Chemical Engineering Dynamics: Modelling with PC Simulation

J. Ingham, I. J. Dunn, E. Heinzle, J. E.Pfenosi1

Chemical Engineering Dynamics

VCH

The included diskette contains the ISIM simulation language as well as simulation examples. It can be run on all DOS-PC’s.

OVCH VerlagsgesellschaftmbH, D-69451Weinheim (Federal Republic of Germany), NY4 Distribution: VCH, P. 0.Box 10 1161, D-69451 Weinheim (Federal Republic of Germany) Switzerland: VCH, P. 0.Box, CH-4020 Basel (Switzerland) United Kingdom and Ireland: VCH, 8 Wellington Court, Cambridge CB11HZ (United Kingdom)

USA and Canada: VCH, 220 East 23rd Street, New York, NY 10010-4606 (USA) Japan: VCH, Eikow Building, 10-9 Hongo 1-chome, Bunkyo-ku, Tokyo 113 (Japan) I

ISBN 3-527-28577-6 (VCH. Weinheim)

ISBN 1-56081-783-6 (VCH, New York)

John Ingham, Irving J. Dunn Elmar Heinzle, Jifi E. Pfenosil

Chemical Engineering Dynamics Modelling with PC Simulation

Weinheim New York - Base1 Cambridge - Tokyo

Dr. J. Ingham Department of Chcmical Engineering University of Bradford Bradford BD7 I D P United Kingdom

Dr. I. I. Dunn, Dr. E. Heinzle, Dr. J.E . Pienosil Department of Chemical Engineering ETH Zurich CH-8092 Zurich Switzerland

This book and thc diskette were carefully produced. Nevcrtheless, authors and publisher do not warrant the information contained therein to be free of errors. Readers are advised to keep in mind that statcments, data, illustrations, procedural details or other items may inadvertently be inaccurate.

Published jointly by VCH Verlagsgesellschaft mbH, Weinheim (Federal Republic of Gcrmany) VCH Publishers, Inc., New York, NY (USA)

Editorial Directors: Philomena Kyan-Bugler, Louise Elsam Production Manager: Claudia Gross1

I.ibrary of Congress Card No. applied for

British Library Cataloguing-in-Publication Data: A catalogue record for this book is available from the British Library

Die Deutsche Bibliothek - CIP-Einheitsaufnahme Chemical engineering dynamics : modelling with PC simulation I John Ingham .. . - Weinheim ; New York ; Basel ; Cambridge ; Tokyo : VCH, 1994 ISBN 3-521-28511-6 NE: Ingham, John

OVCH Verlagsgesellschaft mbH, D-69451 Weinheim (Federal Republic of Germany), 1994 Printed on acid-free and low-chlorine paper All right reserved (including those of translation into other languages). No part of this book may be reproduced in any form -by photoprinting, microfilm, or any other means -nor transmitted or translated into a machine language without written permission from the publishers. Registered namcs, trademarks, etc. used in this book, cvcn when not specifically marked as such, are not to be considered unprotected by law. Data conversion: Hagedornsatz, D-68519 Viernheim Printing: strauss offsetdruck GmbH, D-69509 MGrlenbach Bookbinding: GroRbuchbinderei J. Schaffer, D-67269 Grunstadt Printed in the Federal Republic of Germany

Preface

The aim of this book is to teach the use of modelling and simulation as an aid to the understanding of chemical engineering processes and their dynamics. This is done via a combination of basic modelling theory and computer-based examples, which are used to emphasise basic principles and also cover a wide range of applications. As in our first book, "Biological Reaction Engineering", the examples are based on the use of the ISIM simulation language, which provides a very powerful tool in solution, but which is also very simple in application. The many examples, demonstrate simple modelling procedures, that can be used to represent a wide range of chemical and chemical engineering process phenomena. The study of the examples, by direct computer experimentation, has been shown to lead to a positive improvement in the understanding of physical systems and confidence in the ability to deal with chemical rate processes. Quite simple models can often be shown to give quite realistic representations of process phenomena. The methods, described in the text, are applicable to a wide range of differing applications, including process identification, the analysis and design of experiments, process design and optimisation, process control and plant safety, all of which are essential aspects of modern chemical technology. The book is based on the hands-on use of the computer as an integral part of the learning process. Although computer-based modelling procedures are now commonplace in chemical engineering, our experience is that there still remains a considerable lack of ability in basic modelling, especially when applied to dynamical systems. This has resulted from the traditional steady-state approach to chemical engineering and the past emphasis on flow-sheeting for large-scale continuous processes. Another important contributing factor is the perceived difficulty in solving the large sets of simultaneous differential equations, that result from any realistic dynamic modelling description. With modern trends towards more intensive high value batch processing methods, the need for a better knowledge of the plant dynamics is readily apparent. This is also reinforced by the increased attention, which must now be paid to proper process control, process optimisation and plant safety. Fortunately the PC computer with suitable software, such as the ISIM simulation language, now provides a fast, convenient means of solution.

VI

Preface

Organisation of the Book The book consists of an introduction to basic modelling presented in Chapters 1 through 4 and the simulation examples in Chapter 5 . The first four chapters cover the basic theory for the computer simulation examples and present the basic concepts of dynamic modelling. The aim is not to be exhaustive, but simply to provide sufficient introduction, for a proper understanding of the modelling methodology and computer-based examples. Here the main emphasis is placed on understanding the physical meaning and significance of each term in the resulting model equations. Chapter 5 , constituting the main part of the book, provides the ISIM-based, computer simulation exercises. Each of the examples is self-contained and includes a model description, the model equations, exercises, program listing, nomenclature and references. The combined book thus represents a synthesis of basic theory and computer-based sirnulation examples. The accompanying diskette includes the ISIM simulation language and a copy of all the simulation example programs. Chapter I deals with the basic concepts of modelling, and the formulation of mass and energy balance relationships. In combination with other forms of relationship, these are shown to lead to a systematic development of models. Though thc concepts are simple, they can be applied to very complex problems. Chapter 2 is employed to provide a general introduction to signal and process dynamics, including the concept of process time constants, process control, process optimisation and parameter identification. Other important aspects of dynamic simulation involve the numerical inelhods of solution and the resulting stability of solution; both of which are dealt with from the viewpoint of the simulator, as compared to that of the mathematician. Chapter 3 concerns the dynamic characteristics of stagewise types of equipment, based on the concept of the well-stirred tank. In this, the various types of stirred-tank chemical reactor operation are considered, together with allowance for heat effects, non-ideal flow, control and safety. Also included is the modelling of stagewise mass transfer applications, based on liquid-liquid extraction, gas absorption and distillation. Chapter 4 concerns differential applications, which take place with respect to both time and position and which are normally formulated as partial differential equations. Applications include diffusion and conduction, tubular chemical reactors, differential mass transfer and shell and tube heat exchange. It is shown that such problems can be solved with relative ease, by utilising a finite-differencing solution technique in the simulation approach. Chapter 5 comprises the computer simulation examples. The exercises are by no means mandatory or restrictive. Most instructive is to study the influence of important model parameters, using the interactive and graphical features of ISIM. Working through a particular example will often suggest an interesting variation, such as a control loop, which can then be inserted into the model. ln I

'

Preface

VII

running our courses, the exercises have proven to be very open ended and in tackling them, we hope you will share our conviction that computer simulation is fun, as well as being useful and informative. An Appendix gives a brief instructional guide to the ISIM language, which is intended to be sufficient to work with the simulation examples. We are confident that the book will be useful to all who wish both to obtain a better understanding of chemical engineering dynamics and have an interest in sharpening their modelling skills. We hope that teachers with an interest in modelling will find this to be a useful textbook for chemical engineering and applied chemistry courses, at both undergraduate and postgraduate levels.

ISIM Simulation Software Modern simulation languages are now available that provide the possibility of carrying out the interactive simulation of complex problems at one's own desk. For DOS systems, several languages are currently available. The ISIM simulation language, provided with this book, is the language we have used during the last five years for our continuing education courses and for much of our university teaching. It is especially suitable because of its sophisticated computing power, interactive facility and ease of programming. The use of this PC-based, digital simulation programming language makes it possible for the reader, student and teacher to experiment directly with the model, in the classroom or at the desk. In this way, it is possible to immediately determine the influence of various model parameters on the simulation - a real learning experience. The ISIM software is made available only for the purposes described in this book, and its features are restricted to these examples. An advanced simulation language, ESL, is highly recommended and is also available. Users wishing to purchase the latest version of the ISIM or ESL should contact ISIM International Simulation directly. User manuals for ISIM may also be purchased for &40 from ISIM International Simulation. ISIM International Simulation Limited, Technology House, Salford University Business Park, Lissadel Street, Salford M6 6AP, England, (Tel: +44-(0)61-745 7444; Fax: +44-(0)61-737 7700).

VIII

Preface

Acknowledgements A major acknowledgement should be made to the pioneering texts of Franks

(1966, 1973), Smith, Pike and Murrill (1970), Luyben (1973), Robinson (1 9 7 3 , and Ramirez (1976) for inspiring our interest in digital simulation. The textbook of Russell and Denn (1972) also applied an excellent and disciplined approach to mass balancing, which we have attempted to follow. We are especially grateful to all participants of our post-experience courses, over the last twenty years, for their assistance in the development of the material presented in this book. Continual stimulus and assistance has also been given by a sequence of students, both at the University of Bradford and at the ETHZurich. Special mention should also be given to John Stephenson, Paul Luker, John Ramsay, Stephen King, Carolyn Cable, Ian Dick and many others for their help in running the courses, at the University of Bradford, to Murray Rose, for the times the course went to Dorset and to all past and present collaborators. Dr. Jonathan Snape contributed three months of his time on a postdoctoral grant from the ETH, which permitted the course material to be improved and the course to be offered in Switzerland (see advertisement). The kind collaborations of J. Ramsay and S. King in allowing the use of some parts of their University of Bradford lecture course material on numerical integration is gratefully acknowledged. Our special thanks are again due to Professor John L. Hay of ISIM International Simulation Limited for his agreement to release the TSIM digital simulation programming language, for use with this book. We hope that the book will be useful in drawing attention to his advanced simulation language, ESL, for which we are happy to include a programmed example and an advertisement. This book was done on an Apple Mac, using Microsoft Word and Claris MacDraw Pro. Linotype film was made from the Word file. Special thanks are due to Sabine Meyer and Albert Ochsner, for their able work on the text and drawings. Hermann Stockinger helped to develop many of the simulation example programs. Dr. J. Baldyga critically examined the presentation of the theory in Chapters 1-4 and developed the mixing and segregation example in Sec. 5.5.6. We are grateful to VCH for giving financial assistance for the wordprocessing and especially wish to thank Philomena Ryan-Bugler, Louise Elsam and Claudia Griissl of VCH for their many useful discussions and careful proofreading. Finally, the authors would like to thank Professor John R. Bourne of the ETH for giving the opportunity to develop this area of teaching at the ETH-Zurich and Professor W. L. Wilkinson for encouraging the start of the work at thc University of Bradford.

Table of Contents

Preface

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

Nomenclature for Chapters 1 to 4

1 1.1 1.1.1 1.1.2 1.1.3 1.1.4 1.1.5 1.1.6 1.1.7 1.2 1.2.1 1.2.2 1.2.2.1 1.2.2.2 1.2.2.3 1.2.3 1.2.3.1 1.2.4 1.2.4.1 1.2.4.2 1.2.5 1.2.5.1 1.2.5.2 1.2.5.3 1.2.6 1.2.7 1.2.7.1 1.2.7.2

V

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Basic Concepts . . . . . . . . . . . . . . . . . . . . Modelling Fundamentals . . . . . . . . . . . . . . . . Chemical Engineering Modelling . . . . . . . . . . . . General Aspects of the Modelling Approach . . . . . . . . General Modelling Procedure . . . . . . . . . . . . . . Simulation Tools . . . . . . . . . . . . . . . . . . . ISIM . . . . . . . . . . . . . . . . . . . . . . . . Teaching Applications . . . . . . . . . . . . . . . . . Introductory ISIM Example: BATSEQ - Complex Reaction Sequence . . . . . . . . . . . . . . . . . . Formulation of Dynamic Models . . . . . . . . . . . . . Mass Balance Equations . . . . . . . . . . . . . . . . Balancing Procedures . . . . . . . . . . . . . . . . . Case A . Continuous Stirred-Tank Reactor . . . . . . . . . Case B . Tubular Reactor . . . . . . . . . . . . . . . . Case C . Coffee Percolator . . . . . . . . . . . . . . . Total Mass Balances . . . . . . . . . . . . . . . . . . Case A . Tank Drainage . . . . . . . . . . . . . . . . Component Balances . . . . . . . . . . . . . . . . . Case A . Waste Holding Tank . . . . . . . . . . . . . . Case B . Extraction from a Solid by a Solvent . . . . . . . Energy Balancing . . . . . . . . . . . . . . . . . . . Case A . Continuous Heating in an AgitatedTank . . . . . . Case B . Heating in a Filling Tank . . . . . . . . . . . . Case C. Parallel Reaction in a Semi-Continuous Reactor with Large Temperature Changes . . . . . . . . . . . . Momentum Balances . . . . . . . . . . . . . . . . . Dimensionless Model Equations . . . . . . . . . . . . . Case A . Continuous Stirred-Tank Reactor (CSTR) . . . . . Case B . Gas-Liquid Mass Transfer to a Batch Tank Reactor with Chemical Reaction . . . . . .

1 1 1 5 6 8 10 11 12 15 15 17 17 18 19 29 30 31 32 33 35 41 43 44 46 47 47 SO

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1.3 1.3.1 1.3.2 1.3.3 1.3.4 1.3.5 1.4 1.4.1 1.4.2 1.4.3

Chemical Kinetics . . . . . . . . . Rate of Chemical Reaction . . . . . Reaction Rate Constant . . . . . . Heats of Reaction . . . . . . . . . Chemical Equilibrium and Temperature Yield. Conversion and Selectivity . . Mass Transfer Theory . . . . . . . Stagewise and Differential Mass Transfer Phase Equilibria . . . . . . . . . Interphase Mass Transfer . . . . . .

