glucose transported across the kidneys.

8.10

Consider a nonisothermal reactor. The concentration C and temperature T change with time within the reactor may be described by the following equations:

dC - - c exp ( -10) dT - - lOOOCexp(- ~ ) - lO(Twhere the initial conditions are T(0) = 288.15 K and C(0) = 1.0 gmol/L. Plot the concentration and temperature of the reactor as a function of time. 8.11

Consider the following chemical reaction kinetics equations for the components x, y, and z:

dx --O.031x-lOOOxz dt dt dz dt

- -2500xy - -0.03 lx - 1000xz

-

2500yz

where the initial conditions are x(0) = y(0) = 1 and z(0) = 0. Plot the concentration as a function of time. 8.12

k3 If we consider the back reaction F1 < k4 >F° + P' where k4 is not zero in the reactions of the enzymesubstrate system, modify the Michaelis-Menten kinetics. Show that when equilibrium is established, after a very long time, equilibrium concentrations of substrate and product are related by the following Haldane's relation

eeq kxk3 Seq k2k4 _

References

8.13

451

In an autocatalysis system, the product P of a reaction catalyses its formation from a substrate S. If the initial concentration values of substrate and products are So and P0, respectively, define the rate equations for the concentrations and find their dependence on time. Assume that an intermediate complex is not formed in the reaction.

REFERENCES Zs. Ablonczy, A. Lukacs and E. Papp, Biophys. Chem., 104 (2003) 240. S.R. Caplan and A. Essig, Bioenergetics and Linear Nonequilibrium Thermodynamics." The Steady State, Harvard University Press, Cambridge (1983). E.P. Gyftopoulos and G.P. Beretta, Thermodynamics. Foundations and Applications, Macmillan, New York ( 1991 ). S. Kjelstrup and T.V. Island, Ind. Eng. Chem. Res., 38 (1999) 3051. S. Kjelstrup, E. Sauar, D. Bedeaux and H. van der Kooi, Ind. Eng. Chem. Res., 38 (1999) 3046. D. Kondepudi and I. Prigogine, Modern Thermodynamics. From Heat Engines to Dissipative Structures, Wiley, New York (1999). G. Nicolis and I. Prigogine, Exploring Complexly, Freeman & Company, New York (1989). S.I. Rubinow, Introduction to Mathematical Biology, Wiley, New York (1975). M.T. Suchiya and J. Ross, Proc. Natl. Acad. Sci., 100 (2003) 9691.

REFERENCES FOR FURTHER READING R.A. Alberty, Biophys. Chem., 124 (2006) 11. H. Qian and E.L. Elson, Biophys. Chem., 101-102 (2002) 565. J. Ross and M.O. Vlad,Annu. Rev. Phys. Chem., 50 (1999) 51. S. Sieniutycz, Int. J Eng. Sci., 36 (1998) 577.

9 COUPLED SYSTEMS OF CHEMICAL REACTIONS AND TRANSPORT PROCESSES 9.1

INTRODUCTION

Nonisothermal reaction-diffusion systems control the behavior of many transport and rate processes in physical, chemical, and biological systems. A reaction-diffusion system with appropriate nonlinear kinetics can cause instability in a homogeneous, steady-state system and generate stable concentration patterns. Some of the chemical reactions coupled with the transport of species can lead to pumps and chemical cycles in biological systems, such as a sodiumpotassium pump. A considerable work has been published on reaction-diffusion systems. This chapter discusses mathematically and thermodynamically coupled differential equations of nonisothermal reaction-diffusion systems. Here, the thermodynamic coupling refers that a flow occurs without its primary thermodynamic driving force, or against the direction imposed by its thermodynamic force. The principles of thermodynamics allow the progress of a process without or against its primary driving force only if it is coupled with another process. This is consistent with the second law, which states that a finite amount of organization may be obtained at the expense of a greater amount of disorganization in a series of coupled spontaneous processes. Modeling of spatio-temporal evolution may serve as a powerful complementary tool for studying experimental nonisothermal reaction diffusion systems within a porous catalyst particle and a membrane. The linear nonequilibrium thermodynamics approach may be used in modeling coupled nonisothermal reaction-diffusion systems by assuming that the system is in the vicinity of global equilibrium. In the modeling, the information on coupling mechanisms among transport processes and chemical reactions is not needed.

9.2

NONISOTHERMAL REACTION-DIFFUSION SYSTEMS

The basic equations for an unsteady-state process of one-dimensional (in the y-direction) heat and mass transport with a simultaneous chemical reaction in a porous catalyst pellet are

OCA, o? --D"-~ OvO---(ybOC----~A OV ) (pCp)-~-

ke

--

--

~, O v

Ov

-k-(-AHr)Jr(CA,T)

(9.1)

(9.2)

where ~L/r is the heat of reaction, De and ke the effective diffusivity and thermal conductivity, respectively, and v is stoichiometric coefficient, which is negative for reactants. Here, b describes the shape: b = 0, slab; b = 1, cylinder; and b - 2, sphere. These partial differential equations are mathematically coupled. The initial and boundary conditions with internal and external and resistances across the boundary are Att=0 At v - 0

(surface conditions)

CA=CA~, a n d T = -c ) C- A

--

0 and

~c)T= 0

Ov - - D e OLT----~A" - - k ( C A - -

&, CAO )

Ov

-k~ -°r- = h ( r - r0) 0v

(symmetry conditions) (external mass transfer effect)

(external heat transfer effect)

(9.3)

454

9.

Coupled systems of chemical reactions and transport processes

The linear nonequilibrium thermodynamics approach can provide a quantified description of the fully coupled phenomena for systems in the vicinity of global equilibrium.

9.2.1

Effective Diffusivity

In a multicomponent fluid, a species can be driven not only by its own thermodynamic force (its own concentration gradient) but also by concentration gradients of all the other species. Flow of speciesj in an n multicomponent fluid system is n-1

Nj = - ~ , CDjkVY k + yj k=l

Nk j = 1,2,...,n- 1 k=l

where C is the total concentration. The last term represents the bulk flow of the fluid. To simplify this equation, a common approach is to introduce a mean effective binary diffusivity for species j diffusing through the fluid mixture

Nj = -CDjm Vyj + yj ~ N k

(9.4)

k=l

For ideal fluid mixtures, the Maxwell-Stefan equation yields

-CVyj = ~

1---~-(ykNj -- yjNk) Djk k--/=j

(9.5)

k=l

For a binary mixture, this equation becomes !

-CVy 1 = -__:__.[N 1 - Yl (N1 + N2)] D12

For a multicomponent gas mixture, the effective binary diffusion coefficient for species j diffusing through the mixture may be found by equating the driving forces Ayj in Eqs. (9.4) and (9.5) //

1

~.k=l(1/Djk)(Yk -yj(Nk/Nj)) -

(9.6)

n

Djm

1-- yj Z k=I(Nk/Nj )

This equation reduces to Wilke equation, when it is used with the zero flow rates for k = 2, 3. 1

Dim

_

1

'(-" Yk

1 - Yl k=2Z-"D1k

This equation represents the diffusion of species 1 through stagnant species 2,3..., n in the reacting system, and is mainly suitable for very dilute solutions. When the other species are not stagnant, the steady-state flow ratios are determined by the reaction stoichiometry. For a reaction, N/vj = constant, and Eq. (9.6) becomes n

1 _ ~-~ke=j(1/Djk)(yk - Yj (vk/vj)) n Djm 1- yj Z k = l (l~k/PJ ) or

1

Djm

/

1+ 6jyj k~j Dj---k Yk - Yj ~j [

(9.7)

(9.8)

where j is the reactant species, and 6j = (Evi)/vj. Usually, some average composition Yj,av is used to determine an average value of Dim.

9.2

4.55

Nonisothermalreaction-diffusion systems

For certain applications, we define an effective binary diffusivity with the flow relative to the fixed solid and include only bulk flow in the values of D~m Nj -- -CD)mVy j

Then, the same procedure yields

--

(9.9)

Yk--Yj

k~ i Djk

The effective diffusion coefficient may be obtained from the molecular diffusion coefficient Dj, the catalyst porosity 8p and tortuoisty %. _ ~p

Dej

mDj ?p

(9.10)

Tortuoisty describes the deviation of the pores from an ideal structure.

Example 9.1 Effective diffusivity Consider the following chemical reaction aA + bB ~ rR + sD Equation (9.8) yields

1

DAm

where

-

' [2(

1 + 6 AYA

6 A = (r + s - a - b)/a.

YB--YA

+

l(

DAR

YR +YA

a) l(

')1

-t-1 Ys + Y A - DAS a

The flow of species A is dYA NA =-CDAm---~-y +YA (NA +NB +NR + N s )

or N A = _ C D Am dY AT _F YA N A ( + ~b a- r - s

Here the counter diffusion is used. The flow of species A is N~

z

m

CD Am dY A

l + 6 A y A dy

After integrating this equation between y - 0 and y - L with NA = constant and an average constant value of DAm, we have NA _ CDAm In 1+ 8AYA0 L8 A

1 + 8 A YAL

_

CDAm YA0 - - YAL L

where YfA is the film factor defined by

YfA --

When

•A

--

(1 + 6AYA0)-- (1 + 6AYAL) (1 + 6A YA0 )/(1 + 6AYAL)

0 that represents equimolar counterdiffusion, then YfA = 1.

YfA

456

9.2.2

9.

Coupled systems of chemical reactions and transport processes

Effective Thermal Conductivity

For a heterogeneous solid where one solid phase is dispersed in a second solid phase, or one solid phase contains pores, we introduce an effective thermal conductivity to describe steady-state conduction. The geometry of the dispersed solid or pores affect the thermal conductivity. If we have a material made of spheres with thermal conductivity kl dispersed in a continuous solid phase with thermal conductivity ks, then the effective thermal conductivity ke is ke - 1 + 305 ks (kl + 2ks )/(kl - ks ) - 4'

(9.11)

where 4' is the volume fraction of the embedded material. This equation is called Maxwell's Derivation and assumes that spheres do not interact thermally and the volume fraction 4~ is small. For solids containing gas pockets with thermal conductivity kl, thermal radiation may be important, and an effective thermal conductivity may be approximated by ke_

1

ks

1 + (k 1/ks6 + 4oT3L/ks ) - 4,

(9.12)

where o- is the Stefan-Boltzmann constant and L the total thickness of the material in the direction of the heat conduction. For gas-filled granular beds, the thermal conductivity of the gas may be very low. Since gas phase heat conduction mainly occurs near the points of contact between adjacent solid particles, the distance for heat conduction over the gas phase may approach the mean free path of the gas molecules. This reduces the thermal conductivity of the gas further, since the whole system may become rarefied for evacuated beds of fine powders. In separation processes and chemical reactors, flow through cylindrical ducts filled with granular materials is important. In such systems conduction, convection, and radiation all contribute to the heat flow, and thermal conduction in axial ke,x and radial ke,r directions may be quite different, leading to highly anisotropic thermal conductivity. For a bed of uniform spheres, the axial and radial elements are approximated by ke, x = 0 . 5 p C p v o D p

ke, r = O. l p C p v o O p

(9.13)

where v0 is the superficial velocity, and Dp the diameter of the particles. These equations are used for highly turbulent flow.

9.2.3

Balance Equations

Assuming a steady state, for first-order reaction-diffusion system A ~ B under nonisothermal catalyst pellet conditions, the mass and energy balances are 0 =-V.J-kvC A O -- - V . J q -+-( - , ~ r

(9.14) )kvC A

(9.15)

By using Fick's and Fourier's laws in one-dimensional transport in a slab catalyst pellet (Figure 9.1) with, equimolar counter-diffusion under mechanical equilibrium, Eqs. (9.14) and (9.15) become d2CA O - D e dy 2 - kvC A

(9.16)

O-- ke d2T dy---5- + ( -

(9.17)

AH r)kvCA

9.2

457

Nonisothermal reaction-diffusion systems

I I

y=0

y=L

I

CAs I

II cA

I I

\\,

,, I ] Stagnant A I film I I I

Catalytic surface B

Figure 9.1. Schematic heterogeneous reaction-diffusion system.

Without the external mass and heat transfer resistances, the boundary conditions with the x-coordinate oriented from the centerline (y - 0) to the surface ( y - L) are dC A (0) = 0 and T ( L ) - ~ , c(L) - CAs, ~

d~'

dT(O)

dy

- 0

The values of De tend to be smaller than those of ordinary gas diffusivity, while the values of ke are smaller than those of the thermal conductivity for a similar nonporous solid. The effective reaction rate kvCa is based on the total rate of reaction within any small, representative volume. Eliminating the reaction terms from Eqs. (9.16) and (9.17), and integrating twice with the boundary conditions above, the temperature is related to concentration by q~-1-/3(0-1)

(9.18)

where 0

CA = c-2,'

T

(-AHr)DeCAs '

Equation (9.18) is valid for any particle geometry under steady-state conditions, and can be used to eliminate the 0 or q~ from one of the differential Eqs. (9.16) and (9.17). The nondimensional parameter/3 (positive for exothermic reactions) is a measure of nonisothermal effects and is called the heat generation function. It represents the ratio between the rate of heat generation due to the chemical reaction and the heat flow by thermal conduction. Nonisothermal effects may become important for increasing values of/3, while the l i m i t / 3 - , 0 represents an isothermal pellet. Table 9.1 shows the values of/3 and some other parameters for exothermic catalytic reactions. For any interior points within the pore where the reactant is largely consumed, the maximum temperature difference for an exothermic reaction becomes ( - A H r )OeCAs

ke

= fi~

(9.19)

458

9.

Coupled systems of chemical reactions and transport processes

Table 9.1. Parameters for some exothermic reactions used in the thermodynamically coupled model Reaction systems Synthesis of vinyl chloride from acetylene and HC1 Dissociation of N20 Hydrogenation of benzene Oxidation of SO2 NH3 synthesis Oxidation of CH3OH to CH20 Hydrogenation of ethylene Oxidation of ethylene to ethylene oxide Oxidation of H2

~b

/3~/3'

y

Le

•a

(.oa

0.27 5.0 0.05-19 0.9 1.2 1.1 0.2-2.8 0.08 0.8-2.0

0.25 0.64 0.12 0.012 0.000061 0.0109 0.066 0.13 0.1

6.5 22.0 14-16 14.8 29.4 16.0 23-27 13.4 6.75-7.52

0.1 0.01 a 0.006 0.0415 0.00026 0.0015 0.11 0.065 0.036

0.001 0.001

0.001 0.001

0.001

0.001

aAssumed values.

Source." Hlavecek et al. (1969). Transient forms of Eqs. (9.16) and (9.17) become 00 OT 1 0q~

Le 0~-

-

0expE,/' +/]

020

OZ2

(9.20)

02@

(9.21)

OZ2

after using the following dimensionless parameters

y Z----"

L'

E

Det (!/)2 L2ko exp(E/RTs) T--

--~-'

; Y = ~ s s ' Le .

De

ke/pCp _ a e . De. .

De

(9.22)

where Le is the modified Lewis number and ae the effective thermal diffusivity. The nondimensional group 3' is called the Arrhenius group, and represents a nondimensional activation energy for the chemical reaction. The initial and boundary conditions are 0(0, z) - 1, 0(~-,1) - 1,

dO(r,O) = 0

dz dq~(~-,O) =0 ~(0, z ) - 1, q~(~-,1) = 1, dz After substituting Eq. (9.18) into Eq. (9.20), steady and nonisothermal concentration profiles become

d20 dz 2 - ch2Oexp( - 1yfl(O-1) -/3(0-1) )

(9.23)

Diffusion may reduce average rates relative to those obtained if the concentration everywhere was the surface concentration CAs (Froment and Bischoff, 1979). This limitation is quantified as the effectiveness factor rl defined by 77 - 1 ] Jr(C/)dV V

(9.24)

Jr(Cis)

where V is the volume. For exothermic reactions (/3 > 0) a sufficient temperature rise due to heat transfer limitations may increase the rate constant kv, and this increase may offset the diffusion limitation on the rate of reaction (the decrease in reactant concentrations CA), leading to a larger internal rate of reaction than at surface conditions CAs. This, eventually, leads to rl > 1. As the heat of reaction is a strong function of temperature, Eq. (9.24) may lead to multiple solutions and three possible values of the effectiveness factor may be obtained for very large values of/3 and a narrow range of ¢h values (approximately 0.47-0.49). In common catalytic reactions,/3 is usually <0.1, and therefore, we do not observe multiple values of the effectiveness factor. The criterion fly < 4(1 +/3) may provide a good estimate of the stability condition.

(9.25)

9.2

459

Nonisothermal reaction-diffusion systems

Example 9.2 Maximum temperature difference in the hydrogenation of benzene Consider the hydrogenation of benzene, which is exothermic with a heat of reaction 50 kcal/mol. For the catalyst pellet containing 58% Ni on Kieselguhr Harshaw (Ni-0104P), the effective thermal conductivity and diffusivity are 3.6 × 10 -4 cal/(cm s K) and 0.052 cmZ/s, respectively. For a benzene surface concentration of 4.718 × 10 -6 mol/cm 3, and a surface temperature of 340 K, from Eqs. (9.18) and (9.19), /3--

ATma x -- ( T -

(50,000)(0.052) (4.718× 10-6 ) (3.6×10-4)(340)

=0.1

Ts )max -- (-AHr)DeCAs ke =/3T~ = (0.1)(340) = 34 K

This result shows about a 34°C maximum internal temperature difference for this highly exothermic reaction (Froment and Bischoff, 1979).

Example 9.3 Effectiveness factor for first-order irreversible reaction-diffusion system Consider a first-order reaction occurring on the pore walls of a catalyst with equimolar counter diffusion. Assume that isothermal conditions are maintained, and a catalyst with simple slab geometry is used (Figure 9.1). If the y-coordinate is oriented from the centerline to the surface, the steady-state reaction diffusion equation for reaction A ~ B between reactant A and product B is

d2CA D~ ~ - ( - v A)kvCA = 0

(9.26)

dy 2

with the boundary conditions

CA(L)-CAs dC A (0) __ 0

dv

(surface concentration) (symmetry at center line)

where De is the effective diffusivity and k,, the reaction rate constant based on pellet volume. In terms of surface rate constant ks, we have kv = kspsas, where as is the surface area and Ps the surface density. This linear differential equation has the general solution

C A ( y ) = l , exp "~De

+12exp

~De)

where I1 and 12 are the constants and are estimated from the boundary conditions. The solution is C A(y) _ (cosh x/kv/De )y CAs

(9.27)

(cosh x/kv/De )L

Figure 9.2 shows the concentration profile for various values of Lx/kv/D e . This solution shows that the diffusion resistance causes a concentration profile to exist in the pellet when the reactant cannot diffuse in from the bulk sufficiently rapidly. If the resistance is small due to a large value of De, then the concentration profile becomes flat, while it will behave conversely for a large diffusion resistance. In practice, however, the possible adverse effect of diffusion resistance on the rate of reaction is highly compensated by the enormous increase in surface area of the pores. The effectiveness factor r/is a measure of how much the reaction rate is lowered because of the diffusion resistance, and Eq. (9.24) is defined by

l/Vp)f kvCAdVp

rate of reaction with diffusion resistance m

rate of reaction without diffusion resistance

kvCAs

460

9.

Coupled systems of chemical reactions and transport processes

1 "

¢=0.5

0.9-

_________

~

0.80.7-

_._______/-

0.6<

q9

/

0.50.40.3-

,-3

0.20.1-

,=10

0 0

0.2

0.4

0.6

0.8

1

y/L Figure 9.2. Change and average value of reactant concentration within a catalyst pore as a function of the Thiele modulus.

_

=-

0.1-

0.01 0.01

0.1

1

10

100

4, Figure 9.3. Change of effectiveness factor with modules for a simple-slab geometry.

where kvCAs represents chemical reactions at surface conditions. Substituting the concentration profile into this equation, we have r/-

tanh 4~

(9.28)

where 4~ is the Thiele modulus defined by L~[kv/D e . Therefore, the actual reaction rate is (JrA )obs = rlJrA(CAs )

(9.29)

Figure 9.3 is a plot of Eq. (9.28), which shows that if ~b ~ 0 then r / ~ 1, which means there is no considerable diffusion resistance. As diffusion resistance increases, we have ~b ~ ~ and hence r / ~ 0. The latter can occur not for small diffusivity, for large pellet size L, or for very fast reaction rate, or for all three factors. This regime where the diffusion strongly affects the rate of reaction is called strong pore resistance. For a first-order reaction, a general criterion of Jr,obs L2

DeGAs indicates that there are pore diffusion limitations.

>> 1

(9.30)

9.2

9.2.4

461

Nonisothermal reaction-diffusion systems

Effectiveness Factor for Various Geometries

For a spherical pellet, from the steady-state mass balance

D e l d F 2 dCA -kvC A -0 r- dr

(9.31)

dr

and using the similar boundary conditions, the effectiveness factor becomes 3 b coth b - 1 b

(9.32)

b

where b = Rx/kv/De, and R is the sphere radius. For a cylinder and other geometries, a general modulus for a first-order reaction is

_Vp /kv cb-~

(9.33)

ap ~ De where lip and ap are the volume and external surface area of the pellet. A general modulus for an nth-order irreversible reaction is

__ _Vp //7 + 1 (--V A)kv(CAs) n-I _

lJ) --

.t

ap ~

2

n --> 0

(9.34)

De

The effectiveness factor in terms of/3 and y for an nth-order reaction may be calculated from a finite series for a region of low/3 (Tavera, 2005)

rt - ~ ' 0 5

exp 1+/3

1 - exp(-A)

k=0

k' •

/3 -< 0.1

(9.35)

where

A

z

(1+~ 2)

For a first-order reaction, this equation reduces to

oxp l+t

i

,-

+

/3 --< O. 1

(9.36)

With Eqs. (9.35) and (9.36), overestimation of the nonisothermal effects does not exceed 5% (Tavera, 2005).

9.2.5

External Diffusion Resistance

To consider external diffusion resistance for a first-order reaction, we need to determine the surface concentration of reactant Cas from the boundary condition

kg(Cb-CAs)-De[dCA I dy

s

(9.37)

The solution of this equation in terms of the bulk concentration Cb is

CAs --Cb

( y/L ) cosh ch cosh 05 + (Dedp/Lkg)sinh 05

(9.38)

462

9.

Coupledsystems of chemical reactions and transport processes

This equation, when is used in the definition of the effectiveness factor based on the fluid bulk concentration Cb, leads to the combined resistance of fluid and particle 1 _1+

rig

_

+m

rt kgL/De tanh4~

(9.39)

kg

Example 9.4 Effectiveness for a first-order reversible reaction Consider a first-order reversible reaction

R < kf >P. The rate of reaction is kb Jrv (CR) = ~7Y7[(1 + where K =

kf/kb and C = Cas + Cps = (1 +K)

K)C R -C]

CR, eq. The mass balance equation is

De d2CR = Jrv(CR) dy 2 After the first integration of this equation with constant De, we have

CRs __ kvDe ~s I DeJrv(CR)dCR [ ( I + K ) C R -C]dC R K

CR,eq

CR,eq

K

2

R'eq) -- C(CRs - CR'eq)

]

This equation can be used in the generalized effectiveness factor obtained from Eq. (9.24)

De dCR (L)

7-y

]1/2

,5 ICc

LJrv(CRs ) - L[Jrv(CRs) ] ~ DeJrv(CR)dCR

{ [

L[Jrv (CRs)]

kvD e I + K ( C _ C 2 K

]}12

(9.40)

R,eq ) -- C(CRs - CR,eq )

2

which yields an asymptotic effectiveness factor for Cc = CR, eq. Here, Cc is the reactant concentration at the centerline and may be assumed as the concentration of the reactant at equilibrium

r / ~ ~-7, where

9.2.6

= n

ap

External T e m p e r a t u r e Gradients

The balance Eqs. (9.16) and (9.17) for simple slab geometry are

0--De

d2CA 2 -kvCA dy

d2r + (- a G )kvCA 0=ke dy--T

K

(9.41)

9.2

Nonisothermal reaction-diffusion systems

463

As before these balance equations can be combined as follows

~2/ DeCA + ~keT / - 0

dv 2

( - AH r )

and integrated twice from the pellet center to the surface using the following boundary conditions

De

dCa dT - kg(¢ b - GAs), ke ~ - - hf(Tb -Ts) dv av •

where Cb and Tb are the fluid bulk concentration and temperature. The integration yields the following sum of external and internal temperature differences

kg De T - Tb - (-AH r ) ht----f(Cb--CAs)+-(--~Hr)-f--(CAstQ --CA )

(9.42)

The maximum temperature difference occurs when CR = 0, and after rearranging terms, the equation above becomes

~axTb ~b:~bESh l1 ~CAsi -+-~~ ----7 Nu Cb J The ratio

Cas/Cb is obtained in terms of the observable

rate Jrv, obs

1L De(dCaj kg Jrv,obs = T f 0 JrvdY- T dy L --T(Cb-CAs) where Sh' is the modified Sherwood number Sh' = at bulk fluid conditions is defined by

(9.43)

(9.44)

kgL/De, Nu' the modified Nusselt number Nu' = hfL/ke, and/~b

D e

/~b -- ( - / ~ H r ) ~ C b t%l b

(9.45)

After substituting Eq. (9.44) into Eq. (9.43), we have

1)]

Tmax--]'b=/~b I1 +~b ( 1 Tb

Nu'

Sh'

~946,

where (I) b is the observable Weisz modulus and is defined by

-

with 4~-

L2~DcCb Jrv'°bs

= Sh' ( G 1 -A st~)b

-

~2

L~Jkv/D~ and 1/r/g - l/r/+ 052/Sh'

The internal and external temperature differences can be obtained from Eq. (9.46)

[[

-)

,

Tmax Tb __ /~b I1 -- (I)b(~-7) 1 ~ int

!~axTb ~b)~b,i 1/ ext NUT/"

(9.47)

464

9.

Coupled systems of chemical reactions and transport processes

Example 9.5 Maximum overall temperature difference in the hydrogenation of benzene Consider the hydrogenation of benzene, which is exothermic with a heat of reaction 50 kcal/mol. For a catalyst pellet containing 25% Ni-0104P, 25% graphite, 50% 7-A1203 (Harshaw A1-0104T), the effective thermal conductivity and diffusivity are 3.5 × 10 -3 cal/(cm sK) and 0.035 cm2/s, respectively. The fluid bulk concentration of benzene is 5.655 × 10-6mol/ cm 3, and the fluid bulk temperature is 412 K. The characteristic length of the pellet is 0.31 cm. The observed rate for the reaction is 22.4 × 10 -6 mol/(gcat s) and the density of the catalyst is 1.57 g/cm 3. The modified Sherwood and Nusselt numbers are 401 and 1.35, respectively. From Eqs. (9.6) and (9.7), we find the values of/3b and ~bb /3 = (50,000)(0.035)(5.655 × 10 -6) = 0.007 (3.5×10-3)(412) The reaction rate is

Jrv,obs = 22.4 × 10 -6 mol/(gcat s) × 1.57 gcat/cm 3 = 35.168 × 10 -6 mol/(cm 3 s) L2Jrv obs

~b = ~ '

DeCb

=

(0.31)2(35.168×10 -6 ) (0.035)(5.655×10 -6 )

=16.7

The external temperature difference can be obtained from Eq. (9.46)

( /

,'"(T-ax - Tb ,]ext -- Tb/~b(I)b 1

Nu'

007167(1 /

=

= 36 K

The experimental value is 11 °C. This lower temperature may indicate that the internal pellet concentration may not be exactly zero, contrary to expectations. The internal temperature difference is

[ (ill

(Tmax --Tb)in t = Tb/~b 1--~0b ~

/

=(412)(0.007) 1 - - ~ }

2.8K

This result shows close agreement with the experimental value of 2 K and the low internal temperature gradient, as the value of/3 is small.

9.2.7

Criteria for Negligible Transport Effects

After Dekker et al. (1995), criteria for negligible transport effects in steady-state kinetics are as follows. The criterion for negligible external mass transport resistance in steady-state kinetics is

Jr,obs

g(Sp/Vp)Cb

< 0.05

(9.48)

For negligible intraparticle mass transport resistance, the criterion is rt4~2 =

Jr obsL2 n + 1 ' DeCA

< 0.1

(9.49)

2

For negligible external heat transport resistance, the criterion is

/~y(Cbcb-CA ) < 0.05

(9.50)

The criterion for negligible internal heat transport resistance is (9.51)

9.3

465

Chemical reaction with coupled heat and mass flows

The criterion for external mass transport resistance is

kga

--t

>_ 2.9

(9.52)

where a is the particle surface area per unit volume of packed bed, and c the void fraction of the catalyst bed. For negligible external and internal mass transport resistance, the criterion is

De 8R 2

9.3

t -> 0.25

for Bi m -

kgR De

--> 20

(9.53)

CHEMICAL REACTION WITH COUPLED HEAT AND MASS FLOWS

Nonisothermal reaction-diffusion systems represent open, nonequilibrium systems with thermodynamic forces of temperature gradient, chemical potential gradient, and affinity. The dissipation function or the rate of entropy production can be used to identify the conjugate forces and flows to establish linear phenomenological equations. For a multicomponent fluid system under mechanical equilibrium with n species and Nr number of chemical reactions, the dissipation function q~ is * -- Tq) -

--~-JqVTi=l

J i ( V # i ) + Z JrjA/ j=l

>- 0

(9.54)

where, Jq is the vector representing heat flow, d; the vector representing mass flow, kc; the chemical potential of species i, A the affinity A = -EVil, i, v the stoichiometric coefficients, and Jr the reaction velocity. Equation (9.54) consists of scalar processes of chemical reactions and vectorial processes of heat and mass flows, and excludes viscous flow, electrical, and magnetic effects. Equation (9.54) and the rate of entropy production may be used to identify the conjugate forces and flows to establish linear phenomenological equations. Excluding the coupling of chemical reactions with the heat and mass flows, and using the relationship/X/x a = (OtJ6A/OCA)z~iCA,Eq. (9.54) reduces to 1

- -'--Jq~T-JAAT,pVC

T

A ~ 0

(9.55)

where AT,P --(1--t-(VACA/VBCB))(0[,.I,A/0CA )T,P' and V, is the partial molar volume of component i. Assuming that the reaction-diffusion system is not far from global equilibrium, the linear phenomenological equations based on Eq. (9.55) are 1 - J A - LAAAT.p~'CA + --~ L A q V T - DeVCA + DS,eVT 1

(9.56)

1 - J q - LqAAT,pVC 4 - + - T L q q V T - DD,eVC A -+-keVT

(9.57)

where DD,e is a coefficient in m 2 J/(mol s) related to the effective Dufour effect, and Ds, e a coefficient in mol/(m s K) related to the effective Soret effect (thermal diffusion). When there is no volume flow, the mass flow JA is JA = --LAAAT,p~7CA

(9.58)

and comparing Eq. (9.58) with Fick's law J = -DeVCA, the coefficient LAA is related to the effective diffusion coefficient by LAA --

De AT,P

(9.59)

Using Fourier's law Jq -- -keVT in Eq. {9.57), the primary coefficient Lqq is related to the effective thermal conductivity ke by

Lqq = k~T

(9.60)

466

9. Coupledsystems of chemical reactions and transport processes

The thermal diffusion coefficient for species A is

DT - -

LAq

(9.61)

CAT

For liquids, the diffusion coefficient D is of the order of 10 -5 cm2/s, and the thermal diffusion coefficient D T is of the order of 10-8-10 -1° cm2/(s K). For gases, the order of magnitude for D and D T is 10 -1 cm2/s, and 10-4-10 -6 cm2/(s K), respectively. By using the flows JA and Jq from Eqs. (9.56) and (9.57), respectively, in Eqs. (9.14) and (9.15), we have 0 = V'(DeVC ~ + DS,eVT' ) - kvC'A

(9.62)

0 : ~'(DD,e~C~k nt-kel~T')-+-(-ASr) kvCrA

(9.63)

where T' and C), represent the temperature and concentration in the coupled system. As before, the elimination of the reaction terms from Eqs. (9.62) and (9.63) yields q~'= 1-/3'(0-1)

(9.64)

where

O'

C'A" q9' r' = Cs' = Tss' and

~ , _ (Oe(-AHr)-kDD,e)Cs - (ke + Os'e(-~r-/r) ) ~

(9.65)

and the maximum temperature difference becomes AT' /3' ~ (qgcenter- 1) . . . . .

(9.66)

rs

The modified dimensionless group/3' represents the ratio between the chemical reaction's rate of heat and rate of heat conduction when only the heat and mass flows are coupled (Demirel, 2006). By disregarding the coupling effects, we would have/3'=/3. Using Eqs. (9.62) and (9.63) with the Arrhenius equation kv = koe-E/Rr, we have the transient forms of the coupled heat and mass flows for a single component

OC'A - V.(DeVC~ +Ds,eVT')-(koe-U/Rr')C' a Ot

(9.67)

pCpOT'ot ~'( DD'e~7C~ q-kegT'+[-(AHr)] ) (k°e-E/"r')

C'A'

(9.68)

The initial and boundary conditions are the same as those given for Eqs. (9.20) and (9.21). For a simple plane geometry and one-dimensional unsteady state, Eqs. (9.67) and (9.68) become

00' 0T

D

020' -+E02¢-----~P' 02 Z ~20 exp [T (1 - ~ , ) ] 0Z2

1)]

1 Oq~' O2q)' ~+ ~+13'q520'exp T 12 Le Or OZ2 (.00Z 2 where

Z ~y

L'

Ds,e Ts

DD,eCs

DeC s

keTs

L2ko exp(E/RT s) E k~/pCp _ Ole 7"-- Det 492_ De ' Y - ~ s s ' Le --~-' De De

(9.69)

(9.70)

9.3

467

Chemical reaction with coupled heat and mass flows

Equations (9.69) and (9.70) represent the modeling of reaction-diffusion systems with the thermodynamically coupled heat and mass flows excluding the coupling effects due to reaction. After combining Eqs. (9.64), (9.69), and (9.70) steady-state balance equations with the coupled heat and mass transfer become

de0 ' (y/3'(0'-l) ) ( 1 - fi'e) dT~ = 0520' exp - 1 - / 3 ' ( 0 ' - 1)

(9.71)

(9.72)

Since the dynamic behavior of a reaction-diffusion system may be more apparent with state-space diagrams, the temperature and concentration profiles are replaced with the spatial integral averages obtained from I

1

()

0

(9.73)

Example 9.6 Coupled heat and mass flows in oxidation of CH3OH to CH20 The modeling Eqs. (9.69) and (9.70) are used with parameters for the exothermic catalytic oxidation of CH3OH to CH20 00'

020 ' --+s~-q~20'exp c)z2 Oz2

Or 10q~' Le Or

02~ '

+o)~+ Oz2 Oz2

t, 1 -

[( 1)]

exp y 1 -

-~

where Le is the modified Lewis number. The initial and boundary conditions are

o(o,z)- 1, O ( r , 1 ) =

1,

dO(.~,O)

= 0

dz d p ( r , O) ¢ ( 0 , z ) - 1, ¢ ( r , 1 ) - 1, ~ = 0 & The other parameters are defined by

DDxCs

Ds,~T~ g - - ~ ,

DeCs v L

Z =:z---,

to-

keys

E D~t 052 - - L2k o exp(E/RT~) De ' Y ~ s ' L2 '

T - - ~ -

ke/pCp L e - ~De-

ole D~

Here % is the effective thermal diffusivity. The nondimensional group y is called the Arrhenius group, and represents a nondimensional activation energy for the chemical reaction. MATHEMATICA is used to solve the partial differential simultaneous equations of the mathematically and thermodynamically coupled systems given in Eqs. (9.69) and (9.70). g = 16.0; le = 0.0015; f - 1.1; e = 0.001; w = 0.001; b = 0.0109; da =f'f; Print[" ~b= ",f,"; y = ",g,"; [3 = ",b, ". le = ", le, "; e = ",e,"; m = ",w, "; oxidation of CH3OH to CH20"]; eq 1 = D[c 1 [t,x], t] = = D[c 1 [t,x] ,x,x] + e* D [q[t,x] ,x,x] - d a * c 1 [t,x] *Exp [g* ( 1.0 - 1. O/q[t,x])]; eq2 - D [q[t,x] ,t] --= le*D [q[t,x] ,x,x] + le*weD [c 1 [t,x] ,x,x] + le*b*da*c 1 [t,x]*Exp [g* ( 1.0- 1.0/q[t,x])]; soll -NDSolve[{eql, eq2, cl [0,x]--- 1.0,cl[t,1] 1.0, Derivative[0,1][cl][t,0]-- 0.0, q[0,x] == 1.0,q[t, 1] - - 1.0,Derivative[0,1] [q] [t,0] - - {0.0},{cl,q},{t,0,1 },{x, 1,0},PrecisionGoal --, 0.08,MaxStepSize --, 0.0005, MaxSteps --, Infinity]:

468

9.

Coupled systems of chemical reactions and transport processes

Plot3D [Evaluate [c 1 [t,x]/.soll [[1 ]]] ,{t,0,1 },{x, 1,0},PlotPoints --, 40, A x e s L a b e l --, {"'r"," z "," 0' "}, DefaultFont --, {"Times-Roman", 13}]' Plot3D [Evaluate [q[t,x]/. sol 1 [[ 1]]], {t, 0,1 }, {x, 1,0},PlotPoints --, 40, AxesLabel --, {'"r"," z "," cp' "},DefaultFont --, {"Times-Roman", 13}]' ParametricPlot [{NIntegrate [ Evaluate [c 1 [t,x]/.sol 1 [[ 1]]], {x,0,1 }], NIntegrate [ Evaluate [q[t,x]/.sol 1 [[ 1]]], {x,0,1 }] }, {t,0,1 },PlotRange --, All, Frame --, True, GridLines --, Automatic, PlotStyle --, {PointSize[0.007],Thickness[0.009]}, FrameStyle --, Thickness[0.007], FrameLabel --, { "0' ", "~ '"}, RotateLabel --, True, DefaultFont -, {"Times-Roman", 13}]' Plot[NIntegrate[ Evaluate[c 1 [t,x]/.soll [[1 ]]], {x,0,1 }],{t,0,1 }, Frame --, True, AxesStyle --, Thickness[0.007], FrameStyle --, Thickness[0.007], DefaultFont --, {"Times-Roman", 13}, PlotStyle --, {PointSize[0.007],Thickness[0.009]}, GridLines --, Automatic, F r a m e L a b e l --, {"~-.... 0 '(,r)"}]' Plot[NIntegrate[ Evaluate[q[t,x]/.soll [[1]]], {x,0,1 }],{t,0,1 }, Frame -, True, AxesStyle --, Thickness[0.007], FrameStyle --, Thickness[0.007],DefaultFont --, {"Times-Roman", 13}, PlotStyle --, {PointSize[0.007],Thickness[0.009]},GridLines -, Automatic, F r a m e L a b e l --, {"~-","~ '(~-)"}]' Table 9.1 shows some of the experimental and assumed values of the parameters considered for catalytic oxidation of CH3OH to C H 2 0 with/3 = 0.0109 and hence display relatively fewer nonisothermal effects. The thermal diffusion coefficient is usually smaller by a factor of 102-103 than the ordinary diffusion coefficient for nonelectrolytes and gases. Therefore, for the present analysis the values for e and co are assumed to be 0.001. y = 16.0, Le = 0.0015, ~h = 1.1, /3 = 0.0109, e = 0.001, co = 0.001

1

0.9 0' 0.~ 0.'

(a)

1.00002 1.00002 rp' 1.00001

1

1.00001

0.8

0.6

1

0

"--... 0.2 ~

3.4 0.4

(b)

I:

z

0.2

0.6 0.8 1

0

Figure 9.4. Dynamic behavior of thermodynamically coupled nonisothermal reaction-diffusion system of catalytic oxidation of CH3OH to CH20: (a) concentration surface, (b) temperature surface. The parameters used are in Table 9.1.

9.3

469

Chemical reaction with coupled heat and mass flows

The numerical solutions from Mathematica are obtained with precision goal = 0.08, maximum step size = 0.0005, and maximum steps = infinity. Figure 9.4 shows the dynamic behavior of coupled values of concentrations and temperatures for the catalytic oxidation of CH3OH to CH20. The surface of constant temperature closely follows the change in concentrations. Figure 9.5 shows the changes of the spatial integral averages of concentration and temperatures with time. Figure 9.5c shows the state-space representation of temperature versus concentration when the time changes from 0 to 1. For catalytic oxidation of CH3OH to CH20, the temperature reaches its maximum value when the dimensionless concentration is approximately 0.97, as seen in Figure 9.5c.

1.00002 1.00001 1.00001 1.00001 1 .oooo I

0.8

0.75

0.85

0.9

0.95

1

O' (a)

0.95 0.9 0.85 0.8 i

0.75

,

!. . . . . .

0

0.2

0.4

0.6

0.8

(b) .,

.

.

.

1.00002 1.00001 1.00001

i

1.ooooI 1.00001

S 0

.............

0.2

0.4

0.6

0.8

!

(el Figure 9.5. Spatial integral averages for the catalytic oxidation of CH3OH to CH20; (a) change of concentration with time, (b) change of temperature with time, (c) change of temperature with concentration when the time varies between 0 and 1. The parameters used are in Table 9.1.

470

9.

Coupled systems of chemical reactions and transport processes

This analysis considers the thermodynamic coupling between heat and mass flows in an industrial reactiondiffusion system with a low value of/3. Modeling with the coupling effects of Soret and Dufour opens the path to describing more complex reaction-diffusion systems by adding the two new controlling parameters e and w.

9.4

COUPLED SYSTEM OF CHEMICAL REACTION AND TRANSPORT PROCESSES

Adenosine triphosphate (ATP) as a universal free energy transmitter undergoes the following turnover reaction ADP + Pi = ATP + H20. This reaction represents a simplified synthesis of ATP and hydrolysis of ATE which releases energy utilized in the transport processes. The quoted standard Gibbs free energy AG ° of this reaction is 30.5 kJ/mol at physiological conditions of 37°C, 1 atm, pH 7, pMg = 3, and 0.2 M ionic strength. However, the reported values of AG ° vary between 28 and 37 kJ/mol. The relations AG ° = -RTlnK' with the apparent equilibrium constant K' = {[ATP]c°/[ADP][Pi]}eq with c o representing the standard concentration may not be enough to determine AG ° directly. By introducing a H+-translocating ATP synthase, the coupled reaction cycle above becomes ADP + Pi + nHi+n = ATP + H 20 + nHo+ut where "in" and "out" denote two phases separated by a membrane, and n is the ratio H+/ATP which is the level of transmembrane proton transport for each ATP molecule to be synthesized. The apparent equilibrium constant K ' - {[ATP]c°(H+)out/([ADP][Pi](H+)in}eq is related to the pH level by -AG ° log(K') = ~ + nAPHeq 2.303RT where ApH is the transmembrane difference between outside and inside pH. The H+/ATP coupling ratio of 4 is a widely accepted value and determined from energy balance and proton flow measurements. The respiratory electron transport chain in the inner membrane of mitochondria creates a proton motive force across the membrane, which is used in synthesizing ATE Consequently, the hydrolysis of ATP is coupled in transporting substrates, leading to osmotic work of active transport and other mechanical work. Moreover, the ATP synthesis, in turn, is matched and synchronized to cellular ATP utilization according to the chemiosmotic theory. Enzyme-catalyzed reactions, including the electron transport chain and proton translocation, are composed of series of elementary reactions that proceed forward and backward. One of the methods in describing this thermodynamically and mathematically coupled complex chemical reaction-transport system is the nonequilibrium thermodynamic model, which does not require the detailed knowledge of the system. Lateral gradients of ions, molecules, and macromolecules may occur in an anisotropic medium, such as between mitochondrial compartments. An anisotropic medium or a compartmental structure may support the coupling between a chemical reaction and the transport processes of heat and mass according to the Curie-Prigogine principle. Such a coupling requires interactions between the scalar process of a chemical reaction and the vectorial process of heat and mass transport. In the vicinity of global equilibrium, coupled mass and heat flows and the reaction velocity can be derived by the linear nonequilibrium thermodynamic approach without the need for the detailed mechanisms of coupling phenomena. The modeling equations may be helpful in describing and controlling the evolution of some complex systems, such as reaction-diffusion phenomena with heat effects and active transport in biological systems.

9.4.1

Balance Equations

Consider a reversible homogeneous elementary reaction between a substrate S and a product P S<

kf kb

>P

where kf and kb are the forward and backward reaction rate constants, respectively. This type of reaction system is highly common and plays an important role in chemical and biological systems, such as unimolecular isomerization,

9.4

Coupled system of chemical reaction and transport processes

471

enzyme kinetics, or racemization of molecules with mirror-image structures. For this reaction-transport system, general balance equations of mass and energy under mechanical equilibrium are _

'3Cs

- -V. N s + vsd r

- -

'3Cp _

(9.74)

(9.75)

4- vpd r

-V'Np

ht

pCp

where

~-/r is the heat of reaction.

OT

- -V.q +(-AHr)J r

'3t

(9.76)

The symbols Ns and Np denote the total flows of species S and P defined by Ns - Js + Csv

(9.77)

Np - Jp + Cpv

(9.78)

where v is the molar average velocity. The total heat flow q is

q-Jq+~_~JiHi

(9.79)

i

where H; is the partial molar enthalpy of species i. The reaction velocity is (9.80)

d C s _ dCp _ Jr ~'sdt vpdt

where the parameters Vs and Vp are the stoichiometric coefficients, which are negative for reactants (Vs- - 1 ) . Equations (9.74)-(9.76) show that the changes in concentrations and temperature are due to diffusion and convection, and chemical reaction. Using only the molecular transport in these differential equations, we have '3Cs '3t

(9.81)

- - V. a s + vsd r

'3Cp _

(9.82)

-- --V" Jp + vpd r

- -

'3t OT

p u p (J[ - - - V ' J q

+(-AHr)J

r

(9.83)

where Jq is the vector of the reduced heat flow Jq - q - ~ J;H;. By using Fick's and Fourier's laws in the onei=1

dimensional y-direction, Eqs. (9.81)-(9.83 ) become '3Cs - Ds,e O2Cs + V s J r

,31

0Cp

'3t ~T

0v 2

-

Dp,e 02 Cp

'3v2

+ VpJ r

(9.84)

(9.85)

O:T

p C p -~t -- ke - - + ( - A H r ) J r Ov 2

(9.86)

472

9.

Coupled systems of chemical reactions and transport processes

I I I I I I I

, I I I I

T

"5 Css

Cs

I y=+L

y=-L

I

I y=0

I I I

Figure

9.6. Schematic temperature and concentration profiles in a thin film.

where Di, e is the effective diffusivity for component i, and ke the effective thermal conductivity. Assuming a simple slab geometry (Figure 9.6), the initial and boundary conditions are t = 0

y = ±L

y = 0

C S -- CS0 Cp = Cpo

(9.87)

T = TO

C s -- CSs Cp -- Cps T = Ts (surface conditions)

OC s _ OCp _ OT _ 0 Oy Oy Oy

(symmetry conditions)

(9.88)

(9.89)

where L is the half-thickness of the slab. At stationary state, eliminating the reaction terms from Eqs. (9.84) and (9.85), and integrating twice with the boundary conditions given in Eqs. (9.88) and (9.89), concentrations of the species are related to each other by 0p = a 1 + a2(1- 0s)

(9.90)

with _ Cs 0s - < ,

_ Op

Cp

_ Cps

KCss , a 1

KCss , a 2 -

Ds,e KDp,e

Cps , K-

Css

(9.91)

The value of al determines the direction of the reaction; the net reaction is toward the P if al < 1. Similarly, eliminating the reaction terms from Eqs. (9.84) and (9.86), the temperature is related to the concentration by q~: 1+/3(1 - 0s) where T

(-AHr)Ds,eCss

The value of/3 is a measure of nonisothermal effects. As/3 approaches zero, the system becomes isothermal.

(9.92)

9.4

Coupled system of chemical reaction and transport processes

473

Example 9.7 Diffusion in a liquid film with a reversible homogeneous reaction Consider the reaction

S<

/q-

>P. After relating the two concentrations of species S and P by Eq. (9.91), at stationary state, Eq. (9.84) kb becomes

d20s

(9.93)

(Da s + Dap)0 s - (Dasa 1 + Dap)

-

dz 2

with the boundary conditions

dOs

(9.94)

0s(+_L) = 1, - - 7 ( 0 ) = 0 dz where

krE

Y

z = - - " Da s L'

Dap-

Ds,e

kbE Dp,e

Das and Dap are the Damk6hler numbers, and represent the ratios of the forward and backward reaction rates to the diffusion velocities Di,e/L; they measure the intrinsic rates of the reactions relative to those of the diffusions, and represent an interaction between reaction and diffusion. For the product, P, an expression similar to Eq. (9.94) can also be derived. Figure 9.7 displays the concentration profiles obtained from MATHEMATICA for reactant S and product P for two different sets of Damkohler numbers when 3' = 0 Da s - 5 0 . 0 ,

Dap-40.0,

Da s - l . 0 ,

Dap=0.5,

y=0.0 y=0.0

When the chemical reaction is fast (with large Damkohler numbers or with very low diffusivities) the reactant and product reach their equilibrium concentration throughout most of the film. The concentration gradients are very steep at the nonequilibrium region. The set of parameters Das = 1.0, Dap = 0.5, 3' = 0.0 represent slow reaction and nonequilibrium film.

0.8

"

m -...

m

0.6

0.4

0.2 ~.

•" .

"

"',

,

i

0.2

,

,

,

i

0.4

~

........

.

.

m

.

""

.

.

-

.

0.6

0.8

Figure 9.7. Concentration profiles of reactants and products for diffusion in a stagnant film with reversible homogeneous chemical reaction: Das = 1.0, Dap = 0.5, y = 0; bold dashed line is for 0s; gray dashed line is for 0p; Das = 50.0, Dap = 40.0, y = 0; bold solid line is for es, and gray solid line is for 0p.

474

9.

Coupled systems of chemical reactions and transport processes

MATHEMATICA code: (*a for S and b for P *) al : 50.0;bl = 40.0;g: 0.0; a2 : 1.0;b2 : 0.5; soil = NDSolve[{ ta 1"In] == (a 1+b 1)*tal [n]-(a l*g+b 1),tal [0]== 1.0,ta 1'[1 ] ==0.0, tb 1"[n] ==(a i +b 1)*tb 1 [n]-(a 1*g+b 1),tb 1 [0] ==0.0,tb 1'[1 ] ==0.0, ta2" [n] = = (a2 +b2)*ta2 [n]-(a2*g+b2),ta2 [0] = = 1.0,ta2' [ 1] ==0.0, tb2" [n] ==(a2 + b2)*tb2 [n]-(a2*g+b2),tb2 [0] = =0.0,tb2' [ 1] = =0.0}, {tal,tbl, ta2,tb2},{n,0,1}] Plot [Evaluate [{ta 1 In] ,tb 1 [n], ta2 [n],tb2 [n]}/.sol 1],{n,0,1 }, PlotRange ~ {{0,1 },{0,1 }}, Frarne ~True, GridLines ~Automatic, GridLines--,Automatic, PlotStyle~{{Thickness[0.007], Dashing [{0.0, 0.0}]}, {Thickness[0.005], Dashing [{0.0, 0.0}]}, {Thickness [0.007], Dashing [{0.025, 0.02}]}, {Thickness[0.004], Dashing [{0.025, 0.02}]}}, FrameStyle ~ Thickness [0,005], FrameLabel ~{"z", "0s, 0p"}, RotateLabel ~ True, DefaultFont ~ {"Times-Roman", 14}];

9.4.2 Linear Phenomenological Equations For a multicomponent fluid system under mechanical equilibrium with n species and Nr number of chemical reactions, the entropy production function is

o

(V/Ai)T'P -Jr ~ Aj T j=l --f- Jrj > 0

,q

(9.95)

i=1 where Ji is the vector of mass flows, }tZ i the chemical potential of species i, and A the affinity A = -Evil.Li; ifA > 0, the reaction proceeds toward the right. Equation (9.95) is derived from the general balance equations including the entropy balance and the Gibbs relation, and identifies a set of independent conjugate flows Ji and forces Xk to be used in the linear phenomenological equations when the system is in the vicinity of global equilibrium

Ji - Z LikYk

(9.96)

Equation (9.95) excludes viscous flow, electrical, and magnetic effects. For the chemical system considered in Eq. (9.72), Eq. (9.95) yields ~ = Jq "~7(T1--)- Js "(~7/zS)T'~P T - JP "(V/zP)T'~P T + Jrs A T -> 0

(9.97)

where

n-1 OILi (V/LLi)T,P = 2 - ~ / V C i i=1 By using the Gibbs-Duhem equation at constant temperature and pressure Cs~TIAS -k-Cp~7/.Zp = 0

(9.98)

JsVs + JpVp = 0

(9.99)

and no volume flow condition

9.4

Coupled system of chemical reaction and transport processes

475

where Vi is the partial molar volume of species i, Eq. (9.97) becomes

Op

-

1 -J

1

A

V T - 'Is ~- asVCs + J,s --T >- 0

T2

q

(9.100)

where

"(' +

CS

for(Vs =Vp)

/T,p

Using the net flows and net forces based on Eq. (9.100), we obtain the linear phenomenological equations as follows 1

1

A

1

A

Lsq --~VT + Lsr -~-

'Is - -Lss y AsVCs 1

'I,t = -Lq s -T AsVCs - Lqq 7 VT + Lqr --T 1

Jrs

-

1

-Lrs -~-AsVCs - - Lrq - ~ V T

A

(9.102)

(9.103)

Lrr ~-

+

(9.101)

The above phenomenological equations obey Onsager's reciprocal rules, and hence there would be six instead of nine coefficients to be determined.

9.4.3

Degree of Coupling

The linear phenomenological equations help determine the degree of coupling between a pair of flows; the degree of coupling between heat and mass flows qsq and between the chemical reaction and the transport process of heat and mass flows, qsr and qrq are

-

qsq

-

Lsq (LssLqq)12, qsr ,

-

-

Lsr (LssLrr)l/2 , qrq

_

_

Lrq

(LrrLqq)l/2

(9.104)

Equations (9.101)-(9.103) assume coupling between the vectorial flows of transport processes and the scalar chemical reaction velocity. This type of coupling is possible in an anisotropic medium only according to the Curie-Prigogine principle. Consequently, the nonvanishing values of cross coefficients, Lsr, LrS, Lqr, and Lrq, must have vectorial characteristics. For example, during the active transport of sodium ions, in which the hydrolysis of ATP is coupled with the flow of sodium ions, the direction of flow is determined by the property of the membrane in the mitochondria. The medium may be locally isotropic, although it is not spatially homogenous. In this case, the coupling coefficients are associated with the whole system.

9.4.4

Efficiency of Energy Conversion of a Reaction-Diffusion System

The following entropy production function from Eq. (9.100) shows the input and output energies when there is no heat effect 1

A

1

1

- - ' I s -~ AsVCs + Jrs -~ - output + input -> 0

(9.105)

Applying the entropy production function above to the active transport in a biological cell, the chemical reaction term (Jrs(A/T)) represents the hydrolysis of ATE which can pump an ion in a direction opposite to the direction imposed by its thermodynamic force, and hence we have

,

)

-3 s-~hsvC s < 0

(9.106)

476

9.

Coupled systems of chemical reactions and transport processes

However, the hydrolysis of ATP can pump the ions only if some degrees of coupling exist between the reaction velocity and the mass flow. The efficiency of energy conversion for pumping a substrate with the help of a chemical reaction may be related to the degree of coupling by using Eq. (9.105) -

output input

-

-Js(1/T)AsVCs _ x((Lss/Lrr) 1/2x + qSr) -

Jrs(A/T)

[qsr

+(Lrr/Lss) 1/2 ]

(9.107)

where x is the ratio of thermodynamic forces defined by

(1/T)AsVCs (A/T)

x=

(9.108)

The optimal efficiency would be a function of the degree of coupling.

9.4.5

Phenomenological Coefficients

The diagonal elements of the coefficient matrix [L] may be identified by using Fick's, Fourier's, and the mass action laws. Comparing the first term on the right of Eq. (9.101) with Fick's law, J = -Ds,~VCs, yields Lss -

Ds,eT

As

(9.109)

Similarly, comparing the second term in Eq. (9.102) with Fourier's law, Jq = -keVT , yields

Lqq = keT2

(9.110)

The cross coefficients Lsq and Lqs may be represented by the Soret coefficient ST or the thermal diffusion coefficient DT, which are related to each other by

Lsq = STDs,eT2Cs = DTT2Cs

(9.111)

The Soret coefficient is the ratio of the thermal diffusion coefficient DT to the ordinary diffusion coefficient D

DT ST --

Ds,e

(9.112)

at steady state, and has the dimension of T-1. It changes in the range 10-2-10 -3 1/K for gases, nonelectrolytes, and electrolytes. The term Lqs(= DTT2Cs) is expressed by the Dufour coefficient DD

Lqs

DDCsT =

(9.113)

As

For Lqs = Lsq, we have D D -- DTTAs, which is proved experimentally. We may define two new effective diffusion coefficients, DT,e and DD,e, which are related to the thermal diffusion and the Dufour effect, respectively 1

DT, e --

Lsq --~ = STDs,eC s = DTC s

%

DD,e = L qs--f-= D DCs

(9.114)

(9.115)

With these newly defined primary and cross coefficients, Eqs. (9.101)-(9.103) become A Js - -Ds,eVCs - DT,eVT + Lsr "~-

(9.116)

9.4

Coupledsystem of chemical reaction and transport processes

A

Jq - - D D ~VC s - keV T + Lqr

-

'

Jrs 9.4.6

1

= -Lrs ~ AsV C s - Lrq ~ 1

1

V T nt-

477

(9.117)

-

T

kf Cs,eq A

R

(9.118)

T

Reaction Velocity

For the reaction S <

kf

kb

>P, the affinity is A -/Xs-/Xp. The local rate of entropy production due to chemical reaction is ~_A

~-Jr

(9.119)

where Jr is the reaction velocity. The phenomenological form of Jr is A Jr - L~r -~

(9.120)

where Lrr is the phenomenological coefficient. The reaction velocity is also defined in terms of the forward (f) and backward (b) reaction rates Jr = J r f - J r b

(9.121)

= kfC S - k bCP

The affinity is also related to the forward and backward reaction rates as follows

A - RTln Jrf / or

Jrf _Cskofexp(-Ef/RT)(A)

(9.122)

Cpkobexp(-Eb/RT) = exp

Jrb

where Ef and Eb are the activation energies for the forward and backward reactions, respectively. Using Eqs. (9.12 l) and (9.122), the reaction velocity Jr in terms of affinity A is given by

xpl ¢/)

(9.123)

Far from global equilibrium, the reaction velocity is not only related to affinity but also depends on the concentration of species. If we expand Eq. (9.123) and consider the near global equilibrium state ( A~ Rrl << 1), then we have a linear relationship between the reaction velocity and the chemical affinity for an elementary reaction Jr-

(9.124)

Jrf, eq A

R

T

So, the reaction rate is uniquely defined by the corresponding affinity, since Jf, eq becomes constant due to uniform concentration when a system is in the vicinity of global equilibrium with fast diffusion and heat conduction processes. Comparing Eq. (9.120) with Eq. (9.124), the coefficient Lrr is defined by Lr r _ Jrfieq _ kfCs,eq

R

R

_ No exp(-Ef/RT)

Cs,eq

(9.125)

R

where k0 is the frequency and Ef the activation energy for the forward reaction. Equation (9.125) indicates that the value of Lrr is dependent on the rate constant and consequently on the equilibrium concentration (Cs,eq) and the amount of chemical catalyst.

478

9.

Coupled systems of chemical reactions and transport processes

Linear flow-force relations are valid for chemical reactions not far from global equilibrium at which the Gibbs free energy ranges < 1.5 kJ/mol. However, some selected biological pathways occur at near global equilibrium conditions, and for some chemical reactions the linear flow-force relations can be used in wider ranges than usually expected (Cukrowski and Kolbus, 2005). By conservation of mass, some flow-force relations of enzyme catalyzed and other chemical reactions can be described by a simple hyperbolic tangent function. Therefore, a plot of reaction velocity versus affinity has three regions; the regions at very high positive and negative values of affinity, the reaction velocity is almost independent of affinity. In between these regions of the curve, reaction velocity varies smoothly leading to a quasi-linear region around the inflection point. This region extends the linear flow-force relations over a 7 kJ/mol with an error in the reaction velocity < 15%. This behavior is independent of the reaction rate constants, and mainly occurs due to conservation conditions.

9.4.7

Determination of Cross Phenomenological Coefficients

When we can control the temperature and concentration gradients, the coupling coefficients between the chemical reaction and the flows of mass and heat may be determined experimentally by using Eqs. (9.116)-(9.118)

Lrs = Lsr =

(Js)

(A/T) vcs=0,vr=0

O(A/T) vcs,vr = A(A/T) VCs,Vr

(9.126)

Similarly, we have

a(A/T) vc~,vr

Lrq--Lqr = ~-)VCs=O,VT=O

6(A-7-r) v c ~ , v r

For a stationary state closed system Js = 0, and we have A Js = 0 = -Ds,eVC s - DT,eVT + Lsr -~

LSr = Os'e

VCs

vv ) = LrS

(A/T---~+ oT'e (A/T)

(9.128)

(9.129)

Using A = RTln(Jrf/Jrb ) in Eq. (9.128), we get Lsr =

1

R ln(Jrf/Jrb )

(Ds eVCs + DTeVT)= Lrs '

(9 130)

'

On the other hand, at chemical equilibrium, where A = 0 and Jr = 0, we have

1

Jr = 0 = -Lrs ~

1

AsVC S -

Lrq ~ VT

(9.131)

And the two coupling coefficients are related to each other by

Lrq = - L r s T A

S VCs

VT

_

Lqr

(9.132)

By using the relationship -CSST = V Cs/VT at steady state, the cross coefficient Lrq in terms of the Soret coefficient ST becomes (9.133)

Lrq = Lrs TASC Ss T Using Eq. (9.129) in Eq. (9.133), we have

Lrq--

TasCssr [DseVCs (A/T)

'

+

DT,eVT]

(9.134)

9.4

479

Coupled system of chemical reaction and transport processes

Equations (9.130) and (9.134) suggest that the cross coefficients Lrq and Lrs are related to the gradients concentration and temperature. This reflects the vectorial character of the coupling coefficients Lrq and Lrs, as they relate the vectorial flows of heat and mass with the scalar reaction velocity. 9.4.8

Coupled Chemical Reaction System of Heat and Mass Flows

By substituting Eqs. (9.116) and (9.118) into Eqs. (9.80) and (9.82), we find the thermodynamically and mathematically coupled chemical reaction velocity with heat and mass flows

ocs Ot

(

-- --V" -Ds,eVC s -

OT pCp Ot

-- - - V "

(

DT,eVT +

As

A)

Lsr ~-

Lrq

-L,~Tvc~-TvT+

(

-DD,eVC

s -

k~VT

+

(9.135)

kfCs,eq T) R

A/

Lqr -~

AS

Lrq

+(-~XH~)-L~s7-VCs-TVT

(9.136)

Jr-kf Cs'eqRA/T

Under mechanical equilibrium, we have

v T

-

~;-r~

v

+

r

=-H

i

VT (V~i)T 7 + ~ T

(9.137)

where Hi is the partial enthalpy of species i. Using Eq. (9.137), we obtain

-/)

(-AHr) VT T2

(9.138)

after incorporating the Gibbs-Duhem equation CsV/Xs + CpV/Xp = 0 at constant temperature and pressure and the relations

for Vs = Vp

(9.139)

and (9.140)

G r + T2~S~ = AH~.

Combining Eqs. (9.138), (9.135), and (9.136), we have OCs - Ds,eV2Cs + D T ~V2T + c)l

Lrq + Lsr (- AH r ) V T - kf Cs,eq A

"

T2

R

(9.141)

T

As[(- AHr)Lrs + Lqr ] ( - AHr)kfCs,eq A OT V2 p C p - ~ = DD,e C s + keV2T VCs + -7"

R

T

(9.142)

where the group A/RT is the dimensionless affinity A*, and may be expressed by

A* -

A _lnK(T)+ln

RT

( aa~-Sp) - In ( ~ - ) + In (cC---~sp)

(9.143)

480

9.

Coupled systems of chemical reactions and transport processes

In Eq. (9.143) by disregarding the nonideality effects on the species, the activities of as and ap are assumed to be equal to concentrations Cs and Ce, respectively. Using the Arrhenius equations in Eq. (9.143), we have (9.144) With the dimensionless affinity A* Eqs. (9.141) and (9.142) become OCs - D s e~TZCs-Jr-DT,eVZT+ Ot

Lrq + Lsr ( - AHr)

'

T2

VT (9.145)

-Ef

L RT

+ln{c~)]

OT DD eV2Cs + keV2T - As[(-AHr)LrS +Lqr]vcs pCp -~--= '

T

+(-- AHr) Cs'eqk° exp (--R-T-

(9.146)

RT

One-dimensional forms of Eqs. (9.145) and (9.146) in the y-direction are OCs _ O s , e 02CS +DT, e -O2T - k - ~ - b OT Ot Oy2 Oy2 T 2 0 y

-Ef

OT Ot

Oy2

"t- (-Z~/--/r)pcp Cs'eqk°

RT

+ln(cC--~Sp)]

Asb OCs

OD,e O2Cs _+_oLe 02T pCp Oy2

[

(9.147)

pCpT Oy

exp(-~

(9.148)

[_ e r

+

In ~

where the parameter b in terms of the degree of couplings is

b = Lrq + Lsr ( - AH r ) = Lrr [Lqqqr2q+ Lss qs2~(- ~d-/r ) -- ~

qsr(-AHr)

keTqr2q+

Equations (9.147) and (9.148) use the same boundary conditions given in Eqs. (9.87)-(9.89). With the following parameters L2koexp(Er/RT~)

Ds,et 4)2 z = y/L, "7"-

L2 ,

=

Ds,e

_ Ef

; Ye

Eb

ke/pCp

ole

RTs , Yb = ~ s ' Le = ~ D s , e - D s , e

(9.149)

the nondimendional forms of Eqs. (9.147) and (9.148) become 2 + e ----+T2q- -~A O zOz 00S0r- 020s0z

1 0q~ _ 0 2q~ -k- ¢.o - Le Or Oz2 Oz2

-+- A

q~ Oz

q~s0S'eqexp YY 1-

(9.150)

YY 1.2~sj~0S,eq exp E/1)]

(9.151)

9.4

481

Coupledsystem of chemical reaction and transport processes

where

DT,eTs

DD,eCs

,to-

Ds, eCss

k~ L

3/b -- yf

A*

+ In

q~

bL

,o--

:

Ts Ds,e Css Os K[a 1 + a 2 (1 - 0s )]

[Lrq + LSr(-AHr)] L TsDs,e Css bCssLAs

keTs2

The initial and boundary conditions based on Eqs. (9.87)-(9.89) are 0 S -- OSO 0s -- 1

'r -- 0

Z -- + 1, r > 0 z-0,

r>O

Op -- Opo Op = 1

q? -- q)O q~ - 1

(9.152)

O0s _ o 0 P _ o ~ p _ 0 Oz

Oz

Oz

As an approximation, the nondimensional concentrations 0s and 0p a r e related to each other by the relation 0p = al + a2(1-0s), which is derived for stationary states, via the nondimensional affinity A*. The accuracy of the solutions of Eqs. (9.150) and (9.151 ) depends on the availability of reliable data, such as the effective transport coefficients and cross coefficients. Relating the parameter b to degrees of coupling qqr and qsr, as shown in Eqs. (9.147) and (9.148), may be helpful in solving these equations, since the degrees of coupling vary between - 1 and + 1. According to the Curie-Prigogine principle, a scalar flow, such as the rate of reaction, cannot be coupled with a vectorial flow of a transport process in an isotropic medium where an equilibrium-dividing surface is symmetric with respect to rotations around any local normal vector. However, the symmetry properties alone are not sufficient for identifying physical coupling; the actual physics considered in deriving the entropy production equation and the specific structure, such as anisotropy, are necessary. Some processes will not be dependent on some of the forces when the appropriate cross coefficients naturally vanish. For example, some degrees of imperfections due to parallel pathways of reaction or intrinsic uncoupling within the pathway itself may lead to leaks and slips in mitochondria. This, however, may add complexity to the phenomenological analysis, because failure of models to fit the properties of a system may be the result of unaccounted coupling. Previously, we considered the case where heat and mass flows are coupled in a reaction diffusion system with heat effects, in which the cross coefficients Lrq, Lqr, and Lrs, Lsr have vanished (Demirel, 2006). Here, we consider the other three cases. The first involves the stationary state balance equations. In the second case, there is no coupling between the heat flow and chemical reaction with vanishing coefficients Lrq and Lqr. Finally, in the third, there is no coupling between the mass flow and chemical reaction because of vanishing cross-coefficients of Lrs and Lsr. The thermodynamically coupled modeling equations for these cases are derived and discussed briefly in the following examples.

Example

9.8 Stationary

coupling

of chemical

reactions with heat and mass flows

Stationary forms of Eqs.

(9.150) and (9.151) are

0-- 020-------~S+8~-~ OZ2 ¢OZ2 2 0 Z

0 - 02¢ + ~ o - Oz 2

Oz 2

- --

A*~20S,eq exp 7 1 -

(9.153)

~-A*f~2/30S,eqexp 7 1 -

(9.154)

q~ Oz

The boundary conditions are defined in Eq. (9.152). All the other nondimensional parameters are the same as those defined in Eqs. (9.150) and (9.151). Using the relation q~ - 1 +/3(1 - 0s ) and Op -- a I -[- a 2 (1 -- 0 S )

482

9.

Coupled systems of chemical reactions and transport processes

in Eq. (9.153), the temperature q~is related to the concentration 0s, and the concentrations 0p and 0s are related to each other, and we have

0 -- d20s e,~ d20s -dz 2 -

dz 2

or~

dOs

[1 +/3(1 - 0s )]2 dz

-

[1 + 13(1- 0s)]

[a 1 + a 2 (1 - 0s)]

~ 0S,eqexp y 1 -

1

(9.155)

[1 + 13(1- 0s)]

An equation can be derived in a similar manner for the temperature.

Example 9.9 Chemical reaction velocity coupled to mass flow For the vanishing cross coefficients Lrq and Lqr in Eqs. (9.116)-(9.118), the heat flow and the reaction velocity become A Js = -Ds,eVCs - DT,eVT + Lsr "~-

Jq

(9.156)

= -CsDDVCs - keVT

ko exp(-Ef/RT)Cs,eq A 1 Jrs - -Lrs ~- AsgCs + R T

(9.157)

Still, heat and mass flows are coupled. The new balance equations are

00 S

020S nt_e 02 @ nt- 0" ~0qO _

07"

OZ2

OZ2 qO20Z

1 a ~ ° - 0 2 ~ ° + ~ o020S -Le Or az 2 Oz2

qO OZ

A 4~ Os,eqexp y 1 - 7

(9.158)

~ /~0S,eq exp y 1 -

(9.159)

where

o' ' = [Lsr(-AHr)]L" K'-TsDs,eCss

[Lsr(-AHr)]CssLAs ker2

The cross coefficient Lsr may be eliminated by relating it to the degree of coupling qsr

Lsr 1 qsr / 2(LssLrr = ) = qsr ( Ds'eT 1 ~ 2 kfCs'eq A sR )

(9.160)

and the parameters o-' and K' in terms of the degree of coupling qsr are

Os,eT kfes,e q )1/2 (_AHr) L o-'= qsr

As

R

(9.161)

TsDs,eCss

I Ds,eT kfCs,eq / 1/2 (-A/--/r)CssLA s As R keTs2 K' = qsr

(9.162)

Example 9.10 Chemical reaction velocity coupled to heat flow In this case, Lsr and Lrs vanish. Still, heat and mass flows are coupled. The new phenomenological equations are Js = - D e V C s - CsDTVT

(9.163)

9.4

Coupledsystem of chemical reaction and transport processes

J q _ _Cs DD V Cs _ ke V T + Lqr 1 A T k0 exp ( - E l / R T )

1

Jrs -- -Lrq 7

VT +

483

(9.164)

Cs,eq A

R

(9.165)

T

So, the balance equations become

eqexE,(1 +)]

OOs _ 020 S jr_ 02 q~ + _o-" _ _0q~ _ OT 022 802----T q~2 0z 1 0qo

Le Or

02@ 020 S - --+w-07_2

(9.166)

(9.167)

c)Z 2

where

L L

0""

-

qr

-

--

.

,

K

,,

LqrCssLAs z

keTs2

~DsxCss By relating the cross coefficient to the degree of coupling qqr

Lqr = qqr (LqqLrr) 1/2 = qqr

ke T2

kfCs'eq

)1/2 (9.168)

R

the parameters d ' and K" are expressed in terms of the degree of coupling qqr

o'" -- qqr ( keT2 kfCs'eq ]

(9.169) TsDs,eCss

K"= qqr ( ke T2 kfCs'eq

1/2 CssLas ker 2

R

(9.170)

E x a m p l e 9.11 M o d e l i n g of a n o n i s o t h e r m a l plug flow reactor Tubular reactors are not homogeneous, and may involve multiphase flows. These systems are called diffusion convection reaction systems. Consider the chemical reaction A --, bB described by a first-order kinetics with respect to the reactant A. For a nonisothermal plug flow reactor, modeling equations are derived from mass and energy balances

Ot

-

v

Oz

OCA _ --V OCA Ot Oz c)CB _ Ot

-

-

--

- - V

I ( (,0 exp( pCp

ko exp - ~

OCB + b k o exp Oz

CA

pCp-----~(rrc) 4h

CA

where zX/-/ris the heat of reaction, v the constant velocity, h the heat transfer coefficient, and d the diameter of the tube. Here, the reaction rate constant k is described by the Arrhenius equation. The initial conditions for 0 <- z-< L are

T(2,0) =

TO,

CA(Z,O)--CAo

CB(Z,0)--0

484

9.

Coupledsystems of chemical reactions and transport processes

The boundary conditions for 0 -< t are

CA (O, t) : eAin

T(0, t) = Tin,

where Tin and

9.5

CAi n a r e

CB(0, t) = 0

the temperature and concentration at the inlet.

EVOLUTION OF COUPLED SYSTEMS

During a diffusion-controlled reaction, matter may be transported around an interface, which separates the reactants and the product. The progress of the reaction may be affected by the morphology of the interface with a complicated structure, which controls the boundary conditions for the transport problem. The morphological stability of interfaces with nonequilibrium systems may lead to self-organization or pattern-formation, arising in biological, physical, chemical, and geological systems. Turing (1952) demonstrated that even some simple reaction diffusion systems could lead to spatial organizations due to the instability of the stationary structure, depending on the activator-inhibitor interactions, control parameters, and boundary conditions. If we consider the change of affinity with time at constant temperature and pressure, we have

dt

(9.171)

V,P ~

It is possible to split the dCs into two parts: dCs = deCs + diCs, which describe the part resulting from the exchange with the surrounding and the part due to a chemical reaction. The rate of the second part is the reaction velocity diCs/dt = VsJr. With these relations, Eq. (9.171) yields

(9.172) dt

T,P

dt

T,P

Therefore, affinity changes at the rate of exchanged matter and chemical reaction velocity. Depending on the rate of exchanged matter, the first term in Eq. (9.172) may counterbalance the reaction velocity, and the affinity may become a constant. This represents as system where one of the forces is fixed, and may lead to a specific behavior in the evolution of the whole system. The evolution equation, in general, is expressed by 0Y - f(Y, A) Ot

(9.173)

where Y is the column vector with the elements of the state variables Y1,...,Y/, which are continuously subjected to either internal fluctuations or external perturbations. Thefis mainly a nonlinear space operator, and A denotes a set of controlling parameters affecting the evolution, such as thermal conductivity, diffusivity, chemical rate constants, and initial concentrations of reactants and products. The evolution equations for the dimensionless concentration 0s and the temperature q~in the form of Eq. (9.173) become OCs A* Ot - fs[Css'Cps'Ts'A(~"Ds'e'Dp'e'DT'e'e'°"Tf' 'q~ss'Os'eq'qsq'qrq'qsr)] OT Ot

-- fy[Css , Cps , Ts , A(T,C~e,~,Ds,e,O),K, Tf,T b, A * , Os,eq,qsq,qrq,qsr,-ZkHr,Le)]

(9.174)

(9.175)

These equations suggest that the degrees of coupling besides the other parameters control the evolution and stability of the system. Comparing Eq. (9.174) with a simple rate expression

dCs_ dt

- k f C s + kbC P

(9.176)

9.6

485

Facilitated transport

alone displays the expansion in the number of controlling parameters in the coupled system of an elementary reaction with heat and mass flows. Therefore, induced cross effects due to various coupling phenomena can allow the system to evolve to multiple solutions and diversify its behavior. Using the state variables Y - ~ + y(t)

(9.177)

in Eq. (9.173), and by retaining the linear terms only in the Taylor expansion off, we obtain

Oy - . f ( [ Y ~ + y], A ) - f ( Y ~ , A ) - [M]y i)t

(9.178)

where y shows small perturbations around the stationary and spatially uniform solutions of Ys and [M] is the Jacobian matrix (called the linear stability operator) with the elements (Of/O ~)~ calculated at stationary state. The stability of the stationary states depends on whether the perturbation x grows or decays with time. This depends on the eigenvalues of the Jacobian matrix. The nature of the eigenvalues is determined from the solution of Eq. (9.178) in the form y - ue"

(9.179)

Here, ce is the eigenvalue of the Jacobian matrix, usually a complex quantity, and u is the eigenvector accounting for the structure of y and its dependence on the spatial coordinate.

9.6

FACILITATED TRANSPORT

In facilitated transport in membranes, a carrier agent can interact specifically with a substrate in the feed mixture, the substance-carrier complex diffuses across the membrane, and the carrier dissociates at the end of the membrane, and finally returns to its original position (Figure 9.8). The exterior substrate concentration has no effect on the rate of transport. The transport rate may reach a saturation point, which is its maximum rate. For example, a red blood cell membrane transports oxygen with the hemoglobin as the carrier. Only the specific substrates are transported, depending on the character of the carrier agent. Only specific inhibitors slow down the facilitated transport. In biological systems, the carrier agents are mainly proteins and are called permeases. Carrier-facilitated transport is used successfully to extract various organic and inorganic substances from a feed mixture in liquid membranes. Liquid membranes are employed as bulk liquid membranes, emulsion liquid membranes, and supported liquid membranes. Many biological mass transfer processes occur as a result of the combination of a substance with a membrane constituent to form a complex. For example, myoglobin has a single oxygen-binding site and is present in the muscle cytosol, and it binds to oxygen in a reversible reaction O2(s) + Myoglobin (c) ¢:~ Oxymyoglobin (bc)

(9.180)

The number of binding sites per hemoglobin is 4. Figure 9.9 shows a schematic of oxygen flow facilitated by hemoglobin as the carrier molecule. In facilitated diffusion, the carrier molecules are limited in number, and therefore the transport rate is not controlled by the concentration gradient and shows saturation. The flow is expressed by ,/-

- P (c~ 1 - Cso )

(9.181 )

where the permeability P is dependent on the concentration of substrate and carrier molecules cs1 and cs0. In a simple model, it is usually assumed that the membrane is thin, and a stationary state is established; all flows are determined by the substrate concentrations on both sides, and the reaction takes place on the surface. If k~/k o = k(/kl, the transport is passive. If k~/k o ¢ k(/k 1, the cycle prefers one direction, and the substrate will accumulate on one side. Since the carrier and bound carrier molecules do not leave the membrane, we have (Ccl + CcO + Cbc0 + Cbc1)

-C

(9.182)

486

9.

Coupled systems of chemical reactions and transport processes

4 Cso +

Cco

ko

"~

kI

k;

Cbc o

+cs I

Cc I

_.~

*"'-

kI

Cbc 1

dbc

Figure 9.8. Facilitated transport of a substrate.

Membrane nO 2 + Hb

k'*lk

P1

P2

HbO2n

~-- Jo2

x=0

x=h

Figure 9.9. Facilitated transport of oxygen by hemoglobin.

In membrane transport, one-dimensional models are usually used. If the permeates move independently of one another and with the ideal interface permeability, the simple diffusion, described by Fick's law, across the membrane is given by the boundary value problem OCs -- D 02cs Ot Ox2

(9.183)

cs (0,t) = C~Cs0, cs (l,t)

= O/Csl

(9.184)

where a is the distribution (or partition) coefficient. At steady state, the flow J is constant, and then

(;

J = -a

The permeability is a (

1 dx

oD(x)

/-1

(Csl -- Cs0 )

(9 185)

(1/D(x))dx) -1. IfD is constant, then the permeability is aD/1.

The interaction of particles with each other and with the membrane is a common situation, and there are several models proposed for such interactions. One model assumes a membrane as an energy profile for molecules moving across the barriers of binding sites; the transport takes place in a series of reactions, and the energy profile consists of a series of conformation alterations of the particles. If a membrane has a homogeneous interior, the diffusion coefficient is constant and the energy profile consists of energy barriers of equal size.

9.6

Example

9.12 Steady-state

substrate

487

Facilitated transport

flow in a f a c i l i t a t e d t r a n s p o r t

C o n s i d e r a s c h e m a t i c o f facilitated

transport shown in Figure 9.8. The steady-state facilitated transport may be described by dc (Ccl - Cc0 ) = - k ; C b c 0 -+- kocsoCco

dc (Ccl - Cc0) =

k(Cbc

1 -

klCslCcl

(9.186) (9.187)

dbc (Cbcl -- Cbc O) -- k;CbcO -- kocsoCco

(9.188)

dbc (Cbcl -- Cbc O) -- --k~Cbcl -+- klCslCcl

(9.189)

where dc and dbc are the association and the dissociation constants of the substrate-carrier reaction system, respectively. The flow of oxygen J is expressed as J = - d b c (Cbc I -- Cbc 0 ) = d e (Ccl - Cco )

(9.190)

By solving Eqs. (9.186)-(9.189), we have -CD(o"1 - o.o)

J =

(9.191)

1 + D ( o ' o / k I + o.1/k'o ) + ( D / d ) ( o " o + o"1) + (D/dc)o'oo" ,

where d= 2

(

1 +1

dbc

dc

)' ( , D-

1

1

~ b c + -ko -+

1 -

_ Csl Cs0 , o"1 - Kll , o"o -- Ko

K1 = ~-1' K° - ko

9.6.1

Kinetic Formulation

Since biological membranes act as barriers for hydrophilic and large molecules, a mobile carrier molecule, due to increased mobility of the substrate-carrier complex, may increase the transport of a substrate. Facilitated transport may be described by the jumping mechanism for a fast reaction between the cartier and substrate. Consider a schematic of facilitated transport shown in Figure 9.8. If the transport of substance-carrier across the membrane is not fast enough, then the conventional diffusion-reaction system of Eq. (9.180) is described by ~2Cs -- Ds 7-7- + k2Cbc - klCsCc at Ox"

(9.192)

02Cc -- Dc .-T---7-+ k2Cbc -- klCsCc ht Ox"

(9.193)

Oc+

Oc c

OCbc 31

02Cbc - Dbc ~ -- k2Cbc + klcsCc 3x 2

(9.194)

where Cs, Cc, and Cbc denote the concentrations of substrate, free carrier, and bound carrier, respectively, Ds, Dc, and Dbc are corresponding diffusion coefficients, and k~ and k2 are the chemical reaction constants. Initially, there is no

488

9.

Coupled systems of chemical reactions and transport processes

oxymyoglobin, and the concentration of myoglobin co0 is uniform across the membrane. With a membrane thickness of l, the boundary and initial conditions are Occ(O,t) _ OCc(1,t) _ OCbc(O,t) _ OCbc(1,t) Ot

Ox

Ox

Ox

=0 (9.195)

Cs(0,t) = a0Cs0, Cs(1,t) = °glCsl, Cc(X,0) = ¢c0 where Cs0 and Csl are the concentrations of substrate on the two sides of the membrane, and a0 and a 1 are the partition coefficients. The carrier cannot leave the membrane, and there is no considerable resistance to the diffusion of the substrate at the interface. When a carrier molecule such as hemoglobin is much larger than the substrate, it is often assumed that Dc - Dbc. Since the concentrations inside the membrane are generally not known, we can express the equilibrium oxygen concentration at steady state using the diffusion-reaction relations of Eqs. (9.192)-(9.194); assuming that Dc = Dbc, we have d2 cs _ Ds

dx 2

-k2Cb~ + klCgC~

(9 196)

-k2Cb~ + klcsC ~

(9.197)

kzCbc -- klCsCc

(9.198)

d2cc _ Dc

Dc

dx 2

d2Cbc _ dx 2

By adding Eqs. (9.197) and (9.198), we have d2cc ~ d2Cbc - - 0 dx 2 dx 2

(9.199)

By integrating we find dcc/dx + dCbc/dx--0, since the carrier and the bound carrier molecules do not leave the membrane. Further integration with the boundary condition yields Cc -k- Cbc = Cc0

(9.200)

As expected, the total concentration of the carrier molecules is equal to the summation of the free carrier and the bound carrier molecules. By subtracting Eq. (9.196) from Eq. (9.197), we have D c d2cc - D s ~d2cs = 0 dx 2

(9.201)

dx 2

The first integration yields dcc dcs De --~x - Ds d x = J

(9.202)

Myoglobin does not leave the membrane, and the myoglobin flow is zero dcc/dx - 0; therefore, J represents the flow of oxygen at steady state. The second integration of Eq. (9.202) yields Dcc c - Dsc s = J x + b 1

(9.203)

where the constant b 1 is determined from the boundary conditions. Using Eqs. (9.200) and (9.203), we express Cc in terms of Cbc 1

Cbc( x ) = Coo - - - - < - ( J x + b 1 + Dscs) r, c

(9.204)

9.6

By substituting the concentrations Cc and D~

Cbc ,

d2cs _

dx.2

489

Facilitated transport

defined in Eqs. (9.203) and (9.204), in Eq. (9.196), we have

1 - - - - ( Dsc ~ + Jx + bl)(k 2 + klcs) + k2cc0

D

(9.205)

This single nonhomogeneous and nonlinear equation is called the Wyman equation, and it can represent experimental data of the facilitated oxygen transport by hemoglobin for given boundary concentrations of oxygen, or for given values of the oxygen partial pressures across the membrane.

Example 9.13 Effect of temperature on myoglobin-facilitated transport Myoglobin exists in the muscle cytosol and has a single oxygen-binding site. Myoglobin is a carrier in the facilitated transport of oxygen represented by the following reaction kf

0 2 + Mb <

kb

>MbO 2

The net rate of oxygen (1) binding to myoglobin is Jrl - kt[O2 ][Mb]- kb[MbO2 ] With the reaction equilibrium constant K - kr/kb, this equation becomes

kf

Jrl - - k f [ O : ] [ M b ] - ~-[MbO:] This equation represents the temperature effecl on the reaction rate through the kinetic parameters kf and K. The steady-state one-dimensional energy balance is d2T

]; - - 7 - - ( - A H r ) ( J r l ) - qthl

(a)

dA- ~

where k~ is the effective thermal conductivity, AH,. the heat of reaction, and q the calorific oxygen equivalent, and th 1 t h e oxygen consumption rate, which is assumed constant through out the muscle tissue. The second term is the rate of heat generated by other metabolic processes in the tissue, and it is much larger than the heat of chemical reaction by the myoglobin. The heat of reaction for human myoglobin is reported as - 1.3 × 104 cal/mol. By disregarding the heat of reaction term and with the following dimensionless variables -z - -

x

__

0

m

L

T

rb

Equation (a) becomes d:O

L2

- -~qrh

1- -w

(b)

where L is the half thickness of the tissue (at the center of the tissue z = 1), Tb the boundary value of the temperature, and w a constant. The effective thermal conductivity is assumed to be 0.764 × 10 -3 cal/(cm s °C), while the value of q is assumed to be 4.8 cal/mL 02 for a typical diet, and L - 20 p~m. The oxygen consumption rate is assumed to be 6.0 × 10 -s mol/(cm 3 s). The boundary conditions are 0-1

atz-0,

~d0-0 &

atz=l

Integrating Eq. (b) with the boundary conditions, we have

r-rb

-wz

1-~-

490

9.

Coupled systems of chemical reactions and transport processes

The largest temperature difference occurs at the center of the tissue (z = 1), and for typical tissue fiber conditions, the maximum temperature difference is w/2 = 1.7 × 10-5°C at the tissue core. A similar increase with the effect of the chemical-binding reaction between myoglobin and oxygen is approximately 1.1 × 10-5°C. Equation (2) shows that the temperature difference increases with the square of the fiber thickness. Since the radii of skeletal muscle fibers are approximately 20 Ixm, the temperature difference is not considerable. However, some experiments suggest that there is a temperature effect on the rate of facilitated transport (Dowd et al., 1991). The steady state, one-dimensional mass balance equations for oxygenated myoglobin complex (2) and free oxygen (1) are

De2

d2C2 _

dx 2

Jrl

(C)

De 1 d2C1 dx 2 - - J r l + thl

(d)

where De, i is the effective diffusivity coefficient. Dimensionless concentrations of oxygen and myoglobin are

[MbO2 ] [02 ] 4,=~, ~C [02] b where C is the total myoglobin concentration (= 8.0mg/mL), and the oxygen concentration at boundary [ 0 2 ] b = 4.44 × 10 -7 mol/cm 3. With these definitions, Eqs. (c) and (d) become

dz 2

w

dz 2

----W 3 (1--(~)@--

-+-W2

w4

(e)

(f)

The boundary conditions are q~=latz=0,

d - - ~ = 0 a t z = l , and d 4 ~ = 0 a t z = 0 , 1 dz dz

Here, it is assumed that resistance to oxygen transport through the fiber membrane is negligible at the boundary, and hence the boundary oxygen concentration is equal to the plasma concentration. Since myoglobin as a carrier molecule cannot diffuse through the membrane, its gradient is zero at the boundaries. It is assumed that at the center of the fiber both oxygen and myoglobin concentration gradients are zero. The dimensionless parameters wi in Eqs. (e) and (f) are L2fkf

Del

= W1,

L2thl

Del [02 ]b

= W2,

L2kf [02 ]b CDe2

= W3, K[O2]b -- W4

The numerical value of each of these constants depends on temperature due to the temperature dependence of the diffusion coefficients, chemical reaction rate constant, and equilibrium constant. The Stokes-Einstein equation (D = kBT/(6~rttr)) for spherical particles in liquids suggests that the diffusion coefficient is proportional to the temperature and inversely proportional to the viscosity. At 310.2 K, the diffusion coefficients for oxygen and myoglobin are assumed to be 1.6 × 10 -5 and 1.0 x 10 -6 cmZ/s, respectively. It is often assumed that the diffusion coefficients for oxygen-myoglobin and free myoglobin are the same. By disregarding the viscosity correction, the temperature dependence of the diffusion coefficient may be TCK) D(T) = D ( 3 1 0 . 2 K ) ~ 310.2K

9.6

491

Facilitated transport

The temperature dependence of the forward reaction rate constant can be described by Arrhenius form

where k0 is the Arrhenius frequency factor and E the activation energy. At 293 K, the association rate constant kf and activation energy for horse myoglobin are 1.4 × 10 -7 1/(M s) and 5500 cal/mol, respectively. The temperature dependence of the reaction equilibrium constant is described by the van't Hoff equation

K(T) - K(293K)exp(-ZXH(293K) ( T

293K1) )

At 293 K, the value of K is 8.11 × l05 1/M. Dowd et al. (199 l) solved coupled two-point boundary-value equations. Global temperature changes affect the facilitated transport of oxygen. Elevated temperatures (i) increase local oxygen concentrations and decrease oxymyoglobin concentrations and (ii) increase the fraction of total oxygen carried by myoglobin. A parametric study shows that oxygen transport in muscle fibers is mainly controlled by the diffusion coefficient and the concentration of myoglobin.

9.6.2 NonequilibriumThermodynamic Approach In experimental studies, a membrane composed of a filter soaked in a solution of hemoglobin was used, as shown in Figure 9.8. Oxygen gas, at different pressures P~ > P2, was placed in the two compartments, and the steady-state flow of oxygen across the membrane was measured. The presence of hemoglobin enhanced the flow of oxygen at low oxygen pressure, however, the facilitation of oxygen transfer disappeared at a higher pressure of oxygen. Katchalsky and Curran (1967) used the linear nonequilibrium thermodynamics theory for the facilitated oxygen transport by the hemoglobin based on the linear flow-force relations. The following reaction is used for the facilitated oxygen transport by the carrier hemoglobin nO + H b ~ HbO •

2t~

(9.206)

A representative dissipation function takes the form XF -- J1 ~7(- ~1 )-+- J2V( - ~ 2 ) -Jr-J3V( - / z 3 )

(9.207)

A - n#, + / x - / x

(9.208)

The affinity A is given by

where the subscripts 1, 2, and 3 refer to oxygen, hemoglobin (Hb), and oxyhemoglobin (HbO2n), respectively. We assume that the rate of reactions is more rapid than that of diffusion, so that the reaction is at equilibrium, and hence A- 0

(9.209)

Applying the gradient operator to Eq. (9.208) at equilibrium, we obtain ,V#l + V/~2 - V/~3

(9.210)

This equation relates forces that lead to a coupling of flows. Passing at any point in the membrane are the flows free oxygen J1, hemoglobin J2, and oxyhemoglobin J3. The externally measured flow of oxygen Jl equals to the flows of free oxygen and oxygen carried by hemoglobin

JJ - J1 + n J3

(9.211)

Since no external flows of hemoglobin J2 take place, we have •12 - J2 + J3 - 0

(9.212)

492

9.

Coupled systems of chemical reactions and transport processes

Using Eq. (9.210), we transform the dissipation equation as follows

xt* = J ; V ( - / x 1) + J2V(-/x2 )

(9.213)

From Eq. (9.213) the following linear phenomenological equations are obtained

J( = -L~lVa~- L~2Va2

(9.214)

J2 --0 = - L 2 l V ~ I -

(9.215)

L22V~2

Equation (9.214) can also be expressed in terms of the diffusivity coefficients J1 - -D1 dcl + nD2 dc2 dx dx

(9.216)

By integrating Eq. (9.216) between x = 0 and x = h, and assuming that J(, D1, and D 2 are constant, we get j~ = D1 c o - c~ + n D 2 c~ - c~ h h

(9.217)

The external oxygen pressure determines the first term on the right side of Eq. (9.217). If P1 > P2 and co > c~, then the contribution of the carrier is considerable. The transport of oxygen increases with increasing hemoglobin concentration. When P2 - 0 and c~ = 0 the facilitated oxygen transfer decreases with increasing P1.

Example 9.14 Nonisothermal facilitated transport An approximate analysis of facilitated transport based on the nonequilibrium thermodynamics approach is reported (Selegny et al., 1997) for the nonisothermal facilitated transport of boric acid by borate ions as carriers in anion exchange membranes within a reasonable range of chemical potential and temperature differences. A simple arrangement consists of a two-compartment system separated by a membrane. The compartments are maintained at different temperatures T1 and T2, and the solutions in these compartments contain equal substrate concentrations. The resulting temperature gradient may induce the flow of the substrate besides the heat flow across the membrane. The direction of mass flow is controlled by the temperature gradient. A set of example chemical reactions for boric acid that take place in an anion exchange membrane is HB + OH- ~ B- + H 20 ( a - b)B- + bHB ~- HbBa -(a-b) where HB denotes boric acid, B- the borate, which is the cartier ion, and a and b are the stoichiometric coefficients. Assuming that the system is in mechanical equilibrium and at steady state, the entropy production is

AT A t.tB _+_Z jr i ~. OP=Jq~av -+-JB Tav i ' Tav where A~B =/*B,I- ~B,II, A T - TI - TII, and Tav = (T[ + TII)/2, Jq is the heat flow, which takes into account the enthalpy of mixing and the heat of reactions, Jri the flow for reaction i, A~B the chemical potential difference of H3BO3, AT the temperature difference across the membrane, and JB the absolute flow of boric acid directed from compartment I to compartment II when the absolute flow of water through the membrane is negligible. Equation above shows the three contributions to the rate of entropy production due to heat flow, mass flow, and the chemical reaction, respectively, and excludes the viscous and electrical effects. As the membrane is assumed to be an isotropic medium, there will be no coupling between the vectorial heat and mass flows and scalar chemical reaction, according to the Curie-Prigogine principle. Under these conditions, entropy production equation identifies the conjugate forces and flows, and linear relations for coupled heat and mass flows become AT

Jq = Lqq ~

Tav

+ LqB

A/xB

Tav

(a)

9.6

493

Facilitated transport

AT

JB -- LBq ~

Tav

A/.zB

+ LBB ~

(b)

Tav

According to the Onsager reciprocal rules, we have LqB = LBq. If we replace the chemical potential in terms of the concentration (instead of activity), we transform Eqs. (a) and (b)

AT (CBI ] Jq - Lqq ~ q- LqBRln Tav k CBII )

(c)

AT (CBI ] JB -- LBq ~ + LBBR In Tav ~,CBII )

(d)

To estimate the flow of borate, and assuming that the phenomenological coefficients are dependent on the average concentration (CBI + CBn)/2 linearly, we have

LBB -- aBB if- ]~BBCav LBq -- OLBq-ff ~ BqCav Substituting these relations into JB, we get

JB

_

{

(

1

OlBq'-~ff-~Bq%]~T'+" Tav Ta~)

1

Cav ]

O/BBT--~vq-ffiBB Tav ) ~/J'B

The coefficients ffBB,/~BB may be determined by the isothermal mass transport

J

1

,,so =

Cav)

Therefore,

LBB --

JB iso ' Tav - C~BB-I- ~BBCav '~P-.~B

(e)

On the other hand, the difference between nonisothermal and isothermal mass transport AJB yields a relation to estimate the coefficients aBq, ~Bq

AJB - JB --JB,iso -

C~Bq~

1

Ev

Cav / if-/~Bq ~ AT

ray

After rearranging this relation, we find AJB 2 LBq -- A T Tar - °lBq + [~3qCav

(t)

The linear Eqs. (e) and (f) help in determining the coefficients aBB,/3BB as well as ~Bq,/~Bq. Equation (e) produces a linear plot between (JB,iso/Al~B)Tav and Car, while from Eq. (f) a straight line results when plotting (AJB/AT)T~v versus Cav.The coefficients obtained from the slopes and intercepts of these straight lines are tabulated in Table 9.2 (Selegny et al., 1997).

494

9.

Coupledsystems of chemical reactions and transport processes Table 9.2. The phenomenological coefficients in the facilitated transport of boric acid by borate ions

Coefficient 2.8 × 10-7 K mol2/(m2 s J) 7.8 × 10 -9 K mol m/(s J) -9.15 × 10-3K mol/(m 2 s) --1.0 X 10-14K m/s

O/BB /~BB O/Bq

/~Bq

Source: Selegny et al., 1997.