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51 51 53 55 56 56 58

2 2.1 2.1.1 2.1.1.1 2.1.1.2

Process Dynamics Fundamentals . . . . . . . . . . . . . Signal and Process Dynamics . . . . . . . . . . . . . . Measurement and Process Response . . . . . . . . . . . First-Order Response to an Input Step-Change Disturbance . Case A . Concentration Response of a Continuous-Flow, Stirred Tank . . . . . . . . . . . . . . . . . . . . . Case B . Concentration Response in a Continuous Stirred Tank . . . . . . . . . . . . . . . . with Chemical Reaction Case C . Response o f a Temperature Measuring Element . . . Case D . Measurement Lag for Concentration in a Batch Reactor . . . . . . . . . . . . . . . . . . . . Higher Order Responses . . . . . . . . . . . . . . . . Case A . Multiple Tanks in Series . . . . . . . . . . . . Case B . Response of a Second-Order Temperature Measuring Element . . . . . . . . . . . . . . . . . . . . . . . Pure Time Delay . . . . . . . . . . . . . . . . . . . Transfer Function Representation . . . . . . . . . . . . First-Order Lag Model . . . . . . . . . . . . . . . . . Pure Time Delay . . . . . . . . . . . . . . . . . . . Higher Order Response Curves . . . . . . . . . . . . . Time Constants . . . . . . . . . . . . . . . . . . . . Common Time Constants . . . . . . . . . . . . . . . . Flow Phenomena . . . . . . . . . . . . . . . . . . . Diffusion-Dispersion . . . . . . . . . . . . . . . . . Chemical Reaction . . . . . . . . . . . . . . . . . . Mass Transfer . . . . . . . . . . . . . . . . . . . . Heat Transfer . . . . . . . . . . . . . . . . . . . . . Application of Time Constants . . . . . . . . . . . . . Fundamentals of Automatic Control . . . . . . . . . . . Basic Feedback Control . . . . . . . . . . . . . . . . Types of Controller Action . . . . . . . . . . . . . . . On/Off Control . . . . . . . . . . . . . . . . . . . . Proportional-Integral-Derivative (PTD) Control . . . . . . .

65 65 65 66

2.1.1.3 2.1.1.4 2.1.1.5 2.1.2 2.1.2.1 2.1.2.2 2.1.3 2.1.4 2.1.4.1 2.1.4.2 2.1.4.3 2.2 2.2.1 2.2.1.1 2.2.1.2 2.2.1.3 2.2.1.4 2.2.1.5 2.2.2 2.3 2.3.1 2.3.2 2.3.2.1 2.3.2.2

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60 60

67 69 70 72 73 74 76 78 80 81 82 83 89 90 90 91 91 92 93 93 94 95 96 96 97

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Case A . Operation of a Proportional Temperature Controller Controller Tuning . . . . . . . . . . . . . . . . . . Trial and Error Method . . . . . . . . . . . . . . . Ziegler-Nichols Method . . . . . . . . . . . . . . . Cohen-Coon Controller Settings . . . . . . . . . . . . Ultimate Gain Method . . . . . . . . . . . . . . . . Time Integral Criteria . . . . . . . . . . . . . . . . Advanced Control Strategies . . . . . . . . . . . . . Cascade Control . . . . . . . . . . . . . . . . . . Feed-Forward Control . . . . . . . . . . . . . . . . Adaptive Control . . . . . . . . . . . . . . . . . . Sampled Data or Discrete Control Systems . . . . . . . Numerical Aspects of Dynamic Behaviour . . . . . . . . Optimisation . . . . . . . . . . . . . . . . . . . . . Case A . Optimal Cooling for a Reactor with an Exothermic Reversible Reaction . . . . . . . . . . . . . . . . . Parameter Estimation . . . . . . . . . . . . . . . . 2.4.2 Non-Linear Systems Parameter Estimation . . . . . . . 2.4.2.1 Sensitivity Analysis . . . . . . . . . . . . . . . . . 2.4.2.2 2.4.3 Case B . Estimation of Rate and Equilibrium Constants . . Numerical Integration . . . . . . . . . . . . . . . . 2.4.4 System Stability . . . . . . . . . . . . . . . . . . . . 2.4.5 2.3.2.3 2.3.3 2.3.3.1 2.3.3.2 2.3.3.3 2.3.3.4 2.3.3.5 2.3.4 2.3.4.1 2.3.4.2 2.3.4.3 2.3.4.4 2.4 2.4.1 2.4.1.1

3 3.1 3.2 3.2.1 3.2.2 3.2.2.1 3.2.2.2 3.2.2.3 3.2.2.4 3.2.2.5 3.2.2.6 3.2.3 3.2.3.1 3.2.4 3.2.4.1 3.2.5 3.2.5.1 3.2.6 3.2.7 3.2.8

Modelling of Stagewise Processes . . . . . . . . . . . Introduction . . . . . . . . . . . . . . . . . . . Stirred-Tank Reactors . . . . . . . . . . . . . . . Reactor Configurations . . . . . . . . . . . . . . Generalised Model Description . . . . . . . . . . . Total Mass Balance Equation . . . . . . . . . . . . Component Balance Equation . . . . . . . . . . . . Energy Balance Equation . . . . . . . . . . . . . Heat Transfer to and from Reactors . . . . . . . . . Steam Heating in Jackets . . . . . . . . . . . . . . Dynamics of the Metal Jacket Wall . . . . . . . . . . . . . . . . . . . . . . . . . . The Batch Reactor Case A . Constant-Volume Batch Reactor . . . . . . . The Semi-Batch Reactor . . . . . . . . . . . . . . Case B . Semi-Batch Reactor . . . . . . . . . . . . . . . . . . . . The Continuous Stirred-Tank Reactor Case C . Constant-Volume Continuous Stirred-Tank Reactor Stirred-Tank Reactor Cascade . . . . . . . . . . . . Reactor Stability . . . . . . . . . . . . . . . . . Reactor Control . . . . . . . . . . . . . . . . .

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100 101 101 102 103 103 104 105 105 105 107 107 . 108 108

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114 116 122 126

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129 129 129 129 131 131 131 132 132 136 139 142 143 144 146 147 149 150 151 156

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3.2.9 3.3 3.3.1 3.3.1.1 3.3.1.2 3.3.1.3 3.3.1.4 3.3.1.5 3.3.1.6 3.3.1.7 3.3.1.8 3.3.1.9 3.3.1.10 3.3.1.11 3.3.2 3.3.3 3.3.3.1 3.3.3.2 3.3.3.3 3.3.3.4 3.3.3.5 3.3.3.6 3.3.3.1 3.3.4 4

4 .I 4.1.1 4.1.2 4.2 4.2.1 4.2.2 4.2.3

4.3 4.3.1 4.3.2 4.3.3 4.3.4 4.3.5 4.3.6

Non-Ideal Flow . . . . . . . . . . . . . . . . . . . . Stagewise Mass Transfer . . . . . . . . . . . . . . . . Liquid-Liquid Extraction . . . . . . . . . . . . . . . . Single Batch Extraction . . . . . . . . . . . . . . . . Multisolute Batch Extraction . . . . . . . . . . . . . . Continuous Equilibrium Stage Extraction . . . . . . . . . Multistage Countercurrent Extraction Cascade . . . . . . . Countercurrent Extraction Cascade with Backmixing . . . . Countercurrent Extraction Cascade with Slow Chemical Reaction . . . . . . . . . . . . . . . . . . Multicomponent Systems . . . . . . . . . . . . . . . . Control of Extraction Cascades . . . . . . . . . . . . . Mixer-Settler Extraction Cascades . . . . . . . . . . . . Staged Extraction Columns . . . . . . . . . . . . . . . Column Hydrodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Stagewise Absorption Stagewise Distillation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Simple Overhead Distillation Binary Batch Distillation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Continuous Binary Distillation . . . . . . . . . . . . . . Multicomponent Separations Plate Efficiency . . . . . . . . . . . . . . . . . . . . Complex Column Simulations . . . . . . . . . . . . . . Multicomponent Equilibria . . . . . . . . . . . . . . . Multicomponent Steam Distillation . . . . . . . . . . . . Differential Flow and Reaction Applications . . . . . . . Introduction . . . . . . . . . . . . . . . . . . . . . Dynamic Simulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Steady-State Simulation Diffusion and Heat Conduction . . . . . . . . . . . . Unsteady-State Diffusion Through a Porous Solid . . . . . Unsteady-State Heat Conduction and Diffusion in Spherical and Cylindrical Coordinates . . . . . . . . . . . . . Steady-State Diffusion with Homogeneous Chemical . . . . . . . . . . . . . . . . . . . . . . Reaction Tubular Chemical Reactors . . . . . . . . . . . . . . The Plug-Flow Tubular Reactor . . . . . . . . . . . . . . . . . . . . . . . . Liquid-Phase Tubular Reactors . . . . . . . . . . . . . Gas-Phase Tubular Reactors . . . . . . . . . . . . . . . Batch Reactor Analogy . . Dynamic Simulation of the Plug-Flow Tubular Reactor Dynamics of an lsothermal Tubular Reactor with Axial Dispersion . . . . . . . . . . . . . . . . . . . . . .

159 166 166 167 169 171 175 177 179 180 182 183 192 195 198 201 201 203 208 210 212 213 213 214

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224

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4.3.6.1 Dynamic Difference Equation for the Component Balance Dispersion Model . . . . . . . . . . . . . . . . . . Steady-State Tubular Reactor Dispersion Model . . . . . 4.3.7 Differential Mass Transfer . . . . . . . . . . . . . . 4.4 Steady-State Gas Absorption with Heat Effects . . . . . 4.4.1 Steady-State Design . . . . . . . . . . . . . . . . . 4.4.1.1 Steady-State Simulation . . . . . . . . . . . . . . . 4.4.1.2 Dynamic Modelling of Plug-Flow Contactors: 4.4.2 Liquid-Liquid Extraction Column Dynamics . . . . . . . Dynamic Modelling of a Liquid-Liquid Extractor with 4.4.3 Axial Mixing in Both Phases . . . . . . . . . . . . . 4.5 Heat Transfer Applications . . . . . . . . . . . . . . 4.5.1 Steady-State Tubular Flow with Heat Loss . . . . . . . . Single.Pass. Shel1.and.Tube. Countercurrent-Flow Heat 4.5.2 Exchanger . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5.2.1 Steady-State Applications 4.5.2.2 Heat Exchanger Dynamics . . . . . . . . . . . . . . Difference Formulae for Partial Differential Equations . . 4.6 References Cited in Chapters 1 to 4 . . . . . . . . . . 4.7 Additional Books Recommended . . . . . . . . . . . 4.8

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244 247 250 250 251 . 253

. 253 . 258 . 261 . 261 . . . . .

262 262 264 268 270 273

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5

Simulation Examples of Chemical Engineering Processes

5.1 5.1.1 5.1.2 5.1.3 5.1.4 5.1.5 5.1.6 5.1.7 5.1.8

Batch Reactor Examples . . . . . . . . . . . . . . . BATSEQ -Complex Batch Reaction Sequence . . . . . . BATCHD . Dimensionless Kinetics in a Batch Reactor . . COMPREAC . Complex Reaction . . . . . . . . . . . BATCOM . Batch Reactor with Complex Reaction Sequence CASTOR . Batch Decomposition of Acetylated Castor Oil . HYDROL .Batch Reactor Hydrolysis of Acetic Anhydride . OXIDAT . Oxidation Reaction in an Aerated Tank . . . . RELUY . Batch Reactor of Luyben . . . . . . . . . .

5.2 5.2.1 5.2.2 5.2.3 5.2.4 5.2.5

Continuous Tank Reactor Examples . . . . . . . . . . . . 316 CSTRCOM . Isothermal Reactor with Complex Reaction . . 316 DEACT . Deactivating Catalyst in a CSTR . . . . . . . . 319 TANK and TANKD .Single Tank with Nth-Order Reaction . 323 CSTR . Continuous Stirred-Tank Cascade . . . . . . . . . 327 CASCSEQ . Cascade of Three Reactors with Sequential 330 Reactions . . . . . . . . . . . . . . . . . . . . . . REXT . Reaction with Integrated Extraction of Inhibitory 335 Product . . . . . . . . . . . . . . . . . . . . . . . THERM and THERMPLCrr -Thermal Stability of a CSTR . 340 COOL . Three-Stage Reactor Cascade with Countercurrent Cooling . . . . . . . . . . . . . . . . . . . . . . . 345

5.2.6 5.2.7 5.2.8

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280 280 283 286 . 290 . 293 . 298 . 302 . 306

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Table of Contents

5.2.9 5.2.10 5.2.11 5.2.12 5.2.13

. . . . . . . 350 OSCIL-Oscillating Tank Reactor Behaviour Auto-Refrigerated Reactor . . . 357 REFRIGl and REFRIG2 . Stability of Chemical Reactors with Disturbances . 361 STABTL . Homogeneous Free-Radical Polymerisation . . 366 HOMPOLY . REVTEMP .Reversible Reaction with Variable Heat Capacities . . . . . . . . . . . . . . . . . . . . . . 372

5.3 5.3.1

Tubular Reactor Examples . . . . . . . . . . . . . . . 381 TUBE and TUBED -Tubular Reactor Model for the Steady State . . . . . . . . . . . . . . . . . . . . . 381 TUBTANK . Comparison of Tubular and Tank Reactors . . . 384 BENZHYD . Dehydrogenation of Benzene . . . . . . . . 388 ANHYD .Oxidation of O-Xylene to Phthalic Anhydride . . 393 NITRO .Conversion of Nitrobenzene to Aniline . . . . . . 400 TUBDYN . Dynamic Tubular Reactor . . . . . . . . . . 405 DISRE . Isothermal Reactor with Axial Dispersion . . . . . 410 DISRET . Non-Isothermal Tubular Reactor with Axial Dispersion . . . . . . . . . . . . . . . . . . . . . . 414 VARMOL .Gas-Phase Reaction with Molar Change . . . . 419

5.3.2 5.3.3 5.3.4 5.3.5 5.3.6 5.3.7 5.3.8 5.3.9 5.4

5.4.1 5.4.2 5.4.3 5.4.4 5.5

5.5.1 5.5.2 5.5.3 5.5.4 5.5.5 5.5.6 5.6 5.6.1 5.6.2 5.6.3 5.6.4

Semi-Continuous Reactor Examples . . . . . . . . . . . . 423 Parallel Reactions in a Semi-Continuous SEMIPAR . Reactor . . . . . . . . . . . . . . . . . . . . . . . 423 SEMlSEQ .Sequential Reactions in a Semi-Continuous 426 Reactor . . . . . . . . . . . . . . . . . . . . . . . Semi-Batch Manufacture of Hexamethylenetriamine . 430 HMT . RUN .Relief of a Runaway Polymerisation Reaction . . . . 433 Mixing-Model Examples . . . . . . . . . . . . . . . NOSTR . Non-Ideal Stirred-Tank Reactor . . . . . . . . TUBEMIX-Non-IdealTube-Tank Mixing Model . . . . . Mixed-Flow Residence Time MIXFLOl and MIXFL02 . Distribution Studies . . . . . . . . . . . . . . . . . Gas-Liquid Mixing and GASLIQl and GASLIQ2 . . . . . . . . . . . . . Mass Transfer in a Stirred Tank SPBEDRTD .Spouted Bed Reactor Mixing Model . . . . Mixing and Segregation BATSEG, SEMISEG, COMPSEG . in Chemical Reactors . . . . . . . . . . . . . . . .