Effect of heat of reaction From phenomenological equations AT A]..£B Js = Lsq ~ + LBB

AT A/xB Jq = Lqq ~ + LqB

rav

rav

7av

Vav

we have the isothermal facilitated heat and mass flows Ax/_______~B

Jq,is°=Lq B h~"-""~B Tav '

JB,iso = LBB Tav

Under these assumptions, the stationary heat of transport for formation of the complex per mole of B transported becomes Jq,iso = AH(JB,iso )

This equation shows that heat and mass flow in the same direction if &/-/> 0 (symport) and in opposite directions (antiport) if AH < 0. The two equations above and Onsager's rules yield

LqB -- LBq = L B B Z ~ The AJB is obtained from

A JB = JB -- JB,iso -- LBB ~

-L5-

The direction of AJB is determined by the sign of the product (~-/. AT). The difference of the heat of transport is related to

AJq = Jq -Jq,iso

= ~J-/(AJB)

This relation yields

Lqq = LBB~ 2

Level flow and static head Level flow occurs at zero load, which is at A/z = 0. At level flow, the mass flow is induced by AT and the phenomenological equation becomes

J ,max =

LBq

AT /av

9. 7

495

Active transport

With LqB = LBq = LBBZ~r--/, this equation is transformed into the facilitated form

JB,max

---

L B B A H AT

r:2v

At static head (JB -- Jq = 0), we have the maximum mass load A~max. From equation

AT T,7,,

JB -- LBq -X-V-+ LBBRIn

{CBI ] CBa

we have AT_ LBq To;,v

LBq AT

A/xB

--LBB ~,,, A~B'max . . .LBB . Tar

By using LqB = LBq -- LBBAH, this equation can be transformed into

~/d,B,ma x -- -- AH

9.7

AT

LV

ACTIVE TRANSPORT

Diffusion occurs spontaneously from a region of higher chemical potential/.L 1 to a region of lower chemical potential /x2, and the direction of flow is the same as the direction of decreasing chemical potential. The total Gibbs energy change for such a system is expressed by

- d G - - d N [&l + dN tx 2 - dN (ix 2 - ~ )

(9.218)

However, if a compensating process with a coupling mechanism, is added into the isolated system, causing the total free energy of the complete system to decrease, then it is possible to envisage diffusion against a potential gradient or an electrochemical potential gradient. This type of mass diffusion is called active transport for which the product of the flow d~ being acted upon by the generalized force X, is negative JiXi < 0. This inequality implies that the flow is occurring in the direction opposite to the direction of the force. If the compensating process is a chemical reaction, then the coupling will only be allowed in an anisotropic medium, according to the Curie-Prigogine principle. Since the chemical potential gradient is a vector quantity grad/~ - i ~0/~i + j Ol~i +k O/zi

&r

Oy

Oz

(9.219)

and the chemical reaction rate is a scalar quantity, no chemical reaction can impose directional properties onto the flow of substances unless the spatial gradient of a thermodynamic potential is altered. In biological systems, active transport involves the use of proteins that require the use of cellular energy (usually the energy released by the hydrolysis of ATP) to actively pump substances into or out of the cell (Figure 9.10). For example, protein found in the cell membrane of neurons pumps sodium ions from the inside to the outside of the neuron and pumps potassium ions in the opposite direction. This process sets up a high concentration of sodium ions outside the cell and a high concentration of potassium ions inside the cell. This concentration difference across the membrane is important for the generation of nerve impulses for transmitting information from one end of the neuron to the other. The sodium-potassium pump in red blood cells is operated by an oligomeric protein called Na+-K ÷ ATPase, which is embedded within and transverses the plasma membrane. Almost all cells have an active transport system to maintain nonequilibrium concentration levels of substrates. For example, in the mitochondrion, hydrogen ions are pumped into the intermembrane space of the organelle as part of producing ATR Active transport concentrates ions, minerals, and nutrients inside the cell that are in low concentration

496

9.

Coupled systems of chemical reactions and transport processes

3Na + ~i, Na+_K + ATPase Outside cell

aN~

Membrane

Jr

ATP Inside cell

ADP +

JK

Figure 9.10. Schematic active transport coupled to hydrolysis of ATE outside. Active transport also keeps unwanted ions or other molecules that are able to diffuse through the cell membrane out of the cell. Another common active transport system class pumps molecular nutrients such as glucose and amino acids into the cell at a much faster rate than can be achieved by passive or facilitated transport. For example, in the cells of higher animals, glucose active transport is dependent on the cotransport of Na + into the cell, in which Na + is pumped back out with simultaneous ATP hydrolysis by the Na+-K + ATPase. If an uncharged substrate is pumped from a region where its concentration is 0.001 M to the another region where this substrates concentration is 0.1 M, then we need the following amount of energy at 298 K to drive the transport

AG°=[8"314J/(m°lK)](298K)ln{O'l) =ll'41kJ/m°lO.O01 Free energy for active transport is supplied by a driving system from a high potential state to a low potential state. In primary active transport, the driving system is a chemical reaction away from equilibrium, and in secondary active transport, it is a concentration gradient. A diffusional flow against its conjugate thermodynamic force driven by dissipation of another diffusional process is called an incongruent diffusion, not active transport. Conventional methods for establishing the existence of active transport are to analyze the effects of metabolic inhibitors, to correlate the rate of metabolism with the extent of ion flow or the concentration ratio between the inside concentration and the outside concentration of the cells, and to measure the current needed in a short-circuited system having identical compositions solution on each side. Measurements indicate that the flow contributing to the short-circuited current, and any net flow detected are due to active transport, since the electrochemical gradients of all ions are zero (A~b- 0, Co ci). -

Example 9.15 Long-term asymptotic solution of reversible reaction diffusion systems The formation of dynamic reaction-diffusion fronts occurs when two species A and B are uniformly distributed on opposite sides of an impenetrable barrier, which is removed at time t = 0 at isothermal conditions. The species A and B start to diffuse and react upon mixing, and produce species C. This creates a dynamic reaction front, and the spatio-temporal evolution of this front may exhibit some unique features, which may be valuable in understanding many phenomena in physical, chemical, biological, and geological systems. The long-term behavior of this reversible reaction-diffusion system was studied by Koza (2003). For the reaction system A + B < - - ~ C , kb differential equations

we have the following mathematically coupled nonlinear partial

OA Ot

02A Ox2

DA ~ -

Jr

(9.220)

9. 7

497'

Active transport

DB __02 B _ Jr Ox2

OR _

3t OC

O2C Dc ~ + Jr

-

at

(9.221)

(9.222)

3x 2

where JR is the effective reaction rate with forward kf and backward kb reaction rate constants

kf AB

Jr -

-

k bC

(9.223)

Here, the concentration of species is a function of distance x and time t; A(x,t), B(x,t), C(x,t), and D; show the diffusion coefficient of species i. The initial conditions are

A ( x , O ) - ~ , B ( x , O ) - B o, C ( x , 0 ) - 0

(9.224)

We measure distance, time, and concentration in units o f x/DA/kf A o ,1/kf A 0 ,and A0, respectively. The problem will have four independent control parameters: DB, Dc, B0, and kb by assuming D A - l, A 0 - 1 and kf = 1. Long-term solutions of Eqs. (9.220)-(9.223) are series with ~-= l/t, and coefficients being some functions of r-s = x/qt

A ( x , t ) - ~ ~"a,(e)

(9.225)

n=()

-/

B(x,,;) -- ~ tl

-::

r'b,(e)

(9.226)

(i

C(x,t) - ~ "r'c, (e)

(9.227)

n=0

and

.1~- ~ ~-"r,,(e)

(9.228)

n=O

where r,(e) = Ej=oaj(e)b,,_i(e) - kt, c, (e), and a, b, c, and r are the scaling functions. When collecting coefficients with ~ + l where n - 0, 1, 2 .... , we have c__t2 a ,, ( e___~~-~- s

de 2

DB

Dc

2

d 2b, (e) de 2

da,,(e) +-ha, -- r,+ 1(e) = 0 de e db, (e) 2

de

(9.229)

+nb, - r,,+] (e) - 0

(9.230)

+n G + r , + l ( e ) - O

(9.231)

d2c,,(e) , s dc,(e) de 2

,:.~ de

Therefore, the lowest order terms (at ~.0) satisfy d:a,)(e)

e dao(e) t

d~ 2

2

de

~i(e)- 0

(9.232)

498

9.

Coupled systems of chemical reactions and transport processes

d 2 b 0 (e) e db 0 (e) DB~ -+ de 2 2 de

d 2c O(e)

DC

de 2

t

e d c O(e) 2

de

r1(e) = 0

(9.233)

+-r1(e) = 0

(9.234)

ro(e ) = 0 = ao(e)bo(e ) - kbCo(e)(k f = 1)

(9.235)

lim ao ( e ) = Ao, lima 0 (e) = 0, limbo (s) = 0, and limb o (e) = B o

(9.236)

with the boundary functions

Equations (9.232)-(9.235) have four unknowns functions, a0(e), b0(e), Co(e), and rl(e), which control the asymptotic (long-term) properties. Equation (9.235) shows that in the long term the chemical reaction tends toward a local chemical equilibrium at which the forward and backward rates become asymptotically equal. Because of the nonlinear of form Eq. (9.235), the solutions to these functions can be found for specific cases. One of these specific cases would be to consider a system with equal diffusion coefficients D A = D B = D c = 1, k1 = 1, and Ao = 1 The remaining unknown parameters are Bo and kb. By adding Eq. (9.222) to Eq. (9.221) and (9.220), we have two diffusion relations A ( x , t ) + C ( x , t ) = -~ erfc

(9.237)

B ( x , t ) + C ( x , t ) = ~ Boerfc

(9.238)

These equations have well-known solutions a0(e ) = bo(~) =

*(,)-

kb + J±(~) 2

-*(~)-

kb +

,/±(~)

2

(9.239)

erfc(e/2) + B o e r f c ( - e / 2 ) + 2k b - 2 ~x/-~ Co(~) =

4

~(~) = kbB0 (1 + k b + B o ) e x p ( - e 2 / 2 ) 27r[A(e)] 3/2

where ~(~)

= Aoerfc(e/2 ) - B0erfc(-e/2 ) 2

A ( e ) = [ ( I ) ( e ) - k b ] 2 + 2kberfc(2 /

Since ~(e) decreases monotonically from Ao to B 0, equation ~(e) = 0 has a unique solution denoted by ef. In the case of an irreversible reaction, where kb ~ 0, we have

~¢(s), lim a 0 (e) = [0, kb~0

s < sf} e ~ ef

(9.240)

9. 7

499

Active transport

lira bo(e) -

-0,~ ( e ) ,

f} e8 ~< gef

(9.241)

k b --' 0

lim c o (e) -

~ B° erfc

(9.242)

erfc

lim q (e) = B° exp(-e2/4) 6(e - el) kb~0 x/~erfc(ef/2)

(9.243)

where 6 is the Dirac's delta distribution. Both species A and B will be segregated at el, which may be identified as the position of the reaction front. Beyond this front, the reaction rate approaches zero and the species concentrations satisfy the diffusion equations. The reaction must be restricted to a region narrower than ~7, which is in line with the fact that for irreversible reactions, the width of the reaction front grows as t~ with c~ < 1/2. If we consider the opposite limit where kb--+ % the concentration of species C approaches 0, while the concentrations of species A and B evolve as if there was no reaction, and the scaling functions become

1

l i m a 0 (e) = -;- erfc(e/2) z

k b --,

1

lim bo (e) = -g Boerfc(- e/2) z, kb--+~f_ lira c o (e) = 0

(9.244)

kb-, ~,_

Example 9.16 Nonisothermal heterogeneous autocatalytic reactions-diffusion system Consider an autocatalytic reaction A

k, >B

A+(n-1)B

k2 >nB

where n is the order of the autocatalytic system, and kl and k2 the rate constants of the first- and second-reaction steps, respectively. Here, the product of the reaction acts as a catalyst and hence affects the conversion of reactant to product. The first step is slower than the second autocatalytic step. Effectiveness factors for autocatalytic reactions can be much larger than unity for exothermic and endothermic reactions. One example of such reactions is the catalytic cracking of paraffins on a zeolite catalyst to produce olefins. A mathematical model was suggested by Neylon and Savage (1996) with the assumptions: (i) the effective diffusivities of species A and B are the same, (ii) the catalyst pellet is symmetrical, (iii) effective diffusivities and thermal conductivities are constant, and (iv) external resistances to heat and mass transfer are negligible. It is also assumed that the diffusion is equimolar and heats of reaction for both reaction steps are equivalent to M/r, which is assumed to be constant. With power-law kinetics and the Arrhenius relationships for the reaction rate constants kl and k2, the rate of reaction becomes

- - J r A -- k01 e x p

--~-~ C A -Jr-k02 e x p --~-~- CAC~n-')

(9.245)

Under the conditions assumed and at steady state, the rate of diffusion of A within the pellet is equal to the rate of consumption due to reaction, and we have DA

d 1,, d/ a/

ko exp ' 77

CA

E2 )

~(n-l)

+ ko2 exp -~-~- C A,~B

(9.246)

500

9.

Coupledsystems of chemical reactions and transport processes

where 1 is the spatial coordinate and s the shape factor of the catalyst particle; s = 0 for a slab, s = 1 for a cylindrical shape, and s = 2 for a spherical shape. Similarly, the rate of heat conduction is equal to the rate of heat generation due to the reaction, and we have

lS dl

-~

/

=(-AHr)

exp/

E

k01exp --~-f CA

]

(9.247)

The boundary conditions are Atl=L

CA=CA0

At 1 = 0 dCa = 0

and and

dl

T=T o ~=dT 0

dl

where L is the characteristic length of the particle and CA0 and To the concentration and temperature of the fluid at the surface of the particle, respectively. Equations (9.246) and (9.247) represent mathematically coupled system. Temperature is related to concentration at any internal point by Prater's relation

T_To = _ (~Hr)DA (CA0-C A) ke

which may be used in Eq. (9.246) to reduce the number of variables from two to one by eliminating temperature within the Arrhenius relationships. Once the concentration profile is obtained, then the effectiveness factor is determined by

DAap z

m

--JrA0 where ap is the external surface area and Vp the volume of the particle. The reaction rate at the surface conditions is represented by-JrA0. With n = 2, we have a quadratic expression for the rate of reaction, which can display multiple steady states with high values of effectiveness factor for exothermic autocatalytic reactions.

9.8

NONLINEAR MACROKINETICS IN A REACTION-DIFFUSION SYSTEM

For a reaction S --, R the affinity is A =/Zs - ~p. After substituting the chemical potentials of the substrate and product in an ideal system,/x =/x ° + In C, where/x ° is any reference state, or the component compositions in the form (Sieniutycz, 2004)

Ck=exp

/zk exp -

(9.248)

we have the Marcelin-de Donder form Jr=Jrf-Jrb=kfCs-kbCp-Jro(T)

exp vs ~

- e x p Vp

(9.249)

where Jro is the exchange current and is expressed in terms of forward or backward reaction rate constant

(

(

Jro ( T ) = kf exp - v s ~-f = kb exp -Vp RT)

(9.250)

Here, the stoichiometric coefficients Vs = - 1 and Vp 1 are used. The exchange current Jro satisfies the microscopic reversibility at the state of thermodynamic equilibrium. These relations can be applied to chemical reactions with ionic substances by replacing the chemical potentials with electrochemical potentials. =

501

Problems

9.8.1

Generalized Chemical Kinetics

Consider an elementary generalized chemical reaction step

VofBor + ~ ....

pkt.Bkf--~V0bB0b+ k=l £ VkbBkb

(9.251)

1

where u is the stoichiometric coefficient matrix. Equation (9.251) involves the energy B0 (for k = 0) as well as the chemical species B~. (for k = 1,2,...,s, where s is the number of state coordinates in the generalized system) and therefore includes the heat effects. Here, the forward f and the backward b processes are considered. The net reaction rate is defined by

/ /

Jr--J0 exp --~v/dk=l

--exp --~Vbf--~ k=l

Where F is the potential defined by F-

1 /x)

(9.253)

7'7

The 1/T and tx/T appear in the thermodynamic conjugates of the extensive variables in the Gibbs equation for the system entropy dS = 1 d E T

(9.254)

Ixk dc k T

When the transport process is fast, potentials are equal at forward and backward states. With the total number of system coordinates s comprising the energy ( s - 0) and n components, we have the extended affinity defined by

A

--

(Pkbfkb-- U/'t"F1i")--

k=0 =--1If --[|b

£ Pkf[~/,T /"Of Tf k=l rf

I£Pkb[&kb P0b / k=l

rb

rb

(9.255)

where 1-If and H b are the generalized potentials of forward and backward states, respectively. The second terms within the generalized potentials describe heat effects through the energy barrier. These potentials are state functions, and the reaction proceeds from a higher to a lower potential, which is similar to heat flowing from higher to lower temperature.

PROBLEMS 9.1

Consider the hydrogenation of benzene, which is exothermic with a heat of reaction 50 kcal/mol. For a catalyst pellet containing 25% Ni-0104R 25% graphite, 50% T-AI203 (Harshaw A1-0104T), the effective thermal conductivity and diffusivity are 3.5 × 10 .....3 cal/(cm s K) and 0.035 cm2/s, respectively. For a benzene surface concentration of 4.718 x 10 -6 mol/cm 3, and a surface temperature of 340 K. (a) Estimate the maximum internal temperature difference. The observed rate for the reaction is 22.4 × 10-6mol/(g cat s) and the density of the catalyst is 1.57 g/cm 3. (b) Using the surface concentration as fluid bulk concentration, and the modified Sherwood number 401, estimate the internal temperature difference and compare with the one obtained from part (a).

9.2

Consider the hydrogenation of benzene, which is exothermic with a heat of reaction 50 kcal/mol. For a catalyst pellet containing 58% Ni on Kieselguhr (Harshaw A1-0104T), the effective thermal conductivity and diffusivity are 3.6 × 10 -.4 cal/(cm s Ki and 0.052 cm2/s, respectively. The fluid bulk concentration of benzene is 5.655 × 10-6 mol/cm s, and the fluid bulk temperature is 412 K. The characteristic length of the pellet is 0.296 cm. The observed rate for the reaction is 2.258 x 10 --~ mol/(g cat s) and the density of the catalyst is 1.88 g/cm 3. The modified Sherwood and Nusselt numbers are 215 and 10.8, respectively. Estimate the internal and external

502

9.

Coupled systems of chemical reactions and transport processes

temperature differences. (Note: The experimental values of the internal and external temperatures are 35 and 40 K, respectively. The difference between the surface and bulk fluid temperature is 6-7K, Froment and Bischoff, 1979). 9.3

Derive modeling equations for diffusion with a heterogeneous chemical reaction A ~ B. (a) The reaction is instantaneous. (b) The reaction has a rate NA = kCA. (c) Diffusion with a homogeneous chemical reaction A + B ~ AB, which is a first-order reaction with respect to component A.

9.4

Model a process of diffusion and chemical reaction A ~ B inside a porous catalyst. Assume that the reaction takes place on the catalytic surface within the porous medium.

9.5

Chemical vapor deposition is an important technique in the production of all kinds of solid-state devices. In the process, the active metal organic vapor is swept into a two-dimensional slit reactor by a carrier gas, and deposition occurs at the hot top and bottom plates. The reaction at the plate surfaces can be written MOoM+O Assuming a laminar operation, develop an expression to compute the rate of loss of MO for diffusioncontrolled process.

9.6

A hollow tubular reactor has inside walls that are coated with a catalyst. A feed of reactant A and an inert fluid pass through the reactor. At the tube wall, the irreversible catalytic reaction takes place A~P Assuming a plug-shaped velocity profile, estimate the concentration profile.

9.7

In a nuclear fuel rod, the rate of production of neutrons is proportional to the neutron concentration. Derive the differential equation to describe the flow of neutrons in the fuel rod.

9.8

A compound diffuses in a 4-cm long tube while it reacts. The compound starts from a source with a concentration of 0.2 M at the beginning of the tube. The reaction is a first-order irreversible reaction. Compound A is absorbed completely by an adsorbent at the other end of the tube, so that the concentration is assumed to be zero. Determine the concentration as a function of distance in the tube. Assume that the diffusivity is D = 1.5 × 10 -5 cmZ/s and the reaction rate is constant k = 5 × 10 -61/s.

9.9

A heterogeneous reactionA ~ 2B with nth order kinetics JrA = -kC~ ( n > 0) takes place on a catalyst surface. The component A with initial concentration CA0 diffusses through a stagnant film on the catalyst surface at isothermal and isobaric conditions. Assume one-dimensional diffusion, and determine the concentration profile of component A within the film of thickness 6 if the k is constant.

9.10

Derive a finite difference formulation for a steady-state reaction-diffusion system.

9.11

Consider the antibiotic production of Cycloheximide from streptomyces (Ablonczy et. al., 2003). For the first 24 h, streptomyces grows quickly and produces, a small amount of cycloheximide. After that the mass of streptomyces remains relatively constant and cycloheximide accumulates. Once the cycloheximide reaches a certain level, extracellular cycloheximide is degraded (feedback is initiated). One way to alleviate this problem and to maximize cycloheximide production is to continuously remove extracellular cycloheximide. The rate of growth of streptomyces is

dt

-

~max 1 -

x(O) = 1

Xmax

x

503

References

where x is the mass concentration,/Xma~ the maximum specific growth rate, and Xmax the maximum mass concentration. Experiments show that ]£max 0.3 l/h, and Xmax 10 g/L. (a) For 0 < t <24, solve the initial value problem and plot x vs. t -

-

-

-

The rate of accumulation of cycloheximide is the difference between the rate of synthesis and the rate of degradation

dP -- R s - R d dt R d - kd P

t(-QoEx

1+ P

-

where kd - 5 × 10 -3 I/h, Q0 is the specific enzyme activity (= 0.6 CH/g protein h),/£l the inhibition constant, and E the intracellular concentration of enzyme assumed as constant. For large values of/£l and t, we have x(t) - 10, and (1 + P/KI-1 _ 1) Thus, R~.- 10 QoE, and

dP dt

- 10QoE - kdP

P(24)- 0 (b) Solve the initial value problem above for 24 < t < 1000 and E -

1, and plot P vs. t.

[Source" K.H. Dykstra and H.Y. Wang, Biochemical Engineering V. Annals of the New York Academy of Science, Vol. 56, NYAS (1987), 511--522]. 9.12

Repeat the example 9.14 with LBB -- aBB + / ~ B B C a v

LBq -- OLBq

in the phenomenological equation JB, Eq. (d). 9.13

Use the modeling Eqs. (9.69) and (9.70) with the parameters in Table 9.1 to describe the coupled heat and mass flows by plotting coupled concentration and temperature profiles in NH3 synthesis. Assume the values e - 0.0001 and o) - 0.0001.

9.14

Use the modeling Eqs. (9.69) and (9.70) with the parameters in Table 9.1 to describe the coupled heat and mass flows by plotting coupled concentration and temperature profiles in oxidation of ethylene to ethlylene oxide. Assume the values e - 0.001 and w - 0.001.

REFERENCES Z. Ablonczy, A. Lukacs and E. Papp, Biophys. Chem., 104 (2003) 240. R.B. Bird, W.E. Stewart and E.N. Lightfoot, Transport Phenomena, 2nd ed., Wiley, New York, (2002). A. Burghardt and M. Berezowski, Chem. Eng. Sci., 58 (2003) 2657. A.S. Cukrowski and A. Kolbus, Acta Phys. Pol., 36 (2005) 1485. EH.M. Dekker, A. Bliek, E Kapteijn and J.A. Moulin. Chem. Eng. Sci., 50 (1995) 3573. Y. Demirel, Chem. Eng. Sci., 61 (2006) 3379. Y. Demirel and S.I. Sandler, J Phys. Chem. B, 108 (20041 3 I. M.K. Dowd, R. Murali and R.C. Seagrave, Biophys. J., 60 t 1991 ) 160. EG. Froment and K.B. Bischoff, Chemical ReactorAnah'sis and Design, Wiley, New York, NY, 1979. H. Hlavecek, M. Kubicek, M. Mareek, J. Catalysis, 15 (lC)69). A. Katchalsky and RF. Curran, :~)meqttilibrium Thermod~'nami~'s in Biophysics, Harvard University Press, Cambridge, MA (1967).

504

9.

Coupled systems of chemical reactions and transport processes

Z. Koza, PhysicaA, 330 (2003) 160. M.K. Neylon and P.E. Savage, Chem. Eng. Sci., 51 (1996) 851. E. Selegny, J.N. Ghogoma, D. Langevin, R. Roux and C. Ripoll, J. Memb. Sci., 123 (1997) 147. S. Sieniutycz, Int. J. Heat Mass Transf, 47 (2004) 515. E.M. Tavera, Chem. Eng. Sci., 60 (2005) 907. A. Turing, Phil. Transactions of Royal Society B, 237 (1952) 37. L. Yang and I.R. Epstein, Phys. Rev. Lett., 90 (2003) 178303-1.

REFERENCES FOR FURTHER READING L.P Kholpanov and L.A. Prokudina, Theor. Found. Chem. Eng., 39 (2005) 36. D.H. Kim and J. Lee, Chem. Eng. Sci., 59 (2004) 2253. M. Kohout, I. Schreiber and M. Kubicek, Comp. Chem. Eng., 26 (2002) 517. M. Ozer and I. Provaznik, J. Thoret. Biol., 233 (2005) 237. T. Pan and B. Zhu, Chem. Eng. Sci., 53 (1998) 933. S. Sieniutycz, Catal. Today, 66 (2001) 453. P. Travascio, E. Zito, A. De Maio, C.G.P.H. Schroen, D. Durante, P. De Luca, U. Bencivenga and D.G. Mita, Biotechnol. Bioeng., 79 (2002) 334.

10 MEMBRANE TRANSPORT 10.1

INTRODUCTION

The conventional approach to analyzing membrane transport is based on the specific models of the classical laws of electrostatics (Poisson's equation) and hydrodynamics (Navier-Stokes equations) combined with Fick's diffusion equation. The Nernst-Planck equation is used for the transport of charged particles in an electric field. Specific Fick or Nernst-Planck equations treat all flows as independent, requiring mobility be subject to gradients of electrostatic and chemical potentials, which may not be fulfilled, especially in concentrated solutions. However, interactions (coupling) are a common phenomenon in membrane transport. One of the trends in such analysis is the approach of linear nonequilibrium thermodynamics and phenomenological equations to incorporate the coupled phenomena into membrane transport. Since the interactions between the permeant and membrane may be complex, it may be useful to describe the problem with the phenomenological approach without the need for detailed examination of the mechanism of transport and interactions. The nonequilibrium thermodynamics approach identifies the possible pathways and interactions, such as coupling between the flow of a substance and a reaction, or between two flows. In this chapter, we summarize the formulations for passive, facilitated, and active transport problems, and the degree of coupling.

10.2

MEMBRANE EQUILIBRIUM

The transport of matter across membranes is an important class of biological processes. Membranes may have various physical and chemical structures, and hence are able to restrict transport processes by having different permeabilities for different substrates. Determining properties of equilibrium across and within a membrane may help in understanding the transport phenomena through membranes. Besides thermal equilibrium, the principle of electroneutrality is also satisfied. Because of the absence of mechanical equilibrium, a pressure difference known as osmotic pressure exists between subsystems separated by the membrane. In the case of substrates in ion form, both nonpermeating and permeating ions create an electric potential difference known as membrane potential across the membrane. For the separated parts of A and B, electrochemical equilibrium for permeating species k is

[~;- + RTln(ak/a[; )+ zkFO]a -- [IZ;.+ RTln(ak/a;)+ ZkFO]B

(10.1)

For dilute aqueous solutions, we assume that the standard activity for each species and the electrochemical standard potential for each species are the same in each phase at the same temperature and pressure o

0

o

o

aka -- akB, and #x-A - ~kB -- Vk(PB - PA ) at (T, PIA - (T,P)B

(10.2)

These equations are based on the integration of the partial molar Gibbs energy equation dG- -SdT + VkdP for component k at constant 7" and volume Vx~of pure component. Substituting Eq. (10.2) into Eq. (10.1), we have the equilibrium condition

V k ( ~ - P A )+ RTln( a~Aak----B-B)+zkF (~B -- OA) = 0

(10.3)

506

10.

Membrane transport

If PA = PB, we have the Nernst equation

- R---~Tln( akB ] A'q~' -- ~B -- ~A -= ZkF akA )

(10.4)

Under the condition of electroneutrality, Eq. (10.1) describes the Donnan equilibrium across a membrane, which separates solutions containing nonpermeating ions. With the Donnan equilibrium, differences of pressure and electric potential will appear. If the nonpermeating components are electrically neutral, only the pressure difference occurs. In biological systems with dilute aqueous solutions, the last term in Eq. (10.3) disappears, since Zw = 0 and the activity of the species determines the osmotic pressure (II). For water, we have

(10.5) \ awA J

We may introduce the following approximations. First, for ideal solutions, the activity coefficients are unity (Yk = 1), and concentrations are equal to mole fractions: ak = Xk. Second, using the definitions

XwA =

V

XkA , Vw -

1k= 2

, CkA-

N w

NkA V

and the approximation In(1 - x) - - x for x << 1, we have m

(-~w) 1 lnXwA~-- (_.~__)m ~XkA=-- (_~)

~ NkA ...." - -~_~ CkA

k=2

k=2

N

(10.6)

k=2

where m is the number of solutes and Nw - N for dilute solutions. Substituting Eq. (10.6) into Eq. (10.5), we find the approximation called the van't Hoff equation

n = &

--PA

----- R r

(10.7)

-

k=2

k=2

In a system with one single nonpermeating component dissolved in water, the above equation becomes

II = PB -- PA ~ RT (C2B -- C2A )

(10.8)

If component 2 is present in phase B only, we obtain

H = PB --PA ~ RTC2B

(10.9)

Consider a solution of 0.1 mol/L of sucrose in water (phase B). It is at equilibrium with pure water (phase A) across a membrane, which is impermeable to sucrose. When the temperature is 37°C in both phases, we can determine the osmotic pressure by assuming that equilibrium exists and the solution is ideal. For this system, the electric potential difference is zero, ~B -- qJA = 0, and the solutions are ideal. The water is the only permeating species and from Eq. (10.9), we have H = PB -- PA = RTC2B = 8.314(310)(0.1) = 257.7 kPa. This simple example shows that the osmotic pressure may be considerable depending on the concentration of the permeating species.

10.2.1

Degrees of Freedom

When a membrane system has two phases, m number of permeating components, and zk ionic valences, the thermodynamic state of the composite system is determined uniquely by T, PA, PB, mole fraction xk in the two phases, and the electric potential difference OB -- ~A across the membrane. These all add up to 1 + 2 + 2m + 1 = 4 + 2m variables. These variables are restricted by m equilibrium relations (Eq. (10.1)), so that the degrees of freedom are 4 + m. This is a special form of the Gibbs phase rule for electrochemical or chemical membrane equilibrium.

507

10.2 Membrane equilibrium

Membrane

l Phase A

Phase B

Water + Na* + C I

Water + Na + + C1- +Protein

WA

WB PB

Figure 10.1. Equilibrium between two phases A and B separated by membrane.

Example10.1Membraneequilibrium

An aqueous solution (phase A) of 100mmol/L of NaC1 is in equilibrium across a protein-tight membrane with an aqueous solution (phase B) of NaC1 and protein. The protein concentration is 5 mmol/L with a negative ionic valency of 10. Determine the difference in electric potential and hydrostatic pressure across the membrane when both solutions are assumed to be ideal and the temperature is 25°C. Figure 10.1 shows the membrane system with the phases A and B. There are four components, water (w), sodium ions (Na+), chloride ions (C1-), and protein (P-), and the first three are permeating. After neglecting the pressure difference, the equilibrium conditions for sodium and chloride ions from Eq. (10.3) are c(Na)B RTln c(Na)A

RTln

c(C1)B c(C1)A

i +z(Na)F(OB--OA)=0

(10.10a)

)+z(C1)F(OB -

(10.10b)

O A ) -- 0

For water, where z - 0, we obtain the equilibrium condition

(PB--PA) v. RTln Xw--B-B~ RT 2 CkB--Z CkA XwA k=2 k=2

(10.10c)

The condition of electroneutrality in phase B yields (CZ)Na + (CZ)c1 "]- (CZ)p -- 0

With z(Na) = + 1, z(C1) = - 1, and z(P) = - 10, we have [(C)Na - - (C)c1 - - ( 1 0 C ) p ] B = 0. Therefore, the concentration of sodium and chloride ions cannot be the same in phase B. The condition of electroneutrality in phase A is c(Na +) = c(C1-) = 100 mmol/L. The relations above with the given data determine the differences in electric potential and pressure across the membrane, and the concentration of sodium and chloride ions in phase B. With the concentrations in phase A, the sum of Eqs. (10.10a) and (10.10b) yields [( C)Na ( C)('I ]B -- [( C)Na (C)cI ] --

100 X 100

Using this relation with the electroneutrality condition above, we obtain c(C1)B and c(Na)B" c(C1) B - 78 mmol/L and c(Na)B = 128 mmol/L The electric potential difference is now obtained from Eq. (10.10a) or (10.10b): 8.314(298) ~'B - - 4 ' A

= --

(1)(96, 5001)

128) In 1 - ~ - - 6 . 4 mV

508

10.

Membrane transport

This shows that the algebraic sign of the potential difference is determined by the algebraic sign of the charge of the nonpermeating ion. The pressure difference is calculated from Eq. (10.10c): PB--PA -=8.314(298)[(128+78+5)-(100+100)]= 27.3kPa As expected, this shows that the pressure is largest on the side of the membrane having the largest concentration of nonpermeating components. In the absence of NaC1, the potential difference disappears, and the pressure difference decreases. For a cell membrane in a living animal, a very slight pressure difference will activate transport processes in the membrane, which will effectively eliminate the pressure difference. Introducing this change, there is no longer a state of equilibrium across the membrane, and other transport processes will take place. Such transport often is supported by chemical pumps, which move sodium ions from the protein phase to the aqueous phase. The simple estimation above illustrates that relatively small changes in the concentration are necessary to eliminate the osmotic pressure. In order to force PB - P A - - 0 , c(Na) in phase B must be reduced by 11 mmol/L, or by less than 10% of the previously determined concentration of 128 mmol/L (Garby and Larsen, 1995).

10.3 10.3.1

PASSIVE TRANSPORT Gas Permeation

In gas permeation, a gas species is separated based mainly on its permeability in hollow fiber and spiral wound membranes. The hollow fiber systems can have an inside diameter up to 200 I~m and hence very large surface-to-volume ratios, but high pressure drops inside the tubes. The basic flow equation for a species i is

Ji -- P--J-iAPi zk,c

(10.11)

where Pi is the permeability of the membrane for species i, zk~cis the thickness of the membrane, and z~kP i the partial pressure difference that is the driving force. The permeability is the product of the solubility of the gas in the membrane and the diffusivity of the gas in the membrane. In terms of total pressures and mole fractions, Eq. (10.11) reduces to

Pi Ji = --~ (PH YH,i -- Pp Yp,i )

(10.12)

where Pp is the total permeate pressure, PH the total pressure on the high-pressure side, YH,i the mole fraction of species i on the high-pressure side, while Yp,i is the mole fraction of i in the permeate. The selectivity is the ratio ofpermeabilities

a i j - P~ Pj

(10.13)

The selectivity should be greater than 20 to accomplish significant separation of species i from speciesj. For a completely mixed membrane system, the external mass balance yields JinYin

JpYp

-- JoutYout -

(10.14)

or

Yin =

(1 - 0) Yout

+

Oyp

where 0 = Jp/Jin" If 0 is specified as a design parameter, we can estimate Youtby

Oyp 1-0

Yin - Yout - -

(10.15)

10.3

Passive transport

509

For a binary gas mixture, Eq. (10.12) fi)r species i andj becomes

PiAPi

JP YP -

~x

p/Apj J'p(1 -yp ) .-. . . . . . . . . . . . ..,Xx

(PH Yout--PLYp)

(10.16a)

[ P H ( 1 - Y,.,~t ) - - P:_ ( 1 -

(10.16b)

3),)]

where p is the molar density at the permeate conditions, PL is the total pressure at the low pressure side, and A the area for mass transfer. After dividing Eq. (10.16a) by Eq. (10.16b) and substituting Eq. (10.15), we have

l'p

-

(

Di

--

c~,. i

[{Ym --

Oyp)/(1-

0)] -- (PL/PH) yp

DI (1-[(ym-Oyp)/(1-O)]-(PL/eH)(l-yp)

l--yp

i

(10.17a)

This equation reduces to the following quadratic tbrm once the fractions are eliminated: ,v))',i --X2y p + X3 -- 0

(10.17b)

where xj --

x~

- +-1 .... 0 f]~

-- (1 - %

Pi -

o

/9.;j-

0//!

joi

~ .0+ - - + PL 1-0 Pu

f)/

)----

I

.i i

pj

.vi, ] 1-0)

1 1-0

-~'Jn ']

7o

The root needed is positive and bctwee~ 0 and i

Example 10.2 Gas permeation in a binary gas mixture We want to separate carbon dioxide contaminating methane using a cellulose acetate membrane. The mixture is perfectly mixed on both sides of the membrane. The methane mole fraction in the feed (high-pressure gas) is y(CH4)= 0.90. The permeate pressure is 1.5 atm. At 35°C and 20atm, the permeability of the membrane is p(CO2) = 15 × 10 -1° (cm 3 STP cm)/(cm2 scmHg) and p(CH4) = 0.48 × 10 -1° (cm 3 STP cm)/(cm 2 s cmHg) (Wankat, 1994). The membrane thickness is 1 Ixm, and 0 = 0.4. Estimate the membrane selectivity, permeant mole fraction yp(CO2) , and flows of carbon dioxide and methane J(CO2) and J(CH4). The selectivity is OgCO~-CH4 -

Pco~ Pcna

15x10 -l° 0.48 × 10- 10

= 31.25

The permeate mole fraction yp -yp(CO2) can be determined from Eq. (10.17b) with the constants x1 -

l0~ + 1-0

~

Pn

7

-1

o4

-

1.5 ] (31.25-1) = 22.43 20 )

" 1-0.4

+ - - 1.5 + - - 0 . 1 / _ 1 - -28.91 x~ = ( 1 - ai/) -Pi( - } •01 - -+PL+ - PH - 0 1ym0 - 1 10 - ( - 30.25) { ~ 0.4 1-0.4 20 0.6 0.6 "

p j

Pi

x3 =c~'J 7

Yin i-~

-

-

- 31.25

-

0.1 0-~--5.21

510

10.

Membrane transport

Here, Pco2/PCH4 = 1. With the constants above, Eq. (10.17a) yields yp = YcQ = 0.216. This result displays an uphill transport of carbon dioxide, since the outlet concentration is higher than the carbon dioxide mole fraction in the feed. The pressure difference across the membrane allows for this transport direction. From the mass balance of Eq. (10.15), we have

Yout =

Yin--Oyp

_

-

1-0

0.1-0.4(0.216) 0.6

= 0.022

The flow of carbon dioxide is

JCO2 =

15 X 10-1° (76 cmHg/1 atm) [20(0.022)-- 1.5(0.216)] = 1.322 × 10 -4 cm 3(STP)/(cm 2 s) 1×10 -4

And Yuco2 = Your, since the module is well mixed

JCH4 =

10.3.2

0.48 × 10-10 (76 cmHg/1 atm) [20(0.978)-- 1.5(0.784)] = 6.706 × 10 - 4 cm 3(STP)/(cm 2 s) 1×10 -4

Transport in Liquid Solutions

Consider a homogeneous membrane with thickness Ax separating an outer solution 1 from an inner solution 2. The flow occurs along the x-coordinate perpendicular to the membrane surface. The zero point of x is on the surface in contact with solution 1. The electrolyte solutions are characterized by their electrochemical potentials. Within the membrane, the chemical potential is different from both/.Lil and ]£i2- However, it is widely assumed that the potentials on the membrane surface are equal to those of the solutions they are in contact with. Across the membrane, we identify the mechanical pressure difference (AP = P 1 - P2), the difference between concentration of species (Ac = Cil - ci2), and the drop of electric potentials (AS = $1 - ~P2).These differences are related to the electrochemical potential difference by (10.18)

At~ i = VizMo + R T l n A c i + I i A ~ t

where Vi is the partial molar volume and Ii the electric charge of species i. In passive transport, electrolytes and other substances are transported due to concentration, pressure, and electric potential differences.

10.3.3

Flows and Forces

For identifying conjugate flows and forces, we may use the dissipation function per unit area of membrane. A representative dissipation function is

- T diS dt

JwA#w + JsA#s - hwA~w + hsAtz s (10.19) AP + J a A H

:

cw

+

ca

For a nonelectrolyte binary solution consisting of water (w) and a solute (s), we have

A/xw - Vw AP - R T In Ac w - Vw AP

RTAc s -

Vw A P -

Cw

A/~s = V ~ A P - R T In Ac s = V~AP

RTAc s Cs

HA ---

(10.20a)

Cw

-

+

H - - -A Cs

(10.20b)

10.3

511

Passive transport

where AII = RTAc~ is the osmotic pressure difference. With these equations, the dissipation equation becomes

- T diS - JwA#w +.]~A#~ - nwA/xw + h~A/x~ dt = 'iw( VwAP- A+ I~'{V isAp+AHJ-)vAP+JdA,cI

(10.21)

C~

where dv is the total volume flow and Jd the relative flow of solute versus water defined by .Iv = nwVw + n~v~

Jd--

/;/s

nw

Cs

Cw

Therefore, Jv and NP and Jd and AII are the pairs of conjugate flows and forces. If the solution is sufficiently dilute, then

Jv "~ JwVw. With the identified flows and forces, the linear phenomenological equations are ,J, = Lp A P + LpdA H

(10.22a)

ff d -- LdpAP + L d A I I

(10.22b)

Due to Onsager's relations, Lpd = Ldp , and the membrane could be characterized by three parameters. It is always desirable to characterize a membrane with the minimum number of parameters.

10.3.4

Membrane Transport Coefficients

If we consider a membrane having the same solute concentration on both sides, we have A l I = 0. However, a hydrostatic pressure difference AP exists between the two sides, and we have a flow Jv that is a linear function of ~¢'. The term Lp is called the mechanical filtration coefficient, which represents the velocity of the fluid per unit pressure difference between the two sides of the membrane. The cross-phenomenological coefficient Ldp is called the ultrafiltration coefficient, which is related to the coupled diiNsion induced by a mechanical pressure of the solute with respect to the solvent. Osmotic pressure difference produces a diffusion flow characterized by the permeability coefficient, which indicates the movement of the solute with respect to the solvent due to the inequality of concentrations on both sides of the membrane. The cross-coefficient Lpd relates the flow J, at AP = 0 to AH, and is called the coefficient of osmotic flow. The cross-coefficients are imposed by the nature of the flow in the membrane, for example, Lpd shows the selectivity. If Jv = 0, we have

(AP)j :0

=

-- Lpd Lp A H

(10.23)

Only if --Lpd -- Lp, then ( A P ) j :0 - A I ] . This is the condition for an ideal semipermeable membrane, which blocks the transport of solute no matter what the value of AP and All is. When this is not the case, the membrane allows some solute to pass, -Lpd/L p < 1. The ratio --Lpd/L p is called the reflection coefficient o. The value cr = 1 indicates that all solute is reflected (ideal membrane); the solute cannot cross the membrane. When cr < 1, on the other hand, some of the solute is reflected and the rest crosses the membrane; when o-= 0, the membrane is completely permeable and is not selective. If we introduce cr into Eq. (10.22a), we have ,], = Lp (AP - o-All)

(10.24)

The solute permeabili~ coefficient oo is defined as

w

-~p

(LpL d - Lpd )

(10.25)

512

10.

Membrane transport

Table 10.1. Solute permeability and reflection coefficient for some membranes Membrane

Solute

Solute permeability, o) (× 1015mol/(dyne cm))

Reflection coefficient (or)

Toad skin

Acetamide Thiourea Methanol Ethanol Isopropanol Urea Urea Ethylene glycol Malonamide Methanol Urea Glucose Sucrose Urea Glucose Sucrose

0.0041 0.00057 11 11 7 0.008 17 8 0.04 122 20.8 7.2 3.9 31.6 12.2 7.7

0.89 0.98 0.50 0.44 0.40 1 0.62 0.63 0.53 0.013 0.123 0.163 0.0016 0.024 0.036

Nitella Translucens

Human red Blood cell

Visking dialysis Tubing Dupont "Wet gel"

Source: Katchalsky and Curran (1967).

where Cs,avis the average solute concentration. For an ideal membrane, o- = 1 and co = 0. For nonselective membranes, Lpd - 0 and co = CsLd.The permeability coefficient is a characteristic parameter in both synthetic and natural membranes. Table 10.1 shows the permeability and reflection coefficients for some membranes.

10.3.5

Measurements of Membrane Transport Coefficients

The transport coefficients may be measured based on the equations Jv Lp = ( ~ ) ~ X H = 0

(10.26a)

(10.26b) and w=(~)j

and

Jd

w d =(~--H)~/~=0

(10.26c)

v=0

When 2if' = 0, then flows affect the measured value of Wo. Also, the definition of co requires that Jv = 0 and there is a nonzero mechanical pressure difference. Therefore, experimental measurements of co and cod may be difficult (Kargol and Kargol, 2003), and a correlation may be used cod = ( 1 - or) Lpcav

(10.26d)

Hence, we have the following relation between co and cod: cod(1 + o') = co. Some measured transport parameters are listed in Table 10.2. The solute flow can be expressed in terms of the solute permeability Js - Cs,av(1- °') Jv + cozXH

(10.27)

For a simplified model of the membrane having cylindrical pores of radius r and length I with Np pores per unit surface area, Poiseuille's law expresses the volume flow Q for a pressure difference ~ as follows: Q -

,rrr4 z.~~ 8~1

(10.28a)

513

10.3 Passive transport

Table 10.2. Measured transport parameters Membrane

,Solutior~

'~x (mol/m 3)

~o i / l()l()mo'/'(N s))

wd (× lOl°mol/(N s))

Lp (/lOl2m3/(N s))

Neprophane Neprophane Cellophane Cellophane Dialysis membrane

Ethanol Glucose Ethanol Glucose Glucose

4(1!(i} 2()0 3(i)0 1:50 300

19.0 9.95 6,68 3.3 l 3.21

19.5 9.35 6.55 3.01 2.84

5.0 5.0 2.23 2.23 1.09

0.025 0.065 0.02 0.1 0.13

Sou;re: Kargol and Kargol (2003).

where rl is the viscosity of the liquid. The flow ,], is expressed by ~p 7rr 4 A P r

Jv -

(10.28b)

8rfl

Using Eqs. (10.22a) and (10.28b), we find the filtration coefficient Lp at A I I - 0

N pTTF

Lp --

(10.29)

8v/l Let us assume that k P - 0, but there is a concentration difference and the solute can diffuse. If the solution is ideal, van't Hoff's law" states that A I I - R TAc s and then, from Eq. (10.27), we have J, - wAH - RTooAcs,av

(10.30)

According to Fick's law, the diffusion flow in each pore of transverse area of rrr 2 and length I is given by TrY2 )

,]-D

--~

Acs,av

where D is the diffusion coefficient. With the total number of pores Np, the total flow becomes

- T ~XCs,av

•

(10.31)

Comparing Eqs. (10.30) and ( 10.31 ) yields

~o -

NpD rrr2 RT

(1 - o-')

(10.32)

l

where o-' is the fraction of solute reflected. From a hydrodynamic point of view, the reflection coefficient o- may be defined as

/ 1-o--1

)2 -rs

(10.33)

F

where rs and r are the radii of the solute molecule and the pore, respectively. If rs > r and o- - 1, since the solute molecules do not fit into the pores, all the solute is reflected. Equation (10.33) results from the ratio of the effective

514

lO.

Membranetransport

area of the pores, which is r r ( r - rs)2, to the total area effective area _ ( r - rs)2 ( ~ ) actual area r-------5~= 1 -

2

Generally, the reflection coefficient decreases as the molecular radius increases.

Example 10.3 Time necessary to reach equilibrium in a membrane transport Consider two containers of volumes V1 and V2 and the initial solute concentrations of Cl and c2 (Cl > c2) separated by a permeable membrane. We want to know the time necessary to reach the equilibrium at which cl = c2. The changes in the number of moles of solute in containers 1 and 2 are calculated as a function of the area A of the membrane and the solute flow J~

dN1 _ -Js A, dt We know that Ni

=

Vici,

Js

--

°)RT(cl

--

c2),

dN2 _ ---Js dt

A

and then we have

V1 dc---L1= -A~oRT dt

(C 1 -- C2 )

(10.34a)

V2 ---~ = AwRT dt

(C 1 -- C2)

(10.34b)

After subtracting Eq. (10.34b) from Eq. (10.34a), we obtain

d(Cl - C2) - - A w R T ( ~ +-~2 ) dt -

(Cl -

c2)

(10.35)

Integration of this equation yields

Ac(t)=(c 1 - c 2 ) = Ac (0) e x p ( - t-~-)

(10.36)

V~V2 to - (A,oRT) (Vl + V2)

(10.37a)

where to is defined using Eq. (10.35)

If we assume that V1 >> V2, we have a simplified approximation for to to---

V2

(10.37b)

AooRT In practice, after a time of 4t0, the concentration differences between the two containers will vanish, and to is considered the characteristic time. This approximation may be useful in cellular transport, and artificial kidneys and lungs.

10.3.6

Frictional Forces and Resistance-Type Phenomenological Coefficients

We can express the phenomenological coefficients in terms of the frictional forces; assuming that for a steady-state flow, the thermodynamic forces Xare counterbalanced by a sum of suitable frictional forces F. Thus, for a solute in an aqueous solution, we have X s = -Fsw -Fsm

(10.38)

10.3 Passivetransport

515

where Fsw is the frictional force between solute and water and Fsm the corresponding friction with the membrane. Similarly, the force on water is .V w - - F w s - fwm

(10.39)

The terms fsm and fwm comprise complex hydrodynamic interactions within the membrane matrix and should be regarded as macroscopic averages. For sufficiently swollen membranes, however, Fsw, which indicates the interactions of solute and solvent, may approach free diffusion. The individual frictional forces F O.are assumed to be linearly proportional to the relative velocity vii = v i - vj, and because the proportionality factor.f 7 is the frictional coefficient per mole of ith component, we have Fu - - f , i ( v ; - v i )

(10.40)

C, ji j - C/fji

(10.41)

The S9 obey the reciprocity relations

where C is the concentration in the membrane. Introducing Eqs. (10.41) and (10.40) into Eqs. (10.38) and (10.39), we obtain (10.42)

Xs -- .fsw ( Vs -- Vw ) .or.fsm (Vs -- Vm )

Choosing the membrane as the frame of reference and for vm = 0, we have (10.43)

X s --!'s (,/sw -+- ./4sm)--Ww fsw

A.~ -- - - / '

w

~ Cw

!

+ Uw fwm q-

cs i fsw

(10.44)

Since the flows Js and Jw are expressed as .L - Csv~ and Jw - Cwvw, we may express these equations in terms of the flows X

- , l s w -+- fsm Js

'~ -

Q

•

X w = _ _./;w _ d~ + (-7.... "

"-

fsw

(10.45)

C--7-w dw

Cwfwm -+- Cs fsw (Cw) 2

dw

(10.46)

These equations are the resistance type of formulations X ~ - K s J s + KswJ w

(10.47)

X,, - KswJ s + K w J w

(10.48)

The resistance phenomenological coefficients are given by Ks _ fsw + fsm, Q

10.3.7

K~,, = -fs__ww, Cw

Kw _ Cwfwm nt- C s L w (Cw) 2

(10.49)

Kirkwood's Procedure

In Kirkwood's procedure, the differences of chemical potentials (for nonelectrolyte solutions) are expressed by Abt w

-

vwAP-

R T A In c w - v w A P -

RTAcs = vwAP-~

Cw

A#~ - v s A P + R T A In c s = v s A P +

AH

(lO.5O)

Cw

AH RTAc s - vwAP +

(10.51)

516

10.

Membrane transport

where AH = RTAcs is the gradient of osmotic pressure across the membrane and c denotes the concentrations in the external solutions. If the forces are expressed in terms of the chemical potentials, we have

Xi

-

-

-

d/x___Li dx

(10.52)

Integration of this equation yields II ~c ~ d/xs RTAc s _ 4,sAP + AII I Xsdx = dx = A~s = vsAP-4 Cs Cs dx o I

(10.53)

txs

II

RTAc s _ 4~wAP- AH

ax

I w X- "~sI- dtx,,xw dx = Atx w = v wAP o

cw

I ixs

cw

(10.54)

where ~b = civ i is the mean volume fraction of component i in the external solution. On the other hand, integration of the right sides of Eqs. (10.45) and (10.46) requires the concentration profiles in the membrane and the dependence o f f j on X. For a homogeneous membrane, we may assume that f j are constant, and instead of integrating over the concentration, we can use the mean values given by

(Cs)o Cs'av=

G 2

'

Cw-v w

where ~bw is the mean volume fraction of water in the membrane. Assuming the equality of the chemical potential on the surface of the membrane and in the surrounding solution, we may write (Cs) 0 =

Kc~

'

(Cs)Ax = Kc~ I

and

I nt_C~I Cs,av = X cs 2 =

Kc

s,av

where K is the distribution coefficient of solute between the membrane and the solution in equilibrium with it. The K is taken as a concentration-independent parameter uniform throughout the membrane in an ideal behavior. These assumptions permit the integration of the right of Eqs. (10.45) and (10.46). Since ~bw = CwVw"-~ 1, and ~bs is usually sufficiently small to make ~bsAP < AII, Eqs. (10.45) and (10.46) yield the following expressions: Al-I= z~c fsw +fsm J s - Z~XCs,av fswVw Jw

(10.55)

K AP-AII=

Ax fSW Js + A x ( f w m + K Csfsw ) j

4'w

4'w

(4,w)2

w

(10.56)

where &,c denotes the thickness of the membrane. The permeability coefficient of the solute flow is determined at zero volume flow (Js)Jv=0 -- ~A1--[

(10.57)

In this case, the solute flow at Jw = 0 is practically the same as at Jv = 0. So, Eq. (10.55) at Jw = 0 leads to All = z~oc Aw q- Am Js K

(10.58)

and therefore, w=

K

(10.59)

Ax (fsw + fsm) In the ideal case, K and the frictional coefficients are concentration independent, and o) becomes a constant characterizing the mobility of the solute in the membrane.

10.3

Passive transport

517

Equation (10.59) shows the characteristics of the permeability coefficient co. It is obvious that the total frictional resistance is larger in a membrane than in free solution. Although~w is approximately equal tOfs°wof the solute diffusing freely in the solvent, we have to consider the additional friction j~m due to interactions of the solute with the membrane. If the membrane, for example, has a porous structure and the penetrating molecules are sufficiently large, thenj,~m may be very large and {,) ~v 0. In this case, the system will behave as an ideal semipermeable membrane. If the solute penetrates by dissolving in the lipoid components of the membrane, then a large friction between the solute and membrane will develop, and the value Of/sin will be large. However, if the attraction between the solute and lipoid is very large, K may increase to a larger value thanJ~m. The reflection coefficient G is also defined at ,Iv = 0 using the relation

iT--

~

.1,={}

(10.60)

However, the coefficient G' at Jw = 0 is not exactly equal to the coefficient cr at Jv = 0. The relationship between oand G' is ( r - o r ' - ~°vs Lp

(10.61)

The coefficient o-' is readily obtained from Eqs. ( 10.55 ) and (10.56) at d~ = 0

K,~£SW

G' =:- 1--

(10.62)

(Dw(,£w -+-fsm ) In view of Eq. (10.61 ), we have

(.Ol'S

/~fSW

o-=1-~-

(10.63)

Jsv~ ' /2m)

Substituting Eq. (10.59) in the above equation~ ~,,c ge~ G - 1 ~ov~ oJAg~sw

(10.64)

Lp Equations (10.63) and (10.64) help 1o understand the physical meaning of the reflection coefficient. Ideal semipermeable membranes prevent the permeation of the solute; hence, ~o = 0 and cr = 1. Generally, an increase in f~m (due to large molecular weight) decreases ~o, and hence increases G. On the other hand, if K increases simultaneously withfsm, then ~o may increase, while cr decreases. In some cases, the increase in K is so strong that ~o assumes large values, which make o- negative. This causes negative anomalous osmosis, since ~ , 0 (=o.AH) becomes negative. In coarse nonselective membranes, we have ~o,,~

~oA Xj;w

= 1

(10.65)

If we assume that the solute penetrates only through the water-filled channels in the membrane (with volume fraction &w), then the solute-water frictional coefficient is close to that of free diffusion D O

D°

(10.66)

and the solute flow will be given by

AH/RT ,I. = D Q ~ Ax

(10.67)

518

10.

Membrane transport

m .

x

CiA

CiB

~/A

Ji

CA PA

PB L

Figure 10.2. Diffusion cell with uniform concentration and electric potential chambers connected with a barrier of length L. where D is the diffusion coefficient in the membrane. The relation between to and D is given by to - Dchw RTAx

(10.68)

o- = 1 toYs D Lp D O

(10.69)

Using Eqs. (10.66) and (10.68), we obtain for tr

Example 10.4 Diffusion cell with electrolytes The diffusion cell shown in Figure 10.2 has an aqueous solution of NaC1 with a concentration of 100 mmol/L. Later, 0.1 mmol radioactive Na with a specific activity of 1 x 108 units is added to chamber A, which has a volume V - 1.0 L and is stirred continuously. Measurements show that the radioactivity in reservoir A decreases at a rate of 14 units/min. The process is at steady state. Estimate the flow of sodium ions, the diffusion coefficient, and the mobility at 298.15 K and in a transfer area A - 100 mm 2. At the time of addition, the specific activity is (108)(0.1) (Garby and Larsen, 1995). Initially, the relative decrease in the concentration of radioactive Na + in reservoir A is (~_~)(1 ~_f)]=

107-14_l.4X10_6min_ 1

(10.70)

The initial concentration of labeled Na is 0.1 mmol/L. Since there is no chemical reaction, the following reaction-diffusion equation: drli) dt)

-

E°ut/~i

- Ein

/l i

="

ViJr-

E

out

AJ i

(10.71)

yields

dt From Eqs. (10.70) and (10.72), we find

dt )

-

_ A j

+ ,

+ = c~V nA

) (,4 06/

10~4 (0.1)

60

The diffusion coefficient is estimated from

J+L D ~ ~

J+L

+ + ~ + CA -- CB CA (0)

= 2.33×10 -8 mol/(m 2 s)

(10.72)

519

10.3 Passive transport

In this equation, a quasi-steady state is assumed during the measurement of the initial decrease in radioactivity, and the concentration of labeled Na ÷ is negligible in reservoir B. If we assume that L = 0.005 m, we obtain

D = ---;-

--~

=

1.4 X10 -6 (0.005X10 -3 j 60 10 - 4

1.17 X 10 - 9 m 2/s

--

CA

The mobility is obtained from

b/ m

Di _ 1.173<10 -9

= 4.7× 10 -13 m o l m 2 / ( j s )

8.314(298)

RT

Example 10.5 Diffusion cell and transference numbers The diffusion cell shown in Figure 10.2 has NaC1 mixtures in the two chambers with concentrations C~A- 100mmol/L and ClB = 10mmol/L. The mobilities of Na ÷ and C1- ions are different and their ratio yields their transference numbers b+/b - = t+/t - = 0.39/0.61 (NaC1). The transference number t for an ion is the fraction of the total electric current carried by the ion when the mixture is subjected to an electric potential gradient. For monovalent ions, we have t+/t - = 1. Estimate the diffusion potential of the cell at steady-state conditions at 298 K. Assume that activity coefficients are equal in the two reservoirs (Garby and Larsen, 1995). Due to electroneutrality, the local concentrations ofNa + and C1- ions are the same, and are denoted by cl(x). The flows of ions will be the same and directed from A to B. The Nernst-Planck equation describes the diffusive flow of ions at constant T and P

Ji - -bi(RT) ( dcidx-t ciZiFRTdO

(10.73)

Using the above equation and eliminating fluxes, we have the electrical potential gradient

("/

b+z+-b z

- V c-?

j

(10.74)

After integrating this equation over the diffusion cell, we obtain the diffusion potential b+ - b A O -- ~A -- OB -- --

b+z+ _ b _ z _

--F-- In k ClB }

(10.75)

With the relations t + + t- = 1 and z + = z- - l, we have the difference of diffusion potential RT /ClA] A0 - ( 1 - 2t +) - - ~ In k clB )

(10.76)

With the data given, the potential difference is AO - [1- 2(0.39)]

8.314(298) ln(100 ] _ 96,500 k-i-0--) 13mV

The molar flows of the ions and NaC1 are all the same. In that case, we estimate the flows for Na + and for Cl-, and divide them by b + and b- respectively. By adding these results, we obtain 2b+b-

Jl =

b+ + b -

(10.77)

RT (

dx)

520

lO.

Membrane transport

Comparing this equation with Fick's law for diffusion Jr, = -Di(dci/dx), the diffusion coefficient for NaC1 is obtained by 2b+b D1 = ~=+ + bb- (RT)

(10.78)

2(1-t+)b+(RT)

The mobility of NaC1 is obtained directly from D i = biRT 2b+b ba = ~( 2+-+t +b )- b + =b

(10.79)

For an electrolyte with arbitrary ionic valences, Eq. (10.78) may be generalized by

D1

=

2b+b-(z + - z - ) b+z+-b z

(10.80)

(RT)

Example 10.6 Estimation of flow in a diffusion cell Each chamber of the diffusion cell shown in Figure 10.2 has 100 mmol/L at 298 K. An electric potential difference an aqueous solution of NaC1 with concentrations ClA of 100 mV is established between the two chambers. Estimate the diffusion flow of NaC1 and its direction if "-

ClB

--

D 1 = 1.48×10 -9 m 2/s Here, the externally established potential difference causes a concentration gradient and hence a flow, provided that the mobilities of the ions are different (Garby and Larsen, 1995). Due to electroneutrality, the local concentrations ofNa ÷ and C1- ions are the same, and are denoted by Cl(X). Also, the flows of negative and positive ions are the same, J+ - J-. The flows of ions are oppositely directed. Using the Nernst-Planck equation at constant T and P for Na + and C1- ions and eliminating the flows yields (10.81)

dc 1 _ _ C l ( b + - b - ) F d~b dx b + + bR T dx

This equation shows the magnitude of an externally induced concentration gradient. The gradient is zero if b + - b-, which is nearly true for KC1. The flow of NaC1 can be obtained by substituting Eq. (10.81) into Fick's law

J1 = DlCl b+ + bWith b÷/b - = t+/t - = 0.39/0.61 (NaC1), L = 0.005 m, and D )1+0.39/0.61 J1 =(1"48×10-9)(100)(1-0"39/0"61)t,

R-T i-

(10.82)

1.48

× 10 . 9 m 2 / s ,

the above equation yields

8.314(298)96'500(0.---~)0"1= 2.54×10_5 mol/(m2 s)

For arbitrary ionic valences, the flow may be obtained from J l = D l C l ( ( Z + ) 2 b ++--( zz--)b2-b - )

10.3.8

-R-f F ( ~x

(10.83)

Composite Membranes

For composite membranes (with compartments) the dissipation function • in terms of flows of volume, salt, and electric current, and the corresponding forces are xp = Jv (AP - A~r) + JsA/z c + I~t

(10.84)

where ~ is the difference in hydrostatic pressure across the membrane, A~r the difference in the osmotic pressure, and A/Zs~the concentration-dependent part of the chemical potential differences of the salt; Jv - JwVw + JsVs, where Vwand

521

10.3 Passivetransport

Vs are the partial molar volumes of water and salt, respectively. ~ is the electric potential and I the electric current. Here, Jv, Jw, and J~ represent the virtual flows. Experimentally, Jv is determined by measuring the change in volume of one or both compartments at opposite surfaces of the membranes. Equation (10.84) yields a set of three-flow linear phenomenological equations of conductance type

J' - L l l ( A P - A H ) + L I 2 ~

c~

Js = L21(AP- AH)+ L22 ( AHs ]+ L23~ Cs ) I

L31(AP-AH)+L32

(lo.85)

cs )

where AII~ is the difference in the osmotic pressure due to the permeant solute (salt) and cs the mean concentration of the salt given by %-

(10.86)

As Eq. (10.84) is an appropriately derived dissipation function consisting of the conjugate flows and forces, the Onsager's reciprocal rules states that L!/= LJ; The set of Eq. (10.85) can also be expressed in terms of the flows using the resistance coefficients K!j, and we have the resistance-type formulation (~iP- AII) = KllJ v + K l z J s +KI3I

~II~ Cs

- K21J v + K22J s + K231

(10.87)

dJ = K31Jv + K32J~ + K33I where the coefficients K;j are the inverse of the conductance coefficients L!j and are symmetric, K!j = ~;.

10.3.9

Two-Flow System

The set of Eq. (10.85) is related to various classical studies of electrokinetic phenomena, since the equations describe the coupled processes and yield naturally a number of symmetry relationships, which have been observed experimentally. Therefore, they provide a practical application of the linear nonequilibrium thermodynamic approach. For example, we may consider studies with identical solutions at each surface of the membrane, so that dill = AHs = 0. Then the system has only two degrees of freedom, and we have

= J,.~P + 14,

(10.88)

and the linear phenomenological equations become

.], = L11AP + L12g;

(10.89)

I = L12AP + L22¢)

(10.90)

These equations represent the basis of classical electrokinetics. For example, the magnitude of the electroosmotic volume flow per unit potential at zero pressure difference, • = (Jv/g0zp= o, and the streaming current per unit pressure difference at short circuit, (I/~J~),/,= o, must be identical. Equations (10.89) and (10.90) indicate that the existence of a pressure difference will produce an electric flow if the coupling coefficient is nonvanishing; when no pressure is applied, ~D = 0, the action of the electric force will cause a volume flow of water. We also observe the well-known Saxen relations between the ratios of force and flow ./,

522

10.

Membrane transport

(10.92)

=0 These symmetry relationships do not depend on the specific features of any given model but follow quite generally from the linear phenomenological equations of nonequilibrium thermodynamics. Therefore, any linear model that does not predict these relations is likely to be incorrect. Sometimes, we may consider a significant separation of salt from water in the volume flow. For solutions where L11AHi << AH, so that JvAH = JvAIIs, we have (10.93)

xI* = Jv AP - Jw VwAl-Is +

cwVw = 1, we have approximately

If we have dilute solutions, csVs <<

,w)

xlr= JvAP +

AH s = JvZXP + JDAH

(10.94)

Cw

The contents of the parenthesis show velocity of the salt relative to water and is called the diffusional flow JD. The dissipation function provides a natural basis for the analysis of systems in which mechanical energy derived from the volume flow and the hydrostatic pressure gradient are utilized to produce a separation of salt from water in the face of an adverse concentration gradient. In Eqs. (10.85)-(10.87), a system with three degrees of freedom is characterized by the six independent phenomcould be readily evaluated if it is possible to control the two enological coefficients. The conductance coefficients forces, for example

Lij

or

L12=

i O(AHs/cs) OJv I Ap-AII,4,

(10.95)

For the resistance coefficients K/j, it would be desirable to control the two flows. It may be more useful to consider alternative expressions of phenomenological equations without the need for further transformation of the dissipation function. For example, consider the set of Eq. (10.87) expressing the forces as functions of the flows. For practical purposes, it is desirable to use relations in which the independent variables are readily controlled experimentally. We may rewrite the set of Eq. (10.85) in such a way that AIIs/cs becomes an independent variable, while Js becomes a dependent variable, so that we have

( A [ I - AP) - KllKz2 - KlzK21 Jv q K12 AI-Is + KzzK13 - KlzK23 I K22 K22 Cs K22 Js = - K2------LJv + 1 AIIs + K23 I K22 K22 Cs K22 = K22K31 - K21K32 Jv -+ K32 Al-ls + K22K33 - K32K23 I K22

K22

Cs

(10.96)

K22

It is useful to replace the complex coefficients of Eqs. (10.81)-(10.83) with the practical transport coefficients; they may be evaluated experimentally under conditions in which two of the independent variables, Jv, AIIs/Cs, and/, are set equal to zero. Such a set of coefficients may be identified with six coefficients from the set of Eq. (10.96). Because of the Onsager's reciprocal relations, the remaining three coefficients may be evaluated as follows: = -

APs,I

= cs (1 - o-)

A IIs /Cs

Jv,i

(10.97)

10.3

Passive

)

(,s / T

,Ix,, 1

"

523

transport

Alls,J v

_ (AP-AH)

where cr is the

=_/3

I

IIs./

(10.98)

Vl Z l F

jv,_XUs

(10.99) Lp

reflection coefficient given by o-=[ A P~H)AH~ -

T1 the

(10.100) ,1v, I

•

transport number 7,--v,z~F[

dJ ) AII~Ic~ J

(10.101)

v ,I

/3 the

electroosmotic permeabiliw

[Jv]

/3- -7- -~'-~",~"~ and Lp the filtration

(10.102)

coefficient

Lp /

)

(10.103)

A P - AII arts,I

With these coefficients, the set of Eq. (10.96) can be expressed in a more useful form (AII-AP)-

~

Jv-cs(1-o-)

Cs - ~

AII~ J~ - c, ( 1 - o-)Jv + c~ w ~ •

( r, +

Cs

•

,--~J~-

) I

(10.104)

vlzlF

-

V l Z 1F

I

cs

+

)I

where oJ is the solute permeabili~

O)

and K the

i s/ ~

']v ' l

(10.105)

electric conductance (10.106)

Here, a molecule of the salt dissociates into v~ cations of charge zl and 1;2 anions of charge z2, and F is the Faraday constant. The set of Eq. (10.104) is useful for the treatment of a composite membrane consisting of compartments in series. The practical parameters above were derived long before the linear nonequilibrium thermodynamics formulations

524

10.

Membrane transport

were developed. A combination of these parameters in a self-consistent formulation provides a sound basis for analysis. Table 10.2 shows some of the membrane parameters. When the permeant solute is nonelectrolyte, there is no current flow to be considered, and the set of Eq. (10.104) can be expressed in terms of the flows as a function of the forces Jv = Lp (Ap - A I I ) - o-LpAIls

(10.107)

Js = Cs(1 - o')J v - wAl-~

(10.108)

These equations correspond, respectively, to the set of Eq. (10.87) for the case of dilute solutions. The value of reflection coefficient o-must depend on the nature of both the solute and the membrane. For the case of volume flow in the absence of the concentration gradient in the permeant solute (M-Is = 0), we see that the quantity (1 - o-) is a direct measure of the extent of coupling between the solute flow and the volume flow. If the membrane is completely nonselective, then o- = 0; if the membrane is perfectly selective, permeable only to the solvent, then o- = 1. In most cases, o- will lie between 0 and 1. A diffusion flow against its conjugate gradient driven by the dissipation of another diffusional process would be called "incongruent" diffusion. For example, the flow of the ith component across a membrane may be expressed by

Ji -- LiiA~i

+~ LikAtzk k=l

(10.109)

If A~i = O, but A~k 4= O, a flow of substance i may still be possible.

Example 10.7 Energy conversion in the electrokinetic effect

Electrokinetic effects are the consequence of the interaction between the flow of matter and flow of electricity through a porous membrane. The linear phenomenological equations for the simultaneous transport of matter and electricity are (Eqs. (10.89) and (10.90)) Jv = LllAP

+ L12A~

I = L21AP + Lz2A~ where I is the electric current per unit area, Jv the volume flow of matter, At/, the potential difference, AP the pressure difference, and L/j the phenomenological coefficients defined as

Jv)

(hydraulic permeability, filtration coefficient)

L12--(~)

(10.110)

(electroosmosis)

(10.111)

(streaming current)

(10.112)

(conductance of permeant-filled electroosmotic cell)

(10.113)

AP=O

L21 = | - : - ~ | ~x~-0

L22 "- ( ~ - ~ ) AP=0

Since L12 = L21, the electric current per unit pressure force at A~p = 0 is equal to the volume flow J~ per unit potential difference at AP = 0, (10.114)

10.4

525

Facilitated and active transports in membranes

Table 10.3. Efficiency of electrokinetic energy conversion for various mixed-lipid membranes Type of membrane

fi~o ( >: 10 ~)

/3~p ( X 105 )

v/........ ( X 105 )

~/....... p ( X 105)

Cephalin-serine Cephaline-inositol Lecithin- serine L ecithin-ino sithol Lecithin-cephalin Inositol-serine

2.108 2.235 3.687 3.946 5.855 10.044

1.714 2.255 3.686 3.942 5.852 10.156

0.527 0.558 0.921 0.986 1.463 2.511

0.428 0.563 0.920 0.985 1.463 2.539

Source" Caplan and Essig (1989).

The efficiencies of electrokinetic energy conversion for two operations modes, namely electroosmosis Tieo and streaming potential rlsp, are expressed as Tie°

=

T~sp --

JvzXP IA4t -

output input -

JvzSat' - (A0) ~ 2/R

IAO

output

(AO)2/R

Jv2kP

input

JvAP

(10.115)

(10.116)

where R is the resistance. The maximum values of energy conversions occur when the output forces equal half of their steady-state values. For example, r/eo is maximum when z5,¢'equals half the value of electroosmotic pressure zXP = - _ l ( A p ) j ,

=0

2 The maximum energy conversion efficiency may be related to the m e r i t ~ through the degree of coupling q TImax

(1 "+"/~)1/2 _ (1 -+-/~) 2 + 1

1 (10.117)

with (10.118) where q - Li//(LiiLij) 1/2 with L O. = Lji. When the value of/3 is much smaller than unity, then we have TImax ~ / 3 / 4 . Due to Onsager's reciprocal relations, we have TJmax,eo -- 7~max,sp

(10.119)

Table 10.3 shows the efficiency of electrokinetic energy conversion for mixed-lipid membranes.

10.4

FACILITATED AND ACTIVE TRANSPORTS IN MEMBRANES

Facilitated transport in a membrane involves a chemical agent as a carrier to increase the passive transport. A chemical agent can react reversibly with a permeant, and yields high selectivity and permeability, which makes facilitated transport a very attractive separation technique. The chemical agent carries the substance in the form of a carrier-bound substance; the carrier releases the substance on the other side of the membrane due to chemical conditions (mainly pH and electric charge) and diffuses back. Usually a carrier with high association and dissociation rate constants that are similar in magnitude is desirable. For the transport to be selective, the membrane is permeable to the flow of certain substances and impermeable to the flow of others including the carrier molecule. Therefore, transport is not described by Fick's law and exhibits saturation at a higher concentration of the permeant. Various substances such as amino acids, organic acids, NaOH, NaC1, carbon dioxide, oxygen, and metals, and various ions such as Cd(II), Cu(II), Co0I), and Fe(IIIi, can be separated by using suitable carrier agents in liquid or solid

526

lO.

Membrane transport

composite membranes. Liquid membranes behave like double liquid-liquid extraction systems where the usage of organic solvent is minimized. Such devices are generally prepared as bulk liquid, emulsion liquid, and supported liquid membranes.

10.4.1

Liquid and Composite Membranes

Supported liquid membranes and emulsion liquid membranes are widely used in facilitated transport systems. However, they are not suitable for large-scale industrial applications, and they are not stable, since chemical mediators are easily lost in liquid membranes. One of the methods for overcoming this disadvantage is the use of an ion exchange membrane as the support. Also, solid polymer composite membranes containing a chemical agent are being tested successfully in separation technology. Some solid composite membranes are Nafion-poly(pyrrole) films with silver or sodium, activated composite membrane, and solid polymer electrode composite membranes. Composite membranes are stable and suitable for industrial applications, and are usually made of a support polymer matrix of porous structure to which a chemical carrier is added. The preparation of polymer layers containing different amounts of carrier agents may require special polymerization techniques, such as interfacial polymerization. In ion separation, composite membranes utilize a chemical agent as a carrier dissolved in an organic solvent contained in a polymeric matrix or within the pores of a polymer membrane. For example, pseudo-crown ethers can be used as the fixed site carriers in ion separation. Supported liquid membranes and composite membranes allow wide-scale applications in industrial separation, leading to far less expensive processes. For example, olefins, amino acids, heavy metals, gases, fatty acids, water, and inorganic salts can be separated selectively by facilitated transport. Transport in membranes is mostly a complex and coupled process; coupling between the solute and the membrane, and coupling between diffusion and the chemical reaction may play an important role in efficiency. It is important to understand and quantify the coupling to describe the transport in membranes. Kinetic studies may also be helpful. However, thermodynamics might offer a new and rigorous approach toward understanding the coupled transport in composite membranes without the need for detailed examination of the mechanism of diffusion through the solid structure. Table 10.4 shows some of the applications of facilitated transport.

10.4.2

Active Transport in Artificial Membranes

Some artificial membranes are used for active transport. For example, the active transport of metal ions through synthetic polymer membranes is used; some specific examples are the cation-exchange membranes from 2,3-epithiopropyl methacrylate (ETMA) and 2-acrylamide-2-methyl propane sulfonic acid (AMPS) copolymers for the active transport of alkali and alkali earth metal ions, and ETMA-AMPS copolymer membranes for the active transport of the amino acids glycine, penylalanine, and lycine by using hydrochloric acid as the receiving solution. One possible mechanism for achieving this is that glycine and water are transferred by osmotic pressure into the membrane, and then glycine is protonated with H + released from sulfonic acid groups in the membrane; later, the protonated glycine is transported by means of sulfonic acid groups to the other side of the membrane regardless of the smaller electric potential difference. The transport of amino acids depends on the composition of the membranes and the structure of the amino acids. A membrane with the function of active transport can recover uranyl ions UO2 2+ in the eluate. Uranyl ions form anion complexes with sulfate ions in sulfuric acid, and can be transported against their concentration gradient through a liquid membrane with tertiary amine by using carbonate solution as the receiving solution. Polymeric anion-exchange membranes can also transport uranyl ions selectively from the eluate of sulfuric acid containing alkali earth metal ions or cupric ions. Table 10.5 shows some applications of active transport.

10.5

BIOMEMBRANES

A cell membrane is a fluid mosaic of lipids and proteins. Phosphoglycerides are the major membrane lipids that form a bilayer with their hydrophilic head groups interacting with water on both the extracellular and intracellular surfaces, and their hydrophobic fatty acyl chains in the central and hydrophobic regions of the membrane. Peripheral proteins are embedded at the periphery, while integral proteins span from one side to the other. Biomembranes separate the contents of the cell from the external environment. Some of the proteins are involved in the transport of substances across the membrane, and some other proteins are enzymes that catalyze biochemical reactions. Proteins on the exterior surface can function as receptors and bind external ligands such as hormones and growth factors. Proteins migrate in an electric field; positively charged proteins

10.5

527

Biomembranes

Table 10.4. Some facilitated membrane transport systems

Permeant

Carrier

Membrane

Analysis

References

Amino acids (p-amino benzoic acid) Amino acid phenylalanine

Undeconoic acid

Organic liquid membranes ELM

R

Uegla and Zanoaga (1989)

TR+R

Liu and Liu (1998)

R

LNET TR+ R

Buschman et al. (1999) Calzado et al. (2001 ) Narebska and Staniszewski (1998) Nakano et al. (1996)

Amines and amino acids

Di(2-ethylhexyl) phosphoric acid Macrocyclic ligands

Lactic acid Polyunsaturatedfatty acid esters from sardine oil Binary organic acids Orthoboric acid Boric acid Boric acid Organic residues and inorganic salts Silver Alkali metal ions Cd(II) ions

Neutralization Ag ~

Liquid membrane Composite membrane Anion-exchange membrane ELM

Tri-n-ocytylamine Counterions 1,3-Diols Borate ions Polymeric liquids

SLM Ion-exchange membrane SLM Ion exchange membrane SLM-polymeric

TR+R LNET TR+R TR+R TR+R

Juang et al. (1998) Selegny et al. (1997) Bachelier et al. (1996) Selegny et al. (1994) Ho et al. (1996)

D2EHPA Phenoxy compounds Cyanex-302, DeEPHA

ELM Bulk liquid membrane SLM

TR+ R TR+R TR+R

Carbon dioxide

Monoprotonated ethylene diamine EDAH Tetramethylammonium fluoride tetrahydrate Polymers Diamines Water NaC1 Salt Silver tetrafluoroborate

Ion exchange membrane

TR+R

Lee et al. (1996) Rankumar et al. (1999) Daoud et al. (1998) Gumi et al. (2000) Selegny et al. (1995)

Liquid membrane

TR + R

Quinn et al. (1995)

SLM Ion exchange membrane Nafion 120 Anion-exchange membrane Anion-exchange membrane Composite (poly ethylene oxide) Nafion-poly(pyrrole)

TR + R R LNET LNET LNET

Jeong and Lee (1999) Matsuyama et al. (1996) Narebska and Koter (1997) Narebska et al. (1995) Narebska and Warszawski (1994) Pinnau and Toy (2001 )

Carbon dioxide Carbon dioxide Carbon dioxide Ion HC1 Acid Olefins

Silver-I, sodium-I

Sungpet et al. (2001 )

Note: (ELM) Emulsion liquid membranes; (SLM) supported liquid membranes; (LNET) linear nonequilibrium thermodynamics; (TR) conventional transport equations; (R) conventional rate equations.

Table 10.5. Some active transport systems

Permeant

Carrier

Membrane

Analysis

References

Uranyl ions

2,3-Epithiopropyl methacrylate-dodecyl methacrylate-methacrylamide propyl trimethyl ammonium terpolymer Reversed bienzyme 2,3-Epithiopropyl methacrylate-2-acrylamide-2-methyl propane sulfonic acid (14C)-methylamine, (14C)-thiocyanate

Anion-exchange membranes

R

Nonaka and Kawamoto (1995)

Porous membrane

LNET

Nigon et al. (1998)

Cation-exchange membranes

R

Nonaka et al. (1993)

Inverted membrane vesicles of Escherichia coli

R

Dung and Chen( 1991 )

Anion Amino acids: glycine, phenylalanine, lycine D-Lactate

Note: (LNET) Linear nonequilibrium thermodynamics approach: (R) conventional rate equations.

are cations and m i g r a t e t o w a r d the cathode, while negatively c h a r g e d proteins are anions and m i g r a t e t o w a r d the anode. T h e r e are several m e c h a n i s m s for e x p l a i n i n g h o w biological m e m b r a n e s can transport c h a r g e d or u n c h a r g e d substrates against their t h e r m o d y n a m i c forces. It is w i d e l y accepted that cross-transports by a protein are discrete events. B i o m e m b r a n e s contain e n z y m e s , pores, charges or m e m b r a n e potentials, and catalytic activities associated with the transport o f substrates. It is well established that the electrostatic interactions b e t w e e n the m e m b r a n e and a c h a r g e d

528

10.

Membrane transport

solute may play an important role in transport. Thus, we have to establish reliable links between the membrane's charge effect and the pore size, length distribution, and density to describe the interactions. 10.5.1

Reaction-Diffusion in Biomembranes

From the local conservation of mass, we obtain

OCi --

Ot

dJi t-ViJri

(10.120)

dx

At steady state, the local concentrations do not vary with time, and we have

dJi - PiJri dx

(10.121)

This equation shows that a stationary state imposes a relation between the diffusion and chemical reactions, and is of special interest in isotropic membranes where the coupling coefficients vanish. For a homogeneous and isotropic medium the linear phenomenological equations are

dt~i -- Z KikJk

(10.122)

A=KrJ r

(10.123)

dx

k

where A is the affinity, and Kik represents the resistance coefficients. Here, diffusion and chemical reaction are not coupled since the cross-coefficients between the scalar chemical reaction and vectorial mass flow vanish in an isotropic medium, according to the Curie-Prigogine principle. If the coefficients Kji are independent of position, we can differentiate Eq. (10.122) with respect to x, and insert Eq. (10.121) to obtain

dZld'i - Z dx2 k

Kik -~x -

VkKik

(10.124)

Jr

After multiplying both sides by vi and summing, the above equation becomes

--~_.a~i = Z-PiVkgik i dx2 i,k

Jr

(10.125)

From the definition of affinity A, we find

d2A-(dx2 - ~ffivid2tXk)dx2

(10.126)

After substituting Jr from Eq. (10.123), we have

(10.127)

The expression

Zi.kViVkKie/Krhas the dimension of cm -2 and will be denoted by A -2 d2A dx 2

-A-ZA

to obtain (10.128)

The characteristic parameter A is called the relaxation length of the coupled reaction-diffusion processes within a membrane.

10.5

529

Biomembranes

For a membrane thickness of Ax, dimensionless number A/zXr is closely related to the Thiele modulus used for the characterization of heterogeneous reaction columns. This dimensionless quantity is also related to the relaxation time of chemical reaction rr and the average relaxation time of diffusion processes rd as follows: ( A )2rr -A-TX 2<7"d)

(10.129)

Ifa reaction relaxes faster than the time necessary for diffusion across the membrane % < iTd), then A will be smaller than ~vc, and the reaction will reach equilibrium on the surface after the reactants have diffused only a short distance within the membrane. On the other hand, when Ax is very small, which is the case in biological systems, then % and irdl may be of the same order of magnitude, and hence the system cannot be treated as an equilibrium state.

10.5.2

Compartmental Structure and Dissipation Function

For a simple derivation of the dissipation function, consider an isothermal composite system with three compartments consisting of two external chambers (I and II) and a membrane compartment (m) in between. The volumes of the compartments are constant (dVl = dVll = dVm = 0). The Gibbs relations for the compartments are dUj - TdS l + Z ~i,I dNi, t

(10.130)

i

dUll - T d S n + ~_, I~i,II dNi, II

(10.131)

i

]-JJi,lll dNi, m

d u l l I -- TdSI11 + 2

(10.132)

i

It is assumed that reactions take place only in the membrane, and the net change of the number of moles is expressed by dN,, m - dNi,exch + dNi, react

(10.133)

where dNi, exch is the number of moles of the component i exchanged with the surrounding compartments and dNi,reac t denotes the number of moles of component i produced by the chemical reaction, which is expressed for the kth chemical reaction as follows:

(10.134)

dNi, react = Z v i k d e k k

where vik is the stoichiometric coefficient of the ith component in the kth reaction and e is the extent of advancement of the kth chemical reaction. Combining Eqs. (10.124)-(10.126) we obtain dUm - rdSm + Z I~idNi, exch - Z Ak, mdek i

(10.135)

k

where Ak,m is the affinity of the kth reaction within the membrane. Since the composite system is a closed system, we have dNi. I + dNi,ii + dNi, m - 0

(10.136)

dlj" l + dUii + d U m - 0

(10.137)

where - d N i , i is the number of moles of the ith component gained by the membrane and dNi,ii the number of moles of component i lost by the membrane through the diffusion processes, and the input and output flows are expressed by d N i.l J,

in

•

.

.

.

J i out -

.

dt

'

'

d N i,ii dt

( 10.138)

530

10.

Membrane transport

The macroscopic driving forces for the flows are given by A~i, in -/J"i,I -/'l'i,m

and

A/~i, out -- /-gi,m -- }t/,i,ii

(10.139)

The total rate of the kth chemical reaction within the membrane is (10.140)

Jrk -- dek dt

The entropy contributions of the three compartments are expressed by dS I + dSii -Jr-d S m -- diS I + diSii -Jr-diS m = diS

(10.141)

The contributions from the exchangeable entropy terms deS cancel one another in the above equation. By combining Eqs. (10.130)-(10.132) with Eqs. (10.135)-(10.137), we obtain TdiS + Z (/J,i,i - t'Li,m)dNi,I - £ (~i,m -- I~i,ii)dNi,ii - Z Ak, mdSk : 0 i i k

(10.142)

Dividing the above equation by dt and using Eqs. (10.138)-(10.140), we obtain an expression for the dissipation function xlr = r diS = Z Ji, inAi~i, in + Z Ji, outA~i, out + Z Ak, m Jrk dt i i k

(10.143)

Assuming that under steady-state conditions, the difference between the output and input flows of the ith component is equal to the amount of ith component reacted in all chemical processes, we have (10.144)

Ji, out - Ji, in = Z PikJrk k

The last term in Eq. (10.143) can be expressed by the difference of the output and input flows

Z A k ' m J r k - - - - Z k (~i

i

k

i

(10.145)

Introducing the above equation into Eq. (10.143), we obtain aYlf= Z Ji, in ~i,I - Z Ji, out ~i, II i

(10.146)

i

However, only the difference in chemical potentials is measurable, and the above equation can be further transformed for practical use by introducing the affinity of the kth reaction in compartment II, Ak, c Ak, c -- - - Z l~ikldCi,II i

(10.147)

By introducing Ji,out from Eq. (10.138), we obtain XIt -- Z Ji, in (~i,I - / ~ i , II ) + Z Jrk Ak, c = £ Ji, inZ~l~i + £ Jrk Ak, c i k i k

(10.148)

with

A~i =

].Li,I -- jtLi,II

which is the cross-membrane difference of chemical potentials, and it is measurable.

(10.149)

531

10.5 Biomembranes

10.5.3

Active Transport

Active transport is found in biological desalination and ion separation. Salt excretion by the glands of desert plants and accumulation of potassium in certain bacteria against a very large concentration gradient are some examples of active transport. Active transport is a highly selective process; it can be prevented by specific metabolic inhibitors, and is closely related to facilitated transport. The total turnover per unit membrane per unit time is Jr, tot --

f~

Jr dx 4= 0

(10.150)

X

where Jr is the local reaction coupled to the transport of substrates, x a coordinate normal to the membrane surface, and z£v the membrane thickness. In facilitated transport, Jr, tot-- 0. Biological membranes show anisotropy, as their molecules are preferentially ordered in a definite direction in the plane of the membrane, and the coupling between chemical reactions (scalar) and diffusion flow (vectorial) can take place. Almost all outer and inner membranes of the cell have the ability to undergo active transport. Sodium and potassium pumps operate in almost all cells, especially nerve cells, while the active transport of calcium takes place in muscle cells. The proton pumps operate in mitochondrial membranes, chloroplasts, and the retina. The following representative dissipation function can be used in the thermodynamic formulation of active transport: (10.151)

q, = J / A f , ~ + JrAr

where A/2; is the difference in electrochemical potentials, ~ and Jr the diffusion flow and the rate of reaction, respectively, and A the conjugate affinity within the cell. The phenomenological relations in terms of the forces are expressed by

A£, - £ K!/Jj + K~rd r

(10.152)

/--1

&' .4 r - ~_, Kr/J~ + K,.rJ r

(10.153)

/=1 The cross-coefficient Kir represents the coupling between the diffusion flow and the rate of reaction. Since the diffusion flow is a vector and the rate of reaction is a scalar, the coefficient Kit must be a vector. For nonvanishing values ofKir , biological membranes must have a structural anisotropy in the direction of the diffusion flow. Biomembranes have a topological organization, which determines the pattern of flow across the membrane, and hence Eqs. (10.152) and (10.153) are applicable to the local flows and forces in a biomembrane system. We can integrate the expressions over the membrane thickness at steady state so that the parameters do not change with time. The rate of metabolic reaction Jr may be taken as the rate of oxygen intake by a tissue, and J; may be obtained from Eq. (10.152)

Ji - A fzi - £ Kii

Kii

/=, -~ii J

Kir - ~ii Jr

(10.154)

The last term on the right side represents the active transport of substance i. Equations (10.152) and (10.153) can be used to analyze sodium flow in frog skin. The flow of sodium chloride across the skin comprises the flow of sodium ions JNa, which is coupled to the metabolic process Jr, tot, while the flow of the chloride ions Jcl may be assumed to be passive transport. The driving forces for the ionic flows are the electrochemical potential differences, and are given for a component i in a simple system as follows: A/2i = R T (In ci, 1 - In Ci, II ) -+- Z i F (@i - @II )

(10.155)

where I and II denote the surrounding compartments adjacent to the membrane, z i is the valence of the ion, OI the electric potential, and F the Faraday constant.

532

lO.

Membrane transport

10.5.4 PhenomenologicalEquations Using the dissipation function given in Eq. (10.151), the phenomenological equations for the sodium pump can be expressed by A~Na -- KNa JNa -+- KNarJr,tot

(10.156)

z~/-tc1 -- Kc1Jc1

(10.157)

A = KrNaJNa -k- KrJr, tot

(10.158)

Equation (10.157) implies that no coupling exists between sodium and chloride flows, while the sodium flow JNa is coupled to the metabolic reaction Jr. Equations (10.156)-(10.158) can be applied to the following experimental cases: (i) Short circuit: Consider two electrodes inserted in each compartment. If these electrodes are short circuited, the potential difference (~I - ~bm)is made zero, and an electric current I is allowed to flow across the membrane. If the experiment is carried out at equal salt concentrations in I and II, so that In c i - In cii = 0, and hence we have A/2Na = 0 and A/2Cl - 0, the only remaining driving force is the affinity A of the metabolic reaction in Eq. (10.156), and we have JNa = -- KNa------~rJr tot KNa '

( 10.159)

Since the flow of electricity is determined by the ionic flows I = (JNa -- Jc1) F, from the above equation we have KNar

I =

KNa Jr, totF

(10.160)

Under the short-circuited measurements, the electric current is linearly related to the overall rate of reaction, and the coupling coefficient KNar 4= 0. (ii) Open circuit: In an open-circuited potentiometric experiment where I = 0, both JNa and J¢~ vanish, and Eq. (10.156) becomes ~]~Na -- KNarJr, tot

(10.161)

At steady state, the concentrations and electric potentials in compartments I and II have to be different, and we have RT In CNa'I -+"F (~bI - ~tii ) = KNa r Jr, tot

(10.162)

CNa,II

R T In ccl:I - F (~bI - ~bii) = 0

( 10.16 3)

CCI,II

Adding Eqs. (10.162) and (10.163) yields RT In CNa'ICcI'I = KNar Jr, tot

(10.164)

CNa,IICCI,II

Since CNa = CC1in the compartments, and using the salt concentration denoted by Cs, we have

2RT In

Cs,I

= KNar Jr, tot

(10.165)

Cs,II

The above equation shows the salt distribution due to nonvanishing coupling coefficient KNa r. If the total rate of the chemical reaction is known, short-circuit and open-circuit experiments allow us to determine the straight and cross-coefficients.

10.5 Biomembranes

533

The chemical and electrochemical composition of the extracellular medium is essentially different from the intracellular medium. Equilibrium is achieved when the electrochemical potential of ions is the same on the inside and outside, and is expressed by RT(lnci - lnc0) = F z ( O o - 6 i )

(10.166)

This equation is called Nernst's law and relates the diffusion to electrochemical flow. We can restate Nernst's law in the following form: qJo - ~ / - k--T-Tin c / Q Ca

(10.167)

where k is the Boltzmann constant (k - 8.62 × 10- 5 eV/K) and Q the charge of each ion of the substance (Q is positive for potassium). Potassium leaves the cell, while the net flow of sodium is inward. A nonequilibrium stationary state for the cell at rest is maintained by the sodium and potassium pumps, which pump out the entering sodium ions and pump the leaking potassium ions back into the cell interior, using a certain metabolic output. The sodium transfer is coupled with the chemical reaction. The electrochemical potential difference for sodium ions is expressed as XNa -- /~Na,ext --/'~Na, int

(10.168)

JNa = LNaXNa + LNarA

(10.169)

Jr = LNarXNa + LrA

(10.170)

The related phenomenological equations are

According to the Onsager's relations, three coefficients are to be determined. They are the passive permeability to sodium LNa, the metabolic reaction coefficient if there is no sodium transport Lr, and the cross-coefficient between the chemical reaction and the sodium flow LN~r. The linear nonequilibrium thermodynamics formulation for the active transport of sodium and the associated oxygen consumption in frog skin and toad urinary bladders are studied experimentally. Sodium flow JNa is taken as positive in the direction from the outer to the inner surface of the tissue. The term Jr is the rate of suprabasal oxygen consumption assumed to be independent of the oxygen consumption associated with the metabolic functions. It is essential that the phenomenological coefficients evaluated remain near-constant in the course of perturbations of the external variables. In general, these coefficients and A will be the functions of state, and may be influenced by perturbation, which alters the tissue confiiguration and composition. IfXNa is changed by the perturbation of AO alone, then Eqs. (10.169) and (10.170) become JNa = LNa ( - F A O ) -+-LNar A

(10.171)

Jr =- LNar ( - F A i / / ) -Jr-LrA

(10.172)

where AO = O i - ~0. For constant affinity A, the phenomenological coefficients are (1){dJNa)

(10.173)

(10.174) d(A~))

L~ -

Yro A

(10.175)

where Jro is the rate of oxygen uptake. The quantitative values of the phenomenological coefficients indicate the character of the coupled transport.

534

lO.

Membrane transport

We can express the phenomenological equations (10.169) and (10.170) in terms of the resistance formulation

XNa = KNaJNa +KNarJr

(10.176)

A = KrNaJNa -{-Xr J r

(10.177)

10.5.5 Degreeof Coupling We can define the phenomenological stoichiometry Z and the degree of coupling q as follows: Z=

q=

(10.178)

K~

KNar ( (Jr)JNa =0 ) (KyaKr)l/2= 1 ~o

(10.179)

As for the general case, the degree of coupling is - 1 <- q -< 1. Two processes will be coupled if they are combined in a single overall reaction system. When the coupling protein guides the reaction along a path involving both processes, its specificity and transport properties have to be altered through the substrate binding energy level. The ratio of coupled to uncoupled rates, called the tightness of coupling, is a function of the binding energy; a strong binding may limit the tightness of coupling by reducing the rate of substrate dissociation after the substrate has been transferred through the membrane. From Eqs. (10.176) and (10.177), we can express the flows as functions of the forces and resistance coefficients

JNa = yNa + ( q / Z ) A KNa(1-- q 2)

(10.180)

Jr = (q/Z)Xya +(1/Z)ZA

(10.181)

KNa (1 -- q2 ) The above model is useful; however, biological membranes, which transport various substances, are complex systems. Such membranes are close to composite membranes with series and parallel elements. A value of q < 1 shows an incomplete coupling, where a metabolic energy must be expended to maintain an electrochemical potential difference of sodium even in the absence of active transport, that is (Jr)JNa=0 4: 0.