. 440 . 440 . 445 . 450

. 457 . 466

. 470

Tank Flow Examples . . . . . . . . . . . . . . . . . . 485

CONFLO1. 2 and 3 . Continuous Flow Tank . . . . . . . 485 . . . . . . . . . . 492 TANKBLD .Liquid Stream Blending Ladle Discharge Problem . . . . . . . . . . 496 TANKDIS . TANKHYD .Interacting Tank Reservoirs . . . . . . . . . 501

xv

Table of Contents

5.7 5.7.1 5.7.2 5.7.3 5.7.4 5.7.5

Process Control Examples . . . . . . . . . . . . . . . TEMPCONT . Control of Temperature in a Water Heater . . TWmANK . Two Tank Level Control . . . . . . . . . . CONTUN . Controller Tuning Problem . . . . . . . . . . SEMIEX . Temperature Control for Semi-Batch Reactor . . TRANSIM . Transfer Function Simulation . . . . . . . .

5.8 5.8.1 5.8.2

Mass Transfer Process Examples . . . . . . . . . . . . BATEX . Single Solute Batch Extraction . . . . . . . . TWOEX . Two-Solute Batch Extraction with Interacting Equilibria . . . . . . . . . . . . . . . . . . . . . . KLADYN .Dynamic Method for K, a . . . . . . . . . Simple Equilibrium Stage Extractor . . . . . . EQEX . COLCON .Extraction Cascade with Backmixing and Control . . . . . . . . . . . . . . . . . . . . . Continuous Equilibrium Multistage Extraction EQMULTI . EQBACK .Multistage Extractor with Backmixing . . . . HOLDUP . Transient Holdup Profiles in an Agitated Extractor . . . . . . . . . . . . . . . . . . . . . Differential Extraction Column with Axial AXDISP . Dispersion . . . . . . . . . . . . . . . . . . . . . . AMMONAB .Steady-State Absorption Column Design . . MEMSEP . Gas Separation by Membrane Permeation . . FILTWASH . Filter Washing . . . . . . . . . . . . .

5.8.3 5.8.4 5.8.5 5.8.6 5.8.7 5.8.8 5.8.9 5.8.10 5.8.11 5.8.12

505 505 509 514 518 524

. 527

.

527 530

. 534 . 540

543 . 548 . 552

. 556

.

. .

560 567 572 578

5.9.7

Distillation Process Examples . . . . . . . . . . . . . . 584 BSTILL .Binary Batch Distillation Column . . . . . . . . 584 DIFDIST . Multicomponent Differential Distillation . . . . 589 MUBATCH . Multicomponent Batch Distillation . . . . . . 593 CONSTILL . Continuous Binary Distillation Column . . . . 599 MCSTILL .Continuous Multicomponent Distillation Column . . . . . . . . . . . . . . . . . . . . . . . 605 Bubble Point Calculation for a Batch Distillation BUBBLE . Column . . . . . . . . . . . . . . . . . . . . . . . 610 STEAM . Multicomponent, Semi-Batch Steam Distillation . . 616

5.10 5.10.1 5.10.2 5.10.3

Heat Transfer Examples . . . . . . . . . . . . . . . . 622 HEATEX . Dynamics of a Shell-and-Tube Heat Exchanger . . 622 SSHEATEX . Steady.State, Two-Pass Heat Exchanger . . . 628 ROD . Radiation from Metal Rod . . . . . . . . . . . . 632

5.9

5.9.1 5.9.2 5.9.3 5.9.4 5.9.5 5.9.6

XVI

Table of Contents

5.11.3 5.11.4

Diffusion Process Examples . . . . . . . . . . . . . . . 636 DRY - Drying of a Solid . . . . . . . . . . . . . . . . 636 ENZSPLIT - Diffusion and Reaction: Split Boundary Solution . . . . . . . . . . . . . . . . . . . . . . . 641 ENZDYN - Dynamic Diffusion and Enzymatic Reaction . . 647 BEAD - Diffusion and Reaction in a Spherical Bead . . . . 653

5.12 5.12.1 5.12.2 5.12.3

Dynamic Numerical Examples . . . . . . . . . . . . . . 659 LORENZ - Random Differential Equation Behaviour . . . . 659 HOPFBIF - The Hopf Bifurcation . . . . . . . . . . . . 661 CHAOS - Chaotic Oscillatory Behaviour . . . . . . . . . 663

5.11 5.11.1 5.11.2

Appendix: Using the ISIM Language

1 2 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 3 3.1 3.2 3.3

4

Index

. . . . . . . . . . . . . . . 667

ISIM Installation Procedure . . . Programming with ISIM . . . . . Getting Started . . . . . . . . . . Reading Files from Disk . . . . . Running Simulations . . . . . . . Interacting with ISIM Simulations . Editing ISIM Files . . . . . . . Writing ISIM Models . . . . . . ISIM Statements and Functions . . Output . . . . . . . . . . . . . Useful Sequences of Statements . . Further Information . . . . . . . Summary of ISTM Commands . . . COMMAND and INPUT Modes . ISIM Commands . . . . . . . . ISIM Program Statements . . . . ISJM Error Messages . . . . . .

. . . . . . . . . . . 667 . . . . . . . . . . . 669 . . . . . . . . . . 669

. . . . . . . . . . . 669. . . . . . . . . . . . 669

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

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

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

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

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

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

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

. . . .

670 670 672 674 676 . 677 . 678 . 679 . 679 . 679 . 683 . 687 691

Nomenclature for Chapters 1 to 4 Units

Symbols A A a a B b

C cP CV

D d d, D E E E F F f G g G' H AH H H HG

HL h hi J j

K K k

Area Magnitude of controller input signal Specific interfacial area Various parameters Magnitude of controller output signal Various parameters Concentration Heat capacity at constant pressure Heat capacity at constant volume Diffusivity Differential operator Diameter Energy Activation energy Residence time distribution Residence time distribution Volumetric flow rate Frequency in the ultimate gain method Gas or light liquid flow rate Gravitational acceleration Superficial light phase velocity Enthalpy Enthalpy change Height Henry's law constant Rate of heat gain Rate of heat loss Fractional holdup Partial molar enthalpy Total mass flux Mass flux Constant in Cohen-Coon method Mass transfer coefficient Constant

m2 various m2'm 3 various various various kg/m3, kmol/m3 kJkg K, kJ/mol K kJkg K, kJ/mol K m2/s

m

kJ or k J k g kJ/mol

m3/s 1/s m3/s Ids2 Ids kJ/mol, kJkg kJ/mol, k J k g

rn bar m3/kg kJ/S

kJ/S

kJ/mol kg/s, kmoVs kg/m2 s, mol/m2 s various m/S

various

XVlll -

I<@ kc,a KLa kLa

KLX a

KP

L L L' M N N n n P

P

P Pe

Q Q

q R

R R

r

Ti

rQ S S S S

T t

U U V

v V

w

Nomenclature for Chapters 1 to 4 _ _ _ -

~~

_-

.

Gas-liquid mass transfer coefficient referring to concentration in G-phase Gas film mass transfer coefficient Gas-liquid mass transfer coefficient referring to concentration in L-phase Liquid film mass transfer coefficient Overall mass transfer capacity coefficient based on the aqueous phase mole ratio X Proportional controller gain constant Length Liquid or heavy phase flow rate Superficial heavy phase velocity Mass Mass flux Molar flow rate Number of moles Reaction order Controller output signal Total pressure or pure component vapour pressure Partial Pressure Peclet number (L v/D) Heat transfer rate Total transfer rate Heat flux ldeal gas constant Reaction rate Number of reactions Reaction rate Reaction rate of component i Heat production rate Slope of process reaction curve/A Selectivity Number of compounds Laplace operator Temperature Time Heat transfer coefficient Internal energy Vapour flow rate Volume Flow velocity Rate of work

~

l/s 11s

I Is l/s kmol/m3 s various m m%, mol/s mlS

kg, mol kg/m2 s mol/s various bar bar -

HIS

kgls, molls kJ/m2 s bar m3/K mol kg/s, kmol/s kglm3 s, kmol/m3 s kg i/m3 s, kmol/m3 s kJ/m3 s various

"C, K h, min, s kJ/m2 K s kJ/mol moVs m3 m/S

HIS

XIX

Nomenclature for Chapters 1 to 4

W X X X

X X

Y Y Y Y Y Z Z Z

Mass flow rate Concentration in heavy phase Mole ratio in the heavy phase Conversion Mole fraction in heavy phase Input variable Fractional yield Concentration in light phase Mole ratio in the light phase Mole fraction in light phase Output variable Arrhenius constant Length variable Length variable

kgls kglm3, mollm3 -

various kg/m3, mollm3

-

various various m m

Greek

a a a, P

A &

Q,

?-l

P

V

0

0 P

c z z

z T

?

Backmixing factor Relative volatility Reaction order Difference operator Controller error Thiele modulus Effectiveness factor Plate efficiency Dynamic viscosity Stoichiometric coefficient Dimensionless time Dimensionless temperature Density Summation operator Controller time constant Residence time Shear stress Time constant Time lag Partial differential operator

-

various -

kg m l s kglm3 S

h and s kg m l s 2 h, min, s h, min, s -

Indices 0 1 1, 2, ..., n

Refers to initial, inlet, external, or zero order Refers to outlet or first order Refers to segment, stage, stream, tank or volume element

_____

A

a abs agit aPP avg B C C

D E eq F f G h ht 1 1

inert j L m max mix mt n

6 R r S S

SL ss St t

tot V W

--_____

Nomenclature for Chapters 1 to 4 -

__

~~

Refers to component A Refers to ambient Refers to absorption Refers to agitation Refers to apparent Refers to average Refers to component B, base, backmixing, surface position or boiler Refers to component C or combustion Refers to cross-sectional or cold Refers to derivative control, component D, delay or drum Refers to electrode Refers to equilibrium Refers to formation or feed Refers to final or feed plate Refers to gas or light liquid phase or generation Refers to hot Refers to heat transfer Refers to integral control Refers to component i or to interface Refers to inert component Refers to reaction j or to jacket Refers to liquid phase, heavy liquid phase or lag Refers to metal wall or mixer Refers to maximum Refers to mixer Refers to mass transfer Refers to tank, section, segment or plate number Refers to plug flow or pocket Refers to heat Refers to recycle stream Refers to reactor Refers to settler, steam, solid or surroundings Refers to surface, settler or shell side Refers to liquid film at solid interface Refers to steady state Refers to standard Refers to tube Refers to total Refers to vapour Refers to water or wall Bar above symbol refers to dimensionless variable Refers to perturbation variable, superficial velocity or stripping section Refers to equilibrium concentration

Chemical Engineering Dynami :Modelling

with PC Simulation

John Inph;m. Iruine J. Dunn. Elmar Heinzlc &JiiiE. Picnoul

copyright OVCH VerlagsgesellschaftmbH, 1994

1

Basic Concepts

1.1

Modelling Fundamentals

1.1.1

Chemical Engineering Modelling

The use of models in chemical engineering is well established, but the use of dynamic models, as opposed to the more traditional use of steady-state models for chemical plant analysis, is much more recent. This is reflected in the development of new powerful commercial software packages for dynamic simulation, which has arisen owing to the increasing pressure for design validification, process integrity and operation studies for which a dynamic simulator is an essential tool. Indeed it is possible to envisage dynamic simulation becoming a mandatory condition in the safety assessment of plant, with consideration of such factors as start up, shutdown, abnormal operation, and relief situations. Dynamic simulation can thus be seen to be an essential part of any hazard or operability study, both in assessing the consequences of plant failure and in the mitigation of possible effects. Dynamic simulation is thus of equal importance in large scale continuous process operations as in other inherently dynamic operations such as batch, semi-batch and cyclic manufacturing processes. Dynamic simulation also aids in a very positive sense in gaining a better understanding of process performance and is a powerful tool for plant optimisation, both at the operational and at the design stage. Furthermore steady-state operation is then seen in its rightful place as the end result of a dynamic process for which rates of change have become eventually zero. The approach in this book is to concentrate on a simplified approach to dynamic modelling and simulation. Large scale commercial software packages for chemical engineering dynamic simulation are now very powerful and contain highly sophisticated mathematical procedures which can solve both for the initial steady-state condition as well as for the following dynamic changes. They also contain extensive standard model libraries and the means of synthesising a complete process model by combining standard library models. Other important aspects are the provision for external data interfaces and builtin model identification and optimisation routines, together with access to a physical property data package. The complexity of the software, however, is

such that the packages are often non-user friendly and the simplicity of the basic modelling approach can be lost in the detail of the solution procedures. The correct use of such design software requires a basic understanding of the sub-model blocks and hence of the methodology of modelling. Our simplified approach to dynamic modelling and simulation incorporates no large model library, no attached database and no relevant physical property package. Nevertheless quite realistic process phenomena can be demonstrated, using a very simple approach. Often a simplified approach can also be very useful in clarifying preliminary ideas before going to the large scale commercial package, as we have found several times in our research. Again this follows our general philosophy of starting simple and building in complications as the work and as a full understanding of the process model progresses. Kapur ( I 988) has listed thirty-six characteristics or principles of mathematical modelling. These are very much a matter of common sense, but it is very important to have them restated, as it is often very easy to lose sight of the principles during the active involvement of modelling. They can be summarised as follows: 1. The mathematical model can only be an approximation of real-life processes, which are often extremely complex and often only partially understood. Thus models are themselves neither good nor bad but, as pointed out by Kapur, will either give a good fit or a bad fit to actual process behaviour. Similarly, it is possible to develop several different models for the same process, and these will all differ in some respect in the nature of their predictions. Indeed it is often desirable to try to approach the solution of a given problem from as many different directions as possible, in order to obtain an overall improved description. The purpose of the model also needs to be very clearly defined, since different models of a process, each of which has been developed with a particular purpose in mind, may not satisfy a different aim for which the model was not specifically constructed. Of course, whilst the aim of the modelling exercise is always to obtain as realistic a description of the process phenomena as possible, additional realism often involves additional numerical complexity and will demand additional data, which may be difficult or impossible to obtain. A marginal additional degree of realism can thus become rapidly outweighed by the large amount of extra time and effort needed.

2. Modelling is a process of continuous development, in which it is generally

advisable to start off with the simplest conceptual representation of the process and to build in more and more complexities, as the model develops. Starting off with the process in its most complex form, often leads to confusion. A process of continuous validation is necessary, in which the model theory, data, equation formulation and model predictions must all be examined repeatedly. In formulating any model, it is therefore important to

1.1 Modelling Fundamentals

3

include the potential for change. The final form of the model will lie somewhere between the initial, highly-simplified and unrealistic model and a possible final overcomplicated and over-ambitious model, but should provide a reasonable description of the process and must be capable of being used. Often it is possible to consider the process or plant, as a system of independent sub-sets or modules, which are then modelled individually and combined to form a description of the complete system. This technique is also used in the large scale commercial simulation software, in which various library sub-routines or modules for the differing plant elements, are combined into a composite simulation program.

3. Modelling is an art but also a very important learning process. In addition to a mastery of the relevant theory, considerable insight into the actual functioning of the process is required. One of the most important factors in modelling is to understand the basic cause and effect sequence of individual processes. Often the model itself, by generating unexpected behaviour, will assist in gaining the additional process insight, since the basic cause for the anomalous behaviour must be thought out and a plausible explanation found. The modelling process will often also suggest the need for new data or for experimentation needed to elucidate various aspects of process behaviour that are not well understood. 4. Models must be both realistic and also robust. A model which predicts effects which are quite contrary to common sense or to normal experience is unlikely to be met with confidence. To accord with this, some use of empirical adjustment factors in the model may be needed, in order to represent combinations of relatively unknown unknown factors. The basic stages in the above modelling methodology are indicated in Fig. 1.1. Compared to purely empirical methods of describing chemical process phenomena, the modelling approach attempts to describe performance, by the use of well-established theory, which when described in mathematical terms, represents a working model for the process. In carrying out a modelling exercise, the modeller is forced to consider the nature of all the important parameters of the process, their effect on the process and how each parameter can be defined in quantitative terms, i.e., the modeller must identify the important variables and their separate effects, which in practice may have a very highly interactive effect on the overall process. Thus the very act of modelling is one that forces a better understanding of the process, since all the relevant theory must be critically assessed. In addition, the task of formulating theory into terms of mathematical equations is also a very positive factor that forces a clear formulation of basic concepts. Once formulated, the model can be solved and the behaviour predicted by the model compared with experimental data. Any differences in performance

4

I Basic Concepts -

-..

~

~

G Define model aim

Build In complexities later

Use the model to learn

Models are there to be applled Figure 1.1. Steps in model building.

may then be used to further redefine or refine the model until good agreement is obtained. Once the model is established it can then be used, with reasonable confidence, to predict performance under differing process conditions, and used for process design, optimisation and control. Input of plant or experimental data is, of course, required in order to establish or validate the model, but the quantity of data required, as compared to the empirical approach is considerably reduced. A comparison of the modelling and empirical approaches are given below.

Empirical Approach: Measure productivity for all combinations of plant operating conditions, and make correlations. - Advantage: - Disadvantage:

Little thought is necessary. Many experiments are required.

Modelling Approach: Establish a model and design experiments to determine the model parameters. Compare the model behaviour with the experimental measurements. Use the model for rational design, control and optimisation. ~

-

Advantages: Disadvantage:

Fewer experiments are required and greater understanding is obtained. Time is required for developing models.

1 . 1 Modelling Fundamentals

1.1.2

5

General Aspects of the Modelling Approach

An essential stage in the development of any model, is the formulation of the appropriate mass and energy balance equations (Russell and Denn, 1972). To these must be added appropriate kinetic equations for rates of chemical reaction, rates of heat and mass transfer and equations representing system property changes, phase equilibrium, and applied control. The combination of these relationships provides a basis for the quantitative description of the process and comprises the basic mathematical model. The resulting model can range from a simple case of relatively few equations to models of great complexity. The greater the complexity of the model, however, the greater is then the difficulty in identifying the increased number of parameter values. One of the skills of modelling is thus to derive the simplest possible model, capable of a realistic representation of the process. A basic use of a process model is to analyse experimental data and to use this to characterise the process, by assigning numerical values to the important process variables. The model can then also be solved with appropriate numerical data values and the model predictions compared with actual practical results. This procedure is known as simulation and may be used to confirm that the model and the appropriate parameter values are "correct". Simulations, however, can also be used in a predictive manner to test probable behaviour under varying conditions, leading to process optimisation and advanced control strategies. The application of a combined modelling and simulation approach leads to the following advantages: 1. Modelling improves understanding. In formulating a mathematical model, the modeller is forced to consider the complex cause-and-effect sequences of the process in detail, together with all the complex inter-relationships that may be involved in the process. The comparison of a model prediction with actual behaviour usually leads to an increased understanding of the process, simply by having to consider the ways in which the model might be in error. 2. Models help in experimental design. It is important that experiments be designed in such a way that the model can be properly tested. Often the model itself will suggest the need for data for certain parameters, which might otherwise be neglected. Conversely, sensitivity tests on the model may indicate that certain parameters may be negligible and hence can be neglected in the model.

3. Models may be used predictively for design and control. Once the model has been established, it should be capable of predicting performance under differing process conditions, that may be difficult to achieve experimentally. Models can also be used for the design of relatively sophisticated control

6

1 Ba5ic Concepts -

systems and can often form an integral part of the control algorithm. Both mathematical and knowledge based models can be used in designing and optimising new processes. 4. Models may be used in training and education. Many important aspects of reactor operation can be simulated by the use of simple models. These include process start-up and shut down, feeding strategies, measurement dynamics, heat effects and control. Such effects are easily demonstrated by computer, as shown in the accompanying simulation examples, but are often difficult and expensive to demonstrate in practice.

5 Models may be used for process optimisation. Optimisation usually involves thc influence of two or more variables, with one often directly related to profits and the other related to costs.

1.1.3

General Modelling Procedure

One of the more important features of modelling is the frequent need to reassess both the basic theory (physical model), and the mathematical equations, representing the physical model, (mathematical model), in order to achieve agreement, between the model prediction and actual plant performance (experimental data). As shown in Fig. 1.2, the following stages in the modelling procedure can be identified: (i) The first involves the proper definition of the problem and hence the goals and objectives of the study. All the relevant theory must be assessed in combination with any practical experience, and perhaps alternative physical models need to be developed and examined.

(ii) The available theory must then be formulated in mathematical terms. Most reactor operations involve many different variables (reactant and product concentrations, temperature, rates of reactant consumption, product formation and heat production) and many vary as a function of time (batch, semi-batch operation). For these reasons the mathematical model will often consist of many differential equations. (iii) Having developed a model, the equations must then be solved. Mathematical models of chemical engineering systems, are usually quite complex and highly non-linear and are such that an analytical means of solution is not possible. Numerical methods of solution must therefore be employed, with the method preferred in this text being that of digital

7

1.1 Modelling Fundamentals

simulation. With this method, the solution of even very complex models is accomplished with relative ease. Design, optimization, control

1-

Experimental data

A

+

Physical model

4

Mathematical model

Solution: computer data

Revise ideas, equations and parameters

Figure 1.2. Information flow in model building and its verification.

Digital simulation languages are designed specially for the solution of sets of simultaneous differential equations, based on the use of numerical integration. Many fast and efficient numerical integration routines are now available, such that many digital simulation languages are able to offer a choice of integration routine. Sorting algorithms within the structure of the language enable very simple programs to be written, having an almost one-to-one correspondence with the way in which the basic model equations are originally formulated. The resulting simulation programs are therefore very easy to understand and also to write. A further major advantage is a convenient output of results, in both tabulated and graphical form, obtained via very simple program commands. (iv) The validity of the computer prediction must be checked and steps (i) to (iii) will often need to be revised at frequent intervals. The validity of the solution depends on the correct choice of theory (physical and mathematical model), the ability to identify model parameters correctly and accuracy in the numerical solution method.

In many cases, the system will not be fully understood, thus leaving large areas of uncertainty. The relevant theory may also be very difficult to apply. In such cases, it is then often necessary to make simplifying assumptions, which may subsequently be eliminated or improved as a better understanding is obtained. Care and judgement must be taken such that the model does not become over complex and that it is not defined in terms of immeasurable parameters. Often a lack of agreement can be caused by an incorrect choice of parameter values, which can even lead to quite contrary trends being observed during the course of the simulation. It is evident that these parameters to which the model response is very sensitive have to be chosen or determined with greatest care. It should be noted that often the model does not have to give an exact fit to data as sometimes it may be sufficient to simply have a qualitative agreement with the process.

1.1.4

Simulation Tools

Many different digital simulation software packages are available on the market. Fortunately many, but not all, conform to the standard structure of a Continuous System Simulation Language (CSSL). The programming structures for all CSSL languages are very similar. In addition, all CSSL languages are adjuncts to other high level languages such as FORTRAN, PASCAL or BASIC and thus provide the programmer with all the facilities and all the power of the host language. Other interesting features of simulation tools concern user interfacing, computing power, portability to various computer systems, optimisation and parameter estimation. Some tools allow a graphical set-up of a model. Some languages, for example ACSL and ESL, a development by the ISIM-group as advcrtised in the back of this book, run on PC's and larger machines, and include more powerful numerical algorithms and supply many predefined building blocks, which facilitate such tasks as the modelling of control systems. Mathematical software, such as MATLAB and MATHEMATICA have great computing power and may be portable to almost any available computer system. One useful application of modelling and simulation is optimisation of a process. MATLAB and SIMUSOLV include powerful algorithms for nonlinear optimisation, which can also be applied for parameter estimation. For this latter purpose SIMUSOLV is an excellent tool especially for practical work owing to its flexibility in handling experimental data (Heinzle and Saner, 199 I). Some characteristics of the differing simulation programs are given in Table I . 1. Optimisation and parameter estimation are also discussed in greater detail in Secs. 2.4.1 and 2.4.2. Recent surveys of the differing packages now available include Matko et al. (1992) and Wozny and Lutz (1991).

9

1.1 Modelling Fundamentals

Table 1.1 Characteristics of Simulation Programs. Program

Computer

Characteristics

ISIM

Pc

Enclosed in this book, highly interactive, moderate computational power, easy-touse.

ESL

All types

Graphical interface, powerful, and easyto-use.

ACSL

All types

Powerful mathematics, building blocks especially for control purposes.

MODELWORKS

PC and Mac

Pascal-based simulation environment, very flexible, public domain.

SIMNON

PC and others

Non-CSSL standard but very interactive and allowing system decomposition into both continuous and discrete subsystems.

STELLA

Mac

Graphical set up of model, highly interactive, but direct use of model equations not possible.

SIMUSOLV

VAX, IBM, SUN Simulation same as ACSL, includes powerful and user-friendly optimisation and parameter estimation.

MATLAB

All types

Powerful mathematical package especially for linear problems, includes optimisation routines.

SIMULlNK

Workstations, Mac and PC

Based on MATLAB but with improved integration routines, model building blocks and graphical interface.

MATHEMATTCA

All types

General mathematical software, includes algebraic integration methods, high precision number presentation.

SPEEDUP

Workstations, and Dynamic and steady-state simulation of large processes with data base. larger

10

1 Basic Concepts

~

ISIM and ESL, ISIM lnternational Simulation Ltd., Technology House, Salford University Business Park, Lissadel Street, Salford M6 6AP, UK; STELLA, HighPerformance Systems Inc., 13 Dartmouth College Highway, Lyme, New Hampshire, 03768, USA; MATLAB, The Math Works, Inc., 21 Eliot Street, South Natick, MA 01760, USA; MATHEMATICA, Wolfram Res. Inc., P.O. Box 6059, Champaign, Illinois 61821, USA; ACSL. SIMUSOLV, Mitchell & Gauthier Associates, 73 Junction Square Dr., Concord, MA 0 1742-3096, USA; MODELWORKS, A. Fischlin, Inst. Terrestr. Ecology, ETH, Grabenstr. 11, CH8952 Schlieren, Switzerland; SIMNON, Department of Automatic Control, Lund Institute of Technology, Sweden; SPEEDUP, Aspen Technology, Inc. Ten Canal Park, Cambridge, Mass. 02141, USA.