Example 10.8 Coupled system of flows and a chemical reaction For a specific membrane, the phenomenological equations relating the flows and forces of either vectorial or scalar character may be written. Such flows and forces must be derived from an appropriate dissipation function. Consider the following dissipation function:

-- JwA~w -q-J1A/~I -+-J2A/~2 + Jr A

(10.182)

where the subscripts "w,.... 1,' and "2" refer to water, cation, and anion, respectively, and A and 4 refer to chemical affinity and chemical rate, respectively. Therefore, we have the corresponding linear relations in the resistance-type formulations where the forces are expressed as the function of flows (10.183)

~l~w "-" gwwff w q- gwlff 1 "Jr-Kw2J 2 ~1

-- KlwJw q- K11J1

-+-K12J2 "+"KlrJr

A~2 = K2wJ w Jr-K21J1 -+-K22J 2 +

A = K r l J 1 "+"Kr2 4 "Jr-Krr 4

K2rJ r

(10.184) (10.185) (10.186)

535

10.5 Biomembranes

In systems with several flows interacting, the resistance coefficients can reflect the extent of the interactions directly. Also, the resistance formulation utilizes the flows as independent variables, and often it is easier to measure and control the flows than the forces. The nonzero values of Klr and Kzr indicate coupling between the ionic flows and the reaction. Since both Jr and A are scalars, the coefficients K;r must be vectors. Consider that initially uncharged species M and O move into the cell, where M is reacted and transformed into N. Some of the N flows out of the cell. The transformation is mediated by the action of an enzyme confined to the interior of the cell. The component O does not take part in the transformation. However, the flow of O is coupled with the flow of M. The linear phenomenological equations are AIAM = K M J M + K M o J o

(10.187)

A#N = KNJN

(10.188)

A~o = KMOJ M + KoJ 0

(10.189)

Here, J refers to inward fluxes, while A ~ M -- ~ i e x - ~iin. Eqs. (10.187)-(10.189) indicate the coupling between certain flows and lack of coupling between others. The metabolic reaction occurring in the cell is not coupled to any of the flows. After a certain time has passed the system reaches a state in which the concentrations of M and N, but not that of O, become constant, so that in the stationary state we have J r - JM -- --JN

(10.190)

IfA ex is the affinity of the reaction measured externally, where the requisite enzyme is absent, we have A ex - A in = ( / x ~

_

ex ) - ( ] z ~ /d,N

in ) = A/AM -- A~N - jAN

(10.191)

Ain refers to the affinity of the reaction in the cell. By substituting Eq. (10.190) into Eqs. (10.187)-(10.189), we get &IJ~M -- KM J r + K M o J o

(10.192)

A#N = -KNJ r

(10.193)

A#o = KMOJ r +

Kod o

A in - KriJ r

(10.194) (10.195)

The set of Eqs. (10.192)-(10.195) is combined with Eq. (10.191), and we have /Xtz o = K o J o + K M O J r

(10.196)

Aex = KMOJo +

(10.197)

KreJ r

where Kre = Kri + KM + KN. The only constraint is the fixing of A ex, and the system eventually reaches a state in which Jo = 0. In this state, an accumulation of O occurs, which is given by ~tz o - KMOJ ' -

KM° Aex

(10.198)

Kre

The stationary state of the whole cell, represented by Eq. (10.190), yields a new dissipation function corresponding to Eqs. (10.196) and (10.197) - J o A # o + JrA ex

(10.199)

In this case, a stationary-state coupling occurs between the flow of component O and the reaction. The coupling is generally a property of the membrane, and is associated with enzymes that are an integral part of the membrane. Table 10.1 shows the permeability and reflection coefficients for some biomembranes.

536

10.

Membrane transport

~ | | | 0 | | | | 0 | | | | | 0 | | | | | 0 0 | | | | | | | | | | | | | | | | | | | | | | | | 0 | | | | | 0 | 0 | | | | | |

'.= Glucose i cvl ~ + oxygen | ~ =/ I \

.. ATP cvz , cv3 .

cv411

CV[I; Sj

t

+water =

!i

|! |~.~~ i

~~LTPoot~ ~ o ~ ~

o[] ' I ~ o ~ o o

-q/

Membrane .......

1Jr4

Carbon dioxide

-q2

I i=

Extracellular space Ce qJe cv5

Ci~i • Intracellularspace"

Ii! i

Jr5

-q4

i

l

I |;

~ o ~ l , ~ o ~ ~ ~ o ~

-q3

", i!

~o~o

,,n

-45

Figure 10.3. Active transport of Na + ions from intracellular to extracellular space and from low to high concentration regions (Garby and Larsen, 1995).

10.5.6

Active Transport and Energy Conversions

Figure 10.3 shows a simplified schematic of active transport (Garby and Larsen, 1995). If changes in the kinetic energy are small, then the first law for a control volume (cv) becomes d(n~__~) +

~.~~

_ ~.. ~Ta - gt + m

out

dt

(10.200)

in

where f~ = u + z F q , and h = h + z F q , , z being the valency of the ions, F the Faraday constant, and qJ the electric potential. As shown in Figure 10.3, Na + ions can flow from low electric potential of the intracellular phase to high electric potential of the extracellular phase• In this highly simplified schematic of active transport, the net reactions for energy transformation begin with the oxidation of glucose, which supplies chemical energy mch1 to the synthesis of ATE The hydrolysis of ATP ATP + H 2 0 ~ A D P + Pi can supply chemical energy mch2 to various processes that are coupled to the transport of Na + ions to the extracellular space, while the return flow of Na + ions takes place as passive diffusion. During this complex and coupled energy conversion, various heat flows at different rates occur. Figure 10.3 contains the five chemical reactions taking place in five control volumes. The reaction rates are represented by vertical arrows; Jr1 = ho for glucose oxidation, Jr2 -- nATP for ATP production, Jr3 = nATP for ATP hydrolysis, Jr4 -- tiNa for active transport, and Jr5 = nNa for leakage of sodium ions. The choice of control volumes and mass and energy balances requires that

Jr3 = Jr2,

Jr5 = Jr4,

J/Vchl= ~rch2 -- q2 -- q3

(10.201)

The energy balances for each control volume are

JrlAHr, G = ql -- Wchl

(10.202)

Jr2 ( - AHr, ATP ) = 02 + Wchl

(10.203)

Jr3 (z~-Ir, ATP ) = q3 -- ~rch2

(10.204)

Jr4[ZNa+F(Oe - @ ) ] =

Jr5 [ZNa+ F(~bi

-

t)4 "k-Wch2

Oe)] = 05

(10.205) (10.206)

537

10.5 Biomembranes

Equations (10.205) and (10.206) are based on the assumptions of isothermal and ideal mixtures. Using Eq. (10.201) and summing Eqs. (10.202)-(10.206) yields the energy expenditure ~+ for the control volume cv d~l AH~.(; - c)l + q2 + q3 + q4 + c)5 + I'Vch2 = ~

(10.207)

With JrJ = hc and Jr5 = Jr4 :/~/Na <, summation of Eqs. (10.202)-(10.205) yields nGAHr, G +/tNa [ZNa-f(ff/e - O i )]-- ql -k- q2 + q3 + q4

(10.208)

This equation shows that part of the chemical energy of the oxidation of glucose is consumed for the flow of electric potential energy, and that part is removed as heat. Table B9 lists some of the reaction enthalpies at specified temperatures. The efficiency of the conversion of chemical energy to electric potential energy may be defined by {" nNa=-[ZNa~ff(Oe-- ~i)]} r/= (10.209) -- nG AHr, G Matching the supply of chemical energy W~hl to the chemical energy Wch2 necessary for active transport requires highly complex and organized thermodynamic buffering systems, which may include uncoupling, slips, and leaks.

E x a m p l e 10.9 A representative active transport and energy conversions This example is from Garby and

Larsen (1995). Consider a tissue with active transport of Na + ions from the intracellular to the extracellular space. Assume that 40% of the energy of ATP hydrolysis can be used for the transport and 38 mol of ATP are formed per mole of glucose (G) combusted. At steady state, the extracellular concentration ofNa + is observed to be 145 mM and the intracellular concentration to be 12 mM, and the electric potential of the intracellular fluid is determined to be - 9 0 mV in relation to the extracellular space. The transcellular flux ofNa + ions is 1 mmol/min and the temperature of the tissue is 37°C. Estimate the oxygen consumption, heat output, and energy expenditure of the process when Z~-/r, ATP : - 2 0 kJ/mol for hydrolysis of ATP and M-/r,G = - 2 8 6 7 kJ/mol for glucose combustion. Assume that 02/01 = 1 and £/4/05 = 4/5, and Jr3 = Jr2 and Jr5 -- Jr4 (stationary state). Solving this example requires information about the partition of the total heat power in Eq. (10.207) among the five different parts of the total process Jrl~-/r,G -- ~/1 -'}-02 + 03 -t- £/4 + 05 -'t-~'rch2 --

The unknowns are two reaction rates (Jr1 and Jr2), tWO internal energy flows (mch 1 and Wch2), and the five heat powers. Besides Eqs. (10.202)-(10.206), we have the following additional relations from the data above: mch2 --

0"4~/chl,

Jr2 _ 38, Jrl

q__L2= 1, ql

04

_

05

4

(10.210)

5

Therefore, Eq. (10.206) yields 05 = Using

0.001) 60 [(1)(96,500)(0.090)] = 0.145W

q4/~/5 - 4/5 in Eq. (10.205) yields

Wch2 =

05 = 0.261 W,

Wch1 --

0.261

-- 0.653W

0.4

Using Eq. (10.210) and Table B9, c)1 can be eliminated from Eqs. (10.202) and (10.203); the reaction rates are estimated as Jr~ =

--2Wch , z~r-/r, G + 38AHr.ATp

=

--2(0.653) - 2 8 6 7 - 38(20)

= 0.360 txmol/s

and

Jr2 = 38(0.360) = 13.7 txmol / s

538

lO.

Membrane transport

The oxygen consumption is obtained from the oxidation reaction of glucose C6H120 6(aq) + 60 2(g) ~ 6CO 2(g) + 6H 2O(1) ho2 = 6(hG) = 2.16txmol/s The heat flow rates are estimated by --ql = --02 =

--03 =

0.360 × 10- 3 (-2867) + 0.653 = 0.379 W

13.7×10-3(20) - 0.261 = O.O13W

-04 =(4) (0.145) = 0.116W -05 = 0.145W Therefore, using Eq. (10.207), the total energy expenditure becomes ~/" = 0 . 3 7 9 + 0 . 3 7 9 + 0 . 0 1 3 + 0 . 1 1 6 + 0 . 1 4 5 = 1.032W From Eq. (10.209), we estimate the efficiency ~7=

{[INa+[ZNa+F(~e--$i)]}

0.145 --

-n6AHr, 6

1.032

--0.14

In this highly simplified example, the Na + ion concentrations in the extracellular and intracellular spaces are not used because we are assuming isobaric, isothermal, and ideal mixtures.

PROBLEMS 10.1

An aqueous solution (phase A) of 10mmol/L of NaC1 is in equilibrium across a protein-tight membrane with an aqueous solution (phase B) of NaC1 and protein. The protein concentration is 1 mmol/L with a negative ionic valency of 10. Determine the difference in electric potential and hydrostatic pressure across the membrane when both solutions are assumed to be ideal and the temperature is 25°C.

10.2

We want to separate carbon dioxide contaminating methane using a cellulose acetate membrane. The mixture is perfectly mixed on both sides of the membrane. The methane mole fraction in the feed (high-pressure gas) is y(CH4) = 0.70. The permeate pressure is 1.2 atm. At 35°C and 20 atm, the permeability of the membrane is p(CO2) = 15 x 10-10 (cm 3 STP cm)/(cm2 s cmHg) and p(CH4) = 0.48 × 10-10 (cm 3 STP cm)/(cm 2 s cmHg). The membrane thickness is 0.8 Ixm, and 0 = 0.5. Estimate the membrane selectivity, permeant mole fraction yp(CO2), and flows of carbon dioxide and methane, J(CO2) and J(CH4).

10.3

We want to separate carbon dioxide contaminating methane using a cellulose acetate membrane. The mixture is perfectly mixed on both sides of the membrane. The methane mole fraction in the feed (high-pressure gas) is y(CH4) = 0.70. The permeate pressure is 2 atm. At 35°C and 20 arm, the permeability of the membrane is p(CO2) = 15 x 10-10 (cm 3 STP cm)/(cm 2 s cmHg) and p(CH4) = 0.48 × 10-10 (cm 3 STP cm)/(cm 2 s cmHg). The membrane thickness is 0.8 txm, and 0 = 0.7. Estimate the membrane selectivity, permeant mole fraction yp(CO2), and flows of carbon dioxide and methane, J(CO2) and J(CH4).

10.4

A diffusion cell has an aqueous solution of NaC1 with a concentration of 90mmol/L. Later, 0.08 mmol radioactive Na with a specific activity of 1 X 108 units is added to chamber A, which has a volume of 1.0 and

539

References

is stirred continuously. Measurements show that the radioactivity in reservoir A decreases at a rate of 14 units/rain. The process is at steady state. Estimate the flow of sodium ions, the diffusion coefficient, and the mobility at 298.15 K and in a transfer area of 100 mm 2. 10.5

The diffusion cell shown in Figure 10.2 has NaC1 mixtures in the two chambers with concentrations ClA = 50 mmol/L and ClB -- 4 mmol/L. The mobilities o f N a + and C1- ions are different, and their ratio yields their transference numbers b + / b - - t ~ I t -~ -0.39/0.61 (NaC1). The transference number t for an ion equals the fraction of the total electric current carried by the ion when the mixture is subjected to an electric potential gradient. For monovalent ions, we have t - ~ t - = 1. Estimate the diffusion potential of the cell at steady-state conditions at 298 K. Assume that the activity coefficients are equal in the two reservoirs.

10.6

Each chamber of the diffusion cell shown in Figure 10.2 has an aqueous solution of NaC1 with concentrations ClA- ClB- 80mmol/L at 300K. An electric potential difference of 90mV is established between the two chambers. Estimate the diffusion flow of NaC1 and its direction if D 1 = 1.45 × 10 -9 m2/s.

10.7

A membrane, permeable to hydrogen ions but not to chloride ions, separates two aqueous solutions of HC1. The concentration of solution A is 15 mmol/L, while that of solution B is 2 mmol/L HC1. The two solutions have the same pressure (1 atm) and temperature (25°C). Estimate the electric potential difference between the two solutions at equilibrium, as well as the number of hydrogen ions that have moved across the membrane during the time taken to reach equilibrium.

10.8

The electric potential difference between the interior and exterior of a cell is measured as 90 mV, with the cell interior negative, so that AO = 0B -- OA -- -- 90 InV. The cell's interior and exterior ions and their concentrations are shown in figure below. The activity coefficients for these ions are assumed to be the same in both phases. The temperature is uniform at 37°C. We want to estimate which of the three ions is closest to equilibrium. Membrane Interior: Phase A Na + = 12 mmol/1 C1- = 4 mmol/1 K + = 139 mmol/l qJA P~\

Exterior: Phase B Na+= 145 mmol/1 C1- =116 retool/1 K ÷ = 4 mmol/1 WB PB

10.9

Reconsider Example 10.9 with active transport of Na + ions from the intracellular to the extracellular space. Assume that 25% of the energy of hydrolysis of ATP can be used for the transport and that 38 mol of ATP are formed per mole of glucose (G) combusted. In the steady state, the extracellular concentration of Na + is observed to be 145 mM and the intracellular concentration to be 12 raM, and the electric potential of the intracellular fluid is determined to be -90.5 mV in relation to the extracellular space. The transcellular flux ofNa + ions is 1 mmol/min and the temperature of the tissue is 37°C. Estimate the oxygen consumption, heat output, and energy expenditure of the process when ~/r, aVP = - 2 0 k J / m o l for hydrolysis of ATP and ~¢-/r,G = -2867kJ/mol for glucose combustion. Assume that it2/gll = 1 and c)4/c)5 = 4/5.

10.10

Reconsider Example 10.9 and justify the following assumptions:

(/2/(/1

=

1 and

q4/~/5

--

4/5.

REFERENCES N. Bachelier, C. Chappey, D. Langevin, M. Metayer and J.-E Verchere, J. Membr. Sci., 119 (1996) 285. H.-J. Buschman, L. Mutihac and R. Mutihac, Sep. Sci. Technol., 34 (1999) 331. J.A. Calzado, C. Palet and M. Valiente, Anal. Chim. Acta, 431 (2001) 59. S.R. Caplan and A. Essig, Bioenergetics and Linear" Nonequilibrium Thermodynamics, The Steady State, Harvard University Press, Cambridge (1989). J.A. Daoud, S.A. E1-Reefy and H.E Aly, Sep. Sci. Technol., 33 (1998) 537. H. H. Dung and C.-H. Chen, J Memb. Sci., 56 ( 1991 ) 327. B.J. Eliott, W.B. Willis and C.N. Bowman, J Membt: Sci.. 168 {2000) 109. L. Garby and P.S. Larsen, Bioenergetics. lts Thermodynamic Foundations, Cambridge University Press, Cambridge (1995). T. Gumi, M. Oleinikova, C. Palet, M. Valiente and M. Munoz. Anal. Chim. Acta, 408 (2000) 65.

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10.

Membrane transport

S.V. Ho, P.W. Sheridan and E. Krupetsky, J. Membr. Sci., 112 (1996) 13. S.-H. Jeong and K.-H. Lee, Sep. Sci. Technol., 34 (1999) 2383. R.-S. Juang, S.-H. Lee and R.-H. Huang, Sep. Sci. Technol., 33 (1998) 2379. A. Katchalsky and P.E Curran, Nonequilibrium Thermodynamics in Biophysics, Harvard University Press, Cambridge (1967). S.C. Lee, B.S. Ahn and W.K. Lee, J. Membr. Sci., (1996) 171. X. Liu and D. Liu, Sep. Sci. Technol., 33 (1998) 2597. H. Matsuyama, M. Teramoto, H. Sakakura and K. Iwai, J. Membr. Sci., 117 (1996) 251. A. Narebska and S. Koter, Pol. J. Chem., 71 (1997) 1707. A. Narebska, W. Kujawski, S. Koter and T.T. Le, J Membr. Sci., 106 (1995) 39. A. Narebska and M. Staniszewski, Sep. Sci. Technol., 34 (1998) 1455. A. Narebska and A. Warszawski, J. Membr. Sci., 88 (1994) 167. K. Nakano, S. Kato, H. Noritomi and K. Nagahama, J Memb. Sci., 110 (1996) 219. C. Nigon, E Michalon, B. Perrin and B. Maisterrena, J. Membr. Sci., 144 (1998) 237. T. Nonaka and M. Kawamoto, J. Membr. Sci., 101 (1995) 135. T. Nonaka, T. Takeda and H. Egawa, J. Memb. Sci., 76 (1193) 193. M. Oleinikova, M. Munoz, J. Benavente and M. Valiente, Anal. Chim. Acta, 403 (2000) 91. Y.S. Park, J. Won and Y.S. Kang, J. Membr. Sci., 183 (2001) 163. I. Pinnau and L.G. Toy, J. Membr. Sci., 184 (2001) 39. R. Quinn, J.B. Appleby and G.E Pez, J. Membr. Sci., 104 (1995) 139. J. Rankumar, B. Maiti, B. Nayak and EK. Mathur, Sep. Sci. Technol., 34 (1999) 2069. E. Selegny, J.N. Ghogoma, D. Langevin, R. Roux and M. Metayer, J. Membr. Sci., 108 (1995) 161. E. Selegny, J.N. Ghogoma, D. Langevin, R. Roux and C. Ripoll, J. Membr. Sci., 123 (1997) 147. E. Selegny, J.N. Ghogoma, R. Roux, D. Langevin and M. Metayer, J. Membr. Sci., 93 (1994) 217. A. Sungpet, J.D. Way, C.A. Koval and M.E. Eberhart, J. Membr. Sci., 189 (2001) 271. C. Uegla and C.V. Zanoaga, J. Membr. Sci., 47 (1989) 285. P.C. Wankat, Rate-Controlled Separations, Chapman & Hall, Glasgow (1994). K.M. White, B.D. Smith, P.J. Duggan, S.L. Sheahan and E.M. Tyndall, J. Membr. Sci., 194 (2001) 165.

REFERENCES FOR FURTHER READING S. Freni, S. Cavallaro, S. Donato, V. Chiodo and A. Vita, Mater. Lett., 58 (2004) 1865. A. Gherrou, H. Kerdjoudj, R. Molinari, E Seta and E. Drioli, J Membr. Sci., 228 (2004) 149. S.W. Kang, J.H. Kim, K.S. Oha, J. Won, K. Char, H.S. Kim and Y.S. Kang, J Membr. Sci., 236 (2004) 163. M. Kargol andA. Kargol, Gen. Physiol. Biophys., 22 (2003) 51. S. Koter, J Membr. Sci., 206 (2002) 201. M.A. Lomholt, EL. Hansen and L. Miao, Eur. Phys. J E, 16 (2005) 439. T.E Moodya and H.K. Shepard, Biophys. Chem., 108 (2004) 51. M. Teramoto, S. Kitada, N. Ohnishi, H. Matsuyama and N. Matsumiya, J Membr. Sci., 234 (2004) 83.

11 THERMODYNAMICS AND BIOLOGICAL SYSTEMS 11.1

INTRODUCTION

Systems may exhibit two different types of behavior: (i) the tendency towards maximum disorder or (ii) the spontaneous appearance of a high degree of organization in space, time, and/or function. The best examples of the latter are dissipative systems at nonequilibrium conditions, such as the B6nard cell, the tricarboxylic acid (TCA) cycle, ecosystems, and living systems. As living systems grow and develop, a constant supply of energy is needed for reproduction and survival in changing conditions. Organized structures require a number of coupled metabolic reactions and transport processes that control the rate and timing of life processes. Schrodinger proposed that these processes appear to be at variance with the second law of thermodynamics, which states that a finite amount of organization may be obtained at the expense of a greater amount of disorganization in a series of interrelated (coupled) spontaneous changes. Biochemical reaction cyclic processes maintain the biological cell in nonequilibrium state by controlling the influx of reactants and efflux of products. Biological systems do not decay towards an equilibrium state, but instead increase in size, developing organized structures and complexity. An evolved and adapted biological system converts energy in an efficient manner for the transport of substances across a cell membrane, the synthesis and assembly of proteins, muscular contraction, reproduction, and survival. The source of energy is adenosine triphosphate (ATP), which is produced by oxidative phosphorylation in the inner membrane of the mitochondria. Kinetic equations and statistical models can describe such processes satisfactorily. However, these procedures often require detailed information, which may be unavailable. The nonequilibrium thermodynamics theory may be a useful approach to describe energy pathways and coupling in a quantitative manner, evaluate the stoichiometry in partially coupled systems, and formulate the efficiency of energy conversion in bioenergetics. Nonequilibrium thermodynamics formulations may provide a new approach of analyzing the results of experimental studies and guide the design of new experimental methods relating to biological energy conversions. The linear nonequilibrium thermodynamics theory is valid for systems close to equilibrium, and does not require detailed information about the mechanisms of biological process, although a complete analysis requires a quantitative description of the mechanisms of energy conversion. This chapter starts with a simplified analysis of biological processes using the basic tools of physics, chemistry, and thermodynamics. It provides a brief description of mitochondria and energy transduction in the mitochondrion. The study of proper pathways and multi-inflection points in bioenergetics are summarized. We also summarize the concept of thermodynamic buffering caused by soluble enzymes and some important processes of bioenergetics using the linear nonequilibrium thermodynamics formulation.

11.2

SIMPLIFIED ANALYSIS IN LIVING SYSTEMS

Living systems consist of many subsystems with characteristic functions and outputs. Communication among these various functions and outputs leads to an organized system that can be maintained by a constant supply of energy and matter from the outside. Therefore, living systems represent nonequilibrium open systems with various thermodynamic forces and flows. The second law of thermodynamics suggests that living systems are capable of creating order through various coupled (interrelated) chemical cycles and transport processes. The relations between the flows and forces can provide quantitative information on the characteristics and level of organization without detailed knowledge about the mechanism of coupling. Many chemical cycles and transport processes are the result of controlled energy conversions, and the principles of thermodynamics can describe the efficiency of energy conversions and the exchange of energy and matter between living systems and the environment.

542

11.

Thermodynamicsand biological systems

The analysis of real biological systems may be introduced in idealized simplifications using the principles of physics, chemistry, biology, thermodynamics, and kinetics. The following examples are the simple application of these principles in describing some biological processes.

Example 11.1 Cell electric potentials In living systems, ions in the intracellular phase and the extracellular phase produce a potential difference of about 80 mV between the two phases. The intracellular phase potential is negative (Garby and Larsen, 1995). Determine the difference in electrical potential energy per mole positive monovalent ion, e.g., Na+, between the two phases. Electric potential differences can exist in living systems. Changes in electrical potential energy are produced by ions in solutions whose electric potentials change across cell membranes. The potential energy per mole species i, Epi, at the potential 6 is obtained from Epi =

ziF~

With the valence of the ion z = + 1, and F Faraday constant (F = 96,485 C/mol), we have (EpNa)ext -- (EpNa)int "- z i F ( ~ e x t

- ~tint) --

1(96485)(80 X 10 -3) = 7718.8 J / m o l

This result illustrates one of the nonequilibrium conditions maintained by ion transport.

Example 11.2 Excess pressure in the lungs The lungs have a large surface area (approximately 100 m 2 in adults). During inspiration, the incoming air fills the canals in the bronchial tree that ends with the near-spherical alveoli, across whose walls transport of oxygen and carbon dioxide takes place (Garby and Larsen, 1995). The single alveolus can be considered an elastic membrane covered by a thin film. At equilibrium, between the surface tension crs and the excess pressure AP of the air inside a spherical surface with radius R, we find A P - 2°'s R

(a)

If we assume that O's is constant, then an increasing radius would reduce excess pressure, and the alveoli would collapse when the excess pressure fell below a certain value. This kind of instability is not common, since o-s does not remain constant in a normal lung but increases with increasing radius such that AP always increases with increasing radius. This behavior of O's is caused by a surface-active agent (phospholipids) in the liquid film. The concentration of the surface-active agent decreases in the interface between liquid and air when the surface of the film distends, and therefore both surface tension and excess pressure increase. The hysteresis of this process is controlled by the diffusion of matter between the free interface and interior of the liquid. The surface tension is about 0.06 N/m for water and about 0.05 N/m for blood plasma, and it can vary between 0.002 and 0.04 N/m for the liquid films of alveoli, and the total area changes by a factor of 5. With these data on surface tension and for a maximal area of 100 m 2, the total surface energy (o'sA) varies between the values (o-sA)l = (0.04)(100) = 4 J (o-sA)2 : (0.002)(20): 0.04 J Assuming that the hysteresis over a cycle is 25% of the maximal energy, that the power due to surface tension at a respiration frequency of 12 min -1 becomes

The radii of the --~300 million alveoli vary in the interval R = 0.06--0.15 mm. The excess pressures (Eq. (a)) to distend an alveolus and hence surface tensions can vary between 2(0.04) (~)1

-- O. 15 / 1 0 0 0

= 533Pa

11.2 Simplified analysis in living systems

(AP)2 =

2(0.002) 0.06 / 1000

543

= 66 Pa

These relations show the equality between the work of the excess pressure and the increase in surface energy.

Example 11.3 Enthalpy and work changes of blood due to the pumping work of the heart Calculate the change in enthalpy of blood when subjected to an isothermal increase in pressure of 16 kPa (120 mmHg) (Garby and Larsen, 1995). Assuming that the blood is an incompressible fluid, the density becomes constant, and the second term in the total differential dU= (OU/OS)vdS + (OU/OV)sdVvanishes. From the definition ofenthalpy H= U+ PV, we have dH = CvdT + VdP

(a)

This relation shows that the first term dominates when a liquid is heated, and the second dominates when isothermal pumping takes place. From Eq. (a), assuming the density of water (1000 kg/m 3) is similar to the density of blood, and V= 1/p, we have H 2 - H 1 = V(P2 - P~) = 10- 3 (16000) = 16 J/kg Assuming a pulse rate of 70 beats per minute at rest and a stroke volume of 70 mL, the power is obtained from

/

rn(H 2 - H1) - 0.07 ~

(16) = 1.3 W

This is about 1000 times the power associated with the kinetic energy of the flow. The increase in pressure originates in the heart, and here we calculate the work added to the blood as useful energy. The pump is understood as a system that supplies energy to the fluid in the form of an increase in pressure. So, the blood can circulate through the closed circuit of blood vessels despite the associated pressure drops. The pump power Wp supplied by a muscle can be estimated by th(H 2 -

HI) + AEki n + AEpo t = 0 "+"~"p

Assuming that the process is steady, and kinetic z~kEkin and potential ~ p o t energies are negligible (the same area and velocity in the inflow and outflow), the equation above reduces to rn(H 2 - H, ) = c) + Wp For an isothermal process, we have (b) Therefore, the pump power can be calculated only when the heat flow rate is known. From the general entropy balance equation dS= deS+ diS, we conclude that for an incompressible and isothermal process, we have des = -diS. This relation shows the equality between the dissipated heat flow and internal entropy production and hence the loss of power is c) - ~/'loss.Therefore, Eq. (b) becomes

/

r h ( p (P2-P1)+Eloss =Wp where Elos~is the power loss due to friction and associated internal backflow. If the pump is reversible, ~+loss = 0, and we have

544

11.

Thermodynamicsand biological systems

The isothermal mechanical efficiency for this pump becomes

Tip--

/~rp,rev I//Zp

In turn, by using the metabolic energy expenditures EM needed by the muscle to affect the mechanical motion, we can define the metabolic or chemo-mechanical efficiency of the muscle by

T~ch -- /~M

Combining the two efficiencies, and representing the conversion of chemical energy into useful (reversible) energy received by the fluid, we may define the net chemo-mechanical pump efficiency:

T] : 7~p~ch --

l/~rp,rev /~M

For external muscle work, the chemo-mechanical efficiency is about 0.2-0.25.

11.2.1

Biological Fuels

Biological fuels can be categorized in three groups: carbohydrates (CH), representing a mixture of mono-, di-, and polysaccharides, fats (F), and proteins (Pr) (Garby and Larsen, 1995). The fuel value is equal to the negative reaction enthalpy. Carbohydrates and fats can be completely oxidized, while proteins can only be partially oxidized. Therefore, protein has a lower fuel value. The oxidation enthalpies per unit mass differ only slightly within each group. Table 11.1 shows the reaction enthalpies and stoichiometries of biological fuels. The respiratory quotient shows the ratio of carbon dioxide elimination and oxygen uptake associated with oxidation. The last column in Table 11.1 is the thermal energy equivalency of oxygen, which is a measure of the chemical energy that is liberated per mole of oxygen consumed. It is possible to measure only the uptake of oxygen and the rates of elimination of carbon dioxide and nitrogen. The energy expenditure with the reaction enthalpies is

]E : Z(nAHr)i : (hAHr)cH -F (nAHr) F + (nAHr)Pr : (-0) + ( - W )

(11.1)

i

Example 11.4 Energy expenditure in small organisms A small organism has a heat loss o f - q = 1.52 W and performs external work W = 0.02 N m/s. Calculate that part of the total energy expenditure that originates from its internal circulation, which involves the pumping of 120 mL/min of fluid against a pressure drop of 3.34 kPa with a net chemo-mechanical efficiency of 10%. The energy expenditure of the pump is /~ = - q + (-W) = 1.52 + 0.02 = 1.54 W

Table 11.1 Biological fuel parameters Fuel

CH Fat Pr

Specific reaction enthalpy (kJ/g) - 17 -39 -17

CH: carbohydrate; Pr: protein.

Source: Garby and Larsen, (1995)

Specific turnover 02 (mmol/g)

CO2 (g/g)

N (mmol/g)

33.3 90.6 43.3

33.3 63.8 34.43

0.16

Respiratory quotient = n(CO2)/n(02)

Energy equivalency (kJ/mol 02)

1.00 0.70 0.79

511 431 393

11.2 Simplified analysis in living systems

545

The reversible pump power becomes mp,re v

- ( 0"12X10-3 ] (3340) = 0.0067W 60

Therefore, the net chemo-mechanical pump efficiency becomes

Ep l/Vp,rev/0.1 0 . 0 0 6 7 / 0 . 1 ~ / - EM

EM

1.54

= 0.043

Consider an amphipod with a body weight of l0 p~g consuming 4.0 × 10 - 9 mol oxygen every hour at steady state and eliminating 3.6 x 10-9 mol carbon dioxide, 0.4 x 10 -9 mol N (as ammonia), and 0.1 x 10 -9 mol lactic acid. The external work power is 50 × 10 -9 W. The heat loss of the animal may be calculated when the following four net reactions contribute to the energy expenditure" C6H12O 6 + 6 0 2 ~ 6 C O 2 + 6 H 2 0 - 2 8 7 0 k J / m o l C6H1206

--~ 2 C 3 H 6 0 3

- 100 kJ/mol

C55H1040 6 + 780 2 ~ 55CO 2 + 52H20 - 34300 kJ/mol

C32H48010N 8 + 330 2 ~ 32CO 2 + 8NH 3 + 12H20 - 14744 kJ/mol Therefore, the energy expenditure (Garby and Larsen, 1995) from Eq. (11.1) with the given reaction enthalpies is L- = 2870(h)G + 100(rt)G__.La + 34300(n)F + 14744(n)p r where G refers to the combustion of glucose, G ~ La to the metabolism of glucose to lactic acid (anaerobic process), F to the combustion of fat, and Pr to the combustion of protein. The mass conservations are ho~ = 6n G + 78n F + 33fipr = 4 X10 -9 mol/h _

hL~ -- 2n~_~La -- 0.1X 10 - 9 mol/h

rico;

= 6h~ + 55n F + 32hpr = 3.6X10 -9 mol/h

fin -- 8/)/Pr -- 0 . 4

x 10 - 9 m o l / h

From Eq. (11.1) and solving the above equations, we have ~" = [2870(0.194)~ + 100(0.05)~ ~La + 34300(0.0152)F + 14744(0.05)p r ] X 10 - 9 = 1820 X 10 -9 kJ/h = 0.506 I~W Therefore, the heat loss is ( - q ) = L ' - ( - / ~ ) - 0 . 5 0 6 - 0 . 0 5 0 : 0.461~W

Example 11.5 Energy expenditure in an adult organism An adult organism has an oxygen uptake of about 21.16mol over 24h, and the associated elimination of carbon dioxide and nitrogen is 16.95mol and 5.76g, respectively (Garby and Larsen, 1995). If the adult has performed 0.9 MJ of external work over the same period and his energy expenditure at rest is L'0 = 65 W, estimate his energy expenditure, heat loss, and net efficiency for the external work.

546

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Thermodynamicsand biological systems

The energy expenditure may be calculated from the energy balance. Assume that (i) carbohydrate (CH), fat (F), and protein (Pr) are the only compounds involved in the oxidation process; (ii) the other compounds are stationary; and (iii) the uptake and elimination of oxygen, carbon dioxide, and nitrogen is instantaneous. From the first law of thermodynamics, we have /~ = E ( / ; / ~ r ) i i

= ( / ; / ~ r ) C H + (/:/~-/r)F + ( / ; / ~ r ) P r = ( _ q ) + (__~f/r)

From Table 11.1, we get 17hca + 39h F + 17npr = q + W = / ~

(a)

The conservation of mass requires that ho2 = 33.3hcH + 90.6h F + 43.3hpr mmol/s hco 2 = 33.3hcn +63.8h F + 34.4hpr mmol/s

th N -- 0.16/;/pr g/s

From these equations, we obtain

npr

5.76 0.16

- 36 g/day, hcn = 194 g/day, and tiv = 145 g/day

From Eq. (a), we have 17(194)+ 39(145)+ 17(36) = - q + ( - W ) =/~ = 9566 kJ/day = 110.7W Since the work is - 0 . 9 MJ/day, the heat loss is ( - q ) = 9566 - 900 = 8666 kJ/day = 100 W A net efficiency is defined by

rtnet -- E _ E0

900 = 0.22 9 5 6 6 - 5500

where E0 is the energy expenditure during resting and/~ is the energy expenditure during the performing of external work (-W).

Example 11.6 Oxidation of glucose (a) We wish to estimate the reaction enthalpy for the isothermal and isobaric oxidation of glucose at 310 K and 1 atm. (b) Estimate the energy expenditure for oxidation of 390 g/day glucose at steady conditions. C6H120 6 (aq) + 60 2 (g) ~ 6CO 2 (g) + 6H 2O(1) If the control volume is a piece of tissue, the reaction above may take place in an aqueous solution (aq). Consider an aqueous solution of 0.01 mol/L glucose; the partial pressure of carbon dioxide and oxygen are 0.07 and 0.21 atm, respectively. From Table BS, we obtain the enthalpy of formations for the components of the reaction above at the standard state (298 K and 1 atm) AH ° = 6 ( - 393) + 6(-286) - (1264) - 6(0) = - 2 8 1 0 kJ/mol

547

11.2 Simplified analysis in living systems

From Table B8, we see that the difference in enthalpy between solid and dissolved glucose is 11 kJ/mol, while the differences in enthalpy of formation for gas and dissolved matter is 20 kJ/mol for carbon dioxide and 10 kJ/mol for oxygen. For all the components in aqueous solution, we find from Table B9. AHr° = - 2870 kJ/mol So, the transfer of gaseous components to aqueous solutions is small: 60/2810 - 0.02 or 2%. The reaction enthalpy at 310 K can be estimated from

AHr(310K) -- A H

°

+ Cp,av (T - To ) -- z~kg ° + E piCpi (T - To )

where v is the stoichiometric coefficient. By using the heat capacities from Table B8, we have AH r (310 K)

= AH °

+ Cp,av ( T

- T O) -

-2870 + (0.279)(310 - 298) = -2867 kJ/mol

This value is the same as the one in Table B9, and shows that the temperature correction for the heat of reaction is less than 0.2% and is often negligible. The energy expenditure (E) at a glucose consumption of 390 g/day is 390/180 = 0.025 mmol/s n~ = 24(3600) and =nG ( - A H r ) = 0.025(2867)= 71.7W where the molecular mass of glucose is 180 g/mol.

Example 11.7 Unimolecular isomerization reaction One of the simplest biochemical reactions is a unimolecular isomerization reaction (Qian and Beard, 2005)

S<

kf

kb

>P

where kf and kb are the forward and backward reaction rate constants. A chemical equilibrium is defined by

[P]eq[s]eq_kfkb -

(11.2)

exp (-/x~' -/x~/RT

To simulate a biological metabolic network, consider a special controlling mechanism so that the concentrations of S and P are maintained at prescribed levels. Therefore, the chemical system is at steady state since concentrations remain unchanged with time. However, the system in not at an equilibrium state and the reaction velocity (flow) is not zero: Jr = kfcs - kbCv

= Jrf -Jrb 4:0

We can determine the affinity of the reaction as the driving force for the chemical system

A = -Al~

-

I~p -

tx s -

o

I~p -

o

ix s + RT

Cp Cs

ln~ -

RT

In

[Jrb / ( Jrf )

(11.3)

Introducing chemical potentials for biochemical substrates needs to be done with caution when considering, for example, molecular crowding and signaling molecules with limited copy numbers (Parsegian et al., 2000). This simple chemical system is for cellular metabolic networks, and concentrations replace activities in ideal solutions.

548

11.

Thermodynamicsand biological systems

Equation (11.3) can be transformed to describe the amount of work done by the controlling mechanism, which pumps reactant S and product P to maintain nonequilibrium steady-state conditions

JrA= RT(Jrf -Jrb)ln(Jrf ]>__O

(11.4)

~,Jrb )

The equality of this equation represents a system at equilibrium where Jr = A = 0. The work done by the controlling system dissipates as heat. This is in line with the first law of thermodynamics. The inequality in Eq. (11.4) represents the second law of thermodynamics. The cyclic chemical reaction in nonequilibrium steady-state conditions balances the work and heat in compliance with the first law and at the same time transforms useful energy into entropy in the surroundings in compliance with the second law. The dissipated heat related to affinity A under these conditions is different from the enthalpy difference M-/° = (O(Atx°/T)/O(1/T)). The enthalpy difference can be positive if the reaction is exothermic or negative if the reaction is endothermic. On the other hand, the A contains the additional energy dissipation associated with removing a P molecule from a solution with concentration Cp and adding an S molecule into a solution with concentration Cs. This simple example may be generalized to biochemical reaction cycles in which there are a number of reactions and boundary flows that add and remove substrates. The controlled concentrations and boundary flows maintain the system in a nonequilibrium steady-state condition. A dynamic equation of entropy change is

T --~ : r

-t-T ---~-]

qdis-t-xIt

(11.5)

This equation shows that in isothermal biochemical reaction cycles, the entropy of the system changes because of the heat dissipation rate qdis exchanged with the surrounding and the rate of free energy dissipation • due to entropy production. This equation also indicates the dissipative character of biochemical cycles. Dynamic equations similar to Eq. (11.5) can also be written for enthalpy and Gibbs free energy changes

where Wnowis the flow work (chemical motive force, Qian and Beard, 2005) determined by the controlling boundary flows and/or concentration. At steady state

we have ~ = !)dis = ~/flow ~ 0, in which the equalities represent the first law of thermodynamics and the isothermal Clausius equation, while the inequality represents the second law of thermodynamics.

11.3

BIOENERGETICS

Energy production, conversion, and storage to maintain the nonequilibrium character of living systems form the basis for bioenergetics. Energy is supplied with the intake of food or with solar radiation. Living systems convert part of this energy to produce electrons and protons. The flow of protons across a specific membrane leads to the production of ATE The hydrolysis of ATP is coupled to synthesizing protein molecules, transporting ions and substrates, producing mechanical work, and other metabolic activity. Some of the internal mechanical work involves the pumping of blood by the heart. Continuous chemical cycles and transport processes maintain a stationary state of chemical nonequilibrium and integrity by the regulated and synchronized production, conversion, and utilization of energy. Bioenergetics provides a quantitative description of the transformation of materials and energy in living systems. Most biochemical reactions occur in pathways, in which other reactions continuously add substrates and remove products. The rate of reactions depends on the properties of the enzymes (large proteins produced in cells) that catalyze the reaction. Substrates bind at the active sites of enzymes, where they are converted to products and later released. Enzymes are highly specific for given substrates and products. Inhibitors of enzymes decrease the rate of reaction.

11.3 Bioenergetics

549

Proteins can bind to enzymes and alter their activities. The clusters of orthologous genes database has identified 210 protein families involved in energy production and conversion; the protein families show complex phylogenetic patterns and exhibit diverse strategies of energy conservation. The organized structures of living systems degrade incoming solar radiation and chemical potential through well-controlled chemical cycles.

11.3.1

Mitochondria

Mitochondria are organelles typically ranging in size from 0.5 to 1 txm in length, found in the cytoplasm of eukaryotic cells. Mitochondria contain inner and outer membranes, separated by a space. Both the inner and outer membranes are constructed with tail-to-tail bilayers ofphospholipids into which mainly hydrophobic proteins are embedded. One portion of the lipid molecule is hydrophilic (water-attracting) and the other portion is hydrophobic (lipid-attracting). The selfassembled lipid bilayer is in a dynamic and liquid-crystalline state. The outer membrane contains proteins and lipids and numerous transport proteins, which shuttle materials in and out of the mitochondrion. The outer membrane is 60-70 thick and permeable to small molecules, including salts, adenine and nicotinamide nucleotides, sugars, and coenzymes. The inner membrane contains all the enzymes and fewer lipids than the outer membrane. The inner membrane is permeable to small neutral molecules such as water, oxygen, and carbon dioxide, while its permeability to charged molecules such as proton and ions is limited. Mitochondrial membranes produce two compartments; one of them is called the intermembrane space (C-side) and the space enclosed by the inner membrane is called matrix (M-side) (Figure 11.1). The intermembrane space is usually 60--80 A in width and contains some enzymes. The matrix, however, is very viscous and rich in proteins, enzymes, and fatty acids. The number of mitochondria in a cell depends on the cell's function. Cells with particularly heavy energy demands, such as muscle cells, have more mitochondria than other cells. The inner membrane houses the electron transport chain and ATP synthesis. The inner membrane has numerous folds called cristae, which have a folded structure that greatly increases the surface area where ATP synthesis occurs (Figure 11. lb). Mitochondria contain deoxyribonucleic acid (DNA) and ribosomes, protein-producing organelles in the cytoplasm. The DNA directs the ribosomes to produce proteins as enzymes (biological catalysts) in ATP production. Outer membrane

(a)

Inner membrane

Matrix

Outer membrane

Cristae

2

Cristae

\ (b)

Inner membrane

Figure 11.1. (a) Structure of the mitochondria and (b) inner membrane structure of the mitochondria.

11. Thermodynamicsand biological systems

550

Mitochondria are involved in the transport and regulation of Ca 2+, protein import, cell death and aging, and obesity. Mitochondria from different organ systems, such as the liver, heart, and brain, display morphological and functional differences. Mitochondria are the major source of reactive oxygen species throughout the respiratory chain. These oxygen radicals may affect the function of the enzyme complexes involved in energy conservation, electron transfer, and oxidative phosphorylation, and play an important role in aging. Experimental evidence shows that mitochondria exhibit anisotropy. Three-dimensional images show that inner membrane involutions (cristae) have narrow and long tubular connections to the intermembrane. These openings may lead to lateral gradients of ions, molecules, and macromolecules between the compartments ofmitochondria. This type of structure may influence the magnitude of local pH gradients produced by chemiosmosis and the internal diffusion of adenine nucleotides. The mitochondria have elongated tubes aligned approximately in parallel and are embedded in a multilamellar stack of endoplasmic reticulum, which could be related to the specific function of the mitochondria (Ovadi and Saks, 2004).

11.3.2

Tricarboxylic Acid Cycle in Mitochondria

The TCA cycle, also called the citric acid cycle or the Krebs cycle, is the major energy-producing pathway and occurs in mitochondria. Foodstuffs enter the cycle as acetyl coenzyme A CoA and are oxidized. The cycle starts with the four-carbon compound oxaloacetate, adds two carbons from acetyl CoA, loses two carbons as CO2, and regenerates the four-carbon compound oxaloacetate. Electrons are transferred to nicotinamide adenine dinucleotide NAD + and flavin adenine dinucleotide FAD, and NADH and FADH2 are produced (Marks, 1999). As the electrons are transferred eventually to oxygen, energy is released, and ATP is produced by the process called oxidative phosphorylation. Intermediate products of the TCA cycle are converted to glucose during the fasting state and to fatty acids during the fed state in the liver. Some intermediate products are synthesized to amino acids (Figure 11.2). The change in free energy or available energy AG to perform work at constant temperature and pressure is A G = M 4 - TAS, where M-/is the change in enthalpy and/kS is the change in entropy. If AG is negative, the reaction proceeds spontaneously and releases energy. If AG is positive, the reaction needs an energy supply to proceed. If AG is zero, the reaction is at chemical equilibrium. The rate of a chemical reaction is not related to its free energy change; a reaction with a large negative free energy change may not lead to a fast reaction. Directions of reactions proceeding near equilibrium can be reversed by small changes in the concentrations of substrates or products. The need for energy by the cell regulates the TCA cycle, which acts in concert with the electron transfer chain and the ATPase to produce ATP in the inner mitochondrial membrane. The cell has limited amounts of ATE ADP,

Intermembrane

/

electrons~------7"xzzEZ 2H+ ~~~_TL-NADH

/

- --el tro.s /

~

/

/ /[

/

I [ Proton /

P

~

2H+

Quinone

/

Cytochrome

/ ~

02

l/m°tlV\ \forcee /

/ ADP+P

cristae

ATP ATP Synthesis

Figure 11.2. Electron transport in phosphorylation.

11.3 Bioenergetics

551

and AMP. When ADP levels are higher than ATR the cell needs energy, and hence NADH is oxidized rapidly and the TCA cycle is accelerated. When the ATP level is higher than ADP, the cell has the energy needed; hence, the electron transport chain slows down, and TCA cycle is inhibited.

11.3.3

Mitochondria and Electron Transfer Chain

Mitochondria organize electron transfer and the associated reactions leading to the ATP synthesis called oxidative phosphorylation (Figure 11.2). Synthesis of ATP is an endothermic reaction, and hence conserves the energy released during biological oxidation-reduction reactions. Electron transfer and associated reactions leading to ATP synthesis are completely membrane-bound. NADH and FADH2 are the reduced cofactors of NAD ÷and the FAD÷. The oxidation of one NADH produces approximately three ATR and the oxidation of one FADH2 produces approximately two ATE The hydrolysis of ATP provides energy for all cellular activity. ATP is transported from the mitochondrial matrix to the cytosol in exchange for ADP through the ATP-ADP antiport system. The photosynthetic energy conversion of light energy into Gibbs free energy of protons takes place in plants, algae, and certain species of bacteria. Photosynthesis, driven by light energy, leads to the production of ATP through electron transfer and photosynthetic phosphorylation. The transmembrane electron transfer process occurs in specialized pigment-protein complexes called photosynthetic reaction centers. Photosynthetic energy conservation takes place in the thylakoid membrane of plant chloroplasts; oxidative phosphorylation takes place in the mitochondrial inner membrane. These membranes facilitate the interactions between the redox system and the synthesis of ATR and are referred to as coupling membranes. Electron transport has three major stages: (1) transfer of electrons from NADH to coenzyme Q, (2) electron transport from coenzyme Q to cytochrome c, and (3) electron transport from cytochrome c to oxygen. These stages are briefly described below. (1) Transfer of electrons from NADH to coenzyme Q NADH is produced by (i) the c~-ketoglutarate dehydrogenase, isocitrate dehydrogenase, and malate dehydrogenase reactions within the TCA cycle, (ii) the pyruvate dehydrogenase reaction, (iii) [3-oxidation of fatty acids, and (iv) other oxidation reactions. NADH passes electrons to the flavin mononucleotide FMN using the dehydrogenase complex. NADH produced in the mitochondrial matrix diffuses to the inner mitochondrial membrane where NADH passes electrons to the FMN. The FMN passes electrons through a series of iron-sulfur complexes to coenzyme Q. Coenzyme Q accepts electrons one at a time and forms semiquinone and ubiquinol. The electron transfers produce energy, which is used to pump protons to the cytosolic side of the inner mitochondrial membrane. The protons flow back into the matrix through pores in the ATP synthase complex, and approximately one ATP is produced for each NADH. (2) Electron transport from coenzyme Q to cytochrome c Coenzyme Q passes electrons through iron-sulfur complexes to cytochromes b and cl, which transfer the electrons to cytochrome c. In the ferric Fe 3+ state, the heme iron can accept one electron and be reduced to the ferrous state Fe2÷. Since the cytochromes carry one electron at a time, two molecules on each cytochrome complex are reduced for every molecule of NADH that is oxidized. The electron transfer from coenzyme Q to cytochrome c produces energy, which pumps protons across the inner mitochondrial membrane. The proton gradient produces one ATP for every coenzyme Q-hydrogen that transfers two electrons to cytochrome c. Electrons from FADH2, produced by reactions such as the oxidation of succinate to fumarate, enter the electron transfer chain at the coenzyme Q level. (3) Electron transport from cytochrome c to oxygen Cytochrome c transfers electrons to the: cytochrome a a 3 complex, which transfers the electrons to molecular oxygen, and the oxygen is reduced to water. Cytochromes a and a 3 contain heme a and two different proteins containing copper. The energy released by the transfer of electrons fi'om cytochrome c to oxygen is used to pump protons across the inner mitochondrial membrane. Every two electrons that are transferred from cytochrome c to oxygen produce one ATR

11.3.4

Oxidative Phosphorylation

Oxidative phosphorylation occurs in the mitochondria of all animal and plant tissues, and is a coupled process between the oxidation of substrates and production of ATE As the TCA cycle runs, hydrogen ions (or electrons) are carried by the two cartier molecules NAD or FAD to the electron transport pumps. Energy released by the electron transfer processes pumps the protons to the intermembrane region, where they accumulate in a high enough concentration to phosphorylate the ADP to ATE The overall process is called oxidative phosphorylation. The cristae have the major coupling factors F1 (a hydrophilic protein) and F0(a hydrophobic lipoprotein complex). F1 and F 0 together comprise the ATPase (also called ATP synthase) complex activated by Mg 2+. F 0 forms a proton translocation pathway and F1

552

11.

Thermodynamicsand biological systems

is a catalytic sector. ATP synthesis by F0F 1 consists of three step: (i) proton translocation through F0, (ii) conformation transmission to F1, and (iii) ATP synthesis in the [3 unit. The rotation of a subunit assembly is an essential feature of the mechanisms of ATP synthesis and can be regarded as a molecular motor (Sambongi et al., 2000). ATPase can catalyze the synthesis and the hydrolysis of ATE depending on the change of electrochemical potential of proton A~H. The ratio of ATP production to oxygen consumption P/O can vary according to various physiological processes: (i) maximizing ATP production, (ii) maximizing the cellular phosphate potential, (iii) minimizing the cost of production, and (iv) a combination of these three processes. The values of P/O change within the range of 1-3, and characteristic of the substrate undergoing oxidation and characteristic of the organ's physiological role. In the case of excess oxygen and inorganic phosphate, the respiratory activity of the mitochondria is controlled by the amount of ADP available. In the controlled state called state 4, the amount of ADP is low. With the addition of ADP, the respiratory rate increases sharply; this active state is called state 3. The ratio of the respiratory rates of state 3 to state 4 is called the respiratory control index. The control of the respiration process and ATP synthesis shifts as the metabolic state of the mitochondria changes. In an isolated mitochondrion, control over the respiration process in state 4 is mainly due to the proton leak through the mitochondrial inner membrane. This type of control decreases from state 4 to state 3, while the control by the adenine nucleotide and the dicarboxylate carriers, cytochrome oxidase, increases. ATP utilizing reactions and transport activities also increase. Therefore, in state 3, most of the control is due to respiratory chain and substrate transport. According to the chemiosmotic coupling hypothesis, ATP synthesis decreases the proton electrochemical gradient and hence stimulates the respiratory chain to pump more protons across the mitochondrial inner membrane and maintain the gradient. However, electron supply to the respiratory chain also affects respiration and ATP synthesis. For example, calcium stimulates mitochondrial matrix dehydrogenase, and increases the electron supply to the respiratory chain and hence the rate of respiration and ATP synthesis.

11.3.5

Glycolysis Pathway

The glycolysis pathway occurs in the cytoplasm outside the mitochondria, and requires no oxygen. During glycolysis, glucose is broken down into pyruvate. The initial reactions of the pathway produce triose phosphate, which produces ATP in the second sequence of reactions. Overall, the glycolysis pathway produces ATP, NADH, and pyruvate. NADH cannot directly enter mitochondria. Pyruvate can enter the mitochondria; it is broken down to acetyl CoA by a special enzyme, and carbon dioxide is released. Acetyl CoA enters the TCA cycle, producing additional ATE Only four ATP molecules can be produced by one molecule of glucose.

11.3.6

Transport Processes and Mitochondria

There are many carrier molecules for electrons: one is called the nicotinamide adenine dinucleotide (NAD +) and another is the flavin adenine dinucleotide FAD+. The reduced cofactors NADH and FADH 2 transfer electrons to the electron transport chain. FMN receives electrons from NADH and passes them to coenzyme Q through Fe-S systems. Coenzyme Q receives electrons from FMN and FADH2 through Fe-S systems. Cytochromes receive electrons from the reduced form of coenzyme Q. Each cytochrome consists of a heme group, and the iron of the heme group is reduced when the cytochrome receives an electron: Fe 3+ ~ Fe z+. At the end of the electron transfer chain, oxygen is reduced to water. Membrane proteins transfer material and information between cells and their environment and between the compartments housing the organelles. Some of these proteins selectively transport specific molecules and ions, and some others are receptors for chemical signals from outside the cell. They act as transducers capable of gathering information, processing it, and delivering a response. Their electrical activities are measurable as an electric potential difference across the membrane. Changes in the membrane permeability would yield a change in the potential difference. In a cotransport system, the movement of one permeant is dependent on the simultaneous movement of a different permeant, either in the same direction (called symport) or in the opposite direction (called antiport). The best-known antiport system is the Na+/K+-ATPase pump that is present in the plasma membrane of all animal cells. The pump transports sodium ions out of the cell and potassium ions into the cell through the lipid bilayer against their electrochemical potential gradients, and operates as an antiport. Transmembrane activities are thermodynamically driven by the gradients of chemical and electrochemical potentials and are able to maintain steady nonequilibrium conditions across cell membranes by generating and controlling the flows of ions or electrons. The cytoplasm houses many metabolic cycles and synthetic pathways, as well as protein synthesis. At the cellular level, communication via the membrane is called signal transduction, and is facilitated by ligands or messengers, such as proteins and peptide hormones. These ligands facilitate communication by directly entering the cell,

11.3 Bioenergetics

553

or interacting with a specific receptor situated on or in the lipid bilayer of the membrane. Insulin stimulates glucose transport into muscle and adipose cells. However, it does not significantly stimulate the transport of glucose into tissues such as the liver, brain, and red blood cells (Marks, 1999).

11.3.7

Formulation of Oxidative Phosphorylation

Theoretical approaches applied to oxidative phosphorylation are the kinetic model, metabolic control analysis, and nonequilibrium thermodynamics. These approaches are helpful for quantitatively describing and understanding the control and regulation of oxidative phosphorylation. The metabolic control theory can provide a quantitative description of microbial growth. A kinetic model of oxidative phosphorylation may not be fully completed, because of several assumptions and simplifications associated with it. A proper kinetic approach, however, allows for a deeper insight into the mechanisms related to the control and regulation of oxidative phosphorylation. It may provide a model and methodological approach for describing the dynamic and stationary properties of energy coupling in membranes. The application of nonequilibrium thermodynamics assumes a linear flow-force relationship between the oxidation and phosphorylation flows. Such a linear dependence has been established by measurements taken during the transition on from state 4 to state 3 as a linear part of a more general sigmodial relationship. Nonequilibrium thermodynamics has proved to be useful in describing the energetics aspects of oxidative ATP production and the transport of substrates coupled to ATP hydrolysis without knowledge of the detailed mechanisms of coupling (Stucki, 1980). However, it is not realistic to assume that simple formulations can lead to a complete description of such complex and coupled biological systems. A starting point in linear nonequilibrium thermodynamic formulations is the representative dissipation function given by altt = J p Ap -+-JHA/~H -+-JoAo

(11.7)

Here, the subscripts R H, and O refer to phosphorylation, the H + flow, and substrate oxidation, respectively, and A/~H is the electrochemical potential difference of protons. We consider only systems at steady state. The dissipation function can be transformed as ~r = j p A ~ x -Jr-JHA/~H 4- JoA(e)x

(11.8)

where A ex is the external affinity. When the interior of the mitochondrion is in a stationary state, it suffices to measure the changes in the external solution only. From Eq. (11.8), the linear phenomenological equations with the resistance coefficients are obtained A~,x -- K p J p + KpHJ H + KpoJ O

(11.9)

~/)'H = KpHJP -+-K H J H -+-KOHJo

(11.10)

A~x -- KpoJ P + KOHJ H + K o J o

(11.11)

If A~ is kept constant and A/2H is not controlled, Jn - 0 in the stationary state, which is also called the static h e a d , and Eqs. (11.9) and (11.11 ) become

11.3.8

,4~x - K p d p + Kpod O

(11.12)

~,4~ -- KpoJ p -t-KoJ o

(11.13)

Degree of Coupling

If no cross-coefficients vanish, then we have three degrees of couplings, qPH, qoH, and qPo. Based on the resistance-type phenomenological coefficients, the degree of coupling qpo is Kpo q--

(KpKo)I/2

(11.14)

554

11.

Thermodynamicsand biological systems

When we have levelflow, the force vanishes, A/2H = 0, and Eqs. (11.9)-(11.1 l) reduce to A~,x = Kp (1 - q2 H )Jp - ( K p K O )1/2 (qPo -k-qPHqOH ) J o

(11.15)

A~)X= _(KpKo )1/2 (qPo + qPHqOH )JP + Ko (1 - q2H)Jo

(11.16)

where q=

11.3.9

qPo+qPHqOU (1-- qZH)(1-- qZOH)

(11.17)

Efficiency of Energy Conversion

With the electroosmotic work JpXp compensated by the chemical work JrA, we can define the effectiveness of energy conversion

JpXp

n = - ~ Jr A

(11.18)

where Xp is the force for proton transportation. It is also useful to consider the force developed per given rate of expenditure of metabolic energy, which is called the efficacy of force

exp =

11.3.10

Xp jr A

(11.19)

Photosynthesis

In photosynthesis, energy-rich organic molecules emerge from simple energy-poor molecules, which absorb solar photons. After charge separation occurs, a proton electrochemical gradient, up to 200 mV, is created. In the purple photosynthetic bacterium Rhodobacter sphaeroids, membrane-bound proteins couple electron transfer to proton release into the periplasmic space of the bacterium. In the case of Halobacterium salinarium, photon-free energy is directly converted into the proton electrochemical gradient by the membrane protein bacteriorhodopsin. Using a flow of material JD and flow of energy Go, the linear nonequilibrium formulations can find a simple relation between the efficiency of photosynthesis and common transport properties of the chloroplast for producing ATE Chloroplasts are found only in plants and have double membranes. Inside a chloroplast's outer membrane, a set of thin membranes called hyaloids contain chlorophyll pigments that absorb solar energy. This is the ultimate source of energy for all the plant's needs and for synthesizing carbohydrates from carbon dioxide and water. The chloroplasts first convert the solar energy into ATP-stored energy, which is then used to synthesize carbohydrates, which can be converted back into ATP when energy is needed. Photosynthesizing bacteria called cyanobacteria use chlorophyll bound to cytoplasmic thylakoids. The maximum available energy of a photon that a chlorophyll at temperature T can utilize is (11.20)

where hv is the photon energy and TR is calculated from the assumed Planck distribution of radiation. Steady-state affinity A of a pigment is

/

(11.21)

where P and P* are the ground and excited chlorophyll states, respectively. A part of photon energy can be used to perform charge separation only if an appropriate branched pathway exists; electron acceptors and donors are located so that charge separation through a branched pathway takes place with high efficiency.

555

11.3 Bioenergetics

The thermodynamic force for light reactions is X L -

(l 1.22)

Area x - A ~ 0

The dependence of XL on photochemistry quantum yield 05 = J/I may be described by

(11.23)

X L --kBT{ln(1-ch)-ln[l+chI]}l+kd

where I is the flow of absorbed photons, and kd is the nonradiative relaxation constant, kd = 108 s-1. The second part inside the curly brackets is small compared with the first term and one needs to know the temperature and the photochemical yield to calculate Xc. Equation (11.23) also relates the flow Jr to the force XL when light intensity is regarded as constant

,{1 exp[xL]IB

(11.24)

The corresponding dissipation function qt due to transmitted free energy A and flow Jr is = AJ r

(11.25)

Using Eqs. (11.20)-(11.23), the dissipation as a function of thermodynamic force XL is q~ = a ( 1 - x)

1-exp(-x/b)

(11.26)

1 + c exp(-x/b)

or as a function of photochemical yield 05, we have = ad~[1 + b ln(1- ~b)- b ln(1 + c~b)] where XL X--

Amax

1 ,

a--

IAmax,

A max

I

- - - , b

c-

k BT

I

+ kd

With the approximation in Eq. (11.24), we have (11.27)

A max= A - kBT ln(1 - 4')

Assuming a small quantum yield, O << 1, or an equivalently small thermodynamic force, XL -- kBT, Eq. (11.27) can be linearized Jr=

X

LI kBT

(11.28)

Equation (11.28) resembles the theory of electrical circuits; Areax is the electromotive force, A is the voltage drop on a load, and XL is the voltage drop through internal resistance R i. External R and internal resistances R i are defined by R - / Amax - XL

--~)

and Ri

kBT

(11.29)

The dissipation in Eq. (11.25) is on the external resistor and obtained by using Eq. (11.28) q~ -- Jr( Ama~ -- XL )= I( Amax - X L )

X L

kBT

= RJ )

.3o)

556

11.

Thermodynamicsand biological systems

The functional and morphological heterogeneity of a lamellar system of chloroplasts indicates that pH values in different compartments (in granal and intergranal thylakoids) differ. This type of structure makes it difficult to measure local pH values at different sites. Therefore, mathematical models taking into account the spatial structure of chloroplasts provide a tool for studying the effect of diffusion restrictions on pH distributions over the thylakoid on the rates of electron transport, proton transport, and ATP synthesis. The rate of ATP synthesis depends on the osmotic properties of a chloroplast-incubation medium and, therefore, on topological factors.

Example 11.8 Efficiency of energy conversion of photosynthesis Consider a model process with an energy exchange between a photon and a composite particle. In this over-simplified model, energy is exchanged through an excited state of the chloroplast by which energy-rich electron/proton pairs from the water react with the carbon dioxide. This produces carbohydrate and oxygen molecules, and heat is dissipated away. The linear nonequilibrium thermodynamics formulations start with the rate of entropy production Vn

VT

n

T

* =--JD - - - J q

(11.31)

where n is the atomic density. Based on the rate of entropy production, the phenomenological equations are

JD = -(nD) --Vn _ VT n LDq T

Jq =--LqD

Vn

n

k VT

ks T

(11.32)

(11.33)

where D is the diffusion coefficient, k is the thermal conductivity, ks is the Boltzmann constant, and LDq and Lqo are the cross-phenomenological coefficients. The flows are related by an approximation (Andriesse and Hollestelle, 2001) (11.34)

Only a small part of the initial free energy of photons is available for photosynthesis, and the rest is dissipated. The efficiency of energy conversion in photosynthesis is low and varies in the range 2.4-7.5% (Andriesse and Hollestelle, 2001). The efficiency of energy conversion is defined by r/-

JD ,~G Jq hp

(11.35)

where iXG is the Gibbs energy per molecule and hv is the energy per photon. Some approximate values for these driving forces are AG--~ 7.95 × 10-19 J per unit carbohydrate, and (by) --~ 2.92 × 10-19 J per solar photon (pertaining to red light with a wavelength of 680 nm, which is best absorbed by chlorophyll-a). These approximate values and Eqs. (11.34) and (11.35) yield an approximate relation for the efficiency of energy conversion

( k B n D ) 1/2 r/= __~qD( 7"95 × 10-19 ) 2.92 × 10-19 = 2.72 k

(11.36)

where n --~ 3.3 × 1028 molecules/m 3, which is a typical value for water and condensed matter in general. Based on the thermal conductivity of water, we have k = 0.607 W/(m K). For a chloroplast that is drenched with water, we may assume that the high frequency of motion of water molecules transports energy. However, the diffusion coefficient is linked to the low frequency motions of molecules through the chloroplast. Assuming that the intercellular diffusion of carbon dioxide could be the limiting process in photosynthesis, we have D--~ 1.95 × 10 -9 mZ/s based on carbon dioxide in water or D--~ 0.67 × 10 -9 mZ/s based on glucose in water. Using these values of k and D in Eq. (11.36), the efficiency of energy conversion is estimated and compared with experimental values in Table 11.2. For any plant growing under ideal conditions, the efficiency is expected to be close to 7%.

11.4 Properpathways

557

Table 11.2 Comparison of predicted efficiencies of photosynthesis with measured values at various values of diffusion coefficients (predicted) 2.4 6.1 7.5

D (m-~/s)

r/(measured)

0.1() x 10 '~ 0.67 x 10 '~ 1.0 ~ 10 ~

For a minimum of D, 4.9 C3-plants ~' 6.2 C4-plants ~' 7 ideal crop, for a maximum of D

~ In practice C4-plants converts approximately 14 Ixg CO2/J of intercepted daylight in CH20, while C3-plants converts approximately 11 txg CO?/J of intercepted daylight in CH20. Source: Andriesse and Hollestelle (2001)

11.4

PROPER PATHWAYS

For a steady state far from global equilibrium, there may be pathways in the vicinity of this state along which a linear flow-force relation holds. The linearity of flows observed in experimental studies of active transport in epithelia suggests the existence of proper pathways where the phenomenological coefficients become nearly constant (Caplan and Essig, 1989). Formulating the relationships between forces and flows leads to understanding the change of affinity of a reaction driving the transepithelial active transport, free energy tissue anisotropy (compartmentalization), and activity. Experiments show that biological processes take place in many steps, each of which is thought to be nearly reversible, and exhibit linear relationships between steady-state flows and conjugate thermodynamic forces, such as transepithelial active Na +and H ~-transports and oxidative phosphorylation in mitochondria. The linear relations between the rate of respiration and the specific growth rate are observed for many microbial systems. The forces can be controlled in various ways to find a proper pathway leading to quasi-linear force-flow relationships so that the theory of linear nonequilibrium thermodynamics can be applied. For a first-order reaction S --, R doubling the concentrations of S and P will double the reaction rate for an ideal system, although the affinity remains the same, and a distinction must be made between thermodynamic and kinetic linearity. Proper pathways are associated with thermodynamic linearity. The rate of a process depends not only on the force but also on the reference state; the flow of a solute across a membrane depends on its chemical potential and on its thermodynamic state on both sides of the membrane. The constancy of phenomenological coefficients L may be maintained by applying appropriate constraints to vary the force X in the relationship J = LX. The values of L reflect the nature of the membrane, and can control the force X. If a thin homogeneous membrane is exposed to the same concentrations at each surface, flow is induced solely by the electric potential difference, and L is constant with the variation of X. However, if X is the chemical potential difference, dependent upon the bath solute concentrations, then L becomes L = KmUClm

(11.37)

~Xz where u is the mobility, Km is the solvent-membrane partition coefficient, qm is the logarithmic mean bath concentration (Ac/Aln c), and zXz is the thickness of the membrane. If a value of qm is chosen and the concentrations are constrained to the locus Ac = (qm)Aln c, then L becomes constant. The logarithmic mean concentration can be used in the linear formulation of membrane transport. If the force is influenced by both the concentrations and the electrical potential difference, then L becomes more complex, yet it is still possible to obtain a constant L by measuring J and X in a suitable experiment. For a first order chemical reaction S --+ R the reaction rate is given by Jr - k f c s - kbCp - kbCp( eA/RT --1)

(11.38)

where kf and kb are the rate constants for forward and backward reactions, respectively. At steady state, far from equilibrium, the reaction rate from • - JrA may be L*A Jr = L A RT

where L* = R T L , and can be evaluated my measuring Jr and A

(11.39)

558

11. Thermodynamicsand biological systems

L*=kbCp( A/RT ) -1 eA/RT_ 1

(11.40)

Equation (11.39) shows that for different values of A at various stationary states, the same values of L* will describe the chemical reaction when appropriate concentrations are chosen. For a specified value ofA, Eqs. (11.38) and (11.39) determine Cap and the ratio of cp/cs, respectively, and a constant L can be found by limiting the cp and Cs to an appropriate locus. As the system approaches equilibrium, A tends to vanish and kbcpapproaches the value L*. This procedure can also be used in more complex reaction systems. Proper pathways can be identified in the neighborhood of a reference steady state far from equilibrium by varying the forces X1 and X2 in such a manner as to lead to the linearity of flows and forces. Highly coupled systems show similarity to a single uncoupled flow, and linear dependencies on conjugate and nonconjugate forces exist. In the vicinity of the static head, where the transmembrane flows are zero, linearity would be expected when the degree of coupling is close to unity. Sometimes, kinetic nonlinearity may occur because of a feedback and not due to large affinities, and the sustained oscillations may occur near equilibrium.

11.4.1

Metabolic Control Analysis

The metabolic control analysis determines quantitatively the effects of various metabolic pathway reactions on flows and on metabolic concentrations. The analysis defines two coefficients: (i) the control coefficients, which characterize the response of the system flows, concentrations, and other variables after parameter perturbations; and (ii) the elasticity coefficients, which quantify the changes of reaction rates after perturbations of substrate concentrations or kinetic parameters under specified conditions. Steady-state metabolic flows Jj depend on the total concentrations of the enzymes Ek, and flow control coefficients Cj,E are defined by

= ( E~:AJj )~ _ E k OJj Cj'E ~Jjz2tE---~ Ek-~o JjOEk

(11.41)

The flow control coefficients relate the fractional changes in the steady-state flows to the changes in the total enzyme concentrations. The partial derivatives of reaction rates Jr, i with respect to the substrate concentrations Sj are called the elasticity coefficients e/;, and given by

Sj aJri

eiJ Jri OSj

(11.42)

The flow or concentration control coefficients are related to elasticity coefficients through the conservation relations and connectivity theorems. Besides the forces controlling the pathways, alternatively, flows controlling a certain pathway are also important. The metabolic control analysis can be used to evaluate flow control within biochemical pathways, and provide information on the regulation of pathway flows. Using the metabolic control analysis from the measurements of flows exchanged through the cell membrane, it is possible to quantify pathway flows and alternative pathways to the same metabolite. This methodology is limited only to the analysis of simple two-step pathways, although larger pathways can be lumped into two overall changes. The influence of the individual reaction rates (enzyme activities) on the overall flow through the pathway is called the flow control coefficients Cr, which are expressed by 0 ln(J)

Cri = ~ '

0 ln(Jri )

(i = 1,...,m)

(11.43)

where Jis the steady-state flux through the pathway, Jri is the rate of ith reaction, and m are the enzymic steps. The flux control coefficients are related to the elasticity coefficients e as follows m

8i,jCr, i i=1

-

0

(j = 2,...,m)

(11.44)

11.4

559

Proper p a t h w a y s

where the elasticity coefficients are defined by Oln(dri ) (je;,i = a ln(c i )

2 .... , m ) ( i - 1 , . . . , m )

(11.45)

The determination of flow control coefficients is difficult, and requires the independent variation of the activity of all the enzymes within the pathway. Based on linear nonequilibrium thermodynamics, the kinetics of enzyme reactions can be described by the linear functions of the change in Gibbs free energy. This yields a direct relation between the elasticity coefficients and the change in Gibbs free energy for the reactions in a simple two-step pathway. The control coefficients can be determined by the linear nonequilibrium thermodynamics formulation. Schuster and Westerhoff (1999) provide a simple example for the coupled processes of oxidative phosphorylation with slipping enzymes, for which a representative dissipative function is

- droAo + drpAp + drlAi

(11.46)

Assuming that the system is in the vicinity of equilibrium, the representative linear flow force relations based on Eq. (11.46) are

dro - - LoAo+ LopAp Jrp

--

LpoAo + LpAp

Jrl

i

(11.47)

LIAp

where Jro, Jrp, and Jrl are the reaction velocities for the respiration, ATP production or hydrolysis, and the load, respectively, and A shows the affinities as driving forces in the system. The forces for ATP production and utilization are the same with different signs. A load process may be a chemical pump powered by the hydrolysis of ATP or a proton leak. The linear flow-force relations in Eq. (11.47) indicate that the cross-coefficients Lol, Lpl, Llo, and Lpl vanish, and Lop = Lpo, according to Onsager's reciprocal rules. The reaction flows in Eq. (11.47) represent a steady state attained by the slipping enzymes, with Ao and Ap being constant. However, for a steady state of the whole system, Ap becomes variable (such as the variable proton concentration within the mitochondria), while Ao is constant. At steady state, Jrl = - J r p , and from Eq. (11.47), we have

AP =

L°PA°

(11.48)

L~ +L p

For the steady-state flows of oxygen, ATE and load, we have Jo - Jro and Jp = Jrp - --Jrl" Substituting Eq. ( 1 1 . 4 8 ) in Eq. (11.47), we find

jo-(L

°_

L2°p ~Li + L

dp-

)

A°

LpLop ]

Lop-~ Ao L1+ Lp

(11.49)

(11.50)

The nonnormalized control coefficients are defined by

OJP/OL°

*

-

Lo

*c@ - OJ°/OLP 0-7-#-;Tp

(11.51)

The control coefficient of the load is

OJp/OL 1 Odrl/OL1 The partial differential terms in the control coefficients are obtained from differentiation Eq. (11.47)

(11.52)

560

11.

Thermodynamicsand biological systems

OJro __ Ao, OJr° - 0 , 0Lp OLo

0Jro

=0

OJ~p _ O, OJrp - Ap, OJ rp _ 0 aLp aL 1 OLo OJrl

(11.53)

-

and Eqs. (11.49) and (11.50) yield 2 LopAo OJo _ Ao, O J o _ OLo aLp (L1 + L p ) 2 OJp_

OJ1 -

OLp

OLp

LopL1Ao (L1 + Lp) 2'

OJo __

O/__q

L2opAo

OJp

_

_

(Z,1 + Lp) 2' OLo

OJp_

OJ 1 _

LpLopAo

0L 1

0L1

(L 1 + Lp) 2

OJ1 - 0 OLo

(11.54)

Substituting Eqs. (11.53) and (11.54) in Eq. (11.51), we have *~o

--

1, *~b;

=_

L°P o , (~1 = L 1 + Lp

Lop L 1 + Lp

L 1 + Lp

L 1 + Lp

(11.55)

Here, the coefficients expressing the control by the load process do not carry an asterisk superscript, since they are not overall control coefficients. Multiplying the nonnormalized control coefficients with the ratio of flows Jo/Jp yields the normalized coefficients *CJ° , which satisfy the usual summation conditions (Schuster and Westerhoff, 1999) *CoJ° --1, *CJ° =

2L2opLI(Ll + L p ) , *C -oJp = 0 , Lop - Lo(L1 +Lp)

1 -*C' -Jp

Lp L 1 +Lp

Since Jp = - J b we have the summation condition: *CJ° +*CpJ° +*C1J° = 1. Metabolic systems usually consist of a number of functional units and metabolic pathways. Modular control analyses are developed to streamline the analysis of control and regulation of metabolic systems. The slipping enzyme may be considered a module catalyzing two reactions of exergonic and endergonic processes, providing a biological energy transudation. Control coefficients related to slipping enzymes can be calculated by the linear nonequilibrium thermodynamics approach. The overall control coefficients in the modular approach describe the control exerted by the particular degrees of freedom of a module on the measurable variables at steady state. Using the degree of coupling q (qop = Lop/x/LpLo) and the phenomenological stoichiometry Z (Z = k/Lp/Lo), the reation flows of Jro and Jrp in Eq. (11.47) become Jro = Lo(Ao + qopZAp)

(11.56)

Jrp = Lo ( qop ZAo +Z2Ap)

(11.57)

Using qop and Z, Eq. (11.48) becomes (11.58)

- q°pZL°A°

Ap =

+ LoZ

As before, we can substitute Eq. (11.58) in Eqs. (11.56) and (11.57), and with Jo = Jro and Jp

[ 22 /

Jo =

qop Z Lo 1 - ~ Z - - ~ L ° LoAo

-

Jrp we get (11.59)

561

11.4 Properpathways

Jp - [ 1 -

Z 2Lo

| qopZLo Ao

(11.60)

L1+ Z2Lo )

The control coefficients in terms of q and Z are obtained from the following matrix definition

* o *aSO) *

bo p

-r-p

_

Oj ° Oq OJp Oq

Oj° OZ OJp OZ

OJro OJro Oq OZ OJrp OJrp Oq OZ

-

1

(11.61)

The partial derivatives are obtained from Eqs. (11.56), (11.57), (11.59), and (11.60), and the matrix in Eq. (11.61) yields the control coefficients qopAo + 2 ZAp *6o°=

*~b; =

qopL1"do

ZAp ( + Loz2 ) qop t qopL1 Z Z( L~ + LoZ2 ) (11.62)

• 6op -

Jp(qopAo + 2ZAp) 2qopZ 2Log L 1 ( L o Z 2 _ LI )

2Lo Zz(LI + LoZ2)

(Lo Z2 - L 1 ) L 1 A ° ZZ2LoA + Lo z 2 ) jp

2Z 2LoAp

The metabolic control analysis can be used to study diseases caused by enzyme functions or dysfunctions, and helps us to understand certain pathways. It may be critical in determining the enzymes with the highest flow control coefficient, in order to inhibit or control enzyme functions. This may lead to the quantification of rate limitations in complex enzyme systems.

11.4.2 Complex Systems in Cell Biology Biological systems often reach stationary states, which may have certain characteristic properties, such as being robust when subjected to internal and external fluctuations, and displaying adoptive behavior. The free energy released by the hydrolysis of ATP is also utilized to achieve structuring by lowering entropy as a result of coupling. Biochemical network systems are complex due to the occurrence of multiple branches and cycles. These networks operate with multiple enzymes that sequentially convert different substrates into various products, and are complicated by regulatory interactions including feedback and feed-forward loops, which can be both activating and inhibiting. Biological cells function at the level of (macro)molecules. The cells are composed of interacting low-molecular-weight molecules (metabolites, such as lactate and pyruvate) and macromolecules (enzymes, protein complexes, DNA, and mRNA). Metabolites include substrates, inhibitors, activators, and products. The magnitudes of metabolite concentrations describe the state of a biochemical network at a given time. The cell has a compartmental structure surrounded by a semipermeable lipid-containing membrane, and is composed of networks of interacting microprocesses. The proteins interact either through direct physical interactions or through the binding of metabolites. The amino acid sequence of the proteins is coded by structural genes on the DNA. The genes are transcribed by RNA polymerases into mRNA strands under specific and regulated conditions, and the transcripts are translated into proteins by ribosomes. These macromolecular phenomena underlie cell behavior. We may model complex systems by top-down or bottom-up approaches. In the top-down approach, we describe the components from the systemic behavior of the actual system. For example, from the flow balance analysis in a steadystate bacterium, we learn the input and output flows, topology of the network, and the rates of many metabolic reactions.

562

11.

Thermodynamics and biological systems

On the other hand, the bottom-up approach describes the system's behavior using information about the properties of parts. We rely on the properties of isolated parts. Using kinetic modeling and measuring the parameters that characterize the rates, we can determine the capacities of the parts. The behavior of the whole system is in part a function of these properties. There are two kinds of properties that characterize the parts: (i) intrinsic properties, which are determined by the part itself, such as mass, or the amino acid sequence of a protein, and (ii) relational properties, which are determined not only by individual parts but also by one or more other parts, such as dissociation constants. In complex biochemical systems, aggregative system properties, such as the mass of a bacterium, are a function of only the intrinsic properties of the parts. However, the flow through a biochemical pathway is a nonlinear function of the concentrations of its constituent enzymes. Therefore, the flow is not an aggregative property. To characterize the system, in terms of state-independent properties, we need to impose initial and boundary conditions, as well as concentrations of nutrients, enzymes, metabolites, mRNA, temperature, and pressure. The statedependent properties include rates of free energy dissipation, rates of heat production, nutrient uptake flows, and growth rates. "System biology" requires quantitative predictions on the degree of coupling, metabolic consequences of gene deletion, attenuation, and overexpression.

Example 11.9 A linear pathway Consider a linear metabolic pathway composed of five consecutive reversible reactions (Boogerd et al., 2005) where each reaction is catalyzed by an enzyme E

Xo < 1

)X 1 < 3 )X 2 < 3 )X 3 < 4 )X 4 ( 5 )X 5

A_

(11.63)

i

Metabolite X4 inhibits the rate of enzyme 1. The metabolites X0 and X5 are maintained constant at all times. The kinetics model for this linear pathway yields

d& dt

= Jrl - Jr2

dX 2 dt

= Jr2 - Jr3

(11.64)

dX 3 dt

= Jr3 - Jr4

dX4 -

Jr4 - Jr5

dt

where Jri is the rate of reaction i as the number of product molecules formed per unit time per unit volume. Enzyme 1 E1 has two binding sites" a substrate (product) binding site for Xo or X1 and an allosteric binding site for X4, and Jr1 is a function of the concentrations of Xo, X1 and X4

Jrl

=

1(Jrlf(yO/Kl,yo)-Jrlb(X1/Kl,y,)) 1 + (X4/KI,x4)

(11.65)

1 + (Xo/Kl,xo) + (X1/KI,x,)

The first term on the right represents the inhibition effect, while the second term is the net reaction excluding the inhibition. Kl,Xl is the equilibrium dissociation constant (in mM), and indicates the ratio (E 1 - X1)/(E1X1) , when the binding relaxes to equilibrium. Here, Xi stands for X0, X1, or X4, and (El - X1) is the concentration of the enzyme-substrate X1 complex, and E1X1 is the product of the concentrations of the free enzyme E1 and the free substrate X 1. The reaction velocities Jrlf and Jrlb are the maximal rates of forward and backward rates of catalysis (in mM/min), respectively. As seen from Eq. (11.65), the rate is a nonlinear function of the concentrations of metabolites.

563

11.4 Properpathways

The equilibrium constant of reaction 1 is defined by

/ X') K1-

(11.66)

~ o eq

The equilibrium constant depends only on the properties of the reactants and temperature. The enzyme shortens the time necessary for the reaction to reach equilibrium and does not affect the equilibrium constant. If the actual ratio of product and substrate is X1

then

(11.67)

Q/K1is the extent of displacement from equilibrium. Using the relation

/rlf'lX1/

(11.68)

JrlbKl,xo Equation (11.65) becomes

I

Jr1 =

1 +(X4/KI,x4 )

( Jrlf(X°/KI'x° )(1-(Q/K1))I 1+(Xo/Kl,x,' )+(X1/KI,x, )

(11.69)

The extent of displacement is directly related to the rate Jrl, which vanishes at equilibrium where Q is equal to Keq. The displacement is directly related to the Gibbs free energy difference or the chemical potential difference

/

(11.70)

Modeling the pathway starts with the experimental determination of the rate and the parameters. Equation (11.64) can be integrated to estimate X, at other times.

Example 11.10 Sensitivity of the rate of the enzymatic reaction to substrate concentration Using the reaction J,-1 network in Eq. (11.63), the sensitivity of this enzyme to the substrate X 0 is defined by the elasticity coefficient exo

e'x"-- ~o) ~

- 1--~1 -1-+-(Xo/Kxo)+(X1/Kx,)

(11.71)

J" quantifies the fractional change in the rate of an enzymatic reaction (OJrl/Jrl) upon fractional change in the The ex,, concentrations of any of its substrates (Xo/OXo)or products. We have similar relations for all enzymes with respect to their substrates and products. The first term in Eq. (11.71) shows the change of sensitivity when the systems is approaching thermodynamic equilibrium, while the second term shows the change of sensitivity as the amount of substrate-bound enzyme increases. The elasticity coefficient is a function of the concentrations X0 and X1 as well as the relational properties of enzyme Kxo,Kxl, and the reactants Keq. The displacement from equilibrium [X1/(XoK1)]also affects the elasticity coefficient. As the reaction approaches equilibrium, the first term approaches infinity. Therefore, the values of Kxo and KXl affect the sensitivity of the enzyme to X0 only if the reaction is displaced from equilibrium. For the metabolic pathway in Eq. (11.63), consider the steady-state flow J after changing the concentration of the first enzyme E~. The rate of the first reaction is a linear function of the enzyme concentration and a nonlinear

564

11.

Thermodynamicsand biological systems

function of the concentrations of metabolites X0, XI, and X4. The fractional change in the steady-state rate through the pathway upon a fractional change in E1 is dlnJrl d In E 1

_

OlnJrl + 0~In Jrl d In X 1 t 0 In Jrl d In X 4 = 1 + Jrl CqYl _+. Jrl e xl e x 4 C x4 0 In E 1 d ln X 1 d l n E 1 d l n X 4 d l n E 1

where Cq is the concentration control coefficient. This relationship shows that the effect of the enzyme on its own rate is eE~Jr, = 0 lnJra/O InE 1 = 1 at the start. The effects ofX1 andX4 are shown by the second and third terms, respectively. Changes in component properties describe the changes of a complex system at steady state.

11.4.3

Multiple Inflection Points

A common intermediate metabolite of two enzymes may catalyze consecutive reactions of a pathway, and diffuse from one active center to the other without dissociation. This is called metabolic channeling, which could lead to a decrease in the steady-state concentration of the intermediate metabolite even at constant flux. The existence of a multidimensional inflection point well outside of the global equilibrium in the force-flow space of enzyme-catalyzed reactions may indicate linear behavior between the logarithm of reactant concentrations and enzyme-catalyzed flows (Rothschild et al., 1980). Thus, enzymes, operating near this multidimensional point and leading one to choose particular state variables, may produce some linear coupled biological systems. This range of kinetic linearity may be far from equilibrium. Disregarding the electrical effects, the conditions for the existence of a multidimensional inflection point are (i) each reactant with varying activity influences the transition rates so that only one state remains; (ii) the kinetics of the reaction involving the given reactant are of fixed order with respect to that reactant; and (iii) for various concentrations of reactants, at least a certain cycle is present. The first condition excludes autocatalytic systems; for many biological energy transducers, however, it may well be satisfied. The fact that local asymptotic stability is supported by local symmetry and experimental evidence of the linear behavior of some coupled biological energy-transducing systems suggest that kinetic linearity may lead to thermodynamic linearity and cause a proper pathway to form. Consider an ensemble of enzyme molecules or membrane proteins in the coupled processes of reactions and vectorial flows. Such systems consist of a set of cycles and subcycles of reactions and transport processes. For a flow in cycle k as Jk (k = a, b,..., h), the first two steady-state flows are given by the following relations

Jl = Ja + Jb + J f

(11.72)

J2 = Ja + Jb + Jc + Jg

(11.73)

Expanding these two flows as functions of their conjugate forces in a Taylor series about some reference steady state, and assuming all other forces as constant, yields the finite differences from the first-order terms

•Ji =

axe

¢~Xi -~-

]

OnE k(j) Jk aJj = OXi 6X i

OXj

OXj

I

6Xj +...

(11.74)

6Xj +...

(11.75)

where k(i) and k(j) show the sets of cycles associated with Ji and Jj, respectively. The expansion of flows in a Taylor series about a multidimensional inflection point yields expressions linear in ln(ci) and ln(cj) if changing concentrations of components i andj are the reactant concentrations. In Eqs. (11.74) and (11.75) the proper conjugate forces, X~ and Xj, appear explicitly. The reference state may not be an inflection point except with respect to ci and cj. Caplan and Essig (1989) provided a simple model of active ion transport, having properties consistent with a multidimensional inflection point when one of the variables was the electrical potential difference across the membrane. A multiple inflection point may not be unique; other conditions may exist where flows ,/1 and J2 simultaneously pass through an inflection point of on variation of X1 at constant X2, and vice versa. It is frequently not possible to vary both

565

11.4 Properpathways

forces independently in biological systems. However, if X1 can be controlled experimentally along a proper pathway, while X2 is kept constant, the response of the flows to change in Xl will permit a thermodynamic characterization of the system. Stucki (1980) demonstrated that in mitochondria, varying of the phosphate potential, while maintaining a constant oxidation potential, yields linear flow-force relationships. Extensive ranges of linearity are found in active sodium transport in epithelial membranes, where the sodium pump operates close to a stationary state with zero net flow. In the vicinity of such a stationary state, kinetic linearity to a limited extent simulates thermodynamic linearity at the multidimensional inflection point. There may be a physiological advantage in the near linearity for a highly coupled energy transducer, since local asymptotic stability is guaranteed by these conditions. Also, the thermodynamic regulation (buffering) of enzymes may be interpreted as an indication that intrinsic linearity would have an energetic advantage and may have emerged as a consequence of evolution.

11.4.4

Reaction Diffusion in Biofilms

Many biochemical signaling processes involve the coupled reaction diffusion of two or more substrates. Metabolic biochemical pathways are mainly multicomponent reaction cycles leading to binding and/or signaling and are coupled to the transport of substrates. A reaction-diffusion model can also describe the diffusion of certain proteins along the bacterium and their transfer between the cytoplasmic membrane and cytoplasm, and the generation of protein oscillation along the bacterium (Wood and Whitaker, 2000). Three dominant processes in the reaction diffusion in biofilms and cellular systems are (1) diffusion in a continuous extracellular phase B, (2) transport of solutes across the membrane, and (3) diffusion and reaction in the intracellular phase A. Consider aerobic growth on a single carbon source. The volume-averaged equations of a substrate S and oxygen O (electron acceptor) transport are

0[Cs(eB +c~ leA)]

= V. (Ds, ~ .Vc s ) _ Jrs,maxe A

at

O[Co(eB + K~qle/, )] at

Cs

(11.76)

c s + chK s Co + KeqK o

Cs

-- ~7 • (I)o,e " ~7Co ) -- Jro,max~A

Co

Co

-- ~:A r i o K e q l c o

(11.77)

c s + ol 1K S c o + Keq K O

where rio is the effective oxygen uptake parameter in m3/(kg s). The concentrations above are equilibrium weighted, and defined by CO = Keq (CO,av,A) -- (CO,av,B)

with a volume average concentration of 1

!CO,av, ) = 7 f

dV

Here, V is the averaging volume and VB(t) is the volume of phase B within the averaging volume. The parameters SA and SB are the volume fractions of the phases A (SA = VA/V) and B (SB = VB/V), respectively. The parameter Di,e is the effective diffusivity tensor of component i, J,,max,~ is the maximum reaction rate, and rio is the first-order endogenous respiration rate coefficient. The cell mass per unit volume of biofilm is defined by 1

f p dv

Under thermodynamic equilibrium, the average concentrations of the substrate are related by (CS,av,B) = Oll(CS,av,A ) and (CO,av,B)= Keq (CO,av,A)

where c~1 is the equilibrium coefficient and Keqis the equilibrium coefficient of oxygen portioning between extracellular phase B and intracellular phase A. Even under nonequilibrium condition, the concentrations are close enough, and the equations above can be reasonable approximations. At local mass equilibrium, the solute concentrations on either side of the cell membrane are equal and the parameter a l is equal to 1.

566

11.

Thermodynamicsand biological systems

The oxygen and substrate transport processes are similar. However, through the cell membrane, the flow of oxygen follows the diffusion or permeability model, while the solute flow involves complex transport mechanisms.

11.4.5

Effective Diffusivity of Cellular Systems

The effective diffusivities in Eqs. (11.76) and (11.77) are defined by

DS'e:(eBDsB+eA0/llDsA)I+DSB I nBAbsBda+DSA f V V aBA ( t )

D°'e =(SBDoB

nABbsAda

(11.78)

aAB ( t )

-4-e'AKeqlDoA)I+D°B f nBAboBda+D°A I V V aBA(t )

nABbOA

da

(11.79)

aAB(t)

where I is the unit tensor, nBA = --nAB is the unit normal vector directed from phase B toward phase A, and bsB is the vector field that maps VCs onto the spatial deviation concentration for substrate S in phase B, while bsA is the vector field that maps Vcs onto the spatial deviation concentration for substrate S in phase A times 0/1. The term aAB is the interfacial area contained within the averaging volume. The definitions of effective diffusivity tensors are key parameters in the solution of the transport equations above. For an isotropic medium, the effective diffusivity is insensitive to the detailed geometric structure, and the volume fraction of the phases A and B influences the effective diffusivity. When the resistance to mass transfer across the cell membrane is negligible, the isotropic effective diffusivity, Ds,e = D~,eI may be obtained from Maxwell's equation D~,e _ 3Ks - 2eB(Ks -- 1) DSB 3+eB(K S --1)

(11.80)

where DsB is the mixture diffusivity of S in phase B, and the dimensionless parameter Ks is KS--

DSA

DSB0/1

Maxwell's solution for permeable spheres with 0/1 = 1 is defined by Ds,e _ 2DsB + DSA -- 2e A (DsB -- DSA) DSB

(11.81)

2DsB + DSA -- e A (DsB -- DSA )

Maxwell's solution for impermeable spheres with Ks = 0 is defined by DS,e_ DSB

2~:B

(11.82)

3 - ~3B

These equations are applied to biofilms and cellular systems. If mass transfer across the cell membrane is important, then the following equation is used D's, e _ 2Ks - 2eB(Ks -- 1) -4-28B~3A1/3 (47r/3) 1/3(ys//) 1/3 1/3

DSB

3 + eB (KS --1) + (3 -- eB )e a

(47r/3)

(11.83)

(YS/1)

where 0/2 -+-0/3CSB,av+ 0/4CSA,av nt-0/C 5 SB,avCSA,av TS -- DSA

0/1Eo

The parameter 1is the characteristic length for a unit cell, E0 is the surface concentration of a carrier protein molecule, and c~2,c~3,cq, c~s are the reaction rate parameters analogous to that half saturation constants. Table 11.3 displays the experimental effective diffusion coefficients and the volume fraction of intracellular phase A. In the first four sets

11.5

567

Coupling in mitochondria

Table 11.3 Experimental data of effective diffusivity System

Cell type

Solute

e

D a,e/D A~

Source

Gel immobilized cells

Mammalian cells (ascites tumor)

Glucose

0.980 0.705 0.590

1.05 0.76 0.64

Chresand et al. (1988)

Fermentation media

S. cerevisiae

Oxygen

0.980 0.961 O.922 0.883 0.844 0.822 0.776 0.700

0.990 0.957 0.895 0.900 0.890 0.848 0.805 0.748

Ho and Ju (1988)

Biofilm (natural) on a hollow fiber filter support

E. coli

Nitrous oxide

0.860 0.660 0.640 0.270 0.080 0.040

1.00 0.62 0.65 0.37 0.29 0.28

Libicki et al. (1988)

Biofilm (artificial) on a hollow fiber filter support

E. coli

Nitrous oxide

0.940 0.910 0.835 0.825 0.815 0.810 0.800 0.690 0.630 0.500 0.470 0.400

1.00 0.85 0.77 0.79 0.75 0.74 0.84 0.64 0.61 0.49 0.51 0.50

Libicki et al. (1988)

Gel immobilized cells

S. cerevisiae

Lactose

0.882 0.879 0.770 0.650 0.530

0.96 0.82 0.69 0.62 0.42

Axelsson and Persson (1988)

Gel immobilized cells

S. cerevisiae

Ethanol

0.882 0.770 0.650 0.530

0.88 0.70 0.47 0.36

Axelsson and Persson (1988)

Gel immobilized cells

Mammalian cells (ascites tumor)

Lactate

0.980 0.805 0.705 0.580

0.96 0.76 0.56 0.47

Chresand et al. (1988)

o f data, the substrate was transported inside the cells (B --+ A); the r e m a i n i n g three sets represent the substrate being transported from the cells (A --+ B). The experimental data was satisfactorily r e p r e s e n t e d by Eqs. ( 1 1 . 8 1 ) - ( 1 1 . 8 3 ) , especially for high values o f eB.

11.5

C O U P L I N G IN M I T O C H O N D R I A

A two-flow coupling implies an interrelation b e t w e e n flow i and flow j, so that a flow occurs w i t h o u t a force or against its conjugate driving force. The e n e r g y level o f a reactant m a y c h a n g e due to the c o u p l i n g effect, while the catalyst effect m a y be limited to the lowering o f the reaction barrier for both the forward and b a c k w a r d reactions (Jin and Bethke, 2002). M a n y biological reactions can take place against their o w n affinities b e c a u s e o f the t h e r m o d y n a m i c coupling effect. For e x a m p l e , m a n y transport systems in bacteria are driven by the proton gradient across

568

11.

Thermodynamicsand biological systems

the plasma membrane. The chemiosmotic theory also describes the role of proton flows in bacterial bioenergetics. Proton and lactose are cotransported into the cell by lactose permease. At the same time, protons are transported out of the cell in connection with electron flow through the respiratory chain. Overall, the cell maintains a nonequilibrium level of pH by keeping its interior at a higher pH than its environment. Eucaryotic cells posses a hierarchy of transport systems to maintain nonequilibrium concentration levels of some substrates within organelles than those in the cell's cytoplasm. Still, the cell controls its complex array of chemical reaction cycles so that the supply and demand for substrates, energy, and electrons are balanced and resources are utilized efficiently. In early experimental work, the interior and exterior cell concentrations of K +, Na+, and C1- ions were measured and compared against those obtained from the Nernst potentials, given by

~J =

z - ~ lnt, aj, o )

(11.84)

where ~j is the equilibrium potential, zj the valence, and aj the activity of speciesj on both sides of the membrane (i and o). The measured potentials were different from the Nernst potentials, indicating that a cell maintained the concentration difference at a steady-state diffusion flow. The basis of active transport in animals is the coupled metabolic reaction to external diffusion, while most of the chloride flow in plant cells depends on photosynthesis. One of the conventional methods for establishing the existence of active transport is to analyze the effects of metabolic inhibitors. The second is to correlate the level or rate of metabolism with the extent of ion flow or the concentration ratio between the interior and exterior of cells. The third is to measure the current needed in a short-circuited system having similar solutions on each side of the membrane; the measured flows contribute to the short-circuited current. Any net flows detected should be due to active transport, since the electrochemical gradients of all ions are zero (A~ - 0, Co = ci). Experiments indicate that the level of sodium ions within the cells is low in comparison with potassium ions. The generalized force of chemical affinity shows the distance from equilibrium of the ith reaction X i =

RTln Ki'eq m

I-I c~..~, j=l

(11.85)

where R is the gas constant, K is the equilibrium constant, cj is the concentration of the jth chemical species, and Pji are the stoichiometric coefficients, negative for reactants and positive for products, for the ith reaction. The phosphate potential in mitochondria is expressed as

Xp--AGp-RTln

[ATP] ) [ADP][Pi ]

(11.86)

Stucki (1980, 1984) applied the linear nonequilibrium thermodynamics theory to oxidative phosphorylation within the practical range of phosphate potentials. The nonvanishing cross-phenomenological coefficients Lo.(i 4=j) reflect the coupling effect. This approach enables one to assess the oxidative phosphorylation with H+pumps as a process driven by respiration by assuming the steady-state transport of ions. A set of representative linear phenomenological relations are given by J1 = LalX1 +La2X2

(11.87)

J2 = I-azX1 + L22X2

(11.88)

where J1 is the net flow of ATE J2 is the net flow of oxygen, X1 is the phosphate potential as given by Eq. (11.86), and X2 is the redox potential, which is the difference in redox potentials between electron-accepting and electronreleasing redox couples.

11.5.1

Degree of Coupling in Oxidative Phosphorylation

The degree of coupling is defined as q:

L12 (Zl 1L22) 1/2

0<[q[
(11.89)

11.5 Couplingin mitochondria

569

and it indicates the extent of overall coupling for the various individual degrees of coupling of the different reactions driven by respiration in the mitochondria. With an angle o~ whose sine is q we have c~ - arcsin(q)

(11.90)

By defining the phenomenological stoichiometry Z (11.91)

L22 and by dividing Eq. (11.87) by Eq. (11.88), we can determine that the reduced flow ratioj = reduced force ratio x = X1Z/X2 and the degree of coupling J _

x + q

J1/(J2Z) varies with the

(11.92)

qx+l Assuming that oxidation drives the phosphorylation process, then Xm < 0 and X2 > 0, and Jl/J2 is the conventional P/O ratio, while )(1/)(2 is the ratio of phosphate potential to the applied redox potential. The following relations are from Stucki (1980). At static head (sh), analogous to an open circuited cell, the net rate of ATP vanishes, and the rate of oxygen consumption and the force (the phosphate potential) are expressed in terms of a as follows

(,J2)sh = L22X2 c0S2 a (Xj)sh - -

X2 sin~

(11.93) (11.94)

Z where L22 is the phenomenological conductance coefficient of the respiratory chain. Therefore, energy is still converted and consumed by the mitochondria. The nonequilibrium phosphate value is distant from the equilibrium value 0(1 = -X2 Z) by a factor of q, and is given by

Y2 (Xl)sh -(Xl)eq = - ~ ( 1 - q )

(11.95)

The dissipation at the static head can be obtained from ~ h - Lll( tan2

a)-l(X1)s2h

(11.96)

This equation shows that the energy needed at the static head is a quadratic function of the phosphate potential. At level flow (lf), analogous to a short-circuited cell, the phosphate potential vanishes. Hence, no net work is performed by the mitochondria, and we have

J1

(11.97)

This equation shows the maximal P/O ratio measurable in mitochondria at a zero phosphate potential. Equation (11.97) also indicates that at level flow, the flow ratio does not yield the phenomenological stoichiometry Z but approaches this value within a factor of q. Therefore, if the degree of coupling q is known, it is then possible to calculate Z from the P/O measurements in a closed-circuited cell. Obviously, in state 3, the phosphate potential is not zero, however, for values of q approaching unity, the dependence of the flow ratio on the force ratio is weak, according to Eq. (11.92). Therefore, state 3 is only an approximation of the level flow at values of q close to unity, and the dissipation function to maintain a level flow is given by q~lf - (Jj)~f 2 q Lll

(11.98)

570

11.5.2

11.

Thermodynamicsand biological systems

Efficiency of Energy Conversion

The efficiency of energy conversion as a ratio of output energy to input energy is related to the degree of coupling as follows r/ .

.

J1X1 _

.

.

x +q

.

JzX2

.

(11.99)

q + (1/x)

The efficiency reaches a maximum value between the static head and the level flow, which is the function of the degree of coupling only, and expressed by

r/°pt

[1 + X/1- q2 ]2 = tan2

(11.100)

tan

(11 101)

The value of x at Tlmaxis given by

Xop t .

.

.

.

l+x/l_q2

The dissipation function ~ can be expressed in terms of force ratio x and degree of coupling q xlt = (X 2 + 2qx + 1)L22 X2

(11.102)

If we assume X2 as constant, the dissipation function is at a minimum at the static head force ratio Xsh = - q

(11.103)

Within the region of validity of linear phenomenological equations, the theorem of minimal entropy generation at steady state is a general stability criterion. The static head is the natural steady state where the net ATP flow vanishes and a minimum of • occurs along the loci of the static-head states XI/'sh = COS2 0lL22X2

The dissipation at the state of optimal efficiency is obtained using

Xopt

COS20L L22 X 2 XItopt = 2 ~ 1+ cos a

(11.104) in Eq. (11.102)

(11.105

)

The dissipation for the level flow is given by XIrlf = L22 X2

(11.106)

Without a load, Stucki (1980) suggested the following order Xltsh < xI/'opt < XI/'lf

(11.107)

This inequality means that the minimum dissipation and the natural steady state do not imply the optimal efficiency of oxidative phosphorylation. At the level flow, there is a load and hence a load conductance corresponding to the state of optimal efficiency between the static head and the level flow. The dissipation of oxidative phosphorylation with a coupled process (load) utilizing ATP is given as follows * c = Jl & + J2 X2 + J3 X3

(11.108)

Assuming that the ATP-utilizing processes are driven by the phosphate potential, X3 = X1, and a linear relation between the net rate of ATP utilization and X1, we have

11.5

571

Coupling in mitochondria

J3 = L33X1

(11.109)

Here, L33 is the phenomenological conductance of the load, and the dissipation function in terms of the force ratio x becomes ~c-

L33/

x" l + ~ l

+2qx+l

1L22

(11.110)

Only if the following equation is satisfied L33 - X / 1 - q 2 - cosoL LI l

(11.111)

then Eq. (11.110) is minimal at Xopt. For mitochondria, L33 is an overall phenomenological coefficient lumping together all the conductances of ATPutilizing processes, while Lll shows the conductance of phosphorylation. If these two coefficients match according to Eq. (11.111), then the natural steady state of oxidative phosphorylation is at the optimal efficiency. Stucki called Eq. (11.111) the condition of conductance matching of oxidative phosphorylation, and presented an experimental verification.

11.5.3

Dissipation with Conductance Matching

Dissipation with conductance matching at the static head is given by (q~rc)sh -- ( cOSOI + cOS20l-- COS3 ~)L22222

(11.112)

The dissipation function at the state of optimal efficiency of oxidative phosphorylation is (XIfc)opt -- COS°LL22X2

(11.113)

At the conductance matching state, the dissipation Eq. (11.110) is minimum at the loci of the optimal efficiency states. At the optimal efficiency, the P/O ratio is given by (11.114) "~2 opt

~2-2 If l+cOsa

Equation (11.114) shows that unless q = 1, a maximal P/O ratio is incompatible with the optimal efficiency, and we have the inequality 1

Jl

()opt

(11.115)

Therefore, the low P/O ratios do not necessarily mean a poor performance of the oxidative phosphorylation. Similarly, the net rate of ATP synthesis at the optimal efficiency is given by (')¢1)opt ~---(J1)lf

cos a 1+ cos a

(11.116)

with the boundaries 1

0 < (Jl)opt < 2(J1 )If

(11.117)

This inequality means that a maximal net rate of ATP production is incompatible with the optimal efficiency. Cellular pathways balance the rate and efficiency of ATP production with respect to the energy needs of the cell. For example, heart and brain mitochondrial systems utilize more oxygen and produce ATP at a faster rate than the

572 .

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

11.

.

Thermodynamicsand biological systems

systems in the liver. However, liver mitochondria can produce ATP more efficiently based on a higher P/O ratio and a higher degree of coupling in oxidative phosphorylation.

11.5.4

Variation of Coupling

The degree of coupling depends on the nature of the output required from the energy conversion system in the mitochondria and on metabolic regulation and stability. If the system cannot cope with instabilities, then fluctuations, such as in pH and the pressure of the blood, could irreversibly harm the organism. For an optimal efficiency state, the condition of conductance matching must also be satisfied. Experiments with livers perfused at a metabolic resting state suggest that the conductance matching is satisfied over a time average, and the degree of coupling qec yields an economical oxidative phosphorylation process. Optimization may be based on constraints other than the efficiency, such as the production of thermal energy in the mitochondria, which requires low degrees of coupling. The degrees of coupling are also measured in Na÷transport in epithelial cells, and in growing bacteria where the maximization of net flows are the most important considerations for the system. On the other hand, for a fed rat liver at a metabolic resting state, an economical power output is the priority, while a starved rat liver has to produce glucose, and the maximum ATP production is given priority over energy conservation. In the heart and brain, the experimental value of q for the cellular respiration pathway is close to the value of q~C = 0.953, which suggests that the pathway is optimized to economical ATP for cellular processes. In the brain, the coupling of the acetic acid cycle approaches qp = 0.91, suggesting a maximized cellular energy state. However, in the heart, the acetic acid cycle coupling is 0.786, which is between qp and qf, and consistent with the maximum ATP production necessary for preserving the cellular energy state (Stucki, 1980; Cairns et al., 1998). Optimal flow ratios are also a characteristic of oxidative phosphorylation, and may provide additional information on the relationships between the respiratory response and energy demand stimulation by ADP. Most metabolic processes in living cells are dynamic systems, and the behavior of flows may better reflect complex system mechanisms than do the models dependent on end-point measurements. For example, the ratio of ADP/O describes the state of the end-point capacity of oxidative phosphorylation based on the input flow of ADR Figure 11.3 shows the effect of the degree of coupling on the characteristics of four different output functions f given by f = tanm ( 2 ) c o s ( c 0

0.35

i .........

I

. . . .

I ......

I

.....

I

....

I

(m = 1-4)

I

'

'

I

(11.118)

'

I'

''

I

qf=0.786 0.3

0.25

_

(J1)opt--~ 0.2

0.15

_

q~0=0.953 0.1

0.05

(JlXlq)opt 0

0

0.1

0,2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Degree of Coupling,q

Figure 11.3. Effect of the degree coupling q on the output functions f (Eq. (11.118)) (Demirel and Sandier, 2002).

11.5 Coupling in mitochondria

573

The optimum production functions and the associated constants are described in Table 11.4. The followings are the output functions seen in Figure 11.3: (i)

If the system has to maximize ATP production at optimal efficiency, then f = (J1)opt and the degree of coupling

q f = 0.786. (ii) Conversely, if the system has to maximize power output at optimal efficiency, we have the output function f = (J1X1)opt occurring at the degree of coupling qp = 0.91. (iii) If an efficient ATP synthesis at minimal energy cost is imposed, then the function for economical ATP occurs at the degree of coupling q~C __ 0.953. (iv) Economical power output occurs at the degree of coupling q~C _ 0.972. The optimum output power (J1Xl)opt and the efficiency (J1Xlqq)opt are calculated from the plots of J1X1 vs. x and J1XI~q vs. x, respectively. A transition from qp to q~C causes a 12% drop in output power (J1X1) and a 51% increase in

efficiency. For a favorable ATP production at optimal efficiency of oxidative phosphorylation, we should have q < 1. With the consideration of conductance matching, Stucki (1980) determined four production functions, which are given in Table 11.4. Stucki (1980) analyzed the sensitivity of the force (the phosphate potential) to the fluctuating cellular ATP utilization, and found that the sensitivity is minimal at q - 0.95. This analysis indicates that the phosphate potential is highly buffered with respect to the changing energy demand to maximize kinetic stability and thermodynamic efficiency at the same degree of coupling. For H+-translocating ATPase, the H+/ATP coupling ratio is important for mechanistic, energetic, and kinetic consequences, and a value of 4 has been adopted for the ratio; the standard reaction Gibbs energy of ATP production is obtained as 31.3 kJ/mol at T - 20°C, pH 8.0, pMg 2.5, and 0.08 M ionic strength, and the standard enthalpy of the reaction is 28.1 kJ/mol. 'The differences in the rates of proton pumping, ATPase activities, and degrees of coupling are adjusted by each biological system in order to survive and compete in its environment. Cairns et al. (1998) reported the experimental mean degree of coupling for isolated liver mitochondria as 0.955, which is close to the value of 0.952 found by Soboll and Stucki (1985) using isolated perfused whole livers from fasted rats. Under similar cytoplasmic ATP levels, the ATP utilization for muscular contraction and the ion transport would be much higher in a beating heart than an arrested heart. ADP supply to the mitochondria of the beating heart would lead to higher rates of oxidative phosphorylation and ATP production.

Table 11.4 Production functions with the consideration of conductance matching

Production function

Loci of the optimal efficiency states

1. Optimum rate of ATP production:

From the plot of Jl vs. x:

ZL?gX9 -" -

(11 "123)

_

_

L~X~ .

qp = 0.910

No

(JiX1)op~=tan 2 ~ coso~L22X_~ (11.122)

oz=65.53 °

r/ = constant

From the plot ofJ~r/vs, x:

q~C= 0.953

Yes

(JlT~)opl

. )521.1(

.

=tan3(2)cosc~ZLg~X~ . . . . (11 124)

.

/ ¸) (Y

9

(dlXlq),,pt . . = tan4 ~- cos~L~,X; (11.126)

Reprinted with permission from Elsevier, Demirel and Sandler (2002).

o~=72.38 °

q~C= 0.972

From the plot ofJiX~ Xl vs. x:

4. Optimum output power of oxidative phosphorylation at minimal energy cost: J1Xlrl = - x2(X+qqx +1

r/ = constant

/)

3. Optimum rate of ATP production at minimal energy cost: + q+) 2~ Jl rl = - x ( xxq

c~=51.83 °

(11.120)

cosozZL22X2

From the plot of J1Xi vs. x:

(11.121)

JlXl=x(x+q)Le2X~

No

/)

2. Optimum output power of oxidative phosphorylation:

Energy cost

qt.= 0.786

(Jl)opt =tan ~

(11.119)

JI=(q+x)ZL~,X~

q

c~= 76.34°

Yes

574

11.6

11.

Thermodynamicsand biological systems

REGULATION IN BIOENERGETICS

Regulation implies a physiological outcome as a result of manipulating a mitochondrial function, and hence it is different from control. However, many physiological signals to mitochondria cause large changes in activity associated with control, such as in the shift from state 4 to state 3. Mitochondrial function is regulated by a number of factors over time scales. These include calcium stimulating NADH supply to the respiratory chain, or oxidative phosphorylation complex activities, nitric oxide-inhibiting cytochrome oxidase, and thyroid hormones binding to cytochrome oxidase. The physiological reasons for these regulations of mitochondria are mainly to match ATP supply efficiently to changes in workload, modulate thermogenesis, biogenesis or cell death, as well as respond to external stimuli. According to the chemiosmotic theory, the electrochemical proton gradient across the membrane is one of the important mechanisms for regulating the rate of respiration and ATP synthesis. Respiratory control mainly means stimulation ofmitochondrial respiration by ADP and its decrease because of conversion of ADP to ATP. Various substrates regulate the metabolism of energy; fatty acids may regulate and tune the degree of coupling by inducing uncoupling, leading to optimum efficiency of oxidative phosphorylation. Experiments with incubated rat-liver mitochondria show that the adenylate kinase reaction can buffer the phosphate potential to a value suitable for the optimal efficiency of oxidative phosphorylation in the presence of a very high rate of ATP hydrolysis. Stucki called this class of enzymes, such as adenylate kinase and creatine kinase, thermodynamic buffer enzymes. A fluctuating ATP/ADP ratio and deviations from the optimal efficiency of oxidative phosphorylation are largely overcome by thermodynamic buffering. As a terminal component of the respiratory chain, cytochrome oxidase catalyzes the transfer of electrons from cytochrome c to oxygen, which is coupled with proton pumping. Although the reaction catalyzed by cytochrome oxidase is far from equilibrium, it has been commonly assumed that the flow-force relationship is unique and proportional. The reaction becomes more nonlinear in the thermodynamic branch as the oxygen concentration decreases. This is interpreted as the thermodynamic cost of kinetic regulation. When the oxygen concentration decreases, the cell prefers to optimize the flow through cytochrome oxidase rather than the thermodynamic force of the reaction in order to maintain constant oxygen flow and ATP production. The electrochemical potential of protons may be considered as a universal regulatory factor of cytochrome oxidase. The enzymes are capable of causing certain reaction pathways by catalyzing a conversion of a substance or a coupled reaction. For example, on adding nigericin to a membrane, the system reaches a steady state in which the gradients of H+and K +are balanced. On the other hand, if we add valinomycin and protonophore, both the gradients rapidly dissipate. The mitochondrial creatine kinase is a key enzyme of aerobic energy metabolism, and is involved in buffering, transporting, and reducing the transient nature of the system. This can be achieved: (i) by increasing the enzymatic activities in a pathway, (ii) by the metabolic channeling of substrates, and (iii) by damping oscillations of ATP and ADP flows upon sudden changes in the workload.

11.6.1

Uncoupling

Uncoupling proteins are a subgroup of the mitochondrial anion transporter family, and are identified in prokaryotes, plants, and animal cells. Three mammalian uncoupling proteins are called UCP1, UCP2, and UCP3. The proton electrochemical gradient developed across the inner membrane during electron transport of the respiratory chain is used to phosphorylate ADP to ATP by FoF1-ATP synthase, and hence the respiration is coupled to phosphorylation. However, ATP synthesis is matched to cellular ATP utilization for osmotic work of (downhill and uphill) transport, or mechanical work such as muscle contraction and rotation of bacterial flagellum. The uncoupling of the mitochondrial electron transport chain from the phosphorylation of ADP is physiological and optimizes the efficiency and fine tunes the degree of coupling of oxidative phosphorylation, and prevents reactive oxygen species generation by the respiratory chain in the resting state. Uncontrolled production of reactive oxygen molecules can cause the collapse of mitochondrial energy conservation, loss of membrane integrity, and cell death by necrosis. The respiratory chain is a powerful source of reactive-oxygen molecules, which include oxygen-free radicals, hydroxyl radical hydrogen peroxide, and nitric oxide; they are very reactive and able to damage cellular components and macromolecules, and influence programmed cell death or apoptosis. Cells have developed various strategies to dissipate reactive oxygen molecules and remove their oxidation products. Uncoupling proteins are capable of modulating reactive oxygen molecules. Fatty acids facilitate the net transfer of protons from intermembrane space into the mitochondrial matrix, hence lowering the proton electrochemical potential gradient and mediating weak uncoupling. Uncoupling proteins generally facilitate the dissipation of the transmembrane electrochemical potentials of H +or Na+produced by the respiratory chain, and result in an increase in the H+and Na+permeability of the coupling membranes. They provide adaptive

11.6 Regulation in bioenergetics

575

advantages, both to the organism and to individual cells, and also increase vulnerability to necrosis by compromising the mitochondrial membrane potential. Some uncoupling is favorable for the energy-conserving function of cellular respiration. In oxidative phosphorylation, leaks cause a certain uncoupling of two consecutive pumps, such as electron transport and ATP synthase, and may be described as the membrane potential-driven backflow of protons across the bilayer.

11.6.2

Slippages and Leaks

A slip means a decreased proton/electron stoichiometry of proton pumps. Mainly, slippage results when one of two coupled reactions in a cyclic process proceeds without its counterpart, which is also called intrinsic uncoupling. On the microscopic level, individual enzymes cause slippage by either passing a proton without contributing to ATP synthesis, or hydrolyzing ATP without contributing to proton pumping. On the macroscopic level, the measured degree of coupling may be different from the expected coupling, as these microscopic slips are averaged across a population of enzymes. Slippage is an intrinsic property of the enzyme, and hence is related to an enzyme's mechanism and structure. In terms of the thermodynamic energy conversion, a slip may decrease efficiency; it may, however, allow dynamic control and regulation of the enzyme over the varying ranges of the electrochemical gradient of protons and the chemical potential of ATP in equilibrium with ADP and Pi. Mitochondrial energy metabolism may be regulated by the slippage of proton pumping in cytochrome c oxidase at high proton motive force. It is possible that slips have evolved to enhance the function of particular coupling enzymes in particular conditions. In transportation, leaks can be found in the proton-sugar symport in bacteria where a protein mediates the transport of protons and sugar across the membrane, and adding a protonophore, a parallel pathway occurs, causing a leak in the transport. Leaks and slips may affect the metabolic rate. Schuster and Westerhoff (1999) developed a theory for the metabolic control by enzymes that catalyze two or more incompletely coupled reactions. The control by the coupled reactions is distinguished quantitatively from the control by the extent of slippage using the linear nonequilibrium thermodynamics formulations; here the limits of coupling may be an important parameter, and may be obtained as the ratio of coupled-to-uncoupled rates, which is a function of the binding energy of the substrate and the carrier protein. One other concern in an interconnected biological network is the behavior of a subsystem (e.g., glycolysis), which may become unsteady and chaotic, so that the output of this subsystem (e.g., ATP production) is adversely affected, and becomes external noise for other subsystems, causing inhibition and desynchronization.

11.6.3

Nonequilibrium Thermodynamics Model of a Calcium Pump with Slips

Waldeck et al. (1998a, 1998b) presented a nonequilibrium thermodynamics model for the calcium pump shown in Figure 11.4. This section summarizes this model. During the hydrolysis of ATE a variation of the coupling stoichiometries with the electrochemical gradients is an indication of molecular slips. However, the Ca2+and H+membrane-leak

Ca2+

"~

A

H+ ] H+

Ca2+ • .........

ATP

Membrane

H+--l Ca2+-ATPase

; Ca2+

ADP+Pi+H+

Figure 11.4. Schematics of calcium transport with Ca 2 -ATPase liposome, ionophore (A), and leaks. Ionophore A23187 induces uptake of Ca 2÷iOns. Leaks are shown with dashed arrows. "i" is interior (alkaline) and "e" is exterior (acidic). Scalar flow of ATP hydrolysis drives the uphill transport of Ca 2+and H + .

576

11.

Thermodynamicsand biological systems

conductances may also be a function of their respective gradients. Such leaks yield flow-force relations similar to those that are obtained when the chemical pump slips. Hence, one needs to exercise caution when interpreting data of CaZ+-ATPase mediated flows that display a nonlinear dependence on the electrochemical proton gradient A/2u and/or calcium gradients 2~/2Ca. CaZ+-ATPases exist in the plasma membranes of most cells and in the sarcoplasmic reticulum of myocytes, where they pump Ca 2+out of the cytosol and into the lumen, respectively, while simultaneously counterporting H+ions. Ca2+-ATPase requires MgZ+on the side from which CaZ+is pumped. It is generally established that the Ca2+/ATP stoichiometries for the plasma membrane and sarcoplasmic reticulum are 1 and 2, respectively. Using a nonequilibrium thermodynamics model, the extent of slippage in the plasma membrane Ca 2+-ATPase can be estimated from steadystate H +flow measurements. Molecular slips are intrinsic to the ATPase, while membrane leaks are intrinsic to the membrane proper; thus, slips should be distinguished from leaks. Slips and leaks occur in parallel. Molecular slips in the ATPase may be dependent on the thermodynamic force producing the backpressure, in the absence of the other force. The rate of leakage of the coupling ions depends on the magnitude of the particular thermodynamic force operating within the system in the absence of the other force. In an idealized Ca 2+-ATPase liposome without slips, a representative equation for the dissipation of free energy is ~It -- JpAp - J c a A/d, Ca - JHA/.~,H

(11.127)

where Ap is the affinity of the reaction of the ATP-hydrolyzing activity of the enzyme (-AGp). The Ap, A/.LCa , and A/.~ H are the thermodynamic forces in J/mol. The flow Jp (mol/s) is the rate of reaction of ATP hydrolysis, and Jca and Ju are the respective ATPase mediated transmembrane flow rates for calcium and hydrogen ions. ATP hydrolysis is a highly exergonic (downhill) reaction -AGp. The uphill transport processes of inward Ca2+ (negative flow) and outward H + (positive flow) possess forces with negative signs. Assuming that the system is in the vicinity of equilibrium, the linear phenomenological equations with the conjugate flows and forces based on Eq. (11.127) are Jp = LpAp - LpcaA/~ca - LpHA~H

(11.128)

J c a = Lcap Ap - Lca A/2ca - Lca H A/.LH

(11.129)

J n = LHpAp - L n c a A ~ C a -- LHA/~H

(11.130)

Lii and L~j are the straight and cross-coefficients, respectively. By Onsager's reciprocal rules, we have L O.=

Zji.

The

electrochemical potential differences between internal i and external e regions are defined by A/~ca --/~Cai -/'~Cae = RT in Cai + z F 2 ~

(11.131)

Cae A/'~H -- ~Hi -- ~He = -2.3RTApH + FA~

(11.132)

where subscripts i and e denote interior and exterior, respectively. The Gibbs free energy difference is [ATP] ) AGp = AGp - RT In [ADP][Pi ]([H + ]/10 -p" )

(11.133)

where AG~, is the standard Gibbs free energy for the hydrolysis of ATP at pH 7 ( - 30.5 kJ/mol). Here, R, T, and z are the universal gas constant, absolute temperature, and number of Coulombic charge, respectively. The gradients H ÷and Ca 2÷and the transmembrane voltage may counteract the chemical reaction system of ATP hydrolysis; for example, these forces induce backpressure effects. There are two important conditions with respect to calcium pumping; these are the static head and level flow. At the static head, calcium pumping vanishes (Jca = 0), and at the level flow calcium gradient vanishes (A/2ca = 0). At nearequilibrium and static-head conditions, phenomenological stoichiometry Z and the degree of coupling q are

577

11.6 Regulation in bioenergetics

Zsh = Z H p -

z

qsh -- qup

IL~-~p

(11.134)

LHp N/LHLp

(11.135)

The efficiency of energy conversion becomes JHA/~H = jx ~sh -- TIHP = -- JpAp

(11.136)

In terms of Z and q, the efficiency is T/sh --

T/HP =

Zx( Zx + q ) qZx + 1

(11.137)

where j and x are the flow and force ratio, respectively. For Ca 2+-ATPase, maintaining the thermodynamic force is a priority, and the pumps operate close to static head conditions at close to zero efficiency. During a turnover cycle of the pump, nca calcium ions and nn protons are transported across the membrane per mole of ATP molecule. Therefore, nca and nH represent Ca2+/ATP and H+/ATP coupling stoichiometries, respectively, while nH/nca represents the H+/Ca 2+ stoichiometry. Experimental observations show that both A/XCa and A/2 n can inhibit Jp by exerting a backpressure effect on the rate of AYP hydrolysis for the plasma membranes. Therefore, the actual total thermodynamic force Xp may be Xp - Ap - rtCaA/.ZCa -- r/H A/.~H

Also, Onsager's reciprocal relations suggest that Lpc a - Lca p - ncaLp and LpH = LHp = rtHL p. If AG* denotes the measure of the offset from equilibrium of the Jp - AGp space, an effective driving force may be AGp,ef= AG - AG* ( - -Ap,ef) for the uphill transport of H + and Ca 2+ . As a result, the value of coefficient Lp changes in different proportions with respect to A/Xca and A/2 H, and we need to introduce asymmetry coefficients YCa and VH defined by ~(AGp ).xgc ~,

a(AGp)A/x H

~pCa = 8(~/.~Ca)AGp '

~pH -- 8(A~H)AG p ,

~pCa -- ")/Cap

With the consideration of new forces and the asymmetry coefficients, Eqs. (11.128)-(11.130) reduce to

Jp = Lp (,4 p,~r- nc,a'YpcaA/~ca -

(11.138)

n H ~ p H A]..~H )

Jca - r/Ca Lp ( .4 p,ef-- HCa")/CapA/~Ca -- r/H"YpHA/-~H )

(11.139)

JH -- r/H Lp (.4 p,ef- r/Ca"YpCaA~Ca - r/H'YHpA/'~H )

(11.140)

By assuming, for the sake of simplicity, 7Ca - TH -- 1, we may incorporate ATPase slips and the membrane leaks o f H + and Ca 2+ into the definitions ofphenomenological coefficients Lp,s, Lca,1, and LH,b where the subscripts s and 1 refer to slip and leak, respectively, and we have the following flow-force equations in the matrix form Jp ! Jca,t JH,t

-

Lp + Lp, s

--nca Lp

nHLp

nCa Lp

_[(nca )2 Lp + Lca,1 ]

-- nCa nH Lp

A/~Ca

n H Lp

- n Hnca Lp

_[(nil )2 Lp + LH,1]

A~H

(11.141)

where Lp,s is the slip rate coefficient that does not take into account the kinetic properties of molecular slips, while Lp is the strictly mechanistically coupled ATPase rate coefficient. Jca,1 and Jn,1 are the leak flows of Ca 2+ and H+, respectively,

578

11.

Thermodynamicsand biological systems

while the pump-induced flow plus leaks yields the total value of the flows Ca 2+and H +

Jca,t = Jca 4- Jca,1 JH,t -- JH + JH,1 Equation (11.141) is only a simplified representative model of the ATPase without the leaks of bulk ions Na + and K +. However, it can separate ATPase slips from membrane leaks. In a simple application of Eq. (11.141), Waldeck et al. (1998a, 1998b) estimated nH with and without the ionophore A23187 (see Figure 11.10) from steady-state NMR data using the relation nu

_

Ju

_

Jn,e--JH,s +Ju,1

Jp

(11.142)

Jp

An expression of nH measured in the presence of the ionophore A23187 is

nH

_JH_ __ Z _ _ {JH,e--JH,s Jp

/-

nil,Ca

(11.143)

with

/ pHe/

JH,e = - - / ~ e -

6t

~

ApH e

At

/

JH,s -- g/H,sJp where the subscript e denotes external, fie is the buffering capacity of the extraliposomal medium, and nil,Ca is the number of H + counterported inwards per Ca 2+ extruded by A23187 as a result of Ca 2+ pumping. The leak flow of hydrogen can be estimated from an independent measurement of LH,1and ZX/2H,ss,where subscript ss denotes steady state

JH,1 -- LH,lZ~/-~'H,ss-- LH,1(FA~ss - RTApHss)

(11.144)

From Eqs. (11.142) and (11.143), the estimated value is nH= 1.9 _+0.3 (Waldeck et al., 1998a). Using Eq. (11.141), the extent of slippage Lp,s/L p can be estimated in the plasma membrane CaZ+-ATPase. This is done close to the static head without the ionophore (-), and close to the level flow with the ionophore (+). Using the equation for Jp, the control ratio of the ATPase is obtained from

Jp,+ Lp )

(11.145)

Estimated approximate values of Ap,_ and Ap,+ are 61 and 52kJ/mol, respectively (Waldeck et al., 1998b). Thus, Lp,s/L p <- 0.4, which means for every five coupled turnovers leading to Ca 2+ and H + translocation, there may be as

many as two uncoupled ones or slips. In the case of slips, the energy associated with the hydrolysis of ATP would be dissipated as heat. 11.6.4

Potassium Channels

A class of cardiac potassium channels operates in smooth and skeletal muscle, brain, and pancreatic cells. Potassium channels are activated when intracellular ATP levels decrease, and are an important link between the cellular excitability and the metabolic status of the cell. The ratio of ATP/ADE pH, lactate, and divalent cations determines and modulates the channel activity. The opening of the potassium channels leads to membrane hyperpolarization and a potential decrease as the potassium ions flow out of the cell. Since phosphorylation changes the activity of potassium channels, it modulates cellular excitability.

579

11.6 Regulation in bioenergetics

Potassium channels play an important role in the control of insulin secretion in [3-pancreatic cells. In a resting [3-pancreatic cell, the membrane potential is maintained below the threshold for insulin secretion by an effiux of potassium ions through the open potassium channels. As glucose levels rise within the cell, ATP production increases, and the change in the APT/ADP ratio leads to the closing of potassium channels. This causes the calcium channels to open, and entering calcium ions signal insulin secretion. Factors that modulate the [3-pancreatic potassium channels can fine-tune the insulin secretion.

11.6.5

Aging and Biochemical Cycle Deficiencies

Aging is mainly characterized by a general decline in mitochondrial function and damage to the oxidative chain. Various models and theories on aging are based on the study of aging in a population, an organism, or a single cell, and a large number of parameters changing with aging. The study of aging may produce the survival curve of a genetically uniform population in a controlled environment. Many studies established a correlation between aging and the accumulation of reactive oxygen species-modified molecules such as lipofuscins in various organisms from fungi to humans. Adducts caused by lipid peroximation also accumulate, and may damage DNA and proteins, causing loss o f - S H groups and protein carbonylation. Manipulating the expressions of genes encoding reactive oxygen species scavenging enzymes may help determine the importance of mitochondrial oxidative stress in aging. It is established that mitochondrial genomes accumulate alterations, deletions, rearrangements, or point mutations with age in humans, monkeys, and rats. Reactive oxygen species are produced by the respiration cycle and metabolic activity. Quantifying reactive oxygen species is difficult because of their short life span, low concentrations, and the existence of cellular scavenging systems (Dufour and Larsson, 2004). Mitochondrial uncoupling keeps the proton potential at a lower level and reduces the production of reactive oxygen species. Hence, the regulation of reactive oxygen species production depends on the degree of coupling of oxidative phosphorylation and the efficiency of energy conversion (Lionetti et al., 2004). Reactive oxygen species production can damage the respiratory cycle proteins and may lead to even more production of reactive oxygen species. At least 1% of the total mitochondrial respiration using pyruvate leads to reactive oxygen species and hydrogen peroxide (H202) production. The main sites for producing reactive oxygen species are in the electron transfer chain.

Example 11.11 Approximate analysis of transport processes in a biological cell A typical biological cell and its surroundings are characterized by the following concentrations (Garby and Larsen, 1995) Components

Outer (o) (mmol/L) Inner (i) (mmol/L)

K+

Na+

C1-

HCO3-

P-

3.7 139

145 12

118 -v4

24 -0.8

-1 136

In this table, P- represents anions of protein and organic phosphate. The membrane is permeable to the group represented by P-. The mean values ofthe charge on P- are - 6 . 7 and - 1.08 for the interior and the exterior of the cell, respectively. An electrical potential difference of Atp = Oi - 4'0 = 90 mV is measured, i and o denote the intracellular and extracellular, respectively. The activity coefficients of components inside and outside the cell are assumed to be the same, and pressure and temperature are 1 atm and 310 K. Assume that the diffusion flows in from the surroundings are positive and the diffusion flows out are negative. Using tracers, the unidirectional flows are determined as follows: Components

10l° (J+) (mol/(m2 s)) 10l° (J-) (mol/(m2 s))

K~

Na+

C1-

HCO3-

P-

722

250 -

14.2 -

3.6 -

0 0

Using these approximate flows, we may estimate permeabilities using Jk = --EkPk

% - Cko exp(--Ek)

1 - exp(-Ek)

where E k = z k F A d / R T and the permeability coefficient is

(a)

580

11.

Thermodynamicsand biological systems

DkKm Pk-

where D is the diffusion coefficient and Km is the distribution coefficient defined by Km = c k (x = 0) = c k (x = L) Cko

Cki

The permeability coefficient depends on the characteristics of the membrane and solute, and can vary considerably for various solutes. For example, p = 10 -21 naJs for sucrose and 10 -4 m/s for water in the human red blood cell membrane. Equation (1) may be generalized by including the effect of pressure gradient z ~ m - - P ( 0 ) - P(L), and we have Ek =

zkFAO + gkz~m ~ RT

RT

To use this relation, we need to relate pressure difference in the membrane, z~dgm = P ( 0 ) - P(L), and over the membrane, zkP = P i - Po- This requires the introduction of the concept of osmotic pressure in a nonequilibrium membrane system. From Eq. (a) we obtain the value of E and the permeability for sodium E(Na +) = (+ 1)(96500)(-0.090) = - 3 . 3 7 8.314(310)

p(Na +) =

J + ( N a +){1- e x p [ - E ( N a +)]}

2.5 × 10-811- exp(3.37)]

E(Na + )c o (Na + ) exp [ - E ( N a + )]

( - 3.37)(145)(exp(3.37))

= 4.94 x l 0 -11 m/s

E(Na +) = E(K +) = -3.37. As co(K +) = 0 for tracer, we have P(K+) = 7.22×10-8tl_exp~3.37~j r ~ al = 4.33×10 -9 m/s (-3.37)(145) The net diffusion flows of sodium ions in and potassium ions out can be calculated by J ( N a +) = 3.37(4,94 x 1 0 -11)

1 2 - (145) exp(3.37)

= 2.49 X 10 -8 mol/(m 2 s)

1 - exp(3.37) J ( K +) = 3.37(4.33 ×10 -9)

1 3 9 - (3.7) exp(3.37)

= - 1.63 × 10 -8 mol/(m 2 s)

1 - exp(3.37) The flow of potassium depends on the outside concentration of potassium ions. Under steady-state conditions, the chemical pump creates an active flow ofNa + out of the cell and simultaneously an active flow of K + into the cell: J ( N a + ) = - 2 . 4 9 × 10 -8 mol/(m 2 s) J ( K +) = 1.63 × 10 -8 mol/(m 2 s) The pumping ratio Na+/K + = 2.49/1.63 = 1.53 --~ 3:2. After performing similar estimations, the following values for permeability coefficients and diffusion flows of all components are obtained Components

101°(Pk)(m/s) 101°(Jk) (mol/(m2 s))

K+

Na+

C1-

43.3 163

0.494 249

1.0 0.2

HCO 3-

1.25 0.1

P0

0

11.7

581

Exergy use in bioenergetics

As the estimations above display, the net flows of chloride and bicarbonate ions are negligible, and the transport of ions is passive. Using the van't Hoff equation, the osmotic pressure difference across the membrane is estimated by

( I n a system with one single nonpermeating component dissolved in water, II = P i - P o = RT(cki-Cko).)

k = 2,

and this equation reduces to

P~ - Po = 0.0827(310)[(139 + 12 + 4 +0.8 + 136)-(3.7 + 145 + 118 + 24 + 1)] × 10.3 = 0.0025 atm This result corresponds to an osmotic activity difference of 0.1 mmol/L. The total electrical current density (C/m 2 s) through the membrane is I= ZJ~zk = J(Na+)+Jp(Na+)+J(K+)+Jp(K+)=O

Here, Jk consists of passive diffusion flows and active flows Jp due to chemical pumps, which compensate for the nonequilibrium concentrations of cations and the total charge current becomes zero. The transference number of an ionic component tk

--

Jk Zk and ~_,~-"t k = 1 I

is usually used to describe the passive diffusion of charge. This representative example illustrates transport processes in biological cells using a highly simplified analysis. Biological cells also operate hydrogen and calcium pumps. Some of the concentrations also represent only approximate values.

11.7

EXERGY USE IN BIOENERGETICS

Biological systems extract useful energy from the outside, convert it, store it, and use it for muscular contraction, substrate transport, protein synthesis, and other energy-demanding processes. This useful energy is called exergy, which is lost in every irreversible process because of entropy production. The ATP produced through oxidative phosphorylation is the form of exergy that originates due to oxidation of reduced equivalents of nutrients. A living cell uses ATP for all energy-demanding activities and maintains nonvanishing thermodynamic forces, such as electrochemical potential gradients. However, mitochondria cannot maximize ATP production, and exergetic efficiency at the same time.

11.7.1

Exergy Management

Michaelis-Menten equation shows that the enzyme reactions in certain regions can be approximated by linear kinetics. Stucki (1984) demonstrated that variation of the phosphate potential at constant oxidation potential yields linear flow-force relationships in the mitochondria. Through linear flow-force relationships, cells may optimize their free energy production and utilization by lowering their entropy production and hence exergy losses at stationary states. The second law of thermodynamics states that entropy production or exergy loss as a consequence of irreversibility is always positive. A representative overall dissipation function for oxidative phosphorylation is dS -- it"--i7: __ JoXo +JpYp - i n p u t work+output work -> 0 dt

(11.146)

where the input force Xo is the redox potential of oxidizable substrates, and Xp is the output force representing the affinity A, or the phosphate potential expressed by Eq. (11.86)

- X p = AGp + R T l n

[ATP] ) [ADP][Pi ]

582

11.

Thermodynamics and biological systems

where AGp is the Gibbs free energy at standard conditions. The associated input flow Jo is the net oxygen consumption, and the outflow Jp is the net rate of ATP production. For the sum to be positive in Eq. (11.146), we can have either JoXo > 0 and JpXp > 0, or JoXo >> 0 and JpXp < 0, which requires coupling. The sequence of coupling is controlled at switch points where the mobility, state, and catalysis of the coupling protein can be altered in specific ways, such as sifted equilibria or regulated rates of conversion between one protein state and another. Prigogine showed that the total exergy destruction reaches a minimum in a stationary state, which is a stability criterion. Optimal performance regimes of biological systems are associated with minimum entropy production.

11.7.2 Exergy Efficiency The exergetic efficiency rt is defined as the ratio of dissipations due to output and input powers in oxidative phosphorylation, and from Eq. (11.146) we have output work input work

Jp Xp JoXo

(11.147)

This equation suggests that the efficiency is a function of the state of the system, as both the forces and flows are state dependent. For a coupled system (q < 1), the efficiency is zero at the static head (Jo = 0) and at the level flow (Xo = 0). Therefore, as the process progresses from the level flow to the static head, the phosphorylation, as a linear energy converter, passes through a state of maximal efficiency ~max defined by

~max =

q l+x/1-q 2

(11 "148)

where q is called the degree of coupling which is expressed in terms of the phenomenological cross-coefficients L O. q=

Lop

(LpLo) 1/2

(11.149)

Equation (11.148) shows that 'qmax depends only on the degree of coupling. Here, q is a lump sum quantity for the degrees of coupling of various processes of oxidative phosphorylation driven by respiration. Absolute values of q vary from zero for completely uncoupled systems to unity for completely coupled systems: 0 < ql < 1. It is customary to use the following simplified, representative linear phenomenological equations for the overall oxidative phosphorylation Jp -- Lp Xp + Lpo Y o

( 11.150)

nt- LoX o

( 11.151)

Jo = LopX p

The matrix of the phenomenological coefficients must be positive definite; for example, for a two-flow system, we have L o > 0, Lp > 0, and LoLp-LopLpo > 0. L o shows the influence of substrate availability on oxygen consumption (flow), and

Lp is the feedback of the phosphate potential on ATP production (flow). The cross-coupling coefficient Lop shows the phosphate influence on oxygen flow, while Lpo shows the substrate dependency of ATP production. Experiments show that Onsagers's reciprocal relations hold for oxidative phosphorylation, and we have Lop = Lpo. By dividing Eq. (11.150) by Eq. (11.151), and further dividing the numerator and denominator by Xo(LoLp)1/2, we obtain

"q-ix=-

x+q q+ l/x

(11.152)

where j = (Jp/JoZ), x - (XpZ/Xo). Here, Z is called the phenomenological stoichiometry defined by (11.153) Equation (11.152) shows the exergetic efficiency r/in terms of the force ratio x and the degree of coupling q. The ratio Jp/Jo is the conventional phosphate to oxygen consumption ratio: P/O. Figure 11.5 shows the change of

11.7

1

,

583

Exergy use in bioenergetics

,

,

,

,

,

'

0.9 0.8 0.7

qopt= 0.6195

~~...

~- 0.6

~_q~C

0

" 0.5

/

.m 0

0.4

0.3 0.2 0.1

/o,:/ -0.9

Figure 11.5.

-0.8

. . . . . . -0.7

-0.6 -0.5 -0.4 force ratio, x

-0.3

-0.2

-0.1

0

The change of efficiencies 7, given in Eq. (11.153), in terms of flow ratio x and for the degrees of coupling qf, qp, q~C, and q~C.

efficiencies r/in terms of flow ratio x between-I and 0, and for the particular degrees of coupling qf, q p , q~C, and q~C.As Eq. (11.148) shows, the optimum efficiency values are dependent only on the degrees of coupling, and increase with increasing values of q.

11.7.3

Exergy Losses

For the oxidative phosphorylation described by Eqs. (l 1.151) and (l 1.152), the exergy loss can be obtained from Eq. (11.146) in terms of the force ratio x and the degree of coupling q, and is given by - (x 2 + 2qx + 1)LoXo2

(11 154)

Minimum exergy loss or minimum entropy production at stationary state provides a general stability criterion. There are two important steady states identified in the cell: static head (sh) and level flow (lf). At the static head, where ATP production is zero since Jp =--0, the coupling between the respiratory chain and oxidative phosphorylation maintains a phosphate potential Xp, which can be obtained from Eq. (11.151) as (Xp)sh = --qXo/Z, and the static head force ratio Xsh becomes X~h= --q. The oxygen flow Jo at the static head is obtained from Eqs. (11.151) and (11.152) (Jo)sh

=

LoXo(1-q 2)

(11.155)

where Lo may be interpreted as the phenomenological conductance coefficient of the respiratory chain. If an uncoupling agent, such as dinitrophenol, is used, the ATP production vanishes and hence Xp = 0; then, Eq. (11.151) becomes (Jo)unc -- LoXo

(11.156)

(do)sh -- (do)unc ( 1 - q2)

(11.157)

Combining Eqs. (11.155) and (11.156), we obtain

Using the experimentally attainable static head condition (state 4 in mitochondria) and the uncoupled oxygen flow (Jo)unc, we can determine the degree of coupling q

1/2 (Jo)sh q--

1-- (jo)un

(11.158)

c

At constant Xo, Eq. (11.154) yields the minimum value of exergy loss at x (q~)~h - (q~)mi. - (1- q2 )LoX2o

-q

(11.159)

11.

584

Thermodynamicsand biological systems

1

0.8

0.6 470.4

0.2 ..._...

.

-().9

-0.

8

-().7

-().6

-0'.5 -0'.4 -0'.3 -().2

- .O' 1

0

force ratio,x Figure

11.6. The change of ratio 4'1, given in Eq. (11.160), in terms of flow ratio x and for the degrees of coupling qf, qp, q~C, and q~C.

The ratio of dissipations expressed in Eqs. (11.159) and (11.154) depends only on the force ratio and the degree of coupling, and becomes an exergy distribution ratio over the values ofx ¢hl -

1- q2

a'Itsh __

(11.160)

x 2 + 2qx + 1

Figure 11.6 shows that the values of ¢hl reach unity at various values of x, and the economical degrees of coupling q~C and q~C yield lower values of 4'1 than those obtained for qf and qp. The exergy loss at the static head is relatively lower at the degrees of coupling corresponding to economical ATP production and power output. At level flow, the phosphate potential vanishes, and no net work is performed by the mitochondria, and the flow ratio becomes Jlf -

Lpo Lo

- qZ

(11.161)

Combining Eqs. (11.160) and (11.161) yields an expression for estimating the phenomenological stoichiometry Z from measured Jp/Jo = P/O ratios at level flow Z=

P/O

(11.162)

41-(Jo)sh/(Jo)unc The efficiency expressed in Eq. (11.153) is zero at both the static head and level flow, due to vanishing power at these states. Between the static head and level flow, efficiency passes through an optimum, which is given in Eq. (11.149). The force ratio at optimal efficiency is expressed by _

q

Xopt----Cpt--l+41--q

2

(11.163)

The rate of optimal efficiency of oxidative phosphorylation is not characterized by the exergy decrease, and the exergy loss at optimal efficiency is given by

Xi/,opt __ x 2 )2 _ (1l+x-_------------~LoX2

(11.164)

17.7

585

Exergy use in bioenergetics

1.8

/,///~

1.6 1.4 1.2 O4

e-

1 0.8 0.6 0.4 0.2 0

t._

-0.9

-0.8

i

-0.7

I

I

-0.6 -0.5 -0.4 force ratio, x

I

I

I

-0.3

-0.2

-0.1

Figure 11.7. The change of ratio ~h2,given in Eq. (11.165), in terms of flow ratio x and for the degrees of coupling qf, %, q~C,and q~C.

The ratio of dissipations ~opt and ~ , given in Eqs. ( 1 1.164) and (11.154), respectively, shows the effect of optimal operation on the exergy loss in terms of the force ratio x and the degree of coupling q ~)2 --

opt _

qf

(I-- v2) 2 " (1 + x 2 )(x 2 + 2qx + 1)

(11.165)

Figure 11.7 shows that the values of 052 reach peak values higher than unity, and the exergy loss is not minimized at optimal efficiency of oxidative phosphorylation. The exergy loss is the lowest at the degree of coupling corresponding to the economical power output. For the optimal efficiency to occur at steady state, oxidative phosphorylation progresses with a load. Such a load JL is an ATP-utilizing process in the cell, such as the transport of substrates. A load, which will make the steady state the optimal efficiency state, can be identified through the total exergy loss q~c q/c -- J p X p -Jr-J o X o

+

JLX

(ii.166)

Here, Jk is the net rate of ATP consumed and X is the driving force. If we assume that the ATP-utilizing process is driven by the phosphate potential Xp, and Jc is linearly related to Xp, then we have JL - LXp. Here, L is a phenomenological conductance coefficient. The dissipation caused by the load is JkXp = LX2p, and the total exergy loss becomes

(11.167)

Using Eq. (11.163), and from the extremum of Eq. (11.167), Stucki (1980, 1984) found the condition L _ x/l_q2

(11.168)

Lp

which is called the conductance matching of oxidative phosphorylation. After combining Eqs. (11.23) and (11.168) when the conductance matching is satisfied with a percentage/3, we obtain qZc _ ix211 + fiX/1_ q2 )+ 2qx + 1)LoX,~

(11.169)

The exergy lost at the static head with conductance matching, (Xlrc)sh , and at the state of optimal efficiency, (Xlrc)opt, are expressed by Stucki (1984) as follows

11. Thermodynamicsand biological systems

586

(attc)sh = 4 I - - x 2 (X2 + 4 I - - X 2 )LoX2o

(11.170)

x2 i + x 2 LoX2o

1 -

(atrc)op t

u

I t is now possible to estimate the ratios of exergies 4~3 and q by using Eqs. (11.169)-(11.171)

4'3-

(11.171)

~4 in terms of the force ratio x and the degree of coupling

,,/1- x2 (x2 + x/1- x2 )

(XIYc)sh _ _

(11.172)

x2 (l + ~x/l- q2 )+ 2qx + l

"~II"c

(XItc)opt __

I--X 2

/)4 ~ - ~

4

i

i

I

I

I

I

I

i

2.5 g

2

0.5 0

g

I

-1

I

-0.9-0.8-0.7-0.6-0'.5-o'.4-d.a-d.e

-.1 o'

0

force ratio, x Figure 11.8. The change of ratio (b3, given in Eq. (11.172), in terms of flow ratio x and for the degree of couplings qf, qp, q~C, and q~C with load, and/3 - 1 and/3 = 0.9 conductance matching.

i 1.7

Exergy use in bioenergetics

587

Equations (11.172) and (11.173 ) are analogous to Eqs. (11.160) and (11.165) and show the ratios of exergy with and without the load. Figure 11.8 shows the change of ~3 when a load is attached with the values/3 = 1 and/3 = 0.9 at the static head. With the load, the exergy destruction increases considerably. This increase is larger with decreasing values of/3. Figure 11.9 shows the values of &4 with/3 = 1 and/3 = 0.9. At optimal efficiency, exergy destruction is the lowest at qf, corresponding to the maximum production of ATR while the exergy destruction is relatively higher at the economical power output with a minimal effect of/3. The comparison of energy conversions in linear and nonlinear regions requires a combination of thermodynamic and kinetic considerations to express the exergetic efficiencies of nonlinear TinI and linear r h modes

/ AoX

tin'--

--v-' and TII----

-7-

1

(11.174)

"p o

Denoting y as a measure of the relative binding affinity of a substrate H + on the either side of the membrane, the following inequalities are obtained for y >- 1

i

i

i

!

i

I

i

i

i

i

9=1 _

2.5

e- 1.5

qf 0.5

-1

I

I

i

I

I

I

I

I

I

-0.9

-0.8

-0.7

-0.6

-0.5

-0.4

-0.3

-0.2

-0.1

0

force ratio, x 3 p=0.9 2.5 P

2-

1.5

0.5

0

-1

-0.9

1

1

-0.8

-0.7

I

I

I

-0.6 -0.5 -0.4 force ratio, x

I

I

I

-0.3

-0.2

-0.1

Figure 11.9. The change of ratio ~h4,given in Eq. (11.173), in terms of flow ratio x and for the degrees of coupling qf, q> qp, and q~C with load and/3 = 1 and/3 = 0.9 conductance matching.

588

....

11.

Thermodynamicsand biological systems

~n] > ~7~ for ~'-

T/n1 < T]I

for ~"-

1

(11.175)

q 1

(11.176)

q

where ~"is the reduced stoichiometry Z/n, which is subjected to kinetic limitation (l/q) > ( >_q, and n is the mechanistic stoichiometry. Inequalities (11.175) and (11.176) suggest that at a specified value of (, the efficiencies of nonlinear and linear modes become equal to each other, and there exist values for ~"where the energy converter operates more efficiently. The ratio of efficiencies (called as gain ratio) at linear and nonlinear modes is 'r/1

r/r -

(11.177)

Tlnl

This shows a measure for the efficiency gain in linear mode operation. The efficiency in linear modes depends on only q (Eq. (11.149)), while the efficiency in nonlinear modes depends on input force Xo besides q. In nonlinear regions, the efficiency decreases at high values of input force, and the force ratio at optimum operation Xopt.nl is shifted towards the level flow where x = 0. In oxidative phosphorylation, the input force is the redox potential of the oxidizable substrates and the output force is the phosphate potential. If these two forces are balanced, the system operates close to reversible equilibrium. Experiments show that in mitochondria, q < 1, and the input force is well above 50RT. For a fully coupled system in the nonlinear region of a single force, the phosphate potential Xp would be very small. However, a dissipative structure can only be maintained with a considerable Xp. On the other hand, in the linear mode of operation, optimum force ratio Xop t does not depend on the input force (Eq. (11.163)). Gain ratio T/r can be calculated at a reference force ratio, such as Xopt, which is a natural steady-state force ratio of oxidative phosphorylation. This is seen as a result of the adaptation of oxidative phosphorylation to various metabolic conditions and also as a result of the thermodynamic buffeting of reactions catalyzed by enzymes. The experimentally observed linearity of several energy converters operating far from equilibrium may be due to enzymatic feedback regulations with an evolutionary drive towards higher efficiency. A living organism requires q] < 1. The particular value of the degree of coupling depends on the nature of the output required from the energy converter. For example, fatty acids decrease the degree of coupling and act as uncoupler. Uncoupling is not restricted to thermoregulation; some uncoupling activity is favorable for the performance of the metabolic and even the energy-conserving functions of cellular respiration. Mitochondria can regulate their degree of coupling of oxidative phosphorylation depending on the energy demand of the cell. For example, for fed rats, oxidative phosphorylation operates very close to the conductance matching, i.e., at the state of optimal efficiency with an economical degree of coupling. The load in a living cell fluctuates and compromises the optimal efficiency of oxidative phosphorylation. Some enzymes operate as sensitive thermodynamic buffering to decrease deviations from optimal efficiency. ATP-utilizing reactions act as a load as well as thermodynamic buffers. This regulatory mechanism allows oxidative phosphorylation to operate with an optimal use of the exergy contained in the nutrients. Every reversible ATP-utilizing reaction can, in principle, act as a thermodynamic buffer. For example, adenylate kinase can buffer the phosphate potential Xp to the value permitting optimal efficiency of oxidative phosphorylation in the presence of too high loads. The adenylate kinase reaction is reversible, and acts as a buffer. If this buffer is treated as another load with a conductance La, the overall load conductance L* becomes E =LaO+L

(11.178)

where

0=

c~+ RT ln(1 + e x p - (6 + Xp/RT))

Xp

-1

11.7

Exergy use in bioenergetics

m

589

[AMP] [ATP] + [ADP] + [AMP]

z

6 = kGp - RT ln[Pi ]

Stucki (1984) expressed the dissipation function with buffering from Eq. (11.167) in terms of L*

('qY'c)b-- X2 1%"

+2qx+l

Lp

Lo X2

(11.179)

Dividing Eq. (11.179) with Eq. (11.167) shows the effect of thermodynamic buffering on the exergy loss

!

!

'

'--1--

'

'

,

i

i

!

i

|

1 L=0.9 0.95 0.9 0.85 0.8 -C- 0.75 0.7

qo

0.65 0.6

<__q~c 0.55 0 . 5

i

-1

,

-0.9

i

1

i

1. . . .

-0.8

i

-0.7

i

i,

,, i

i

,,,

-0.6 -0.5 -0.4 force ratio,x i

r

1

i

i

-0.3

-0.2

-0.1

'1

|

......

i

0

i

L=0.5

0.9 0.8 0.7 0.6 ~-

0.5 0.4 0.3

0.2 0.1 0

I ........

-1

i,,

-0.9

I

,I

I

I

I

-0.8

-0.7

-0.6

-0.5

-0.4

,,

I

-0.3

,,

I

I

-0.2

-0.1

0

force ratio,x Figure 11.10. The change of ratio 65, given in Eq. (11.180), in terms of the flow ratio x and for the degrees of coupling qf, qp, q~C,and q~C. The plots were normalized by Xp = (x/Z)Xo with Xo = 209,200 J/moi, Z = 3, 2~G°k = 630/mol, 2~G~ = 35560J/mol, M = 0.005, Pi = 8 mM, T - 310 K with load conductance of L = 0.9 and L = 0.5D.

590

11.

Thermodynamicsand biological systems

4,5 -

(XI/'c)b

(11.18o)

q~c Figure 11.10 shows the change of 4'5 in terms ofx between 0 and-1 for the degrees of coupling qf, qp, q~Cand q;C with the load conductance L - 0.5 and L = 0.9. The values of 4~5are lowest at the economical power output and highest at the maximum output of ATE Therefore, the exergy loss is relatively lower in the economical power output than in maximum ATP flow. Within the framework of the theory of dissipative structures, thermodynamic buffering represents a new bioenergetics regulatory principle for the maintenance of a nonequilibrium conditions. Due to the ATP production in oxidative phosphorylation, the phosphate potential is shifted far from equilibrium. Since hydrolysis of ATP drives many processes in the cell, the shift in Xp to far from equilibrium results in a shift of all the other potentials into the far from equilibrium regime.

Example 11.12 Exergy efficiency Assume that the oxidation of glucose (G) in living cells produces 38 mol of ATP per mole of glucose (Garby and Larsen, 1995). Estimate the maximum theoretical production and the exergy efficiency. C6H1206 +602(g)~6CO2 +6H20 The steady composition of the mixture contains 0.01 mol/L glucose and the partial pressures of carbon dioxide and oxygen are 0.07 and 0.21 atm, respectively. The state of the mixture is characterized by the activity ratios (ai/ai°) = 0.07, 1.0, 0.01, and 0.21 for carbon dioxide, water, glucose, and oxygen, respectively. The entropy of reaction for oxidation of glucose at standard conditions is obtained from Table B8 AS~°G = 6(121) + 6(70) -- 151-- 6(115) = 305 J/(mol K) The entropy of reaction at actual state should be estimated. Assuming that the entropy of mixing is negligible, we have (a)

For a reaction at 310 K, entropy may be obtained from

Si(T,Po)--Si(To,Po)+Cp

In (~-0-o)

Here, Cp is assumed to be a constant. Substituting Eqs. (a) and (b) into the relation of reaction entropy ASr = ~ we have

(b)

viSi,

(c)

The heat capacities from Table B8 show that Cp = 0.037, 0.075, 0.219, and 0.029kJ/(mol K) for carbon dioxide, water, glucose, and oxygen, respectively. At 310 K, this equation with the data from the Appendix and above becomes

: 30' + (0

× 1.06 ] 1000'1hi310)' 3141n[0.076 0.01×0.216

= 305 + 11 - 16.5 = 299.5 J/(mol K) On the other hand, the heat of reaction is obtained from

z~J--/r,G -- AHr°G at-(Z piCp i ) ( T - To) -- -2870 + 0.279(310- 298) = -2867 kJ/mol

11.7

591

Exergy use in bioenergetics

The Gibbs energy of the reaction at 310 K is obtained from - 2 8 6 7 ( 1 0 0 0 ) - 310(299.5) = - 2 9 6 0

AGr, G = A H r , G - T ~ S r , G -

kJ/mol

For the reaction ATP ~ ADP + Pi at the standard state and each component at a concentration of 1 mol/L, at 310 K Table B9 provides the heat of reaction and the Gibbs energy of reaction AHr°ATP = --

20 kJ/mol ATP, and A Gr°ATP -- -- 31 kJ/mol ATP

With the values above, we find (-20)-(-31) ASr°,ATP =

= 0.0355 kJ/(mol ATP K)

310

With concentrations ofATR ADR and Pi of 4, l, and 10 mmol/L, respectively, in the aqueous solution, we find

AGr,ATp=AGr°ATp+RTEPiln( °a--L) a

AGrATP = - - 3 1 + 8.314(310) × 10-3 lnI10-3 ' 4 × (0"01) 10 .3 ] = --46 kJ/mol ATP The maximum yield is AGr,c~

-2960

AGr,ATP

--46

- 64.3 mol ATP mol glucose

Therefore, the process has an approximate efficiency of 38/64.3 --~ 0.6 to transfer useful energy. The total dissipation, ~ = T~, is the summation of the product reaction rate and affinity (the Gibbs energy of reaction), and then we have q.r = _ ( j r A Gr ) a -- ( J r A Gr ) ATP -- --( 1 )(--

2960) - (38)(46) = 1212 kJ/mol glucose

From the energy balance, we have 0 = (JrZ2tHr)a + (JrAHr)ATP = --2867 + 38(20) = --2107 kJ/mol glucose Assuming that

qG/£/ATP

--

1, we have qG __ qATP __

de

do

We can find the chemical work of glucose

1 2 ( - 2107) = - 1053.5 kJ/mol glucose oxidation

f~/ch from

7ch _ 6)G

- - -- AHr, G - - 1 0 5 3 . 5 - ( - 2867) - 1814 kJ/mol glucose

JG

JG

ch alt G = - - J G A G r , G . . . .

JG

- - ( 1 ) - (2960) - 1814 - 1146 kJ/mol glucose

For the ATP synthesis, we have

XIrATP = --JATPZ~Gr ATP -+'

Wch ....

JG

38(-46) + 1814 = 66 kJ/mol glucose

592

11. Thermodynamicsand biological systems

The exergy efficiency is ~rch -- ~ A T P __ 1 ~/rch

'0 --

66 _ 0.96 1814

This result shows somewhat high exergy efficiency.

Example 11.13 Approximate exergy balances in a representative active transport Consider the representative active transport in Example 10.9. Estimate the distribution of exergy and the exergy efficiency of the active transport system shown in Figure 10.3. The dissipation equations for control volumes are -- ~rch1

XI/"1 -- - J r l ~ G r G

for cvl

XI~2 -- Jr2Z~Gr,ATP -1-mch 1

for cv2 for cv3

XI)'3 -- --Jr3Aar,ATP -- Wch 2

• 4=

i +Wch

for cv4

for cv5 The total exergy loss is related to the oxidation of glucose XIf -- --JrGAGr,G

The exergy efficiencies are defined by J'/t/chl " 0 1 - XI$1 _+. ~rch 1

'02 =

~rchl--XIt2 mchl ' '03

J/Vch2 * 3 "t-~rch 2

'04 --

~rch2 -- XI)'4 ~rch 2

For the control volume cv5, there is no yield and no efficiency. For the total control volume, the yield is measured in terms of the capacity of the system to maintain its nonequilibrium character in concentration and electrical potential across the membrane. The maintenance of nonequilibrium conditions is expressed quantitatively as the product of a generalized potential difference and generalized flow represented by ~5. Taking the values from Example 10.9, and for Jr3 -- Jr2 and Jr5 = Jr4 = 1 mmol/min (stationary state), we have xIg'l = - J r l A G r ,

a -}Vchl

=0.360×10-3(2960) -0"653=0"413w

XI)"2 -- Jr2AGr,ATP nt-~fch 1 -- 13.7×10-3(-46)+0.653 = 0 . 0 2 3 W

xIt3 = --Jr3AGr,ATP

-- ~'~ch2 -

0.629 - 0.261 = 0.368 W

for c v l

for c v 2

for cv3

• 4

....

(0.0011[ 60

8.314(310) In

r,4 1 k, 12

]

+ 1(96500)(0.090) + 0.261 = 0.009 W

for cv4

11.9

q.r 5 = J r s R T l n

593

Molecular machines

Ce + J,:sz(Na+)F(Oe - q t i ) = 0.252W

for cv5

The total dissipation is = --JrGAGr,G -- --0.360 × 10-3 (--2960) = 1.066W The total exergy efficiency becomes .

xP5 0.252 . . . q~ 1.066

0.24

The other efficiencies are • Wchl

=

0.61, "112

T]I = XI'rl -Jr- I/Vchl

11.8

];~7chl -- ~ 2 __

=

--

~l/chl

__

~'~rch2

0.44, r/3 -

.

XP3 -+- Wch2

__

- 0.41, 774

__ ~/ch2 -- ~J'4 --

.

=0.96

Wch2

MOLECULAR EVOLUTION

Living systems utilize a set of genetic instructions and develop physical characteristics. Quantitative theories for describing the information transfer generally assume that the organization and transfer of information, while constrained by the principles of chemistry and physics, may not necessarily be a consequence of these principles. Proteins are synthesized as linear polymers with the covalent attachments of successive amino acids, and many of them fold into a three-dimensional structure defined by the information contained within the characteristic sequence. The folding results largely from an entropic balance between hydrophobic interactions and configurational constraints. The information content of a protein structure is essentially equivalent to the configurational thermodynamic entropy of the protein relating the shared information between sequence and structure. From the perspective of the fluctuation-dissipation approach, Dewey (1996) proposed that the time evolution of a protein depends on the shared information entropy S between sequence and structure, which can be described with a nonequilibrium thermodynamics theory of sequence-structure evolution. The sequence complexity follows the minimal entropy production resulting from a steady nonequilibrium state 0 (_~t)= 0 OXj

(11.181)

A statistical mechanical model of thermodynamic entropy production in a sequence-structure system suggests that the shared thermodynamic entropy is the probability function that weighs any sequence average. The sequence information is defined as the length of the shortest string that encodes the sequence. The connection between sequence evolution and nonequilibrium thermodynamics is that the minimal length encoding of specific amino acids will have the same dependence on sequence as the shared thermodynamic entropy. Dewey and Donne (1998) considered the entropy production of the protein sequence-structure system based on linear nonequilibrium thermodynamics. The change of composition with time is taken as the flow, while the sequence information change with the composition is treated as the thermodynamic affinity, which can be interpreted as the chemical potential of the sequence composition. Since the change of entropy with time is a positive quadratic expression in forces, Eq. (11.181 ) shows the regions of the sequence that are conserved; the rest of the sequence is driven to a minimum entropy production, hence toward the lower complexity seen in the protein sequence, creating a stable state away from equilibrium with a specific arrow of time. At steady state, a system decreases its entropy production and loses minimal amounts of free energy (the concept of least dissipation). A restoring and regulating force acting in any fluctuation from steady state may be one of the principles in the evolution of biological systems.

11.9

MOLECULAR MACHINES

Either the hydrolysis of ATP or the draining of an ion gradient is converted into an osmotic work by a pump or a mechanical work by a motor protein. Substrate binding sites in pumps or motor proteins occur in two sections: (i) one of them binds the substrate and (ii) the other converges on the bound substrate molecule, and the protein changes into an

594

11.

Thermodynamicsand biological systems

altered conformation. Some important biological processes resemble macroscopic machines governed by the action of molecular complexes. Pumps are commonly used for the transport of ions and molecules across biological membranes, while the word "motor" is used for transducing chemical energy into mechanical work in the form of rotatory or translationary by proteins or protein complexes. Some identified motor proteins such as kinesins and dyneins move along tubulin filaments, and myosin moves along filaments. These motor protein families play a major role in muscle contraction, cell division, and the transport of substrates in and out of the cell. Molecular motors are isothermal and hence the Carnot efficiency concept is not applicable. The internal states are in local equilibrium; they operate with a generalized force for the motor/filament system. This may be the external mechanical forcefext applied to the motor and the affinity A, which measures the free-energy change per ATP molecule consumed ATP.~ ADP + Pi

(11.182)

A = -A/~ = ~ATP --/J'ADP -- ~Pi

(11.183)

The external forces may be optical tweezers, microneedles, or the viscous load of the substance that is carried. These generalized forces create motion, characterized by an average velocity v, and average rate of ATP consumption Jr. Molecular motors mostly operate far from equilibrium, and the velocity and rate of ATP consumptions are not linear functions of the forces. However, in the vicinity of the linear region, where A << kBT, linear relations hold xp = Vfext + Jr A > 0

(11.184)

v = ~lfext +L12A

(11.185)

Jr = L21fext + L22A

(11.186)

Here, L 11 is the mobility coefficient, while L22 is a generalized mobility relating ATP consumption and the chemical potential difference, and L12 and L21 are the mechano-chemical coupling coefficients. A given motor/filament system can work in different regimes, and in a regime where the work is performed by the motor, efficiency is defined by r / = - ~ Vfext

(11.187)

Jr A

For nonlinear motors operating at far from equilibrium, velocity reversal allows direction reversal without a change in the microscopic mechanism. Molecular motors are classified in two groups depending on whether they operate in groups or individually. Those that operate in groups are relevant to muscle contraction. In principle, muscle fibers can oscillate in appropriate conditions. Skeletal muscle myofibrils can oscillate spontaneously; for example, spontaneous oscillations of asynchronous muscles are common in the wings of many insects. The relationship between the Gibbs energy of ATP hydrolysis and the performance of muscle is sigmoidal, and the normal operating domain is in the quasi-linear region of the curve. This corresponds with the magnetic resonance spectroscopy results obtained from studies of the finger flexor muscle. Chemiosmotic potentials are coupled to rotation in multiprotein subunit systems of bacterial flagellar motors and FoFl-ATPases. Rotation of a subunit assembly of the ATP synthase is considered an essential feature of the ATPase enzyme mechanism and of FoF1 as a molecular motor generating a torque. The bacterial flagellar motor and FoF1 are macromolecular assemblies and utilize six to eight distinct protein components to affect chemiosmotic energy transduction. Both assemblies have integral membrane modules and extensive cytoplasmic modules. With these systems, the work is accomplished outside the membrane, whereas the chemiosmotic pumps and transports take place within the membrane. The flagellar motor rotates an external filament, and according to the intracellular signals, it modulates the direction of rotation. FoF1 system activity is modified by the intracellular phosphate potential. The energy transduction reactions are reversible for FoF1 and partly reversible for the flagellar motor. (The tightly coupled and efficient operation of these machines may be achieved by alternating access of protonable residues to the adjacent bulk phases linked to the conformational motions, which generate force.) ATP is synthesized by mitochondrial oxidative phosphorylation (Jp,op) and glycolysis. The latter contribution may be disregarded for a muscle engaging little or moderate exercise. Exerted power by a muscle P is directly related to the rate ofATP hydrolyzed by the myosin ATPases Jp, and we have P - aJp

(11.188)

11.10

595

Evolutionary criterion

where a is a constant and is independent of the muscle operation. Some of the energy released by ATP hydrolysis is not associated with the muscle operation Jp,1, and may be considered a leak. For the ATP balance for steady-state muscle operation, Eq. (11.188) yields

P = °t(-Jp,op - Jp,1)

(11.189)

The quasi-linear variation of power with ATP hydrolysis is observed experimentally, as the contraction is being activated at the level of actinomyocin activity. The kinetic approach suggests that the muscle power output varies hyperbolically with the ADP concentration. Both the ADP control and the Gibbs energy of ATP hydrolysis control are similar, and when muscle power is varied voluntarily, muscle energetics may be represented by the linear flow-force relationships. 11.10.

EVOLUTIONARY CRITERION

Tellegen's theorem can be used in an evolving network, where the forces are allowed to change with time, and after a time interval dt, the forces become XI + dX,/dt. Since, according to Tellegen's theory, the flows and forces lie in the orthogonal spaces, we have (11.190)

~_~ J; dX; _ 0 dt

Since dt is an arbitrary time interval, we get

"

d[

--

~_~

J i - - ~ irrev

+E

Ji

dt )rev

=0

(ll.191)

This equation comprises both the reversible and irreversible contributions, and also reduces to

dX i ] = - - E Ji--~-]rev

dX i

The reversible part obeys the constitutive relations C i - dNi and Ji - Ci d l~i >- 0 d#; dt

(11.192)

Since the capacity of an ideal capacitor C; >> 0, we have ci dX; _ J~ >_0 dt C;

(11.193)

Therefore, the first term of Eq. (11.191 ) is negative definite

Ji dt

)irrev~ 0

(11.194)

This equation is an evolutionary criterion. The change of dissipation function with time yields d~ dt

dXi ] (Xi dJi -- E (Ji --~)irrev -+-E -~)irrev

(11.195)

In the linear region of the thermodynamic branch and with constant phenomenological coefficients, we have

dX i

dJ

(11.196)

596

11.

Thermodynamicsand biological systems

Combining Eqs. (11.195) and (11.196), we have

d*-2~_~(j i dXi dt

TJirrev

(11.197)

From Eqs. (11.194) and (11.197), we obtain

d~ dt

~-<0

(11.198)

This equation is valid only in the linear region, which may be rare in biology. Equation (11.194) may be used for the evolution of all biological networks, which can be characterized by thermodynamic considerations. Equation (11.194) is valid for both linear and nonlinear constitutive relations, and can be used for quasi-equilibrium and far-from-equilibrium regions of the thermodynamic branch.

PROBLEMS 11.1

In living systems, ions in the intracellular phase and the extracellular phase produce a potential difference of about 85 mV between the two phases. The intracellular phase potential is negative. Determine the difference in electrical potential energy per mole positive monovalent ion, e.g., Na+, between the two phases.

11.2

Calculate the change in the enthalpy of blood when it is subjected to an isothermal increase in pressure of 20 kPa.

11.3

The enthalpy change of a blood stream flow at 5.2 L/rain has an isothermal increase of pressure of 21 kPa. Calculate the work added to the blood as useful energy.

11.4

A small organism has a heat loss o f - q = 1.65 W and performs external work W = 0.025 N m/s. Calculate that part of the total energy expenditure that originates from its internal circulation, which involves the pumping of 122 mL/min of fluid against a pressure drop of 3.4 kPa with a net chemo-mechanical efficiency of 12%.

11.5

Estimate the energy expenditure for a steady process involving oxidation of 425 g/day glucose at 310 K and 1 atm. C6H1206 (aq) 4- 602 (g)~6CO2 (g) + 6H20(1 )

11.6

An adult male has an oxygen uptake of about 21.16 mol over 24 h, and the associated elimination of carbon dioxide and nitrogen is 16.95 mol and 5.76 g, respectively. The male has performed 0.12 MJ of external work over the same period and his energy expenditure at rest is E0 = 70 W. Estimate his energy expenditure, heat loss, and net efficiency for the external work.

11.7

An amphipod with a body weight of 9 txg consumes 3.5 × 10 - 9 mol oxygen every hour at steady state and eliminates 3.5 × 10 -9 mol carbon dioxide, 0.4 x 10 -9 mol N (as ammonia), and 0.1 × 10 -9 mol lactic acid. The external work power is 47 x 10 -9 W. Estimate the heat loss of the animal when the following four net reactions contribute to the energy expenditure. C6H120 6 4- 602(g)--*6CO 2 + 6H20

C6H1206---.2C3H603 C 5 5 H 1 0 4 0 6 4-

7802 ~

- 2870 kJ/mol

- 100kJ/mol

5 5 C 0 2 4-

52H20

- 34300 kJ/mol

C32H48010N8 4- 330 2 --. 32CO 2 4- 8NH 3 4-12H20

- 14744 kJ/mol

597

References

11.8

A small organism has a heat loss of 1.52 W and performs external work of 1.2 N m/min. Calculate that part of the total energy expenditure that originates from its internal circulation, which involves the pumping of 120 mL/min of fluid against a pressure drop of 25 mmHg (3.34 kPa) with a net chemo-mechanical efficiency of 10%.

11.9

The intensity of the sun's radiation on a clear day is observed to be 50 mW/cm 2. Calculate the accumulation of glucose during 8 h of exposure in a green leaf. Its surface area is 10 cm 2 and the heat loss from the leaf is 49 mW/cm 2. It is assumed that the temperature of the leaf is steady at 25°C and that only the net process proceeds (the reaction enthalpy is 478 kJ/mol CO2 at 25°C).

11.10

Consider the total energy available from the oxidation of acetate. What percentage is transferred through the TCA cycle to NADH, FADH2, and GTP? Acetate + 202 -~ 2CO 2 + 2 H 2 0

- 243 kJ/mol

1 N A D H + H +- + - 02 ~ NAD + + H 20 2

1

FADH2 + 2- 02 --* FAD + H 2 0 Z

11.11

GTP---, GDP + Pi

- 8 kJ/mol

ATP ~ ADP + Pi

- 8 kJ/mol

- 41 kJ/mol

- 41 kJ/mol

Consider the oxidation of acetate to produce ATE What percentage of the energy is available from the oxidation? Use the chemical reactions in Problem 11.10.

REFERENCES C.D. Andriesse and M.J. Hollestelle, Biophys. Chem., 90 (2001) 249. A. Axelsson and B. Persson, Appl. Biochem. Biotechnol., 18 (1988) 231. EC. Boogerd, F.J. Bruggeman, R.C. Richardson, A. Stephen and H.V. Westerhoff, Synthase, 145 (2005) 131. C.B. Cairns, J. Walther, A.L. Harken, A. Banerjee, Am. J. Phvsiol. Regul. Integi: Comp. Physiol., 433 (1998) R1376. S.R. Caplan and A. Essig, Bioenergetics and Linear Nonequilibrium Thermodynamics, The Steady State, Harvard University Press, Cambridge (1983). J. Castresana, Biochim. Biophys. Acta, 45098 (2001) 1. T.J. Chresand, B.E. Dale, S.L. Hanson and R.J. Gillies, Biotechnol. Bioeng., 32 (1988) 1029. Y. Demirel and S.I. Sandler, Biophys. Chem., 97 (2002) 87. T.G. Dewey, Phys. Rev. E, 54 (1996) R39. T.G. Dewey, Phys. Rev. E, 56 (1997) 4545. T.G. Dewey and M.D. Donne, J. Theo~: Biol., 193 (1998) 593. A.M. Diehl and J.B. Hoek, J. Bioenerg. Biomemb., 31 (1999) 493. E. Dufour and N.-G. Larsson, Biochim. Biophys. Acta, 1658 (2004) 122. L. Garby and P.S. Larsen, Bioenergetics. Its Thermodvnamic Foundations, Cambridge University Press, Cambridge (1995). C.S. Ho and L.-K. Ju, Biotechnol. Bioeng., 32 (1988) 313. J.L. Howland and M. Needleman, Biochem. Mol. Biol. Educ., 28 (2000) 301. R Jezek, J. Bioenerg. Biomemb., 31 (1999)457. Q. Jin and C.M. Bethke, Biophys. J., 83 (2002) 1797. B. Kadenbach, M. Huttemann, S. Arnold, I. Lee and E. Bender, Free Radic. Biol. Med., 29 (2000) 211. O. Kedem and S.R. Caplan, ~'ans. Faraday Soc., 61 (1965) 1897. S.B. Libicki, P.M. Salmon and C.R. Robertson, Chem. Eng. Sci., 32 (1988) 68. L. Lionetti, R. Crescenzo, M.R Mollica, R. lasso, A. Barletta, G. Liverini and S. Iossa, Cell Mol. Life Sci., 61 (2004) 1366. C.A. Mannella, J Bioenerg. Biomemb., 32 (2000) 1. D.B. Marks, Biochemistly, 3rd ed. Lippincott Williams & Wilkins, New York (1999). M.P. Murphy, Biochim. Biophys. Acta Bioenel~g., 1504 (2001) 1. J. Ovadi and V Saks, Mol. Cell. Biochem., 256/257 (2004) 5. V.A. Parsegian, R.R Rand and D.C. Rau, Proc. Natl. Acad. Sci. U.S.A., 97 (2000) 3987. T. Pfeiffer, S. Schuster and S. Bonhoeffer, Science, 292 (2001) 504. H. Qian and D.A. Beard, Biophys. Chem., 114 (2005) 213.

598

11.

Thermodynamicsand biological systems

M. Rigoulet, A. Devin, P. Espie, B. Guerin, E. Fontaine, M.-A. Piqiet, V. Nogueira and X. Lverve, Biochim. Biophys. Acta, 1365 (1998) 117. K.J. Rothschild, S.A. Ellias, A. Essig, H.E. Stanley, Biophys. J., 30 (1980) 209-239. Y. Sambongi, I. Ueda, Y. Wada and M. Futai, J Bioenerg. Biomemb., 32 (2000) 441. S. Schuster and H.V. Westerhoff, Biosystems, 49 (1999) 1. S. Soboll and J.W. Stucki, Biochim. Biophys. Acta, 807 (1985) 245. J.W. Stucki, Eur. J. Biochem., 109 (1980) 269. J.W. Stucki, Adv. Chem. Phys., 55 (1984) 141. J.W. Stucki, Proc. R. Soc. Lond. Ser. Biol. Sci.,244 (1991) 197. J.W. Stucki, M. Compiani and S.R. Caplan, Biophys. Chem., 18 (1983) 101. J.W. Stucki, L.H. Lehmann and P. Mani, Biophys. Chem., 19 (1984) 131. P.R. Territo, V.K. Mootha, S.A. French and R.S. Balaban, Am. J. Physiol. Cell Physiol., 278 (2000) 423. J.J. Tomashek and W.S.A. Brusilow, J. Bioenerg. Biomemb., 32 (2000) 493. A.R. Waldeck, K. van Dam, J. Berden and P.W. Kuchel, Eur. Biophys. J., 27 (1998a) 255. A.R. Waldeck, A.S.-L. Xu, B.D. Roufogalis, and EW. Kuchel, Eur. Biophys J., 27 (1998b) 247. B.D. Wood and S. Whitaker, Chem. Eng. Sci., 55 (2000) 3397.

REFERENCES FOR FURTHER READING M.A. Bianchet, EL. Pedersen and L.M. Amzel, J Bioenerg. Biomemb., 32 (2000) 517. Y. Evron, E.A. Jhonson and R.E. McCarty, J. Bioenerg. Biomemb., 32 (2000) 501. R.H. Fillingame, W. Jiang and O.Y. Dmitrev, J. Bioenerg. Biomemb., 32 (2000) 433. W.D. Frasch, J Bioenerg. Biomemb., 32 (2000) 539. P.M. Haggie and A.S. Verkman, J. Bio. Chem., 227 (2002) 40782. D. Juretic and P. Zupanovic, Comp. Bio. Chem., 27 (2003) 541. Y. Kagawa, T. Hamamoto and H. Endo, J. Bioenerg. Biomemb., 32 (2000) 471. M.V. Mesquita, A.R. Vasconcellas, R. Luzzi and S. Mascarenhas, Braz. J. Phys., 34 (2004) 459. M. Ozer and I. Provaznik, J. Theor. Biol., 233 (2005) 237. EL. Pedersen, Y.H. Ko and S. Hong, J. Bioenerg. Biomemb., 32 (2000) 423. H.R. Petty and A.L. Kindzelskii, PNAS, 98 (2001) 3145. P. Reimann, Phys. Rep., 361 (2002) 57. H. Rottenberg, Biochim. Biophys. Acta, 549 (1979) 225. M. Schmid, T. Vorburger, K.M. Pos and P. Dimroth, Eur. J. Biochem., 269 (2002) 2997. A.E. Senior, J. Weber and S. Nadanaciva, J. Bioenerg. Biomemb., 32 (2000) 523. B.J. van Rotterdam, H.V. Westerhoff, R.W. Visschers, D.A. Bloch, K.J. Hellingwerf, M.R. Jones and W. Crielaard, Eur. J. Biochem., 268 (2001) 958. A.V. Vershubskii, VT Priklonskii and A.N. Tikhonov, Biochemistry, 69 (2004) 1016.

12 STABILITY ANALYSIS 12.1

INTRODUCTION

Thermodynamics plays an important role in the stability analysis of transport and rate processes, and the nonequilibrium thermodynamics approach in particular may enhance and broaden this role. This chapter reviews the stability analysis based on the conventional Gibbs approach and the nonequilibrium thermodynamics theory. It considers the stability of equilibrium, near equilibrium, and far from equilibrium states with some case studies. The entropy production approach for nonequilibrium systems appears to be more general for stability analysis. One major implication of the nonequilibrium thermodynamics theory is the introduction of distance from global equilibrium as a constraint for determining the stability of nonequilibrium systems. When a system is far from global equilibrium, the possibility of new organized structures of matter arise beyond an instability point. As the nonequilibrium thermodynamics theory considers the implications of distance from global equilibrium, it may play a critical role in our understanding of the stability of nonequilibrium systems. Nonequilibrium conditions may occur with respect to disturbances in the interior of a system, or between a system and its surroundings. As a result, the local stress, strain, temperature, concentration, and energy density may vary at each instance in time. This may lead to instability in space and time. Constantly changing properties cannot be described properly by referring to the system as a whole. Some averaging of the properties in space and time is necessary. Such averages need to be clearly stated in the utilization and correlation of experimental data, especially when their interpretations are associated with theories that are valid at equilibrium. Components of the generalized flows and the thermodynamic forces can be used to define the trajectories of the behavior of systems in time. A trajectory specifies the curve represented by the flow and force components as functions of time in the flow-force space. 12.2

THE GIBBS STABILITY THEORY

The stability of equilibrium and nonequilibrium systems is continuously tested by internal fluctuations and external perturbations. Hydrodynamic instabilities develop mainly by the two competing mechanisms of destabilizing and stabilizing effects, such as (i) kinematic nonlinearity working against viscosity, and (ii) gravity competing with a temperature gradient. Light scattering experiments have confirmed that a temperature gradient causes nonequilibrium fluctuations in liquid mixtures and polymer solutions due to coupled phenomena between viscous flows and heat and mass flows. Fluids in nonequilibrium states are capable of exhibiting long-range fluctuations, which are studied by linearized Boussinesq approximations. The classical Gibbs stability theory considers the stability of isolated systems in which energy is totally randomized, and entropy reaches its maximum value~ acting as Lyapunov function. The Gibbs free energy is also a Lyapunov function for specified boundary conditions. By contrast, an equilibrium state characterized by a thermodynamic potential is a global attractor and asymptotically stable for near nonequilibrium states as illustrated in Figure 12.1. At near global equilibrium, irreversible processes reduce perturbations and drive the system back to equilibrium by producing entropy. However, at far from global equilibrium, perturbations do not tend to decay, and the system evolves to metastable or stable coherent behavior stabilized through the exchange of energy and matter with the environment. Such states might be highly organized and called dissipative structures, which are created and controlled by hydrodynamic and kinetic parameters~ The thermodynamic forces (gradients) imposed on a system measure the distance from global equilibrium, leading to multiple solutions that appear at a bifurcation point. Complex behavior in transport phenomena is usually associated with spatial instabilities, while chemical systems may sustain temporal instabilities by enhancing the effect of slight perturbations, such as the Belousov-Zhabotinsky reaction scheme, which is one of the early examples of chemical oscillations.

600

12. Stabilityanalysis

~2S =/kS = S - Seq _<0

y

t

~>

dt - 0

Figure 12.1. Near equilibrium/kS is a Lyapunov function and equilibrium is a global attractor.

~28 =/kS = S - Seq _<0

y

t

t -0

Stable

Unstable

Figure 12.2. Stability at far from equilibrium. Due to the two terms, deS and diS, the second law does not impose the sign of entropy variation: dS = ~S + deS. Therefore, there is no universal Lyapunov function, which creates ambiguity in the stability of states far from equilibrium.

Stable equilibrium has minimum Gibbs free energy. The necessary (but not sufficient) condition is that the first derivative of the Gibbs energy is zero at the possible equilibrium states, and the second derivative of the Gibbs energy is positive

¢~2G=

/

d x 2 ) (~x) 2 > 0

(12.1)

Here x is a parameter that characterizes the state of the system (for example, the concentration of a gas). Stability in equilibrium plane requires that the second derivative of the entropy be negative a2S < 0

(12.2)

Equations (12.1) and (12.2) represent the energy minimum principle and the entropy maximum principle at equilibrium (Figure 12.2). The Gibbs stability theory provides necessary and sufficient conditions to investigate stability problems with well-defined boundary conditions in equilibrium state. Some examples of this are

12.2

601

The Gibbs stability theory

constant)

(12.3)

d G - - T d iS

<- 0 (T, P = constant)

(12.4)

d H = - T d iS

<- 0 (S, P = constant)

(12.5)

d A - - T d i S <- 0 ( T , V -

Fluctuations in thermodynamic properties determine the entropy change, which can be expanded in certain fluctuations. For an isolated system, the power series expansion of entropy in terms of fluctuation x is

AS

-

S-

Seq --

aS

dx +

Ox

/

( d x ) 2-+- . . . .

~

2

Ox 2

aS-+--

S+...<0

(12.6)

2

The change in entropy is due to the second-order term, as the first-order term vanishes since the entropy reaches its maximum value at equilibrium. Therefore, the system at equilibrium will be stable to perturbations when the entropy decreases and the change is negative. However, the characteristics of perturbations play an important role in driving the system toward stable or unstable state. Decaying perturbations lead to stable equilibrium; otherwise, instability occurs. More interestingly, when the magnitude of perturbations is very large, the system may move to the nonlinear region, which is far from global equilibrium on the thermodynamic branch (Nicolis and Prigogine, 1977), where the instability may cause a system to evolve into an organized structure.

12.2.1

Thermal Stability

For a system with parts 1 and 2, consider a flow of energy d U from part 1 causing a small fluctuation in temperature Expansion of the total entropy of the parts (S) in terms of U1 and U2 yields

aT.

Tx - Tj )(aU) - 0, and as-

Cv(aT)2

a2S -

-r,~

-

r 2

< 0

(12.7)

at equilibrium (T= T 1 = T2). Therefore, fluctuations where d U 1 - - d U 2 - d U , and ( O S / O U ) t w - l/T, a U - C v a T , can only decrease the entropy, and an entropy-producing irreversible process drives the system back to the original state. Equation (12.7) shows that the state at equilibrium is stable to thermal fluctuations because the heat capacity Cv is positive.

12.2.2

Mechanical Stability

For a system similar to the one described above, the second-order term of Eq. (12.6) due to fluctuations in the volume of a subsystem is a2S = _ ( a v ) 2

< 0

(12.8)

T~crV

where

Kr

is the isothermal compressibility defined by 1

OV

, and

i

~

T

(12.9)

We have a S = 0, as P 1 / T I - P2/T2 - P / T at equilibrium, and the second derivative becomes negative. Equation (12.8) shows that the state of equilibrium is stable to fluctuations in volume because the isothermal compressibility is positive.

12.2.3

Chemical Stability

Chemical reactions may cause fluctuations in the number of moles of various components of a system. Equation (12.6) provides expansion in terms of affinity A (A = - £ U i l X i ) . At equilibrium, aS = 0 and A - 0. So, the condition

602

12.

Stability analysis

for the stability of equilibrium state versus the fluctuations in the numbers of moles in terms of the extent of reaction e is

62S=_1(0~) T

12.2.4

(6e)2< 0

(12.10)

eq

Stability in Diffusion

When the fluctuations are due to the exchange of matter between the parts, Eq. (12.6) yields

~2S---~ik( O . ONk I~li-tOT ONk 1~2-----Li ( ~ N)i 6 N k ) < O T

(12.11)

Since (OS/ONk)- -iz#/T, 6Nlk = -~Nzk = ~Nk, and 6S = 0, as the chemical potentials of the two parts are equal, Eq. (12.11) becomes

(~2S =--~i k ( ONk I~li) (6Ni6Nk) <

(12.12)

This is the stability condition of an equilibrium state when the numbers of moles fluctuate. For an isothermal binary mixture, Eq. (12.12) becomes 0/~1 (6N1) 2 + 0~2 (6N2) 2 0/22 (6N1)(6N2)+ ON1 ~2 + -~1

0~1 (6N1)(6N2) > 0

(12.13)

where

01~1 _ 0 O G ON2 ON2 aM

_

O/J,2

ON1

By relating the fluctuations to the reaction coordinate vi de = 6Ni, Eqs. (12.10) and (12.12) imply that ifa system is stable to fluctuations in diffusion, it is also stable to fluctuations in chemical reactions, which is called the DuhemJougeut theorem (Kondepudi and Prigogine, 1999). However, a nonequilibrium steady state involving chemical reactions may be unstable even if the system is stable with respect to diffusion.

12.2.5

General Stability Condition

The general condition for the stability of equilibrium state with respect to thermal, volume, and fluctuations in the numbers of moles is obtained by combining Eqs. (12.7), (12.9), and (12.12)

-

T2

TKT

V

- ..

6Nj

6Ni6Nj < 0

(12.14)

A process is spontaneous if it obeys the following conditions" (AS)u,v > - 0 at constant U and V (A U)s, v -< 0 at constant S and V (M-~s,p <- 0 at constant S and P (&4)v,v-< 0 at constant T and V (AG)v,p <- 0 at constant T and P Equation (12.14) shows that the second-order differential of entropy assumes a quadratic form, and the stability conditions are directly related to the behavior and signs of the coefficients. It is interesting to see that Eq. (12.14) resembles the Gibbs equation T~S = 3U + P~ V - ~,i~i6Ni in the following form: T6 2S -- - 6

T~S + 3P6 V - ~_~6 l~i6N i i

where ~;S = 3q/T= Cv6T/T, 6P = -6 V/(KvV), and 6/~i = ~i (Ot~JONj)dNj.

(12.15)

12.2

603

The Gibbs stability theory

Table 12.1 Necessary and sufficient conditions for the stability of equilibrium state

Boundary conditions: constant properties

Stability criterion

Equilibrium criterion

Infinitesimal perturbations

Finite amplitude perturbations

S, V

((~U)e q - - 0

(~U:> 0

(AU)eq > 0

U, V

(6S)e q -- 0

(~S ~ 0

(AS)e q < 0

(62U')eq > 0 (a2S)eq < 0

S, P

(6H)cq = 0

~ H :> 0

(Z~)e q > 0

(62/-/)eq ~> 0

T, V

(aA)e q = 0

6A > 0

(z~l)e q > 0

(a2A)eq > 0

T, P

(6G)~q = 0

6G > 0

(AG)e q > 0

(32G)eq > 0

For a closed system at uniform temperature and pressure, we have TdiS = T 6 s - d U - p d V >- 0

(12.16)

Equation (12.16) results from the entropy balance di S = d S - deS >__O

with deS = 6q/T and 6q = dU + PdV, and leads to a stability criterion for thermodynamic equilibrium; the inequality shows that equilibrium state is stable. For a small and arbitrary increment 6, Eq. (12.16) provides the stability criterion 6 U + ['6 V - T6S >- O. At constant S and V, the stability condition becomes 6 U >- 0, indicating that the internal energy assumes its minimum value for stable equilibrium state: (6U)e q = 0. Stability after a perturbation with finite amplitude is (A g)e q > 0, or after an infinitesimal perturbation is (6 2 U)e q > 0. Table 12.1 shows the equilibrium and stability criteria for various boundary conditions; the last column is not always satisfied for metastable systems, although we often describe both stable and metastable systems as stable states.

12.2.6

Configurational Heat Capacity

Heat absorbed by a system existing in two isomeric forms changes pressure, temperature, and extent of transformation between isomers 6q = - V d P + dH = (hr, ~ - V ) d P + Cp,~dT + hr,pde

(12.17)

where

where Cp,~ is the heat capacity at constant composition and with very slow relaxation of the transformation. The heat capacity at constant pressure is

/

/

The second term on the right is called the configurational heat capacity, and is due to the relaxation of system to the equilibrium configuration (Kondepudi and Prigogine, 1999)

C t,~, = ( OH

(12.18)

For a transformation at equilibrium (A = 0), we have OH

(12.19)

604

12.

Stability analysis

The configurational heat capacity for a transformation at equilibrium and constant pressure is defined by combining Eqs. (12.18) and (12.19) OA

(12.20) , ~.OT}p,A=O

Since the stability condition for a chemical reaction is (OA/Oe)< 0, the heat capacity at constant composition is always less than the heat capacity of a system that remains in equilibrium as it receives heat. Certain fluid molecules, such as supercooled liquid glycerin, can vibrate but not rotate freely, which is called libration. As the temperature increases, more molecules rotate, and the variable e becomes the extent of libration-rotation transformation. If the transformation equilibrium is reached rather slowly, the heat capacity (Cp,~) will be lower than the heat capacity measured in slow heating. 12.2.7

Phase Stability

If a system is in the thermodynamic equilibrium (dS/dt >- 0), instabilities can occur only at phase transition points, and the new phase may be in a more ordered state (vapor ~ liquid), which is a self-sustaining structure. Phase splitting due to thermodynamic instability and hence symmetry breaking in equilibrium in a feed mixture affect problems associated with the simulation and design of distillation and extraction. Knowing the exact number of phases contributes considerably toward successfully predicting phase equilibria. For ternary mixtures, for example, feed points located within the binodal curve split only into two liquid phases; therefore, knowing the position of the feed leads to the exact number of phases. For multicomponent mixtures, the distance between the tangent plane and the Gibbs energy of mixing surface is used for phase stability analysis. When the distance D for a composition x is negative, a phase with feed mole fractions z is unstable, and the molar Gibbs energy of the mixing surface Gm = AGmix/RT falls below a plane tangent to the surface at z. The distance D is obtained from

D(x)=Gm(x)-Gm(z)-~( OGm] ( x i - z i =1 ~',

i)

(12.21)

OXi Jz

The subscript z indicates the evaluation of partial differentials at x = z, and n is the number of components. The Gibbs energy of mixing Gin, and the reduced excess Gibbs energy g~ are n

GE

Gm (x) = ~_, x i In x i + gE (x) and gE _

(12.22)

RT

i=1

The tangent plane distance analysis minimizes the D subject to the mole fractions by solving the following system of nonlinear equations, which provide the stationary points (Gecegormez and Demirel, 2005)

with

Lk ax,

k - .Jj

=

ox.

0 i = 1,...,n-1

(12.23)

z

i=l Xi -- 1.