1.1.5

ISIM

Modern simulation languages are now available that provide the possibility of carrying out the interactive simulation of complex problems at one's own desk. For DOS compatible systems, several types of language are currently available. The ISIM simulation language, provided with this book, is also the language we have used during the last five years in our continuing education courses, for undergraduate and postgraduate education and as a powerful research tool. The simulation examples in this book are programmed with the ISIM digital simulation language, which is supplied on the diskette. This language is a development from the Electrical Engineering Department of the University of Salford. ISIM is FORTRAN-based and conforms to the basic structure of a CSSL in having three regions; the INITIAL, DYNAMIC and TERMINAL regions, together with an optional CONTROL region. The INITIAL region is used for calculating and setting initial values of constants. The DYNAMIC region contains the model equations, and the TERMINAL region may be used lor end-of-run calculations. During execution, data values are produced i n tabular form by the command OUTPUT, and a graphical output of any one variable by the command PLOT. Post-mortem plots of all variables, stored by the command PREPARE, can be shown using the command GRAPH. The interactive feature of ISIM is particularly valuable. This allows the simulations to be interrupted during execution by pressing any key. Values of constants or parameters can be changed and the simulation continued. This means that changing process conditions can be easily effected without the need for complex programming statements. ISIM is also one of the very few digital simulation programming languages that can be used with PCs. This makes it possible for the reader, student and teacher to experiment directly with the model, in the classroom or at the desk. Probably one of the best ways to learn to use ISIM is by reference to existing programs as provided i n Chapter 5 , rather than by an initial detailed

1.1 Modelling Fundamentals

11

consultation of the condensed program manual. The introductory example, in Sec. 1.1.7, is specially formulated to explain the main points of program structure and execution. The simulation examples serve to enforce the learning process in a very effective manner and also provide hands-on confidence in the use of a simulation language. Readers can program their own examples, by formulating new mass balance equations or by modifying an existing example to a new set of circumstances. Thus by working directly at the PC, the no-longer-passive reader is able to experiment directly on the system in a very interactive way. A true degree of interaction is possible with ISIM because at the stroke of a few keys a simulation run can be stopped, parameters changed, and the run restarted from the point of interruption. Plotting the variables in any configuration is easy at any time during a run. Runs can be repeated with new parameters, and the results from multiple runs can be plotted together for comparison. Full details of the ISIM digital simulation programming language can be found in the appendix, and by reference to the ISIM programs associated with the simulation examples of Chapter 5.

1.1.6

Teaching Applications

The solution technique, based on the use of a simulation language, is particularly beneficial in teaching, because it permits easy programming, easily comprehensible program listings, a rapid and efficient solution of the model equations, interactive variation of the model parameters, and a convenient graphical output of the computed results. In our experience, digital simulation has proven itself to be the most effective way of introducing and reinforcing new concepts which involve multiple interactions. For effective teaching, the introduction of computer simulation methods into modelling courses can be achieved in various ways, and the method chosen will depend largely on how much time can be devoted, both inside and outside the classroom. The most time-consuming method for the student is to assign modelling problems to be solved outside the classroom on any available computer. If scheduling time allows, computer laboratory sessions are effective, with the student working either alone or in groups of up to three on each monitor or computer. This requires the availability of many computers, but has the advantage that pre-programmed examples, as found in this text, can be used to emphasise particular points related to a previous theoretical presentation. This method has been found to be particularly effective when used for short, continuing-education, professional courses. Working through the computer examples, the student can vary parameters and make program alterations, as well as working through the suggested exercises at their own

12

1 Basic Concept\ -.-

-

... .

~

pace. Demonstration of a particular simulation problem via a single personal computer and video projector is also an effective way of conveying the basic ideas, since students can still be active in suggesting changes and in anticipating the results.

1.1.7

Introductory ISIM Example: BATSEQ-Complex Reaction Sequence

The complex chemical reaction, shown below, is carried out in an isothermal, constant-volume, batch reactor. All the reactions follow simple first-order kinetic rate relationships, in which the rate of reaction is directly proportional to concentration (Fig. 1.3).

c

C

A

B

D

Figure 1.3. Example of a complex reaction sequence.

The model equations for each component of this system are ~dCA

dt

- - k l CA

where CA, CB, C c and CD are the concentrations of component, A, R, C and D, respectively and kl, k2, k3 and k4 are the respective kinetic rate coefficients.

13

1 . 1 Modelling Fundamentals

If the number of equations is equal to the number of unknowns, the model is complete and a solution can be obtained. The easiest way to demonstrate this is via an information flow diagram as shown in Fig. 1.4.

d CA = - ~ C A

cA

dt

r I

I

1

Figure 1.4. Information flow diagram for the complex reaction model equations.

It is seen that all the variables, required for the solution of any one equation block are obtained as results from other blocks. The information flow diagram thus emphasises the complex interrelationship of this very simple problem.

ISIM Program BATSEQ The following ISIM program solves the complex reaction problem 1 2

3 4 5 6

PROGRAM BATSEQ COMPLEX BATCH REACTION CONSTANT K1 = 0.01 CONSTANT K2 = 0.01 CONSTANT K3 = 0.05 CONSTANT K4 = 0.02 : :

SEQUENCE

14

1 Basic Concepts

CONSTANT A0 = 1 CONSTANT BO = 0 CONSTANT CO = 0 CONSTANT DO = 0 CONSTANT CINT = 10 CONSTANT TFIN = 500 1 S I M ; R E S E T ; I N T E R A C T ; G O T O~ INITIAL A=AO B=BO

7 8 9

10 11 12 13

14 15 16 17 18 19 20 21 22 23 24 25 26 27

c=co

D=DO DYNAMIC A'=-Kl*A B1=K1*A-K2*B+K3*C-K4*B C1=K2*B-K3*C D'=K4*B MOLE=A+B+C+D OUTPUT T, A, B, C, D, MOLE PREPARE T,A,B,C,D,MOLE PLOT T,A,O,TFIN,O,1

Line numbers are shown here only for reference and are not intrinsic to the functioning of the program. Line number 1

is the program file name for the program is the program title. In each case the colon sign , : , is used to indicate a non-executionable, i.e., reference only, statement 3 to 6 set the values for the four rate constants, kl , k2 , k3 and kq 7 to 10 set the initial values for concentrations CA, CB, Cc and CD 11 to 12 set the values of the control parameters, CINT, for the print interval and, TFIN, for the end of the simulation 13 SIM is a mandatory statement when control parameters are specified. The full significance of the other statements in line 13 are explained later 14 INITIAL specifies the start of the INITIAL region 15 to 18 specify the initial concentrations DYNAMIC specifies the start of the DYNAMIC region 19 20 to 23 represent the model equations is a check on the total mass balance 24 25 Values of the concentrations, A, B, C and D are tabulated, at intervals of time, CINT

2

15

1.1 Modelling Fundamentals

26

27

Values of all the variables, stored by the PREPARE statement can be plotted at the end of the simulation or after an interrupt, using the GRAPH command. The GRAPH T,A,B,C,D, command, for example, plots a combined post-mortem graph of all the concentrations with respect to the independent variable time Only one variable can be plotted during program execution. In this case, concentration, A, is plotted versus time, using appropriate scaling commands of the PLOT statement

Graphical output obtained by the GRAPH command, for the above example is shown in Chapter 5. Line 13, of the program listing, makes use of a special feature in ISIM. This allows those constants and initial conditions which are declared in a CONSTANT statement, to be varied, following the end of each run by using the statement VAL. Thus, the statement 1 SIM; INTERACT; RESET; G O T 0 1 resets the integration to time T=O and permits interaction at this point, so that constant values can be changed and the results of different runs can be compared using the GRAPH command.

1.2

Formulation of Dynamic Models

1.2.1

Mass Balance Equations

Steady-State Balances One of the basic principles of modelling is that of the conservation of mass. For a steady-state flow process, this can be expressed by the statement: Rate of mass flow

( into the system )

=

Rate of mass flow

(out of the system

Dynamic Total Mass Balances Most real situations are, however, such that conditions change with respect to time. Under these circumstances, a steady-state mass balance is inappropriate and must be replaced by a dynamic or unsteady-state mass balance, expressed as of (Ratemassof accumulation in the system )

Rate of

= (mass flow in) -

(m a ~ E ~ f o u t )

Here the rate of accumulation term represents the rate of change in the total mass of the system, with respect to time, and at steady state this is equal to zero. Thus the steady-state mass balance is seen to be a simplification of the more general dynamic balance. At steady state Rate of (accumulation of mass

)

= 0 = (Mass flow in) - (Mass flow out)

hence, when steady state is reached (Mass flow in) = (Mass flow out)

Component Balances The previous discussion has been in terms of the total mass of the system, but most process streams, encountered in practice, contain more than one chemical species. Provided no chemical change occurs, the generalised dynamic equation for the conservation of mass can also be applied to each chemical component of the system. Thus for any particular component Rate of accumulation of mass] of component in the system

=

thecomponent^ Mass flow of

into the system

-

[

Mass flow of the component out of the system

Component Balances with Reaction Where a chemical reaction occurs, the change, due to reaction, can be taken into account by the addition of a reaction rate term in the component balance equation. Thus in the case of material produced by the reaction [c;;;$on)Rate of of component in the system

=

[ ][ ][ :Mass ; :0

flow

component the system

Mass o f t hflow e

-

component out of the system

+

pr;E;m Rate of

]

component by the reaction

The principle of the component mass balance can also be extended to the atomic level and can also be applied to particular elements. Thus for the case of carbon, in say a fuel combustion process

1.2 Formulation of Dynamic Models

[

Rate of accumulation of carbon mass in the system

]

Mass flow

= [rate of the

carbon into the system

1

-

[

Mass flow rate of the carbon out of the system

1

17

Note that the elemental balances do not involve additional reaction rate terms since the elements are not changed by chemical reaction. While the principle of the mass balance is very simple, its application can often be quite difficult. It is important therefore to have a clear understanding of the nature of the system (physical model) which is to be modelled by the mass balance equations and also of the methodology of modelling.

1.2.2

Balancing Procedures

The methodology described below, outlines six steps I through VI, to establish the model balances. The first task is to define the system by choosing the balance or control region. This is done using the following procedure:

I.

Choose the balance region such that the variables are constant or change little within the system. Draw boundaries around the balance region

The balance region can vary substantially, depending upon the particular area of interest of the model, ranging from say the total reactor, a region of a reactor, a single phase within a reactor, to a single gas bubble or a droplet of liquid. The actual choice, however, will always be based on a region of assumed uniform composition, or on another property as in the case of population balances. Generally the modelling exercises will involve some prior simplification of the real system. Often the system being modelled will be considered in terms of a representation, based on systems of tanks (stagewise or lumped parameter systems) or systems of tubes (differential systems), or even combinations of tanks and tubes.

1 . 2 . 2 . 1 Case A.

Continuous Stirred-Tank Reactor

If the tank is well-mixed, the concentrations and density of the tank contents are uniform throughout. This means that the outlet stream properties are identical with the tank properties, in this case concentration CA and density p. The balance region can therefore be taken around the whole tank (Fig. 1.5).

18

1 Basic Concepts

Total mass = p V Mass of A = C A V

Balance region Figure 1.5. The balance region around the continuous reactor.

The total mass in the system is given by the product of the volume of the tank contents V (m3) multiplied by the density p (kg/m3), thus V p (kg). The mass of any component A in the tank is given in terms of actual mass or number of moles by the product of volume V times the concentration of A, CA (kg of A/m3 or kmol of A/m3), thus giving V CA in kg or kmol.

1.2.2.2 Case B. Tubular Reactor In the case of tubular reactors, the concentrations of the products and reactants will vary continuously along the length of the reactor, even when the reactor is operating at steady state. This variation can be regarded as being equivalent to that of the time of passage of material as it flows along the reactor and is equivalent to the time available for reaction to occur. Under steady-state conditions the concentration at any position along the reactor will be constant Balance region

Figure 1.6. The tubular reactor concentration gradients.

19

1.2 Formulation of Dynamic Models

with respect to time, though not with position. This type of behaviour, obtained with tubular reactors, can be approximated by choosing the incremental volume of the balance regions sufficiently small so that the concentration of any component within the region can be assumed to be approximately uniform. Thus in this case, many uniform property sub-systems (well-stirred tanks or increments of different volume but all of uniform concentration) comprise the total reactor volume. This situation is illustrated in Fig. 1.6. The basic concepts of the above lumped parameter and distributed parameter systems are shown in Fig. 1.7.

Lumped parameter

Homogeneous (well-mixed) No spatial variations Concentration = f(time only)

Control volume

Distributed parameter

Spatial variations Concentration = f(time and position)

Figure 1.7. Choosing balance regions for lumped and distributed parameter systems.

1 . 2 . 2 . 3 Case C. Coffee Percolator A coffee percolator operates by circulating a stream of boiling coffee solution from the reservoir in the base of the coffee pot up through a central rise-pipe to the top of a bed of coffee granules, through which the solution then percolates,

20

1 Basic Concepts _.

before returning in a more concentrated state to the base reservoir, as shown in Fig. 1.8.

Coffee grounds packed bed Liquid circulation

Well-mixed liquid reservoir Figure 1.8. Conceptual of coffee percolator.

The above system can be thought of as consisting of two parts with 1) the base reservoir acting effectively as a single well-stirred tank and 2) a fixed bed system of coffee granules irrigated by the flowing liquid stream. Solute coffee is removed from the granules by mass transfer under the action of a concentration driving force and is extracted into the liquid. The concentrations of the coffee, both in the granules and in the liquid flowing through the bed, will vary continuously both with distance and with time. The behaviour of the packed bed is therefore best approximated by a series of many uniform property subsystems. Each segment of solid is related to its appropriate segment of liquid by interfacial mass transfer, as shown in Fig. 1.9. The resulting model would therefore consist of component balance equations for the soluble component written over each of the many solid and liquid subsystems of the packed bed, combined with the component balance equation for the coffee reservoir. The magnitude of the recirculating liquid flow will depend on the relative values of the pressure driving force generated by the boiling liquid and the fluid flow characteristics of the system. The concept of modelling a coffee percolator as a dynamic process comes from a problem suggested by Smith et al. (1970) and extended by Ramsay (Bradford University).