12.3

STABILITY AND ENTROPY PRODUCTION

The Gibbs stability theory condition may be restrictive for nonequilibrium systems. For example, the differential form of Fourier's law together with the boundary conditions describe the evolution of heat conduction, and the stability theory at equilibrium refers to the asymptotic state reached after a sufficiently long time; however, there exists no thermodynamic potential with a minimum at steady state. Therefore, a stability theory based on the entropy production is more general. The change of total entropy is: dS = deS + diS. The term deS is the entropy exchange through the boundary, which can be positive, zero, or negative, while the term diS is the rate of entropy production, which is always positive for irreversible processes and zero for reversible ones. The rate of entropy production is diS/dt = EJeXk. A near equilibrium system is stable to fluctuations if the change of entropy production is negative z~iS < 0. For isolated systems,

I2.3

605

Stability and entropy production

dS/dt >- 0 shows the tendency toward disorder as deS/dt = 0, and dS = diS >- O. For nonisolated systems, diS/dt > 0

shows irreversible processes, such as chemical reactions, heat conduction, diffusion, or viscous dissipation. For states near global equilibrium, diS is a bilinear form of flows and forces that are related in linear form. The second law for isolated systems shows that the excess entropy, AS = S - Seq -< 0, increases monotonically in time, d(AS)/dt >_ O. Therefore, it plays the role of a Lyapunov function, and defines a global stability. So, diS/dt is a Lyapunov function that guarantees the global stability of stationary states that are close to global equilibrium. For nonequilibrium systems far from global equilibrium, the second law does not impose the sign of entropy variation due to the terms deS and diS, as illustrated in Figure 12.2. Therefore, there is no universal Lyapunov function. For a multicomponent fluid system with n components, entropy production in terms of conjugate forces X~, flows J,., and I number of chemical reactions is

+ =

J;x, = a . . v

-

a;. v v

i

- F, +

(W)

i=1 1

(12.24)

l

+ - - E A/Jr/>- 0

T j=l

'

where F i is the force per unit mass of component i. Here, the rate of entropy production is the sum of contributions from heat, mass, momentum transfer, and chemical reactions, excluding electrical, magnetic, and other effects. Equation (12.24) identifies a set of the conjugate flows and forces to be used in the phenomenological equations with the coefficients satisfying Onsager's reciprocal rules in the linear regime of the thermodynamic branch. For a chemical reaction, entropy production is diS _ A de dt

(12.25)

T dt

An approximation of A for a given small fluctuation in the extent of reaction e, c~ = ( e - geq), is expressed by eeq), and used in Eq. (12.25) to obtain the following stability condition

A = (OA/Oe)(e

-

~ A d e = [ a~ OA

AiS =

aO

7

oldol -

eq

< 0

eq

(12.26)

where de = dc~. For I number of chemical reactions, Eq. (12.26) becomes

AiS--

~

1 (04.)a<&j c)g,----7eq

(12.27)

Equations (12.26) and (12.27) do not depend on the boundary conditions imposed on the system. For a thermal fluctuation of Teq +/3, we have

diS=

~eq +/3

(12.28)

Teq a q = - l e q

Using 6q = C, d[3, the condition for thermal stability becomes Cv (6T)2 < 0

(12.29)

As before, a general thermodynamic theory of stability formulation is quadratic in the perturbations of 6T, 6 V, and 6Nk, because the forces and flows vanish at equilibrium

=- ~

2T

~

T

+ ........

KTV

+

.. ~

) ] 6NiBNj

"

< 0

(12.30)

606

12. Stability analysis

Equations (12.30) and (12.14) show that the contributions to the entropy change due to fluctuations in the equilibrium state are of second order, expressed by 62S/2 (12.31)

~2S<0

1 d62S 2 dt - ~ 6Jk6Xk > 0

(12.32)

k

The term to the right of the equal sign in Eq. (12.32) is the excess entropy production. Equations (12.31) and (12.32) describe the stability of equilibrium and nonequilibrium stationary states. The term 62S is a Lyapunov functional for a stationary state.

Example 12.1 Distance of a chemical reaction from equilibrium Consider the synthesis ofhydroiodic acid in the perfect gaseous phase:

H2 + i 2 ., kf '2HI kb The affinity A and the reaction velocity Jr are

A=IXH2+IXI2-2tXHI=RTln(~,Q(T)) K(T)] J r = J r f - J r b = k f c I 2 CH 2 -- kbC~tI = J r f

I

1-

exp - ~-~

(12.33)

(12.34)

where kf and kb are the forward and backward reaction rate constants, respectively, and Q is defined by

Q - cZi CI2CH 2

Equation (12.33) is a measure of the distance of a nonequilibrium chemical reaction from equilibrium. At equilibrium, the affinity A vanishes. Equation (12.34) shows a nonlinear relation between the reaction velocity Jr (flow) and the affinity (thermodynamic force). If the chemical system is close to equilibrium, that is, IA/(RT)] << 1, then the contents of the square parentheses of Eq. (12.34) are approximated as A/(RT), and we have the following linear flow-force relation between the reaction velocity and the affinity Jr -

Jrf,eq A -LXr R T

(12.35)

The phenomenological coefficient L is dependent on the partial rate of reaction at equilibrium. If, however, A/(RT) ~ % then (ci2 CH2/c~i) ~ ~ or Cu~ ~ 0. This shows a sort of saturation effect with respect to the affinity.

Under this condition, entropy production becomes a linear function of the affinity.

12.3.1

Stability of Chemical Reactions

If a system is stable with respect to diffusion, a chemical reaction is also stable in the vicinity of equilibrium. The reaction rate is: Jr = de/dt, where e is the extent of a chemical reaction. The time change in the number of molecules N due to an elementary reaction is d N = v de. In terms of the small fluctuations 6e of the extent of the reaction, the stability condition becomes (6~3)2 > 0 or

OA eq

This condition is valid for equilibrium systems.

< 0 eq

(12.36)

607

12.4 Thermodynamicfluctuations

However, nonequilibrium steady states may be unstable even if the system is stable with respect to diffusion. For a nonequilibrium state, the stability condition for a chemical reaction in terms of excess entropy production is 1 d62S 2 dt

1 T E6Jr6A

>

0

(12.37)

k

where 6Jr and 6A are the perturbation of the chemical affinities and the corresponding reaction rates. The sign of Eq. (12.37) for the linear flow-force region is positive. However, for chemical reactions often the reaction velocities are not linear in terms of affinities. Therefore, it is expected that there exist mechanisms that may lead to a negative contribution.

Example 12.2 Stability of chemical systems Consider the following chemical reaction system:

X+Y

kf ' C + D kb

Assuming that the chemical system is far from equilibrium, and for kf = 1 and kb = 0 (disregarding the backward reaction), the reaction velocity and the affinity become Jr = CxCy, and A = R T l n CxCy CcC D

The excess entropy production due to a fluctuation in the concentration of X around a steady-state value is 6Jr6A = R T c--x-Y(6c x )2 > 0 CX

Such a fluctuation would not violate the stability condition, and the inequality above would have a positive sign. Consider the autocatalytic reaction, X + Y , ks "2X. Under the same conditions, the reaction velocity and kb the affinity become

r=CXC

and

So, the contribution to the excess entropy production becomes negative, and the system becomes unstable 6jr6 A = _ C__y__Ry T ( 6 c x )2 < 0 CX

Both in hydrodynamics and in chemical kinetics, instability may occur due to nonlinear conditions far from equilibrium. In hydrodynamic systems, nonlinear conditions are produced by the inertia terms, such as the critical Reynolds number or Rayleigh number. However, nonequilibrium kinetic conditions may lead to a variety of structures. In chemical systems, some autocatalytic effect is required for instability.

12.4

THERMODYNAMIC FLUCTUATIONS

Equation (12.24) does not describe the dynamics of thermodynamic fluctuations, although the nonequilibrium thermodynamics theory and the thermodynamic fluctuations are connected in deriving the Onsager rules. However, the introduction of the notion of internal degrees of freedom into nonequilibrium thermodynamics implicitly accounts for the fluctuations. Einstein related thermodynamic entropy to the probability of a fluctuation q~by q~ - Ze -xs/k~

(12.38)

608

12. Stabilityanalysis

where AS is the entropy change due to fluctuations from the state of equilibrium, Z the normalization constant ensuring that the sum of all probabilities is equal to one, and kB the Boltzmann constant (= 1.38 × 10 -23 J/K). In a closed system, substituting Eq. (12.30) in Eq. (12.38), the probability q~of fluctuations in terms of T, V, and Nbecomes

¢(6T,6V,6N)=Z exp { -

l[Cv(aT)

]}

(~V) 2

kBT2 q TkBKvV ~ +~i

ONj)

39,

In a more general form, the probability is expressed in terms of the products of pairs of variables (Kondepudi and Prigogine, 1999)

q~= Zexp 2---~-B) Zexp

)]

[-2kBT 1 ( 6T6S-6P6V+~-"6t'*k6Nkk

If we define a fluctuation in a property Y as (Yk- Yk,eq)--Cek, and using (•2S)/2 = -(1/2)£go.ceic~ ments g0 are the appropriate coefficients of matrix [g], then the probability is

/ det[g_] exp (OLI'~2"'"OLm) -- ~/(2rrkB) m

_

1 m ) ~-£ gijoeioQ B i,j=l

(12.40)

j, where the ele(12.41)

where det[g] is the determinant. For a single variable of a, we have q~(a) =

exp - ~-7-_

(12.42)

Example 12.3 Stability under both dissipative and convective effects In some cases, both dissipative as well as convective effects determine the stability of a system. Some examples of such stability are the onset of free convection in a layer of fluid at rest, leading to B6nard convection cells, and the transition from laminar to turbulent flow. For stability considerations, two limiting cases exist: (i) in the case of ideal fluids, dissipative processes are neglected, and (ii) in purely dissipative systems, no convection effects occur. For B6nard convection cells to occur, instability appears at the minimum temperature gradient at which a balance can be steadily maintained between the entropy produced through heat conduction by the temperature fluctuation and the corresponding entropy flow carried away by the velocity fluctuations. Fluctuations may lead to an instability, which appears as the difference between two opposite positive effects: dissipative effects-convective effects. So, stability occurs as the result of competition between two opposite tendencies, stabilizing dissipative effects and destabilizing convective effects (Glansdorff and Prigogine, 197I). In the range of small Rayleigh numbers, dissipative effects produced by the temperature fluctuations are dominant. This region belongs to the linear region on the thermodynamic branch. Effects of velocity fluctuations become dominant at higher values of Rayleigh numbers. At the critical Rayleigh number, however, entropy production changes suddenly due to the occurrence of an unstable mode, at which a new mechanism occurs due to viscous dissipation caused by convection effects. This leads to a dissipative structure, in which the system uses part of its thermal energy to create the kinetic energy necessary to maintain the macroscopic stationary cellular convection current. For a two-component B6nard problem, the convection current includes the effects of thermodiffusion due to coupling effect between heat and mass flows.

12.5

STABILITY IN NONEQUILIBRIUM SYSTEMS

For a nonequilibrium state, the stability condition is

d(a2S) _~2(dS)

dt

--& : 6cop > 0

(12.43)

609

12.5 Stabifity in nonequilibrium systems

Since a definite function 62S leads to the stability condition, it operates as a Lyapunov function, and assures the stability of a stationary state. As the entropy production is the sum of the products of flows J and forces X, we have

82~ = 8J 8X where 6J and 8)( are the perturbations of the J = LX, with L > 0, we get from Eq. (12.44)

(12.44)

flows and forces, respectively. For a linear phenomenological law

82~ = L (8X) 2 --> 0

(12.45)

Since L(•X) 2 is always positive, we always consider the stationary states described by the linear phenomenological equations as thermodynamically stable states. For systems not far from equilibrium, the total entropy production reaches a minimum value and also assures the stability of the stationary state. However, for systems far from equilibrium, there is no such general criterion. Far from equilibrium, we may have order in time and space, such as, appearance of rhythms, oscillations, and morphological structurization. If the stability criterion for a nonequilibrium state (T62S < 0) is violated, a certain class of nonlinear system may appear and is maintained beyond a critical distance from thermodynamic equilibrium. Such new nonlinear systems may be structured states and can be maintained only based on a continuous exchange of energy and matter with their surroundings. For this reason, Glansdorff and Prigogine separated the change in the entropy production into two parts: one due to the change in forces and the other due to the change in flows, which are (12.46)

d * - d x * + djd~ - E Ji d X i + E xi dJi

They have shown that the first term on the right is negative definite even for cases for which the linear phenomenological equations do not hold. By introducing the linear phenomenological equations Ji = LikXk with constant coefficients Lik, we get

dx* - ~_~Ji dXi = ~_~LikXk dXi i

From the reciprocity relations

Lik =

(12.47)

i

Lki, we have u×, - E

(L,, dx; - Z xk dJk = d j ,

i

(12.48)

i

This relation shows that the contribution of the time change of forces to the entropy production is equal to that of the time change of flows. In the domain of validity of linear thermodynamics of irreversible processes, the contribution of the time change of forces to the entropy production is negative or zero dxq~ <- 0

(12.49)

This inequality holds whenever the boundary conditions used are time independent, and it can be extended to include the flow processes as well (12.50) where the forces X~ and the flows Jk include transport processes, such as convective terms. This inequality is regarded as a general criterion of the evolution in macroscopic physics. However, d ~ is not a total differential, and therefore, it is not a thermodynamic potential, although it leads to the concept of local potential. If P is the entropy production in a nonequilibrium stationary state, the change of P due to small changes in the forces 8)(,- and in the flows 8Ji is

dP

(

dX~

)

(

dJi )

dx P

dj P

E - y J-; dv+I, .... E x ; - 2 dv- dt + dt

(12.51)

610

12. Stability analysis

In the nonlinear regime and for time-independent boundary conditions we have

axe

~<-0 dt

(12.52)

dxP is not a differential of a state function, so Eq. (12.52) does not indicate how the state will evolve; it only indicates that the dxP can only decrease. So stability must be determined from the properties of that particular steady state. This leads to the decoupling of evolution and stability in the nonlinear region, and it permits the occurrence of new organized structures beyond a point of instability. The time-independent constraints may lead to oscillating states in time, such as the well-known Lotka-Volterra interactions, where the system rotates irreversibly. A general criterion for stability of a state is given by the Lyapunov function. A physical system x may be defined by an m dimensional vector with elements of xi (i = 1, 2, ..., m) and parameters aj, and we have dxi dt - f i ( x i ' a j )

(12.53)

The stationary states Xsi are obtained using dxi/dt = 0. We define a small perturbation 6xi and a positive function L(6x) called the distance. If this distance between xi and the perturbed state (Xsi + 6xi) steadily decreases, in time, the stationary state is stable d t ( axi )

L (¢~Xi ) > O, ~

dt

< 0

(12.54)

A function L satisfying Eq. (12.54) is called a Lyapunovfunction. The second variation of entropy L = - 62S may be used as a Lyapunovfunctional if the stationary state satisfies E 3XiaJi > 0; hence, a nonequilibrium stationary state is stable if

d 62S dt 2

- ~ aXiaJ i

(12.5 5)

A functional is a set of functions that are mapped to a real or complex value. Equation (12.55) indicates that the quantity d(62S)/dt has the same form for the perturbations from the equilibrium state as well as the nonequilibrium state. In the vicinity of equilibrium, the quantity ~i,6Xi6Ji is called the excess entropy production, and it shows the increase in entropy production. The quantities 6Ji and 6Xi denote the deviations of Ji and X~.from the values at the nonequilibrium steady state. The increase in entropy production for a perturbation from a nonequilibrium state is

6P=6xP+6~P

(12.56)

Since 3S < 0 under both the equilibrium and nonequilibrium conditions, the stability of a stationary state is accomplished if

d 62S dt 2

-- E ~Yi~Ji > 0

(12.57)

Example 12.4 Stability of an autocatalytic reaction For a simple example, we may consider the following autocatalytic reaction, which appears in the reaction scheme of the Brusselator 2X+Y, We have the forward Jrf of reaction Jr are

=

kf "3X ku

kfc2Cyand backward Jrb = kbC3Xreaction rates, respectively. The affinity A and the flow A = RT In Jrf , Jrb

Jr = Jrf - Jrb

12.5 Stability in nonequilibrium systems

611

The excess entropy production is written in terms of 6X = 6A/T and 6Jr

1 da2S _ _R(2kfcxsCys _ 3kbC~) (6Cx)_____~ 2 2 dt Cxs The excess entropy production can become negative if kf >> kb, and hence the stationary state may become unstable. Coupling between chemical kinetics and transport may also lead to dissipative structures, which are caused by auto- and cross-catalytic processes with positive and negative feedback, influencing their own rates of reaction. For example, the Belousov-Zhabotinsky reaction exhibits a wide variety of characteristic nonlinear phenomena. In the nonlinear region, the possible instabilities in chemical and biological systems are (i) multisteady states, (ii) homogeneous chemical oscillations, and (iii) complex oscillatory phenomena. The thermodynamic buffer enzymes may represent a bioenergetics regulatory principle for the maintenance of a far from equilibrium state.

12.5.1

Stationary States

The study of the behavior of the stationary state ofjth order after a disturbance may be helpful for stability considerations. For this purpose, we designate the index s for all forces and flows in the stationary state. Assuming that the force Xk has been disturbed by 6Xk, and keeping the remaining forces constant, we have

XI,. = X,.t,. +tXk, n >-k >- j + l

(12.58)

dk = J,,k + LkktXk

(12.59)

The new value of the flow Jk becomes

According to the Prigogine theorem, d,..k = 0, and we have

Yk = LkI,-tXI,-

(12.60)

After the disturbance, the rate of entropy production is

• = ~ _ J i Y i - ~ L i / X , . i Y , . / + [ ~ _ ~ ( L i k + L l , q)YitYk]-t-Lkk(tYk)2>O i=1

i,j=l

(12.61)

i=1

The first summation term on the right is the minimum entropy production corresponding to the stationary state. The second sum on the right is zero, according to the Onsager reciprocal relations and the Prigogine principle. Therefore, we have Lkk(~Xk) 2 > 0

(12.62)

or

JktXk > 0 This inequality is identical to the Le Chatelier principle for nonequilibrium stationary states, since the disturbance

6Xk has the same sign as the flow dk, indicating a decrease in the disturbance. For example, an increase in the gradient of the chemical potential will cause the mass flow and diminish the gradient. Hence, the stationary state will return to its original status. The change of entropy production can be measured by a sensitive calorimeter during the growth of bacterial colonies. The heat dissipated by the system is directly related to the entropy production. The high-precision calorimetric measurements indicate that the entropy increases sharply and reaches a maximum, and finally decreases to a minimum. This is a stationary state, in which the colony no longer grows. The colony is then limited to maintaining itself. If no growth occurs, the system can be described by means of linear phenomenological laws, and tends toward the minimum entropy production. Similarly, the resting states of biological cells can often be interpreted as stationary states described by linear laws; the active phases or growth are mainly examples of nonlinear phenomena.

612

12.5.2

12. Stability analysis

Stability of Stationary States

In the linear nonequilibrium thermodynamics theory, the stability of stationary states is associated with Prigogine's principle of minimum entropy production. Prigogine's principle is restricted to stationary states close to global thermodynamic equilibrium where the entropy production serves as a Lyapunov function. The principle is not applicable to the stability of continuous reaction systems involving stable and unstable steady states far from global equilibrium. For chemical reactions, the change of entropy production with time P is

dt

T dt

--f- Jrk

(12.63)

The affinity and the reaction velocity are expressed by A = A(P,T,e) and Jrk = £kLki(Ai/T). At constant pressure and temperature, we have

dP dt

2T~i/l~)pTJriJrj

(12.64)

Equations (12.27) and (12.64) show the stability of the nonequilibrium stationary states in light of the fluctuations

6ei. The linear regime requires P > 0 and dP/dt < 0, which are Lyapunov conditions, as the matrix (OAi/Oej)is negative definite at near equilibrium. Dissipative structures can sustain long-range correlations. The temperature and chemical potential are well defined with the assumption of local equilibrium, and the stationary probability distribution is obtained in the eikonal approximation; so the fluctuation-dissipation relation for a chemical system with one variable is

Jr(x) =

2D(x)tanh -2kBT

( 1

= 2D(x)tanh -2kBT VOx

/

(12.65)

where Jr(x) is the net reaction rate representing the flux, D(x) is a probability diffusion coefficient and shows the strength of chemical fluctuations, A(x) is a species-specific affinity representing the thermodynamic force, and q~(x) is the stochastic potential. Equation (12.65) shows a nonlinear relationship between the flux and force, and due to the hyperbolic tangent, the reaction rate approaches toward a constant value for large values of the specific affinity.

12.5.3

Evolution Criterion

An evolution criterion can be obtained from the rate of change of volumetric entropy production P = fEJXdV> 0 as follows:

dt - I v

J k - - ~ j d V + ~.v

/

X k--~ dV= dt ' dt

(12.66)

This equation is independent of the type of phenomenological relations between fluxes and forces. In contrast, linear phenomenological equations and the Onsager reciprocal relations yield

E Jk dXk = E LjkXj dXk = 2 Xj d (LjkX k ) = E Xj dJ k

(12.67)

So, in the linear region, we have

dxP - djP _ 1 dP dt dt 2dt

(12.68)

For time-independent boundary conditions, we have the general conditions for the stability of a state

dP - 2 dxP <-0 dt dt

(12.69)

613

12.5 Stability in nonequilibrium systems

Here, the equality sign is for stationary states. Unfortunately, Eq. (12.69) does not indicate how a state evolves in time. The relations P > 0 and dxP < 0 are called the Lyapunov conditions. The fundamental quantity, which determines the stability, is the sign of excess volumetric entropy production (12.70)

P(6S)=IE6JaXdV>-O

where 6S is the perturbation in entropy. The stability condition for chemical reactions is (Glansdorff and Prigogine, 1971) P(6S)

=

Ei(~'Jric~Ai -->0 T

(12.71)

Consider the following rate of entropy production:

P= ~

T dt

(12.72)

. T dt

and

d P= ZJkdXk7d k

-

(12.73)

7d

with the following approximately defined differentials:

@_d -- - - ( hA(~CA + hB6cB + Cp(~T) dt dt d--~=R~-

ci~,

RE,,

# )

= _(d(6T)

--,

...........

T,

Substituting these into Eq. (12.64), we have

d×P = -R(d(6CA) d(6ca ) f d(6CB) d(6cB) dt cA,, dt CB,,~

Cp d(6T) d(T)) dt L

(12.74)

Equation (12.74) can be used in Eq. (12.69) for stability analysis at near global equilibrium. The Lyapunov function resembles the thermodynamic entropy production function and the asymptotic stability principle. If the eigenvalues of the coefficient matrix of the quadratic form of the entropy production are very large, then the convergence to equilibrium state will be rapid.

Example 12.5 Macroscopic behavior in systems far from equilibrium Consider the nonequilibrium chemical

system R---~ X --~ p X--"-

X----

Concentrations of both R and P are maintained at constant values, while the concentration of the intermediate component X may vary with time. Assume that Xs denotes a steady state (stable or not). The behavior of such a system may be controlled by the position of the steady state: (i)

If the steady state is close to equilibrium, then the system is stable; linear nonequiiibrium thermodynamics can be used. Considering the entropy production below p=l 7EJr/~ i

"

614

12. Stability analysis

The evolution criterion becomes 1 1 11 d x P = --_ '~, Jri ( dAi ) = ~_ d, P = - -- d P < 0 i 1'-2T

This equation is the result of minimum entropy production at stable steady state and shows that if the system is disturbed by a small perturbation, it will return to the steady state. (ii) However, if the steady state is far away from equilibrium, the system may be stable or unstable. A perturbation may lead to multiple states, since the system may enhance the fluctuations instead of damping them, and the system may choose one of the states according to the hydrodynamic and kinetic conditions the system is in. Even if the system is stable, the behavior of the system may vary; the path to the steady state may be spiral or the system may rotate around the steady state. A larger variety of possibilities may exist for the unstable steady-state case. For far from equilibrium conditions, the overall stability is no longer a consequence of the stability with respect to the diffusion, as is the case for conditions in the vicinity of equilibrium.

12.6

LINEAR STABILITY ANALYSIS

Hamiltonian dynamics show that classical mechanics is invariant to ( - t) and (t). In a macroscopic description of dissipative systems, we use collective variables of temperature, pressure, concentration, and convection velocity to define an instantaneous state. The evolution equations of the collective variables are not invariant under time reversal Chemical reaction: dcA -- --kCACB, A + B

k >D

(12.75)

dt

Heat conduction:

OT

= c~V2T, ol > 0

(12.76)

Ot

Diffusion: v--7-~= D V 2 c , D > 0

(12.77)

Ot

Here, T and c are called the even variables whose signs do not change upon time reversal, while convection velocity, and momentum of a particle are called the odd variables whose signs change with time reversal. The general form of a dissipative system with macroscopic variables X1, ..., Am, space r, and time t, may be defined by OX i

- fi(X1,...,Xn,r,t,

tx)

Ot

(12.78)

The evolution of the variables X/is influenced by the variation of some control parameters represented by/x that can be modified by the environment. The control parameters may be the diffusion coefficient, thermal conductivity, chemical rate constants, or initial and final concentrations of reactants and products. Stability analysis has to consider a variety of variables characterizing problems of transport and rate processes. The variables often are functions of time and space. The functionf has the following properties:

f/([Yj, eq ],/Zeq ) = 0, at equilibrium

(12.79)

fi ([Xj, s ],/x s) = 0, at nonequilibrium steady state

(12.80)

The components [X~s] represent the stationary and spatially uniform solution. These relations are associated with certain restrictions, such as T > 0 and c > 0; detailed balance is achieved, and hence physical systems are highly unique (Nicolis and Prigogine, 1989).

615

12.6 Linear stability analysis

The state variables X1, ..., if,., which are continuously subjected to either internal fluctuations or external perturbations, are represented by a column vector X 0X

Ot

- F(X,/z)

(12.81)

Here, F is an operator, and acts on the space in which X is defined. Stability analysis determines if the stationary solutions will remain stable to small perturbations of x(t) X(t) = X~ + x(t)

(12.82)

The stationary state Xs is a particular solution of Eq. (12.81 ): F(X,~) = 0

(12.83)

Using Eq. (12.82) in Eq. (12.81), and by retaining the linear terms only in the Taylor expansion ofF, we obtain 0X

at

- F([X, + x],/x)- F(X,,/x) = Jx

(12.84)

where J is the Jacobian matrix with the elements (Ofi/O~.)~ calculated at stationary state. Equation (12.84) presents an eigenvalue problem: Jq~ = hq~. The solution in terms of each eigenvector (q~) and its eigenvalue (h) becomes x - y_~c k exp(A~t) q~k

(12.85)

k

The coefficients cx. are determined by the initial conditions. Stability depends on whether the perturbation x grows or decays with time. A perturbation may be due to the interference of the environment with the intrinsic dynamics of the system or intrinsic internal deviations called fluctuations that the system generates spontaneously. The property of stability refers to several responses of systems to various types of perturbations (Nicolis and Prigogine, 1989): (i) Perturbations remain smaller than a critical value for all times, and the state X~ is stable in the sense of Lyapunov. Then we can define the notion of orbital stability as the distance between the reference and perturbed trajectories as the whole sequence of possible states. (ii) Perturbations decay in time, and % is asymptotically stable, which implies irreversibility. (iii) State X(t) does not remain in the vicinity of %, and x(t) cannot remain less than a critical value for all times. Then the reference state X,. is unstable: the system experiences the rapid growth of perturbations leading to orbital instability. (iv) State X(t) remains in some vicinity of X, for x(t) <_ critical, and moves away from X~ for x( t) >- critical. This represents a locally stable but globally unstable state %. How the perturbations affect the state of the system depends on the eigenvalues hk. If any eigenvalue has a positive real part, then the solution x grows exponentially, and the corresponding eigenvectors are known as unstable modes. If, on the other hand, all the eigenvalues have negative real parts then a perturbation around the stationary state exponentially decays and the system returns back to its stable state. The linear stability analysis is valid for small perturbations (1x I/1%1 << 1) only.

Example 12.6 Evolution in chemical systems Consider the following set of reactions: A + 2X.

kfl '

X,

kb 1

~~ "B

kb2

'

3X

616

12. Stability analysis

At fixed concentrations of A and B, X is the only variable. At equilibrium, the detailed balance yields kflAX 2 =

kblX3, and

kf2X-- kb2B

Therefore, equilibrium concentrations yield

Xeq

kflAeq _ kb2Beq kbl

(a)

kf2

and impose a condition on A and B B)

_ kflkf2 eq

kblkb2

At nonequilibrium stationary state, however, we have dX dt

-

f ( X i , k f , i,kb,iA, B ) = --kbl X3 + kflAX 2 - kf2X -Jr-kb2B

Using this equation in Eq. (12.83), we obtain F([X s -t- x],/~) - F(Xs, ~) -- -kbl ( Y s -t- x) 3 -t- k f l A ( Y s -k- x) 2 - kf2 ( y s -t- x ) -+-kb2B - (--kbl X3 + kflAX 2 - k f 2 Y s -Jr-kb2B )

After expanding the cubic and quadratic binomial, we have the linearized equation in terms of perturbation x and the stationary solution X~ Ox

Ot

----kbl x3 - + - ( k f l A - 3 k b l X s ) x

2 -+-(--3kblXs2 + 2 k f l A X s - k f 2 ) x

This equation has multiple solutions for some values of A and hence bounded values of B. So, nonequilibrium state can reveal the true properties that are disguised at equilibrium and near equilibrium; nonlinearity combined with nonequilibrium constraints may allow the diversification of the behavior of a system. Macroscopic systems are composed of large numbers of interacting particles, and the state variables represent either averages of instantaneous states over a long time interval, or the most probable states. Most systems communicate with the environment by exchanging small quantities of matter, momentum, or energy, which are treated as experimental error and, confidence level. So, the instantaneous state of a system is not stationary state X~ but rather nearby state Xrelated to Xs through the perturbation x(t); X(t) = Xs + x(t).

12.7

OSCILLATING SYSTEMS

Some autocatalytic chemical reactions such as the Brusselator and the Belousov-Zhabotinsky reaction schemes can produce temporal oscillations in a stirred homogeneous solution. In the presence of even a small initial concentration inhomogeneity, autocatalytic processes can couple with diffusion to produce organized systems in time and space.

12.7.1

The Brusselator System

A well-known oscillating reaction scheme is the Brusselator system, representing a trimolecular model given by A~X B+X~Y+E 2X+Y~3X X~F

(12.86)

617

12.7 Oscillating systems

The net reaction is: A + B ~ E + F. This reaction scheme has been developed by the Brussels School of Thermodynamics, and consists of a trimolecular collision and an autocatalytic step. This reaction may take place in a well-stirred medium leading to oscillations, or the diffusions of the components A and B may be considered. In the latter case, the system may produce Turing structures.

Example 12.7 Linear stability analysis: Brusselator scheme This example is from Kondepudi and Prigogine (1999). Consider the chemical reaction scheme in Eq. (12.86). Assume that the concentration values of A, B, and E, F are maintained at uniform values, and X and Y are the only remaining variables. Then the kinetic equations are

dX - k~A- kzBX + k3xZY - k4X = fl dt dY

dt

(12.87)

- k2BX-k3 X2Y = f2

(12.88)

Here the concentrations A and B are specified. The stationary solutions for which j] =j~ - 0 are

X ~ - k,_~A, and Ys - k4 k2B k4

(12.89)

k 3 kl

Using Eqs. (12.87) and (12.88) in Eq. (12.84), we have

Ot

(12.90)

L-k2Bx - k3 (kl/k4 )2 A2y

where the quadratic perturbations [(B/A)x 2 + 2Axy + x2y] are neglected, since ]xl/IX~] << 1 and lyl/I Ysl << 1. The Jacobian matrix at stationary state is obtained by

J

z

Of1

Of1

0X

0Y =

Of2

Of2

OX

OY

k2B-k 4

k3(klA] 2 [,W)

[_k2 B

(12.91)

-kg(klA] 2 --~-4 )

The eigenvalues are obtained from the characteristic equation: d e t [ J - 1/] = 0, or for a 2 × 2 matrix we have the following quadratic form: A2 - ( J l l

(12.92)

+ J 2 2 ) A + J l l J 2 2 - J21J12 = 0

where Jo are the elements of the Jacobian matrix. The eigenvalues are obtained from Eq. (12.92). We need the trace of the matrix tr < 0 and the determinant det > 0 for a stable stationary state. If the real parts of the eigenvalues are negative, then we have the stable stationary state klA/2 '~1 + '~2 -- J l l -+- J22 -

k2B- k 4 - k 3 U

>0

(12.93)

For a specified concentration of A, the system becomes unstable and oscillates if the condition in terms of the concentration value B becomes B > - k4 --~--k3(klA]

2

This kind of instability is space independent and leads to order in time.

(12.94)

618

12. Stabilityanalysis

Example 12.8 Linear stability analysis with two variables Consider the following reaction scheme previously used: A

kl >X,

B+X

k2 >Y+E,

k2 = 1.0

k3 >3X,

k3 =1.1

2X+Y X

k1 =1.0

k4 >F,

k4 =1.1

The initial values of A and B are maintained at CA = 1 and CB = 1.6, while the products E and F are removed. Calculate: (a) The particular (stationary) solution (b) The Jacobian matrix at steady state (c) The homogeneous solution Solution: (a) the particular solution can be obtained by starting with the rate of change of the intermediate components, Cx and Cy dCx - k,c,~ - k ~ c ~ c ~ + L c ~ c ~ dt dCy k2C~Cx - k3C2Cy dt

- qCx

(a)

-

At steady-state condition and with CA = 1 and CB = 1.6, Eq. (a) becomes 0 = (1)(1)- (1)(1.6)Cx + (1.1)C~Cy - 1.1C× 0 = (1)(1.6)Cx - (1.1) C2Cy

(b)

These simultaneous equations yield the particular solutions or steady-state solutions

X,

~, 1.600 )

(b) The Jacobian matrix is

J

df~(Cx,Cy) dl(cx,Cy) dCy dCx d~ (Cx,Cy ) df2(Cx,Cy) dCy dCx

= [-2.7 + 2.2CxCy k 1.6 - 2.2CXCv

1"1C2 ] -1.1CZx

By substituting the steady state into the above matrix, the Jacobian matrix at steady state becomes

J=

0.5 -1.6

0.9091 1 -0.9091

(c) With MATLAB we have the following eigenvalues matrix and eigenvectors matrix:

A=

[-0.2045 + 0.9789i 0

0 ] - 0 . 2 0 4 5 - 0.9789i

[-- 0°3516 - 0o4885i w0.3516 + 0o4885i1 ('P-0,7985 0°7985

(c)

12.7

619

Oscillatingsystems

Using the eigenvalues and eigenvectors the homogeneous solutions becomes

-

-

cos (0.9789t) [

sin (0.9789t)

x2(t )

0 e -0"20450t

+12

sin(0.9789t)l

(d)

cos(0.9789t) -0.48850

+ e -020450t

Cx (0) - 1 and Cr (0) - 1.

The constants 11 and 12 are obtained from the initial conditions:

I l - -0.7514, [2 - 0.3547 The real parts of the eigenvalues are negative, and the perturbations will decay in time as Figure 12.3 illustrates. When the value of B is 2.4 then the oscillations are sustainable. Figure 12.3b and 12.3d show the state-space plot of concentrations Cx and Cy for the different values of B. Regardless of whether the eigenvalues are real or complex, the steady state is stable to small perturbations if the two conditions tr[J] < 0 and det[J] > 0 are satisfied simultaneously. Here, tr is the trace and det is the determinant of the square matrix J.

12.7.2

The Brusselator Model with Diffusion

The Brusselator reaction-diffusion system is capable of sustaining spatial and temporal structure. When the concentrations of A and B are controlled, the one-dimensional approach to complex reaction-diffusion systems under isothermal conditions yields the kinetic equations for X and Y. Consider the following set of equations representing space-independent evolution: 0Z

1.6

- F(Z, ~)

(12.95)

N,..j--~__

1 >. 1.4 r.j

......

1~

!

G'I ---...

..... .

0.8 6 ....

l0

~0

(a)

3o . . . .

0.7

;0

t

.

.

.

2

2

. .

0 (c)

1

0.9 Cx

.

0

0.8

(b)

.

.

.

.

.

.

10

.

.

.

.

.

.

20 t

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

30

0.5

40 (d)

0.75

1

1.25 Cx

1.5

1.75

2

Figure 12.3. (a) Oscillations of concentrations of Cx and Cy with kl - 1.0, k2 = 1.0, k 3 - 1.1, k4 = 1.1, A = 1.0, B = 1.6. Bold line indicates the concentration of Cy. (b) State-space plot, (c) oscillations of concentrations of Cx and Cy with kl - 1.0, k2 = 1.0, k3 = 1.1, k4 - 1.1, A - 1.0, B = 2.4. (d) State-space plot in Example 12.8.

620

12. Stability analysis

where/x represents some controlling parameters. The perturbations z around a steady state Zs are defined by Z = Zs + z with Izl/Iz, I << 1. We can convert Eq. (12.95) into the following linear form: O Z - F ( Z , + z , / x ) m F(Z,/x) -- ( O ~ )

Ot

Z

(12.96)

z~

The stationary solutions are X~-

k~A

k4

, and Y~-

ka k2B

k3k 1

Here, X~ and Ys are the elements of the vector Z. Using the perturbation equations Z = Zs + z, Eq. (12.96) becomes

7

]

-k2gx-k3(kl/k4)2A2y

(12.97)

Here, x and y are the elements of vector z. Considering the unequal diffusion of X and Y under isothermal conditions, with the spatial coordinate r, Eqs. (12.87) and (12.88) become OX _ kl A _ kzBX + k3X2y_ k4X + Dx O2X

at

(12.98)

Or 2

0Y 02y 0--7 = k z B X - k3xZY + Dy ~ Or2

(12.99)

where Dx and Dy are the respective diffusion coefficients and r is the distance to the region. The boundary conditions are

0 and/°Y) (0Y) r=-L

0 r=+L

As before, x andy are the small perturbations around the stationary state. The stability of the stationary states depends on whether the perturbations ofx and y grow or decay with time. We have the Jacobian matrix of Eqs. (12.98) and (12.99)

(kzB-k4)+DxV2 J=

k

)2 A 2

k3 -~4)

3( 1)212 ~4)

-0

(12.100)

+DyV2

This equation shows that the Jacobian acts as a matrix in the space of two-dimensional vectors and as a differential operator for space-dependent functions. The linearized form of Eqs. (12.98) and (12.99) with the function consisting of time and space variables becomes F', and is F'= F+D'Vzz

(12.101)

Linearized operator (L) for this function is L-D'V2+

~OF) " z,

(12.102)

621

12.7 Oscillating systems

or

2 02

0(~)

Dx --

0

c)r2 0

{x)

DY--oF2

k2B - k4 k3

I k4)(~)

_Jr_

-k2g

=L

{;)

-k3 U

The spatial part of Eq. (12.102) is the eigenvalue problem of the Laplacian

V2O(r) = _er2{~)

(12.103)

where ® denote the eigenfunctions and - e r denote the eigenvalues, which depend on the boundary conditions: er = m r r / L and ® = sin err where m is any integer. The solution of linear Eq. (12.102) admits the form

(12.104)

z = c®(r)exp(wt)

where c denotes the difference between the state variables, and oo is the eigenvalue of the linearized operator L. The corresponding characteristic equation becomes

k3 ( kl ]2 A 2

( k 2 B - k 4 ) - (Dxer 2 + a))

-o

(12.105)

k ]2 12 -(DY er2 ff74)
-k2B

or with kl = k2 = k3 = k4 = 1, we have w 2 - [ B - A 2 _ 1 - (D x + Dy ) 0 -2 ] 09 -Jr- A 2 + [_(B - 1)Dy 0 -2 q- A 2Dx 0.2 ] _jr_Dx Dy 0 -4 If the eigenvalues of Eq. (12.105) are real and at least one of them is positive due to the controlled parameters A, B, and the diffusion coefficients, only the spatial patterns of the Turing structure of sin err will arise without temporal oscillations. If, on the other hand, the eigenvalues are a complex conjugate pair (,Oreal-+-i(.Oirn,then the solutions represent spatiotemporal instabilities, and lead to propagating waves, in which e '°"ira' represents the oscillations in time. If the real part ((.Oreal)is positive, the perturbation grows. The corresponding imaginary part is obtained from the determinant of Eq. (12.105) (ierim) 2 -- A 2 + [A2D× - ( B c - 1)Dy ]

(12.106)

+ DxD Y

For the stability, we must have

det - (k2B - k 4 - o-2 D x ) ( - k 3X 2 - o-2Dy ) + k z k g B , ~ )

klA]

> 0

(12.107)

and tr = k2 B - k 4

- A2Dx - k3X2 - A2Dy < 0

(12.108)

Since the det is the product of the eigenvalues, if both eigenvalues are negative, then det > 0 and tr < 0, and hence the system will be stable. However, for some variation of B, if one of the eigenvalues becomes positive, then det < 0 and the system becomes unstable, and the propagating wave or the Turing structure occurs when the value of B satisfies the following condition: 1

(

B-->~(k 4 k 2 +er2Dx) 1+

k3(klA/k4)2 I er2DY

(12.109)

622

12.

Stability analysis

By plotting the fight of Eq. (12.109) against o2, a minimum value of B is obtained. At this minimum, stationary patterns with the wave number O'min arise. Eq. (12.109) corresponds to Turing bifurcation, in which stable, a spatially inhomogeneous pattern forms. Such spontaneous states are common in biological, chemical, and ecological systems. Wittenberg and Holmes (1997) reviewed the spatiotemporal complexity and richness occurring from the variation of the parameters A, B, Dx, and Dy independently, and complex behaviors, such as two-frequency spatiotemporal dynamics and chaos. A specific and very narrow parameter domain may play a role in initiating instability. Because of their extreme sensitivity to controlling parameter values, the relevance of reaction-diffusion systems to biological systems has been an issue. Linear stability analysis does not provide information on how a system will evolve when a state becomes unstable. It does not distinguish between metastable and stable states when multiple local states are possible for given boundary conditions. Boundary conditions affect the value of the Lyapunov functional, and cause changes between stable and metastable states, hence altering the relative stability. An unstable state corresponds to the saddle points of the functional and defines a barrier between the attractors. Approximate solutions of nonlinear evolution equations may help us to understand how the system will behave in time and space. 12.7.3

The Brusselator under Nonisothermal Conditions

Under nonisothermal conditions, Eqs. (12.98) and (12.99) become OX 0 . . . . (Jx)+klA-k2BX+k3X2y-k4 Ot Or OY Ot

(12.110)

X

0

- - - - ( J r ) + kzBX- k3xZY

(12.111)

Or

OT 0 - - - - ( J q ) + (-zXHr) kAB pCp Ot Or

(12.112)

where k is the overall reaction A + B ---,E + F rate constant. Using the coupled heat and mass flow relations given in Eqs. (9.56) and (9.57), we have 02T

~

0X _ Dx 02X+D Y 0 2 Y + D s q x ~ + k l A _ k 2 B X + k 3 x z Y _ k 4 Ot

Or2

Or2

X

Or2

0Y _ Dy 02Y+D x o z X + D s q Y ~02T +k2BX-k3XZY

(12 114)

~

Ot OT at

Or2

-

a

02T ~ Or2

Or2

Or2

DDxq 02X -t

pCp

Or2

(12 113)

DDYq 02y t

pCp

Or2

(-AHr) (k0e_E/Rr t -

-

)

pCp

AB

(12.115)

Here, DsqX and DsqY are the cross coefficients representing the temperature gradient-induced mass flows (thermal diffusion) of X and Y, respectively, and DDyq and DDXq are the cross coefficients representing the Dufour effects. Under steady-state conditions, the temperature is related to concentration by Eq. (9.18), we have 0-D x 1-

0-Dy

1-

(_AH r ) Dsqx

ke

2X + D r 02Y +klA Or2

(_AH r ) DsqY o2Y + D x O2X + kzBX

ke

Or2

kzBX+k3X2y

Or2

Or2

k3X2y

k4X

(12.116)

(12.117)

These equations display the spatial order with the thermodynamically coupled heat and mass flows. Here, the coupling between chemical reactions and transport processes of heat and mass is excluded. The analysis of reaction-diffusion systems would be more complete if we consider heat effects and coupling among fluxes of mass and heat. The nonequilibrium thermodynamics approach is useful for incorporating the coupling phenomena into reaction-diffusion systems.

12. 7

623

Oscillating systems

Example 12.9 Chemical instability Consider the following set of reactions" S=X 2X + Y = 3X B+X=Y+D X=P

(a)

Here, the initial and final concentrations of S, B, D, and P are maintained constant so that only the concentrations of X and Y are the independent variables. The autocatalytic step 2X + Y - 3X involves a trimolecular reaction. The overall reaction in Eq. (a) is S+B-E+P We have two distinct reactions between the reactant R and the final product P S~P,

B--,D

The equilibrium concentrations of X and Y are klfS Xeq -- ~ , klb

Yeq =

k2bklfS klbk2f

and the ratios of constant concentrations are

P_ klfk4f

D _ k2fk3f

S

]~

klbk4b '

k2bk3b

Assuming that all the forward reaction rate constants equal unity (ki,f- 1), and the backward reaction rate constants equal k (ki,b- k), we have the following kinetic relations: dX

- S + X2y - BX- X + k(YD+ P-X-X

3)

(b)

dt dY

- BX + X 2 y + k (X 3 - YD)

dt

(c)

Steady-state concentrations are

X~.-

S -nt-kP

l+k

,

kX 2 + r D

Y , - -S - X , X 2 +kD

(d)

where r - B/D. Since the reaction steps S - X and X - P cannot compromise the stability or the asymptotic stability, we may then assume the values of S and P obtained from the law of mass action, for example, we have S - kZE Therefore, Eq. (d) becomes S(S 2 + krD)

S

x ,', - k ,

r•, -

s2 +k3D

Also, the linearized perturbation equations around a steady-state yield the following dispersion relation: A2 + [ X 2 + r D + l -

2X,.Y, + k ( 3 X [ + D + I ) ] A + X 2 + k ( X 2 + D ) - O

(e)

624

12.

Stability analysis

In this equation, the values of r as coefficients of A vanish and correspond to a transition point. Beyond this point, the real part Ar of the roots A 1 and ~2 changes its sign and hence the system becomes unstable, and at the marginal state, we have A1 + A2 = 0. Using Eq. (e), the marginal stability condition becomes

I

I

k k3D + k2S + (1 + k) 282 + k 2 + rc = 82 -- k3D

s2s2+k2]} k3D

(f)

Or in terms of the critical affinity, we have the marginal stability condition Ac = RT

ln~

1

(g)

k2rc

As a consequence of Eq. (f), rc > 0 implies that 82 ~>D>0 k3

For D = 0 or D = ~, rc approaches to infinity, while rc reaches its minimum value at {82[2(82 + k 2 ) ( 2 8 2 +k2)]1/2 - 82(82 + k 2 ) } k 3 (382 - k 2 )

When the affinity is smaller than the critical affinity, the steady-states lie along the thermodynamic branch (Glansdorff and Prigogine, 1971). If we assume that backward reaction rate constants are all zero (corresponding to the limiting case of IAI ~ ~), then Eqs. (b) and (c) become dX

- S+X2y-BX-X

dt

dY

- BX + X 2 y

dt

The characteristic equation is A2 + ( X 2 + B + I - 2 X ~ Y ~ ) A + X 2 = 0

(h)

The steady-state values with k - 0 are

x,=s,

Y,-

rD

B

S

S

Then the values ofeigenvalues show that the solution ofEq. (g) becomes unstable when B > Bc with Bc = 1 + S 2.

Example 12.10 Multiple steady states Multiple steady states and dissipative structures may play an important role in nerve excitations. Consider the following simple set of reactions" S + X ~ 2X X + E ,-~--C C,-~- E + B The overall reaction is S~B

(a)

12. 7 Oscillating systems

625

In Eq. (a), there is the autocatalytic production of X and its enzymatic consumption. The total amount of enzyme is constant and, we have E0 -constant

E + C ~--

Assuming that all the kinetic rate constants are equal to one, the kinetic equations are dX

-

SX - X 2 - XE + C

dt dE

(b) -

-XE-

BE + 2C

dt With E0, the steady-state equations are f(X)-- X 3 +(2+B-S)X

2 +[E0 - S ( 2 + B ) ] X - B E

E-

0 =0

2E°

(c) (d)

X+B+2

With B = 0.2 and E0 = 30.0, Figure 12.4 shows the (X,S) space where multiple steady states occur. Only the upper and lower values of X are stabile, while intermediate values are unstable. It would be interesting to know when the system will jump from one branch to another. This stability problem may be controlled by the direction and magnitude of fluctuations. Figure 12.4 is obtained from the following MATHEMATICA code: (*Multiple steady states*) b = 0.2;eo = 30.0;

solx = Solve [(x A3) + (2,0+b - s)* (x A2) + (eo - s* (2 + b))*x- b*eo = = 0, x]; Plot [Evaluate[x/,solx], {s, 8,3, 8,65}, Frame->True, GridLines->Automatic, PlotStyle->{Thickness [0,008]}, FrameStyle->Thickness [0,004], FrameLabel->{"S", "X"}, RotateLabel->True, DefaultFont- > {' 'Times-Roman", t 2 }]:

12.7.4

Belousov-Zhabotinsky Reaction Scheme

Sometimes, even spatially homogeneous chemical systems can cause bistability and show complex behavior in time. For example, autocatalysis may occur due to the particular molecular structure and reactivity of certain constituents, and reactions may evolve to new states by amplifying or repressing the effect of a slight concentration perturbation.

3.5

X 2.5

1.5

,

8.3

,

,

i

.

8.35

.

.

.

~

.

8.4

.

.

.

,

.

.

.

.

8.45

,

.

8.5

.

.

.

,

.

8.55

.

.

.

.

.

.

.

.

.

8.6

S Figure 12.4. Multiple steady states in Example 12.10.

8.65

626

12. Stability analysis 0.0002 0.0002 0.00015

0.00015

×

0.0001

0.0001

0.00005

0.00005

500 (a)

1000 t

1500

2000

500

1000

(b)

1500

2000

t

F i g u r e 12.5. C h a n g e of composition of X with time, kl = 1.28, k2 = 8.0, k3 = 8 × 105, k4 = 2 × 103, k 5 -- 1.0, H = 0.06, B = 0.02.

(a) f - 0.51319711709999 and (b) f= 0.5131971170999999.

The Belousov-Zhabotinsky reaction system is one example leading to such chemical oscillations. One of the interesting phenomena is the effect of the very narrow range of controlling parameter ~ on the stability of the BelousovZhabotinsky reaction system. The following reactions represent the Belousov-Zhabotinsky reaction scheme: H+Y

k, >X+P

H+X X +Y

k~ >2X+2Z 'k3 >2P

2X

(12.118)

k4 >H+P

The evolution equations for the Belousov-Zhabotinsky system are dX

- klHY+k2HX-k3XY-2k4

X2

dt

dY--klHY-dt k3XY + ( f ) kSBZ dZ dt

(12.119)

(12.120)

- 2kzHX- ksBZ

(12.121)

Figure 12.5 shows the effect of the kinetic and controlling parameterfon the evolution of concentration of X estimated from Eqs. (12.119) to (12.121).

Example 12.11 Reaction-diffusion model The linear stability analysis (Zhu and Li, 2002) may be used to investigate the evolution of a reaction-diffusion model of solid-phase combustion (Feng et al., 1996). The diffusion coefficients of the oxygen and magnesium (g) are the two controlling parameters besides kinetics Mg(s) + 02 bMg(g) + 0 2

kl >MgO(s) + aMg(g)

k2 >MgO(s) + cMg(g)

Mg(g) + 02env

k3 >MgO(s)

The flows of Mg and 02 are"

Mg(g) k4

> MgO(g)env

and 0

2 ( k5

02env

12.7

by assuming that X denotes

02, and Y

627

Oscillatingsystems

denotes Mg(g), the mass-action law yields

dX

k2XY b + ks(X 0 - X)

-- -klX-

dt

dY

- a k l X + ( c - b) k2XY b - k3Y + k 4 (Yo - Y)

dt

For the combustion, the linearized equations become dx - - w x dr

x y b -~ 1 - x to

dy _ a w x + ( c _ b ) x y b - uy-~ Yo - Y t1 dr

where X --

X

Y

k2xbt,

w--

k,

Xo ' y = ~-0-0, r =

k2 Xb

k3

k2xb, u-k2xb

, t o --

ks

k2Xob , t,-

With the following numerical values: a = 1, b = 2, c - 3, w = 1/650, v = 1/20, and Y0 = 0.006 (Feng et al., 1996), the linearized equations reduce to ----

x-- xy 2 + ~ to

dr

+ -- (6+o)x + x y 2 --

y+

0006,

dr

t~

From the eigenvalue problem for these equations, the two controlling parameters to and t~ are obtained. As the parameters to and t 1 contain kinetics and transport coefficients, they represent a combined effect, and make the study more interesting and complex.

Example 12.12 Adiabatic stirred flow reactor Consider the following reaction: A.

k, "B k, _

The reaction occurs in an adiabatic stirred flow reactor with feed flow rate F, transient compositions CA, and CB and reaction rate Jr, and total mass of reacting mixtures M. For small perturbations around the stationary state(s), the following expansions are used: CA (t) -" CA,s -Jr-~CA (t), CB (t) = CB,s Jr- ~CB (t), T ( t ) = T, + 6 T ( t ) Jr (CA, CB, T) -- Jrs (CA,s,CB,s,Ts) + ~Jr (CA,CB, T ) to find the differential model equations of heat and mass balances d(~CA) . . . . dt

d(,~CB) dt

F M

6c A - 6 J r

F - - - ~ a C B + a Jr M

628

12.

Stability analysis

d ( 6 T ) - - F - - - 6 T + Q ~Jr dt M Cp In stable systems, such disturbances vanish in time and the stationary values are restored. For very small perturbations, the reaction rate disturbance may be expanded with negligible second order and higher terms as follows:

~Jr : ( OJr ] ~CA"k-(OJr ) OJr) OCA)s ~~CB)s aCBq-( OT .)s ~r This expansion may lead to the linearization of differential equations above.

PROBLEMS 12.1

Using the truncated virial equation of state with the second virial coefficient B(T)

P V - 1 + ~(TB,________~ RT V Obtain the thermodynamic stability condition based on the constraint on B(T). 12.2

If we have a fluid in a closed system at constant entropy and pressure, prove that the stability condition of the fluid is Cp > O.

12.3

Using the truncated virial equation of state

PV

B C -1-+----+-~ RT V V2

and the constant-volume heat capacity C v = a + b T. For what values of B and C this fluid undergo a vapor-liquid phase transition? 12.4

Solve the following initial value problem as an eigenvalue problem, and prepare a state-space plot.

dXl__ - 9 x 1 + 4x 2 dt

dx2_

dt X 1 (0)

12.5

_ _ 2x 1 + 2x 2

--

1; x 2 (0) = - 1

Solve the following initial value problem as an eigenvalue problem, and prepare a state-space plot of xl versus x2 and x2 versus x3.

dXl -

8x 1 - 5x 2 + 10x 3

dt

dx2 _ 2 x 1 + x 2 + 2 x 3 dt

dx3_

dt

_ - 4 x 1 + 4x 2 - 6x 3

The initial conditions are x 1( 0 ) = x 2 ( 0 ) = 2; x 3 ( 0 ) - - 3

References

12.6

629

We have a first-order homogeneous reaction, taking place in an ideal stirred tank reactor. The volume of the reactor is 20 × 10-3 m 3. The reaction takes place in the liquid phase. The concentration of the reactant in the feed flow is 3.1 kmol/m 3 and the volumetric flow rate of the feed is 58 × 10-6 m3/s. The density and specific heat of the reaction mixture are constant at 1000 kg/m 3 and 4.184 kJ/(kg K). The reactor operates at adiabatic conditions. If the feed flow is at 298 K, investigate the possibility of multiple solutions for conversion at various temperatures in the product stream. The heat of reaction and the rate of reaction are 2U-/,. - -2.1 × 108 J/kmol Jr = k C -

4.5 × 106 C (kmol/m 2) exp[(-62800/(RT)] kmol/(m 3 s)

REFERENCES G. Captained, lnt. J. Heat Mass Transfer, 46 (2003) 3927. Y. Demirel, Int. J. Thermodynamics, 8 (2005) 1. C.-G. Feng, Q.-X. Zeng, L.-Q. Wang and X. Fang, J. Chem. Soc. Faraday Trans., 92 (1996) 2971. H. Gecegormez and Y. Demirel, Fluid Phase Equilibria, 237 (2005) 48. E Glansdorff and I. Prigogine, Thermodvnamic Theoo, oj'Structure, Stability and Fluctuations, Wiley, New York ( 1971 ). W Horsthemke and EK. Moore, J. Phys. Chem. A., 108 (2004) 2225. G. Izfis, R. Deza, O. Ramirez, H.S. Wio, D.H. Zanette and C. Borzi, Phys. Rev. E, 52 (1995) 129. D. Kondepudi and I. Prigogine, Modern Thermodynamics, Firm Heat Engines to Dissipative Structures, Wiley, New York (1999). W.B. Li, K.J. Zhang, J.V. Sengers and R.W. Gammon, J. Chem. Phys., 112 (2000) 9139. G. Nicolis and I. Prigogine, Self-Organization in Nonequilibrium Systems, Wiley, New York (1977). G. Nicolis and I. Prigogine, Exploring Complexity, Freeman & Company, New York (1989). J.M. Ortiz de Zarate, R.P. Cordon and J.V. Sengers, Phys. A. 291 (2001 ) 113. M. Tsuchiya and J. Ross, Proc. Natl. Acad. ScL, 100 (2003) 9691. A. Turing, Phil. Trans. R. Soc. B, 237 (1952) 37. M.O. Vlad, A. Arkin and J. Ross, Proc. Natl. Acad. Sci., 101 (2004) 7223. L. Yang and I.R. Epstein, Phys. Re~: Lett., 90 (2003) 178303-1. R. Zhu and Q.S. Li, Theot: Chem. Acc., 107 (2002) 357.

REFERENCES FOR FURTHER READING I.D. Epstein, An Introduction to Nonlinear Chemical Dynamics: Oscillations, Waves, Patterns, and Chaos, Oxford University Press, Oxford, 1998. I.R. Epstein and V.K. Vanag, Chaos, 15 (2005) 047510-1. A. Goldbeater, Biochemical Oscillations and Cellular Rhythms: The Molecular Bases of Periodic and Chaotic Behavior, Cambridge University Press, Cambridge 1996. I.A. Halatchew and J.P. Denier, Int. J. Heat Mass Transfer, 46 (2003) 3881. T. Kolokolnikov, T. Erneux and J. Wei, Phys. D, 214 (2006) 63. I.S. Kovacs, Nonlinear Anal., 59 (2004) 567. M. Mincheva and D. Siegel, Nonlinear Anal., 56 (2004) 1105. M.G. Newbert, H.Caswell and J.D. Murray, Math. Biosci.. 175 (2002) 1. J.M. Ortiz de Zarate and J.V. Sengers, J Stat. Phys., 115 (2004) 1341. S. Petrovski, B.-L. Li and H. Malchow, Bull. Math. Biol., 65 (2003) 425. I. Schreiber, E Hasal and M. Marek, Chaos, 9 (1999) 43. R. Sureshkumar, J. Non Newtonian Fhtid Mech., 97 (2001) 125. M. Vendruscolo, Trends Biotechnol., 20 (2002) 1. R. Wittenberg and P. Holmes, Phvs. D, 100 ( 1<)97) 1.

13 ORGANIZED STRUCTURES 13.1

INTRODUCTION

In physics, irreversibility and dissipation are interpreted as the loss of available energy and hence randomness, while biological evolution is associated with increased complexity and organization in time and space. In principle, we observe the destruction of structure at global equilibrium, and the creation of structure far from global equilibrium. Hence, the distance from global equilibrium appears as a new organizing factor, similar to the lowering of temperature, both of which can induce long-range correlations leading to phase transitions or self-organization phenomena. Some well-known autocatalytic reaction systems are periodic in time and sometime are called biological clocks. Turing's instabilities, published in 1952, suggested that the morphogens transported by diffusion and other mechanisms from a production site may cause the patterning of biological host tissues during the growth phase. When the uniformly distributed morphogens are disturbed and become unstable, the system may evolve to a stable nonuniform steady state from which may be observed the patterns of growth, differentiation of tissue, and pigmentation. Some reaction-diffusion systems represent many biological signaling processes and pharmacokinetic applications, such as growth promoter/inhibitor factors, extravascular drug delivery, and polymeric controlled-release drug codelivery design. The modeling of spatiotemporal evolution may serve as a powerful complementary tool to experimental medical analyses for assessing disease states and drug efficacy, enzyme systems, and two interacting biological cells. Spatial gradients may facilitate self-organization in time and space, and complicated kinetics, irreversible reactions, and different diffusivity coefficients are not prerequisite for this. Even simple reaction-diffusion systems with unequal diffusions, flows, or autocatalytic systems may lead to multiple solutions and organized structures at stable and metastable states. Turing instabilities, which lead to pattern formations demonstrated that transport processes such as diffusion, which ordinarily leads to uniform concentration in time, may cause unexpected differentiation and organization of species. Some examples of self-organization of diverse phenomena in biology are: (i) Spontaneous folding of proteins and other biomacromolecules, (ii) Formation of lipid bilayer membranes, (iii) Homeostasis {the self-maintaining nature of systems from the cell to the whole organism), (iv) Morphogenesis, or how the living organism develops and grows, and (v) Flocking behavior (the formation of flocks by birds, schools of fish, etc.). This chapter briefly discusses some of the structured states in physical, chemical, and biological systems. 13.2

EQUILIBRIUM AND NONEQUILIBRIUM STRUCTURES

With the Carnot-Clausius principle the second law of thermodynamics is interpreted as a law describing the evolution of the continuous disorganization of a system leading to the disappearance of the structure introduced by the initial conditions. On the other hand, biology is closely associated with an increase of organization leading to high level and complex structures. For example, if we mix two different liquids, diffusion occurs with a system's progressive forgetting of its original state and increases entropy. In biological system, however, heterogeneity is required. Nonequilibrium conditions are maintained by chemical cycles and chemical pumps, which lead to coherent behavior. For these two types of evolution, the same physical principles are still valid. These different types of evolution are the results of different thermodynamic situations that are near and far from global equilibrium (Glansdorff and Prigogine, 1971). Mainly, disorganization and the loss of structure occur in the vicinity of thermodynamic equilibrium. On the other hand, order may occur beyond the, stability limit of the thermodynamic branch that is in the far from equilibrium region under specific nonlinear kinetics and transport processes. The type of solutions changes drastically, since the system behaves quite differently against the perturbations and fluctuations before and after the instability limit. The second law of thermodynamics remains valid for both situations. The existence of order in biological systems may

632

13.

Organized structures

be maintained under nonequilibrium conditions and far from equilibrium. The occurrence of dissipative structures depends mainly on the boundary conditions. The boundary conditions of living systems are far less arbitrary and constitute the specific domain of space-time organization characteristics of dissipative structures. Equilibrium structures may be formed and maintained through reversible transformations. A crystal is an example of an equilibrium structure. In equilibrium state, all parts possess the same physical properties, both locally and globally. However, dissipative structures are formed and maintained through the continuous exchange of matter and energy between the system and its surroundings in nonequilibrium conditions. B6nard convection cells and living organisms are some examples of dissipative structures. Irreversible processes may promote disorder at near equilibrium, and promote order at far from equilibrium known as the nonlinear region. For systems at far from global equilibrium, flows are no longer linear functions of the forces, and there are no general extremum principles to predict the final state. Chemical reactions may reach the nonlinear region easily, since the affinities of such systems are in the range of 10-100 kJ/mol. However, transport processes mainly take place in the linear region of the thermodynamic branch.

13.2.1

Self-Assembly and Self-Organization

Living cells display internal structures that are controlled and regulated dynamically These structures may occur at various lengths and scales. Some examples of such structures are the organization of nerve cells onto axon, body, and dendrites, and the appearance of short-lived signaling patches in the membrane. The structures occur mainly because of an uneven distribution of biochemical molecules as a result of spontaneous symmetry breaking through local fluctuations. Self-assembly and self-organization may contribute toward the variety of time and length scales in cellular structures. Self-assembly refers to spatial structuring caused by the minimization of the free energy in a closed system. In principal, therefore, a self-assembled structure corresponds to a thermodynamic equilibrium. One of the examples of such phenomena is the phase separation of lipids and proteins due to macromolecular interactions in which the interaction energy dominates the entropy contribution. Liquid crystalline ordering and molecular association are other examples of self-assembly. On the other hand, self-organization occurs when an open system is far from equilibrium. For example, rotating spiral waves and patterns form in some reaction-diffusion systems. A self-organized system can be maintained under nonequilibrium conditions by exchanging energy and matter between itself and its environment. The dissipation of chemical energy provides the thermodynamic force for self-organization. Experimental patterns of calcium waves and protein distributions are modeled satisfactorily by reaction-diffusion equations (Falcke, 2004; John and B/Jr, 2005).

13.3

BIFURCATION

The stability of transport and rate systems is studied either by nonequilibrium thermodynamics or by conventional rate theory. In the latter, the analysis is based on Poincare's variational equations and Lyapunov functions. We may investigate the stability of a steady state by analyzing the response of a reaction system to small disturbances around the stationary state variables. The disturbed quantities are replaced by linear combinations of their undisturbed stationary values. In nonequilibrium thermodynamics theory, the stability of stationary states is associated with Progogine's principle of minimum entropy production. Stable states are characterized by the lowest value of the entropy production in irreversible processes. The applicability of Prigogine's principle of minimum entropy production is restricted to stationary states close to global thermodynamic equilibrium. It is not applicable to the stability of continuous reaction systems involving stable and unstable steady states far from global equilibrium. The steady-state deviation of entropy production serves as a Lyapunov function. When a system is sufficiently far from equilibrium, it may arrive at a bifurcation of states and move to ordered structures. Transitions between different modes of dynamic organizations are called bifurcation. If the system continues to move away from equilibrium, the structures become more complex, leading to a chaotic situation in the macroscopic sense. Some regularity may involve in such chaotic behavior. To study the behavior of systems far from equilibrium, a new interdisciplinary field called synergetics was developed. Synergetics is concerned with the cooperation of individual parts of the system that produce macroscopic spatial and temporal structures, which are mainly dissipative. Let us define the distance of a system from global equilibrium by parameter fi (e.g., a temperature or a concentration gradient). After a value of ill is reached, the system displays ordering characterized by a certain frequency or a wavelength. Figure 13.1 shows the bifurcation in the velocity in Bdrnard convection cells. If the parameter/3 is

13.4

633

Limit cycle

Right

(T2-Tj)~.,.I ~.~

T2-T1 Left

Figure 13.1. Bifurcation phenomena in Benard's cells. increased further, we may reach a value of/32 at which the system is characterized by two frequencies. This behavior can be repeated for the values of/33,/34, leading to increasingly complicated states. In 1975 Mitchell Feigenbaum described a general characteristic among the critical values of/3i given as

/3, -/3,_~ ~4.669201...

Limlarge" ]3.+1 -- fin

(13.1)

and Limlargen

~ 2.5029...

(13.2)

8n+l

where el is the separation between the two branches coming from the ( i - 1)th bifurcation. Therefore, the scheme of the bifurcation displays a series of universal characteristics independent of a specific system. These universal numbers were observed for example, in the transition of laminar flow to turbulent flow, and in chemical and electrical systems. These Feigenbaum numbers characterize the order-chaos transition, and reveal a certain regularity in chaos or in one of the series of processes that leads to chaos. Bifurcation and chaos can be expressed in finite difference equations of the form (13.3)

xi+ 1 = f ( x i )

If we assume that the functionfis a quadratic function, we have xi_ 1 = axi(1-

xi)

(0 < a < 4)

(13.4)

As parameters change in finite difference equations, bifurcation (e.g., changes in qualitative dynamics) is found. At steady state, Xs is a value for which x~ =f(xs). For Eq. (13.4) there are two steady states: x~ = 0 and x~ = (a - 1)/a. A cycle of period m is defined by xi+ , = x i

and

xi= j 4: x i

forj = 1,...,m- 1

(13.5)

The stability of steady states and cycles implies the restoration of steady states and cycles, respectively, following a small perturbation. One type of bifurcation is period-doubling bifurcation, in which a stable cycle of period n becomes unstable and a new stable cycle of period 2n is generated with a new parameter. Equation (13.4), for example, produces successive period-doubling as a increases. For 3 < a < 3.57, stable cycles of lengths 1, 2, 4, 8, 16, 32, and 64 are generated. Bifurcation, instability, multiple solutions, and symmetry-breaking states are all related to each other. Chemical cycles in living systems show asymmetry. The bifurcation of a solution indicates its instability, which is a general property of the solutions to nonlinear equations.

13.4

LIMIT CYCLE

The theory of nonlinear oscillations can describe the periodic solution that appears beyond the instability of the steady state. Stable states exist before the instability. The perturbations correspond to complex values of the normal mode frequencies and spiral toward the steady state to a focus. As soon as the steady state becomes unstable, a stable periodic

634

13.

Organized structures

process called the limit cycle occurs. This behavior is independent of the initial conditions, and the system approaches in time the same periodic solution determined by the nonlinear differential equations. The periodic solution is characterized by its period and amplitude. The limit cycle is unique and stable with respect to small fluctuations. For a chemical reaction system, the characteristics of the periodic solutions are uniquely determined by the kinetic constants as well as by the concentrations of the reactants and final products. Starting from the neighborhood of steady state as an initial condition, the system asymptotically attains a closed orbit or limit cycle. Therefore, for long times, the concentrations sustain periodic undamped oscillations. The characteristics of these oscillations are independent of the initial conditions, and the system always approaches the same asymptotic trajectory. Generally, the further a system is in the unstable region, the faster it approaches the limit cycle.

13.5

ORDER IN PHYSICAL STRUCTURES

There are two types of macroscopic structures: equilibrium and dissipative ones. A perfect crystal, for example, represents an equilibrium structure, which is stable and does not exchange matter and energy with the environment. On the other hand, dissipative structures maintain their state by exchanging energy and matter constantly with environment. This continuous interaction enables the system to establish an ordered structure with lower entropy than that of equilibrium structure. For some time, it is believed that thermodynamics precludes the appearance of dissipative structures, such as spontaneous rhythms. However, thermodynamics can describe the possible state of a structure through the study of instabilities in nonequilibrium stationary states.

13.5.1

Order in Convection

If a system is far from equilibrium, then a dissipative structure associated with the initiation of macroscopic organization such as a motion can appear. The kinetic energy of the motion accounts for the lower entropy of the system relative to the equilibrium value

(13.6)

AS = --Z~k

One of the best-known physical ordering phenomena is the Benard cells, which occur during the heating a fluid held between two parallel horizontal plates separated by a small distance. The lower plate is heated, and the temperature is controlled. The upper plate is kept at a constant temperature. When the temperature difference between the two plates reaches a certain critical value, the elevating effect of expansion predominates, and the fluid starts to move in a structured way; the fluid is divided into horizontal cylindrical convection cells, in which the fluid rotates in a vertical plane. At the lower hot plate, the hot fluid rises; later, it is cooled at the upper plate, and its density increases again; this induces a movement downward, as seen in Figure 13.2. The Benard cells are one of the best-known physical examples of spontaneous structurization as a result of sufficient distance from equilibrium, which is the large temperature difference between the plates. The critical temperature difference (T2 - T1)c can be determined from the dimensionless Rayleigh number Ra

Ra

gH30l( T 2 -

=

T 1)c

(13.7)

17k

Thermal flow

1'

1'

1'

1'

1'

1'

1'

Heat

Figure 13.2.

Thermal flow in Benard's cells.

13.5 Order in physical structures

635

where g is the acceleration of gravity, r/and k the viscosity and the thermal conductivity of the fluid, respectively, a the thermal expansion coefficient, and H the distance between the plates. The critical value of Rayleigh number is Rac = 1707. Therefore, the critical temperature difference is given as 1707~/k (~ - ~ )c

(gH3a)

(13.8)

Here, the dissipative coefficients r/or k and the expansion coefficient a represent two opposing effects in the B6nard phenomenon. The stability occurs as the result of a competition between the stabilizing dissipative effects and the destabilizing convective effects. In the :range of the small Rayleigh numbers, dissipative effects caused by the temperature fluctuations are dominant, while the velocity fluctuations become more effective at higher values of the Rayleigh number. This transition leads to the appearance of a dissipative structure; at the critical point, the system uses part of its thermal energy to create the kinetic energy necessary to maintain the macroscopic stationary cellular motion. This structure needs a continuous supply of energy, and disappears as soon as the heating stops. According to the hydrodynamics analysis, the approximate velocity distribution in the B6nards cells is given by

v(x)

--

C

COS

(13.9)

where h is the repetition length of the horizontal cells, and C a constant. If we plot the velocity versus the temperature difference, we have the bifurcation phenomena as seen in Figure 13.1. When the temperature difference is above a critical level, the resting fluid becomes unstable and it rotates in two structural states: one rotating toward the right and the other rotating toward the left as seen in Figure 13.2.

E x a m p l e 13.1 L o r e n z e q u a t i o n s : T h e s t r a n g e a t t r a c t o r The Lorenz equations (published in 1963 by Edward N.

Lorenz a meteorologist and mathematician) are derived to model some of the unpredictable behavior of weather. The Lorenz equations represent the convective motion of fluid cell that is warmed from below and cooled from above. Later, the Lorenz equations were used in studies of lasers and batteries. For certain settings and initial conditions, Lorenz found that the trajectories of such a system never settle down to a fixed point, never approach a stable limit cycle, yet never diverge to infinity. Attractors in these systems are well-known strange attractors. The Lorenz equations may produce deterministic chaos because we know how it will instantaneously change. However, for high enough Rayleigh numbers, the system becomes chaotic. Small changes in the initial conditions can lead to very different behavior after long time interval, since the small differences grow nonlinearly with feedback over time (known as the Butterfly effect). These equations are fairly well behaved, and the overall patterns repeat in a quasi-periodic fashion. Let us solve the Lorenz equations below with MATLAB between t = 0 and 20 and prepare plots of yl versus t, and a state-space representation of y2 versus Yl, and Y3 versus Y2 by using two different sets of initial conditions: y l ( 0 ) = y 2 ( 0 ) = y 3 ( 0 ) = 5 . 0 and y l ( 0 ) -- y 2 ( 0 ) -- y 3 ( 0 ) = 5 . 0 . dYl

dt

_ _ 1 0 y 1 + 10y 2

_

dy2 - 28yl - Y2 - YlY3 dt dy3_ - -2.666667y3 + YlY2 dt

(13.10)

Figure 13.3 shows the solutions to the set of Eq. (13.10) with the initial conditions yl(0)--y2(0)--y3(0)= 1.0 with the time span 0 -< t -< 20 using the following m-file function lorenz(time) tspan= [0:0.01 :time] ;yo= [5.0;5.0;5.0]; [t,y] =ode4 5 (@f,tspan,yo); figure; plot(t,(y (:,1)), xlabel ('t'),ylabel('yl')

636

13.

Organized structures

20 15 10 5 :g,

0

-10 -15 -20

' 0

'

5

10 t

(a)

15

20

30 20 10 >,,

0 -10 -20 -30 -20

. -15

.

. -10

. -5

(b)

.

. 0

. 5

10

15

20

Yl 30 20 10

~,

0 -10 -20 -30 0

10

(c)

20

30

40

50

Y3

Figure 13.3. (a) Plot of Yl versus time, (b) state-space plot yl versus Y2, (c) state-space plot Y2 versus Y3 obtained from Lorenz equations with the initial conditions: y~(O) = Y2(O) = Y3(O) = 1.0.

figure; plot(y(:, 1),y(:,2)),xlabel ('yl '),ylabel('y2') figure; plot(y(:,3),y(:, 2)),xlabel ('y3'),ylabel('y2') function dz = f(t,y) d z : [- 10. O'y( 1) + 10. O'y(2); 28. O'y( 1)-y(Z)-y( 1) *y(3); ...-2.666667*y(3)-y(1)*y(2)];

13.5

637......

Order in physical structures

2.5 2 1.5 1

0.5 :7,

0 -0.5 -1 -1.5 -2 -2.5

5

10

(a)

15

20

25

30

t

3

04

>,

0

-2

-3 -3 _

t

I

t

t

I

-2

-1

0

1

2

Yt

(b)

Figure 13.4. (a) Plot of Yl versus time, (b) state-space plot of Yl versus Y2 Van der Pol's equations with yl(O) = 0.1, y2(O), and a = 1.0.

Example 13.2 Van der Pol's equations Van der Pol's equations provide a valuable framework for studying the important features of oscillatory systems. It describes self-sustaining oscillations in which energy is fed into small oscillations and removed from large oscillations. Consider the following system of ordinary differential equations called Van der Pol's equations

dYl_ _ Y2 dt dy2_ - a (1 - y2) Y2 - Yl dt

m

(13.11)

The initial conditions are yl(0) = 0.1 and y2(0) =0.5, and a is constant. Figure 13.4 shows the solution for Van der Pol's equations (Eq. 13.11) with the time span 0 -< t-< 30 using a MATLAB code presented below. Figure 13.4a shows the periodic plots, while Figure 13.4b displays the limit cycle plots. function vdpol(time) tspan: [0:0.01 :time] ;yo: [0.1 ;0.5]; [t,y] =ode45(@f,tspan,yo); figure; plot(t, (y(:, 1))),xlabel('t'),ylabel('y 1') figure;

638

13.

Organized structures

plot(y(:, 1),y(:,2)),xlabel ('yl '),ylabel('y2') function dz - f(t,y) a=l.O' dz = [y(2); a* (1. O-y(1 )*y(1 ))*y(2)-y( 1)]'

13.6

ORDER IN CHEMICAL SYSTEMS

The dynamics of chemically reacting systems far from equilibrium allow them to self-organize in time and space. The entropy of the universe must increase during a spontaneous process, and the Gibbs free energy, G = H - TS, approaches a global minimum during a spontaneous chemical reaction. An oscillating reaction converts reactants to products and the products back to reactants, and hence it requires the free energy to decrease and then increase, which may appear to violate the second law. In fact, however, when a chemical reaction oscillates, it is far from equilibrium and cannot pass through its equilibrium point. For example, the concentrations of the intermediates in a chemical reaction can oscillate as the free energy monotonically decreases because of the conversion of reactants with high free energy to products with low free energy. Any decrease in entropy caused by the periodic concentration changes is compensated by an entropy increase from the other processes (Epstein et al., 2006). So, the oscillation of concentrations about the nonequilibrium value of the extent of reaction is consistent with the second law.

Example 13.3 The Brusselator system and oscillations One theoretical model displaying order in time is the Brusselator scheme, which was introduced in Section 12.7.1. In the limit of irreversible reactions the Brusselator scheme is A B+X 2X+Y X

kl >X k2 > Y + E k3 >3X k4 >F

When the concentrations of A and B are controlled, the kinetic equations for intermediate components Xand Y are

OX Ot

- k l A - kzBX + k3xZY - k4X

and

OY = k z B X - k3xZY Ot

(13.12)

These equations have the following stationary(s) solutions

X s=klA k4

and

Y~- k4k2 B k3k1 A

Equation (12.93) shows that Det > 0 as long as B > k4/k2 + (k3/kz)X2 and the system remains stable, otherwise instability arises, and the system undergoes a Hopfbifurcation at

k2 k2 The following MATHEMATICA code produces the oscillations in the concentrations X and Ywith the Brusselator scheme. With the parameters kl = 1.3, k2 = 1.0, k3 = 1.0, k4 = 1.0, and the concentrations A = 1.1, B = 3.0, we have the Hopfbifurcation point BH as given in Eq. (12.94)

k2

10+10(,

k2 --~--4)=1.---6

1.----6

1.0

=3.0449>3.0=B

Since B < BH, the oscillations are not sustainable and die out after some time. Figure 13.5a and b shows the oscillations and the limit cycle with these parameters.

13.6

3.5 3 2.5 >. 2 1.S 1

0.5 0

Order

in chemical

639

systems

3.5

A^

_

~ ............

/ !///7\ A A/'x r t j ~ _ i IIIkJ ~ k.i y w ~ w

2.5

"<,'-

-d + E l :

1.5 1

,

0

10

20

(a)

30

t

.

.

.

.

0.5

40

.

.

.

.

.

.

.

.

1

.

A

1.5

2

.

.

.

.

.

.

2.5

X

(b)

[

'A

3 >. >/2

.

'

I l lt_ 1 It~ 17 Itlltlh [ II IX IX II

b

L Y ' \,)' Yx' YW"~J' \ .,

0

....

,

....

10

(c)

i

20

t

....

i

30

. . . . .

.

0.5

40

(d)

1

1.5

.

.

.

.

.

2

.

.

.

.

.

2.5

.

.

.

.

.

.

.

3

.

3.5

X

Figure 13.5. The oscillations and the limit cycle obtained from the Brusselator scheme in Example 13.3: (a) X and Yversus time, where the bold line displays the concentration of Y, and (b)limit cycle with kl = 1.3, k2 = 1.0, k3 = 1.0, k4 = 1.0, A = 1.1, B = 3 . 0 (c) X and Y versus time, where the bold line displays the concentration of Y, and (d)limit cycle with kl = 1.0, k2 = 1.0, k3 = 1.0, k4-- 1.0, A = 1.0, B = 3.0.

With the parameters kl = 1.0, k2 = 1.0, k3 - 1.0, k4 = 1.0, A = 1.0, B = 3.0, we have the Hopfbifurcation point B H = 2.0 < B = 3.0 Since B > BH, the oscillations are sustainable. Figure 13.5c and d shows the oscillations and the limit cycle. Starting from the neighborhood of steady state as an initial condition, the system asymptotically attains a limit cycle in (X,Y) space. Therefore, for long times, X(t) and Y(t) exhibit periodic undamped oscillations, and the system always approaches the same asymptotic trajectory regardless of the initial conditions. The M A T H E M A T I C A code is: kl=l.3;k2=l.O; k3=l,O; k4=l,O; a=l.1; b=3.0; sol 1-NDSolve [{x' [t] - = k 1* a-k2*b*x[x] + k3* (x[t] ~ 2) *y [t]-k4*x[t] ,x[0] = - 1.0, y' It] = =k2*b*x[t]-k3*(x [t] A2)*y[t] ,y[0] - - 1.0},{x,y},{t,0,40}, MaxSteps-> 1000] Plot [Evaluate [{x It] }/.sol 1], {t, 0,40},Frame->True, GridLines->Automatic, PlotStyle-> {PointSize[0.03],Thickness[0.02]}, FrameStyle->Thickness[0.004], FrameLabel->{"t", "X"}, RotateLabel->True, DefaultFont-> {"Times-Roman", 12}]; Plot [Evaluate [{y[t] }/.sol 1], {t,0,40},Frarne->True, GridLines->Automatic, PlotStyle->{PointSize [0.03] ,Thickness [0.02] }, FrameStyle->Thickness[0.004] ,FrameLabel->{"t", "Y"}, RotateLabel->True, DefaultFont-> {"Times-Roman", 12}]; Plot [Evaluate [{x [t] ,y[t] }/.sol 1], {t, 0,40},Frame- >True, GridLines->Automatic, PlotStyle->{PointSize[0.02],Thickness[0.01 ]}, FrameStyle->Thickness[0.004],FrameLabel->{"t", "X,Y"}, RotateLabel->True, DefaultFont-> {"Times-Roman", 12}]; ParametricPlot [Evaluate [{x[t] ,y[t] }/.sol 1],{t,0,40}, PlotRange->All,Frame->True, GridLines- >Automatic, PlotStyle->{PointSize[0.03],Thickness[0.02] }, FrameStyle->Thickness[0.004],FrameLabel->{"X", "Y"}, RotateLabel->True, DefaultFont-> {"Times-Roman", 12 }];

640 13.6.1

13.

Organized structures

Limit Cycle in the Brusselator Model

In dissipative systems, we frequently encounter behavior called the limit cycle, in which, the system tends toward a certain orbit or an oscillation regardless of the initial conditions. If we assume the following perturbations X = X s + ceexp(At)

(13.13)

Y = Ys +/3exp(At)

(13.14)

and introduce these perturbations into Eq. (13.12), we obtain (13.15)

(13.16)

Here, we have disregarded the terms quadratic in ce and/3. For nonzero, initial perturbations, the determinant of Eqs. (13.15) and (13.16) should be equal to zero, which leads to the relation A 2 nt- (a 2 + 1 - b) A + a 2 = 0

(13.17)

where a = A(kl/k4)(k3/k4) 1/2 and b = B(k2/k4) If the solutions for Eq. (13.17) have a real and positive part, then the perturbations grow and make the system unstable. If the real parts of A1 and A2 are negative, then the perturbations decrease, and the system becomes stable. The solutions o r a l and A2 are given by A1,2 = _ ( 2 1 _ . ) ( a 2 + l _ b ) _ + _ [ ( a 2 + l _ b ) 2 4 a 2

]1/2

(13.18)

The system will be stable if b < (a 2 -+- 1), and it will be unstable if b > (a 2 + 1). When b - a 2 + 1, and A 1 and A2 are purely imaginary, we have an undamped oscillation similar to the Lotka-Volterra model. I f (a2+ 2a + 1 ) > b > (a 2 -- 2a + 1), and A1 and A2 have a nonzero imaginary part, then we have oscillatory behavior. These oscillations damp out in the stable z o n e : (a 2 -+- 2a + 1) > b > (a 2 + 1). When b > (a 2 -+- 2a + 1), limit cycle oscillations occur, which are independent of the initial values of the perturbations in X and Y. Figures 13.5b and d display the limit cycles obtained in Example 13.3 under the two set of parameters.

Example 13.4 Order in time and space with the Brusselator system The Brusselator model with unequal diffusion may produce order in time and space. When the concentrations of A and B are controlled, the one-dimensional approach to complex reaction-diffusion systems with the spatial coordinate r under isothermal conditions yields the kinetic equations for Xand Y(Eqs. (12.98) and (12.99)) OX

02X

Ot - k l A - kzBX + k 3 x Z Y - k4X + Dx Or2

OY Ot

- k 2 B X - k3xZY + Dy

02y Or2

where Dx and Dy are the respective diffusion coefficients. The boundary conditions are

( r3 r=-L

r=+L

=0

13.6

Order in chemical systems

641

For a specified value for A and some variation of B if one of the eigenvalues becomes positive, then Det < 0, and the system becomes unstable, and the propagating wave or the Turing structure occur

B >_ 1-L(k4 +o-2Dx) (l k3(klA/ Yk4)2 The following is a slightly modified version of the MATLAB demo program the BRUSSODE, and displays order in time and space (see Figure 13.6). The parameter N -> 2 is used to specify the number of grid points; the resulting system consists of 2N equations. The problem becomes increasingly stiff and increasingly sparse as Nis increased. The Jacobian for this problem is a sparse constant matrix (banded with bandwidth 5). The property 'Jpattern' is used to provide the solver with a sparse matrix of l's and O's showing the locations ofnonzeros in the Jacobian. By default, the stiff solvers of the ODE Suite generate Jacobians numerically as full matrices. However, when a sparsity pattern is provided, the solver uses it to generate the Jacobian numerically as a sparse matrix. Providing a sparsity pattern can significantly reduce the number of function evaluations required to generate the Jacobian and can accelerate integration. [Source: E. Hairer and G. Wanner,

Solving OrdinaryDifferentialEquations lI, Stiff and Differential-AlgebraicProblems, Springer-Verlag, Berlin, 1991 .] The Brusselator for N = 6 0

The Brusselator for N = 6 0

I

--"g

3d¢. " "

2.1.-" "

.-rI

".

-- ~

,

I

,

i

,--

~l ~ ~

......

- - , _

,

i

i

I i

,

1

~'1-~

I

.0

'"

6

i

. .~'~ "

.t

1

0.2 0

time

0

time

0

The Brussdator for N = 60 I"

/

~

i

0

space

space

The Brusselatcr ...~

~~.~

,

,...-- ""

,

i

for N = 60

~ ~ ....

I

i ~ "" ~

-

~

7

t

0

time

0

space

o

time

The Brusselator for N = 8 0 ~ -i" ..- ,-- " •-- ~-

t

T h e Brusselator for N = 6 0

_

i i

o

.~ "T" ~ .

'"i ~ ~ ~ i ~ ~ ~

"-

._8

""

6

"5 •-6

== -~

co

1

time

0

0

space

1

time

0

0

space

Figure 13.6. Brusselator reaction scheme and order in time and space produced by Brussode demo of matlab.

642

13.

Organized structures

function b r u s s o d e ( N ) if nargin< 1 N = 50; end tspan = [0; 30]; y0 = [1 +sin((2*pi/(N+ 1))*(I:N)); repmat(3,1,N)]; options = odeset('Vectorized','on','JPattern',jpattern(N)); [t,y] = ode 15s(@f, tspan,y0,options,N); u = y(:,l:2:end); x = (1 :N)/(N+ 1); figure; surf(x,t,u); view(-40,50); xlabel('space'); ylabel('time'); zlabel('solution u'); title(['The Brusselator for N = ' num2str(N)]); /0 .

.

.

.

.

.

function dydt - f(t,y,N) c - 0.000005 * (N+ 1)"2; dydt - zeros(2*N, size(y,2)); % preallocate dy/dt i=l; dydt(i,:) = 1 + y(i+ 1,:).*y(i,:)."2- 4*y(i,:) + c*(1-2*y(i,:)+y(i+2,:)); dydt(i+ 1,:) = 3*y(i,:) - y(i+ 1,:).*y(i,:)."2 + c*(3-2*y(i+ 1,:)+y(i+3,:)); % Evaluate the 2 c o m p o n e n t s of the function at all interior grid points. i = 3:2:2"N-3; dydt(i,:) = 1 + y(i+ 1,:).*y(i,:).A2- 4*y(i,:) + ' c* (y(i- 2, :)- 2*y(i, :) +y(i+ 2,:)); dydt(i+ 1,:) = 3*y(i,:) - y(i+ 1,:).*y(i,:).A2 + '.. c*(y(i- 1, :)-2*y(i+ 1,:)+y(i+3,:)); i - 2*N-l; dydt(i,:) = 1 + y(i+ 1,:).*y(i,:)."2 - 4*y(i,:) + c*(y(i-2,:)-2*y(i,:)+ 1); dydt(i+ 1,:) = 3*y(i,:) - y(i+ 1,:).*y(i,:)."2 + c*(y(i-l,:)-2*y(i+ 1,:)+3); % function S = jpattern(N) B = ones (2 *N, 5) ; B(2:2:2*N,2) = zeros(N, 1); B(l:2:2*N-1,4) = zeros(N, 1); S = spdiags(B,-2:2,2*N,2*N);

13.6.2

The Belousov-Zhabotinsky Reaction Scheme

One of the best-known oscillatory reactions is the Belousov-Zhabotinsky reaction scheme, which contains a set of oxidation-reduction steps; the oxidizing agent is bromate (BROW), the reducing agent is malonic acid [H2C(COOH)2], and cesium ions are used as the catalyst. The concentrations of Ce+3 and Ce +4 vary periodically with a frequency on the order of 0.01 Hz. The Belousov-Zhabotinsky reaction also organizes itself into bands. The Belousov-Zhabotinsky reaction is a chemical oscillatory reaction, and originally consisted of a one-electron redox catalyst, an organic substance that is easily brominated and oxidized, and a bromate ion is dissolved in acid. The typical catalyst ferroin, in its oxidized state, has a blue color, while in its reduced state, ferroin is red. As the Belousov-Zhabotinsky reaction alternates between the oxidized state, and the reduced state, the solution changes its color. The overall reaction is complex; however, its oscillatory effects can be understood by studying the following reaction steps (Field et al., 1972) Ce+3

BrO3' H+ >Ce+4

(oxidation)

(13 19)

13.6 Order in chemical systems

Ce+4

malonicacid

>Ce+3 (reduction)

643 (13.20)

At the start, the cycle begins with a certain amount of Ce+4 ions. The second reaction provides Br- ions, which inhibit the first reaction. This leads to an increase in concentration of Ce +3. After reaching a certain amount of Ce +3, the oxidation reaction starts, since little Ce+4 remains. The system can no longer produce sufficient Br- to inhibit the reaction, and Ce +3 decreases rapidly, producing Ce +4 until the cycle is completed. It is possible to maintain indefinite oscillations with constant frequency in a continuous flow stirred reactor into which bromate, malonic acid, and cerium catalyst are being supplied at a uniform rate. The Belousov-Zhabotinsky reaction scheme can also produce moving spatial inhomogeneties in unstirred solutions. Spatial waves develop as an oxidizing region advances into a region of low but finite bromide ion concentration that falls below a critical value. The autocatalytic production ofbromous acid at the interface advances the wave faster than the diffusion of any other molecules proceeds (Field et al., 1972). Nagy-Ungvarai and Hess (1991) used the electrochemical method to produce experimental data on the two-dimensional concentration profile of three variables in distributed Belousov-Zhabotinsky solutions. Biological systems might be oscillatory, which may be the direct result of biological evolution on earth. The earth moves around the sun and rotates on its own axis. These periodicities induce rhythms in the changes of temperature, light, humidity, and their effects are reflected in the physiology of living systems. The periodicity of day and night is reflected in the characteristics of living systems. Related to these rhythmic changes are the concepts of biological clocks and circadian rhythms for oscillations with a time span of approximately 24 h. The circadian rhythms are generated internally, since their periods are practically independent of environmental factors. Periodic self-oscillatory processes are characteristics of the processes of glycolsis (anaerobic catabolism of glucose) and in the conversion of ADP to ATP. Some reactions display oscillatory behavior in the transport of substrates across membranes, such as facilitated transport.

Example 13.5 The Belousov-Zhabotinsky reaction scheme Field et al. (1972) explained the qualitative behavior of the Belousov-Zhabotinsky reaction, using the principles of kinetics and thermodynamics. A simplified model with three variable concentrations producing all the essential features of the Belousov-Zhabotinsky reaction was published by Field and Noyes (1974). Some new models of Belousov-Zhabotinsky reaction scheme consist of as many as 22 reaction steps. With the defined symbols X = HBrO2, Y = Br-, Z = Ce 4+, B = organic, A = BrO3 (the rate constant contains H+), FKN Model (Field et al., 1972) consists of the following steps summarized by Kondepudi and Priogogine (1999): (i) Production of HBrO2 BrO 3 + BrA+Y

2H +

>HBrO 2 + HBrO k, > X + P

(13.21)

(ii) Autocatalytic production of HBrO2 BrO3 + HBrO2 A+X

H + ' Ce 3+

>2HBrO2 + 2Ce4+ k2 >2X+2Z

(13.22)

(iii) Consumption of HBrO2 HBrO 2 + Br-

H+

>2HBrO

2HBrO 2 -, BrO 3 + HBrO + H + X +Y k3 >2P 2X

xr4 > A + P

(13.23)

(13.24)

(iv) Oxidation of the organic reactants Ce4+ + ( 2 ) C H 2 ( C O O H ) 2 + BrCH(COOH)2--+(@) B r - + Ce3+ + Products

(13.25)

644

13.

Organized structures

The oxidation step is approximated with the reaction above. Concentration of the organic compounds (B) is assumed constant. Effective stoichiometry (f) is a variable, and the oscillations occur whenfvaries in the range 0.5-2.4. Representative kinetic equations of the Belousov-Zhabotinsky reaction scheme based on Eqs. (13.21)(13.25) are

dX - klAY + k z A X - k 3 X Y - 2k4 X2 dt dY - - k , A Y - k 3 X Y + ( f ) ksBZ dt dZ - 2 k z A X - ksBZ dt

(13.26)

By using the following data and maintaining the concentrations of A and B constant, the oscillatory solutions of concentrations can be obtained: kl = 1.28, k2 = 8.0, k3 = 8.0 × 105, k4 = 2 x 103, ks = 1.0 L(mol s); A = 0.06 M, B = 0.02M, f = 1.55. Figure 13.7 displays the oscillations in the concentrations obtained by the following MATHEMATICA code (* The Belousov-Zhabotinsky reaction*) (*FKN Model*) (*X-HBrO2, Y=Br-, Z=Ce4+, B=org, A=BrO3-*) Clear kl = 1.28;k2-8.0;k3=8.0* 10 ^ 5;k4=2" 10 ^ 3;k5= 1.0; A=0.06;B=0.02 ;f= 1.55; Soln3=NDSolve[{X' [t] ==kl *A*Y[t] +k2*h*X[t]-k3*X[t]*Y[t]-2*k4*X[t] ^2, Y' [t]==-kl *A*Y[t]-k3*X [t]Y[t] + (f/2)*k5*B*Z [t], Z' [t] = = 2*k2*A*X [t]-k5*B*Z [t], X[0] = =2" 10 ^-7,Y[0] = =0.00002,Z[0] ==0.0001 },{X,Y,Z}, {t,0,800},MaxSteps-> 10000] Plot[Evaluate[{X[t]}/.Soln3],{t,0,800},Frame->True, FrameLabel->{"t", "X"}, DefaultFont->{"TimesRoman", 14}, PlotStyle->{PointSize[0.03],Thickness[0.02]}, FrameStyle->Thickness[0.0075], PlotRange->{0.0,10^-4}] Plot[Evaluate[{Y[t]}/.Soln3],{t,0,800}, Frame->True, FrameLabel->{"t", "Y"}, DefaultFont->{"TimesRoman", 14},PlotStyle->{PointSize[0.03],Thickness[0.02]}, Frame Style- > Thickness [0.0075]] Plot[Evaluate[{Z[t]}/.Soln3],{t,0,800},Frame->True, FrameLabel->{"t", "Z"}, DefaultFont->{"TimesRoman", 14},PlotStyle->{PointSize[0.03],Thickness[0.01 ]}, Frame Style-> Thickness [0.0075]] Plot[Evaluate[{X[t],Y[t]}/.Soln3],{t,0,800}, Frame->True, FrameLabel->{"t", "X,Y"}, DefaultFont- > {"TimesRoman", 14},PlotStyle-> {PointSize [0.03] ,Thickness [0.01 ]}, FrameStyle->Thickness[0.0075], PlotRange->{0.0,0.0001 }] ParametricPlot[Evaluate[{X[t],Y[t]}/.Soln3],{t,0,800}, Frame->True, FrameLabel->{"Y", "Z"}, DefaultFont->{"TimesRoman", 14},PlotStyle->{PointSize[0.03],Thickness[0.01]}, FrameStyle->Thickness[0.005], PlotRange->{0.0,10"-4}]

Example 13.6 Order in time: Thermodynamic conditions for chemical oscillations Consider the following set of reactions S + X < klf ,,') 2 X X+Y < Y<,

ktb k2f

k2b k3f k3b

....>2Y >D

(13.27)

645

13.6 Order in chemical systems

0.0001

l

0.00008

]

0.00006

t

t

1 0.00004 0.00002

26o

4oo -~

(a)

800

600

t

o.oool 0.00008 0.00006 0.00004 0.00002 ,

,

,

,

,

0

200

400

600

800

(b)

t

0.0012 0.001 0.0008 N

0.0006 0.0004 0.0002 0 0

2()0'

(c)

'

'4()0'

'600'

'

'800

t

0.0001 0.00008

>.. X-

0.00006 0.00004 0.00002

0 (d)

200

400

600

80(

t

Figure 13.7. Concentration diagrams for Belousov-Zhabotinsky reaction with parameters: k1=1.28, k2 = 8.0, k3 = 8.0 × 105, k4 = 2 × 103 , k5-- 1.0, A - O . O 6 M , B--O.O2M, f - 1.55; (a) X - H B r 0 2 versus time, (b) Y - - B r - v e r s u s time, (c) Z - C e 4÷, (d) X, Y versus time, Y is displayed with the bold line.

The concentrations of initial and final products S and D are maintained at constant values. So, there are two independent variables Xand Y. The kif and k,~ denote the forward and backward chemical reaction rate constants, respectively. The overall affinity characterizes the thermodynamic state of the chemical system, and is found form

A_ tx, +ljc2+tJc3_ RTln( klfk2fk3fklbk2bk3b S

(13.28)

646

13.

Organized structures

At chemical equilibrium, we have

F

eq

klfk2fk3f

'

-- klf S, Xeq - "klb

Yeq -- klf k2f S klbk2b

(13.29)

Within the vicinity of a nonequilibrium (A 4: 0), the magnitude of the affinity may determine the time behavior of the system: (i) If (S/D) is close to its equilibrium value, then affinity and the reaction velocity are related linearly. Using the subscript s for steady-state values, we have

Jrl,s = klf SXeq AR 1T '

Jr2,s = k2 f Xeq Yeq RA2T '

Jr3,s -- k3f Yeq RA3T

(13.30)

So that the system will be stable to small disturbances around the steady state if it satisfies the inequality p=l T ZJriAi

>- 0

(13.31)

i For the chemical system, the stability condition in Eq. (13.31) becomes T 2 p = klfSXeq ((~A1) 2 -nt- k2f XeqYeq ((~A2) 2 q- k3f Yeq ((~A3) 2 ~ 0

(13.32)

(ii) Assume that the system is far from equilibrium. If the reverse reactions in the system (13.27) are negligible, the kinetic relations become dX dt dY dt

-- klfS x - k2fX Y

(13.33)

- k2f X Y -

(13.34)

k3f Y

These kinetic relations yield the following steady-state solutions

Xs - k3f k2f'

Ys = klf' S k2f

(13.35)

This model resembles a Lotka-Volterra model, which may be used in studying the evolution of systems in time, such as biological clocks or the time-dependent properties of neural networks. In the vicinity of the steady state, X(t) and Y(t) may be X ( t ) = X s + xe at,

Y(t) = X s + ye at

with small enough magnitudes of perturbations of x and y

I+01 L0L <<1,

<<1

(13.36)

647

13.6 Order in chemical systems

Substituting Eq. (13.36) into Eqs. (13.33) and (13.34), and considering only the first-order terms in the perturbations, we have the following linearized equations ASX + k3f8 Y = 0

(13.37)

-k~f.SSX + A8 Y = 0

(13.38)

Disregarding the higher order terms in the perturbations, the corresponding "characteristics equation" is

a 2 + klfk3rS - 0

(13.39)

This equation indicates that small fluctuations around the steady state are periodic with the frequency

Ai - _+(klfk3fS) 1/2,

Ar -- 0

where Ai and Ar are the imaginary and real parts of the eigenvalue, respectively. From Eqs. (13.33) and (13.34), we determine that 8xP vanishes around the steady state

~xP=

k2f-~s- k i t . ~ ,

(~X)2 + ~,Ys - k 2 f ~

(~y)2 - 0

(13.40)

This equation shows that the system will be at the steady state if the equality holds. (iii) Consider the intermediate values of the overall affinity 1 << IA/RT I << oc. Assuming that k i f - O and kib = k, the steady-state solutions of Eqs. (13.33 )and (13.34) become

X~-l+kY~-~

krS

(13.41)

<

k3Ys 4 + ( 1 - kS q- 2k2 ) ys3 + ( k - S - k r S - 2k3 rS) Y2 + (krS2 - 2 k 2 r S ) Ys +k3r2S2 - 0

(13.42)

where r = D/S, and is a measure of distance from equilibrium; Eq. (13.28) now becomes

('}

(13.43)

A - - [..ZI -ff l,.Z2 +123 -- R T ln -~r

The characteristics equation is

A2 +(Ys - Xs + 2 k X S +2kYs)a+ X~ + 2 k X ~ - 1 - 2 k X 2 - 2 k Y s

+Ys q-4k2XsYs - 0

(13.44)

For the whole range of overall affinity, one has the inequality Ar < 0. Therefore, the thermodynamic branch is stable, and the fluctuations will disappear. For values of A > 9.2 RT, however, A becomes complex, and the regression is not monotonic in time.

13.6.3

The LengyeI-Epstein Model

The Lengyel-Epstein model is a more realistic chemical reaction scheme. The Lengyel-Epstein model is a twovariable model for the chlorite-iodide-malonic acid (CIMA) reaction scheme and its variant, the chlorine dioxide-iodine-malonic acid (CDIMA} reaction scheme. In the model, the oscillatory behavior is related with: (i) The iodination of malonic acid [CH2(COOH)2] (MA) MA + 12 ---+IMA + I- + H + d[I2] _ kl[mA][I2]

dt

w I + [I 2 ]

(13.45)

648

13.

Organized structures

(ii) The oxidation of iodide ions by chlorine dioxide C102 + I- ~ C10 2 + _1 H+ 2 d[C102 ] _ ~ = k2[C102][I- ] dt

(13.46)

(iii) The reaction between the chlorite and iodide ions to produce iodine C10 2 + 4 I - 4 H + ~ C 1 - + 2I 2 + 2H20 _ d[C102] _ k3a[C102][I-][H+]+ k3b[C102][12][I- ] dt w3 + [ I - ] 2

(13.47)

where ki denotes the reaction rate constants, and Wl and w3 describe saturation phenomena. The last term on the right side of Eq. (13.47) represents the autocatalytic effect of I2 and the inhibition of I-. This term vanishes when no iodide is available ([I-] --, 0) and when very strong self-inhibition ([I-] --, oo) occurs. The above rate equations indicate a five-variable model involving the concentrations of [I-], [C102], [CH2(COOH)2], [C102], and [I2]. During an oscillation, the concentrations of malonic acid, chloride dioxide, and iodine remain almost constant while the concentrations of iodide and chlorite ions vary. For the CDIMA reaction scheme in a constantly stirred tank reactor, the kinetic equations of the two-variable Lengyel-Epstein model become (Horsthemke and More, 2004). cr

d[I-] dt

dU - ~r~= dt

4UV a - U - ~ 1+ U 2

d[C102] dV _ b (U UV ) d-------~ - dt - --------~ 1+ U

(13.48)

(13.49)

where U and V are the dimensionless concentrations of [I-] and [C102], respectively. The constants a is proportional to [MA]/[C102] while the constant b is proportional to [I2]/[C102]. The value of the parameter o- depends on the concentration of the complexing agent and lies in the range 1 < o- < 1000. If o- = 1 there is no complexing agent. The CDIMA reaction scheme has the following steady-state solutions 2

Us_a

5

and

Vs - l + a

(13.50)

25

This steady state becomes oscillatory after a Hopfbifurcation bH b
2-125 5ao"

fora>

(13.51) 3

Figure 13.8 display the oscillations for the Lengyel-Epstein model with different sets of parameters: o - - 2 . 0 , a - 30.0, b = 8.0 < bH; c r - 2.0, a -- 30.0, b = 11.0 > bH, the time interval is 0 < t < 20. c=2.0;a=30.0;b=8.0; h=20; Print[" c -",c,"; a = ",a,"; b - " , b , "; d - " , d, "; LE-model"]; s - 1.0+u[t]*u[t] eq 1-u' It] =--(a/c)- ( 1/c) *u [t]-(4.0*u [t] *v [t] / (c* s)); eq2--v' [t] ==b* (u [t]-u [t] *v [t]/s); soll =NDSolve[{eql, eq2,u[0] ==0,v[0] ==0},{u,v},{t,0,h}, MaxSteps -> 5000]; Plot[Evaluate[ u[t]/.soll],{t,0,h},PlotPoints -> 40, Frame- > True, FrameLab el- > {"t", "U" }, AxesLabel ->{"t"," U "}, DefaultFont->{"TimesRoman", 14},PlotStyle->{PointSize[0.03],Thickness[0.0075]}, Frame Style- > Thickness [0.0075]]; Plot[Evaluate[ v[t]/.soll],{t,0,h},PlotPoints->40,Frame->True, FrameLabel->{"t","V"},

649

13.6 Order in chemical systems

5O

50

4O 40

>- 30

23

20

30

10 20

0

.

0

5

10 t

(a)

15

20

4

.

.

.

.

.

6

(b)

60

.

.

.

.

.

7

50

40

40

D- 30

30

2O

20

10

10 0

5

10

15

t

(c)

0

20

.

))

60

0

.

8

U

5O >

.

5

//

0

z

2

(d)

6

8

10

U

Figure 13.8. Oscillations of concentrations and the limit cycle obtained from the LengyeI-Epstein model with the parameters: o-= 2.0, a = 30.0, b = 11.0 > bH, h = 20, which indicates time; (a) U, Y versus time, the bold line is for V, (b) limit cycle: V versus U with the initial conditions of U(O)= O, and V(O)= O, (c) U, Y versus time, the bold line is for V, (d)limit cycle: V versus U with the initial conditions of U(O) = O, and V(O) = 0 with the parameters: o- = 2.0, a = 30.0, b = 8.0 < bH, h = 20.

AxesLabel-> {"t"," V "},DefaultFont- > {"TimesRoman", 14},PlotStyle-> {PointSize [0.03] ,Thickness [0.001 ]}, FrameStyle->Thickness[0.0075]]; ParametricPlot [Evaluate [{u [t] ,v[t] }/.sol 1 [[ 1]]], {t, 0,h}. PlotPoints- > 40 ,Frame-> True, FrameLabel->{"U"," V "},DefaultFont->{"TimesRoman", 14},PlotStyle->{PointSize[0.01 ],Thickness[0.01 ]}, FrameStyle->Thickness [0.0075], GridLines- >Automatic]; Plot [Evaluate [{u[t] ,v[t] }/. sol 1]],{t,O,h},PlotPoints->40,Frarne->True,FrameLabel->{"t""U, V"}, AxesLabel->{"t"," V "},DefaultFont->{"TirnesRoman", 14},PlotStyle->{PointSize[0.0 i ],Thickness[0.01 ]}, FrameStyle->Thickness[0.008]; When a one-dimensional diffusion is considered with the Lengyel-Epstein model, we have d[I-]

o-

dt

-

o

dU dt

-

a - U -

4UV U2 1+

+

Du

O2U dx 2

d[C102 ] dV _ b (U - UV ) 02V d~--[---- dt I+U---------~ + D f . , dx ~ 2

(13 52)

(13.53)

A homogeneous steady state undergoes a Turing bifurcation at the following critical value of b (Rudovics et al., 1999) d br - -5-~a(13ae - 4 x/~a~/25 + a2 +125)

(13.54)

where d is the ratio of diffusion coefficients: d - DI~/D u -Dclov/DI_" The critical value b T is independent of the parameter o-. The Turing pattern occurs before the Hopfbifurcation when sufficiently large amount of complexing agent is used for a specified value of a (Horsthemke and Moore, 2004). For example, starch (S) forms a complex with iodide S+I- +I2--SI 3 The SI 3 complex is practically immobile in the gel.

650

13.7

13.

Organized structures

BIOLOGICAL STRUCTURES

The fields related to spatiotemporal organizations and the stability and robustness of chemical and biological systems keep attracting a growing number of scientists. Mainly, the Turing instability leads to steady patterns and the Hopf instability leads to oscillation. Some examples of Turing instabilities in coupled systems occur in inhomogeneous arrays of diffusively coupled reactors, and the other results with two coupled layers; one of them supports oscillatory Turing patterns, while the other supports the stationary Turing structure. Biological cells exchange not only species but also signals in a highly nonlinear manner. Continuous models are applicable only to systems where the cell-cell interaction is well approximated by diffusion. For diffusion driven models, the stationary states are stable as long as the number of cells is small, and they destabilize when the number of cells in the lattice increases beyond a certain bifurcation value. Different phase space perturbations from the homogeneous state lead the system to completely different patterned states. The pattern of the final state is influenced by the initial perturbation along an unstable lattice vector. For signal-driven models, the homogeneous state is mainly unstable and independent of the number of cells in the lattice. Linear Turing analysis may predict patterns, while nonlinear analysis can be useful in explaining them. In discrete cellular systems, the eigenvector and eigenvalue analysis of the homogeneous state shows that the set of lattice vectors provides a natural basis for describing the final spatial patterns for each species. Linear analysis gives at best a prediction of the final pattern, and the effects of boundary and initial conditions on the stability of patterns need more research. In many cases, there is no obvious resemblance between the final, unsteady state and the pattern of the unstable mode, or the mode corresponding to the initial perturbation, in the case of multiple unstable modes. The Turing mechanism can only predict a qualitative resemblance of the final state to the lattice vector resulting from the linear analysis. However, the linearization of evolution equations using very small perturbations for biological and chemical systems has limitations due to large experimental errors, and the "response approaches" that avoid linearization developed by Vlad et al. (2004) may be promising in that respect. To solve highly nonlinear differential equations for systems far from global equilibrium, the method of cellular automata may be used (Ross and Vlad, 1999). For example, for nonlinear chemical reactions, the reaction space is divided into discrete cells where the time is measured, and local and state variables are attached to these cells. By introducing a set of interaction rules consistent with the macroscopic law of diffusion and with the mass action law, semimicroscopic to macroscopic rate processes or reaction-diffusion systems can be described. Living objects are self-reproducing and are capable of creating information that influences their evolution and self-reproduction. The ability to create new information gives a meaning to structuring. We see this structuring in living objects at all levels of organization.

13.7.1

Reaction-Diffusion Systems:Turing Structures

In reaction-diffusion systems, chemical reactions are the source terms for both mass balance and energy balance equations. Therefore, the heat of reaction may lead to coupling between mass and heat fluxes. The coupling depends on whether the system is isotropic or anisotropic. About 50 years ago, Turing demonstrated that even simple reaction-diffusion systems could lead to spatial differentiation due to the instability of the homogeneous equilibrium depending on the activator-inhibitor interactions and boundary conditions. The Rayleigh-B6nard instability shows that the maintenance of nonequilibrium might be the source of order in fluids subjected to a thermodynamic force above a critical value. Therefore, the distance from global equilibrium in the form of magnitude of a thermodynamic force emerges as another constraint of stability; some systems may enhance perturbations, and evolve to highly organized states called dissipative structures after a critical distance on the thermodynamic branch. Although the kinetics and transport coefficients represent short-range interactions, chemical instabilities may lead to Hopf bifurcation, which is long-range order and coherent time behavior, such as a chemical clock. Stability analyses of linear and nonlinear modes for stationary homogeneous systems are useful in understanding the formation of organized structures. A differential flow of activator and inhibitor species at different rates can have a destabilizing effect of periodic traveling waves on the homogeneous steady state of a reaction-diffusion advection system similar to unequal diffusion in the Turing system. Internal fluctuations on the oscillatory reaction-diffusion systems at a mesoscopic level may also affect the long-range order controlled by short-range interactions. The Turing analysis had two important conclusions: (1) Transport processes such as diffusion normally decrease inhomogeneties, and may promote greater inhomogeneity and organization in the presence of unequal diffusion, flow, or/and an autocatalytic reaction system. (2) The analysis of the approximate linearized system of equations is sufficient to determine the conditions for the onset of instability.

13. 7

651

Biological structures

The spatial coupling of local chemical reaction systems to transport processes of diffusion, convection, and electromigration can lead to spatial organization. Spatial organization and pattern forming may occur in the large class of reaction-diffusion systems. These systems are modeled by an appropriate set of thermodynamically and mathematically coupled partial differential equations (Demirel, 2006). The set of equations describes the evolution of the concentration as well as the temperature of the system. For example, stable homogeneous chemical reaction systems may become unstable because diffusion and inhomogeneous steady states of Turing structures arise. This is a symmetrybreaking process and can lead to stable spatial patterns, if the global stability of the system is maintained. A general Turing model for morphogenesis in a two-component (A and B) system may be formulated by Oc A

Ot

-- f ( C A , C B , k ) + D A V c A

OCB

Ot

-- f (c A

,cB,k) + DBVC B

where D A and D B are the diffusion coefficients of the components A and B, respectively. The analytical form of the functionfcontains kinetics relations and kinetic constants denoted by k. A simple inhibitor-activator model can be represented by the above relations. Another important application of the model above is to regulated gap junctions. In two coupled cells, two substrates can cross the membranes through gap-junction protein channels. When one of the substrates controls the movement through the gap-junctions, nonlinear diffusion arises. This nonlinear diffusion can lead to Turing structures (Klein and Seelig, 1995).

13.7.2 Chiral Symmetry Breaking Many species living in an environment changing with a daily rhythm have evolved a biological internal clock known as the circadian clock. Many genes and proteins take part in the biochemical process dynamics producing this stable rhythm. The basic mechanism involves one or more genes whose products may enter into the nucleus and then suppress the transcription of their own gene(s) (Kurosawa et al., 2002). Circadian rhythm occurs in cells that are compartmentalized into the nucleus and the cytosol, and also in prokaryotes (exemplified by cyanobacteria) in which the cells are not compartmentalized (Goldbeter, 1996; Merrow and Roenneberg, 2001). A molecule whose geometrical structure is not identical to its mirror image possesses chirality. For example, enantiomers are mirror-image structures of a chiral molecule. Two mirror-image molecules are identified as L- and D-enantiomers. Amino acids and deoxyribose in DNA are chiral molecules. Asymmetry in biochemistry requires the constant catalytic production of the preferred enantiomer in the reactions between enantiomers, a process known as racemization. In systems with appropriate chiral autocatalysis, instability may appear. Due to random fluctuations, the instability occurs accompanying the bifurcation of asymmetric states in which one enantiomer dominates. These states of broken symmetry can be observed in the following simple model reaction scheme with chiral autocatalysis (Kondepudi and Prigogine, 1999) S + T ~.--~--X L

(1)

S + T + X L ~ 2X L

(2)

S+T~X

(3)

D

S + T + XD ~,--~-2XD (4) XL+ XD~

P

(5)

Enantiomers of XL and XD are produced from the reactants S and T, as shown in reactions (1) and (3), respectively They are also produced by the autocatalytic reactions (2) and (4). The reaction rate constants in reactions (1) and (3) and in reactions (2) and (4) are identical. In reaction (5), the two enantiomers react to produce component R Obviously, at equilibrium, XL = XD, and the system will be in a symmetric state. If we control the incoming flows of T and S and outgoing flow of P, and assume that the reverse reaction in (5) can be ignored, then we have the following kinetic equations dX dt

k _ k f l S T _ kblX L + k f 2 X L S T _ kb2X2 _ k 3 X L X D

(13.55)

652

13.

Organized structures

d X D - k f l S T - k b l X D + k f 2 X D S T - kb2 X 2 - k 3 X L X D

(13.56)

dt

With the following parameters, the symmetric and asymmetric states may become more explicit a--ST;

og-- XL - X D ;

/~ = XL if- XD

2

(13.57)

2

Using these parameters in Eqs. (13.55) and (13.56), we have dez dt

d3 dt

-- -kblOl - kf 2 A ol - 2kb2Ol ~

(13.58)

__ kfla _ kbl/~ _+_kf2A/~ _ kb 2 (/~2 + ~2 ) _ k3 (/~2 _ 2 )

(13.59)

For small values of A, the steady-state solutions of Eqs. (13.58) and (13.59) yield

c~s = 0;

/3s =

2kb2m + 4(2kb2m) 2 + 4(kb2 + k3) kfl/~.

(13.60)

2(kb2 +k3)

where m =

2kf2A- kbl 2kb2

Linear stability analysis indicates that the symmetric solution in Eq. (13.60) becomes unstable when A is greater than the following critical value 1c

AC

S-+-x/S 2 -- 4k22kl~l

(13.61)

2G

where 4kl~2kbl s = 2kf2kbl + ~ k 3 --kb2

The asymmetric stationary solutions are I

4kfla

and

O~a = _+_ /~2 _ k3 _ kb2

kf2'~- kbl

/~a =

(13.62)

2kb2

where the subscript a denotes asymmetry.

Example 13.7 Chiral symmetry breaking The symmetry breaking property of Eqs. (13.55) and (13.56) can be seen by the following simple MATHEMATICA code. Figure 13.9 displays the Chiral symmetry breaking using the parameters kfl = 0.5, kba 0.1, kf2 = 0.1, kb2 = 0.2, k3 = 0.5, S = 0.4, T = 0.4. (*Chiral symmetry breaking*) kfl =0.5;kb 1=0. I ;kf2 =0.1 ;kb2=0.2;k3=0.5;S=0.4;T=0.4; Solnl =NDSolve [{XL'[t] = =kfl *S*T-kb 1*XL It] +kf2*S*T*XL It] -kb 2* (XL[t] ^ 2)-k3*XZ [t] *XD [t], XD' [t] = =kfI *S*T-kb 1*XD [t] +kf2*S*T*XD It]

13. 7

Biological structures ,

,

,

r

,

,

,

i

653 ,

,

,i

i

r

0.28335

\

m

0.28334 ×

0.28333

X 0.28332

0.28331 i

,

,

J

,

J

20

0

Figure 13.9. Chiral breaking with the parameters:

~

i

i

i=

40

60

k l f -- 0 . 5 ,

k b l -- 0 . 1 ,

i

i

80

kf2 --- 0 . 1 ,

=

i

i

100

kb2 -

i

i

J

i

120

0.2, k3 = 0.5, S = 0.4, T = 0.4.

-k2b 2 * (XD [t] A2 )-k3* XL[t] *XD [t], XL[0]==0.002,XD[0]-=0.0},{XL,XD},{t,0,120}, MaxSteps->500] Plot[Evaluate[{XL[t],XD It] }/.Solnl ],{t,0,120}, Frame->True, FrameLabel->{"t", "XL, XD"}, PlotStyle->{{GrayLevel[0], Dashing[{0.02,0.025}]},Thickness [0.008], GrayLevel[0.3]}, DefaultFont->{"TimesRoman", ]2}] 13.7.3

Lotka-Volterra Model

In biological dissipative structures, self-organization may be related to the attractors in the phase space, which correspond to ordered motions of the involved biological elements (De la Fuenta, 1999). When the system is far from equilibrium, ordering in time or spontaneous rhythmic behavior may occur. The L o t k a - V o l t e r r a m o d e l of the predator-prey interactions is a simple example of the rhythmic behavior. The interactions are described by the following kinetics dX J1 -

- klX-k2XY

(13.63)

- - k 3Y - k 4 Y X

(13.64)

dt dY J~ -

dt

where the terms X and Y represent the number of individuals of species, and k 1 and k4 are the biological potentials, which are the difference between the birth and death rates, respectively. The terms k2 and k3 are the interactions between both populations. The flows shown by Jl and J2 have two stationary solutions; the first of these is X = Y = 0, and the second is at the stationary values o f % and Y~, which are given by

k4

k:

(13.65)

To see whether this state is stable or not, we add small perturbations of 6 X and 6 Y to Xs and Ys, so that X-A'~+aX

fix

and

Y-Y~+aY

(13.66)

We can introduce these expressions into Eqs. (13.63) and (13.64) and disregard the terms containing products of and 6 Y, and we obtain d(aX) dt

-- k l r ~ X - k 2

(Xs6Y + Y~aX)

d(ar) dt

-- k 3 ( X s ~ Y -4 Y s ~ X ) - k 4 ~ Y

(13.67)

(13.68)

654

13.

Organized structures

With Eq. (13.65), we can rewrite Eqs. (13.67) and (13.68) as follows d(aX)

(13.69)

- -kzXsaY

dt

d(aY) -

dt

(13.70)

k3YsaX

We now differentiate Eq. (13.69) with respect to time, and after combining with Eq. (13.70), we have d2(6X) dt 2

=

-k2k3Xs

aX -

-k &ax

(13.71)

This equation has the form of motion of a harmonic oscillator and the solution yields a harmonic oscillation 6X = 6X(O) cos (2'n'fi)

(13.72)

where f is the frequency given by

{1

( 1)

(13.73)

Therefore, the stationary states given by Eq. (13.65) are not stable. This means that after small perturbations, the system does not return to the original state. Instead, it oscillates. This oscillatory behavior can be explained with the following example: As X increases, species 2 has more food and tends to increase its population. As Y increases, the amount of species 1 consumed by species 2 also increases. Therefore, X begins to decrease, and hence the amount of food available for species 2 decreases; this leads to a decrease of Y. As the number of predators Y decreases, the population of prey X recovers. This causes the start of a new cycle. The amplitude and period of oscillation depend on the initial state.

Example 13.8 Prey-predator system" Lotka-Volterra model The Lotka-Volterra predator and prey model provides one of the earliest analyses of population dynamics. In the model's original form, neither equilibrium point is stable; the populations of predator and prey seem to cycle endlessly without settling down quickly. The Lotka-Volterra equations are dX dt dY dt

-¢¢Y-aYY

- -6 Y+ faXY

When the birth rate equals the death rate in the prey (host) population, so that/3 = 0, we have dX -

-olXY

dt dY dt

- Y(-6 + faX)

If we take the ratio of these equations, we have dY dX

--f+~

aX

13. 7

655

Biological structures

This equation shows that dY/dX may be negative, zero, or positive, according to the values and signs of (8laX). Because dX/dt is negative, Xdecreases with time. Even though the host population is dying out, the predator population can increase until X decreases to the critical value 6/fa; after that, the host population also begins to die out. If none of the predator eggs hatches, then f - O, and the Lotka-Volterra equations become dX dt dY -

-BY

dt

Here, the trajectories are determined by dY

dY w

dx

x(~-~Y)

The trajectories show that the prey population is minimum at Y-/3/ce.

13.7.4

Stability Properties of Lotka-Volterra Equations

To investigate the stability properties of the Lotka-Volterra equation in the vicinity of the equilibrium point (X, Y) = (0, 0), we linearize the equations ofXand Y appearing on the right side of Eqs. (13.63) and (13.64). These functions are already in the form of Taylor series in the vicinity of the origin. Therefore, the linearization requires only that we neglect the quadratic terms in XY, and the Lotka-Volterra equations become dX dt

dY

- fiX,

dt

- -8 Y

(13.74)

The first equation shows that X increases exponentially, and hence the equilibrium point (0, 0) is unstable. A relationship between the X and Y can be obtained by the method of separation of variables, which yields dX m

-

=

x ( ~ - ,~Y)

dY

(13.75)

Y (-~ + . f ~ x )

Furthermore this equation in separable form becomes dt-

(-8 + faX) dX

(fi - c~Y) d Y

X

Y

(13.76)

Integration of Eq. (13.76) yields f o~X-81nX - ~lnY-o~Y +C

(13.77)

X '~e./~x - C'Y~e -~r

(13.78)

In exponential form, Eq. (13.77) is

Here C' is the constant of integration, and is obtained from the initial condition. Equation (13.78) describes a family of closed trajectories in the X, Y space. Each trajectory is determined by a particular initial condition. Figure 13.10 displays such trajectories obtained with an m-file with the following MATLAB code. function lotka(time) tspan- [0:0,01 :time]; zo--[2;1]; kl=l; k2-1; k3=l; k4-1; [t, z] =ode4 5 (@fz,tsp an, zo): zo=[1,8;1];

656

13.

Organized structures

[t,y] =ode4 5 (@fy,t sp an, zo); zo:[1.6;1]; [t,x] : o d e 4 5 (@fx,tspan, zo); figure; plot(z(:, 1),z(:,2)),xlabel ('X'),ylabel('Y') figure; plot(y(:, 1),y(:,2)),xlabel ('X'),ylabel('Y') figure; plot(z(:, 1),z(:, 2),y(:, 1),y(:, 2),x(:, 1),x(:, 2)) xlabel ('X'),ylabel('Y') function dz = fz(t,z) dz- [k 1*z(1 )-k2*z ( 1)*z (2);-k3* z (2)+k4* z ( 1) *z (2)]; function dy = fy(t,y) dy= [k 1*y( 1)-k2 *y( 1) *y(2 );-k3 *y( 2) + k4*y( 1) *y( 2) ]; function dx = fx(t,x) dx= [k 1*x(1 )-k2*x (1)*x (2);-k3*x (2)+ k4*x (1)*x (2) ];

Example 13.9 Sustained oscillations of the Lotka-Volterra type An open system far from equilibrium could exhibit spontaneous self-organization by dissipating energy to the surroundings to compensate for the entropy decrease in the system. Prigogine called such systems dissipative structures. Sustained oscillations, therefore, require an open system capable of exchanging energy and matter with its surroundings. Since it cannot exchange matter with its environment, a closed system can exhibit transitory oscillations only, as it must approach equilibrium. These studies are helpful in understanding the spontaneous self-organization, independent from genes and natural selection, in biological systems. The Lotka-Volterra type of equations provides a model for sustained oscillations in chemical systems with an overall affinity approaching infinity. Perturbations at finite distances from the steady state are also periodic in time. Within the phase space (Xvs. Y), the system produces an infinite number of continuous closed orbits surrounding the steady state

dX dt

-

- X ( Y - 1)

and

dY = Y ( X - 1) dt

After dividing one of these equations by the other, we have the following equation for the trajectories in the X, Yspace

dY

Y ( X - 1)

dx

x ( r - 1)

Integration of this equation yields X+Y-lnX-lnY=

C

(13.79)

where C is an arbitrary constant, which is determined by the initial conditions. The critical point of marginal stability is reached in the limiting situation of infinite overall affinity. There is no mechanism for the decay of fluctuations. Equation (13.79) represents a family of cycles each corresponding to a given value of constant C around a steady state (Figure 13.10). Each cycle appears as a state of marginal stability where even a small perturbation can change the motion of the system to a new cycle, corresponding to a different frequency (Glansdorff and Prigogine, 1971). There is no average orbit in the vicinity where the system is maintained. The Lotka-Volterra model has properties similar to those of unstable systems at the marginal state. Only the orbits infinitesimally close to the steady state may be considered stable, according to Liyapunov's theory of stability. However, at a finite distance from the steady state, two neighboring points belonging to two distinct cycles tend to be far apart from each other because of differences in the period. Such motions are called stable in the extended sense of orbital stability. The average concentrations of X and Y over an arbitrary cycle are equal to their steady-state values (Xs = 1 and Ys = A = 1). Under these conditions, the average entropy production over one period remains equal to the steady-state entropy production.

13. 7

657

Biological structures

2.4 2.2

1.8 1.6 >- 1.4 1.2

1 0.8 0.6 0.4

0

0.5

1

1.5

2

2.5

X

Figure 13.10. State-space plots with various initial values of Xo for the Lotka-Volterra model.

Diffusive instability can appear in simple predator-prey models. Bartumeus et al. (2001) used the linear stability analysis and demonstrated that a simple reaction-diffusion predator-prey model with a ratio-dependent functional response for the predator can lead to Turing structures due to diffusion-driven instabilities.

Example 13.10 Lotka-Volterra model Solve the following equations and prepare a state-space plot where Xis plotted against Y using the solution. dX dt dY dt

- 0 . 2 X - 1.2XY - -0.5Y + 0.25XY

Initial conditions" at t = 0, X = 1.0, and Y = 0.01. Figure 13.11 display the trajectories of X and Y, and the state-space plot produced by the following MATHEMATICA code (*Lotka-Volterra Model*) k1-0.2;k2-1.2 ;k3-0.5;k4-0.25; Solnl =NDSolve [{X' It]-=kl *X[t]-k2*X [t]*Y[t], Y' It]- --k3*Y[t] + k4* X [t] *Y It], X[0]-= 1.0,Y[0]--0.01 },{X,Y},{t,0,100}, MaxSteps->4000] Plot[Evaluate[{X[t],Y[t]}/.Soln 1],{t,0, IL00}, Frame->True, FrarneLabel->{"t", "X,Y"}, PlotStyle->{{GrayLevel[0], Dashing[{0.02,0.025}]}, GrayLevel[0.3]}, DefaultFont->{"TimesRoman", 14}]; PararnetricPlot[Evaluate[{X[t],Y[t]}/.Soln 1],{t,0,100}, Frame->True, FrameLabel-> {"X", "Y"}, RotateLabel-> True,DefaultFont-> {"TimesRoman", 14}]

Example 13.11 Enzymatic reactions" Oscillations in the glycolytic cycle Biochemical chains are highly likely to exhibit limit cycles and dissipative structures. Oscillations appear in living systems in a variety of ways with

658

13.

6

'

'

;I

. . . . . . . . . . . . . . . Ii

I

5 >-

X-3

I

I

II

III

III

II

I

I

I Ii

I

II

f ~

,,,,"

I

i"

.....

20

ii

11

I

I

" ....

40

>" 0.4 0.2

|

,ll 7 .....

60

0.8 0.6

I

iI

'

I

I

l

0

Figure 13.11.

I

I

I

0

(a)

I

'

I I ii

I

I I

II

2 1

i'

iI

I

I

4

Organized structures

~i

80

. . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

100

t

2

3

4

.-

5

6

x

(b)

Plots from prey-predator in Example 13.10: (a) Dashed line displays X, (b) state space plot with kl = 0.2, k2 = 1.2, k3 = 0.5, k4 = 0.25, initial values: X(0) = 1.0, Y(0) = 0.01.

very different properties. Glycolysis is one the best examples of temporal structures, which show the richness and variety of self-organization far from thermodynamic equilibrium in a metabolic pathway. The temporal structures result when an instability-producing multienzymatic mechanism experiences variations in the domains of the initial functions (De la Fuenta, 1999). For example, oscillations of the concentrations of some metabolites of enzymatic reactions are possible at the molecular level or at the supracellular level (circadian ryttmas) (Goldbeter, 1996). Some characteristic examples of sustained chemical oscillations are (i) substrate- and product-inhibited enzymatic reactions and (ii) the product activated enzymatic reaction of phosphofructokinase in the glycolytic cycle. In a model by De la Fuenta (1999), the activity of three enzymes, namely hexokinase, phosphofructokinase, and pyruvatekinase, is considered, and the chaotic dynamics of a dissipative glycolytic subsystem is suggested. For a single range of the control parameter, this model predicts the coexistence of chaos with different periodic regimes and limit cycles. The enzyme phosphofructokinase is allosteric, that is, it is made up of equivalent units that possess specific reaction sites for the fixation of the substrate and product. Each unit exists in two conformational states: one active with more affinity for the substrate, and one inactive. The reaction products of phosphoffuctokinase (FDP and ADP) displace the conformational equilibrium in favor of the active form of the enzyme. This may create a destabilizing effect on the excess entropy production. In the glycolytic cycle, the allosteric properties of the phosphofructokinase may lead to oscillations. Consider the following simple model A--~C 1 C 1 Jr-D 1--* D 2 D 2 --~ C 2 + D 1

(13.80)

C 2 + D 3 ~ D1

C2-*F Where A is the initial product glucose, F the final product glyceraldehyde 3P, D1 and D 3 the active and inactive forms of the enzyme, respectively, D 2 the enzymatic complex, C1 and C2 the fructose 1P and fructose 2P, respectively. In the conformational equilibrium step (C2 + D3 ~ D1) FDP activates the enzyme. Experimental results indicate that under physiological conditions, the enzyme is controlled mainly by the ATP and ADP. Jrl

A 3 + D~ D2

,YA2 + D3 < A2

>A3

kfl >D2

(a) (b)

kbl

kf2 > A 2 + D 1 kf

(c)

3 )D1 kb3

(d)

>

(e)

k4

(13.81)

The reactant A3 is the ATP and enters the system at constant rate OfJr~ at step (a). The A2 is the ADP, which is the product of reaction and also the activator in step (d). The factor y (3' > 1) shows that the fixation of A2 on the enzyme activates more than one reacting site because of the concerted conformational equilibrium. The kinetics of the model above is

13. 7

659

Biological structures

OA3ot

02A3 - Yrl - kflA3D1 + kblD2 + DA 3 Or 2

OA2

at -

kfaD2 -

Nf3A~D3+ kb3D1 -

Old___.= .!'-" -kflA3D 1+(kbl Ot

+ kfz)D 2 +

02A2

k4A2 + DA2 Or"~

kf3A~D3 -

kb3D1

(13.82)

0 0 3 = - k f 3 A ~ D 3 + kb3D1 Ot

OD2 _ Ot

kflA3O 1 --(kbl

n t-

kfa)D 2

This model neglects the diffusion of the enzyme since it is much slower than for ATP and ADP. In agreement with experiments, the following conditions are assumed to hold (Glansdorff and Prigogoine, 1971)

kfl, kbl, kf2 , kf------~3kb3 >>1 A3 A~'

(13.83)

A3 , A2 >>1,Do = D 1 + D 2 + D 3 Do Do where D o is a constant. OA30t - Jrl - °llDoC~

°A2at-

A3A~ + DA3

02A3 Or 2

o24 a, Doc~~ A 3A~ - k 4A 2 + DA2 Or2

(13.84)

(13.85)

with kflkf2 , oe~ = kf3 oe5 = alD0C~~ kbl --}-kf2 kb3'

t9ll - - - ~

At steady state, we have

A2s- Jrl ,A3~ = j(1-rlY)~(kblq-kf2)k~kb3 k4

kflkf2kf3O 0

(13.86)

The linearization of the perturbation equations around the steady state yields the characteristic equation

A2 +[°es(A2s)'Y+ k4-T°esA3s(A2s)Y-I +DA2A+ DA3 2 ] (13.87) + ~(A~s) '

k4 + - - / -

+

< + - - r - - ~ A 3 s ( A ~ s ) '-~ = 0

The condition for homogeneous perturbations becomes

b7+l

_

J ~ < (Jrlc)7 - kbl -ff kf2 kb3 '~4

/~flkf2 kf3 Do

(13.88)

This condition corresponds to the vanishing of the coefficient A in Eq. (13.87). The frequency becomes

Ai - k4 ( T - 1)1/2 and a focus is followed by a limit cycle for the phosphofructokinase reaction.

(13.89)

660

13.

Organized structures

The condition for inhomogeneous perturbations becomes bY+l J~ < JYrlc -- kbl "+-kf2 kb3 ~4 DA3 (~-~_ 1)2

k lkf2 kf3 Do DA2

(13.90)

The critical wavelength (~Oc)is 2 % =

DA2

(13.91)

Comparing the conditions in Eqs. (13.88) and (13.89) indicates that both instabilities occur when DA3 ~.~ x ~ +1 D A2

(13.92)

xf-Y - 1

An approximate value of k4 is obtained from Eq. (13.86) as k4 = 4 x 10 -2 1/s (Glansdorff and Prigogine, 1971). The values of Jr1 and the steady-state value of ADP concentration are Jrl -- 6 X 10 -6 mole / s, A2s = 1.5 × 10 -4 mole, and k4 -- 4 × 10 -2 S- 1(approximate)

Setting y = 2 in Eq. (13.89), the frequency of oscillations is A = 2.4 min- 1 and the period is 2.6 min. This value is in agreement with the experimental value of 3-5 min. The spatial differentiation is related to k4 and the diffusion coefficient DA. The critical wavelength changes in the range 10 -4 cm < ~oc< 10 -2 cm.

Example 13.12 Long-wavelength instability in bacterial growth Growth conditions may lead to different morphologies. A nonlinear diffusion coefficient may cause the generation of patterns and a long-wavelength instability. Consider a two-dimensional reaction-diffusion system for the bacteria density B(r,t) with a nonlinear diffusion term, and nutrient density N(r,t) with a linear diffusion term OB - f B ( B , N ) + VDB(B)VB Ot ON Ot

- - f y (B,N) + DN V2N

(13.93)

(13.94)

where DN is the diffusion coefficient of the nutrient, and DB the bacteria-dependent diffusion coefficient of the bacteria, defined by

DB(B ) = Do Bk

(13.95)

Here, k describes the nonlinearity. For simplicity, it is assumed that

f B ( B , N ) = f N ( B , N ) = BN

(13.96)

In a simplified model, this representation reflects a bilinear autocatalytic reaction: B + N ~ 2B, and simply means that the bacteria needs nutrient (assume no shortage of nutrient) to double themselves (Muller and Saarloos, 2002). Any instability observed for k > 0 is due to the nonlinearity in the diffusion coefficient. With the relations above, the reaction-di~sion system becomes

OB _ BN + D ~2Bk+l Ot k+l

(13.97)

13.7

661

Biological structures

ON --BN+V2N Ot

(13.98)

where D is the rescaled diffusion coefficient: D Do/O N. The model in Eqs. (13.97) and (13.98) has two homogeneous states: a stable solution in which only bacteria are present, and an unstable solution with only nutrients. The propagation of stable state to unstable state can be studied. For k > 0 and beyond a critical diffusion (Dc) D < Dc(k), the planar front is unstable and has a long wavelength instability (Muller and Saarloos, 2002). In a more realistic model, one needs to be able to incorporate the bacteria growth properties into the effective diffusion coefficient. =

Example 13.13 Instability in a simple metabolic pathway Instability can occur in a system composed of interacting subsystems. The subsystems may be stable in isolation. Consider the following simple pathway consisting of three enzymes (e 1, e2, and e3). ÷ I

I

Xo< 1 >XI< 2 ~'X2 < 3 ~'X3

(13.99)

and the metabolites of Xo and X3, which are maintained at constant values. The other metabolites X and X2 have varying concentrations. The metabolites Xo and)(3 constitute the boundary conditions that keep the system under nonequilibrium conditions. The rate equation of enzyme i is denoted by Jri. The kinetics of the system is

dX1 ]

I

dX2

(13.100)

Jr2 -- Jr3

k--g-J The system dynamics is obtained by integrating these equations with some initial conditions. The system in Eq. (13.99) may be decomposed into two subsystems + I

Xo< 1 )~)(1( 2 )x2

(1)

X l ( 2 )x2 ( 3 )x3

(2)

(13.101)

The behavior of the subsystems can be described in isolation. In the first subsystem, X2 is kept constant, while X1 is constant in the second subsystem. If the whole system is at steady state then dX1/dt = 0 and dXz/dt = 0. If the system is stable, then any small perturbations in)(1 and X2 are corrected, and the system returns to its original state. We assume that the two subsystems are stable in isolation. The enzymes in the subsystems are sensitive to the metabolites and hence interact with the relevant subsystem. The stability of the subsystems assumes the following relationships

OJrl OX1

OJr2 < 0 OX 1

and

OJr2 OX2

OJr3 < 0 OX2

(13.102)

The first inequality indicates that the sensitivity of enzyme 1 to X1 is less than the sensitivity of enzyme 2 to X1. The second inequality indicates that the sensitivity of enzyme 2 to X2 is less than the sensitivity of enzyme 3 to X2. These partial differentials known as unscaled elasticities, are the component properties and quantify the sensitivity of a rate to a metabolite. The whole system is stable if the following conditions (tr < 0 and Det > 0) are satisfied

OJrl OX 1 I OJrl OX~

OJr2 OX 1

OJr2 OX2

-

OJr 3 < 0 (1) OX2

(13.103)

> 0

(2)

(13.104)

662

13.

Organized structures

Since we assumed that the subsystems are stable the first condition is satisfied. The condition is violated if we have

'r2) OY 1

>o

o ]

&

'

J~

'r3/ OX 2

>0

(13.105)

OX 2

&

The left side term indicates the interactions between the two component subsystems, while the fight term shows the interactions within the subsystem of each component. Thus, even with stable subsystems (in isolation), the system can be unstable if the interactions among subsystems are more significant than the interactions within subsystems. So, the enzymatic parameters and the boundary conditions can be controlled in such a way that systemic instability occurs. This particular phenomenon is known as a saddle-node bifurcation. Under slightly different conditions, oscillations appear when the first condition is violated. However, this would be unexpected because the subsystems are stable in isolation and contain only one variable metabolite. These conditions show that: (i) when one subsystem contains two variable metabolites arid the other subsystem has one variable metabolite, the system with dynamic subsystems in isolation oscillates, and (ii) instability is impossible with two dynamically stable subsystems if the influence is one-directional; they must influence each other mutually to become unstable. This simple example also displays the importance of nonlinear dynamics. The unscaled elasticities are constant only for a linear system and hence result in the same behavior: if a linear system is stable, it will remain stable when boundary conditions change, and if it is unstable, it will remain unstable. Under nonlinear equations, however, the values of elasticity depend on the system state, which varies with the boundary conditions. This may lead to a transition from a stable to unstable system, known as bifurcation or symmetry breaking. This means that nonlinearity causes a variety of new behaviors in a system.

Example 13.14 A model for an enzyme reaction inhibited by the substrate and product Consider the enzymatic reaction inhibited by the substrate and product Jri ) S 1 + E < S1E

kfl >$1E

kbl

kf2 )E_[_S2

Jrf )

kr3 >$1S1E Sl+SlE< k~

(a) (b)

(C)

(13.106) kf4 )

S2 + E <

kb4

82 +S1E<

ES 2

kf5 )S1ES2

kb5

S 2 "F S1S1E < kf6 > S1S1ES2

(d)

(e) (f)

kb6

where $1 is the substrate, $2 the product, E the enzyme, and S1E the active enzyme-substrate complex. ES2, S1S1E, S1ES2, and S1SaES2 are inactive enzymatic complexes. Jri is the rate at which $1 enters the system and is given by Jri = Jro -- koSI, Jrf is the rate at which $2 disappears and is defined by the Michaelis-Menten kinetics Jrf --

Jr max82 K m -Jr-S2

(13.107)

where Jrmax is the maximum rate when $2 --+ ~ and Km the Michaelis Menten constant The kinetic equation for the substrate $1 is

) 0281 OS1 -- Jri kf 2E° ( Sl - - -/K -' + Ds, Or2 Ot (1 + $2/Ks2)(1 + S1/Ks~ + Ks1/K's1(S1/Kst )2)

(13.108)

663

Problems

082 = Ot

- J r f nt-

kf2E°(S1/KS, ) + Ds~ 02S2 ( I + S 2 / K s 2 ) ( I + S 1 / K s ' nt_(Ks /KPs,)(S1/KS,)2) " 0?.2

(13.109)

where E0 is the total quantity of the enzyme, and the other definitions are

KS, =

kbl + kf2 K' - kb---)-3 --kb4 kr 1 , S, -- kf 3 , K s 2 kf 4

The following conditions are assumed $ 1 ~ $ 2 ~1, Eo<>1 By defining the a as a = Ks, ~Ks, the instability with respect to diffusion may be determined in terms of a. The system becomes unstable beyond the following critical value

//vo-/lJ2( / /lJ2))-1

/~2

O/c =-

1+

1+

(v0 -/3) 2

Ds, K s

3

/3 v0 - / 3

(13.110)

with critical wavelength Wc:

2 Vo--~+[~(Vo--~)]l/2 (Ds, Ks, I e ( 2 / 3 - V0)

6% E--

koKs, , 1 )

kf2E 0

Jr0 0 --

kf2E0

(13.111)

Jrmax ,

/~-

kf2Eo

kf2Eo

Equations (13.110) yield acceptable values if (1/2)u0
PROBLEMS 13.1

Solve the following evolution equations and prepare a state-space plot where x is plotted against y using the solution.

dt

-0.36x-l.55xy,

--=-O.11y+O.O4xy dt

Initial conditions: at t = 0, x(0) = 1.0, and y(0) = 0.05, the time interval: 0.0 < t < 100. 13.2

Solve the following system of first-order ordinary differential equations and prepare a state-space plot where x is plotted against y using the solution. & /

dz =

dt

z,

-

- 8 z -

dt

1200x

The initial conditions are x(0) = 0.0 and z(0) = 0.5, and time interval is 0.0 < t < 50. 13.3

Solve the following initial value problem as eigenvalue problem, and prepare state-space plot. dxl _ dt

--9x 1 + 4x~ "'

dx2 dt

-- --2x 1 + 2x 2

The initial conditions are: xl(0) = 1: x2(O) = - 1 and time interval is 0.0 < t < 5.0.

664 13.4

13.

Organized structures

Let us solve the Lorenz equations below with MATLAB between t = 0 and 20 and prepare plots ofyl versus t, and a state-space representation ofy2 versus Yl, and Y3 versus Y2 by using two different sets of initial conditions: yl(0) = y2(0) = y3(0) = 5.0 and yl(0) = y2(0) = y3(0) - 5.0. dyl

_

_ _ 10y 1 +

dt

10y 2

- 28yl - Y2 - YlY3 dt dy3_ - - 2.666667y3 + YlY2 dt

dY2

the initial conditions yl(0) = y2(0) - y3(0) = 1.0 with the time span 0 -< t -< 20 13.5

Van der Pol's equations provide a valuable framework for studying the important features of oscillatory systems. It describes self-sustaining oscillations in which energy is fed into small oscillations and removed from large oscillations. Consider the following system of ordinary differential equations called Van der Pol's equations

dYl_

~--

dt

Y2

dY2_ - a (1 -

y 2 ) Y2 - Yl

dt

The initial conditions are yl(0) = 1.0 and yl(0) = 0.5, and a is constant. With the time span 0 -- t prepare the trajectories of yl and Y2 and state-space plot. 13.6

In the limit of irreversible reactions the Brusselator scheme is A B+X 2X+Y X

kl > X

k2 > Y + E k3 >3X k4 )F

k2 k2 With the parameters kl = 1.3, k2 = 1.0, k3 = 1.0, k4 = 1.0, and the concentrations A = 1.1, B - 3.0, Prepare the trajectories of X and Y and state-space plot. 13.7

Consider the following chemical reaction system: A

B+X 2X+Y X

kl )X,

kl = 1.0

k2 ) Y + E , k3 >3X, k4 >F,

k2 = 1 . 0 k3 = 1 . 1

k4=l.1

The initial values of A and B are maintained at A - 0 . 6 M and B = 1.6 M, while the products E and F are removed. Prepare the trajectories and state-space plot of X and Y. 13.8

When the concentrations of A and B are controlled, the one-dimensional approach to complex reactiondiffusion systems with the spatial coordinate r under isothermal conditions yields the kinetic equations for Xand Y

OX -

Ot

k A- k BX + k3x r - k4X + Dx

02X Or

665

Problems

OY 02y c)t - k 2 B X - k 3 x Z Y + Dy Or2

where Dx and Dy are the respective diffusion coefficients. The boundary conditions are

Or

Or r--+L

r=-L

and

-0

For a specified value for A and some variation of B if one of the eigenvalues becomes positive, then Det < 0, and the system becomes unstable, and the propagating wave or the Turing structure occur

B-> 1l (k4 + +k°'2Dx 2 )(

k3(klA/k4)2)0 -2Dy

Solve the above equations and assess the structuring conditions. 13.9

The Belousov-Zhabotinsky reaction scheme is A+Y

k, > X + P

A+Y

k2 >2X+2Z

X+Y

k3 >2P

2X

k4 > A + P

B+Z Representative kinetic equations of the Belousov-Zhabotinsky reaction scheme based on Eqs. (13.21)-(13.25) are dX

- k l A Y + k 2 A X - k 3 X Y - 2k4 X2

dt dY

- -k~AY-

dt dZ dt

k3XY + ( f ) ksBZ

- 2k2A X _ ksBZ

By using the following data and maintaining the concentrations of A and B constant, find the oscillatory solutions of concentrations of AT, Y, and Z: k l - 1 . 2 8 , k 2 - 8 . 0 , k 3 - 8 . 0 × l0 s, k 4 = 2 × 103, ks - 1.0 L/(mol s); A - 0.06 M, B - 0.02 M , f = 1.6, and f = 1.4. 13.10

The kinetic equations of the two-variable Lengyel-Epstein model become (Horsthemke and Moore, 2004)

o

d[I-] dt

-0-

~d[ClO _ ] dt

du

- a-lJ-~

dt dv dt

_

b u-

4uv 1+ u 2

uv 1+ u 2

666

13.

Organized structures

where u and v are the dimensionless concentrations of [I-] and [C102, respectively. The constants a is proportional to [MA]/[C102] while the constant b is proportional to [I2]/[C102]. Display the oscillations for the Lengyel-Epstein model with different sets of parameters: o-= 2.0, a = 30.0, b = 6.0, the time interval is 0
Consider one-dimensional diffusion is considered with the Lengyel-Epstein model

o

d[I-] dt

du =o'~=a-udt

+D u 1+ u 2

d[ClOz]_dV_b dt

02U

4uv

UV ) + D v 02V

u-

dt

dx 2

1 -+- U 2

dx 2

A homogeneous steady state undergoes a Turing bifurcation at the following critical value of b (Rudovics et al., 1999) d bT = 7-- ( 13a2 -- 4 ffi-Oa ff25 + a 2 + 125) 3a

where d is the ratio of diffusion coefficients" d = D v / D U - Dclo2/DI. The Turing pattern occurs before the Hopf bifurcation when sufficiently large amount of complexing agent is used for a specified value of a (Horsthemke and Moore, 2004). Solve the above equations and assess the structuring conditions. 13.12

Following simple model reaction scheme with chiral autocatalysis (Kondepudi and Prigogine, 1999) (1)

S+T~---X L

S + T + X L ~ 2X L

(2)

S -+-T ~.-~-X D

(3)

S + T +XD ~,-~-2XD (4) X L + X D~ P

(5)

Enantiomers OfXL and XD are produced from the reactants S and T, as shown in reactions (1) and (3), respectively. They are also produced by the autocatalytic reactions (2) and (4). The reaction rate constants in reactions (1) and (3) and in reactions (2) and (4) are identical. In reaction (5), the two enantiomers react to produce component P. If we control the incoming flows of T and S and outgoing flow of P, and assume that the reverse reaction in (5) can be ignored, then we have the following kinetic equations dX L

- knST-

kblX L + k f 2 X L S T -

kb2 X 2 -- k 3 X L X D

- kflSr-

kblXD + k f 2 X D S r -

kb2 X 2 - k 3 X L X D

dt

dX D dt

Using the parameters kfl = 0.5, kbl = 0.1, kf2 = 0.1, kb2 = 0.2, k3 = 0.5, S = 0.6, T = 0.6 plot trajectories of XL and XD. 13.13

With the following parameters, the symmetric and asymmetric states may become more explicit A - - ST;

oL=

XL- XD

;

j~ __

X L "-I-X

2

Using these parameters in Eqs. (13.55) and (13.56), we have dog. dt

-- - k b l O e - k f 2 A o l - 2kb2Ol ~

2

D

667

Problems

dfi

dt Using the parameters c~ and/3. 13.14

kfl -

__ kfl/~ _ kblJ ~ _ff kf2,,~j~_

0.5,

kbl =

0.1, kf2 = 0.1,

kb 2 (/~2 _+_2 ) _ k3 (/~2 _ 2 ) kb2 --

0.2, k3 = 0.5, S = 0.4, T = 0.4 plot trajectories of

The Lotka-Volterra model of the predator-prey interactions is a simple example of the rhythmic behavior. The interactions are described by the following kinetics dX

J~ -

dt

- k~x-

~2xY

dY J2 -

dt

- - k 3 r - k4 ~ ' x

where the terms X and Y represent the number of individuals of species, and kl and k4 are the biological potentials, which are the difference between the birth and death rates, respectively. The terms k2 and k3 are the interactions between both populations. Solve the equations above and prepare a state-space plot where x is plotted against y using the solution with kl = 0.5, k2 = 2.0, k3 = 0.6, k4 = 0.1. The initial conditions: at t = 0, x = 1.0, and y = 0.01. The time interval is 0 < t < 30. 13.15

A nonlinear diffusion coefficient may cause the generation of patterns and a long-wavelength instability. Consider a two-dimensional reaction-diffusion system for the bacteria density B(r,t) with a nonlinear diffusion term, and nutrient density N(r,t) with a linear diffusion term OB _ at ON

-

BN

+

D

V2Bk+l

k+l

-BN

+ V 2N

Ot where D is the rescaled diffusion coefficient: D = Do/DN, where DN is the diffusion coefficient of the nutrient, and DB the bacteria-dependent diffusion coefficient of the bacteria, defined by DB(B) = D0Bk. Assess the propagation of stable state to unstable state. (For k > 0 and beyond a critical diffusion (Dc) D < Dc(k), the planar front is unstable and has a long wavelength instability, Miiller and Saarlos, 2002). 13.16

(a) Instability can occur in a system composed of interacting subsystems. Consider the following simple pathway consisting of three enzymes (e 1, e2, and e3). 4-

X°(

1 )kl(

I 2 >X2<

3 )x3

and the metabolites ofXo andX3, which are maintained at constant values. The other metabolites X1 and X2 have varying concentrations. The metabolites Xo and X3 constitute the boundary conditions that keep the system under nonequilibrium conditions. The rate equation of enzyme i is denoted by Jrg. The kinetics of the system is

dX, ] dXz

r2/ ~,Jr2 - Jr3

dt J Write the reaction velocities Jr; on the right of equation above.

Xo(

i 1 )Xl(

+ 2 )x2

1

(1)

668

13.

Organized structures

Xl ( 2 ) x 2 ( 3 ) x 3

(2)

Using the following relationships assess the stability of subsystems above: OJrl

OJr2 <

OX 1

OY 1

0

and

OJr2

OJr3 < 0

OX 2

OX 2

The first inequality indicates that the sensitivity of enzyme 1 to)(1 is less than the sensitivity of enzyme 2 to X1. The second inequality indicates that the sensitivity of enzyme 2 to X2 is less than the sensitivity of enzyme 3 to X. (b) The whole system is stable if the following conditions (tr < 0 and Det > 0) are satisfied OJrl

OJr2 _~ OJr2

OJr 3

OYl

ax~

ox2

oY2

<0

Or3)(Orl

(1)

-

OX 1

OX 1

OX 2

Oy 2

OX 2

Oy 2

~ OXl

)

>0

(2)

Using the kinetics of the reaction system and the relations above assess the overall stability.

13.17

There is a possibility that during embryogenesis, the angle of branching of a small artery is determined according to the principle of minimization of energy dissipation. Assume that the kinetic energy loss of the blood in going from the inlet to the outlet is neglected. For steady flow in a tube, the dissipated power P is equal to the work done by the pressure forces at the inlet and outlet. Assume that a symmetric bifurcation in an artery with the distance d fixed but the angle a considered to be variable. Derive the total power dissipated in the arterial, and by minimizing dissipated power determine the origin x and angle a of the bifurcation for specified values of Q1/Q2 and ra/r2, where Q1 and Q2 are the blood flows in the upstream branch and each of the downstream branches, respectively.

REFERENCES E Bartumeus, D. Alonso and J. Catalan, Physica A, 295 (2001) 53. I.M. De la Fuenta, BioSystems, 50 (1999) 83. Y. Demirel, Chem. Eng. Sci., 61 (2006) 3379. I.R. Epstein, J.A. Pojman and O. Steinbock, Chaos, 16 (2006) 037101-1. M. Falcke, Adv. Phys., 53 (2004) 255. R.J. Field, E. K6r6s and R.M. Noyes, J. Am. Chem. Soc., 94 (1972) 8649. R.J. Field and R. M. Noyes, J. Chem. Phys., 60 (1974) 1877. P. Glansdorff and I. Prigogine, Thermodynamic Theory of Structure, Stability and Fluctuations, Wiley, New York, NY (1971). A. Goldbeter, Biochemical Oscillations and Cellular Rhythms: The Molecular Bases of Periodic and Chaotic Behavior, Cambridge University Press, Cambridge, MA (1996). W. Horsthemke and EK. Moore, J. Phys. Chem. A, 108 (2004) 2225. K. John and M. Bar, Phys. Rev. Let., 95 (2005) 198101-1. C.T. Klein and EE Seelig, Biosystems, 35 (1995) 15. D. Kondepudi and I. Prigogine, Modern Thermodynamics, From Heat Engines to Dissipative Structures, Wiley, New York, NY (1999). G. Kurosawa, A. Mochizuki and Y. Iwasa, J. Thor. Biol., 216 (2002) 193. M. Merrow and T. Roenneberg, Trends. Genet., 17 (2001) 1207. J. Muller and W.V. Saarloos, Phys. Rev. E., 65 (2002) 61111. Z. Nagy-Ungvarai and B. Hess, Physica D, 49 (1991) 33. O. Nekhamkina and M. Sheintuch, Phys. Rev. E., 68 (2003) 36207. G.M. Neuter, H. Caswell and J.D. Murray, Math. Bioscience, 175 (2002) 1. G. Nicolis and I. Prigogine, Self-Organization in Nonequilibrium Systems, Wiley, New York (1977). S. Petrovskii, B.-L. Li and H. Malchow, Bull. Math. Biol., 65 (2003) 425. E. Plahte, J. Math. Biol. 43 (2001) 411. J. Ross and M.O Vlad, Annu. Rev. Phys. Chem., 50 (1999) 51. B. Rudovics, E. Barillot, EW. Davies, E. Dulos, J. Boussonade and P. De Kepper, J. Phys. Chem. A, 103 (1999) 1790. R.A. Satnoianu, J.K. Merkin and S.K. Scott, Chem. Eng. Sci., 55 (2000) 461. I. Schreiber, E Hasal and M. Marek, Chaos, 9 (1999) 43. A.M. Turing, Philos. Trans. R. Soc. Lond. B, 237 (1952) 37. V.K. Vanag and I.R Epstein, Proc. Natl. Acad. Sci., 100, (2003) 14635.

References W. Vance and J. Ross, Phys. Rev. Lett., 62 (2000) 3303. M.O. Vlad, A. Arkin and J. Ross, Proc. Natl. Acad. Sci., 101 (2004) 7223. L. Yang and I.R. Epstein, Phys. Rev. Lett., 90 (2003) 178303-1. L. Yang and I.R. Epstein, Phys. Rev. E, 69 (2004) 26211.

REFERENCES FOR FURTHER READING M. Bar, A.K. Bangia and I.G. Kevrekidis, Phys. Rev. E, 67 (2003) 056126. EC. Boogied, F.J. Bargeman, R.C. Richardson, A. Stephen and H.V. Westerhoff, Synthase, 145 (2005) 131. M.A.J. Chaplain, M. Ganesh and I.G. Graham, J. Math. Biol., 42 (2001) 387. S. Cortes, N. Glade, I. Chartier and J. Tabony, Biophys. Chem., 120 (2006) 168. M. Dolnik, A.M. Zhabotinsky, A.B. Rovinsky and I.R. Epstein, Chem. Eng. Sci., 55 (2000) 223. T. Kirner and J. Ackermann, J. Theor. Biol., 224 (2003) 539. J. Ovadi and V. Saks, Mol. Cell. Biochem., 256/257(2004) 5. M. Perc and M. Marhl, Bioelectlvchemistly, 62 (2004) 1. S. Shima andY. Kuramoto, Phys. Rev. E, 69 (2004)036213. E. Smith, Phvs. Rev. E, 72 (2005)036130-1.

669

14 NONEQUILIBRIUM THERMODYNAMICS APPROACHES 14.1

INTRODUCTION

One needs to describe nonequilibrium phenomena by the simultaneous consideration of mass, temperature, and time of the local states while accounting for the given time and energy dissipation due to temperature changes. The time scale over which microscopic changes occur is much smaller than the time scale associated with macroscopic changes. Temperature fluctuations in a microstate will be different from those in a macroscopic state in which the properties are the averages of many microstate values. Linear nonequilibrium thermodynamics has some fundamental limitations: (i) it does not incorporate mechanisms into its formulation, nor does it provide values for the phenomenological coefficients, and (ii) it is based on the local equilibrium hypothesis, and therefore it is confined to systems in the vicinity of equilibrium. Also, properties not needed or defined in equilibrium may influence the thermodynamic relations in nonequilibrium situations. For example, the density may depend on the shearing rate in addition to temperature and pressure. The local equilibrium hypothesis holds only for linear phenomenological relations, low frequencies, and long wavelengths, which makes the application of the linear nonequilibrium thermodynamics theory limited for chemical reactions. In the following sections, some of the attempts that have been made to overcome these limitations are summarized. 14.2

NETWORK THERMODYNAMICS WITH BOND GRAPH METHODOLOGY

Highly structured, organized, and coupled systems cannot be reduced to simple systems without failing to capture their unique behavior. For such complex systems, network thermodynamics combines classical and nonequilibrium thermodynamics along with the electrical network theory and kinetics to provide a practical formulation. The formulations contain information that is vital for describing the organization of the system. Therefore, the network thermodynamics method can provide insight into the system's topology and permits a systematic analysis of the dynamics of the system. It acts as a bridge between classical thermodynamics and the general dynamics theory of modern physics, and allows for the introduction of thermodynamic concepts into the approach for analyzing the system. Combined with the bond graph methodology, network thermodynamics provides a graphic representation of the processes that control the system's behavior. The governing equations can be formulated from a bond graph description of the system to evaluate the perturbations in system configurations and compositions (Mikulecky, 1994, 2001 ). In order to define the governing equations, the bond graph method identifies the cause (force) and effect (flow) relation for the energy exchange. This model can be modified easily to account for changes in the system or its environmental perturbations. Initial and boundary conditions can be related to one another within the formulations. Network thermodynamics can be used in the linear and nonlinear regions of nonequilibrium thermodynamics, and has the flexibility to deal with complex systems in which the transport and reactions occur simultaneously. The results of nonequilibrium thermodynamics based on Onsager's work can be interpreted and extended to describe coupled, nonlinear systems in biology and chemistry. The bond graph method defines the structure and constitutive equations of the system. Standard bond graph elements are used to build a model of the structure of the system. Suitable computer programs are available to generate the governing equations, and alternative methods have also been developed for deriving equivalent block diagrams, which can represent nonlinear systems.

672

14.2.1

14. Nonequilibrium thermodynamics approaches

Transport Processes

Figure 14.1 shows a typical membrane system through which a nonelectrolyte substance flows. The elements of the membrane system comprise of two reservoirs 1 and 2 containing the same substance with different levels of the chemical potentials Ix1 and IX2.In this system, the inner (1) and outer (2) compartments communicate through the membrane by the exchange of substances. The system consists of the energetic flow between two regions with different chemical potentials, and results in a flow of power. For such a system, the dissipation function and accompanying linear phenomenological equations are given by = JA A~A + J B A ~ B

(14.1)

JA = LAA A/J'A -1- LABA[/"B

(14.2)

JB = LBAA/J'A + LBBA/J'B

(14.3)

The bond graph of the transport across the membrane is shown in Figure 14.2 by a two-port resistance R element. The basic element of the bond graph is the ideal energy bond transmitting power without loss. A bond graph illustrates the system components and their interconnections with arrows, which indicate the positive direction of power flow associated with the transport processes. All time-dependent processes and all dissipative transformations are localized conceptually as capacity and resistance elements. Two ideal junctions are used in the method; the 0-junction is defined

I I I

JA

~I

JB ~1

I I I

Membrane

1

I I

I I

~;

I

I I I I I

I I I I I

~- JA ~

JB

~2

Figure 14.1. Membrane flow system.

RA ~.IA,1

XA

~-IA,2

1

~

XA

JA

TD(rA) JA/rA 0

l

~ RAB

JB/rB

TD(rB)

~tB,l

XB

~tB,2

1

~

JB

XB

l

RB Figure 14.2. Bond graph of steady-state membrane transport of two substances A and B.

14.2 Networkthermodynamics with bond graph methodology

673

in a way that all forces connected to the junction are equal, so that the sum of all the flows over a 0-junction is zero. At the 1-junction, all flows entering or exiting the junction are equal, no power accumulates, and the sum of all the forces is zero. The flow into the membrane is supplied by reservoir i with a chemical capacity Ci (14.4)

Ci - dNi dtxi

The flow can be defined for the capacitative elements as follows: d txi _

Ci - - &

dX i Ci ~ d t

dN i dt

= Ji -

(14.5)

The membrane is a resistor and transmits the flow in a dissipative process. A steady-state membrane relates the thermodynamic force X to the conjugate flow J through a resistance function R, and we have OX; _ OAtx OJ i

(14.6)

-R i

OJ i

Resistive modules represent the irreversible dissipative processes in the system. This simple model illustrates the general dissipative nature of the flows between the chambers 1 and 2 without any specific indication of the mechanisms involved. The flows J1 and J2 are due to the differences in chemical potentials AtxA = tXA,1 -- IXA,2and AIxB = IXB,1-- IXB,2. The bond graph in Figure 14.2 contains a dissipative coupling between flows A and B, in which only an interacting fraction is involved in the process. Therefore, the linear transducer TD, which converts energy from one form to another, thereby conserving power, is introduced into the bond graph. The operation of transducer is characterized by a modulus r, which may be a function of the parameter of state, such as temperature or concentration, and is independent of flows and forces. The scaling of flows and forces by the transducer gives X 2 = r X 1 and

J1

= rJ2

(14.7)

In Figure 14.2, there are two transducers, which convert the flows of A and B, and we have JA/rA and JBIrB, respectively. At the 0-junction, the coupled flow Jc is given by Yc -

Ja

./

+ v--Ab rA rB

(14.8)

The relation between the force X c and the flow Jc may be expressed by (14.9)

Xc -- RAB ( J c )

Similarly, nonlinear relations are assumed for the dissipative elements R A and RB and

X~ - RA(JA)

X~ = RB(JB)

(14.10)

The summation of the forces around the 1-junction yields !

--/'LA,I q- XA -Jr-/'~A,2 +

X

C -- 0

(14.11)

rA --/'LB,I q-)(t3 q-/J'B,2 -[- Xc -- 0 rB

(14.12)

Equations (14.11) and (14.12) may be rearranged using Eqs. (14.7), (14.9), and (14.10) A#A -- R A J A + RABJc

rA

(14.13)

674

14. Nonequilibrium thermodynamics approaches

A].I,B = RBJ B -Jr

RABJc

(14.14)

rB Equations (14.13) and (14.14) represent the nonlinear phenomenological relations between the external driving forces of permeation flows A/t£A and 2~/xB and the conjugate flows JA and JB. Linear phenomenological equations obey the Onsager reciprocal relations. For the nonlinear region, from the symmetry of the Jacobian of forces versus flows, we have (14.15) OJB

JA

JB

As RAJA and rA are independent of flow JB, from the above equation we obtain

( aZ~.k~AI _ 1 aRAB agB

JA

(14.16)

Elk oJB

Combining Eqs. (14.7) and (14.16), we obtain aAl&A ) _ 1 dRAB OJB JA EArB dgc

(14.17)

aA/xB ] _ 1 dRAB OJA )JB rBEA dgc

(14.18)

From Eqs. (14.17) and (14.18), we have (14.19)

OJB JA

JB

The symmetry of coupled membrane transport holds in a wide range of network structure applications and for the general behavior of biological networks. A thermodynamic flow system may be fully described in n-dimensions of flow and n-dimensions of conjugate force. According to Tellegen's theorem, we have jT'x = EJiX-> o

(14.20)

The above equation also shows that the general space consists of two orthogonal subspaces: the subspace of the flow (vector of flow) and the subspace of forces orthogonal to those flows. One of the consequences of this consideration is the network equivalent of the evolutionary principles of Glansdorff and Prigogine (1971). Tellegen's theorem can be used to demonstrate the existence of Onsager reciprocity and topological justification in the network formulation. The network thermodynamics model has been applied to understand the effects of diffusion coupling in the membrane transport of binary flows. In the formalism of network thermodynamics, a membrane is treated as a sequence of discrete elements called lumps, where both dissipation and storage of energy may occur. These lumps are joined in the bond graphs, and have a resistance Ri and capacitance (volume) C, which are defined by /~. -

L

DiAn

C - aL

(14.21)

(14.22)

n

where D i is the diffusion coefficient, which may change from lump to lump, n the number of lumps, and a and L the membrane area and thickness, respectively.

14.2 Networkthermodynamics with bond graph methodology

R1

C 1 R_~ C~

C~----~1--~0----~

R,,

1---~0----~

675

C,, R,,+I

.......... - ' ~ 1 - ' ~ 0 " ' ~ 1 " - ~

C,,+l

Figure 14.3. Network thermodynamic model with bond graph for a single-component flow system.

Analogous to electrical circuits, the energy flow in the bond graph can be determined. Diffusion flow J and force Xcorrespond to current and voltage, respectively, as seen in Figure 14.3, which shows the bond graph for the diffusion of a single component. Using the concept of parallel (0) and series (1) junctions, we may derive equations showing the dynamics of the transport process dJo _ J0 _ J1 dt CoR l C1R l

(14.23)

_I J2 d J1 _ J_______0__0 J______L__I J_____L__~ dt CoR 1 C1R 1 C1R2 C2R 2

(14.24)

dJn_

Jn-i

.

Jn . . CnR n

Jn Jn+l . + ~ CnRn+1 Cn+lRn+1

dt

C,,_ 1R n

dJn+l _

Jn

Jn+l

C,R,÷ 1

Cn+lRn+l

dt

(14.25)

(14.26)

There are n equations for the n lumps of the membrane system, and two equations for the adjacent reservoirs (0 and n + 1). All the lump equations have the following common form: dJ i

(14.27)

- l, Ji-1 - miJi + ~Ji+l

dt

where the indices 1, m, and r simply mean left, middle, and right, respectively. The first and the last terms on the right side incorporate the boundary conditions. If there are additional flows ofJ~_ 1 and Jn+2, then corresponding coefficients l0 and r~+l vanish, and we obtain /o=0,

li-

1

Ci_IR 1

,

mo--~,

mi-

1

ln+1 - - ,

C, Rn-~I

1

C/R/

1

r0=-~,

CoR1

C1R1

1 + ~ ,

i=0

1 r i = ~ ,

CiRi+ 1

m,,+l=~,

1

Ci+IR/+ 1

1

Cn+IR~+I

rn+l = 0 ,

(14.28)

l<-i<-n

i=n+l

(14.29)

(14.30)

With these assumptions, Eqs. (14.23)-(14.26) may be written in a compact form, and Eqs. (14.27) and (14.28)-(14.30) can be solved numerically. When there is a two-component flow with coupling, then we have two flows J,,l and J,,2 for each lump i, and a matrix of R,jk coefficients. The bond graph is modified additively to accommodate the two coupled flows, and the two-component coefficients 1,./k,m!/k, and r!/~.are expressed in terms of R,.ik (Figure 14.4)

/0,ik=0,

m0,ik = -

r,/~. =

1

Co Rl,ik

1 ro,jk = - ~ ,

,

1

Ci+I Ri+l,jk

C1 Rl,jk

,

1 -< i -< n

i=0

(14.31)

(14.32)

676

14. Nonequilibriumthermodynamicsapproaches

RI,ll C1 R2,11 C2

t Co'--~

t

1 "-~ 0"-~

Rn,ll Cn Rn+l,ll

t

t

t

1 ~

0"-~

.........

t

"-~ 1 "-~ 0"-~

-'~

Cn+ l

(Rn,12,Rn,21)c (Rn+l,12,Rn+l,21)c

(RI,12,R1,21)c (R2,12,R2,21)c

TT

II Co---~ 1 ~

t

0"-=" 1 "-~ 0---~

m

--~ 1 ~

R1,22 C1 R2,22C2

0"~

1 "-~

Cn+1

Rn,22 Cn Rn+l,22

Ngure 14.4. Networkthermodynamic model with bond graph for a two-component flow system.

1 ln+l,jk=~, CnRn+ l,jk

1

mn+l,jk=

,

rn+l,nk=O,

Cn+ l Rn+ l, jk

i=n+l

(14.33)

The formulation of the network thermodynamics bond graph can be used in modeling coupled nonlinear diffusion in two-component transport through a membrane. The linear nonequilibrium thermodynamics formulation is used in the network approach to describe the coupled diffusion of water and the cryoprotectant additive in cryopreservation of a living multicellular tissue during cell freezing, and in pancreatic islets. Standard membrane transport parameters and interstitial diffusion transport properties have been calculated for the transport of water and cryoprotective agent in pancreatic islets. Assuming that the living tissue is a porous medium, Darcy's law with temperature-dependent viscosity is used to model the flows of water and cryoprotective agent. The three independent phenomenological coefficients are expressed in terms of the water and solute permeability and the reflection coefficient. The network thermodynamics model is able to account for interstitial diffusion and storage, the transient osmotic behavior of cells and interstitium, and chemical potential transients in the tissue compartments. The bond graph method of network thermodynamics is widely used in studying homogeneous and heterogeneous membrane transport. Electroosmosis and volume changes within the compartments are the critical properties in the mechanism of cell membrane transport, and these properties can be predicted by the bond graph method of network thermodynamics. In another study, a network thermodynamics model was developed to describe the role of epithelial ion transport. The model has four membranes with series and parallel pathways and three transported ions, and simulates the system at both steady-state and transient transepithelial electrical measurements. Network thermodynamics has also been applied to nonstationary diffusion through heterogeneous membranes; concentration profiles in the composite membrane and change of the osmotic pressure have been calculated with the modified boundary and experimental conditions.

14.2.2

Chemical Reaction Processes

Chemical reactions are dissipative processes, and can be easily adapted to a network structure. When there is no diffusion in the system, dNi/dt is related to the flow of reaction Jr, which is measured by the time derivative of the advancement of reaction de/dt, and for a reaction VAA.~VBB

(14.34)

we have de : ~ dt

VA

dN__~ A : 1 dt

dN B

(14.35)

v B dt

The rate of flows in terms of components A and B can be expressed based on the forward kr and backward kb rate constants

677

14.2 Networkthermodynamics with bond graph methodology

dN A _

-]A --

--VA ( kfCA \ -- NbCB L~) -- - - P A J r

(14.36)

dt dNB _

J B --

dt

VB

vB(kfCA:' -- kbC B ) -- PBJr

(14.37)

The relationship between the capacitive and the resistive flow is expressed by (14.38)

,l i = v i J r

The above equation shows that in the bond graph structure, all flow contributions will center on the 1-junction. The driving force of the chemical reaction is the affinity A A-

-Z

(14.39)

pi [&i

The resistor function of the reaction is given by OA

(14.40)

Rr /)Jr

The resistor function is mostly nonlinear and approaches a constant value only in the vicinity of equilibrium. Combining Eqs. (14.5) and (14.38), the constitutive relation for the capacitative element Ci is (14.41)

dIJ, i _ Pi Jr dt Ci

After multiplying the above equation by (-v/), and summing over i, we obtain

-Zui

-- --

dt

. C~ )

Jr -

(14.42) dt

The change in affinity is related to the progress of the reaction as follows: dA

dA d J r

dt

d J r dt

dt

-

(14.43)

Combining this with Eq. (14.40), we have

~i Ci ) J r

(14.44)

From the above equation, we may express a typical relaxation time -r d Jr m dt

Jr

(14.45)

1"r

where the relaxation time is expressed by

Rr

(14.46)

As the above equation shows, the network structure relates the relaxation time to capacity C and resistance R, which is similar to what occurs in electrical circuits, and also provides information on any reaction far from equilibrium. The bond graph structure can be extended for multiple coupled reactions. For example, the change of ith substance in the kth chemical reaction is expressed by

678

14. Nonequilibriumthermodynamics approaches

(d__xNi dt )k

-

VikJrk

(14.47)

The total chemical transformation of the ith component is expressed by dNi - E VikJrk k

dt

(14.48)

Ci

The above equation represents a 0-junction on the capacitor of the component i, since it is a summation of various flows. Such a 0-junction divides the flow of a substrate for various chemical reactions, and retains the same chemical potential of this substrate in all chemical reactions. A model of a biphasic enzyme membrane reactor for the hydrolysis of triglycerides has been formulated according to the bond graph method of network thermodynamics, and the kinetics, the permeabilities of fatty acids and glycerides, the rates of inhibition of the immobilized enzyme, and the concentration of enzyme in a reaction zone are studied.

14.3.

MOSAIC NONEQUILIBRIUMTHERMODYNAMICS

Another attempt to overcome the phenomenological character of nonequilibrium thermodynamics is called mosaic nonequilibrium thermodynamics. In the formulation of mosaic nonequilibrium thermodynamics, a complex system is considered a mosaic of a number of independent building blocks. The species and each process are separately described and hence the biochemical and biophysical structures of the system are included in the description. The mosaic nonequilibrium thermodynamics model can be expanded to complex physical and biological systems by adding the well-characterized steps. These steps obey the thermodynamic laws and kinetic principles. The theory of mosaic nonequilibrium thermodynamics has been applied to the following biological free-energy converters (Westerhoff and Dam, 1987): (i) Bacteriorhodopsin liposomes use light as the energy source to pump proton besides receiving protons through passive permeability K +and C1-. Flow-force relations for each of the elemental processes are formulated, and by adding the flows of each chemical substance, a set of equations is obtained based on the proposed structure of the system. Verification of the mosaic nonequilibrium thermodynamics relations can be used to test the applicability of the proposed structure. If the verification is not realized, then either the formulation is in error or the proposed structure is not appropriate. In testing the formulation experimentally, some certain states, such as steady states, are assumed. The effect of the addition of ionophores on the predicted rate of light-driven proton uptake is experimentally tested; the light-driven pump is inhibited by the electrochemical gradient of protons developed by the system itself. (ii) The mosaic nonequilibrium thermodynamics formulation of oxidative phosphorylation uses the chemiosmotic model as a basis, besides assuming that the membrane has certain permeability to protons, and that the ATP synthase is a reversible H+pump coupled to the hydrolysis of ATE It is assumed that the reversibility of the reactions allows the coupled transfer of electrons in the respiratory chain for the synthesis of ATE and the proton gradient across the inner mitochondrial membrane is the main coupling agent. The following flow-force relations are used in the mosaic nonequilibrium thermodynamics formulation: Jn = LHA/XH

(14.49)

Jo = Lo (AGo + YnnnA/Xn )

(14.50)

Jp = Lp (AGp + Ynnn A/XH)

(14.51)

The terms Ln, Lo, and Lp are the transport coefficients for proton, oxygen, and ATP flows, respectively. The ~/factors describe the enzyme-catalyzed reactions with the rates having different sensitivities in the change of free energy for the proton pump and other reactions. This differential sensitivity is a characteristic of the enzyme and is reflected by the mosaic nonequilibrium thermodynamics formulation of the flow-force relationships of that enzyme. The term nu shows the number of protons translocated per ATP hydrolyzed, while Ju, Jo, and Jp indicate the flows of hydrogen, oxygen, and ATP, respectively. The mosaic nonequilibrium thermodynamics approach was also used for studying microbial growth. In a simple configuration, aerobic microbial metabolism is considered a combination of three elemental steps that are mutually

14.4 Rationalthermodynamics

679

dependent through the intracellular phosphate potential. The first is catabolism, which is the conversion of the growth-supporting energy source with the concurrent generation of ATP. The catabolic substrates are glucose and oxygen, while the catabolic products are carbon dioxide and water. The second is anabolism, where the ATP produced in catabolism is utilized for the conversion of low-molecular-weight anabolic substrates into biomass. Some of the anabolic substrates are sulfate, phosphate, glucose, and ammonia. The third is leakage, which encompasses all the processes that utilize ATP without coupling to anabolism. For example, the passive proton flow through the bacterial membrane is a leak. From these three steps, the working equations can be derived in terms of free-energy differences in the system. In mosaic nonequilibrium thermodynamics formulations not all the flows are dependent on all the free-energy differences, mainly because only a subset of catalytic components affects each flow relation. In this respect, the models differ from classical nonequilibrium thermodynamics where all flows are a function of all forces. The mosaic nonequilibrium thermodynamics model uses linear relationships between the rate of anabolism and the rate of catabolism, and the coupling is quantified through the stoichiometric coupling constant. In these linear relations, the empirical microbiological constants, such as growth rate dependency and growth rate maintenance, and maximal and theoretical growth can be projected. The mosaic nonequilibrium thermodynamics approach can accommodate biochemically known mechanisms as well as the microbial growth in more complex environments with certain simplifications. For example, the anabolic reactions can be subdivided into various distinctive sections, such as protein and lipids. Conductivities, force asymmetry factors, and stoichiometry numbers are treated as real constants. However, in practice, the composition of the microbial cell is dependent on the environmental conditions and the growth rate. Considering all these complexities, however, increases the mathematical involvement, which is a disadvantage. Efforts are being made to overcome the shortcomings of the mosaic nonequilibrium thermodynamics by developing an expanded version of it.

14.4

RATIONAL THERMODYNAMICS

Rational thermodynamics provides a method for deriving the constitutive equations without assuming local equilibrium. In this formulation, absolute temperature and entropy do not have a precise physical interpretation. It is assumed that the system has a memory, and the behavior of the system at a given time is determined by the characteristic parameters of both the present and the past. However, the general expressions for the balance of mass, momentum, and energy are still used. Rational thermodynamics is formulated based on the following hypotheses: (i) absolute temperature and entropy are not limited to near-equilibrium situations, (ii) it is assumed that systems have memories, their behavior at a given instant of time is determined by the history of the variables, and (iii) the second law of thermodynamics is expressed in mathematical terms by means of the Clausius-Duhem inequality. The balance equations were combined with the Clausius-Duhem inequality by means of arbitrary source terms, or by an approach based on Lagrange multipliers. The Clausius-Duhem equation is the fundamental inequality for a single-component system. The selection of the independent constitutive variables depends on the type of system being considered. A process is then described by solving the balance equations with the constitutive relations and the Clausius-Duhem inequality. Studies on thermodynamic restrictions on turbulence modeling show that the kinetic energy equation in a turbulent flow is a direct consequence of the first law of thermodynamics, and the turbulent dissipation rate is a thermodynamic internal variable. The principle of entropy generation, expressed in terms of the Clausius-Duhem and the ClausiusPlanck inequalities, imposes restrictions on turbulence modeling. On the other hand, the turbulent dissipation rate as a thermodynamic internal variable ensures that the mean internal dissipation will be positive and the thermodynamic modeling will be meaningful. Rational thermodynamics is not limited to linear constitutive relations, and when the constitutive equations are expressed in terms of functionals, generally a vast amount of information is necessary. Rational thermodynamics may be useful in the case of memory effects; nonequilibrium processes may approach equilibrium in a longer time than is generally assumed; as a result, natural systems have a much longer memory of irreversible processes. There are efforts to combine thermodynamic theories such as nonequilibrium thermodynamics, rational thermodynamics, and theories using evolution criteria and variational principles into a bracket formalism based on an extension of Hamiltonian mechanics. One result of this bracket approach is a general equation for nonequilibrium reversible-irreversible coupling (GENERIC) formalism for describing isolated discrete systems of complex fluids. The foundation of rational thermodynamics is the Clausius-Planck inequality defining the change of entropy between two equilibrium states, 1 and 2

680

14. Nonequilibriumthermodynamicsapproaches

AS > - i 6q T 1

(14.52)

In the rational thermodynamics formulations, the above equation becomes

ds

1

r

P ~O + Vd' -T- q -t P-T ->

(14.53)

where r is a specific rate of energy supply or energy lost and q the transport of internal energy due to conduction. Introducing the Helmholtz energy, A = u - Ts, and the following energy balance equation: du p ~ = - V.q - P : Vv + pr

(14.54)

dt

Equation ( 14.53) becomes

-P

d(A+sT)_p.v_ dt

1

T q VT -> 0

(14.55)

Here, P is the pressure tensor, and the velocity gradient splits into a symmetric (V) and an antisymmetric (W) part Vv = V + W

(14.56)

where V = 1 [Vv + (Vv)T ]. 2 m

w

1 = = [Vv - (Vv)T ]

2

Elements of the symmetric part of the velocity gradient are V/j = (l/2)[0vj/0xi + Ovi/Oxj]. Equation (14.55) is known as the Clausius-Duhem or the fundamental inequality for a single-component system. The selection of the constitutive independent variables depends on the type of system considered. For example, the density, velocity, and temperature fields in hydrodynamics are customarily chosen. A process is then described by solving the balance equations with a consideration of constitutive relations and the Clausius-Duhem inequality. For simplicity, a set of constitutive equations for a Stokesian fluid without memory is (14.57)

4~ = 4~(v,v,T, Vv, VT)

The dependence of + is expressed by ordinary functions instead of functionals. With Eq. (14.57) the Clausius-Duhem inequality becomes

/

/

-P ~Oa +s T - P ovOCt" v - P

oa

1

o(vT).(VT)-~q.VT-

U+P

) "V->0

(14.58)

where the mass conservation is expressed as

pv = V'v = V :U

(14.59)

When the constitutive equations are expressed in terms of functionals representing the whole history of the variables, a vast amount of information may be necessary.

14.5

EXTENDED NONEQUILIBRIUMTHERMODYNAMICS

Chemical process rate equations involve the quantity related to concentration fluctuations as a kinetic parameter called chemical relaxation. The stochastic theory of chemical kinetics investigates concentration fluctuations (Malyshev, 2005). For diffusion of polymers, flows through porous media, and the description liquid helium, Fick's and Fourier's laws are generally not applicable, since these laws are based on linear flow-force relations. A general formalism with the aim to go beyond the linear flow-force relations is the extended nonequilibrium thermodynamics. Polymer solutions are highly relevant systems for analyses beyond the local equilibrium theory.

681

14.5 Extended nonequilibrium thermodynamics

Extended nonequilibrium thermodynamics is concerned with the nonlinear region and deriving the evolution equations with the dissipative flows as independent variables, besides the usual conserved variables. Typical nonequilibrium variables such as flows and gradients of intensive properties may contribute to the rate of entropy generation. When the relaxation time of these variables differs from the observation time they act as constant parameters. The phenomenon becomes complex when the observation time and the relaxation time are of the same order, and the description of system requires additional variables. To coordinate components, the generalized flows and the thermodynamic forces can be used to define the trajectories of the evolution of nonequilibriun systems in time. A trajectory specifies the curve represented by the flow and force components as a function of time in the flow-force space. A useful trajectory can be found and analyzed by a variation principle. In thermodynamics, the variation principles lead to the least energy dissipation and minimum entropy generation at steady states. According to the most general evolutionary criterion, open chemical reaction systems are dissipative, and evolve toward an asymptotic state in time. Extended nonequilibrium thermodynamics is not based on the local equilibrium hypothesis, and uses the conserved variables and nonconserved dissipative fluxes as the independent variables to establish evolution equations for the dissipative fluxes satisfying the second law of thermodynamics. For conservation laws in hydrodynamic systems, the independent variables are the mass density, 9, velocity, v, and specific internal energy, u, while the nonconserved variables are the heat flux, shear and bulk viscous pressure, diffusion flux, and electrical flux. For the generalized entropy with the properties of additivity and convex function considered, extended nonequilibrium thermodynamics formulations provide a more complete formulation of transport and rate processes beyond local equilibrium. The formulations can relate microscopic phenomena to a macroscopic thermodynamic interpretation by deriving the generalized transport laws expressed in terms of the generalized frequency and wave-vector-dependent transport coefficients.

14.5.1

Extended Nonequilibrium Thermodynamics of Polymer Solutions

Extended nonequilibrium thermodynamics theory is often applied to flowing polymer solutions. This theory includes relevant fluxes and additional independent variables in describing the flowing polymer solutions. Other contemporary thermodynamic approaches for this problem are GENERIC formalism, matrix method, and internal variables (Jou and Casas-Vfizquez, 2001), which are summarized in the following sections. Polymer solutions may have the memory effects observed in viscoelastic phenomena. This requires additional relaxation terms in the constitutive equations for the viscous pressure tensor, which may be affected by the changes in the velocity gradient. Besides this, the orientation and stretching of the macromolecules may have an influence on the flow. The following summary is from Jou and Casas-Vfizquez (2001). In the extended nonequilibrium thermodynamics for a binary liquid mixture, the viscous pressure tensor pv and the diffusion flux J are considered as additional independent variables. The viscous pressure tensor, pv, by the simplest Maxwell model, is defined by the following constitutive equation:

d P " _ . .1p,, . . 2r/v dt

r

(14.60)

"r

where rl is the shear viscosity and r is the viscoelastic relaxation time for the viscous pressure tensor. In extended nonequilibrium thermodynamics of polymer solutions, the generalized extended Gibbs equation for a fluid characterized by internal energy U and viscous pressure pv is d, -:

1 T

1

1

T

T

d. + :- Pd,,-

= a.

-

dJ -

dpv

(14.61)

where T, P, u, and V are the temperature, pressure, internal energy, and volume, respectively, wl the mass fraction of the solute, A~ = tx~- ~2 the difference between the specific chemical potentials of the solute and the solvent, c~1 and OL2 the coefficients, and J the diffusion flux. In Eq. (14.61), T, P, and Atx depend on the fluxes J and Pv. Various formalisms on rheology indicate the connection between thermodynamics and dynamics. To integrate Eq. (14.61) T, P, and Atx depend also on the fluxes pv and J, while the corresponding definitions of values based on the local equilibrium quantities are defined as functions of only u, V, and wl. Assume the following entropy flux:

1 1 J, - -~-q- = A/~ J +/3P v .J 1

1

(14.62)

682

14. Nonequilibriumthermodynamics approaches

where [3 is a coupling coefficient for coupling between viscous pressure tensor and diffusion, and q the energy flux. The energy and mass balance equations are du p--=-V.q-p(V.v)-pv: dt

p

(Vv)

(14.63)

dw 1 = - V . J

(14.64)

dt

Combining Eqs. (14.62)-(14.64) yields the time derivative of the entropy 1

ds--1--V.qdt

T

(A/~)~7. J - - -

T

1 pv

"V

dJ

_

o/1J'---

T

o/2 P v "

dt

dP v

(14.65)

dt

Substituting Eqs. (14.62) and (14.65) in the general form of the balance equations of entropy ds p - - + V . J ~ =(P dt

we obtain the entropy production -o/1-d-7+V'(/3Pv) +

( P = q .V 1 + j . - V

• ---V-o/2

+/3VJ

(14.66)

For an isothermal process the simplest evolution equations for J and pv that are compatible with the positive chracter of the entropy production are --V

-- O/1 ~

1

- - -T - V - o/2

-+" V ' ( ~ e v ) -- [~1J

dP v - - ~ + flVJ = flzP v

(14.67)

(14.68)

where [31 and ~2 are the positive phenomenological coefficients which may be identified by comparing Eqs. (14.67) and (14.68) in the steady state and uncoupled forms with the Newton-Stokes' and Fick's equations PV =-2,qV,

J = -/)V(A/z)

(14.69)

where/) is related to the diffusion coefficient D by

Then, the parameters [3j and ~2 are 1

/31 - / ) T '

1

/32

2vtT

(14.70)

The respective relaxation times of diffusion flux and viscous pressure tensor are ~'1 -

o/1

~

- o/1 ( D T ) ,

T2 --

°/2

-- O/2 (2z/T)

(14.71)

Therefore, the evolution equations (14.67) and (14.68) for diffusion flux and viscous pressure tensor become dJ

'7"1 - - ~ --

- [ J +/)V(A/.,)] +/3/)TV. pv

(14.72)

683

14.6 Generic formulations

dp v 7" 2 ~

_ _(pv + 2~V) + 2/3T~VJ

(14.73)

_

dt

These equations describe the coupled phenomena between diffusion and viscous stresses, existing, for example, in diffusion of small molecules in polymer matrix. Other possible couplings occur in shear-induced diffusion and shear-induced separation. If pv is the sum of various viscous pressure contributions P(, each of them with its own relaxation time "r2i, Eq. (14.61) becomes

1 1 ds = l du +-- Pdv - -- A/x T T T

dw

vT.1 "~ V7.2i v. ~ J'dJPi dP~v DT ~i 2"qiT

1 -

(14.74)

By neglecting diffusion effects, integration of the above equation yields

s(u,v, p V ) _ Seq(u v ) _ VJ pv. pv 4T

(14.75)

'

where subscript "eq" refers to local equilibrium value and J (=~27q) is the steady-state flux, n2 being an averaged relaxation time. This equation is only a second-order expansion of the entropy in the viscous pressure. Often extended nonequilibrium thermodynamics with maximum-entropy formalism leads to more general expression for the entropy not limited to second order in fluxes.

14.6

GENERIC FORMULATIONS

GENERIC was formulated by Grmela and Ottinger (1997). The time evolution of the physical systems may be written in terms of two generators E and S and two matrices L and M to represent the essential features of the dynamics of the system

dx 6E 6S -L. +M.~ dt 6x 6x

(14.76)

where x represents a set of independent variables to describe the nonequilibrium system completely (namely hydrodynamic fields and additional structural variables), E and S are the total energy and entropy, respectively, and L and M the linear functional operators. The g/gx indicates functional derivatives, and the dot shows the multiplication of a vector by a matrix. The first term on the right of Eq. (14.76) describes the reversible part of the time evolution equations of x generated by the energy E and entropy S, while the second term represents the irreversible contributions. Equation (14.76) requires the following general and essential degeneracy conditions:

6E L.--= 6x

0,

6S M.~-0 6x

(14.77)

The first condition is for the reversible contribution of L to the time evolution of the system and requires that the functional form of the entropy is unaffected by the operator L responsible for the reversible dynamics. The second term is the conservation of the total energy by the contribution of the dynamics. In the GENERIC derivation, the following three brackets are defined:

{A, B I -

6A

6B)

6A

6B>

[A,B]- 7x,n. x

(14.78)

where {,) denotes the scalar product, the bracket {,} is the extension of the usual Poisson brackets of classical mechanics, and [,] describes the dissipative behavior. Using the chain rule and the brackets defined above, the evolution equation of an arbitrary function A becomes

dA dt

-{A,E}+[A,S]

(14.79)

684

14. Nonequilibriumthermodynamicsapproaches

The conditions of L and M become clear with the following properties of the brackets: {A,B} = - { B , A }

(antisymetric) (Jacobi identity)

{A,{B,C} } + {B,{C,A} } + {C,{A,B} } = 0

(14.80)

These properties indicate that L is antisymmetric, and restrict the possible forms of the connection mechanisms for the structural variables. The antisymmetric requirements of L guarantee the consistency of L with the structure of the equivalent Poisson bracket. The following properties:

[A,B]=[B,A],

[A,A]_>0

(14.81)

require that M is symmetric (assuming that all the variables x have the same time-reversal parity) and definite positive which leads to ds/dt >- O. The symmetry o f M is directly related to Onsager's reciprocal rules. Using two generators, E and S, provides more flexibility in the choice of variables. The behavior of the variables x under space transformation determines the matrix L. The information related to the dynamics of material describes the friction matrix M which is related to the transport coefficients. The GENERIC formulation describes the consistency between the generalized entropy and the corresponding evolution equation of the system. It suggests how to generalize entropy of extended nonequilibrium thermodynamics containing the terms beyond the second order in the viscous pressure tensor. The GENERIC formulations may be applied to polymer solutions, emulsions and blends, and polymer melts (Ottinger and Grmela, 1997; Jou and Casas-Vfizquez, 2001). The rational nonequilibrium thermodynamics and GENERIC are compared by Muschik et al. (2000).

14.7

MATRIX MODEL

In the matrix model (Jongschaap, 1990), the global thermodynamic system is composed of two separate physical parts, which are called the environment and the intemal variables. For the polymer solutions, for example, the pressure tensor pv may be the internal variables, and the classical variables density, velocity, and internal energy are the environment variables. In the matrix model, the power supplied to the system is characterized by a set of controllable extemal forces Fex and rate variables F and is given by dF pe = Fex @ ~ dt

(14.82)

where Q is the full-contracted product. The internal subsystem is characterized by a set of variables x i and the fundamental equation of the rate of change of energy is

(14.83)

Pw = I I @ dx~ dt

where II is the thermodynamic force conjugated to the respective variables xi. The total dissipation rate is the difference between the rate of power supplied to the system and the internal rate of storage of energy dF dx i A - Pwe -Pw - Fex @ - - ~ - - I I @ dt

(14.84)

When there are no internal variables, the whole supplied power would dissipate. The matrix model is derived from Eq. (14.84)

kdt)

/

(14.85)

where A = - 1/z. Here, it is assumed that the variables x i and the dissipation A remain unchanged under a reversal of sign of the rate variables dF/dt. This is called the principle of macroscopic time reversal. The upper part in Eq. (14.85) represents the flux-force relations and the bottom part is the time evolution equations for the internal variables.