21

1.2 Formulation of Dynamic Models

*

----- - - _ _b -----* -----* ----- - - _'_ b ----- - - _ _b

Liquid flow

Packed

* *

grounds

I t

Finite difference elements for the solid coffee grounds

lnterphase mass transfer Liquid phase between finite difference the bed elements elements

Figure 1.9. Modelling concepts for the packed bed solid-liquid extraction. Process of coffee percolation.

II.

Identvy the transport streams which flow across system boundary

the

Having defined the balance regions, the next task is to identify all the relevant inputs and outputs to the system (Fig. 1.10). These may be well-defined physical flow rates (convective streams), diffusive fluxes, but may also include interphase transfer rates. It is important to assume transfer to occur in a particular direction and to specify this by means of an arrow. This direction may reverse itself, but the change will be accommodated by a reversal in sign of the transfer rate term.

22

I Basic Concepts ~-

-~

Out by diffusion

Convective flow in

Convective flow

-

/---

/

-F

_--

l p , \ \

\

\

d b y diffusion Figure 1.10. Balance rcgion showing convective and diffusive flows in and out.

III.

Write the mass balance in word form

This is an important step because it helps to ensure that the resulting mathematical equation will have an understandable physical meaning. Just starting off by writing down equations is often liable to lead to fundamental errors, at least on the part of the beginner. All balance equations have a basic logic, as expressed by the generalised statement of the component balance given below, and it is very important that the model equations also retain this. Thus Rate of accumulation in the system

Mass flow

Mass flow of the

~

into the system

out of the system

+[

Rate of production ofthe component by the reaction

]

This can be abbreviated as (Accumulation)

IV.

=

(In)

-

(Out)

+

(Production)

Express each balance term in mathematical form with measurable variables

A. Rate of Accumulation Term This is given by the rate of change of the mass of the system, or the mass of some component within the system, with changing time and is expressed as the derivative of the mass with respect to time. Hence

23

1.2 Formulation of Dynamic Models

Rate of accumulation of mass (of component i within the system)

=

dMi

(T)

where M is in kg or mol and time is in h, min or s. Volume, concentration and, in the case of gaseous systems, partial pressure are usually the measured variables. Thus for any component i

where Ci is the concentration of component i (kmol/m3 or kg/m3). In the case of gases, the Ideal Gas Law can be used to relate concentration to partial pressure and mol fraction. Thus,

where pi is the partial pressure of component i, within the gas phase system, and R is the Ideal Gas Constant, in units compatible with p, V, n and T. In terms of concentration,

where yi is the mol fraction of the component in the gas phase and P is the total system pressure of the system. The accumulation term for the gas phase can be therefore written in terms of number of moles as dni __ dt For the total mass of the system

- d(V ___ p) dt dt

_dM _

with units

24

1 Basic Concepts

B. Convective Flow Terms Total mass flow rates are given by the product of volumetric flow multiplied by density. Component mass flows are given by the product of volumetric flow rates multiplied by concentration. (Convective mass flow rate) = (Volumetric flow rate) for the total mass flow

M

(Volume Mass 1

= Fp

and for the component mass flow with units

A stream leaving a well-mixed region, such as a well-stirred tank, has the identical properties as in the system, since for perfect mixing the contents of the tank will have spatially uniform properties, which must then be identical to the properties of the fluid leaving at the outlet. Thus, the concentrations of component i both within the tank and in the tank effluent are the same and equal to Cil, as shown in Fig. 1.1 1 .

Figure 1.11. Convective flow terms for a well-mixed tank reactor.

C. Diffusion of Components As shown in Fig. 1.12, diffusional flow contributions in engineering situations are usually expressed by Fick's Law for molecular diffusion

25

1.2 Formulation of Dynamic Models

where ji is the flux of any component i flowing across an interface (kmol/m2 s or kg/m2 s) and dCi/dZ (kmol/m3 m) is the concentration gradient and Di is the diffusion coefficient of component i (m2/s) for the material.

Figure 1.12. Diffusion flux ji driven by concentration gradient (Cil - Cio)/AZ through surface area A.

In accordance with Fick’s Law, diffusive flow always occurs in the direction of decreasing concentration and at a rate, which is proportional to the magnitude of the concentration gradient. Under true conditions of molecular diffusion, the constant of proportionality is equal to the molecular diffusivity of the component i in the system, Di (m2/s). For other cases, such as diffusion in porous matrices and for turbulent diffusion applications, an effective diffusivity value is used, which must be determined experimentally. The concentration gradient may have to be approximated in finite difference terms (finite differencing techniques are described in more detail in Secs. 4.2 to 4.4). Calculating the mass diffusion rate requires a knowledge of the area, through which the diffusive transfer occurs, since Mass rate ( c o r n b e n t i) =

Diffusivity -

( c o m g i e n t i)

(

Concentration gradient of i

)(

Area perpendicular to transport

)

The concentration gradient can often be approximated by difference quantities, where

with units

D. Interphase Transport Interphase mass transport also represents a possible input to or output from the system. In Fig. 1.13., transfer of a soluble component takes place across the interface which separates the two phases. Shown here is the transfer from phase G to phase L, where the separate phases may be gas, liquid or solid.

Phase L

Phase G

Figure 1.13. Transfer across an interface of area A from phase G to phase L.

When there is transfer from one phase to another, the component balance equations must take this into account. Thus taking a balance for component i around the well-mixed phase G, with transfer of i from phase G to phase L, gives Rate of kcu:;pon]

=

-

in phase G

[

Rate of interfacial mass transfer ofi from phase G into phase L

]

This form of the transfer rate equation will be examined in much more detail in Sec. 1.4. Suffice it to say here that the rate of transfer can be expressed in the form shown below Rate of (mass transfer) =

(

Mass transfer coefficient

Concentration

) (thAEzr%e) ( driving force

27

1.2 Formulation of Dynamic Models

J = KAAC The units of the transfer rate equation (with appropriate molar quantities) are kmol

~

S

-

m ,2

-

S

kmol m3

where, J is the total mass flux, A is the total interfacial area for mass transfer (m2), AC is the concentration driving force (kmol/m3) and K is the overall mass transfer coefficient (m/s). It is important to note that the concentration driving force is represented as a difference between the actual concentration and the corresponding equilibrium value and is not a simple difference between actual phase concentrations. Mass transfer rates can be converted to mass flows (kg/s), by multiplying by the molar mass of the component.

E. Production Rate The production rate term allows for the production or consumption of material by chemical reaction and can be incorporated into the component balance equation. Thus, Rate of

Mass flow

Mass flow

Rate of

in the system

the system

the system

the reaction

Chemical production rates are often expressed on a molar basis but can be easily converted to mass flow quantities (kg/s). The mass balance equation can then be expressed as

where RA is the total reaction rate. The units are

Equivalent molar quantities may also be used. The quantity rA is positive when A is formed as product, and rA is negative when reactant A is consumed.

V.

Introduce other relationships and balances such that the number of equations equals the number of dependent variables

The system mass balance equations are often the most important elements of any modelling exercise, but are themselves rarely sufficient to completely formulate the model. Thus other relationships are needed to complete the model in terms of other important aspects of behaviour in order to satisfy the mathematical rigour of the modelling, such that the number of unknown variables must be equal to the number of defining equations. Examples of this type of relationships, which are not based on balances, but which nevertheless form a very important part of any model are: Reaction rates as functions of concentration and temperature. Reaction stoichiometry. Equations of state or Ideal Gas Law behaviour. Physical property correlations as functions of concentration, temperature, etc. Hydraulic flow equations. Pressure variations as a function of flow rate. Dynamics of measurement instruments, as a function of instrument response time. Equilibrium relationships (e.g., Henry's law, relative volatilities, etc.). Controller equations with an input variable dependent on a measured variable. Correlations for mass transfer coefficients, gas holdup volume, and interfacial area, as functions of system physical properties and agitation rate or flow velocity, etc. How these and other relationships are incorporated within the development of particular modelling instances are illustrated, throughout the text and in the simulation examples.

VI.

For additional insight with complex problems, construct an information flow diagram

Information flow diagrams can be useful in understanding complex interactions (Franks, 1967). They help to identify missing relationships and provide a graphical aid to a full understanding of system interaction. An example of such a diagram is shown in Fig. 1.4.

29

1.2 Formulation of Dynamic Models

1.2.3

Total Mass Balances

In this section the application of the total mass balance principles is presented. Consider some arbitrary balance region, as shown in Fig. 1.14 by the shaded area. Mass accumulates within the system at a rate dM/dt, owing to the competing effects of a convective flow input (mass flow rate in) and an output stream (mass flow rate out). Mass flow rate out

------+

Mass flow rate in

Figure 1.14. Balancing the total mass of an arbitrary system.

The total mass balance is expressed by Rate of accumulation in the system of mass dM dt

Mass flow

) = ( into the system)

Mass flow out

-

(of the system

= Mass rate in - Mass rate out

or in terms of volumetric flow rates F, densities p, and volume V,

When densities are equal, as in the case of water flowing in and out of a tank

The steady-state condition of constant volume in the tank (dV/dt = 0) occurs when the volumetric flow in, Fo, is exactly balanced by the volumetric flow out, F1. Total mass balances therefore are mostly important for those modelling situations in which volumes are subject to change, as given in simulation examples CONFLO, TANKBLD, TANKDIS and TANKHYD.

30

1 Basic Concept\ ~-

. .

1 . 2 . 3 . 1 Case A. Tank Drainage A tank of diameter D, containing liquid of depth H, discharges via a short base connection, of diameter d, as shown in Fig. 1.15 (Brodkey and Hershey, 1988).

Figure 1.15. Liquid discharge from the bottom of a tank

In this case, the problem involves a combination of the total mass balance with a hydraulic relationship, representing the rate of drainage. For zero flow of liquid into the tank and assuming constant density conditions, the total mass balance equation, becomes

Assuming the absence of any frictional flow effects, the outlet flow, velocity v, is related to the instantaneous depth of liquid within the tank, by the relationship v = (2 g h)”* where

and

31

1.2 Formulation of Dynamic Models

v is the discharge pipe velocity, V is the volume of liquid in the tank, h is the depth of liquid in the tank and g is the constant of gravitational acceleration. The above equations are then sufficient to define the model which has the following simple analytical solution

where H is the liquid depth at time t = 0. However, with a time variant flow of liquid into the tank, then analytical solution is not so simple. The above problem is treated in more detail by the simulation example TANKDIS.

1.2.4

-

Component Balances

Each chemical species, in the system, can be described by means of a component balance around an arbitrary, well-mixed, balance region, as shown in Fig. 1.16.

Species i inflow

d(

Speciesi outflow

v ci 1

Figure 1.16. Component balancing for species i

In the case of chemical reaction, the balance equation is represented by Rate of accumulation in the system

Mass flow of

Mass flow of

Rate of

the system

the system

component i by reaction

Expressed in terms of volume, volumetric flow rate and concentration, this is equivalent to

32

1 Basic Concepts

with units of massltime

In the case of an input of component i to the system by interfacial mass transfer, the balance equation now becomes: Rate of accumulation

Mass flow of

Mass flow of

the system

the system

in the system

(’

dt

=

(Fo Cio) - (F1 Cil)

+

Rate of interfacial into the system

Qi

where Qi, the rate of mass transfer is given by

1 . 2 . 4 . 1 Case A. Waste Holding Tank A plant discharges an aqueous effluent at a volumetric flow rate F. Periodically, the effluent is contaminated by an unstable noxious waste, which is known to decompose at a rate proportional to its concentration. The effluent must be diverted to a holding tank, of volume V, prior to final discharge, as in Fig. 1.17 (Bird et al. 1960). This situation is one involving both a total and a component mass balance, combined with a kinetic equation for the rate of decomposition of the waste component. Neglecting density effects, the total mass balance equation is

33

1.2 Formulation of Dynamic Models

F F

Figure 1.17. Waste holding tank.

The rate of the decomposition reaction is given by

and the component balance equation by

The tank starts empty, so that at time t = 0, volume V = 0, and the outlet flow from the tank F1 = 0. At time, t = V/Fo, the tank is full; then F1 = Fo =F, and the condition that dV/dt = 0 also applies. The above model equations can be solved analytically. For the conditions that at time t = 0, the initial tank concentration CA = 0, the tank is full and overflowing and that both F and CAOare constant, the analytical solution gives

where

z=

F F+kV

When the flow and inlet concentration vary with time, a solution is best however obtained, by numerical integration.

1 . 2 . 4 . 2 Case B. Extraction from a Solid by a Solvent An agitated batch tank, is used to dissolve a solid component from a solid matrix into a liquid solvent medium, as in Fig. 1.18, Geankoplis (1983).

34

1 Basic Concepts

0

0 0

0

0

0 0

0

0

O

0 O

0 0

0

Figure 1.18. Agitated tank for dissolving solids.

For a batch system, with no inflow and no outflow, the total mass of the system remains constant. The solution to this problem, thus involves a liquid-phase, component mass balance for the soluble material, combined with an expression for the rate of mass transfer of the solid into the liquid. The component mass balance is then Rate of accumulation of the material in the solvent

Rate of transfer of the solid to the solvent

I

where, VL is the volume of the liquid, CL is the concentration of the component in the liquid, k L is the liquid phase mass transfer coefficient, A is the total interfacial area for mass transfer and CL* is the equilibrium value. The analytical solution to the above equation, assuming constant VL, kL, A and equilibrium concentration, CL*, is given by cL*-cL = ,-kLAt/VL CL* - CLO For the case, where the soluble component is leaching from an inert solid carrier, a separate solid phase component balance would be required to establish the solute concentration in the solid phase and hence the time dependent value of the equilibrium concentration, CL*. If during this extraction the volume and area of the solid remains approximately constant, the balance for the component in the solid phase is

35

1.2 Formulation of Dynamic Models

Vs%

= - k L A (CL*-CL)

where

CL" = feq ( CS> and the subscript "eq" refers to the equilibrium condition.

1.2.5

Energy Balancing

Energy balances are needed whenever temperature changes are important, as caused by reaction heating effects or by cooling and heating for temperature control. For example, such a balance is needed when the heat of reaction causes a change in reactor temperature. This is seen in the information flow diagram for a non-isothermal continuous reactor as shown in Fig. 1.19.

t Component mass

("91")equation

I

Sk

Rate constant k=f (T)

Energy balance Figure 1.19. Information flow diagram for modelling a non-isothermal, chemical reactor, with simultaneous mass and energy balances.

36

1 Basic Concepts ~-

~~

Energy balances are formulated by following the same set of guidelines as those given in Sec. 1.2.2 for mass balances. Energy balances are however considerably more complex, because of the many processes which cause temperature change in chemical systems. The treatment considered here is somewhat simplified, but is adequate to understand the non-isothermal simulation examples. The various texts cited in the reference section, provide additional advanced reading in this subject. Based on the law of conservation of energy, energy balances are a statement of the first law of thermodynamics. The internal energy depends, not only on temperature, but also on the mass of the system and its composition. For that reason, mass balances are almost always a necessary part of energy balancing. For an open system with energy exchange across its boundaries, as shown in Fig. 1.20, the energy balance can be written as Rate of accumulation ofenergy

[

)

Rate of =

dE dt -

to flow

Rate of energy

Rate of energy output due to flow

i= 1

i= 1

to transfer

Cil Eil +

Q

Rate of work done by the system on the surroundings

- w

Here, E is the total energy of the system, Ei is the energy per mole of component i, F is the volumetric flow rate, C is the molar concentration and S is the number of components (reactants and inerts). The work term can be separated into flow work and other work Ws, according to S

s

i= 1

i= 1

The energy Ei is the sum of the internal energy Ui, the kinetic energy, the potential energy and any other energies. Of these various forms of energy, changes of internal energy are usually dominant in chemical systems. The other terms are usually neglected. Internal energy, U, can be expressed in terms of enthalpy, H Ei = Ui = Hi

-

(PV)j

The accumulation term is given in terms of enthalpies

37

1.2 Formulation of Dynamic Models Combining these equations and assuming the other work Ws = 0, yields S

S

S

i= 1

i= I

i=l

Here ni the number of moles of component i, hi the partial molar enthalpy and Q is the rate of energy loss to the environment.

tQ Figure 1.20. A continuous reactor showing only the energy-related variables.

Enthalpies are generally dependent on temperature as follows: hi = hi0

+

T [cPi dT TO

The temperature dependency of heat capacity can usually be described by a polynomial expression, e.g., cp = a

+

bT

+

c T2

where a, b and c are empirical constants. A detailed derivation of the energy balance is given in various textbooks (e.g., Aris, 1989 and Fogler, 1992). The accumulation term in the energy balance equation can be rewritten as

i=l

i= 1

i= 1

For the solution of the energy balance it is necessary that this is combined with material balance relationships. Using a general mass balance for component i

38

I Basic Concepts

multiplying this equation by hi1 and summing over S components gives

Introducing the reaction enthalpy AH S

and allowing R reactions to occur gives together for the general energy balance

where Cpi is the partial molar heat capacity of component i , rij the reaction rate of component i in reaction j, Vij the stoichiometric coefficient of component i in reaction j and AHj(T1) is the reaction enthalpy of reaction j at temperature T1. The heat of reaction AH is defined by

where AHFi is the heat of formation of component i. Considering the above temperature dependencies, the complete heat balance can then be written in the following form:

This equation can be used directly for any well-mixed, batch, semi-batch or continuous volume element. The term on the left-hand side represents the rate of energy accumulation. The first term on the right-hand side depicts the energy needed to raise the temperature of the incoming reactants, including inert material, to the reactor temperature. The second term describes the heat

39

1.2 Formulation of Dynamic Models

released by the chemical reactions. Since AH is a function of state, the energy balance could also be formulated such that the reaction is considered to take place at the inlet temperature To, followed by heating the reactor contents to temperature T i . The general energy balance equation is applied in simulation example REVTEMP and in Case C, Sec. 1.2.5.3. The general heat balance can often be simplified for special situations. The individual terms are discussed in what follows:

Accumulation Term At moderate temperature changes, Cpi may be assumed to be independent of temperature and therefore

Then

with units

K

mol _ _ (J/mol _ _ _ K) _ ^ . ~ S

m3 (kg/m3) (J/kg K) K __ S

-

kJ S

Thus the accumulation term has the units of (energy)/(time), for example J/s.

Flow Term At moderate temperature changes cpi is again assumed constant and therefore the flow term is

with the units:

40

1 Basic Concepts ~

~~

volume

energy

This term actually describes the heating of the stream entering the system with temperature To to the reaction temperature T I , and is only needed if streams are entering the system.

Heat Transfer Term The important quantities in this term are the heat transfer area A, the temperature driving force or difference (Ta-Tl), where Ta is the temperature of the heating or cooling source, and the overall heat transfer coefficient U. The heat transfer coefficient, U, has units of (energy)/(tirne)(area)(degree), e.g., J/s m2 K. The units for U A AT are thus (heat transfer rate) = U A (Ta - TI) energy energy time - area time degree (area) (degree)

~

The sign of the temperature difference determines the direction of heat flow. Here if Ta > T I , heat flows into the reactor.

Reaction Heat Term As already shown above, the heat of reaction is defined as n

where v, represents the stoichiomelric coefficient for component i. Heats of reaction are conventionally taken as positive for the reaction products and as negative for the reactants. AHF, represents the heat of formation for component i. For exothermic reactions, the value of AH is by convention negative and for endothermic reactions positive. For a set of R individual reactions, the total rate of heat production by reaction is given by

1.2 Formulation of Dynamic Models

41

with the units: (SuZgime

moles energy ) volume = volume ( s) ( x)

Other Heat Terms The heat of agitation may be important, depending on the relative magnitudes of the other heat terms in the general balance equation and especially with regard to highly viscous reaction mixtures. Other terms, such as heat losses from the reactor, by convection, by evaporation, by radiation or by mixing, can also be important in the overall energy balance equation.

Simplified Energy Balance If specific heat capacities can be assumed constant and the mechanical energy input by the stirrer, Wagit, is significant, the energy balance equation simplifies to

where the units of each term of the balance equation are energy per unit time (kJ/s). This equation can be applied to batch, semi-batch and continuous reactors. Most of the non-isothermal simulation examples in Chapter 5 use the above form of the energy balance.

1 . 2 . 5 . 1 Case A.

Continuous Heating in an Agitated Tank

Liquid is fed continuously to a stirred tank, which is heated by internal steam coils (Fig. 1.21). The tank operates at constant volume conditions. The system is therefore modelled by means of a dynamic heat balance equation, combined with an expression for the rate of heat transfer from the coils to the tank liquid. With no heat of reaction and neglecting any heat input from the agitator, the heat balance equation becomes

1 Basic Concepts _.

steam in

(

1

__

TS

4 -

Condensate

Figure 1.21. Continuous stirred tank heated by internal steam coil

where Ts is the steam temperature. The above equation can be made dimensionless (See Sec. 1.2.7.) by substituting 0 = ( t Fo /V) and 81 = (TI - To)/To. Consequently, dO = dt Fo/V and d81 = dTl/To. Then

and after rearrangement

where

1 a = ~U A~I +--P c p Fo

1.2 Formulation of Dynamic Models

43

and

This equation is actually equivalent to a first-order delay (Sec. 2.1.1). With the initial conditions 01 = O1,init and 0 = 0, it has the solution

1 . 2 . 5 . 2 Case B. Heating in a Filling Tank Liquid flows continuously into an initially empty tank, containing a full-depth heating coil. As the tank fills, an increasing proportion of the coil is covered by liquid. Once the tank is full, the liquid starts to overflow, but heating is maintained. A total mass balance is required to model the changing liquid volume and this is combined with a dynamic heat balance equation. Assuming constant density, the mass balance equation is

where for time t less than the filling time, V/Fo, the outlet flow F1, equals zero, and for time t greater than V/Fo, F1 equals Fo. The heat balance is expressed by

Assuming A0 is the total heating surface in the full tank, with volume, Vo, and assuming a linear variation in heating area with respect to liquid depth, the heat transfer area may vary according to the simple relationship

More complex relationships can of course be derived, depending on the particular tank geometry.

44

1 Basic Concepts

___

_______-.

1 . 2 . 5 . 3 Case C. Parallel Reaction in a Semi-Continuous Reactor with Large Temperature Changes Let us assume an adiabatic, semi-continuous reactor (see Sec. 3.2.4) with negligible input of mechanical energy (Fig. 1.22).

Two reactions are assumed to occur in parallel

A+B->C A

+

2 B ->

D

The total energy balance from Sec. 1.2.5 is given by

In this case the number of components, S, is 4 and the number of reactions, R, is 2. The reaction enthalpies at standard temperature, Tst, are then

45

1.2 Formulation of Dynamic Models

All heats of formation, AHFi, are at standard temperature. Assuming that the temperature dependencies for the specific heats are given by

and the reaction enthalpies, AH1 and AH2, at temperature T are AH1 = AHlst

+

(ac - aA -as) (T - Tst)

AH2 = AH2st+ (ac-aA -2 aB) (T-Tst)

b c - bA -bB

2 ( T2 - Tst2) + b c - bA - 2 bB + 2 (T2 Tst2> ~

-

With stoichiometric coefficients, V A l = -1 and V.42 = -1, the total heat of reaction is then

The total heat capacity in the accumulation term is

With only component B in the feed, the flow term in the energy balance becomes

Substitution into the energy balance then gives

46

I Basic Concepts .___

1.2.6

Momentum Balances

Momentum balance equations are of importance in problems involving the flow of fluids. Momentum is defined as the product of mass and velocity and as stated by Newton's second law of motion, force which is defined as mass times acceleration is also equal to the rate of change of momentum. The general balance equation for momentum transfer is expressed by Rate of

Rate of

into the system

Rate of ) generation of momentum)

or alternatively

[

Rate of change of momentum with respect to time

)

=

Rate of

Sum of the forces acting on the system

into the system

I

Force and velocity are however both vector quantities and in applying the momentum balance equation, the balance should strictly sum all the effects in three dimensional space. This however is outside the scope of this text and the reader is referred to more standard works in fluid dynamics. As for the mass and energy balance equations, steady-state conditions are obtained when the rate of change of momentum in the system is zero and Rate of momentum

Rate of momentum

Sum of the forces

Three forms of force, important in chemical engineering flow problems, are pressure forces, shear or viscous forces and gravitational forces. The pressure force is given as the product of pressure and applied area. There are usually taken to be positive when acting on the system surroundings. Shear or viscous forces are also usually taken to be positive when acting on the surroundings and shear force is again the product of shear stress and the applied area. The gravitational forces consist of the force exerted by gravity on the fluid and is equal to the product of the mass of fluid in the control volume times the local acceleration due to gravity. The simulation example TANKHYD utilizes a simple momentum balance to calculate flow rates.

1.2 Formulation of Dynamic Models

1.2.7

47

Dimensionless Model Equations

The model mass and energy balance equations will have consistent units, throughout, i.e., kg/h, kmol/h or kJ/h. and corresponding dimensions of masdtime or energykime. The major system variables, normally concentration or temperature will also be expressed in terms of particular units, e.g., kmol/m3 or "C, and the model solution will also usually be expressed in terms of the resultant concentration or temperature profiles, obtained with respect to either time or distance. Each variable will have some maximum value, which is usually possible to establish by simple inspection of the system. Time, as a variable, usually does not have a maximum value, but some characteristic value of time can always be identified. Using these values, a new set of dimensionless variables can be obtained, simply, by dividing all the variables by the appropriate maximum value or by the characteristic time value. This leads to a model composed of dependent variables that vary only between the limits of zero and unity. This means that the solution may, for example, be in terms of the variation of dimensionless concentration versus dimensionless time, and now has a much greater significance, since the particular units of the problem are no longer relevant to the model formulation. The model can now be used much more generally. Furthermore, the various parameters in the original model can be grouped together such that each group of terms also becomes dimensionless. As an added result, the model equations can now be formulated in terms of a fewer number of dimensionless groups than the original number of single parameters. In addition to extending the utility of the mathematical models, the resulting dimensionless groups are especially valuable in correlating experimental data. The above procedure is best illustrated in the simulation examples (BATCHD, TANKD, HOMPOLY, TUBED, TUBDYN, DISRE, DISRET, ENZSPLIT, ENZDYN, and BEAD).

1 . 2 . 7 . 1 Case A. Continuous Stirred-Tank Reactor (CSTR) The mass balance for a continuous-flow, stirred-tank reactor with first-order reaction is

and this is treated in more detail in Chapter 3. The dimensions for each term, in the above equation, are those of mass per unit time and the units would normally be kmol/h or kg/h. Dividing the balance equation by the volume of reactor, V, leads to the equation in the form

This equation has two parameters 2 , the mean residence time (z = V/F) with dimensions of time and k, the reaction rate constant with dimensions of reciprocal time, applying for a first-order reaction. The concentration of reactant A in the reactor cannot, under normal circumstances, excsed the inlet feed value, CAO,and thus a new dimensionless concentration, C A I , can be defined as

such that CAI normally varies in the range from zero to one. The other variable time, t, can vary from zero to some undetermined value, but the system is also represented by the characteristic time, z. Note that the value of l/k also represents a characteristic time for the process. A new dimensionless time variable is defined here as

t t =: Alternatively the dimensionless time variable

t = kt could be employed, but is not used here. In terms of the dimensionless variables, the original variables are

dt = T dT When substituted into the model equation, the result is

49

1.2 Formulation of Dynamic Models

This equation can now be rearranged such that the parameter for the time derivative is unity. Thus dividing by CAOand multiplying by ‘I: gives

dt The parameter term (k z) which is called Damkohler Number I, is dimensionless and is now the single governing parameter in the model. This results in a model simplification because originally the three parameters, z,k and CAO,all appeared in the model equation. The significance of this dimensionless equation form is now that only the parameter (k z) is important; and this alone determines the system dynamics and the resultant steady state. Thus, experiments to prove the validity of the model need only consider different values of the combined parameter (k z). For this, the dimensionless reactant concentration, CA1 , should be plotted versus dimensionless time, 7, for various values of the dimensionless system parameter, ( k z ) . Although, k is not an operating variable and cannot be set independently, this type of plot may be useful to estimate its value from experimental data, as illustrated below. At the steady state

Knowing

z thus permits finding k from the data.