685

14.8 Internal variables

For polymer solutions, and taking the p, v, and u as environment variables and pv as internal variable, the dissipation becomes

(')

dp v

6 - - ( V v ) . pv + Tq.V ~ - H . ~

->0

(14.86)

dt

The first two terms on the right are the dissipation (entropy production times the absolute temperature), same as in the linear nonequilibrium thermodynamics. The last term is the contribution of the internal variables. The pv acts as an internal variable and as a rate variable. The evolution equation for pv is dp v

1

,lq~)

"r

~"

_ _ _- ev _ - . , (Vv)

dt

(14.87)

Jongschaap et al. (1994) provided detailed examples for rheological problems with the matrix method by using the configuration function as variables. 14.8

INTERNAL VARIABLES

The theories with internal variables provide detailed description of microstructure by introducing additional variables relevant to the microstructure of the system, and enlarge the domain of application of thermodynamics. The theories of internal variables are applied in rheology, dielectric, and magnetic relaxation where the structure of the macromolecules plays a relevant role. In the theories of internal variables, it is usual to propose purely relaxational equations for the internal variables and associate the additional variables with some structure of underlying molecules. If we assume the configuration tensor W = (RR) as an independent variable, the Gibbs equation is ds -

-~ du + P

d v - ol

W'W

(14.88)

Therefore, the time derivative of the entropy becomes

It)

(14.89)

Combining mass and energy balances dv

--=V.v

P dt

du

'

p--=-V.q-P dt

v'v-pU'V

(14.90)

with Eq. (14.89), we have p-~-=V. Tq

-q.V

(14.91)

T -

Here, q / T i s the entropy flux and the first term on the right is the entropy production. The simplest constitutive equations satisfying the requirement of the positive entropy production are 1 1

dt

-~o

1 )paW

(Vv) - Lll

(14.92)

p~W

Here, the first equation is the usual Fourier law, the second relates the viscous pressure tensor to the internal variable W, and the last is the evolution of the internal variable. The matrix of the transport coefficients Lij is positive definite

686

14. Nonequilibriumthermodynamicsapproaches

with L10 = -L01 due to Onsager-Casimir reciprocal rules. Jou and Casas-Vfizquez (2001) compared the various formulations and the extended nonequilibrium thermodynamics and underlined the connection between thermodynamics and dynamics. Rubi and Perez-Madrid (2001) derived some kinetic equations of the Fokker-Planck type for polymer solutions. These equations are based on the fact that processes leading to variations in the conformation of the macromolecules can be described by nonequilibrium thermodynamics. The extension of this approach to the mesoscopic level is called the mesoscopic nonequilibrium thermodynamics, and applied to transport and relaxation phenomena and polymer solutions (Santamaria-Holek and Rubi, 2003). REFERENCES A.N. Beris and S.J. Edwards, Thermodynamics of Flowing Fluids with Internal Microstructure, Oxford University Press, New York (1994). K.R. Diller (Ed.), Biotransport, Heat and Mass Transfer in Living Systems, 858. Annals of the New York Academy of Sciences, New York Academy of Sciences, New York (1998). P. Glansdorff and I. Prigogine, Thermodynamics Theory of Structure, Stability and Fluctuations, Wiley, New York (1971). M. Grmela, D. Jou and J. Casas-Vfizquez, J Chem. Phys., 108 (1998) 7937. M. Grmela and H.C. Ottinger, Phys. Rev. E, 56 (1997) 6620. R.J.J. Jongschaap, Rep. Prog. Phys., 53 (1990) 1. R.J.J. Jongschaap, K.H. de Haas and C.A.J. Damen, J. Rheol., 38 (1994) 769. D. Jou, Extended Irreversible Thermodynamics, Springer-Verlag, New York (1996). D. Jou and J. Casas-Vfizquez, J. Non-Newton. Fluid Mech., 96 (2001) 77. D. Jou, J. Casas-Vfizquez and M. Criado-Sancho, Physica A, 262 (1999) 69. V.A. Malyshev, J. Stat. Phys., 119 (2005) 997. D.C. Mikulecky, Comp. Chem., 19 (1994) 999. D.C. Mikulecky, Comp. Chem., 25 (2001) 369. W. Muschik, S. Gumbel, M. Kroger, H.C. Ottinger, Physica A, 285 (2000) 448. H.C. Ottinger and M. Grmela, Phys. Rev. E, 56 (1997) 6633. R. Paterson, Network Thermodynamics, in: E.E. Bitter, Ed., Membrane Structure and Function, Vol. 2, Wiley, New York (1980). L. Peusner, Studies in Network Thermodynamics, Elsevier, Amsterdam (1986). H.V. Westerhoff and K.V. Dam, Thermodynamics and Control of Biological Free-Energy Transduction, Elsevier, Amsterdam (1987).

REFERENCES FOR FURTHER READING N. Despireux and G. Lebon, J. Non-Newton. Fluid Mech., 96 (2001) 105. B.J. Edwards, J. Non-Equilb. Thermodyn., 27 (2002) 5. E. Hernandez-Lemus and L.S. Garcia-Colin, J. Non-Equilib. Therm., 31 (2006) 397. G. Lebon, T. Desaive and P. Dauby, J. Appl. Trans.ASME, 73 (2006) 16. W. Muschik, C. Papenfuss and H. Ehrentraut, J. Non-Newton. Fluid Mech., 96 (2001) 255. H.C. Ottinger, Beyond Equilibrium Thermodynamics, Wiley, New York (2003). J.M. Rubi and A. Perez-Madrid, Physica A, 298 (2001) 177. I. Santamaria-Holek and J.M. Rubi, Physica A, 326 (2003) 284. ST Serdyukov, Phys. Lett. A, 324 (2004) 262.

APPENDIX APPENDIX A Tensors Scalars are specified by a single numerical value. Vectors come with directions as well as numerical values. In a threedimensional space, a tensor of rank n is determined by 3" elements. A scalar is a tensor with the rank zero, hence 30 = 1. A vector is tensor with the rank 1, hence 31 = 3. For a tensor with n = 2, we have 32 _ 9 elements. Differentiation of a tensor with respect to a scalar does not change its rank. The spatial differentiation of a tensor raises its rank by unity, and identical to multiplication by the vector V, called del or Hamiltonian operator or the nabla

V-

Ox

The gradient of a scalar field a is a vector Oa

grad a = m = Va Ox The derivative of a scalar a with respect to a vector is a vector. The gradient of a vector field v is a tensor of rank two 0V div v = m = Vv 0x

W h e n contraction is performed once (summation over repeated indices), the divergence is obtained instead of the gradient. The divergence of a vector field v is a scalar O divv---.vOx

V.v

The divergence of a tensor field T is a vector divT---.0 T=V.T Ox The Laplace operator or Laplacian is a scalar

V 2 - V. V - div grad -

0 ~

°

0 ~

Ox Ox Nonequilibrium thermodynamics often uses the Gauss-Ostrogradsky theorem, which states that the flux of a vector through a surface a is equal to the volume integral of the divergence of the vector v for the space of volume Vbounded by that surface

f v. da - f div vdV - I V. vdV a

l~

f"

Appendix

688 APPENDIX B

Table B1 Lennard-Jones (6-12) potential parameters and critical properties Species

MW

MW O)

Lennard-Jones parameters or (A)

e/K (K)

H2

2.016

-0.216

2.915

38.0

He

4.003

-0.390

2.576

10.2

Ne

20.183

2.789

35.7

Critical constants

Tc (K)

Pc atm

Vc (cm 3 mol)

33.3

12.80

65.0

34.7

2.26

57.8

25.4

5.26 44.5

26.9

/xc X 106 (g/(cm s))

41.7

156

kc X 106 (cal/(cm s K)) 79.2

Ar

39.948

0.000

3.432

1224.4

150.7

48.0

75.2

264

71.0

Kr

83.80

0.000

3.675

170.0

209.4

54.3

92.2

396

49.4

Xe

131.30

0.000

4.009

234.7

289.8

58.0

118.8

490

40.2

Air

28.9,7

0.035

3.617

97.0

132

36.4

86.6

193

90.8

N2

28.01

0.038

3.667

99.8

126.2

33.5

90.1

180

02

32.00

0.022

3.433

154.4

49.7

74.4

250.0

113

CO

28.01

0.048

3.590

110

132.9

34.5

93.1

190

CO2

44.01

0.224

3.996

190

304.2

72.8

94.1

343

86.8 105.3 86.5 122

NO

30.1

0.583

3.470

119

180

64

57

258

118.2

N20

44.01

0.141

3.879

220

309.7

71.7

96.3

332

131

SO2

64.06

0.245

4.026

363

430.7

77.8

F2

38.00

3.653

112

-

-

C12

70.91

4.115

357

417

Br2

159.82

4.268

520

584

102

I2

253.81

4.982

550

800

-

CH 4

16.04

C2H2

26.04

C2H 4

28.05

C2H 6

30.07

0.069

0.012

76.1

122

411

98.6

124

420

97

144

-

-

3.780

154

191.I

45.8

98.7

159

158

4.114

212

308.7

61.6

112.9

237

-

0.087

4.228

216

282.4

50.0

124

215

-

0.100

4.388

232

305.4

48.2

148

210

203

C3H 4

40.06

4.742

261

394.8

-

C3H 6

42.08

0.140

4.766

275

365.0

45.5

181

233

-

C3H8

44.10

0.152

4.934

273

369.8

41.9

200

228

-

n-C4Hlo

58.12

0.200

5.604

304

425.2

37.5

255

239

-

i-C4Hlo

58.12

0.181

5.393

295

408.1

36.0

263

239

-

n-CsH~2

72.15

0.252

5.850

326

469.5

33.2

311

238

-

i-CsH12 C(CH3) 4

72.15 72.15

5.812 5.759

327 312

460.4 433.8

33.7 31.6

306 304 370 432

248 254

-

492

n-C6HI4

86.18

0.301

6.264

342

507.3

29.7

n-CvH16 n-CsH18

100.20 114.23

0.350 0.400

6.663 7.035

352 361

540.1 568.7

27.0 24.5

259

-

n-C9H20 Cyclohexane

128.26 84.16

0.444 0.2t0

7.463 6.143

351 313

594.6 553

22.6 40.0

548

265

-

308

284

-

Benzene

78.11

0.210

5.443

387

562.6

48.6

260

312

-

CH3C1

50.49

0.153

4.151

355

416.3

65.9

143

338

-

CH2C12

84.93

0.199

4.748

398

510

60

CHC13

119.38

0.222

5.389

340

536.6

54

240

410

-

CC14

153.82

0.193

5.947

323

556.4

45.0

276

413

-

C2N2

52.04

4.361

349

400

59

COS

60.07

CS2

76.14

CClzF 2

0.111

120.92

4.130

336

378

61

4.483

467

552

78

170

404

-

5.116

280

384.7

39.6

218

-

-

Table B2 Collision integrals for predicting transport properties of gases at low densities KT/e or KT/gAB

~

= g~k for viscosity and

~'~D, AB = for diffusivity

thermal conductivity

KT/e or KT/SAB

f~. = fl~ for viscosity and

~-~D,AB = f o r diffusivity

thermal conductivity

0.30

2.840

2.469

2.7

1.0691

0.9782

0.35

2.676

2.468

2.8

1.0583

0.9682

0.40

2.531

2.314

2.9

1.0482

0.9588

(Contd:)

Appendix

689

Table B2 ( C o n t i n u e d )

KT/s or

{)u = ~k for viscosity and

KT/SAB

thermal conductivity

D.D.AB = for diffusivity

KT/e, or KT/eAB

flu = fla for viscosity and

~-'~D.AB =

for diffusivity

thermal conductivity

0.45

2.401

2.182

3.0

1.0388

0.9500

0.50

2.284

2.066

3.1

1.0300

0.9418

0.55

2.178

1.965

3.2

1.0217

0.9340

0.60

1.084

1.877

3.3

1.0139

0.9267

0.65

1.999

1.799

3.4

1.0066

0.9197

0.70 0.75

1.922 1.853

1.729

3.5 3.6

0.9996 0.9931

0.9131 0.9068

0.80 0.85

1.790 1.734

0.90

1.667 1.612

3.7

0.9868

0.9008

1.562

3.8

0.9809

0.8952

1.682

1.517

3.9

0.9753

0.8897

0.95

1.636

1.477

4.0

0.9699

0.8845

1.00

1.593

1.440

4.1

0.9647

0.8796

1.05

1.554

1.406

4.2

0.9598

0.8748

1.10

1.518

1.375

4.3

0.9551

0.8703

1.15

1.485

1.347

4.4

0.9506

0.8659

1.20

1.455

1.320

4.6

0.9462

0.8617

1.25

1.427

1.296

4.6

0.9420

0.8576

1.30

1.401

1.274

4.7

0.9380

0.8537

1.35

1.377

1.253

4.8

0.9341

0.8499

1.40

1.234

4.9

0.9304

0.8463

1.45

1.355 1.334

1.216

5.0

0.9268

0.8428

1.50 1.55

1.315 1.297

1.199

6.0

0.8962

0.8129

7.0

0.8727

0.7898

1.60

1.280

1 183 1 168

8.0

0.8538

0.7711

1.65

1.264

1 154

9.0

0.8380

0.7555

1.70

1.249

1141

10.0

0.8244

0.7422

1.75

1.235

1 128

12.0

0.8018

0.7202

1.80

1.222

1117

14.0

0.7836

0.7025

1.85

1.209

1 105

16.0

0.7683

0.6878

1.90 1.95

1 198

1.095

18.0

0.7552

0.6751

1 186

20.0

0.7436

0.6640

2.00

1 176

1.085 1.075

25.0

0.7198

0.6414

2.10 2.20 2.30 2.40 2.50

1 156

1.058 1.042 1.027 1.013 1.0006 0.9890

30.0

0.7010

0.6235

35.0 40.0 50.0 75.0

0.6854 0.6723 0.6510 0.6140

0.6088 0.5964 0.5763 0.5415

100.0

0.5887

0.5180

2.60

1 138 1 122 1.107 1.0933 1.0807

Table B3 Heat capacities of gases in the ideal-gas state a C'//R = a + b E + c r 2 + dT -2 T(K) from 298 to Tmax Chemical species

Tm,,~

C'/,.2~)sK/R

a

10 3 b

106 c

1O- 5 d

Paraffins Methane

CH 4

1500

4.217

1.702

9.081

Ethane

C2H~

1500

6.369

1.131

19.225

-2.164 - 5.561

Propane

C3Hs

1500

9.011

1.213

28.785

- 8.824 - 11.402

n-Butane

C4H~o

1500

11.928

1.935

36.915

Iso-butane

C4H~o

1500

11.901

1.677

37.853

- 11.945

n-Pentane

CsH~ 2

1500

14.731

2.464

45.351

- 14.111

n-Hexane

C~,H j4

1500

17.350

3.025

53.722

- 16.791

n-Heptane

CTH~(~

1500

20.361

3.570

62.127

- 19.486

CsH~ ~

1500

23.174

4.108

70.567

-22.208

C2H4

1500

5.325

1.424

14.394

- 4.392

n-Octane 1-Alkenes Ethylene

(Contd.)

Appendix

690 Table B3 ( C o n t i n u e d ) Tmax

Cp,298K/R

a

103 b

C3H6 C4H 8 CsH10 C6H12 C 7H14 C8H16

1500 1500 1500 1500 1500 1500

7.792 10.520 13.437 16.240 19.053 21.868

1.637 1.967 2.691 3.320 3.768 4.324

22.706 31.630 39.753 48.189 56.588 64.960

-6.915 -9.873 -12.447 -15.157 -17.847 -20.521

C2H40 C2H2 C 6H6 C4H6 C6H 12 C2H60 CsH 10 C2H40 CH20 CH40 CsHs CyH8

1000 1500 1500 1500 1500 1500 1500 1000 1500 1500 1500 1500

6.506 5.253 10.259 10.720 13.121 8.948 15.993 5.784 4.191 5.547 15.534 12.922

1.693 6.132 - 0.206 2.734 - 3.876 3.518 1.124 -0.385 2.264 2.211 2.050 0.290

17.978 1.952 39.064 26.786 63.249 20.001 55.380 23.463 7.022 12.216 50.192 47.052

-6.158

NH3 Br2 N20 NO NO 2 N204 O2 SO2 SO3 H20 CO CO2 CS2 C1z H2 HzS HC1

2000 1800 3000 2000 2000 2000 2000 2000 2000 2000 2000 2500 2000 1800 3000 3000 2300 2000

3.509 4.269 4.337 4.646 3.590 4.447 9.198 3.535 4.796 6.094 4.038 3.507 4.467 5.532 4.082 3.468 4.114 3.512

3.355 3.578 4.493 5.328 3.387 4.982 11.660 3.639 5.699 8.060 3.470 3.376 5.457 6.311 4.442 3.249 3.931 3.156

0.575 3.020 0.056 1.214 0.629 1.195 2.257 0.506 0.801 1.056 1.450 0.557 1.045 0.805 0.089 0.422 1.490 0.623

Chemical species Propylene 1-Butene 1-Pentene 1-Hexene 1 -Heptene 1 -Octene Organics Acetaldehyde Acetylene Benzene 1,3-Butadiene Cyclohexane Ethanol Ethylbenzene Ethylene oxide Formaldehyde Methanol Styrene Toluene Inorganics Air Ammonia Bromine Nitrous oxide Nitric oxide Nitrogen dioxide Dinitrogen tetroxide Oxygen Sulfur dioxide Sulfur trioxide Water Carbon monoxide Carbon dioxide Carbon disulfide Chlorine Hydrogen Hydrogen sulfide Hydrogen chloride

106 c

10-5d

- 1.299 -13.301 -8.882 -20.928 -6.002 -18.476 -9.296 -1.877 -3.450 -16.662 -15.716 -0.016 -0.186 -0.154 -0.928 0.014 -0.792 -2.787 • -0.227 -0.1015 -2.028 0.121 -0.031 -1.157 -0.906 -0.344 0.083 -0.232 0.151

aH.M. Spencer, Ind. Eng. Chem., 40 (1948) 2152; K.K. Kelly, US. Bur. Mines Buff., (1960) 584; L.B. Pankratz, U.S. Bur. Mines Buff., (1982) 672.

Table B4 Heat capacities of solids a C°p/R = a + bT 4- d T -2 T(K) from 298 to/max Chemical species

Tmax

C°p,298K/R

a

103 b

10-5 d

CaO CaCO3 Ca(OH)2 CaC2 CaCI2 C (graphite) Cu CuO Fe(O) FezO3 Fe304

2000 1200 700 720 1055 2000 1357 1400 1043 960 850

5.058 9.848 11.217 7.508 8.762 1.026 2.959 5.087 3.005 12.480 18.138

6.104 12.572 9.397 8.254 8.646 1.771 2.677 5.780 -0.111 11.812 9.594

0.443 2.637 5.435 1.429 1.530 0.771 0.815 0.973 6.111 9.697 27.112

- 1.047 - 3.120 - 1.042 -0.302 -0.867 0.035 -0.874 1.150 - 1.976 0.409

(Contd.)

Appendix Table B4

691

(Continued)

Chemical species

Tm~x

C'/,.29sx/R

a

103 b

10- 5 d

FeS

411

6.573

2.612

13.286

I2 LiC1

386.8 800

6.929 5.778

6.481 5.257

1.502 2.476

NHzC1

458

10.741

5.939

16.105 4.688

Na

-0.193

371

3.386

1.988

NaC1

1073

6.111

5.526

1.963

NaOH

566

7.177

0.121

16.316

NaHCO3

400

10.539

5.128

18.148

S (rhombic)

368.3

3.748

4.114

- 1.728

-0.783

SiO2 (quartz)

847

5.345

4.871

5.365

- 1.001

aK.K. Kelly,

1.948

US. But: Mines Bull., (1960) 584; L.B. Pankratz, US. Bul: Mines Bull., (1982) 672.

Table B5 Heat capacities of liquids a C~/R= a + bT + cT 2 T from 273.15 to 373.15 K Chemical species

C~_9~x/R

a

Ammonia

103 b

106 c

9.718

22.626

- 100.75

192.71

23.070

15.819

29.03

- 15.80

Benzene

16.157

-0.747

67.96

-37.78

1,3-Butadiene

14.779

22.711

- 87.96

205.79

Carbon tetrachloride

15.751

21.155

- 48.28

101.14

Chlorobenzene

18.240

11.278

32.86

-31.90

Aniline

Chloroform

13.806

19.215

-42.89

83.01

Cyclohexane

18.737

-9.048

141.38

- 161.62

Ethanol

13.444

33.866

- 172.60

349.17

Ethylene oxide

10.590

21.039

- 86.41

172.28

9.798

13.431

- 51.28

131.13

n-Propanol Sulfur trioxide

16.921 30.408

41.653 -2.930

-210.32 137.08

427.20 - 84.73

Toluene

18.611

15.133

6.79

16.35

9.069

8.712

1.25

- 0.18

Methanol

Water aJ.W. Miller, Jr., G.R. Schorr and C.L. Yaws,

Chem. Eng., 83 (1976) 129.

Table B6 Properties of some c o m m o n liquids a Substance

Boiling at 1 atm Tb (°C)

Ammonia

- 33.3

~ff-/, (kJ/kg) 1357

Freezing

Liquid properties

T~-(°C)

Z~r-/v (kJ/kg)

T (°C)

- 77.7

322.4

- 33.3

682

4.43

-20

665

4.52

639

4.60

602 1394

4.80 1.14

0 Argon Benzene Brine (20% NaC1 by mass) n-Butane Carbon dioxide

-185.9

161.6

28

5.5

126

20

879

1.72

-

20

1150

3.11

80.3

-0.5

601

2.31

0

298

0.59 2.46

80.2

394 -

-17.4

385.2

- 138.5

-0.5

230.5 (at 0°C)

C/, (kJ/(kg °C))

-189.3

103.9 -78.4

25 -185.6

p (kg/m 3)

-56.6

Ethanol

78.2

838.3

- 114.2

109

25

783

Ethyl alcohol

78.6

855

- 156

108

20

789

2.84

- 10.8

181.1

20

1109

2.84

18.9

200.6

20

1261

Ethylene glycol Glycerine

198.1

800.1

179.9

974

Helium

-268.9

22.8

Hydrogen

- 252.8

445.7

- 259.2

2.32

-

-268.9

146.2

22.8

59.5

- 252.8

70.7

10.0

(Contd.)

Appendix

692 Table B6

(Continued)

Substance

Boiling at 1 atm Tb (°C)

M-/v (kJ/kg)

Freezing

Liquid properties

Tf (°C)

M-/v (kJ/kg)

T (°C)

p (kg/m 3)

367.1 251 294.7 510.4

- 160 - 24.9 -38.9 - 182.2

105.7 11.4 58.4

- 11.7 20 25 - 161.5 - 100 25 - 195.8 -160 20 25 - 183 20 -42.1 0 50 -50 -26.1 0 25 0 25 50 75 100

593.8 820 13560 423 301 787 809 596 703 910 1141 640 581 529 449 1443 1374 1294 1206 1000 997 988 975 958

Iso-butane Kerosene Mercury Methane

- 11.7 204-293 356.7 - 161.5

Methanol Nitrogen

64.5 - 195.8

1100 198.6

-97.7 -210

99.2 25.3

124.8

306.3

-57.5

180.7

-218.8

13.7

- 187.7

80.0

Octane Oil (light) Oxygen Petroleum Propane

Refrigerant-134a

Water

ay.A. Cengel and M.A. Boles,

- 183 -42.1

-26.1

100

212.7 230-384 427.8

216.8

2257

-96.6

-

0.0

333.7

Cp (kJ/(kg °C)) 2.28 2.00 0.139 3.49 5.79 2.55 2.06 2.97 2.10 1.80 1.71 2.0 2.25 2.53 3.13 1.23 1.27 1.34 1.42 4.23 4.18 4.18 4.19 4.22

Thermodynamics.An EngineeringApproach, 4th ed., McGraw-Hill, New York (2002).

Table B7 Standard enthalpies and Gibbs energies of formation at 298.15 K Chemical species

State

M-/F (J/mol)

AGF (J/mol)

-46,110

- 16,450 -26,500 -64,900 - 1,128,790 - 748,100 - 8,101,900

Ammonia Ammonia Calcium carbide Calcium carbonate Calcium chloride Calcium chloride Calcium chloride Calcium hydroxide Calcium hydroxide Calcium oxide Carbon dioxide Carbon monoxide Hydrochloric acid Hydrogen cyanide

NH3 NH3 CaC 2 CaCO3 CaClz CaCI2 CaClz.6H20 Ca(OH)2 Ca(OH)2 CaO CO2 CO HC1 HCN

(g) (aq) (s) (s) (s) (aq) (s) (s) (aq) (s) (g) (g) (g) (g)

Hydrogen sulfide Iron oxide Iron oxide (hematite) Iron oxide (magnetite) Iron sulfide (write) Lithium chloride Lithium chloride Lithium chloride Lithium chloride

HzS FeO Fe203 Fe304 FeS2 LiC1 LiC1.H20 LiC1.2HzO LiC1.3H20

(g) (s) (s) (s) (s) (s) (s) (s) (s)

- 59,800 - 1,206,920 - 795,800 -2,607,900 -986,090 -635,090 - 393,509 - 110,525 -92,307 135,100 -20,630 -272,000 -824,200 - 1,118,400 - 178,200 -408,610 - 712,580 - 1,012,650 - 1,311.300

- 898,490 - 868,070 -604,030 - 394,359 137,169 -95,299 124,700 - 33,560 -742,200 - 1,015,400 - 166,900

(Contd.)

Appendix Table B7

693

(Continued)

Chemical species

State

Nitric acid

HNO~

Nitric acid Nitrogen oxides

HNO3 NO NO2 N20 N204 Na2CO3 Na2CO3.10H20 NaC1 NaC1 NaOH NaOH SO2 SO3 SOs H2SO 4 H2SO 4 H20 H20

(aq) (g) (g) (g) (g) (s) (s) (s) (aq) (s) (aq) (g) (g) (1) (1) (aq) (g) (1)

CH 4 C2H(, C3H~ C4Hi0 CsHI2 C6H14 CTHI~, CsHIs

Sodium carbonate Sodium carbonate Sodium chloride Sodium chloride Sodium hydroxide Sodium hydroxide Sulfur dioxide Sulfur trioxide Sulfur trioxide Sulfuric acid Sulfuric acid Water Water

(1)

~¢-/~ (J/mol) - 174,100 90,250 33,180 82,050 9,160 - 1,130,680 -4,081,320 -411,153 - 425,609 - 296,830 - 395,720 -441,040 - 813,989

AG~: (J/mol) - 80,710 - 111,250 86,550 51,310 104,200 97,540 - 1,044,440 -384,138 - 393,133 - 379,494 -419,150 - 300,194 - 371,060

- 241,818 - 285,830

-690,003 - 744,530 - 228,572 - 237.129

(g) (g) (g) (g) (g) (g) (g) (g)

- 74,520 - 83,820 - 104,680 - 125,790 - 146,760 - 166,920 - 187,780 - 208,750

- 50,460 - 31,855 -24,290 - 16,570 - 8,650 150 8,260 16,260

C2H4 C3H~ C4H ~ CsH~0 C6H12 C7H14

(g) (g) (g) (g) (g) (g)

52,510 19,710 - 540 -21,280 -41,950 -62,760

68,460 62,205 70,340 78,410 86,830

1,2-Ethanediol

C2H40 C2H402 C2H2 C6H~ C6H~, C4H(~ C6HI~ C6HI: C 2H(~O2

(g) (1) (g) (g) (1) (g) (g) (1) (1)

- 166,190 -484,500 227,480 82,930 49,080 109,240 - 123,140 - 156,230 - 454,800

- 128,860 - 389,900 209,970 129,665 124,520 149,795 31,920 26,850 - 323,080

Ethanol Ethanol Ethylbenzene Ethylene oxide Formaldehyde

C2H(~O C2H~,O CsHi0 C2H40 CH20

(g) (1) (g) (g) (g)

- 235,100 -277,690 29,920 -52,630 - 108,570

- 168,490 - 174,780 130,890 - 13,010 - 102,530

Methanol Methanol Methylcyclohexane Methylcyclohexane Styrene Toluene Toluene

CH40 CH40 C7H14 CTHt4 CsH~ CTHs CTH~

(g) (1) (g) (1) (g) (g) (1)

-

- 161,960 - 166,270 27,480 20,560 213,900 122,050 113,630

Paraffins Methane Ethane Propane n-Butane n-Pentane n-Hexane n-Heptane n-Octane

1-Alkenes Ethylene Propylene 1-Butene 1-Pentene 1-Hexene 1-Heptene Miscellaneous organics Acetaldehyde Acetic acid Acetylene Benzene Benzene 1,3-Butadiene Cyclohexane Cyclohexane

200,660 238,660 154,770 190,160 147,360 50,170 12,180

694

Appendix

Table B8 Selected state properties a

Compound (state) Glycerol (aq) Triglycerol-palmitate (s) Hydrogen H2 (g) Oxygen 02 (g) Oxygen 02 (aq) Water (1) Water (g) Hydrogen ion (aq) Hydroxyl ion (aq) Oxonium ion H 2 0 + (aq) Phosphoric acid (aq) Dihydrogen phosphate ion (aq) Hydrogen phosphate ion (aq) Phosphate ion (aq) Carbamide (urea) (aq) Carbon dioxide (g) Carbon dioxide (aq) Acetic acid (1) Acetic acid (aq) Acetate anion (aq) Ethanol (1) Ethanol (aq) Carbonic acid (aq) Bicarbonate ion (aq) Lactic acid (1) Lactic acid (aq) Lactate anion (aq) Pyruvic acid (aq) Pyruvate anion (aq) a-D-glucose (s) a-D-glucose (aq) Sucrose (s) Sucrose (aq) Palmitic acid (s) Glycerol (1)

M-/F (kJ/mol)

2~GF (kJ/mol)

S (J/(mol K))

Cp (J/(mol

K))

-488 - 2457 0 0 - 10 -286 -242 0 -230 -286

- 1277 - 319 -393 -413 -487 - 385 -485 -278 -266 -699 -690 -675 -687

- 1275 - 1264 - 1221 -882 -669

0 0 17 -237 -229 0 - 157 -237 - 1142 - 1130 - 1089 - 1019 - 204 -394 -386 -392 - 396 -369 - 175 - 182 -623

131 205 115 70 189 0 - 11 70

29 29 75 34 0 - 149 75

-222 174 214 121 160

37 123

87 161 222 191

- 518 -540 - 518 -489 -475 -910 -917 - 1544 - 1552 -305 -477

114

192

211 151 360

219 425

452

aGarby and Larsen (1995).

Table B9 Approximate standard reaction enthalpy and standard reaction Gibbs energy for some selected reactions at standard state T = 25°C, P = 1 atm a Reaction ATP and glucose reactions ATP ~ ADP + P (310K, pH 7.0, l m m o l Mg ++ Glucose to ethanol 298 K Glucose to ethanol 308 K Glucose to lactic acid (298 K, pH 7.0) Oxidation reactions (kJ/mol substrate) Ethanol 293 K Acetic acid 293 K Glucose 298 K, all components (aq) Glucose 310 K, all components (aq) Lactose Palmitic acid, all components (aq) Glycerol Triglycerides, average Protein urea, per gram, average

AH~ (kJ/mol)

AG~ (kJ/mol)

-20 -67 -66

-31 -235

-

100

-

198

-1371

-876 -2870 -2867 -5646 -9982 -1659 -34300

-2930 -9791

-17

(Contd.)

Appendix Table B9

695

(Continued)

Reaction

M-/I (kJ/mol)

Protein ---, a m m o n i a , per g r a m , average

AG,~ (kJ/mol)

- 19

Lactic acid

- 1363

Alanine

- 1621

A l a n i n e --, urea

- 1303

Urea

-634

G l u t a m i c acid --, urea

- 1930

Citric acid

- 1986

P r o t o n dissociation reactions N e u t r a l i z a t i o n by intracellular buffers

- 25

Acetic acid

-0.4

27

G l u t a m i c acid

25

Water, 298 K

56

D i s s o c i a t i o n o f strong electrolytes S o d i u m chloride

-407

"Garby and L a r s e n (1995).

APPENDIX C Table C1 Parameters for the thermal conductivity of alkanes in chloroform k(W/(mK))= Solute ( c o m p o n e n t 1)

Ei4oaixli

ao

a,

a2

a3

n-Hexane

0.11076

- 0 . 0 3 777

0.04917

0.01362

-0.02051

0.22

n-Heptane

0.11079

- 0.04462

0.11050

- 0.07782

0.02156

0.05

a4

%E ~

n-Octane

0.11081

-0.04457

0.14569

-0.14027

0.05211

0.16

3-Methylpentane

0.11079

-0.06830

0.10737

-0.04171

-0.00121

0.11

2,3-Dimethylpentane

0.11080

-0.04081

0.04925

-0.00950

-0.00453

0.17

2,2,4-Trimethylpentane

0.11080

-0.04350

0.01113

0.05298

-0.03542

0.08

"E = ~-Z,=, 1 ,,v

( f ....,~ - .[i.~xpt#;.~,,.

Table C2 Parameters for the mutual diffusion coefficients of alkanes in chloroform Solute ( c o m p o n e n t 1)

ao

al

D(10-9 m 2/s)= ~4i=o aixli

a2

a3

%E

n-Hexane

2.4394

- 1.0243

3.9930

-0.1670

-0.7710

n-Heptane

2.2368

- 0.7534

2.3321

2.7348

- 3.0324

0.08

n-Octane

2.0367

- 1.0923

3.2118

0.6468

- 1.7149

0.06 0.08

0.14

3-Methylpentane

2.2851

-0.2589

2.7260

- 1.3769

1.2063

2,3-Dimethylpentane

2.0936

- 1.1928

8.4556

- 11.7748

6.0184

0.08

2,2,4-Trimethylpentane

1.9529

-0.4573

1.2118

1.4582

- 1.0782

0.00

3 i Table C3 Parameters for the heats of transport of alkanes in chloroform - Q ( ' * ( k J / k g ) = ~_~i=oaiX~

Solute ( c o m p o n e n t 1)

a0

al

a2

n-Hexane

69.7765

- 89.6873

127.5539

- 78.3904

0.21

n-Heptane

63.8333

- 71.3086

68.3994

- 31.5108

0.21

n-Octane

61.3523

- 79.0694

110.1027

- 73.4022

0.10

3-Methylpentane

66.8969

- 75.9630

90.5502

-47.8497

0.08

2,3-Dimethylpentane

62.2974

- 71.8306

80.2944

- 47.6851

0.14

2,2,4-Trimethylpentane

59.7674

- 76.4213

99.8976

- 66.2341

0.45

%E

696

Appendix

4 Table C4 Parameters for the thermal conductivity of alkanes in carbon tetrachloride k(W/(mK)) = ~/--,;--0 aix]i

Solute (component 1)

a0

al

a2

a3

a4

%E

n-Heptane n-Octane 3-Methylpentane 2,3-Dimethylpentane 2,2,4-Trimethylpentane

0.0.9977 0.09979 0.09980 0.09969 0.09981

- 0 . 0 i 821 -0.04871 -0.06310 -0.03883 -0.07541

-0.02666 0.16253 0.15139 0.07447 0.18953

0.15875 -0.14561 -0.12861 -0.02811 -0.21637

-0.09150 0.05612 0.04580 0.0 0.10198

0.05 0.01 0.01 0.11 0.02

Table C5 Parameters for the mutual diffusion coefficients of alkanes in carbon tetrachloride D(10 -9 m2/s)-- Z4i = 0 Solute (component 1) n-Hexane n-Heptane n-Octane 3-Methylpentane 2,3-Dimethylpentane 2,2,4-Trimethylpentane

ao

al

1.5886 t.5027 1.4127 1.4940 1.3964 1.2280

1.9757 1.4798 0.3.903 1.7198 0.9422 0.4974

a2

a3

- 2.7016 -0.5978 2.3090 - 1.3199 1.1393 2.3803

6.6258 1.9214 - 1.0357 4.6769 -0.7047 - 1.9014

aixli

a4

%E

- 3.6101 - 1.1210 -0.5235 -2.5842 0.4628 0.4919

3

0.08 0.08 0.15 0.06 0.02 0.01

i

Table 06 Parameters for the heats of transport of alkanes in carbon tetrachloride -Q('* (10-9 (kJ/kg)= ~,;--oaiX~ Solute (component 1)

a0

al

a2

a3

%E

n-Hexane n-Heptane n-Octane 3-Methylpentane

68.5474 65.8806 63.0431 70.7849

-90.6118 - 85.9901 - 80.5948 - 122.0631

118.5839 111.5736 83.2332 221.6t95

-63.0289 -69.6799 - 50.3668 - 138.4118

0.01 0.01 0.13 0.07

2,3-Dimethylpentane 2,2,4-Trimethylpentane

61.3236 58.4977

- 75.3408 - 106.5203

88.1979 t 79.0495

- 55.9357 - 122.3974

0.40 0.01

APPENDIX D

Table D1 Saturated water-temperature table Temperature, T (°C)

Sat. press., /:'sat

Specific volume (m3/kg) Sat. liquid, VI

Sat. vapor, Vg

Internal energy (kJ/kg)

Enthalpy (kJ/kg)

Sat. liquid, U~

Sat. vapor, Ug

Sat. liquid, HI

Sat. vapor, Hg

Entropy (kJ/(kg K)) Sat. liquid, S~

Sat. vapor, Sg

(kPa) 0.01 5

0.6113 0.8721

0.001000 0.001000

206.14 147.12

0.0 20.97

2375.3 2382.3

0.01 20.98

2501.4 2510.6

0.000 0.0761

9.1562 9.0257

10 15

1.2276 1.7051

0.001000 0.001001

106.38 77.93

42.00 62.99

2389.2 2396.1

42.01 62.99

2519.8 2528.9

0.1510 0.2245

8.9008 8.7814

20 25 30 35

2.339 3.169 4.246 5.628

0.001002 0.001003 0.001004 0.001006

57.79 43.36 32.89 25.22

83.95 104.88 125.78 146.67

2402.9 2409.8 2416.6 2423.4

83.96 I04.89 125.79 146.68

2538.1 2547.2 2556.3 2565.3

0.2966 0.3674 0.4369 0.5053

8.6672 8.5580 8.4533 8.3531

40

7.384

0.001008

19.52

167.56

2430.1

167.57

2574.3

0.5725

8.2570

45 50 55 60

9.593 12.349 15.758 19.940

0.001010 0.001012 0.001015 0.001017

15.26 12.03 9.568 7.671

188.44 209.32 230.21 251.I1

2436.8 2443.5 2450.1 2456.6

188.45 209.33 230.23 251.13

2583.2 2592.1 2600.9 2609.6

0.6.387 0.7038 0.7679 0.8312

8.1648 8.0763 7.9913 7.9096

65

25.03

0.001020

6.197

272.02

2463.1

272.06

2618.3

0.8935

7.8310

70 75 80

31.19 38.58 47.39

0.001023 0.001026 0.001029

5.042 4.131 3.407

292.95 313.90 334.86

2469.6 2475.9 2482.2

292.98 313.93 334.91

2626.8 2635.3 2643.7

0.9549 1.0155 1.0753

7.7553 7.6824 7.6122

(Contd.)

Appendix Table D1

(Continued)

Temperature, T (°C)

Sat. press., P~,

Specific volume (m3/kg) Sat. liquid, V~

85 90 95

697

57.83 70.14 84.55

0.001033 0.001036 0.001040

Sat. vapor,

Internal energy (kJ/kg)

V,:

2.828 2.361 1.982

Enthalpy (kJ/kg)

Entropy (kJ/(kg K))

Sat. liquid, U~

Sat. vapor, ~

Sat. liquid, HI

Sat. vapor, ~

Sat. liquid, S~

Sat. vapor, Sg

355.84 376.85 397.88

2488.4 2494.5 2500.6

355.90 376.92 397.96

2651.9 2660.1 2668.1

1.1343 1.1925 1.2500

7.5445 7.4791 7.4159

418.94 440.02 461.14 482.30 503.50 524.74 546.02 567.35 588.74 610.18 631.68 653.24 674.87 696.56 718.33 740.17 762.09 784.10 806.19 828.37 850.65 873.04 895.53 918.14 940.87 963.73 986.74 1009.89 1033.21 1056.71 1080.39 1104.28 1128.39 1152.74 1177.36 1202.25 1227.46 1253.00 1278.92 1305.2 1332.0 1359.3 1387.1 1415.5 1444.6 1505.3 1570.3 164.19 1725.2 1844.0 2029.6

2506.5 2512.4 2518.1 2523.7 2529.3 2534.6 2539.9 2545.0 2550.0 2554.9 2559.5 2564.1 2568.4 2572.5 2576.5 2580.2 2583.7 2587.0 2590.0 2592.8 2595.3 2597.5 2599.5 2601.1 2602.4 2603.3 2603.9 2604.1 2604.0 2603.4 2602.4 2600.9 2599.0 2596.6 2593.7 2590.2 2586.1 2581.4 2576.0 2569.9 2563.0 2555.2 2546.4 2536.6 2525.5 2498.9 2464.6 2418.4 2351.5 2228.5 2029.6

419.04 440.15 461.30 482.48 503.71 524.99 546.31 567.69 589.13 610.63 632.20 653.84 675.55 697.34 719.21 741.17 763.22 785.37 807.62 829.98 852.45 875.04 897.76 920.62 943.62 966.78 990.12 1013.62 1037.32 1061.23 1085.36 1109.73 1134.37 1159.28 1184.51 1210.7 1235.99 1262.31 1289.07 1316.3 1344.0 1372.4 1401.3 1431.0 1461.5 1525.3 1594.2 1670.6 1760.5 1890.5 2099.3

2676.1 2683.8 2691.5 2699.0 2706.3 2713.5 2720.5 2727.3 2733.9 2740.3 2746.5 2752.4 2758.1 2763.5 2768.7 2773.6 2778.2 2782.4 2786.4 2790.0 2793.2 2796.0 2798.5 2800.5 2802.1 2803.3 2804.0 2804.2 2803.8 2803.0 2801.5 2799.5 2796.9 2793.6 2789.7 2785.0 2779.6 2773.3 2766.2 2758.1 2749.0 2738.7 2727.3 2714.5 2700.1 2665.9 2622.0 2563.9 2481.0 2332.1 2099.3

1.3069 1.3630 1.4185 1.4734 1.5276 1.5813 1.6344 1.6870 1.7391 1.7907 1.8418 1.8925 1.9427 1.9925 2.0419 2.0909 2.1396 2.1879 2.2359 2.2835 2.3309 2.3780 2.4248 2.4714 2.5178 2.5639 2.6099 2.6558 2.7015 2.7472 2.7927 2.8383 2.8838 2.9294 2.9751 3.0208 3.0668 3.1130 3.1594 3.2062 3.2534 3.3010 3.3493 3.3982 3.4480 3.5507 3.6594 3.7777 3.9147 4.1106 4.4298

7.3549 7.2958 7.2387 7.1833 7.1296 7.0775 7.0269 6.9777 6.9299 6.8833 6.8379 6.7935 6.7502 6.7078 6.6663 6.6256 6.5857 6.5465 6.5079 6.4698 6.4323 6.3952 6.3585 6.3221 6.2861 6.2503 6.2146 6.1791 6.1437 6.1083 6.0730 6.0375 6.0019 5.9662 5.9301 5.8938 5.8571 5.8199 5.7821 5.7437 5.7045 5.6643 5.6230 5.5804 5.5362 5.4417 5.3357 5.2112 5.0526 4.7971 4.4298

(MPa) 100 105 110 115 120 125 130 135 140 145 150 155 160 165 170 175 180 185 190 195 200 205 210 215 220 225 230 235 240 245 250 255 260 265 270 275 280 285 290 295 300 305 310 315 320 330 340 350 360 370 374.14

0.10135 0.12082 0.14327 0.16906 0.19853 0.2321 0.2701 0.3130 0.3613 0.4154 0.4758 0.5431 0.6178 0.7005 0.7917 0.8920 1.0021 1.1227 1.2544 1.3978 1.5538 1.7230 1.9062 2.104 2.318 2.548 2.795 3.060 3.344 3.648 3.973 4.319 4.688 5.081 5.499 5.942 6.412 6.909 7.436 7.993 8.581 9.202 9.856 10.547 11.274 12.845 14.586 16.513 18.651 21.03 22.09

0.001044 0.001048 0.001052 0.001056 0.001060 0.001065 0.001070 0.001075 0.001080 0.001085 0.001091 0.001096 0.001102 0.001108 0.001114 0.001121 0.001127 0.001134 0.001141 0.001149 0.001157 0.001164 0.001173 0.001181 0.001190 0.001199 0.001209 0.001219 0.001229 0.001240 0.001251 0.001263 0.001276 0.001289 0.001302 0.001317 0.001332 0.001348 0.001366 0.001384 0.001404 0.001425 0.001447 0.001472 0.001499 0.001561 0.001638 0.001740 0.001893 0.002213 0.003155

1.6729 1.4194 1.2102 1.0366 0.8919 0.7706 0.6685 0.5822 0.5089 0.4463 0.3928 0.3468 0.3071 0.2727 0.2428 0.2168 0.19405 0.17409 0.15654 0.14105 0.12736 0.11521 0.10441 0.09479 0.08619 0.07849 0.07158 0.06537 0.05976 0.05471 0.05013 0.04598 0.04221 0.03877 0.03564 0.03279 0.03017 0.02777 0.02557 0.02354 0.02167 0.019948 0.018350 0.016867 0.015488 0.012996 0.010797 0.008813 0.006945 0.004925 0.003155

698

Appendix

Table D2 Superheated steam T (°C)

V (m3/kg)

S~. 50 100 150 200 250 300 400 500 600 700 800 900 1000 1100 1200 1300

14.674 14.869 17.196 19.512 21.825 24.136 26.445 31.063 35.679 40.295 44.911 49.526 54.141 58.757 63.372 67.987 72.602

U (kJ/kg)

H (kJ/kg)

S (kJ/(kg K))

p = O.O01MPa(45.81°C) 2437.9 2584.7 2443.9 2592.6 2515.5 2687.5 2587.9 2783.0 2661.3 2879.5 2736.0 2977.3 2812.1 3076.5 2968.9 3279.6 3132.3 3489.1 3302.5 3705.4 3479.6 3928.7 3663.8 4159.0 3855.0 4396.4 4053.0 4640.6 4257.5 4891.2 4467.9 5147.8 4683.7 5409.7

8.1502 8.1749 8.4479 8.6882 8.9038 9.1002 9.2813 9.6077 9.8978 10.1608 10.4028 10.6281 10.8396 11.0393 11.2287 11.4091 11.5811

V (m3/kg)

3.240 3.418 3.889 4.356 4.820 5.284 6.209 7.134 8.057 8.981 9.904 10.828 11.751 12.674 13.597 14.521

U (kJ/kg)

H (kJ/kg)

P = O.05MPa(81.33°C) 2483.9 2645.9 2511.6 2585.6 2659.9 2735.0 2811.3 2968.5 3132.0 3302.2 3479.4 3663.6 3854.9 4052.9 4257.4 4467.8 4683.6

2682.5 2780.1 2877.7 2976.0 3075.5 3278.9 2488.7 3705.1 3928.5 4158.9 4396.3 4640.5 4891.1 5147.7 5409.6

S (kJ/(kg K))

7.5939 7.6947 7.9401 8.1580 8.3556 8.5373 8.8642 9.1546 9.4178 9.6599 9.8852 10.0967 10.2964 10.4859 10.6662 10.8382

Sat. 150 200 250 300 400 500 600 700 800 900 1000 1100 1200 1300

0.8857 0.9596 1.0803 1.1988 1.3162 1.5493 1.7814 2.013 2.244 2.475 2.705 2.937 3.168 3.399 3.630

P = 0.20 MPa 2529.5 2576.9 2654.4 2731.2 2808.6 2966.7 3130.8 3301.4 3478.8 3663.1 3854.5 4052.5 4257.0 4467.5 4683.2

(120.23°C) 2706.7 2768.8 2870.5 2971.0 3071.8 3276.6 3487.1 3704.0 3927.6 4158.2 4395.8 4640.0 4890.7 5147.5 5409.3

7.1272 7.2795 7.5066 7.7086 7.8926 8.2218 8.5133 8.7770 9.0194 9.2449 9.4566 9.6563 9.8458 10.0262 10.1982

0.6058 0.6339 0.7163 0.7964 0.8753 1.0315 1.1867 1.3414 1.4957 1.6499 1.8041 1.9581 2.1121 2.2661 2.4201

P = 0.30 MPa (133.55°C) 2543.6 2725.3 2570.8 2761.0 2650.7 2865.6 2728.7 2967.6 2806.7 3069.3 2965.6 3275.0 3130.0 3486.0 3300.8 3703.2 3478.4 3927.1 3662.9 4157.8 3854.2 4395.4 4052.3 4639.7 4256.8 4890.4 4467.2 5147.1 4683.0 5409.0

6.9919 7.0778 7.3115 7.5166 7.7022 8.0330 8.3251 8.5892 8.8319 9.0576 9.2692 9.4690 9.6585 9.8389 10.0110

S~. 200 250 300 350 400 500 600 700 800 900 1000 1100 1200 1300

0.3749 0.4249 0.4744 0.5226 0.5701 0.6173 0.7109 0.8041 0.8969 0.9896 1.0822 1.1747 1.2672 1.3596 1.4521

p = 0.50MPa(151.86°C) 2561.2 2748.7 2642.9 2855.4 2723.5 2960.7 2802.9 3064.2 2882.6 3167.6 2963.2 3271.9 3128.4 3483.9 3299.6 3701.7 3477.5 3925.9 3662.1 4156.9 3853.6 4394.7 4051.8 4639.1 4256.3 4889.9 4466.8 5146.6 4682.5 5408.6

6.8213 7.0592 7.2709 7.4599 7.6329 7.7938 8.0873 7.3522 8.5952 8.8211 9.0329 9.2328 9.4224 9.6029 9.7749

0.3157 0.3520 0.3938 0.4344 0.4742 0.5137 0.5920 0.6697 0.7472 0.8245 0.9017 0.9788 1.0559 1.1330 1.2101

P = 0.60MPa (158.85°C) 2567.4 2756.8 2638.9 2850.1 2720.9 2957.2 2801.0 3061.6 2881.2 3165.7 2962.1 3270.3 3127.6 3482.8 3299.1 3700.9 3477.0 3925.3 3661.8 4156.5 3853.4 4394.4 4051.5 4638.8 4256.1 4889.6 4466.5 5146.3 4682.3 5408.3

6.7600 6.9665 7.1816 7.3724 7.5464 7.7079 8.0021 8.2674 8.5107 8.7367 8.9486 9.1485 9.3381 9.5185 9.6906

S~. 100 150 200 250 300

1.6940 1.6958 1.9364 2.172 2.406 2.639

p = O.lOMPa(99.63°C) 2506.1 2675.5 2506.7 2676.2 2582.8 2776.4 2658.1 2875.3 2733.7 2974.3 2810.4 3074.3

7.3594 7.3614 7.6134 7.8343 8.0333 8.2158

(Contd.)

Appendix Table D2

(Continued)

T (°C)

V (m3/kg)

U (kJ/kg)

H (kJ/kg)

S (kJ/(kg K))

400

3.103

2967.9

3278.2

8.5435

500

3.565

600 700 800 900 1000 1100 1200

4.028 4.490 4.952

3131.6 3301.9 3479.2 3663.5 3854.8 4052.8

3488.1 3704.4 3928.2 4158.6 4396.1 4640.3

4257.3 4467.7 4683.5

4891.0 5147.6 5409.5

8.8342 9.0976 9.3398 9.5652 9.7767 9.9764 0.1659

1300

5.414 5.875 6.337 6.799 7.260

699

V (m3/kg)

U (kJ/kg)

H (kJ/kg)

S (kJ/(kg K))

6.5865 6.6940 6.9247

0.16333 0.16930 0.18234

P = 1.20 MPa (187.99°C) 2588.8 2784.8 2612.8 2815.9 2704.2 2935.0

7.1229

0.2138

2789.2

3045.8

7.0317

0.2345 0.2548

2872.2 2954.9

3153.6 3260.7

7.2121 7.3774

0.3463 0.5183

p = 0.40MPa(143.63°C) Sat. 150 200 250 30O 40O 500 600 7O0 800 900 1000 1100 1200 1300 Sat. 200 250 300 350 400 500 600 700 800 900 1000 1100 1200 1300 Sat. 200 250 300 350 400

0.4625 0.4708 0.5342 0.5951 0.6548 0.7726 0.8893 1.0055 1.1215 1.2372 1.3529 1.4685 1.5840 1.6996 1.8151 0.2404 0.2608 0.2931 0.3241 0.3544 0.3843 0.4433 0.5018 0.5601 0.6181 0.6761 0.7340 0.7919 0.8497 0.9076 0.1944 0.2060 0.2327

2553.6 2564.5

2738.6 2752.8

2646.8 2726.1 2804.8 2964.4 3129.2 3300.2 3477.9 3662.4 3853.9 4052.0

2860.5 2964.2 3066.8 3273.4 3484.9 3702.4 3926.5 4157.3 4395.1 4639.4

6.9299 7.1706 7.3789 7.5662 7.8985 8.1913 8.4558 8.6987 8.9244 9.1362 9.3360

4256.5 4467.0 4682.8

4890.2 5146.8 5408.8

9.5256 9.7060 9.8780

= 170.43°C) 2769.1 2839.3 2950.0 3056.5 3161.7 31267.1 3480.6 3699.4

6.6628 6.8158 7.0384 7.2328 7.4089 7.5716 7.8673 8.1333

3924.2 4155.6 4393.7 4638.2 4889.1 5145.9 5407.9

8.3770 8.6033 8.8153 9.0153 9.2050 9.3855 9.5575

P = 0.80MPa(T 2576.8 2630.6 2715.5 2797.2 2878.2 2959.7 3126.0 3297.9 3476.2 3661.1 3852.8 4051.0 4255.6 4466.1 4681.8

p = 1.00MPa(179.91°C) 2583.6 2778.1 2621.9 2827.9 2709.9

2942.6

0.2579 0.2825

2793.2 2875.2

3051.2 3157.7

0.3066

2957.3

6.8959

6.5233 6.5898 6.8294

3263.9

7.3011 7.4651

500 600

0.3541

3124.4

3478.5

7.7622

0.2946

3122.8

3476.3

7.6759

0.4011

3296.8

3697.9

8.0290

0.3339

3295.6

3696.3

7.9435

700

0.4478

0.3729

3474.4

3922.0

8.1881

0.4943

3923.1 4154.7

8.2731

800

3475.3 3660.4

8.4996

0.4118

3659.7

4153.8

8.4148

900 1000 1100 1200

0.5407 0.5871 0.6335 0.6798

3852.2 4050.5 4255.1 4465.6

4392.9 4637.6 4888.6 5145.4

1300

0.7261

4681.3

5407.4

8.7118 8.9119 9.1017 9.2822 9.4543

0.4505 0.4892 0.5278 0.5665 0.6051

3851.6 4050.0 4254.6 4465.1 4680.9

4392.2 4637.0 4888.0 5144.9 5407.0

8.6272 8.8274 9.0172 9.1977 9.3698

(Contd.)

700

Appendix

Table D2

(Continued)

T (°C)

V (m3/kg)

Sat. 225 250 300 350 400 500 600 700 800 900 1000 1100 1200 1300

0.12380 0.13287 0.14184 0.15862 0.17456 0.19005 0.2203 0.2500 0.2794 0.3086 0.3377 0.3668 0.3958 0.4248 0.4538

S~. 225 250 300 350 400 450 500 600 700 800 900 1000 1100 1200 1300

0.07998 0.08027 0.08700 0.09890 0.10976 0.12010 0.13014 0.13993 0.15930 0.17832 0.19716 0.21590 0.2346 0.2532 0.2718 0.2905

S~. 200 250 300 350 400 500 600 700 800 900 1000 1100 1200 1300

0.14084 0.14302 0.16350 0.18228 0.2003 0.2178 0.2521 0.2860 0.3195 0.3528 0.3861 0.4192 0.4524 0.4855 0.5186

Sat. 225 250 300 350 400 500 600

0.09963 0.10377 0.11144 0.12547 0.13857 0.15120 0.17568 0.19960

U (kJ/kg)

H (kJ/kg)

p = 1.60MPa(201.41°C) 2596.0 2794.0 2644.7 2857.3 2692.3 2919.2 2781.1 3034.8 2866.1 3145.4 2950.1 3254.2 3119.5 3472.0 3293.3 3693.2 3472.7 3919.7 3658.3 4152.1 3850.5 4390.8 4049.0 4635.8 4253.7 4887.0 4464.2 5143.9 4679.9 5406.0 p = 2.50MPa(223.99°C) 2603.1 2803.1 2605.6 2806.3 2662.6 2880.1 2761.6 3008.8 2851.9 3126.3 2939.1 3239.3 3025.5 3350.8 3112.1 3462.1 3288.0 3686.3 3468.7 3914.5 3655.3 4148.2 3847.9 4387.6 4046.7 4633.1 4251.5 4884.6 4462.1 5141.7 4677.8 5404.0 p = 1.40MPa(195.07)°C 2592.8 2790.0 2603.1 2803.3 2698.3 2927.2 2785.2 3040.4 2869.2 3149.5 2952.5 3257.5 3121.1 3474.1 3294.4 3694.8 3473.6 3920.8 3659.0 4153.0 3851.1 4391.5 4049.5 4636.4 4254.1 4887.5 4464.7 5144.4 4680.4 5406.5 P = 2.00 MPa (212.42) °C 2600.3 2628.3 2679.6 2772.6 2859.8 2945.2 3116.2 3290.9

2799.5 2835.8 2902.5 3023.5 3137.0 3247.6 3467.6 3690.1

S (kJ/(kg K))

V (m3/kg)

6.4218 6.5518 6.6732 6.8844 7.0694 7.2374 7.5390 7.8080 8.0535 8.2808 8.4935 8.6938 8.8837 9.0643 9.2364

0.11042 0.11673 0.12497 0.14021 0.15457 0.16847 0.19550 0.2220 0.2482 0.2742 0.3001 0.3260 0.3518 0.3776 0.4034

6.2575 6.2639 6.4085 6.6438 6.8403 7.0148 7.1746 7.3234 7.5960 7.8435 8.0720 8.2853 8.4861 8.6762 8.8569 9.0291

U (kJ/kg)

H (kJ/kg)

S (kJ/(kg K))

0.06668

P = 1.80MPa(207.15°C) 2598.4 2797.1 2636.6 2846.7 2686.0 2911.0 2776.9 3029.2 2863.0 3141.2 2947.7 3250.9 3117.9 3469.8 3292.1 3691.7 3471.8 3918.5 3657.6 4151.2 3849.9 4390.1 4048.5 4635.2 4253.2 4886.4 4463.7 5143.4 4679.5 5405.6 P = 3.00MPa(233.90°C) 2604.1 2804.2

6.1869

0.07058 0.08114 0.09053 0.09936 0.10787 0.11619 0.13243 0.14838 0.16414 0.17980 0.19541 0.21098 0.22652 0.24206

2644.0 2750.1 2843.7 2932.8 3020.4 3108.0 3285.0 3466.5 3653.5 3846.5 4045.4 4250.3 4460.9 4676.6

6.2872 6.5390 6.7428 6.9212 7.0834 7.2338 7.5085 7.7571 7.9862 8.1999 8.4009 8.5912 8.7720 8.9442

2855.8 2993.5 3115.3 3230.9 3344.0 3456.5 3682.3 3911.7 4145.9 4385.9 4631.6 4883.3 5140.5 5402.8

6.3794 6.4808 6.6066 6.8226 7.0100 7.1794 7.4825 7.7523 7.9983 8.2258 8.4386 8.6391 8.8290 9.0096 9.1818

6.4693 6.4975 6.7467 6.9534 7.1360 7.3026 7.6027 7.8710 8.1160 8.3431 8.5556 8.7559 8.9457 9.1262 9.2984 6.3409 6.4147 6.5453 6.7664 6.9563 7.1271 7.4317 7.7024

(Contd.)

Appendix

701

Table D2

(Continued)

T (°C)

V (m3/kg)

U (kJ/kg)

H (kJ/kg)

S (kJ/(kg K))

700 800 900 1000 1100

0.2232 0.2467 0.2700 0.2933 0.3166

3470.9 3657.0 3849.3 4048.0 4252.7

3917.4 4150.3 4389.4 4634.6 4885.9

7.9487 8.1765 8.3895 8.5901 8.7800

1200 1300

0.3398 0.3631

4463.3 5142.9 4679.0 5405.1 p = 3.50MPa(242.60°C)

8.9607 9.1329

Sat. 250 300 350 400 450 500 600 700 800 900 1000 1100 1200 1300

0.05707 0.05872 0.06842 0.07678 0.08453 0.09196 0.09918 0.11324 0.12699 0.14056 0.15402 0.16743 0.18080 0.19415 0.20749

2603.7 2803.4 2623.7 2829.2 2738.0 2977.5 2835.3 3104.0 2926.4 3222.3 3015.3 3337.2 3103.0 3450.9 3282.1 3678.4 3464.3 3908.8 3651.8 4143.7 3845.0 4384.1 4044.1 4630.1 4249.2 4881.9 4459.8 5139.3 4675.5 5401.7 P = 4.0 MPa (250.40°C)

6.1253 6.1749 6.4461 6.6579 6.8405 7.0052 7.1572 7.4339 7.6837 7.9134 8.1276 8.3288 8.5192 8.7000 8.8723

Sat. 275 300 350 400 450 500 600 700 800 900 1000 1100 1200 1300

0.04978 0.05457 0.05884 0.06645 0.07341 0.08002 0.08643 0.09885 O. 11095 0.12287 0.13469 0.14645 O. 15817 0.16987 0.18156

6.0701 6.2285 6.3615 6.5821 6.7690 6.9363 7.0901 7.3688 7.6198 7.8502 8.0647 8.2662 8.4567 8.6376 8.8100

0.04406 0.04730 0.05135 0.05840 0.06475 0.07074 0.07651 0.08765 0.09847 0.10911 0.11965 0.13013 O. 14056 0.15098 0.16139

P = 4.5 MPa (257.49°C) 2600.1 2798.3 2650.3 2863.2 2712.0 2943.1 2817.8 3080.6 2913.3 3204.7 3005.0 3323.3 3095.3 3439.6 3276.0 3670.5 3459.9 3903.0 3648.3 4139.3 3842.2 4380.6 4041.6 4627.2 4246.8 4879.3 4457.5 5136.9 4673.1 5399.4

6.0198 6.1401 6.2828 6.5131 6.7047 6.8746 7.0301 7.3110 7.5631 7.7942 8.0091 8.2108 8.4015 8.5825 8.7549

Sat. 300 350 400 450 500

0.03244 0.03616 0.04223 0.04739 0.05214 0.05665

2602.3 2801.4 2667.9 2886.2 2725.3 2960.7 2826.7 3092.5 2919.9 3213.6 3010.2 3330.3 3099.5 3445.3 3279.1 3674.4 3462.1 3905.9 3650.0 4141.5 3843.6 4382.3 4042.9 4628.7 4248.0 4880.6 4458.6 5138.1 4674.3 5400.5 P = 6.0 MPa (275.64°C) 2589.7 2784.3 2667.2 2884.2 2789.6 3043.0 2892.9 3177.2 2988.9 3301.8 3082.2 3422.2

5.8892 6.0674 6.3335 6.5408 6.7193 6.8803

0.02737 0.02947 0.03524 0.03993 0.04416 0.04814

P = 7.0 MPa (285.88°C) 2580.5 2772.1 2632.2 2838.4 2769.4 3016.0 2878.6 3158.1 2978.0 3287.1 3073.4 3410.3

5.8133 5.9305 6.2283 6.4478 6.6327 6.7965

V (m3/kg)

U (kJ/kg)

H (kJ/kg)

S (kJ/(kg K))

550

0.06101

3174.6

3540.6

7.0288

0.05195

3167.2

3530.9

6.9486

600

0.06525

3266.9

3658.4

7.1677

0.05565

3260.7

3650.3

7.0894

700

0.07352

3453.1

3894.2

7.4234

0.06283

3448.5

3888.3

7.3476

800 900 1000

0.08160 0.08958 0.09749

3643.1 3837.8 4037.8

4132.7 4375.3 4622.7

7.6566 7.8727 8.0751

0.06981 O.07669 0.08350

3639.5 3835.0 4035.3

4128.2 4371.8 4619.8

7.5822 7.7991 8.0020

1100 1200

0.10536 0.11321

4243.3 4454.0

4875.4 5133.3

8.2661 8.4474

0.09027 0.09703

4240.9 4451.7

4872.8 5130.9

8.1933 8.3747

1300

0.12106

4669.6 51396.0 P = 9.0 MPa (303.40°C) 2557.8 2742.1 2646.6 2856.0

8.6199

0.10377

4667.3

5393.7

8.5475

0.018026

P = IO.OMPa (318.06°C) 2544.4 2724.7 2610.4 2809.1

Sat. 325

0.02048 0.02327

5.6772 5.8712

0.019861

5.6141 5.7568

(Contd.)

Appendix

702 Table D2

(Continued)

T (°C)

V (m3/kg)

350 400 450 500 550 600 650 700 800 900 1000 1100 1200 1300

0.02580 0.02993 0.03350 0.03677 0.03987 0.04285 0.04574 0.04857 0.05409 0.05950 0.06485 0.07016 0.07544 0.08072

S~. 275 300 350 400 450 500 600 700 800 900 1000 1100 1200 1300

0.03944 0.04141 0.04532 0.05194 0.05781 0.06330 0.06857 0.07869 0.08849 0.09811 0.10762 0.11707 0.12648 0.13587 0.14526

S~. 300 350 400 450 500 550 600 700 800 900 1000 1100 1200 1300

0.02352 0.02426 0.02995 0.03432 0.03817 0.04175 0.04516 0.04845 0.05481 0.06097 0.06702 0.07301 0.07896 0.08489 0.09080

Sat. 350 400 450 500 550 600 650 700 800 900

0.013495 0.016126 0.02000 0.02299 0.02560 0.02801 0.03029 0.03248 0.03460 0.03869 0.04267

U (kJ/kg)

H (kJ/kg)

2724.4 2956.6 2848.4 3117.8 2955.2 3256.6 3055.2 3386.1 3152.2 3511.0 3248.1 3633.7 3343.6 3755.3 3439.3 3876.5 3632.5 4119.3 3829.2 4364.8 4030.3 4614.0 4236.3 4867.7 4447.2 5126.2 4662.7 5389.2 p = 5.0MPa(263.99°C) 2597.1 2794.3 2631.3 2838.3 2698.0 2924.5 2808.7 3068.4 2906.6 3195.7 2999.7 3316.2 3091.0 3433.8 3273.0 3666.5 3457.6 3900.1 3646.6 4137.1 3840.7 4378.8 4040.4 4625.7 4245.6 4878.0 4456.3 5135.7 4672.0 5398.2 P = 8.0 MPa (295.06°C) 2569.8 2758.0 2590.9 2785.0 2747.7 2987.3 2863.8 3138.3 2966.7 3272.0 3064.3 3398.3 3159.8 3521.0 3254.4 3642.0 3443.9 3882.4 3636.0 4123.8 3832.1 4368.3 4032.8 4616.9 4238.6 4870.3 4449.5 5128.5 4665.0 5391.5 P = 12.5 MPa (327.89°C) 2505.1 2673.8 2624.6 2826.2 2789.3 3039.3 2912.5 3199.8 3021.7 3341.8 3125.0 3475.2 3225.4 3604.0 3324.4 3730.4 3422.9 3855.3 3620.0 4103.6 3819.1 4352.5

S (kJ/(kg K))

V (m3/kg)

U (kJ/kg)

H (kJ/kg)

S (kJ/(kg K))

6.0361 6.2854 6.4844 6.6576 6.8142 6.9589 7.0943 7.2221 7.4596 7.6783 7.8821 8.0740 8.2556 8.4284

0.02242 0.02641 0.02975 0.03279 0.03564 0.03837 0.04101 0.04358 0.04859 0.05349 0.05832 0.06312 0.06789 0.07265

2699.2 2832.4 2943.4 3045.8 3144.6 3241.7 3338.2 3434.7 3628.9 3826.3 4027.8 4234.0 4444.9 4460.5

2923.4 3096.5 3240.9 3373.7 3500.9 3625.3 3748.2 3870.5 4114.8 4361.2 4611.0 4865.1 5123.8 5387.0

5.9443 6.2120 6.4190 6.5966 6.7561 6.9029 7.0398 7.1687 7.4077 7.6272 7.8315 8.0237 8.2055 8.3783

5.9734 6.0544 6.2084 6.4493 6.6459 6.8186 6.9759 7.2589 7.5122 7.7440 7.9593 8.1612 8.3520 8.5331 8.7055 5.7432 5.7906 6.1301 6.3634 6.5551 6.7240 6.8778 7.0206 7.2812 7.5173 7.7351 7.9384 8.1300 8.3115 8.4842 5.4624 5.7118 6.0417 6.2719 6.4618 6.6290 6.7810 6.9218 7.0536 7.2965 7.5182

(Contd.)

Appendix Table D2

(Continued)

T (°C)

V (m3/kg)

U (kJ/kg)

H (kJ/kg)

S (kJ/(kg K))

1000

0.04658

4021.6

4603.8

7.7237

1100

O.O5O45

4228.2

4858.8

7.9165

1200 1300

0.05430 0.05813

4439.3

.5118.0

8.0937 8.2717

Sat. 350 400 450 500 550

0.010337 0.011470 0.015649 0.018445 0.02080 0.02293

4654.8 5381.4 p = 15.0MPa(342.24°C) 2455.5 2610.5 2520.4 2692.4 2740.7 12975.5 2879.5 13156.2 2996.6 13308.6 3104.7 13448.6

600 650 700 800 900

0.02491 0.02680 0.02861 0.03210 0.03546

3208.6 3310.3 3410.9 3610.9 3811.9

1000 1100 1200 1300

0.03875 0.04200 0.04523 0.04845

375 400 425 450 500 550 600 650 700 800 900 1000 1100 1200 1300

0.0019731 0.006004 0.007881 0.009162 0.011123 0.012724 0.014137 0.015433 0.016646 0.018912 0.021045 0.02310 0.02512 0.02711 0.02910

375 400 425 450 500 550 600

0.0016407 0.0019077 0.002532 0.003693 0.005622 0.006984 0.008094

650

0.009063

700 800 900 1000

0.009941 0.011523 0.012962 0.014324

1100 1200 1300

703

V (m3/kg)

5.3098 5.4421

0.007920

5.8811 6.1404 6.3443 6.5199

0.012447 0.015174 0.017358 0.019288

13582.3 13712.3 13840.1 4092.4 4343.8

6.6776 6.8224 6.9572

0.02106 0.02274 0.02434

7.2040 7.4279

4015.4 4596.6 4222.6 4852.6 4433.8 5112.3 4649.1 5376.0 P = 25.0MPa 1798.7 11848.0 2430.1 2580.2 2609.2 2806.3 2720.7 2949.7 2884.3 3162.4 3017.5 3335.6 3137.9 3491.4 3251.6 3637.4 3361.3 3777.5 3574.3 4047.1 3783.0 4309.1 3990.9 4568.5 4200.2 4828.2 4412.0 5089.9 4626.9 5354.4 P = 40.0 MPa 1677.1 1742.8 1854.6 1930.9 2096.9 2198.1 2365.1 2512.8 2678.4 2903.3 2869.7 3149.1 3022.6 3346.4 3158.0 3520.6

U (kJ/kg)

H (kJ/kg)

P = 17.5 MPa (354.75°C) 2390.2 2528.8

0.02738 0.03031

2685.0 2844.2 2970.3 3083.9 3191.5 3296.0 3398.7 3601.8 3804.7

2902.9 3109.7 3274.1 3421.4 3560.1 3693.9 3824.6 4081.1 4335.1

7.6348 7.8283 8.0108 8.1840

0.03316 0.03597 0.03876 0.04154

4009.3 4216.9 4428.3 4643.5

4589.5 4846.4 5106.6 5370.5

4.0320 5.1418 5.4723 5.6744 5.9592 6.1765 6.3602 6.5229 6.6707 6.9345 7.1680 7.3802 7.5765 7.7605 7.9342

0.0017892 0.002790 0.005303 0.006735 0.008678 0.010168 0.011446 0.012596 0.013661 0.015623 0.017448 0.019196 0.020903 0.022589 0.024266

3.8290 4.1135 4.5029 9.9459 5.4700 5.7785 6.0144 6.2054

0.0015594 0.0017309 0.002007 0.002486 0.003892 0.005118 0.006112 0.006966

6.3750 6.6662 6.9150 7.1356

0.007727 0.009076 0.010283 0.011411

3283.6 3517.8 3739.4

3681.2 3978.7 4257.9

0.015642 0.016940

3954.6 4167.4 4380.1

4527.6 4793.1 5057.7

7.3364 7.5224

0.012496 0.013561

0.018229

4594.3

5323.5

7.6969

0.014616

Sat.

0.005834

p = 20.OMPa(365.81°C) 2293.0 2409.7

400

0.009942

2619.3

2818.1

4.9269 5.5540

450

0.012695

500 550

0.014768 0.016555

2806.2 2942.9

3060.1 3238.2

5.9017 6.1401

3062.4

3393.5

6.3348

P = 30.0 MPa 1737.8 1791.5 2067.4 2151.1 2455.1 2614.2 2619.3 2821.4 2820.7 3081.1 2970.3 3275.4 3100.5 3443.9 3221.0 3598.9 3335.8 3745.6 3555.5 4024.2 3768.5 4291.9 3978.8 4554.7 4189.2 4816.3 4401.3 5079.0 4616.0 5344.0 P = 50.0 MPa 1638.6 1716.6 1788.1 1874.6 1959.7 2060.0 2159.6 2284.0 2525.5 2720.1 2763.6 3019.5 2942.0 3247.6 3093.5 3441.8 3230.5 3479.8 3710.3 3930.5 4145.7 4359.1 4572.8

S (kJ/(kg K))

5.1419 5.7213 6.0184 6.2383 6.4230 6.5866 6.7357 6.8736 7.1244 7.3507 7.5589 7.7531 7.9360 8.1093 3.9305 4.4728 5.1504 5.4424 5.7905 6.0342 6.2331 6.4058 6.5606 6.8332 7.0718 7.2867 7.4845 7.6692 7.8432 3.7639 4.0031 4.2734 4.5884 5.1726 5.5485 5.8178 6.0342

3616.8 3933.6 4224.4 4501.1

6.2189 6.5290 6.7882 7.0146

4770.5 5037.2

7.2184 7.4058

5303.6

7.5808

(Contd.)

Appendix

704 Table D2 (Continued) T (°C)

V (m3/kg)

U (kJ/kg)

600 650 700 800 900 1000 1100 1200 1300

0.018178 0.019693 0.02113 0.02385 0.02645 0.02897 0.003145 0.03391 0.03636

375 400 425 450 500 550 600 650 700 800 900 1000 1100 1200 1300

0.0017003 0.002100 0.003428 0.004961 0.006927 0.008345 0.009527 0.010575 0.011533 0.013278 0.014883 0.016410 0.017895 0.019360 0.020815

375 400 425 450 500 550 600 650 700 800 900 1000 1100 1200 1300

0.0015028 0.0016335 0.0018165 0.002085 0.002956 0.003956 0.004834 0.005595 0.006272 0.007459 0.008508 0.009480 0.010409 0.011317 0.012215

H (kJ/kg)

S (kJ/(kg K))

3174.0 3537.6 3281.4 3675.3 3386.4 3809.0 3592.7 4069.7 3797.5 4326.4 4003.1 4582.5 4211.3 4840.2 4422.8 5101.0 4638.0 5365.1 P = 35.0MPa 1702.9 1762.4 1914.1 1987.6 2253.4 2373.4 2498.7 2672.4 2751.9 2994.4 2921.0 3213.0 3062.0 3395.5 3189.8 3559.9 3309.8 3713.5 3536.7 4001.5 3754.0 4274.9 3966.7 4541.1 4178.3 4804.6 4390.7 5068.3 4605.1 5333.6 P = 60.OMPa 1609.4 1699.5 1745.4 1843.4 1892.7 2001.7 2053.9 2179.0 2390.6 2567.9 2658.8 2896.2 2861.1 3151.2 3028.8 3364.5 3177.2 3553.5 3441.5 3889.1 3681.0 4191.5 3906.4 4475.2 4124.1 4748.6 4338.2 5017.2 4551.4 5284.3

V (m3/kg)

U (kJ/kg)

H (kJ/kg)

S (kJ/(kg K))

6.5048 6.6582 6.7993 7.0544 7.2830 7.4925 7.6874 7.8707 8.0442 3.8722 4.2126 4.7747 5.1962 5.6282 5.9026 6.1179 6.3010 6.4631 6.7450 6.9386 7.2064 7.4037 7.5910 7.7653 3.7141 3.9318 4.1626 4.4121 4.9321 5.3441 5.6452 5.8829 6.0824 6.4109 6.6805 6.9127 7.1195 7.3083 7.4837

APPENDIX E Table E1 Saturated refrigerant-134a properties-Temperature

T (°C) Psat(MPa)

-40 -36 -32 -28 -26 -24 -22

0.05164 0.06332 0.07704 0.09305 0.10199 0.11160 0.12192

Entropy (kJ/(kg K))

Enthalpy (kJ/kg)

Specific volume (m3/kg)

Internal energy (kJ/kg)

Sat. liquid

Sat. vapor

Sat. liquid

Sat. vapor

Sat. liquid

Evap.

Sat. vapor

Sat. liquid

Sat. vapor

0.0007055 0.0007113 0.0007172 0.0007233 0.0007265 0.0007296 0.0007328

0.3569 0.2947 0.2451 0.2052 0.1882 0.1728 0.1590

-0.04 4.68 9.47 14.31 16.75 19.21 21.68

204.45 206.73 209.01 211.29 212.43 213.57 214.70

0.00 4.73 9.52 14.37 16.82 19.29 21.77

222.88 220.67 218.37 216.01 214.80 213.57 212.32

222.88 225.40 227.90 230.38 231.62 232.85 234.08

0.0000 0.0201 0.0401 0.0600 0.0699 0.0798 0.0897

0.9560 0.9506 0.9456 0.9411 0.9390 0.9370 0.9351

(Contd.)

Appendix

705

Table E1 (Continued) T (°C)

-20 -18 -16 -12 -8 -4 0 4 8 12 16 20 24 26 28 30 32 34 36 38 40 42 44 48 52 56 60 70 80 90 100

Psat(MPa)

0.13299 0.14483 0.15748 0.18540 0.21704 0.25274 0.29282 0.33765 0.38756 0.44294 0.50416 0.57160 0.64566 0.68530 0.72675 0.77006 0.81528 0.86247 0.91168 0.96298 1.0164 1.0720 1.1299 1.2526 1.3851 1.5278 1.6813 2.1162 2.6324 3.2435 3.9742

Specific volume (m3/kg) Sat. liquid

Sat. vapor

Sat. liquid

Sat. vapor

0.0007361 0.0007395 0.0007428 0.0007498 0.0007569 0.0007644 0.0007721 0.0007801 0.007884 0.0007971 0.0008062 0.0008157 0.0008257 0.0008309 0.0008362 0.0008417 0.0008473 0.0008530 0.0008590 0.0008651 0.0008714 0.0008780 0.0008847 0.0008989 0.0009142 0.0009308 0.0009488 0.0010027 0.0010766 0.0011949 0.0015443

0.1464 0.1350 0.1247 0.1068 0.0919 0.0794 0.0689 0.0600 0.0525 0.0460 0.0405 0.0358 0.0317 0.0298 0.0281 0.0265 0.0250 0.0236 0.0223 0.0210 0.0199 0.0188 0.0177 0.0159 0.0142 0.0127 0.0114 0.0086 0.0064 0.0046 0.0027

24.17 26.67 29.18 34.25 39.38 44.56 49.79 55.08 60.43 65.83 71.29 76.80 82.37 85.18 88.00 90.84 93.70 96.58 99.47 102.38 105.30 108.25 111.22 117.22 123.31 129.51 135.82 152.22 169.88 189.82 218.6(i)

215.84 216.97 218.10 220.36 222.60 224.84 227.06 229.27 231.46 233.63 235.78 237.91 240.01 241.05 242.08 243.10 244.12 245.12 246.11 247.09 248.06 249.02 249.96 251.79 253.55 255.23 256.81 260.15 262.14 261.34 248.49

Entropy (kJ/(kg K))

Enthalpy (kJ/kg)

Internal energy (kJ/kg)

Sat. liquid 24.26 26.77 29.30 34.39 39.54 44.75 50.02 55.35 60.73 66.18 71.69 77.26 82.90 85.75 88.61 91.49 94.39 07.31 100.25 103.21 106.19 109.19 112.22 118.35 124.58 130.93 137.42 154.34 172.71 193.69 224.74

Evap.

Sat. vapor

Sat. liquid

Sat. vapor

211.05 209.76 208.45 205.77 203.00 200.15 197.21 194.19 191.07 187.85 184.52 181.09 177.55 175.73 173.89 172.00 170.09 168.14 166.15 164.12 162.05 159.94 157.79 153.33 148.66 143.75 138.57 124.08 106.41 82.63 34.40

235.31 236.53 237.74 240.15 242.54 244.90 247.23 249.53 251.80 254.03 256.22 258.35 260.45 261.48 262.50 263.50 264.48 265.45 266.40 267.33 268.24 269.14 270.01 271.68 273.24 274.68 275.99 278.43 279.12 276.32 259.13

0.0996 0.1094 0.1192 0.1388 0.1583 0.1777 0.1970 0.2162 0.2354 0.2545 0.2735 0.2924 0.3113 0.3208 0.3302 0.3396 0.3490 0.3584 0.3678 0.3772 0.3866 0.3960 0.4054 0.4243 0.4432 0.4622 0.4814 0.5302 0.5814 0.6380 0.7196

0.9332 0.9315 0.9298 0.9267 0.9239 0.9213 0.9190 0.9169 0.9150 0.9132 0.9116 0.9102 0.9089 0.9082 0.9076 0.9070 0.9064 0.9058 0.9053 0.9047 0.9041 0.9035 0.9030 0.9017 0.9004 0.8990 0.8973 0.8918 0.8827 0.8655 0.8117

Table E2 Saturated refrigerant-134a properties-Pressure P

T~.t

(MPa)

(°C)

0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20 0.24 0.28 0.32 0.36 0.4 0.5 0.6 0.7 0.8 0.9

-37.07 -31.21 -26.43 -22.36 -18.80 -15.62 -12.73 -10.09 -5.37 -1.23 2.48 5.84 8.93 15.74 21.58 26.72 31.33 35.53

Specific volume (m3/kg) Sat. liquid 0.0007097 0.0007184 0.0007258 0.0007323 0.0007381 0.0007435 0.0007485 0.0007532 0.0007618 0.0007697 0.0007770 0.0007839 0.0007904 0.0008056 0.0008196 0.0008328 0.0008454 0.0008576

Internal energy (kJ/kg)

Enthalpy (kJ/kg)

Sat. vapor

Sat. liquid

Sat. vapor

Sat. liquid

Evap.

0.3100 0.2366 0.1917 0.1614 0.1395 0.1229 0.1098 0.0993 0.0834 0.0719 0.0632 0.0564 0.0509 0.0409 0.0341 0.0292 0.0255 0.0226

3.41 10.41 16.22 21.23 25.66 29.66 33.31 36.69 42.77 48.18 53.06 57.54 61.69

206.12 209.46 212.18 214.50 216.52 218.32 219.94 221.43 224.07 226.38 228.43 230.28 231.97

70.93 78.99 86.19 92.75 98.79

235.64 238.74 241.42 243.78 245.88

3.46 10.47 16.29 21.32 25.77 29.78 33.45 36.84 42.95 48.39 53.31 57.82 62.00 71.33 79.48 86.78 93.42 99.56

221.27 217.92 215.06 212.54 210.27 208.18 206.26 204.46 201.14 198.13 195.35 192.76 190.32 184.74 179.71 175.07 170.73 166.62

Entropy (kJ/(kg K)) Sat. vapor 224.72 228.39 231.35 233.86 236.04 237.97 239.71 241.30 244.09 246.52 248.66 250.58 252.32 256.07 259.19 261.85 264.15 266.18

Sat. liquid

Sat. vapor

0.0147 0.0440 0.0678 0.0879 0.1055 0.1211 0.1352 0.1481 0.1710 0.1911 0.2089

0.9520 0.9447 0.9395 0.9354 0.9322 0.9295 0.9273 0.9253 0.9222 0.9197 0.9177

0.2251 0.2399 0.2723 0.2999 0.3242 0.3459 0.3656

0.9160 0.9145 0.9117 0.9097 0.9080 0.9066 0.9054

(Contd.)

Appendix

706 Table E2 (Continued) P

Tsa t

Specific volume (m3/kg)

(MPa)

(°C)

Sat. liquid

1.0 1.2 1.4 1.6 1.8 2.0 2.5 3.0

39.39 46.32 52.43 57.92 62.91 67.49 77.59 86.22

Sat. vapor

0.0008695 0.0008928 0.0009159 0.0009392 0.0009631 0.0009878 0.0010562 0.0011416

0.0202 0.0166 0.0140 0.0121 0.0105 0.0093 0.0069 0.0053

Internal energy (kJ/kg)

Enthalpy (kJ/kg)

Entropy (kJ/(kg K))

Sat. liquid

Sat. vapor

Sat. liquid

Evap.

Sat. vapor

104.42 114.69 123.98 132.52 140.49 148.02 165.48 181.88

247.77 251.03 253.74 256.00 257.88 259.41 261.84 262.16

105.29 115.76 125.26 134.02 142.22 149.99 168.12 185.30

162.68 155.23 148.14 141.31 134.60 127.95 111.06 92.71

267.97 270.99 273.40 275.33 276.83 277.94 279.17 278.01

Sat. liquid

Sat. vapor

0.3838 0.4164 0.4453 0.4714 0.4954 0.5178 0.5687 0.6156

0.9043 0.9023 0.9003 0.8982 0.8959 0.8934 0.8854 0.8735

Table E3 Superheated refrigerant-134a T (°C)

S~. -20 -10 0 10 20 30 40 50 60 70 80 90

V (m3/kg)

0.31003 0.33536 0.34992 0.36433 0.37861 0.39279 0.40688 0.42091 0.43487 0.44879 0.46266 0.47650 0.49031

u (kJ/kg)

h (kJ/kg)

P = 0.06MPa (Tsat 206.12 217.86 224.97 232.24 239.69 247.32 255.12 263.10 271.25 279.58 288.08 296.75 305.58

--

-37.07°C) 224.72 237.98 245.96 254.10 262.41 270.89 279.53 288.35 297.34 306.51 315.84 325.34 335.00

s (kJ/(kg K))

V (m3/kg)

u (kJ/kg)

0.9520 1.0062 1.0371 1.0675 1.0973 1.1267 1.1557 1.1844 1.2126 1.2405 1.2681 1.2954 1.3224

0.19170 0.19770 0.20686 0.21587 0.22473 0.23349 0.24216 0.25076 0.25930 0.26779 0.27623 0.28464 0.29302

P = 0.10 MPa 212.18 216.77 224.01 231.41 238.96 246.67 254.54 262.58 270.79 279.16 287.70 296.40 305.27

0.9273 0.9362 0.9684 0.9998 1.0304 1.0604 1.0898 1.1187 1.1472 1.1753 1.2030 1.2303 1.2573

0.09933 0.09938 0.10438 0.10922 0.11394 0.11856 0.12311 0.12758 0.13201 0.13639 0.14073 0.14504 0.14932

0.9197

0.06322

0.9238 0.9566 0.9883 1.0192 1.0494 1.0789 1.1079 1.1364 1.1644 1.1920 1.2193

0.06576 0.06901 0.07214 0.07518 0.07815 0.08106 0.08392 0.08674 0.08953 0.09229

h (kJ/kg) (Tsa t -"

-26.43°C) 231.35 236.54 244.70 252.99 261.43 270.02 278.76 287.66 296.72 305.94 315.32 324.87 334.57

s (kJ/(kg K))

0.9395 0.9602 0.9918 1.0227 1.0531 1.0829 1.1122 1.1411 1.1696 1.1977 1.2254 1.2528 1.2799

100 SN. -10

0 10 20 30 40 50 60 70 80 90 100

0.10983 0.11135 0.11678 0.12207 0.12723 0.13230 0.13730 0.14222 0.14710 0.15193 0.15672 0.16148 0.16622

S~.

0.07193

0 10 20 30 40 50 60 70 80 90 100

0.07240 0.07613 0.07972 0.08320 0.08660 0.08992 0.09319 0.09641 0.09960 0.10275 0.10587

P = 0.18MPa(Tsat = -12.73°C) 219.94 239.71 222.02 242.06 229.67 250.69 237.44 259.41 245.33 268.23 253.36 277.17 261.53 286.24 269.85 295.45 278.31 304.79 286.93 314.28 295.71 323.92 304.63 333.70 313.72 343.63 P = 0.28 MPa (~at = - 1"23°C) 226.38 246.52 227.37 235.44 243.59 251.83 260.17 268.64 277.23 285.96 294.82 303.83 312.98

247.64 256.76 265.91 275.12 284.42 293.81 303.32 312.95 322.71 332.60 342.62

P = 0.20MPa(Tsat = -10.09°C) 221.43 241.30 221.50 241.38 229.23 250.10 237.05 258.89 244.99 267.78 253.06 276.77 261.26 285.88 269.61 295.12 278.10 304.50 286.74 314.02 295.53 323.68 304.47 333.48 313.57 343.43 P = 0.32 MPa (Tsat = 2.48°C) 228.43 248.66 234.61 242.87 251.19 259.61 268.14 276.79 285.56 294.46 303.50 312.68

255.65 264.95 274.28 283.67 293.15 302.72 312.41 322.22 332.15 342.21

0.9253 0.9256 0.9582 0.9898 1.0206 1.0508 1.0804 1.1094 1.1380 1.1661 1.1939 1.2212 1.2483 0.9177 0.9427 0.9749 1.0062 1.0367 1.0665 1.0957 1.1243 1.1525 1.1802 1.1076

(Contd.)

Appendix Table E3 T (°C)

707

(Continued) V (m3/kg)

u (kJ/kg)

h (kJ/kg)

352.78 363.08

s (kJ/(kg K))

1.2461 1.2727

V (m3/kg)

0.09503 0.09774

u (kJ/kg)

h (kJ/kg)

322.00 331.45

352.40 362.73

s (kJ/(kg K))

1.2345 1.2611

110

0.10897

322.27

120 130 140

0.11205

331.71

Sat.

0.13945

-20 -10 0 10

0.14549 0.15219 0.15875

223.03 230.55 238.21

243.40 251.86 260.43

0.9606 0.9922 1.0230

20 30

0.16520 0.17155

246.01 253.96

269.13 277.97

1.0532 1.0828

40 50 60 70 80 90 100

0.17783 0.18404 0.19020 0.19633 0.20241 0.20846 0.21449

262.06 270.32 278.74 287.32 296.06 304.95 314.01

286.96 296.09 305.37 314.80 324.39 334.14 344.04

1.1120 1.1407 1.1690 1.1969 1.2244 1.2516 1.2785

Sat. -10 0 10 20 30 40 50 60 70 80 90 100

0.08343

224.07

244.09

(t.9222

0.08574 0.08993 0.09339 0.09794 0.10181 0.10562 0.10937 0.11307 0.11674 0.12037 0.12398

228.31 236.26 244.30 252.45 260.72 269.12 277.67 286.35 295.18 304.15 313.27

248.89 257.84 266.85 275.95 285.16 294.47 303.91 313.49 323.19 333.04 343.03

0.9399 0.9721 1.0034 1.0339 1.0637 1.0930 1.1218 1.1501 1.1780 1.2055 1.2326

Sat.

0.05089

231.97

252.32

0.9145

0 10 20 30 40 50 60 70 80 90

0.05119 0.05397 0.05662 0.05917 0.06164 0.06405 0.06641 0.06873 0.07102

232.87 241.37 249.89 258.47 267.13 275.89 284.75 293.73 302.84

253.35 262.96 272.54 282.14 291.79 301.51 311.32 321.23 331.25

0.9182 0.9515 0.9837 1.01448 1.0452 1.0748 1.1038 1.1322 1.1602

100 110

0.07327 0.07550

312.07 321.44

341.38 351.64

1.1878 1.2149

120

0.07771

330.94

362.03

1.2417

130

0.07991

340.58

372.54

1.2681

140

0.08208

350.35

383.18

1.2941

Sat. 20

0.04086 0.04188

253.64 239.40

256.07 260.34

0.9117 0.9264

0.03408

238.74

259.19

0.9097

30 40 50

0.04416 0.04633 0.04842

248.20 256.99 265.83

270.28 280.16 290.04

0.9597 0.9918 1.0229

60

0.05043

274.73

299.95

1.0531

0.03581 0.03774 0.03958 0.04134

246.41 255.45 264.48 273.54

267.89 278.09 288.23 298.35

0.9388 0.9719 1.0037 1.0346

P = O.14MPa(C~. = -18.80°C) 216.52 236.04

0.9322

P = 0.24 MPa (T~ = -5.37°C)

P = 0.40MPa(~=

8.93°C)

P = 0 . 5 0 M P a ( ~ t = 15.74°C)

P = 0.60 MPa (Tsat = 21.58°C)

(Contd.)

Appendix

708 Table E3

(Continued) V(m3/kg)

(kJ/kg)

h (kJ/kg)

70 80 90 100 110 120 130 140 150 160

0.05240 0.05432 0.05620 0.05805 0.05988 0.06168 0.06347 0.06524

283.72 292.80 302.00 311.31 320.74 330.30 339.98 349.79

309.92 319.96 330.10 340.33 350.68 361.14 371.72 382.42

1.0825 1.1114 1.1397 1.1675 1.1949 1.2218 1.2484 1.2746

0.04304 0.04469 0.04631 0.04790 0.04946 0.05099 0.05251 0.05402 0.05550 0.05698

Sat. 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180

0.02547 0.02691 0.02846 0.02992 0.03131 0.03264 0.03393 0.03519 0.03642 0.03762 0.03881 0.03997 0.04113 0.04227 0.04340 0.04452

p = 0.80MPa(Tsa t 243.78 252.13 261.62 271.04 280.45 289.89 299.37 308.93 318.57 328.31 338.14 348.09 358.15 368.32 378.61 389.02

= 31.33°C) 264.15 273.66 284.39 294.98 305.50 316.00 326.52 337.08 347.71 358.40 369.19 380.07 391.05 402.14 413.33 424.63

0.9066 0.9374 0.9711 1.0034 1.0345 1.0647 1.0940 1.1227 1.1508 1.1784 1.2055 1.2321 1.2584 1.2843 1.3098 1.3351

0.02255 0.02325 0.02472 0.02609 0.02738 0.02861 0.02980 0.03095 0.03207 0.03316 0.03423 0.03529 0.03633 0.03736 0.03838 0.03939

Sat. 50 60 70 80 90 100 110 120 130 140 150 16O 170 180 190 200

0.01663 0.01712 0.01835 0.01947 0.02051 0.02150 0.02244 0.02335 0.02423 0.02508 0.02592 0.02674 0.02754 0.02834

P = 1.20MPa(Tsat 251.03 254.98 265.42 275.59 285.62 295.59 305.54 315.50 325.51 335.58 345.73 355.95 366.27 376.69

= 46.32°C) 270.99 275.52 287.44 298.96 310.24 321.39 332.47 343.52 354.58 365.68 376.83 388.04 399.33 410.70

0.9023 0.9164 0.9527 0.9868 1.0192 1.0503 1.0804 1.1096 1.1381 1.1660 1.1933 1.2201 1.2465 1.2724

0.01405

422.16

1.2980

0.02472 0.02541 0.02608

Sat.

0.02918

20 30 40 50 60 70 80 90 100 110 120

0.02979 0.03157 0.03324 0.03482 0.03634 0.03781 0.03924 0.04064 0.04201 0.04335

T (°C)

0.02912

387.21

P = 0.70MPa(~at = 26.72°C) 241.42 261.85 244.51 253.83 263.08 272.31 281.57 290.88 300.27 309.74 319.31 328.98

265.37 275.93 286.35 296.69 307.01 317.35 327.74 338.19 348.71 359.33

s (kJ/(kg K))

V (m3/kg)

0.01495 0.01603 0.01701 0.01792 0.01878 0.01960 0.02039 0.02115 0.02189 0.02262 0.02333 0.02403

u (kJ/kg)

h (kJ/kg)

282.66 291.86 301.14 310.53 320.03 329.64 339.38 349.23 359.21 369.32 P = 0.90 MPa ( T s a t 245.88 250.32 260.09 269.72 279.30 288.87 298.46 308.11 317.82 327.62 337.52 347.51 357.61 367.82 378.14 388.57 P = 1.40 MPa ( T s a t 253.74

308.48 318.67 328.93 339.27 349.70 360.24 370.88 381.64 392.52 403.51 = 35.53°C) 266.18 271.25 282.34 293.21 303.94 314.62 325.28 335.96 346.68 357.47 368.33 379.27 390.31 401.44 412.68 424.02 - " 52.43°C) 273.40

s (kJ/(kg K)) 1.0645 1.0938 1.1225 1.1505 1.1781 1.2053 1.2320 1.2584 1.2844 1.3100 0.9054 0.9217 0.9566 0.9897 1.0214 1.0521 1.0819 1.1109 1.1392 1.1670 1.1943 1.2211 t.2475 1.2735 1.2992 1.3245 0.9003

262.17 272.87

283.10 295.31

0.9297 0.9658

283.29 293.55 303.73 313.88 324.05 334.25 344.50 354.82 365.22 375.71 386.29 396.96 407.73

307.10 318.63 330.02 341.32 352.59 363.86 375.15 386.49 397.89 409.36 420.90 432.53 444.24

0.9997 1.0319 1.0628 1.0927 1.1218 1.1501 1.1777 1.2048 1.2315 1.2576 1.2834 1.3088 1.3338

0.9080 0.9197 0.9539 0.9867 1.0182 1.0487 1.0784 1.1074 1.1358 1.1637 1.1910

(Contd.)

Appendix Table E3 T (°C)

709

(Continued) V (m3/kg)

130

0.04468

140 150 160

0.04599 0.04729 0.04857

Sat. 40 50 60 70 80 90 100 110 120 I30 140 150 160 170 180

0.02020 0.02029 0.02171 0.02301 0.02423 0.02538 0.02649 0.02755 0.02858 0.02959 0.03058 0.03154 0.03250 0.03344 0.03436 0.03528

S~. 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200

0.01208

u (kJ/kg)

h (kJ/kg)

338.76

370.04

1.2179

380.86 391.79 402.82 - 39.39oc)

1.2444 1.2706 1.2963

267.97 268.68 280.19 291.36 302.34 313.20 324.01 334.82 345.65 356.52 367.46 378.46 389.56 400.74 412.02 423.40 P = 1.60MPa(~at = 57.92°C) 256.00 275.33

0.9043 0.9066 0.9428 0.9768 1.0093 1.0405 1.0707 1.1000 1.1286 1.1567 1.1841 1.2111 1.2376 1.2638 1.2895 1.3149

348.66 358.68 368.82 P = 1.OOMPa(~at 247.77 248.39 258.48 268.35 278.11 287.82 297.53 307.27 317.06 326.93 336.88 346.92 357.06 367.31 377.66 388.12

0.01233 0.01340 0.01435 0.01521 0.01601 0.01677 0.01750 0.01820 0.01887 0.01953 0.02017 0.02080 0.02142 0.02203 0.02263

258.48 269.89 280.78 291.39 301.84 312.20 322.53 332.87 343.24 353.66 364.15 374.71 385.35 396.08 406.90

278.20 291.33 303.74 315.72 327.46 339.04 350.53 361.99 373.44 384.91 396.43 407.99 419.62 431.33 443.11

s (kJ/(kg K))

V(m3/kg)

u

(kJ/kg)

h (kJ/kg)

s (kJ/(kg K))

0.8982 0.9069 0.9457 0.9813 1 0148 1 0467 1 0773 1 1069 1 1357 1 1638 1 1912 1.2181 1.2445 1.2704 1.2960 1.3212

Table E4 Ideal-gas properties of air T (R) 360 380 400 420 440 460 480 500 520 537 540 560 580

H (Bm/lbm) 85.97 90.75 95.33 100.32 105.11 109.90 114.69 119.48 124.27 128.10 129.06 133.86 138.66

P,. 0.3363 0.4061 0.4858 0.5760 0.6776 0.7913 0.9182 1.0590 1.2147 1.3593 1.3860 1.5742 1.7800

U (Btu/lbm) 61.29 64.70 68.11 71.52 74.93 78.36 81.77 85.20 88.62 91.53 92.04 95.47 98.90

S (Btu/(lbm R)) 396.6 346.6 305.0 270.1 240.6 215.33 193.65 174.90 158.58 146.34 144.32 131.78 120.70

0.50369 0.51663 0.52890 0.54058 0.55172 0.56235 0.57255 0.58233 0.59173 0.59945 0.60078 0.60950 0.61793

(Contd.)

710 Table E4 T (R) 600 620 640 660 680 700 720 740 760 780 800 820 840 860 880 900 920 940 960 980 1000 1040 1080 1120 1160 1200 1240 1280 1320 1360 1400 1440 1480 1520 1560 1600 1650 1700 1750 1800 1850 1900 1950 2000 2050 2100 2150 2200 2250 2300 2350 2400 2450 2500 2550 2600 2650 2700

Appendix (Continued) H (Btu/lbm) 143.47 148.28 153.09 157.92 162.73 167.56 172.39 177.23 182.08 186.94 191.81 196.69 201.56 206.46 211.35 216.26 221.18 226.11 231.06 236.02 240.98 250.95 260.97 271.03 281.14 291.30 301.52 311.79 322.11 332.48 342.90 353.37 363.89 374.47 385.08 395.74 409.13 422.59 436.12 449.71 463.37 477.09 490.88 504.71 518.71 532.55 546.54 560.59 574.69 588.82 603.00 617.22 631.48 645.78 660.12 674.49 688.90 703.35

P~ 2.005 2.249 2.514 2.801 3.111 3.446 3.806 4.193 4.607 5.051 5.526 6.033 6.573 7.149 7.761 8.411 9.102 9.834 10.61 11.43 12.30 14.18 16.28 18.60 21.18 24.01 27.13 30.55 34.31 38.41 42.88 47.75 53.04 58.78 65.00 71.13 80.89 90.95 101.98 114.0 127.2 141.5 157.1 174.0 192.3 212.1 223.5 256.6 281.4 308.1 336.8 367.6 400.5 435.7 473.3 513.5 556.3 601.9

U (Btu/lbm) 102.34 105.78 109.21 112.67 116.12 119.58 123.04 126.51 129.99 133.47 136.97 140.47 143.98 147.50 151.02 154.57 158.12 161.68 165.26 168.83 172.43 179.66 186.93 194.25 201.63 209.05 216.53 224.05 231.63 239.25 246.93 254.66 262.44 270.26 278.13 286.06 296.03 306.06 316.16 326.32 336.55 346.85 357.20 367.61 378.08 388.60 399.17 409.78 420.46 431.16 441.91 452.70 463.54 474.40 485.31 496.26 507.25 518.26

Vr

S (Btu/(lbm R))

110.88 102.12 94.30 87.27 80.96 75.25 70.07 65.38 61.10 57.20 53.63 50.35 47.34 44.57 42.01 39.64 37.44 35.41 33.52 31.76 30.12 27.17 24.58 22.30 20.29 18.51 16.93 15.52 14.25 13.12 12.10 11.17 10.34 9.578 8.890 8.263 7.556 6.9294 6.357 5.847 5.388 4.974 4.598 4.258 3.949 3.667 3.410 3.176 2.961 2.765 2.585 2.419 2.266 2.125 1.996 1.876 1.765 1.662

0.62607 0.63395 0.64159 0.64902 0.65621 0.66321 0.67002 0.67665 0.68312 0.68942 0.69558 0.70160 0.70747 0.71323 0.71886 0.72438 0.72979 0.73509 0.74030 0.74540 0.75042 0.76019 0.76964 0.77880 0.78767 0.79628 0.80466 0.81280 0.82075 0.82848 0.83604 0.84341 0.85062 0.85767 0.86456 0.87130 0.87954 0.88758 0.89542 0.90308 0.91056 0.91788 0.92504 0.93205 0.93891 0.94564 0.95222 0.95919 0.96501 0.97123 0.97732 0.98331 0.98919 0.99497 1.00064 1.00623 1.01172 1.01712

(Contd.)

Appendix

711

Table E4 (Continued)

T(R)

P,.

H (Btu/lbm)

2750

717.83

2800 2850 2900 2950 3000 3050 3100 3150 3200 3250 3300 3350 3400 3450 3500 3550 3600 3650 3700 3750 3800 3850 3900 3950 4000 4050 4100 4150 4200 4300 4400 4500 4600 4700 4800 4900 5000 5100 5200 5300

732.33 746.88 761.45 776.05 790.68 8O5.34 820.03 834.75 849.48 864.24 879.02 893.83 908.66 923.52 938.40 953.30 968.21 983 15 998 11 10131 10281 1043 1 10581 1073 2 1088.3 1103.4 1118.5 1133.6 1148.7 1179.0 1209.4 1239.9 1270.4 1300.9 1331.5 1362.2 1392.9 1423.6 1454.4 1485.3

U (Btu/lbm)

650.4 702.0 756.7 814.8 876.4 941.4 1011 1083 1161 1242 1328 1418 1513 1613 1719 1829 1946 2068 2196 2330 2471 2618 2773 2934 3103 3280 3464 3656 3858 4067 4513 4997 5521 6089 6701 7362 8073 8837 9658 10.539 11.481

S (Btu/(lbm R))

V,.

529.31

1.566

1.02244

540.40 551.52 562.66 573.84 585.04 596.28 607.53 618.82 630.12 641.46 652.81 664.20 675.60 687.04 698.48 709.95 721.44 732.95 744.48 756.04 767.60 779.19 790.80 802.43 814.06 825.72 837.40 849.09 860.81 884.28 907.8l 931.39 955.04 978.73 1002.5 1026.3 1050.1 1074.0 1098.0 1122.0

1.478 1.395 1.318 1.247 1.180 1.118 1.060 1.006 0.955 0.907 0.8621 0.8202 0.7807 0.7436 0.7087 0.6579 0.6449 0.6157 0.5882 0.5621 0.5376 0.5143 0.4923 0.4715 0.4518 0.4331 0.4154 0.3985 0.3826 0.3529 0.3262 0.3019 0.2799 0.2598 0.2415 0.2248 0.2096 0.1956 0.1828 0.1710

1.02767 1.03282 1.03788 1.04288 1.04779 1.05264 1.05741 1.06212 1.06676 1.07134 1.07585 1.08031 1.08470 1.08904 1.09332 1.09755 1.10172 1 10584 1 10991 111393 1 11791 1 12183 1 12571 1.12955 1.13334 1.13709 1 14079 1 14446 114809 1 15522 1 16221 1 16905 1 17575 1.18232 1.18876 1.19508 1.20129 1.20738 1.21336 1.21923

Table E5 Ideal-gas properties of carbon dioxide, C02 T (K) 0 220 230 240 250 260 270 280 290 298 300

H (kJ/kmol)

U (kJ/kmol)

S (kJ/(kmol K))

T (K)

H (kJ/kmol)

U (kJ/kmol)

S (kJ/(kmol K))

0 6.601 6.938 7.280 7.627 7.979 8.335 8.697 9.063 9.364 9.431

0 4.772 5.026 5.285 5.548 5.817 6.091 6.369 6.651 6.885 6.939

0 202.966 204.464 205.920 207.337 208.717 210.062 211.376 212.660 213.685 213.915

600 610 620 630 640 650 660 670 680 690 700

22.280 22.754 23.231 23.709 24.190 24.674 25.160 25.648 26.138 26.631 27.125

17.291 17.683 18.076 18.471 18.869 19.270 19.672 20.078 20.484 20.894 21.305

243.199 243.983 244.758 245.524 246.282 247.032 247.773 248.507 249.233 249.952 250.663

(Contd.)

Appendix

712 Table E5

(Continued)

T (K)

H (kJ/kmol)

U (kJ/kmol)

S (kJ/(kmol K))

r (K)

H (kJ/kmol)

U (kJ/kmol)

S (kJ/(kmol K))

310 320 330 340 350 360 370 380 390 400 410 420 430 440 450 460 470 480 490 500 510 520 530 540 550 560 570 580 590 1000 1020 1040 1060 1080 1100 1120 1140 1160 1180 1200 1220 1240 1260 1280 1300 1320 1340 1360 1380 1400 1420 1440 1460 1480 1500 1520 1540 1560

9.807 10.186 10.570 10.959 11.351 11.748 12.148 12.552 12.960 13.372 13.787 14.206 14.628 15.054 15.483 15.916 16.351 16.791 17.232 17.678 18.126 18.576 19.029 19.485 19.945 20.407 20.870 21.337 21.807 42.769 43,859 44.953 46.051 47.153 48.258 49.369 50.484 51.602 52.724 53.848 54.977 56.108 57.244 58.381 59.522 60.666 61.813 62.963 64.116 65.271 66.427 67.586 68.748 66.911 71.078 72.246 73.417 74.590

7.230 7.526 7.826 8.131 8.439 8.752 9.068 9.392 9.718 10.046 10.378 10.714 11.053 11.393 11.742 12.091 12.444 12.800 13.158 13.521 13.885 14.253 14.622 14.996 15.372 15.751 16.131 16.515 16.902 34.455 35.378 36.306 37.238 38.174 39.112 40.057 41.006 41.957 42.913 43.871 44.834 45.799 46.768 47.739 48.713 49.691 50.672 51.656 52.643 53.631 54.621 55.614 56.609 57.606 58.606 59.609 60.613 61.620

215.146 216.351 217.534 218.694 219.831 220.948 222.044 223.122 224.182 225.225 226.250 227.258 228.252 229.230 230.194 231.144 232.080 233.004 233.916 234.814 235.700 236.575 237.439 238.292 239.135 239.962 240.789 241.602 242.405 269.215 270.293 271.354 272.400 273.430 274.445 275.444 276.430 277.403 278.361 297.307 280.238 281.158 282.066 282.962 283.847 284.722 285.586 286.439 287.283 288.106 288.934 289.743 290.542 291.333 292.114 292.888 292.654 294.411

710 720 730 740 750 760 770 780 790 800 810 820 830 840 850 860 870 880 890 900 910 920 930 940 950 960 970 980 990 1760 1780 1800 1820 1840 1860 1880 1900 1920 1940 1960 1980 2000 2050 2100 2150 2200 2250 2300 2350 2400 2450 2500 2550 2600 2650 2700 2750 2800

27.622 28.121 28.622 29.124 29.629 20.135 30.644 31.154 31.665 32.179 32.694 33.212 33.730 34.251 34.773 35.296 35.821 36.347 36.876 37.405 37.935 38.467 39.000 39.535 40.070 40.607 41.145 41.685 42.226 86.420 87.612 88.806 90.000 91.196 92.394 93.593 94.793 95.995 97.197 98.401 99.606 100.804 103.835 106.864 109.898 112.939 115.984 119.035 122.091 125.152 128.219 131.290 134.368 137.449 140.533 143.620 146.713 149.808

21.719 22.134 22.522 22.972 23.393 23.817 24.242 24.669 25.097 25.527 25.959 26.394 26.829 27.267 27.706 28.125 28.588 29.031 29.476 29.922 30.369 30.818 31.268 31.719 32.171 32.625 33.081 33.537 33.995 71.787 72.812 73.840 74.868 75.897 76.929 77.962 78.996 80.031 81.067 82.105 83.144 84.185 86.791 89.404 92.023 94.648 97.277 99.912 102.552 105.197 107.849 110.504 113.166 115.832 118.500 121.172 123.849 126.528

251.368 252.065 252.755 253.439 254.117 254.787 255.452 256.110 256.762 257.408 258.048 258.682 259.311 259.934 260.551 261.164 261.770 262.371 262.968 263.559 264.146 264.728 265.304 265.877 266.444 267.007 267.566 268.119 268.670 301.543 302.217 302.884 303.544 304.198 304.845 305.487 306.122 306.751 307.374 307.992 308.604 309.210 310.701 312.160 313.589 314.988 316.356 317.695 319.011 320.302 321.566 322.808 324.026 325.222 326.396 327.549 328.684 329.800

(Contd.)

Appendix Table E5 T (K) 1580 1600 1620 1640 1660 1680 1700 1720 1740

713

(Continued) H (kJ/kmol) 76.767 76.944 78.123 79.303 80.486 81.670 82.856 84.043 85.231

U (kJ/kmol)

S (kJ/(kmol K))

62.630 63.741 64.653 65.668 66.592 67.702 68.721 69.742 70.764

295.161 295.901 296.632 297.356 298.072 298.781 299.482 300.177 300.863

T (K) 2850 2900 2950 3000 3050 3100 3150 3200 3250

H (kJ/kmol)

U (kJ/kmol)

152.908 156.009 159.117 162.226 165.341 168.456 171.576 174.695 177.822

129.212 131.898 134.589 137.283 139.982 142.681 145.385 148.089 150.801

S (kJ/(kmol K)) 330.896 331.975 333.037 334.084 335.114 336.126 337.124 338.109 339.069

APPENDIX F: THE LEE/KESLER GENERALIZED CORRELATION TABLES Table F1 Values of Z o Tr

Pr 0.0100

0.0500

0.1000

0.2000

0.4000

0.6000

0.8000

1.0000

0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.93 0.95 0.97 0.98

0.0029 0.0026 0.0024 0.0022 0.0021 0.9804 0.9849 0.9881 0.9904 0.9922 0.9935 0.9946 0.9954 0.9959 0.9961 0.9963 0.9965

0.0145 0.0130 0.0119 0.0110 0.0103 0.0098 0.0093 0.9377 0.9504 0.9598 0.9669 0.9725 0.9768 0.9790 0.9803 0.9815 0.9821

0.0290 0.0261 0.0239 0.0221 0.0207 0.0195 0.0186 0.0178 0.8958 0.9165 0.9319 0.9436 0.9528 0.9573 0.9600 0.9625 0.9637

0.0579 0.0522 0.0477 0.0442 0.0413 0.0390 0.0371 0.0356 0.0344 0.0336 0.8539 0.8810 0.9015 0.9115 0.9174 0.9227 0.9523

0.1158 0.1043 0.0953 0.0882 0.0825 0.0778 0.0741 0.0710 0.0687 0.0670 0.0661 0.0661 0.7800 0.8059 0.8206 0.8338 0.8398

0.1737 0.1564 0.1429 0.1322 0.1236 0.1166 0.1109 0.1063 0.1027 0.1001 0.0985 0.0983 0.1006 0.6635 0.6967 0.7240 0.7360

0.2315 0.2084 0.1904 0.1762 0.1647 0.1553 0.1476 0.1415 0.1366 0.1330 0.1307 0.1301 0.1321 0.1359 0.1410 0.5580 0.5887

0.2892 0.2604 0.2379 0.2200 0.2056 0.1939 0.1842 0.1765 0.1703 0.1656 0.1626 0.1614 0.1630 0.1664 0.1705 0.1779 0.1844

0.99 1.00 1.01 1.02 1.05 1.10 1.15 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 2.20 2.40 2.60 2.80 3.00 3.50 4.00

0.9966 0.9967 0.9968 0.9969 0.9971 0.9975 0.9978 0.9981 0.9985 0.9988 0.9991 0.9993 0.9994 0.9995 0.9996 0.9997 0.9998 0.9999 1.0000 1.0000 1.0000 1.0001 1.0001

0.9826 0.9832 0.9837 0.9842 0.9855 0.9874 0.9891 0.9904 0.9926 0.9942 0.9954 0.9964 0.9971 0.9977 0.9982 0.9986 0.9992 0.9996 0.9998 1.0000 1.0002 1.0004 1.0005

0.9648 0.9659 0.9669 0.9679 0.9707 0.9747 0.9780 0.9808 0.9852 0.9884 0.9909 0.9928 0.9943 0.9955 0.9964 0.9972 0.9983 0.9991 0.9997 1.0001 1.0004 1.0008 1.0010

0.9277 0.9300 0.9322 0.9343 0.9401 0.9485 0.9554 0.9611 0.9702 0.9768 0.9818 0.9856 0.9886 0.9910 0.9929 0.9944 0.9967 0.9983 0.9994 1.0002 1.0008 1.0017 1.0021

0.8455 0.8509 0.8561 0.8610 0.8743 0.8930 0.9081 0.9205 0.9396 0.9534 0.9636 0.9714 0.9775 0.9823 0.9861 0.9892 0.9937 0.9969 0.9991 1.0007 1.0018 1.0035 1.0043

0. 7471 0.7574 0.7671 0.7761 0.8002 0.8323 0.8576 0.8779 0.9083 0.9298 0.9456 0.9575 0.9667 0.9739 0.9796 0.9842 0.9910 0.9957 0.9990 1.0013 1.0030 1.0055 1.0066

0.6138 0.6355 0.6542 0.6710 0.7130 0.7649 0.8032 0.8330 0. 8764 0.9062 0.9278 0.9439 0.9563 0.9659 0.9735 0.9796 0.9886 0.9948 0.9990 1.0021 1.0043 1.0075 1.0090

0.1959 0.2901 0.4648 0.5146 0.6026 0.6880 0.7443 0.7858 0. 8438 0.8827 0.9103 0.9308 0.9463 0.9583 0.9678 0.9754 0.9865 0.9941 0.9993 1.0031 1.0057 1.0097 1.0115

Appendix

714 Table F2 Values of Z 1

rr

0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.93 0.95 0.97 0.98 0.99 1.00 1.01 1.02 1.05 1.10 1.15 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 2.20 2.40 2.60 2.80 3.00 3.50 4.00

er 0.0100

0.0500

O.1000

0.2000

0.4000

0.6000

0.8000

1.0000

-0.0008 -0.0009 -0.0010 -0.0009 -0.0009 -0.0314 -0.0205 -0.0137 -0.0093 -0.0064 -0.0044 -0.0029 -0.0019 -0.0015 -0.0012 -0.0010 -0.0009 -0.0008 -0.0007 -0.0006 -0.0005 -0.0003 0.0000 0.0002 0.0004 0.0006 0.0007 0.0008 0.0008 0.0008 0.0008 0.0008 0.0008 0.0007 0.0007 0.0007 0.0006 0.0006 0.0005 0.0005

-0.0040 -0.0046 -0.0048 -0.0047 -0.0045 -0.0043 -0.0041 -0.0772 -0.0507 -0.0339 -0.0228 -0.0152 -0.0099 -0.0075 -0.0062 -0.0050 -0.0044 -0.0039 -0.0034 -0.0030 -0.0026 -0.0015 0.0000 0.0011 0.0019 0.0030 O.0036 0.0039 0.0040 0.0040 0.0040 0.0040 0.0039 0.0037 0.0035 0.0033 0.0031 0.0029 0.0026 0.0023

-0.0081 -0.0093 -0.0095 -0.0094 -0.0090 -0.0086 -0.0082 -0.0078 -0.1161 -0.0744 -0.0487 -0.0319 -0.0205 -0.0154 -0.0126 -0.0101 -0.0090 -0.0079 -0.0069 -0.0060 -0.0051 -0.0029 0.0001 0.0023 0.0039 0.0061 0.0072 0.0078 0.0080 0.0081 0.0081 0.0079 0.0078 0.0074 0.0070 0.0066 0.0062 0.0059 0.0052 0.0046

-0.0161 -0.0185 -0.0190 -0.0187 -0.0181 -0.0172 -0.0164 -0.0156 -0.0148 -0.0143 -0.1160 -0.0715 -0.0442 -0.0326 -0.0262 -0.0208 -0.0184 -0.0161 -0.0140 -0.0120 -0.0102 -0.0054 0.0007 0.0052 0.0084 0.0125 0.0147 0.0158 0.0162 0.0163 0.0162 0.0159 0.0155 0.0147 0.0139 0.0131 0.0124 0.0117 0.0103 0.0091

-0.0323 -0.0370 -0.0380 -0.0374 -0.0360 -0.0343 -0.0326 -0.0309 -0.0294 -0.0282 -0.0272 -0.0268 -0.1118 -0.0763 -0.0589 -0.0450 -0.0390 -0.0335 -0.0285 -0.0240 -0.0198 -0.0092 0.0038 0.0127 0.0190 0.0267 0.0306 0.0323 0.0330 0.0329 0.0325 0.0318 0.0310 0.0293 0.0276 0.0260 0.0245 0.0232 0.0204 0.0182

-0.0484 -0.0554 -0.0570 -0.0560 -0.0539 -0.0513 -0.0487 -0.0461 -0.0438 -0.0417 -0.0401 -0.0391 -0.0396 -0.1662 -0.1110 -0.0770 -0.0641 -0.0531 -0.0435 -0.0351 -0.0277 -0.0097 0.0106 0.0237 0.0326 0.0429 0.0477 0.0497 0.0501 0.0497 0.0488 0.0477 0.0464 0.0437 0.0411 0.0387 0.0365 0.0345 0.0303 0.0270

-0.0645 -0.0738 -0.0758 -0.0745 -0.0716 -0.0682 -0.0646 -0.0611 -0.0579 -0.0550 -0.0526 -0.0509 -0.0503 -0.0514 -0.0540 -0.1647 -0.1100 -0.0796 -0.0588 -0.0429 -0.0303 -0.0032 0.0236 0.0396 0.0499 0.0612 0.0661 0.0677 0.0677 0.0667 0.0652 0.0635 0.0617 0.0579 0.0544 0.0512 0.0483 0.0456 0.0401 0.0357

-0.0806 -0.0921 -0.0946 -0.0929 -0.0893 -0.0849 -0.0803 -0.0759 -0.0718 -0.0681 -0.0648 -0.0622 -0.0604 -0.0602 -0.0607 -0.0623 -0.0641 -0.0680 -0.0879 -0.0223 -0.0062 -0.0220 0.0476 0.0625 0.0719 0.0819 0.0857 0.0864 0.0855 0.0838 0.0814 0.0792 0.0767 0.0719 0.0675 0.0634 0.0598 0.0565 0.0497 0.0443

1.0000

1.2000

1.5000

2.0000

3.0000

5.0000

7.0000

10.000

0.2892 0.2604 0.2379 0.2200 0.2056 0.1939 0.1842 O. 1765

0.3479 0.3123 0.2853 0.2638 0.2465 0.2323 0.2207 0.2113

0.4335 0.3901 0.3563 0.3294 0.3077 0.2899 0.2753 0.2634

0.5775 0.5195 0.4744 0.4384 0.4092 0.3853 0.3657 0.3495

0.8648 0.7775 0.7095 0.6551 0.6110 0.5747 0.5446 0.5197

1.4366 1.2902 1.1758 1.0841 1.0094 0.9475 0.8959 0.8526

2.0048 1.7987 1.6373 1.5077 1.4017 1.3137 1.2398 1.1773

Table F3 VaLues of Z °

0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65

2.8507 2.5539 2.3211 2.1338 1.9801 1.8520 1.7440 1.6519

(Contd.)

Appendix Table F3

715

(Continued)

Tr

/Or 1.0000

1.2000

1.5000

2.0000

3.0000

5.0000

7.0000

0.70

0.1703

0.2038

0.2538

0.3364

0.75

0.1656

0.1981

0.2464

0.3260

0.80

0.1626

0.1942

0.2411

0.85

0.1614

0.1924

0.90

0.1630

0.93

10.000

0.4991

0.8161

1.1341

1.5729

0.4823

0.7854

1.0787

1.5047

0.3182

0.4690

0.7598

1.0400

1.4456

0.2382

0.3132

0.4591

0.7388

1.0071

1.3943

0.1935

0.2383

0.3114

0.4527

0.7220

0.9793

1.3496

0.1664

0.1963

0.2405

0.3122

0.4507

0.7138

0.9648

1.3257

0.95

0.1705

0.1998

0.2432

0.3138

0.4501

0.7092

0.9561

1.3108

0.97

0.1779

0.2055

0.2474

0.3164

0.4504

0.7052

0.9480

1.2968

0.98

0.1844

0.2097

0.2503

0.3182

0.4508

0.7035

0.9442

1.2901

0.99

0.1959

0.2154

0.2538

0.3204

0.4514

0.7018

0.9406

1.2835

1.00

0.2901

0.2237

0.2583

0.3229

0.4522

0.7004

0.9372

1.2772

1.01

0.4648

0.2370

0.2640

0.3260

0.4533

0.6991

0.9339

1.2710

1.02

0.5146

0.2629

0.2715

0.3297

0.4547

0.6980

0.9307

1.2650

1.05

0.6026

0.4437

0.3131

0.3452

0.4604

0.6956

0.9222

1.2481

1.10

0.6880

0.5984

0.4580

0.3953

0.4770

0.6950

0.9110

1.15

0.7443

0.6803

0.5798

0.4760

0.5042

0.6987

0.9033

1.2232 1.2021

1.20

0.7858

0.7363

0.6605

0.5605

0.5425

0.7069

0.8990

1 1844

1.30

0.8438

0.8111

0.7624

0.6908

0.6344

0.7358

0.8998

1 1580

1.40

0.8827

0.8595

0.8256

0.7753

0.7202

0.7661

0.9112

1 1419

1.50

0.9103

0.8933

0.8689

0.8328

0.7887

0.8200

0.9297

1 1339 1 1320

1.60

0.9308

0.9180

0.9000

0.8738

0.8410

0.8617

0.9518

1.70

0.9463

0.9367

0.9234

0.9043

0.8809

0.8984

0.9745

1 1343

1.80

0.9583

0.9511

0.9413

0.9275

0.9118

0.9297

0.9961

1 1391

1.90

0.9678

0.9624

0.9552

0.9456

0.9359

0.9557

1.0157

1 1452

2.00 2.20

0.9754 0.9856

0.9715 0.9847

0.9664 0.9826

0.9599 0.9806

0.9550 0.9827

0.9772 1.0094

1.0328 1.0600

1.1516 1 1635

2.40

0.9941

0.9936

0.9935

0.9945

1.0011

1.0313

1.0793

1 1728

2.60

0.9993

0.9998

1.0010

1.0040

1.0137

1.0463

1.0926

1 1792

2.80

1.0031

1.0042

1.0063

1.0106

1.0223

1.0565

1.1016

1 1830

3.00

1.0057

1.0074

1.0101

1.0153

1.0284

1.0635

1.1075

1 1848

3.50

1.0097

1.0120

1.0156

1.0221

1.0368

1.0723

1.1138

1 1834

4.00

1.0115

1.0140

1.0179

1.0249

1.0401

1.0747

1.1136

1 1773

Table F4 Values of Z 1

~

Pr 1.0000

1.2000

1.5000

2.0000

3.0000

5.0000

7.0000

10.000

0.30

-0.0806

- 0.0966

- O. 1207

- O. 1608

- 0.2407

-0.3996

-0.5572

- 0.7915

0.35 0.40

-0.0921 -0.0946

-0.1105 -0.1134

-0.1379 -0.1414

-0.1834 -0.1879

-0.2738 -0.2799

-0.4523 -0.4603

-0.6279 -0.6365

-0.8663 -0.8936

0.45

-0.0929

-0.1113

-0.1387

-0.1840

-0.2734

-0.4475

-0.6162

-0.8608

0.50

-0.0893

-0.1069

-0.1330

-0.1762

-0.2611

-0.4253

-0.5831

-0.8099 -0.7521

0.55

-0.0849

-0.1015

-0.1263

-0.1669

-0.2465

-0.3991

-0.5446

0.60

-0.0803

-0.0960

-0.1192

-0.1572

-0.2312

-0.3718

-0.5047

-0.6928

0.65

-0.0759

-0.0906

-0.1122

-0.1476

-0.2160

-0.3447

-0.4653

-0.6346

0.70

-0.0718

-0.0855

-0.1057

-0.1385

-0.2013

-0.3184

-0.4270

-0.5785

0.75

-0.0681

-0.0808

-0.0996

-0.1298

-0.1872

-0.2929

-0.3901

-0.5250 -0.4740

0.80

-0.0648

-0.0767

-0.0940

-0.1217

-0.1736

-0.2682

-0.3545

0.85

-0.0622

-0.0731

-0.0888

-0.1138

-0.1602

-0.2439

-0.3201

-0.4254

0.90

-0.0604

-0.0701

-0.0840

-0.1059

-0.1463

-0.2195

-0.2862

-0.3788 -0.3516

0.93

-0.0602

-0.0687

-0.0810

-0.1007

-0.1374

-0.2045

-0.2661

0.95

-0.0607

-0.0678

-0.0788

-0.0967

-0.1310

-0.1943

-0.2526

-0.3339

0.97

-0.623

-0.0669

-0.0759

-0.0921

-0.1240

-0.1837

-0.2391

-0.3163

(Contd.)

716 Table F4

Appendix

(Continued)

rr 1.0000

1.2000

0.98

-0.0641

- 0.0661

0.99

-0.0680

-0.0646

1.00

-0.0879

-0.0609

1.01

-0.0223

-0.0473

1.02

-0.0062

1.5000

2.0000

3.0000

5.0000

7.0000

-0.0740

-0.0893

-0.0715

-0.0861

-0.0678 -0.0621

-0.0227

10.000

-0.1202

-0.1783

-0.2322

-0.3075

-0.1162

-0.1728

-0.2254

-0.2989

-0.0824

-0.1118

-0.1672

-0.2185

-0.2902

-0.0778

-0.1072

-0.1615

-0.2116

-0.2816

-0.0524

-0.0722

-0.1021

-0.1556

-0.2047

-0.2731 -0.2476

1.05

0.0220

0.1059

0.0451

-0.0432

-0.0838

-0.1370

-0.1835

1.10

0.0476

0.0897

0.1630

0.0698

-0.0373

-0.1021

-0.1469

-0.2056

1.15

0.0625

0.0943

0.1548

0.1667

0.0332

-0.0611

-0.1084

-0.1642

1.20

0.0719

0.0991

0.1477

0.1990

0.1095

-0.0141

-0.0678

-0.1231

1.30

0.0819

0.1048

0.1420

0.1991

0.2079

0.0875

0.0176

-0.0423

1.40

0.0857

0.1063

0.1383

0.1894

0.2397

0.1737

0.1008

0.0350

1.50

0.0854

0.1055

0.1345

0.1806

0.2433

0.2309

0.1717

0.1058

1.60

0.0855

0.1035

0.1303

0.1729

0.2381

0.2631

0.2255

0.1673

1.70

0.0838

0.1008

0.1259

0.1658

0.2305

0.2788

0.2628

0.2179

1.80

0.0816

0.0978

0.1216

0.1593

0.2224

0.2846

0.2871

0.2576

1.90

0.0792

0.0947

0.1173

0.1532

0.2144

0.2848

0.3017

0.2876

2.00

0.0767

0.0916

0.1133

0.1476

0.2069

0.2819

0.3097

0.3096

2.20

0.0719

0.0857

0.1057

0.1374

0.1932

0.2720

0.3135

0.3355

2.40

0.675

0.0803

0.0989

0.1285

0.1812

0.2602

0.3089

0.3459

2.60

0.0634

0.0754

0.0929

0.1207

0.1706

0.2484

0.3009

0.3475

2.80

0.0598

0.0711

0.0876

0.1138

0.1613

0.2372

0.2915

0.3443

3.00

0.0535

0.0672

0.0828

0.1076

0.1529

0.2268

0.2817

0.3385

3.50

0.0497

0.0591

0.0728

0.0949

0.1356

0.2042

0.2584

0.3194

4.00

0.0443

0.0527

0.0651

0.0849

0.1219

0.1857

0.2378

0.2994

Table F5 Values of

(HR)°/RTc

Tr

Pr 0.0100

0.0500

0.1000

0.2000

0.4000

0.6000

0.8000

1.0000

0.30

-6.045

-6.043

-6.040

-6.034

-6.022

-6.011

- 5.999

- 5.987

0.35

-5.906

-5.904

-5.901

-5.895

-5.882

-5.870

-5.858

- 5.845

0.40

- 5.763

- 5.761

- 5.757

- 5.751

- 5.738

- 5.726

- 5.713

- 5.700

0.45

- 5.615

- 5.612

- 5.609

- 5.603

- 5.590

- 5.577

- 5.564

- 5.551

0.50

- 5.465

- 5.463

- 5.459

- 5.453

- 5.440

- 5.427

- 5.414

- 5.401

0.55

-0.032

-5.312

-5.309

-5.303

- 5.290

-5.278

-5.265

-5.252

0.60

-0.027

-5.162

- 5.159

-5.153

-5.141

-5.129

-5.116

- 5.104

0.65

-0.023

-0.118

-5.008

-5.002

-4.991

-4.980

-4.968

-4.956

0.70

-0.020

-0.101

-0.213

-4.848

-4.838

-4.828

-4.818

-4.808

0.75

-0.017

-0.088

-0.183

-4.687

-4.679

-4.672

-4.664

-4.655

0.80

-0.015

-0.078

-0.160

-0.345

-4.507

-4.504

-4.499

-4.494

0.85

-0.014

-0.069

-0.141

-0.300

-4.309

-4.313

-4.316

-4.316

0.90

-0.012

-0.062

-0.126

-0.264

-0.596

-4.074

-4.094

-4.108

0.93

-0.011

-0.058

-0.118

-0.246

-0.545

-0.960

-3.920

-3.953

0.95

-0.011

-0.056

-0.113

-0.235

-0.516

-0.885

-3.763

-3.825

0.97

-0.011

-0.054

-0.109

-0.225

-0.490

-0.824

- 1.356

-3.658

0.98

-0.010

-0.053

-0.107

-0.221

-0.478

-0.797

- 1.273

-3.544

0.99

-0.010

-0.052

-0.105

-0.216

-0.466

-0.773

- 1.206

-3.376

1.00

-0.010

-0.051

-0.103

-0.212

-0.455

-0.750

- 1.151

-2.584

1.01

-0.010

-0.050

-0.101

-0.208

-0.445

-0.721

- 1.102

- 1.796

1.02

-0.010

-0.049

-0.099

-0.203

-0.434

-0.708

- 1.060

- 1.627

1.05

-0.009

-0.046

-0.094

-0.192

-0.407

-0.654

-0.955

- 1.359

1.10

-0.008

-0.042

-0.086

-0.175

-0.367

-0.581

-0.827

- 1.120

1.15

-0.008

-0.039

-0.079

-0.160

-0.334

-0.523

-0.732

-0.968

(Contd.)

Appendix Table F5

717

(Continued)

Tr

Pr 0.0100

0.0500

0.1000

0.2000

0.4000

0.6000

0.8000

1.0000

1.20

-0.007

-0.036

-0.073

-0.148

-0.305

-0.474

-0.657

1.30

-0.006

-0.031

-0.063

-0.127

-0.259

-0.399

-0.545

-0.857 -0.698

1.40

-0.005

-0.027

-0.055

-0.110

-0.224

-0.341

-0.463

-0.588

1.50

-0.005

-0.024

-0.048

-0.097

-0.196

-0.297

-0.400

-0.505

1.60

- 0.004

- 0.021

- 0.043

- 0.086

- 0.173

- 0.261

- 0.350

- 0.440

1.70

-0.004

-0.019

-0.038

-0.076

-0.153

-0.231

-0.309

-0.387

1.80

-0.003

-0.017

-0.034

-0.068

-0.137

-0.206

-0.275

-0.344 -0.307

1.90

-0.003

-0.015

-0.031

-0.062

-0.123

-0.185

-0.246

2.00

-0.003

-0.014

-0.028

-0.056

-0.111

-0.167

-0.222

-0.276

2.20

- 0.002

- 0.012

- 0.023

- 0.046

- 0.092

- 0.137

- 0.182

- 0.226

2.40

-0.002

-0.010

-0.019

-0.038

-0.076

-0.114

-0.150

-0.187

2.60

-0.002

-0.008

-0.016

-0.032

-0.064

-0.095

-0.125

-0.155

2.80

-0.001

-0.007

-0.014

-0.027

-0.054

-0.080

-0.105

-0.130

3.00

-0.001

-0.006

-0.011

-0.023

-0.045

-0.067

-0.088

-0.109

3.50

-0.001

-0.004

-0.007

-0.015

-0.029

-0.043

-0.056

-0.069

4.00

- 0.000

-0.002

-0.005

-0.009

-0.017

-0.026

-0.033

-0.041

Table F6 Values of

(HR)I/RTc

Tr

Pr 0.0100

0.0500

0.1000

0.2000

0.4000

0.6000

0.8000

1.0000

0.30

- 11.098

- 11.096

- 11.095

- 11.091

- 11.083

- l 1.076

- 11.069

- 11.062

0.35

- 10.656

- 10.655

- 10.654

- 10.653

- 10.650

- 10.646

- 10.643

- 10.640

0.40

- 10.121

- 10.121

- 10.121

- 10.120

- 10.121

- 10.121

- 10.121

- 10.121

0.45

-9.515

-9.515

-9.516

-9.517

-9.519

-9.521

-9.523

-9.525

0.50

- 8.868

- 8.869

- 8.870

- 8.872

- 8.876

- 8.880

- 8.884

- 8.888

0.55

-0.080

-8.211

-8.212

-8.215

-8.221

-8.226

-8.232

-8.238 - 7.596

0.60

-0.059

- 7.568

- 7.570

- 7.573

- 7.579

- 7.585

- 7.591

0.65

-0.045

-0.247

-6.949

-6.952

-6.959

-6.966

-6.973

-6.980

0.70

- 0.034

- 0.185

- 0.415

- 6.360

- 6.367

- 6.373

- 6.381

- 6.388

0.75

-0.027

-0.142

-0.306

-5.796

-5.802

-5.809

-5.816

-5.824

0.80

-0.021

-0.110

-0.234

-0.542

-5.266

-5.271

-5.278

-5.285

0.85

-0.017

-0.087

-0.182

-0.401

-4.753

-4.754

-4.758

-4.763

0.90

- 0.014

- 0.070

- 0.144

- 0.308

- 0.751

-4.254

-4.248

-4.249

0.93

-0.012

-0.061

-0.126

-0.265

-0.612

- 1.236

-3.942

-3.934

0.95

-0.011

-0.056

-0.115

-0.241

-0.542

-0.994

-3.737

-3.712

0.97

-0.010

-0.052

-0.105

-0.219

-0.483

-0.837

- 1.616

-3.470

0.98

-0.010

-0.050

-0.101

-0.209

-0.457

-0.776

- 1.324

-3.332

0.99

-0.009

- 0.048

-0.097

-0.200

-0.433

-0.722

- 1.154

-3.164 - 2.471

1.00

- 0.009

- 0.046

- 0.093

- 0.191

- 0.410

- 0.675

- 1.034

1.01

-0.009

-0.044

-0.089

-0.183

-0.389

-0.632

-0.940

- 1.375

1.02

-0.008

-0.042

-0.085

-0.175

-0.370

-0.594

-0.863

- 1.180

1.05

-0.007

-0.037

-0.075

-0.153

-0.318

-0.498

-0.691

-0.877

1.10

-0.006

-0.030

-0.061

-0.123

-0.251

-0.381

-0.507

-0.617

1.15

-0.005

-0.025

-0.050

-0.099

-0.199

-0.296

-0.385

-0.459

1.20

-0.004

-0.020

-0.040

-0.080

-0.158

-0.232

-0.297

-0.349

1.30

-0.003

-0.013

-0.026

-0.052

-0.100

-0.142

-0.177

-0.203

1.40

-0.002

-0.008

-0.016

-0.032

-0.060

-0.083

-0.100

-0.111

1.50

-0.001

-0.005

-0.009

-0.018

-0.032

-0.042

-0.048

-0.049

1.60

- 0.000

-0.002

-0.004

-0.007

-0.012

-0.013

-0.011

-0.005

-0.009

-0.017

-0.027

1.70

- 0.000

- 0.000

- 0.000

- 0.000

-0.003

1.80

0.000

0.001

0.003

0.006

0.015

0.025

0.037

0.051

1.90

0.001

0.003

0.005

0.011

0.023

0.037

0.053

0.070

(Contd.)

Appendix

718 Table F6

(Continued)

Tr

Pr 0.0100

0.0500

0.1000

0.2000

0.4000

0.6000

0.8000

1.0000

2.00

0.001

0.003

0.007

0.015

0.030

0.047

0.065

2.20

0.001

0.005

0.010

0.020

0.040

0.062

0.083

0.085 0.106

2.40

0.001

0.006

0.012

0.023

0.047

0.071

0.095

0.120

2.60

0.001

0.006

0.013

0.026

0.052

0.078

0.104

0.130

2.80

0.001

0.007

0.014

0.028

0.055

0.082

0.110

0.137

3.00

0.001

0.007

0.014

0.029

0.058

0.086

0.114

0.142

3.50

0.002

0.008

0.016

0.031

0.062

0.092

0.122

0.152

4.00

0.002

0.008

0.016

0.032

0.064

0.096

0.127

0.158

3.0000

5.0000

7.0000

Table F7 Values of

(HR)°/RTc

r~

Pr 1.0000

1.2000

1.5000

2.0000

10.000

0.30

- 5.987

- 5.975

- 5.957

- 5.927

- 5.868

- 5.748

- 5.628

- 5.446

0.35

-5.845

-5.833

-5.814

-5.783

- 5.721

-5.595

-5.469

-5.278

0.40

- 5.700

- 5.687

- 5.668

- 5.636

- 5.572

- 5.442

- 5.311

- 5.113

0.45

- 5.551

- 5.538

- 5.519

- 5.486

- 5.421

- 5.288

- 5.154

- 5.950

0.50

- 5.401

- 5.388

- 5.369

- 5.336

- 5.279

- 5.135

-4.999

-4.791

0.55

- 5.252

- 5.239

- 5.220

- 5.187

- 5.121

-4.986

-4.849

-4.638

0.60

- 5.104

- 5.091

- 5.073

- 5.041

-4.976

-4.842

-4.794

-4.492

0.65

-4.956

-4.949

-4.927

-4.896

-4.833

-4.702

-4.565

-4.353

0.70

-4.808

-4.797

-4.781

-4.752

-4.693

-4.566

-4.432

-4.221

0.75

-4.655

-4.646

-4.632

-4.607

-4.554

-4.434

-4.393

-4.095

0.80

-4.494

-4.488

-4.478

-4.459

-4.413

-4.303

-4.178

-3.974

0.85

-4.316

-4.316

-4.312

-4.302

-4.269

-4.173

-4.056

-3.857

0.90

-4.108

-4.118

-4.127

-4.132

-4.119

-4.043

-3.935

-3.744

0.93

-3.953

-3.976

-4.000

-4.020

-4.024

-3.963

-3.863

-3.678

0.95

- 3.825

- 3.865

- 3.904

- 3.940

- 3.958

- 3.910

- 3.815

- 3.634

0.97

-3.658

-3.732

-3.796

-3.853

-3.890

-3.856

-3.767

-3.591

0.98

- 3.544

- 3.652

- 3.736

- 3.806

- 3.854

- 3.829

- 3.743

- 3.569

0.99

- 3.376

- 3.558

- 3.670

- 3.758

- 3.818

- 3.801

- 3.719

- 3.548

1.00

- 2.584

- 3.441

- 3.598

- 3.706

- 3.782

- 3.774

- 3.695

- 3.526

1.01

- 1.796

-3.283

-3.516

-3.652

-3.744

-3.746

-3.671

-3.505

1.02

- 1.627

-3.039

-3.422

-3.595

-3.705

-3.718

-3.647

-3.484

1.05

- 1.359

-2.034

-3.030

-3.398

-3.583

-3.632

-3.575

-3.420 - 3.315

1.10

-

- 1.487

- 2.203

- 2.965

- 3.353

- 3.484

- 3.453

1.15

-0.968

1.120

- 1.239

- 1.719

- 2.479

- 3.091

- 3.329

- 3.329

- 3.211

1.20

-0.857

- 1.076

- 1.443

- 2.079

- 2.801

- 3.166

- 3.202

- 3.107

1.30

-0.698

-0.860

- 1.116

- 1.560

-2.274

-2.825

-2.942

-2.899

1.40

-0.588

-0.716

-0.915

- 1.253

- 1.857

-2.486

-2.679

-2.692

1.50

-0.505

-0.611

-0.774

- 1.046

- 1.549

-2.175

-2.421

-2.486

1.60

-0.440

-0.531

-0.667

-0.894

- 1.318

- 1.904

-2.177

-2.285

1.70

-0.387

-0.446

-0.583

-0.777

- 1.139

- 1.672

- 1.953

-2.091

1.80

-0.344

-0.413

-0.515

-0.683

-0.996

- 1.476

- 1.751

- 1.908

1.90

-0.307

-0.368

-0.458

-0.606

-0.880

- 1.309

- 1.571

- 1.736

2.00

-0.276

-0.330

-0.411

-0.541

-0.782

- 1.167

- 1.411

- 1.577

2.20

-0.226

-0.269

-0.334

-0.437

-0.629

-0.937

- 1.143

- 1.295

2.40

-0.187

-0.222

-0.275

-0.359

-0.513

-0.761

-0.929

- 1.058

2.60

-0.155

-0.185

-0.228

-0.297

-0.422

-0.621

-0.756

-0.858

2.80

-0.130

-0.154

-0.190

-0.246

-0.348

-0.508

-0.614

-0.689

3.00

-0.109

-0.129

-0.159

-0.205

-0.288

-0.415

-0.495

-0.545

3.50

-0.069

-0.081

-0.099

-0.127

-0.174

-0.239

-0.270

-0.264

4.00

-0.041

-0.048

-0.058

-0.072

-0.095

-0.116

-0.110

-0.061

Appendix

719

Table F8 Values of (HR)I/RTc

Tr

Pr 1.0000

1.2000

1.5000

2.0000

3.0000

5.0000

7.0000

10.000

0.30

- 11.062

- 11.055

- 11.044

- 11.027

- 10.992

- 10.935

- 10.872

- 10.781

0.35

- 10.640

- 10.637

- 10.632

- 10.624

- 10.609

- 10.581

- 10.554

- 10.529

0.40

- 10.121

- 10.121

- 10.121

- 10.122

- 10.123

- 10.128

- 10.135

- 10.150

0.45

-9.525

-9.527

-9.531

-9.537

-9.549

-9.576

-9.611

-9.663

0.50

-8.888

-8.892

-8.899

-8.909

-8.932

-8.978

-9.030

-9.111

0.55

- 8.238

- 8.243

- 8.252

- 8.267

- 8.298

- 8.360

- 8.425

- 8.531

0.60

- 7.596

- 7.603

- 7.614

- 7.632

- 7.669

- 7.745

- 7.824

- 7.950

0.65

-6.980

-6.987

-6.997

- 7.017

-7.059

- 7.147

- 7.239

- 7.381

0.70

-6.388

-6.395

-6.407

-6.429

-6.475

-6.574

-6.677

-6.837

0.75

- 5.824

- 5.832

- 5.845

- 5.868

-5.918

-6.027

-6.142

-6.318

0.80

-5.285

- 5.293

- 5.306

- 5.330

- 5.385

- 5.506

- 5.632

-5.824

0.85

-4.763

-4.771

-4.784

-4.810

-4.872

- 5.000

- 5.149

- 5.358

0.90

-4.249

-4.255

-4.268

-4.298

-4.371

-4.530

-4.688

-4.916

0.93

-3.934

- 3.937

-3.951

-3.987

-4.073

-4.251

-4.422

-4.662

0.95

-3.712

- 3.713

-3.730

- 3.773

-3.873

-4.068

-4.248

-4.497

0.97

- 3.470

- 3.467

- 3.492

- 3.551

- 3.670

- 3.885

-4.077

-4.336

0.98

- 3.332

- 3.327

- 3.363

- 3.434

- 3.568

- 3.795

- 3.992

-4.257

0.99

- 3.164

- 3.164

- 3.223

- 3.313

- 3.464

- 3.705

- 3.909

- 4.178

1.00

- 2.471

- 2.952

- 3.065

- 3.186

- 3.358

- 3.615

- 3. 8 2 5

- 4.100

1.01

- 1.375

- 2. 5 9 5

- 2. 8 8 0

- 3.051

- 3.251

- 3. 5 2 5

- 3. 7 4 2

- 4.023

1.02

- 1.180

- 1. 7 2 3

- 2.650

- 2.906

- 3.142

- 3.435

- 3.661

- 3.947

1.05

-0.877

-0.878

- 1.496

-2.381

-2.800

- 3.167

- 3.418

-3.722

1.10

- 0.617

- 0.673

- 0.617

- 1.261

-2.167

- 2. 7 2 0

- 3.023

- 3.362

1.15

-0.459

-0.503

-0.487

-0.604

- 1.497

- 2.275

- 2.641

- 3.019

1.20

-0.349

-0.381

-0.381

-0.361

-0.934

- 1.840

-2.273

-2.692

1.30

-0.203

-0.218

-0.218

- 0 . 178

-0.300

- 1.066

- 1.592

-2.086

1.40

-0.111

-0.115

-0.128

-0.070

-0.044

-0.504

- 1.012

- 1.547

1.50

-0.049

-0.046

-0.032

1.60

-0.005

0.004

0.023

0.008

0.078

0.065

0.151

-0.142 0.082

-0.556

- 1.080

-0.217

-0.689

1.70

0.027

0.040

0.063

0.109

0.202

0.223

0.028

- 0.369

1.80

0.051

0.067

0.094

0.143

0.241

0.317

0.203

-0.112

1.90

0.070

0.088

0.117

0.169

0.271

0.381

0.330

0.092

2.00

0.085

0.105

0.136

0.190

0.295

0.428

0.424

0.255

2.20

0.106

0.128

0.163

0.221

0.331

0.493

0.551

0.489

2.40

0.120

0.144

0.181

0.242

0.356

0.535

0.631

0.645

2.60

0.130

0.156

0.194

0.257

0.376

0.567

0.687

0.754

2.80

0.137

0.164

0.204

0.269

0.391

0.591

0.729

0.836

3.00

0.142

0.170

0.211

0.278

0.403

0.611

0.763

0.899

3.50

0.152

0.181

0.224

0.294

0.425

0.650

0.827

1.015

4.00

0.158

0.188

0.233

0.306

0.442

0.680

0.874

1.097

0.4000

0.6000

0.8000

1.0000

Table F9 Values of

(Sn)°/R

Tr

Pr 0.0100

0.0500

0.1000

0.2000

0.30

- 11.614

- 10.008

-9.319

-8.635

-7.961

-7.574

-7.304

-7.099

0.35

- 11.185

-9.579

-8.890

-8.205

-7.529

-7.140

-6.869

-6.663 -6.275

0.40

- 10.802

-9.196

- 8.506

- 7.821

- 7.144

-6.755

-6.483

0.45

- 10.453

- 8.847

- 8.157

- 7.472

-6.794

-6.404

-6.132

- 5.924

0.50

- 10.137

- 8.531

- 7.841

- 7.156

-6.479

-6.089

- 5.816

- 5.608 - 5.324

0.55

-0.038

- 8.245

- 7.555

-6.870

-6.193

- 5.803

- 5.531

0.60

- 0.029

- 7.983

- 7.294

- 6.610

- 5.933

- 5.544

- 5.273

- 5.066

0.65

- 0.023

- 0.122

- 7.052

- 6.368

- 5.694

- 5.306

- 5.036

-4. 830

(Contd.)

Appendix

720 Table F9

(Continued)

T~

/:'r 0.0100

0.0500

0.1000

0.2000

0.4000

0.6000

0.8000

1.0000

0.70

-0.018

-0.096

-0.206

-6.140

- 5.467

- 5.082

-4.814

-4.610

0.75

-0.015

-0.078

-0.164

- 5.917

- 5.248

-4.866

-4.600

-4.399

0.80

-0.013

-0.064

-0.134

-0.294

- 5.026

-4.694

-4.388

-4.191

0.85

-0.011

-0.054

-0.111

-0.239

-4.785

-4.418

-4.166

-3.976 - 3.738

0.90

-0.009

-0.046

-0.094

-0.199

-0.463

-4.145

-3.912

0.93

-0.008

-0.042

-0.085

-0.179

-0.408

-0.750

-3.723

-3.569

0.95

-0.008

-0.039

-0.080

-0.168

-0.377

-0.671

- 3.556

- 3.433

0.97

-0.007

-0.037

-0.075

-0.157

-0.350

-0.607

- 1.056

-3.259

0.98

-0.007

-0.036

-0.073

-0.153

-0.337

-0.580

-0.971

- 3.142

0.99

-0.007

-0.035

-0.071

-0.148

-0.326

-0.555

-0.903

- 2.972

1.00

-0.007

-0.034

-0.069

-0.144

-0.315

-0.532

-0.847

- 2.178

1.01

-0.007

-0.033

-0.067

-0.139

-0.304

-0.510

-0.799

- 1.391

1.02

-0.006

-0.032

-0.065

-0.135

-0.294

-0.491

-0.757

- 1.225

1.05

-0.006

-0.030

-0.060

-0.124

-0.267

-0.439

-0.656

-0.965

1.10

-0.005

-0.026

-0.053

-0.108

-0.230

-0.371

-0.537

-0.742

1.15

-0.005

-0.023

-0.047

-0.096

-0.201

-0.319

-0.452

-0.607

1.20

-0.004

-0.021

-0.042

-0.085

-0.177

-0.277

-0.389

-0.512

1.30

-0.003

-0.017

-0.033

-0.068

-0.140

-0.217

-0.298

-0.385

1.40

-0.003

-0.014

-0.027

-0.056

-0.114

-0.174

-0.237

-0.303

1.50

-0.002

-0.011

-0.023

-0.046

-0.094

-0.143

-0.194

-0.246

1.60

-0.002

-0.010

-0.019

-0.039

-0.079

-0.120

-0.162

-0.204

1.70

-0.002

-0.008

-0.017

-0.033

-0.067

-0.102

-0.137

-0.172

1.80

-0.001

-0.007

-0.014

-0.029

-0.058

-0.088

-0.117

-0.147

1.90

-0.001

-0.006

-0.013

-0.025

-0.051

-0.076

-0.102

-0.127

2.00

-0.001

-0.006

-0.011

-0.022

-0.044

-0.067

-0.089

-0.111

2.20

-0.001

-0.004

-0.009

-0.018

-0.035

-0.053

-0.070

-0.087

2.40

-0.001

-0.004

-0.007

-0.014

-0.028

-0.042

-0.056

-0.070

2.60

-0.001

-0.003

-0.006

-0.012

-0.023

-0.035

-0.046

-0.058

2.80

- 0.000

-0.002

-0.005

-0.010

-0.020

-0.029

-0.039

-0.048 -0.041

3.00

- 0.000

-0.002

-0.004

-0.008

-0.017

-0.025

-0.033

3.50

- 0.000

-0.001

-0.003

-0.006

-0.012

-0.017

-0.023

-0.029

4.00

-0.000

-0.001

-0.002

-0.004

-0.009

-0.013

-0.017

-0.021

Table FIO Values of

(SR)I/R

rr

er 0.0100

0.0500

0.1000

0.2000

0.4000

0.6000

0.8000

1.0000

0.30

- 16.782

- 16.774

- 16.764

- 16.744

- 16.705

- 16.665

- 16.626

- 16.586

0.35

- 15.413

- 15.408

- 15.401

- 15.387

- 15.359

- 15.333

- 15.305

- 15.278

0.40

- 13.990

- 13.986

- 13.981

- 13.972

- 13.953

- 13.934

- 13.915

- 13.896

0.45

- 12.564

- 12.561

- 12.558

- 12.551

- 12.537

- 12.523

- 12.509

- 12.496

0.50

- 11.202

- 11.200

- 11.197

- 11.092

- 11.082

- 11.172

- 11.162

- 11.153

0.55

-0.115

-9.948

-9.946

-9.942

-9.935

-9.928

-9.921

-9.914

0.60

-0.078

- 8.828

- 8.826

- 8.823

- 8.817

- 8.811

- 8.806

- 8.799

0.65

-0.055

-0.309

-7.832

-7.829

-7.824

-7.819

-7.815

-7.510

0.70

-0.040

-0.216

-0.491

-6.951

-6.945

-6.941

-6.937

-6.933

0.75

-0.029

-0.156

-0.340

-6.173

-6.167

-6.162

-6.158

-6.155

0.80

-0.022

-0.116

-0.246

-0.578

- 5.475

- 5.468

- 5.462

- 5.458

0.85

-0.017

-0.088

-0.183

-0.400

-4.853

-4.841

-4.832

-4.826

0.90

-0.013

-0.068

-0.140

-0.301

-0.744

-4.269

-4.249

-4.238

0.93

-0.011

-0.058

-0.120

-0.254

-0.593

- 1.219

-3.914

-3.894

0.95

-0.010

-0.053

-0.109

-0.228

-0.517

-0.961

-3.697

-3.658

0.97

-0.010

-0.048

-0.099

-0.206

-0.456

-0.797

- 1.570

-3.406

(Contd.)

Appendix Table FIO

721

(Continued)

Tr

Pr 0.0100

0.0500

0.1000

0.2000

0.4000

0.6000

0.8000

1.0000

0.98

- 0.009

- 0.046

- 0.094

- 0.196

- 0.429

- 0.734

- 1.270

- 3.264

0.99

- 0.009

- 0.044

- 0.090

- 0.186

- 0.405

- 0.680

- 1.098

- 3.093

1.00

-0.008

-0.042

-0.086

-0.177

-0.382

-0.632

-0.977

-2.399

1.01

-0.008

-0.040

-0.082

-0.169

-0.361

-0.590

-0.883

- 1.306

1.02

-0.008

-0.039

-0.078

-0.161

-0.342

-0.552

-0.807

- 1.113

1.05

-0.007

-0.034

-0.069

-0.140

-0.292

-0.460

-0.642

-0.820

1.10

-0.005

-0.028

-0.055

-0.112

-0.229

-0.350

-0.470

-0.577

1.15

-0.005

-0.023

-0.045

-0.091

-0.183

-0.275

-0.361

-0.437

1.20

-0.004

-0.019

-0.037

-0.075

-0.149

-0.220

-0.286

-0.343

1.30

-0.003

-0.013

-0.026

-0.052

-0.102

-0.148

-0.190

-0.226

1.40

-0.002

-0.010

-0.019

-0.037

-0.072

-0.104

-0.133

-0.158

1.50

-0.001

-0.007

-0.014

-0.027

-0.053

-0.076

-0.097

-0.115

1.60

-0.001

-0.005

-0.011

-0.021

-0.040

-0.057

-0.073

-0.086

1.70

-0.001

-0.004

-0.008

-0.016

-0.031

-0.044

-0.056

-0.067

1.80

-0.001

-0.003

-0.006

-0.013

-0.024

-0.035

-0.044

-0.053

1.90

-0.001

-0.003

-0.005

-0.010

-0.019

-0.028

-0.036

-0.043

2.00

- 0.000

-0.002

-0.004

-0.008

-0.016

-0.023

-0.029

-0.035

2.20

- 0.000

-0.001

-0.003

-0.006

-0.011

-0.016

-0.021

-0.025

2.40

- 0.000

-0.001

-0.002

-0.004

-0.008

-0.012

-0.015

-0.019

2.60

- 0.000

-0.001

-0.002

-0.003

-0.006

-0.009

-0.012

-0.015

2.80

- 0.000

-0.001

-0.001

-0.003

-0.005

-0.008

-0.010

-0.012

3.00

- 0.000

-0.001

-0.001

-0.002

-0.004

-0.006

- 0.008

-0.010

3.50

- 0.000

- 0.000

- 0.001

- 0.001

- 0.003

- 0.004

- 0.006

- 0.007

4.00

- 0.000

- 0.000

-0.001

-0.001

-0.002

-0.003

-0.005

-0.006

Table F l l

Values of

(SR)°/R

Tr

Pr 1.0000

1.2000

1.5000

2.0000

3.0000

5.0000

7.0000

10.000

0.30

- 7.099

-6.935

-6.740

-6.497

-6.180

- 5.847

- 5.683

- 5.578

0.35

-6.663

-6.497

-6.299

-6.052

- 5.728

- 5.376

-5.194

- 5.060

0.40

-6.275

-6.109

- 5.909

- 5.660

- 5.330

-4.967

-4.772

-4.619

0.45

- 5.924

- 5.757

- 5.557

- 5.306

-4.974

-4.603

-4.401

-4.234

0.50

- 5.608

- 5.441

- 5.240

-4.989

-4.656

-4.282

-4.074

- 3.899

0.55

- 5.324

- 5.157

-4.956

-4.706

-4.373

- 3.998

-3.788

-3.607

0.60

-5.066

-4.900

-4.700

-4.451

-4.120

-3.747

-3.537

-3.353

0.65

- 4.830

-4.665

-4.467

-4.220

- 3.892

- 3.523

- 3.315

- 3.131

0.70

-4.610

-4.446

-4.250

-4.007

-3.684

-3.322

-3.117

-2.935

0.75

- 4.399

- 4.238

- 4.045

- 3.807

- 3.491

- 3.138

- 2.939

- 2.761

0.80

-4.191

-4.034

-3.846

-3.615

-3.310

-2.970

-2.777

-2.605

0.85

- 3.976

- 3.825

-3.646

- 3.425

- 3.135

-2.812

-2.629

-2.463

0.90

-3.738

-3.599

- 3.434

- 3.231

-2.964

-2.663

-2.491

-2.334

0.93

- 3.569

- 3.444

- 3.295

- 3.108

-2.860

-2.577

-2.412

-2.262

0.95

- 3.433

- 3.326

- 3.193

- 3.023

-2.790

-2.520

-2.362

-2.215

0.97

-3.259

-3.188

- 3.081

-2.932

-2.719

-2.463

-2.312

-2.170

0.98

- 3.142

-3.106

- 3.019

-2.884

-2.682

-2.436

-2.287

-2.148

0.99

- 2.972

- 3.010

- 2.953

- 2.835

- 2.646

- 2.408

- 2.263

- 2.126

1.00

-2.178

-2.893

-2.879

-2.784

-2.609

-2.380

-2.239

-2.105

1.01

- 1.391

-2.736

-2.798

- 2.730

-2.571

-2.352

-2.215

-2.083

1.02

- 1.225

-2.495

- 2.706

-2.673

-2.533

-2.325

-2.191

-2.062

1.05

-0.965

- 1.523

-2.328

-2.483

-2.415

-2.242

-2.121

-2.001

1.10

-0.742

- 1.012

- 1.557

-2.081

-2.202

-2.104

-2.007

- 1.903

1.15

-0.607

-0.790

- 1.126

- 1.649

- 1.968

- 1.966

- 1.897

- 1.810

(Contd.)

Appendix

722 Table F l l

(Continued)

Tr

Pr

1.0000

1.2000

1.5000

2.0000

3.0000

5.0000

7.0000

10.000

1.20

-0.512

-0.651

-0.890

- 1.308

- 1.727

- 1.827

- 1.789

- 1.722

1.30

-0.385

-0.478

-0.628

-0.891

- 1.299

- 1.554

- 1.581

- 1.556

1.40

-0.303

-0.375

-0.478

-0.663

-0.990

- 1.303

- 1.386

- 1.402

1.50

-0.246

-0.299

-0.381

-0.520

-0.777

- 1.088

- 1.208

- 1.260

1.60

-0.204

-0.247

-0.312

-0.421

-0.628

-0.913

- 1.050

- 1.130

1.70

-0.172

-0.208

-0.261

-0.350

-0.519

-0.773

-0.915

- 1.013

1.80

-0.147

-0.177

-0.222

-0.296

-0.438

-0.661

-0.799

-0.908

1.90

-0.127

-0.153

-0.191

-0.255

-0.375

-0.570

-0.702

-0.815

2.00

-0.111

-0.134

-0.167

-0.221

-0.625

-0.497

-0.620

-0.733

2.20

-0.087

-0.105

-0.130

-0.172

-0.251

-0.388

-0.492

-0.599

2.40

-0.070

-0.084

-0.104

-0.138

-0.201

-0.311

-0.399

-0.496

2.60

-0.058

-0.069

-0.086

-0.1 I3

-0.164

-0.255

-0.329

-0.416

2.80

-0.048

-0.058

-0.072

-0.094

-0.137

-0.213

-0.277

-0.353

3.00

-0.041

-0.049

-0.061

-0.080

-0.116

-0.181

-0.236

-0.303

3.50

-0.029

-0.034

-0.042

-0.056

-0.081

-0.126

-0.166

-0.216

4.00

-0.021

-0.025

-0.031

-0.041

-0.059

-0.093

-0.123

-0.162

Table F12 Values of (S R)1/R

rr

er 1.0000

1.2000

1.5000

2.0000

3.0000

5.0000

7.0000

10.000

0.30

- 16.586

- 16.547

- 16.488

- 16.390

- 16.195

- 15.837

- 15.468

- 14.925

0.35

- 15.278

- 15.251

- 15.211

- 15.144

- 15.011

- 14.751

- 14.496

- 14.153 - 13.144

0.40

- 13.896

- 13.877

- 13.849

- 13.803

- 13.714

- 13.541

- 13.576

0.45

- 12.496

- 12.482

- 12.462

- 12.430

- 12.367

- 12.248

- 12.145

- 11.999

0.50

- 11.153

- 11.143

- 11.129

- 11.107

- 11.063

- I0.985

- 10.920

- 10.836

0.55

-9.914

-9.907

-9.897

-9.882

-9.853

-9.806

-9.769

-9.732

0.60

- 8.799

- 8.794

- 8.787

- 8.777

- 8.760

- 8.736

- 8.723

- 8.720

0.65

-7.810

-7.807

-7.801

-7.794

-7.784

-7.779

-7.785

-7.811

0.70

-6.933

-6.930

-6.926

-6.922

-6.919

-6.929

-6.952

-7.002

0.75

-6.155

-6.152

-6.149

-6.147

-6.149

-6.174

-6.213

-6.285

0.80

- 5.458

- 5.455

- 5.453

- 5.452

- 5.461

- 5.501

- 5.555

- 5. 6 4 8

0.85

-4.826

-4.822

-4.820

-4.822

-4.839

-4.898

-4.969

- 5.082

0.90

-4.238

-4.232

-4.230

-4.236

-4.267

-4.351

-4.442

-4.578 -4.300

0.93

- 3.894

- 3.885

- 3.884

- 3.896

- 3.941

-4.046

-4.151

0.95

- 3.658

- 3.647

- 3.648

- 3.669

- 3.728

- 3.851

- 3.966

-4.125

0.97

- 3.406

- 3.391

- 3.401

- 3.437

- 3.517

- 3.661

- 3.788

- 3.957

0.98

- 3.264

- 3.247

- 3.268

- 3.318

- 3.412

- 3.569

- 3.701

- 3.875

0.99

-3.093

-3.082

-3.126

-3.195

-3.306

-3.477

-3.616

-3.796

1.00

-2.399

-2.868

-2.967

-3.067

-3.200

-3.387

-3.532

-3.717

1.01

- 1.306

-2.513

-2.784

-2.933

-3.094

-3.297

-3.450

-3.640

1.02

- 1.113

- 1.655

-2.557

-2.790

-2.986

-3.209

-3.369

-3.565

1.05

-0.820

-0.831

- 1.443

- 2.283

- 2.655

- 2.949

- 3.134

- 3.348

1.10

-0.577

-0.640

-0.618

- 1.241

-2.067

-2.534

-2.767

- 3.013

1.15

-0.437

-0.489

-0.502

-0.654

- 1.471

-2.138

-2.428

-2.708

1.20

-0.343

-0.385

-0.412

-0.447

-0.991

- 1.767

-2.115

-2.430

1.30

-0.226

-0.254

-0.282

-0.300

-0.481

- 1.147

- 1.569

- 1.944

1.40

-0.158

-0.178

-0.200

-0.220

-0.290

-0.730

- 1.138

- 1.544

1.50

-0.115

-0.130

-0.147

-0.166

-0.206

-0.479

-0.823

- 1.222

1.60

-0.086

-0.098

-0.112

-0.129

-0.159

-0.334

-0.604

-0.969

1.70

-0.067

-0.076

-0.087

-0.102

-0.127

-0.248

-0.456

-0.775

1.80

-0.053

-0.060

-0.070

-0.083

-0.105

-0.195

-0.355

-0.628

1.90

-0.043

-0.049

-0.057

-0.069

-0.089

-0.160

-0.286

-0.518

(Contd.)

Appendix Table F12

(Continued)

T~

2.00 2.20 2.40 260 2.80 3.00 3.50 4.00

723

Pr 1.0000

1.2000

1.5000

2.0000

3.0000

5.0000

7.0000

10.000

-0.035 -0.025 -0.019 -0.015 -0.012 -0.010 -0.007 -0.006

-0.040 -0.029 -0.022 -0.018 -0.014 -0.012 -0.009 -0.007

-0.048 -0.035 -0.027 -0.021 -0.018 -0.015 -0.011 -0.009

-0.058 -0.043 -0.034 -0.028 -0.023 -0.020 -0.015 -0.012

-0.077 -0.060 -0.048 -0.041 -0.025 -0.031 -0.024 -0.020

-0.136 -0.105 -0.086 -0.074 -0.065 -0.058 -0.046 -0.038

-0.238 -0.178 -0.143 -0.120 -0.104 -0.093 -0.073 -0.060

-0.434 -0.322 -0.254 -0.210 -0.180 -0.158 -0.122 -0.100

Table F13 Values of ~h° Tr

Pr 0.0100

0.0500

0.1000

0.2000

0.4000

0.6000

0.8000

1.0000

0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.93 0.95 0.97 0.98 0.99 1.00 1.01 1.02 1.05 1.10 1.15 1.20 1.30 1.40 1.50 1.60

0.0002 0.0034 0.0272 O. 1321 0.4529 0.9817 0.9840 0.9886 0.9908 0.9931 0.9931 0.9954 0.9954 0.9954 0.9954 0.9954 0.9954 0.9977 0.9977 0.9977 0.9977 0.9977 0.9977 0.9977 0.9977 0.9977 0.9977 1.0000 1.0000

0.0000 0.0007 0.0055 0.0266 0.0912 0.2432 0.5383 0.9419 0.9528 0.9616 0.9683 0.9727 0.9772 0.9795 0.9817 0.9817 0.9817 0.9840 0.9840 0.9840 0.9840 0.9863 0.9886 0.9886 0.9908 0.9931 0.9931 0.9954 0.9954

0.0000 0.0003 0.0028 0.0135 0.0461 0.1227 0.2716 0.5212 0.9057 0.9226 0.9354 0.9462 0.9550 0.9594 0.9616 0.9638 0.9638 0.9661 0.9661 0.9683 0.9683 0.9705 0.9750 0.9795 0.9817 0.9863 0.9886 0.9908 0.9931

0.0000 0.0002 0.0014 0.0069 0.0235 0.0625 0.1384 0.2655 0.4560 0.7178 0.8730 0.8933 0.9099 0.9183 0.9226 0.9628 0.9290 0.9311 0.9333 0.9354 0.9376 0.9441 0.9506 0.9572 0.9616 0.9705 0.9772 0.9817 0.9863

0.0000 0.0001 0.0007 0.0036 0.0122 0.0325 0.0718 0.1374 0.2360 0.3715 0.5445 0.7534 0.8204 0.8375 0.8472 0.8570 0.8610 0.8650 0.8690 0.8730 0.8770 0.8872 0.9016 0.9141 0.9247 0.9419 0.9550 0.9638 0.9727

0.0000 0.0001 0.0005 0.0025 0.0085 0.0225 0.0497 0.0948 0.1626 0.2559 0.3750 0.5188 0.6823 0.7551 0.7709 0.7852 0.7925 0.7980 0.8035 0.8110 0.8166 0.8318 0.8531 0.8730 0.8892 0.9141 0.9333 0.9462 0.9572

0.0000 0.0001 0.0004 0.0020 0.0067 0.0176 0.0386 0.0738 0.1262 0.1982 0.2904 0.4018 0.5297 0.6109 0.6668 0.7112 0.7211 0.7295 0.7379 0.7464 0.7551 0.7762 0.8072 0.8318 0.8531 0.8872 0.9120 0.9290 0.9441

0.0000 0.0000 0.0003 0.0016 0.0055 0.0146 0.0321 0.0611 0.1045 0.1641 0.2404 0.3319 0.4375 0.5058 0.5521 0.5984 0.6223 0.6442 0.6668 0.6792 0.6902 0.7194 0.7586 0.7907 0.8166 0.8590 0.8892 0.9141 0.9311

1.70 1.80 1.90 2.00 2.20 2.40 2.60 2.80 3.00 3.50 4.00

1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000

0.9977 0.9977 0.9977 0.9977 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000

0.9954 0.9954 0.9954 0.9977 0.9977 1.0000 l. 0000 1.0000 1.0000 1.0000 1.0000

0.9886 0.9908 0.9931 0.9954 0.9977 0.9977 1.0000 1.0000 1.0000 1.0023 1.0023

0.9772 0.9817 0.9863 0.9886 0.9931 0.9977 1.0000 1.0000 1.0023 1.0023 1.0046

0.9661 0.9727 0.9795 0.9840 0.9908 0.9954 0. 9977 1.0000 1.0023 1.0046 1.0069

0.9550 0.9661 0.9727 0.9795 0.9886 0.9931 0. 9977 1.0023 1.0046 1.0069 1.0093

0.9462 0.9572 0.9661 0.9727 0.9840 0.9931 0.9977 1.0023 1.0046 1.0093 1.0116

Appendix

724 Table F14 Values of ~h1

rr

0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.93 0.95 0.97 0.98 0.99 1.00 1.01 1.02 1.05 1.10 1.15 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 2.20 2.40 2.60 2.80 3.00 3.50 4.00

er 0.0100

0.0500

0.1000

0.2000

0.4000

0.6000

0.8000

1.0000

0.0000 0.0000 0.0000 0.0002 0.0014 0.9705 0.9795 0.9863 0.9908 0.9931 0.9954 0.9977 0.9977 0.9977 0.9977 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000

0.0000 0.0000 0.0000 0.0002 0.0014 0.0069 0.0227 0.9311 0.9528 0.9683 0.9772 0.9863 0.9908 0.9931 0.9931 0.9954 0.9954 0.9954 0.9977 0.9977 0.9977 0.9977 1.0000 1.0000 1.0023 1.0023 1.0046 1.0046 1.0046 1.0046 1.0046 1.0046 1.0046 1.0046 1.0046 1.0023 1.0023 1.0023 1.0023 1.0023

0.0000 0.0000 0.0000 0.0002 0.0014 0.0068 0.0226 0.0572 0.9036 0.9332 0.9550 0.9705 0.9795 0.9840 0.9885 0.9908 0.9908 0.9931 0.9931 0.9931 0.9954 0.9977 1.0000 1.0023 1.0046 1.0069 1.0069 1.0069 1.0069 1.0093 1.0069 1.0069 1.0069 1.0069 1.0069 1.0069 1.0069 1.0069 1.0046 1.0046

0.0000 0.0000 0.0000 0.0002 0.0014 0.0068 0.0223 0.0568 0.1182 0.2112 0.9057 0.9375 0.9594 0.9705 0.9750 0.9795 0.9817 0.9840 0.9863 0.9885 0.9908 0.9954 1.0000 1.0046 1.0069 1.0116 1.0139 1.0163 1.0163 1.0163 1.0163 1.0163 1.0163 1.0139 1.0139 1.0139 1.0116 1.0116 1.0023 1.0093

0.0000 0.0000 0.0000 0.0002 0.0014 0.0066 0.0220 0.0559 0.1163 0.2078 0.3302 0.4774 0.9141 0.9354 0.9484 0.9594 0.9638 0.9683 0.9727 0.9772 0.9795 0.9885 1.0023 1.0116 1.0163 1.0257 1.0304 1.0328 1.0328 1.0328 1.0328 1.0328 1.0304 1.0304 1.0280 1.0257 1.0257 1.0233 1.0209 1.0186

0.0000 0.0000 0.0000 0.0002 0.0014 0.0065 0.0216 0.0551 0.1147 0.2050 0.3257 0.4708 0.6323 0.8953 0.9183 0.9354 0.9440 0.9528 0.9594 0.9638 0.9705 0.9863 1.0046 1.0186 1.0280 1.0399 1.0471 1.0496 1.0496 1.0496 1.0496 1.0496 1.0471 1.0447 1.0423 1.0399 1.0375 1.0352 1.0304 1.0280

0.0000 0.0000 0.0000 0.0002 0.0013 0.0064 0.0213 0.0543 0.1131 0.2022 0.3212 0.4654 0.6250 0.7227 0.7888 0.9078 0.9225 0.9332 0.9440 0.9528 0.9616 0.9840 1.0093 1.0257 1.0399 1.0544 1.0642 1.0666 1.0691 1.0691 1.0666 1.0666 1.0642 1.0593 1.0568 1.0544 1.0496 1.0471 1.0423 1.0375

0.0000 0.0000 0.0000 0.0002 0.0013 0.0063 0.0210 0.0535 0.1116 0.1994 0.3168 0.4590 0.6165 0.7144 0.7797 0.8413 0.8729 0.9036 0.9311 0.9462 0.9572 0.9840 1.0163 1.0375 1.0544 1.0716 1.0815 1.0865 1.0865 1.0865 1.0840 1.0815 1.0815 1.0765 1.0716 1.0666 1.0642 1.0593 1.0520 1.0471

Table F15 Values of ~o

Tr

0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65

/'r 1.0000

1.2000

1.5000

2.0000

3.0000

5.0000

7.0000

0.0000 0.0000 0.0003 0.0016 0.0055 0.0146 0.0321 0.06 i 1

0.0000 0.0000 0.0003 0.0014 0.0048 0.0127 0.0277 0.0527

0.0000 0.0000 0.0003 0.0012 0.0041 0.0107 0.0234 0.0445

0.0000 0.0000 0.0002 0.0010 0.0034 0.0089 0.0193 0.0364

0.0000 0.0000 0.0002 0.0008 0.0028 0.0072 0.0154 0.0289

0.0000 0.0000 0.0002 0.0008 0.0025 0.0063 0.0132 0.0244

0.0000 0.0000 0.0002 0.0009 0.0027 0.0066 0.0135 0.,0245

10.000 0.0000 0.0000 0.0003 0.0012 0.0034 0.0080 0.0160 0.0282

(Contd.)

Appendix Table F15

(Continued)

Tr

0.70 0.75 0.80 0.85 0.90 0.93 0.95 0.97 0.98 0.99 1.00 1.01 1.02 1.05 1.10 1.15 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 2.20 2.40 2.60 2.80 3.00 3.50 4.00

725

Pr 1.0000

1.2000

1.5000

2.0000

3.0000

5.0000

7.0000

0.1045 0.1641 0.2404 0.3319 0.4375 0.5058 0.5521 0.5984 0.6223 0.6442 0.6668 0.6792 0.6902 0.7194 0.7586 0.7907 0.8166 0.8590 0.8892 0.9141 0.9311 0.9462 0.9572 0.9661 0.9727 0.9840 0.0031 0.9977 1.0023 1.0046 1.0093 1.0116

0.0902 0.1413 0.2065 0.2858 0.3767 0.4355 0.4764 0.5164 0.5370 0.5572 0.5781 0.5970 0.6166 0.6607 0.7112 0.7499 0. 7834 0.8318 0.8690 0.8974 0.9183 0.9354 0.9484 0.9594 0.9683 0.9817 0.9908 0.9977 1.0023 1.0069 1.0116 1.0139

0.0759 0.1188 0.1738 0.2399 0.3162 0.3656 0.3999 0.4345 0.4529 0.4699 0.4875 0.5047 0.5224 0.5728 0.6412 0.6918 0.7328 0.7943 0.8395 0.8730 0.8995 0.9204 0.9376 0.9506 0.9616 0.9795 0.9908 0.9977 1.0046 1.0069 1.0139 1.0162

0.0619 0.0966 0.1409 0.1945 0.2564 0.2972 0.3251 0.3532 0.3681 0.3828 0.3972 0.4121 0.4266 0.4710 0.5408 0.6026 0.6546 0.7345 0.7925 0.8375 0.8710 0.8995 0.9204 0.9376 0.9528 0.9727 0.9886 0.9977 1.0069 1.0116 1.0186 1.0233

0.0488 0.0757 0.1102 0.1517 0.1995 0.2307 0.2523 0.2748 0.2864 0.2978 0.3097 0.3214 0.3334 0.3690 0.4285 0.4875 0.5420 0.6383 0.7145 0.7745 0.8222 0.8610 0.8913 0.9162 0.9354 0.9661 0.9863 1.0023 1.0116 1.0209 1.0304 1.0375

0.0406 0.0625 0.0899 0.1227 0.1607 0.1854 0.2028 0.2203 0.2296 0.2388 0.2483 0.2576 0.2673 0.2958 0.3451 0.3954 0.4446 0.5383 0.6237 0.6966 0.7586 0.8091 0.8531 0.8872 0.9183 0.9616 0.9931 1.0162 1.0328 1.0423 1.0593 1.0666

0.0402 0.0610 0.0867 0.1175 0.1524 0.1754 0.1910 0.2075 0.2158 0.2244 0.2328 0.2415 0.2506 0.2773 0.3228 0.3690 0.4150 0.5058 0.5902 0.6668 0.7328 0.7907 0.8414 0.8831 0.9183 0.9727 1.0116 1.0399 1.0593 1.0740 1.0914 1.0990

10.000 0.0453 0.0673 0.0942 0.1256 0.1611 0.1841 0.2000 0.2163 0.2244 0.2328 0.2415 0.2500 0.2582 0.2844 0.3296 0.3750 0.4198 0.5093 0.5943 0.6714 0.7430 0.8054 0.8590 0.9057 0.9462 1.0093 1.0568 1.0889 1.1117 1.1298 1.1508 1.1588

Table F16 Values of ~bl Tr

0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.93 0.95 0.97

Pr 1.0000

1.2000

1.5000

2.0000

3.0000

5.0000

7.0000

0.0000 0.0000 0.0000 0.0002 0.0013 0.0063 0.0210 0.0536 0.1117 0.1995 0.3170 0.4592 0.6166 0.7145 0.7798 0.8414

0.0000 0.0000 0.0000 0.0002 0.0013 0.0062 0.0207 0.0527 0.1102 0.1972 0.3133 0.4539 0.6095 0.7063 0.7691 0.8318

0.0000 0.0000 0.0000 0.0002 0.0013 0.0061 0.0202 0.0516 0.1079 0.1932 0.3076 0.4457 0.5998 0.6950 0.7568 0.8185

0.0000 0.0000 0.0000 0.0002 0.0012 0.0058 0.0194 0.0497 0.1040 0.1871 0.2978 0.4325 0.5834 0.6761 0.7379 0.7998

0.0000 0.0000 0.0000 0.0001 0.0011 0.0053 0.0179 0.0461 0.0970 0.1754 0.2812 0.4093 0.5546 0.6457 0.7063 0.7656

0.0000 0.0000 0.0000 0.0001 0.0009 0.0045 0.0154 0.0401 0.0851 0.1552 0.2512 0.3698 0.5058 0.5916 0.6501 0.7096

0.0000 0.0000 0.0000 0.0001 0.0008 0.0039 0.0133 0.0350 0.0752 0.1387 0.2265 0.3365 0.4645 0.5470 0.6026 0.6607

10.000

0.0000 0.0000 0.0000 0.0001 0.0006 0.0031 0.0108 0.0289 0.0629 0.1178 0.1954 0.2951 0.4130 0.4898 0.5432 0.5984

(Contd.)

Appendix

726 Table F16

(Continued)

/'r

0.98 0.99 1.00 1.01 1.02 1.05 1.10 1.15 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 2.20 2.40 2.60 2.80 3.00 3.50 4.00

er 1.0000

1.2000

1.5000

2.0000

3.0000

5.0000

7.0000

0.8730 0.9036 0.9311 0.9462 0.9572 0.9840 1.0162 1.0375 1.0544 1.0715 1.0814 1.0864 1.0864 1.0864 1.0839 1.0814 1.0814 1.0765 1.0715 1.0666 1.0641 1.0593 1.0520 1.0471

0.8630 0.8913 0.9204 0.9462 0.9661 0.9954 1.0280 1.0520 1.0691 1.0914 1.0990 1.1041 1.1041 1.1041 1.1015 1.0990 1.0965 1.0914 1.0864 1.0814 1.0765 1.0715 1.0617 1.0544

0.8492 0.8790 0.9078 0.9333 0.9594 1.0186 1.0593 1.0814 1.0990 1.1194 1.1298 1.1350 1.1350 1.1324 1.1298 1.1272 1.1220 1.1143 1.1066 1.1015 1.0940 1.0889 1.0789 1.0691

0.8298 0.8590 0.8872 0.9162 0.9419 1.0162 1.0990 1.1376 1.1588 1.1776 1.1858 1.1858 1.1858 1.1803 1.1749 1.1695 1.1641 1.1535 1.1429 1.1350 1.1272 1.1194 1.1041 1.0914

0.7962 0.8241 0.8531 0.8831 0.9099 0.9886 1.1015 1.1858 1.2388 1.2853 1.2942 1.2942 1.2883 1.2794 1.2706 1.2618 1.2503 1.2331 1.2190 1.2023 1.1912 1.1803 1.1561 1.1403

0.7379 0.7674 0.7962 0.8241 0.8531 0.9354 1.0617 1.1722 1.2647 1.3868 1.4488 1.4689 1.4689 1.4622 1.4488 1.4355 1.4191 1.3900 1.3614 1.3397 1.3183 1.3002 1.2618 1.2303

0.6887 0.7178 0.7464 0.7745 0.8035 0.8872 1.0186 1.1403 1.2474 1.4125 1.5171 1.5740 1.5996 1.6033 1.5959 1.5849 1.5704 1.5346 1.4997 1.4689 1.4388 1.4158 1.3614 1.3213

10.000 0.6266 0.6546 0.6823 0.7096 0.7379 0.8222 0.9572 1.0864 1.2050 1.4061 1.5524 1.6520 1.7140 1.7458 1.7620 1.7620 1.7539 1.7219 1.6866 1.6482 1.6144 1.5813 1.5101 1.4555

SUBJECT INDEX

A Absolute activity, 6 Active transport, 495-497, 525-527, 531,533-534, 592 Activity coefficient, 32 Actual cogeneration plant, 218-219 Actual regenerative Rankine cycle, 204-207 Actual reheat Rankine cycle, 198-201 Actual reheat regenerative cycle, 211 Adiabatic mixer, 108-109 Affinity, 28, 90 Alveoli, 542 Anisotropic diffusion, 357 Annular packed bed, 166-167 Antiport, 551-552 Antoine equation, 34 Approach temperature, 254 Arrhenius group, 458 Arrow of time, 7 Atomic volumes, 82 ATP synthase, 551 Availability, 286-287 Average velocity, 115 Azeotropes, 38 B Bacterial growth, 660 Balance equations, 14-15, 112-113,456, 470-471 Bejan number, 165 Belousov-Zhabotinsky reaction scheme, 625,642, 643 B6nard cells, 634 Bifurcation, 632-633 Bioenergetics, 548-549 Biomembranes, 526-529 Biological, fuels, 544 gels, 359 structures, 650-651 Boltzmann's H-theorem, 56 Bronsted work principle, 20 Brownian motion, 84 Brusselator system, 616, 638,640 Butterfly effect, 635 C Caloric equation, 21 Caratheodory's theory, 13 Cash flow diagram, 314-315 Cell electric potentials, 542 Chapman-Enskog formula, 70 Chemical, affinity of real gases, 28 equilibrium, 8

exergy, 243-244 potential, 6, 46 stability, 601 Chiral symmetry breaking, 651-652 Clapeyron equation, 28-29 Claude process, 227 Cogeneration, 215 Collision integral, 60 Column grand composite curves, 260 Combined energy flow, 63 Combined momentum flow, 59 Compartmental structure, 529 Composite, curves, 246-248 membranes, 520 wall, 64 Compressibility factor, 45 Conductance coefficients, 132 Conductance matching, 571-573 Configurational heat capacity, 603-604 Conjugate flows and forces, 123 Conservation law, 114-115 Conservation of mass, 3 Conservation of energy, 3, 118 Corresponding state correlation, 60 Couette flow, 161-166 Coupled processes, 90-92, 564 Coupled reactions, 447-448, 677 Cristae, 549-551 Cross relations, 47-48 Cumulative degree of thermodynamic perfection, 280-281 Cumulative exergy consumption, 280-282 Curie-Prigogine principle, 143 Cyclic process, 11 D Damk6hler number, 473 Dead state, 184-185 Decrease of exergy principle, 187 Degree of coupling, 371,475-476, 534, 553,568-570 Degrees of freedom, 506 Depletion number, 244-245 Diabatic distillation configuration, 298 Diffusion, barrier, 83 coefficients, 75, 78 in mixtures, 329 potential, 91 velocity, 73 Discontinues systems, 131 Dissipation function, 124-125,379, 529-532 Dissipation with conductance matching, 571

728 Dissipative processes, 7 Dissipative structures, 55,634 Distillation column targets, 260 Distribution coefficient, 349 Distribution of entropy production, 171 Dufour effect, 91,365 Dynamic equilibrium, 9 E

Ecological cost, 285-286 Effective diffusivity, 453-455 of cellular systems, 566 Effectiveness factor, 458-462 Effective thermal conductivity, 456 Efficiency of coupling, 376 Efficiency of energy conversion, 377, 475-476, 541,554, 570 Elasticity coefficient, 558 Electrical conductance, 340-343 Electric field intensity, 88 Electrochemical affinity, 105 Electrochemical potential, 337 Electrokinetic effect, 90 Electron transport in mitochondria, 549 Electrosmotic pressure, 91 Energy, balance, 15, 117 dissipation, 106-110 expenditure, 544-547 of transport, 370 Enthalpy, 45-48 deficit, 261-263 of blood, 543 Entropy, 3 balance 109, 110, 156 distribution, 14 of melting, 100 of vaporization, 100-102 of transport, 370 production, 16-19, 104-113, 121,426-443 production in distillation, 237 source strength, 16-17 Equation of state, 22 Equilibrium, 6-14 constant of reaction, 563 Equipartition principle, 176-177, 236-237, 289 Essergy, 287 Eucken formula, 70 Euler equations, 21 Evolutionary criterion, 595,681 Evolution of couple systems, 481 Excess, enthalpy, 32, 37 entropy, 32, 33 Gibbs energy, 32 volume, 32 Exergy, 184-195 analysis, 184-185 analysis of distillation, 234 analysis of power plant, 194 analysis of refinery, 242 analysis of refrigeration cycle, 225 analysis procedure, 192 balance, 186-187 cost, 275-277 destruction number, 285 efficiency, 235-236, 582, 590

Subject index factor, 187-188 loss, 184-192, 581 loss profiles, 234-242 use in bioenergetics, 581 Exhaustion of nonrenewable resources, 284 Expansivity, 4-5 Expression of Kohlrausch, 343 Extended exergy, 275-276 Extended nonequilibrium thermodynamics, 680-681 Extensive properties, 2 Extent of reaction, 90, 415 Extremum principles, 48 F

Facilitated transport, 485-487,489-494, 496, 525-526 Faraday constant, 84 Fick's first law, 73 First law of thermodynamics, 11-12 Flow control coefficients, 558-559 Flow exergy, 189 Fokker-Planck equation, 395 Fourier's law, 63 Fugacity, 30-31 Fugacity coefficient, 31, 46 Fundamental equations, 10 G

Gas permeation, 508-509 General stability condition, 602 Generic cubic equation of state, 25 Generic equation of state, 25-26 GENERIC formulations, 683-684 Generalized chemical kinetics, 501 Generalized correlations, 45 Generalized matrix method, 328 Geothermal power plants, 222 Gibbs, -Duhem relation, 20 equation, 20-21 free energy, 46-48 stability theory, 599-601 Global warming potential, 286 Glycolysis pathway, 552 Gradient of chemical potential, 126 Grand composite curves, 248 H

Heat capacities for real gases, 22 Heat exchanger network synthesis, 248 Heat of transport, 369-371,383 Helmholtz free energy, 46-48 Henry's law, 34-35 Heterogeneous azeotrope, 43 Hittorf transference number, 341 Hydrostatic equilibrium, 8 I

Ideal regenerative Rankine cycle, 201-202 Ideal reheat regenerative cycle, 208 Increase of entropy principle, 156 Infinitesimal processes, 2 Information capacity, 245-246 Intensive properties, 2 Internal, energy, 5, 47 variables, 681,684-685 Irreversible process, 3, 6-8

Subject index Isentropic efficiency, 156 Isomerization reaction, 547 Isothermal compressibility, 3-5 d Joule Thomson coefficient, 26

machines, 593 Momentum, balance, 116-119 flow, 57-59 Mosaic nonequilibrium thermodynamics, 678-679 Multidimensional inflection points, 564 Mutual diffusion coefficients, 85,324

K

Kinematic viscosity, 57 Kinetic energy, 2 Kirkwood's procedure, 515 Knudsen coefficient, 347 Kxonecker delta, 58 L Leaks, 537, 575-578 Lengyel-Epstein model, 647 Level flow, 371 Lewis number, 87 Limit cycle, 633-635,637-640 Linear nonequilibrium thermodynamics, 54-55 Linear pathway, 562 Linear stability analysis, 614-615 Lineweaver-Burk plot, 444, 445 Liquid-like root, 25 Liquid-liquid equilibrium, 33 Local entropy production, 123-124 Local equilibrium, 54-55, 98 Local gross exergy loss, 283 Lorenz equations, 635-636 Lotka-Volterra model, 657 Lyapunov function, 599-600 M

Macroscopic behavior, 613 Macroscopic state, 2, 14 Mass action law, 89 Mass average velocity, 73, 115 Mass balance, 15, 116 Matrix model, 684 Maxwell-Cattaneo equation, 89 Maxwell relations, 47 Maxwell-Stefan equations, 86, 319 Mechanical, filtration coefficient, 511 stability, 601 Membrane equilibrium, 505-507 Metabolic control analysis, 558 Metabolic pathway, 661 Michaelis-Menten equation, 444, 445, 581 Microbial growth, 445-446 Microscopic domain, 56 Microscopic reversibility, 7, 97, 421-423 Minimum, entropy production, 146-149, 442 separation work, 182 temperature level, 246 utilities, 250 Mitochondria, 549-552 Mixing functions, 37 Mixing rules, 42 Modified Graetz problem, 390 Modified Raoult's Law, 34 Molar average velocity, 73 Molecular, evolution, 593

N

Natural processes, 2 Nernst-Haskell equation, 84 Nernst-Planck equation, 349 Network thermodynamics, 671 Newtonian fluids, 56-57 Newton's law of viscosity, 56 Nonequilibrium systems, 53-55,629 Nonlinear transport and rate processes, 394 Non-Newtonian fluids, 56-57 Normal stress, 58 Nozzle flow, 102 NRTL model, 36-37 Nusselt number, 463-464 O

Ohm's law, 87-88 Onsager's reciprocal relations, 132 Open circuit, 532 Optimization problem, 279, 287 Oscillating systems, 616-617 Osmotic, diffusion, 83 equilibrium, 43-44 pressure, 44-45,404 temperature, 404 Oxidation of glucose, 546 Oxidative phosphorylation, 550-551 P

Packed duct flow, 168 Partial excess properties, 37 Peltier effect, 91,407 Peltier heat, 409-410 Phase, diagrams, Tyx, Pyx, 39, 41 rule, 33 stability, 604 Phenomenological, coefficients, 127-128, 132-135, 379-383,476 equations, 127-129, 397, 474-475, 532-535 stoichiometric coefficient, 371 Photosynthesis, 551,554 Pinch analysis, 246-249 Polytropic compression, 104 Potassium channels, 578-579 Potential energy, 2 Power plant analysis, 229 Prandtl number, 64 Prefractionation arrangements, 183 Principle of detailed balance, 419 Production functions, 573 Proper pathways, 557-559 Pseudocritical properties, 61 R

Raoult's law, 34 Rational thermodynamics, 679-680

729

730 Rayleigh number, 634-635 Reaction-diffusion systems, 453,622, 650 Reaction velocity, 425,475-479, 547 Reduced, phenomenological coefficients, 343 pressure, 60 temperature, 60 viscosity, 60 Reflection coefficient, 511-514, 523-524 Regulation in bioenergetics, 574-575 Reheat Rankine cycle, 196 Relaxation length, 528 Relaxation times, 89 Residual enthalpy, 45,529 Residual entropy, 45 Resistance coefficients, 132 Retrofit of distillation, 299 Reverse diffusion, 83 Reversible process, 6-8 Rheopectic fluids, 57 S Saxen relations, 521 Scalar flows, 129 Schmidt number, 87 Second law analysis, 155 Seebeck effect, 406 Self-assembly, 632 Self-diffusivity, 76 Self-organization, 631-632 Sensitivity of enzymatic reaction, 563 Separation work, 182 Second law efficiency, 181-182 Second law of thermodynamics, 13-14, 98-99 Shear stresses, 58 Short Circuit, 532 Slippage, 575-576 Solute permeability coefficient, 511 Sonic velocity, 72 Soret effect, 91,363-364 Spurious drift term, 355 Stability of chemical reactions, 606 Stability of stationary states, 612 State function, 2-3 Static head, 371 Stationary states, 111,430-431,609-613 Stokes-Einstein equation, 83 Streaming current, 91 Stress tensors, 58 Substantial differential derivative, 113 Sylvester expansion, 334 Symport, 552 Synergetics, 632

T Temperature interval method, 257-258 Thermal, conductivity, 63-64, 66-72, 379-380 diffusion coefficient, 364 diffusion factor, 365 diffusion ratio, 365-367, 381 diffusion regime, 398 diffusivity, 64 effusion, 403 equation, 21 expansion coefficient, 26, 51

Subject index field-flow fractionation, 387 stability, 601 Thermodynamic, branch, 54-56 cost, 275 coupling, 19 efficiency, 191-192 factor, 82 fluctuations, 607 forces, 131 optimum, 287 potentials, 46 Thermoeconomics, 275-276 of distillation, 298 of extraction, 291 of latent heat storage, 307 Thermoelectric effect, 91 Thermomechanical coupling, 161, 164 Thermomolecular pressure, 404 Thermoosmosis, 404 Thiele modulus, 460 Thixotropic fluids, 57 Thomson heat, 407 Throttling, 157-159 Time variation of affinity, 440-441 Total differentials, 47 Total energy flow, 389 Transference numbers, 344, 519 Transport coefficients, 87 Transport in biological cells, 581 Transport number, 523 Tricarboxylic acid cycle, 550 Turing structures, 650-651 U Ultrafiltration coefficient, 511 Uncoupling, 574-575 V Van der Pol's equation, 637 van der Waals, equation of state, 49 isotherms, 23, 49 van Kampen's hopping model, 355 van't Hoff equation, 418 Vapor-like root, 25 Vapor-liquid equilibrium, 33 Variation of coupling, 572 Vectorial flows, 129 Virial equation, 27, 40, 42, 49, 628 Viscosity of gases, 59-60 Viscous dissipation, 123 Volume average velocity, 115 Volumetric rate of entropy production, 124 W Wall friction factors, 348 Weisz modulus, 463 Wetted wall column, 330, 332-334 Wilke-Chang equation, 82 Wilson equation, 36 Work, 1-2 Z Zeroth law of thermodynamics, 11

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