In a batch system

or In

CAI =

-

(k z)t

+ constant.

Plotting the data as In CAI versus 7 will give - (k z) as the slope and permit calculating k. It should be noted that calculating k from data involves solving the equations, but establishing (k z) as the only important dimensionless group requires only the dimensionless model.

50

1 Basic Concepts

Consider an nth-order reaction; the equivalent dimensionless model for the stirred-tank reactor becomes

The variables are defined as previously. Thus if, for example, experimental gats is to be tested for second-order reaction behaviour, then data plotted as CAI versus t should be examined from experiments, during which (k 1' : CAO)is kept cons tan t.

1 . 2 . 7 . 2 Case B. Gas-Liquid Mass Transfer to a Batch Tank Reactor with Chemical Reaction A second-order reaction takes place in a two-phase batch system. Reactant A is

supplied by gas-liquid transfer and reactant B supplied by liquid feed. The model equations are

The saturation value, CA", will be assumed constant. The dimensionless variables can be defined as

The equations become

51

1.2 Formulation of Dynamic Models

where z is the residence time (=V/F). The number of parameters is reduced, and the equations are in dimensionless form. Alternatively other quantities, for example CA* and T, could have been used to define the dimensionless variables. An equivalency can be demonstrated between the concept of time constant ratios and the dimensionless parameters obtained from differential equations. The concept of time constants is discussed in Sec. 2.2. Thus variables in this example can be interpreted as follows

_1_ - Transfer time constant KLa z - Residence time constant

-

Convection rate Transfer rate

and k_ _ CBO _ _ -- Transfer time constant - Reaction rate KLa Reaction time constant - Transfer rate Further examples of the use of dimensionless terms in dynamic modelling applications are given in Sec. 1.2.5.1, Sec. 4.3.6.1 and 4.3.7 and in the simulation examples KLADYN, DISRET, DISRE, TANKD and TUBED.

1.3

Chemical Kinetics

1.3.1

Rate of Chemical Reaction

By simplifying the general component balance of Sec. 1.2.4, the mass balance for a batch reactor becomes

[

]

[.reduction

Rate of Rate of a c c ~ ~ ~ = ~ ~ component o n of) i

of component i in the system

by reaction

Expressed in terms of volume V and concentration Ci, this is equivalent to

52

I Basic Concepts

__~_____

-

with units of moles/time. Here the term ri is the rate of chemical reaction, expressed as the change in the number of moles of a given reactant or product per unit time and per unit volume of the reaction system. Thus for a batch reactor. the rate of reaction for reactant i can be defined as

ri =

1 dni

moles of i volume time

..

vdt

where ni = V Ci and is the number of moles of i present at time t. The reactants and products are usually related by a stoichiometric equation, which for the case of components A and B reacting to form product C has the form Vp,A+VgB -3 Vcc

vi is the stoichiometric coefficient for species i in the reaction. By convention,

the value of v is positive for the products and negative for the reactants. The stoichiometric coefficients relate the simplest ratio of the number of moles of reactant and product species, involved in the reaction. The individual rates of reaction, for all the differing species of a reaction are related via their stoichiometric coefficients according to ri = rj

(%)

The value of ri is therefore negative for reactants and positive for products. For the reaction

A+2B

+

3P

the individual reaction rates are therefore 1

1

- r A = - - 2 rg= 3 rp Thus in defining the rate of reaction, it is important to state the particular reaction species. The reaction rate must be determined by experiment; the temperature and concentration dependencies can usually be expressed as separate functions, for example

53

1.3 Chemical Kinetics

The exact functional dependence of the reaction rate with respect to concentration must be found by experiment, and for example may have the form

Here, k is the reaction rate constant, which is a function of temperature only; CA, CB are the concentrations of the reactants A, B (molesholume); a is the order of reaction, with respect to A; p is the order of reaction, with respect to B; ( a + p) is the overall order of the reaction. It is important to realize that the reaction rate may represent the overall summation of the effect of many individual elementary reactions, and therefore only rarely represents a particular molecular mechanism. The orders of reaction, a or p, can not be assumed from the stoichiometric equation and must be determined experimentally. For heterogeneous catalytic reactions, the rate of reaction is often expressed as the number of moles of component reacting per unit time, per unit mass of catalyst. For a batch reactor 1 'A=M

dnA dt ~

moles mass time

where M is the mass of catalyst. Sometimes the surface area is used as a reference for solid surface reactions. In vapour phase reactions, partial pressure units are often used in place of concentration in the rate equation, for example

where PA and PB are the gas phase partial pressures of reactants A and B. In this case, k would be expressed in terms of pressure units.

1.3.2

Reaction Rate Constant

The reaction constant, k, is normally an exponential function of the absolute temperature, T, and is described by the Arrhenius equation

The exponential term gives rise to the highly non-linear behaviour in reactor systems which are subject to temperature changes.

54 _____

1 Basic Concepts ~~

The parameters Z and E, the activation energy, are usually determined by measuring k, over a range of temperatures, and plotting ln(k) versus the reciprocal of absolute temperature, 1/T, as shown in Fig. 1.23.

Figure 1.23. Arrhenius plot for determining activation energy.

High reaction temperatures can cause numerical overflow problems in the computer calculation of k, owing to the very large values generated by the exponential term. This can often be eliminated by defining a value of the rate constant, ko, for some given temperature, To. Thus k g = Z e - E/RTO The logarithmic form of the Arrhenius relationship, is then

which permits the calculation of k at any temperature T. The above procedure is used in the simulation examples THERM and THERMPLOT.

55

1.3 Chemical Kinetics

1.3.3

Heats of Reaction

The heat of reaction, A H , is defined as

i= 1

i=l

where A H F i is the heat of formation of component i, A H c i is the heat of combustion of component i and v i is the stoichiometric coefficient for component i. If heats of formation are not available, heats of combustion can easily be determined from calorimetric heats of combustion data, as illustrated in Fig. 1.24. The resulting heat of reaction, AH, can be calculated and is negative for exothermic reactions and positive for endothermic reactions.

H

A 1

----

AHF,R

P I

AHF,P Elements '"C,R

AHC,P ustion products

Figure 1.24. Heat of reaction determination. AHF,R and A H F , ~are the heats of formation of reactants and products, respectively. AHc,R and AHc,p are the heats of combustion of reactants and products, respectively.

Standard heat data are usually compiled at 298 K, and to calculate the heat of reaction at an arbitrary temperature, the temperature dependency of enthalpies of reaction species have to be considered. These are generally dependent on temperature as follows T H i = Hi0

+

lcpi dT

To

where Hi and H i 0 are enthalpies of i at temperatures T and To, respectively, and cpi is the specific heat of component i. The temperature dependency of heat

56

1 Basic Concepts .

~

capacity is usually described by a quadratic polynomial cp = a

+

bT

+

cT2

where a, b, and c are empirical constants. For complex reaction systems, the heats of reaction of all individual reactions have to be estimated and the dynamic heat balance equations must include the heats of all the reactions.

1.3.4

Chemical Equilibrium and Temperature

Chemical equilibrium depends on temperature as described by the van't Hoff equation d(ln K) ~ dT -

AH_ R T2

_

Here K is the thermodynamic chemical equilibrium constant. If AH is constant, direct integration yields an explicit expression. If AH is a function of temperature, as described in Sec. 1.3.3, then its dependancy on cp can be easily included and integration is again straight forward. A calculation with varying AH and cp being functions of temperature is given in the simulation example REVTEMP.

1.3.5

Yield, Conversion and Selectivity

The fractional conversion of a given reactant, XA, is defined for a batch system as moles of A reacted XA = moles of A initially present giving

or

and at constant volume

57

1.3 Chemical Kinetics

where, nAO is the initial number of moles of A, nA is the number of moles of A at fractional conversion XA and (nA0 - nA) is the number of moles of A reacted. From this it follows for a constant-volume batch system that

For a well-mixed flow system at steady state, the fractional conversion XA is the ratio of the number of moles of A converted to the moles A fed to the system

where for equal volumetric flow rates at inlet and outlet (Fo = F1)

This definition is identical to that of the batch case. Fractional yield is defined by YCIA =

Moles of A transformed into a given product C Total moles of A reacted

Again it is important that both the particular reactant and product, concerned, should be stated, when defining a fractional yield. Another definition of fractional yield is based on the reaction rates

where A is the key reactant and C the product. Multiple reaction selectivity can be defined similarly as the ratio of the rate of formation of the desired product to the formation rate of an undesired product as in a parallel reaction A

where

zB -c

58

1 Basic Concepts ~~~

.

Here SB,Cis the selectivity of the desired product B to unwanted product in this parallel reaction.

1.4

Mass Transfer Theory

1.4.1

Stagewise and Differential Mass Transfer Contacting

Mass transfer separation processes, e.g., distillation, gas absorption, etc., are normally treated in terms of a stagewise procedure, in which concentration changes are taken to occur in distinct jumps, as, for example, between the neighbouring plates of a distillation column, or the concentrations are assumed to vary continuously throughout the total length or volume of the contactor, as in say, a packed gas absorption column. The two different types of operation lead to two quite distinct design approaches, namely a stagewise design and a differential design. Both can be handled to advantage by simulation methods. Fig. 1.25 shows a countercurrent stagewise mass transfer cascade and the resulting staged profile of the two streams, owing to the mass transfer between the streams. I

I

I

7-i

I

I

I

I I

I I

I I

I

I

I

I

I I I

I

t

I

I

I

I

I I I

I

I

I

I

I : I

I

I

I I I

I I

I I

I

I I I

Figure 1.25. Concentration profiles for countercurrent, stagewise contactor.

In the stagewise simulation method, the procedure is based on the assessment of the separation achieved by a given number of equilibrium contacting stages. The concept of the equilibrium stage is illustrated, for a particular stage n of the cascade, as shown in Fig. 1.26. According to this assumption the two streams are so well mixed that the compositions of each phase within the stage are

1.4 Mass Transfer Theory

59

Figure 1.26. Equilibrium mass-transfer stage.

uniform. Further, the mass transfer is so efficient that the compositions of the streams leaving the stage are in equilibrium.

The actual stage can be a mixing vessel, as in a mixer-settler used for solvent extraction applications, or a plate of a distillation or gas absorption column. In order to allow for non-ideal conditions in which the compositions of the two exit streams do not achieve full equilibrium, an actual number of stages can be related to the number of theoretical stages, via the use of a stage-efficiency factor. Fig. 1.27 shows a differential type of contactor as in, say a countercurrent flow packed gas absorption column, together with the resulting approximate continuous concentration profiles.

Figure 1.27.

Concentration profiles for countercurrent, differential contactor.

60

1 Basic Concepts -~

~

In this type of apparatus, the two phases do not come to equilibrium, at any point in the contactor and the simulation approach is based, therefore, not on a number of equilibrium stages, but rather on a consideration of the relative rates of transport of material through the contactor by flow and the rate of interfacial mass transfer between the phases. For this, a consideration of mass transfer rate theory becomes necessary.

1.4.2

Phase Equilibria

Knowledge of the phase equilibrium is essential for any mass transfer process, since this is, by definition, implicit in the idea of a theoretical stage, It is also important, however, in determining the concentration driving-force term in the mass-transfer rate expression. At phase equilibrium conditions, the driving force for mass transfer is Lero and therefore further concentration changes via a mass transfer mechanism become impossible. The equilibrium, is therefore also important in determining the maximum extent of the concentration change, possible by mass transfer. Equilibrium data correlations can be extremely complex, especially when related to non-ideal multicomponent mixtures, and in order to handle such real life complex simulations, a commercial dynamic simulator with access to a physical property data-base often becomes essential. The approach i n this text, is based, however, on the basic concepts of ideal behaviour, as expressed by Henry's law for gas absorption, the use of constant relative volatility values for distillation and constant distribution coeficients for solvent extraction. These have the advantage that they normally enable an explicit method of solution and avoid the more cumbersome iterative types of procedure, which would otherwise be required. Simulation examples in which more complex forms of equilibria are employed are STEAM and BUBBLE.

1.4.3

Interphase Mass Transfer

Actual concentration profiles (Fig. 1.28) in the very near vicinity of a mass transfer interface are complex, since they result from an interaction between the mass transfer process and the local hydrodynamic conditions, which change gradually from stagnant flow, close to the interface, to more turbulent flow within the bulk phases.

61

1.4 Mass Transfer Theory

Interface Phase-G

I I

Phase4

Figure 1.28. Concentration gradients at a gas-liquid interface.

According to the Whitman Two-Film theory, the actual concentration profiles, as shown in Fig. 1.28 are approximated for the steady state with no chemical reaction, by that of Fig. 1.29. Interface Phase-G

CG

I I I I

Phase-L

n

--

I I I I I I I

CL

I

‘G

‘L

Figure 1.29. Concentration gradients according to the Two-Film theory.

The above theory makes the following assumptions: 1 . A thin film of fluid exists on either side of the interface. 2. Each film is in stagnant or laminar flow, such that mass transfer across the films is by a process of molecular diffusion and can therefore be described by Fick‘s Law. 3 . There is zero resistance to mass transfer at the interface, itself, and therefore the concentrations at the interface are in local equilibrium. 4. Each of the bulk phases, outside the films, are in turbulent flow. Concentrations within the bulk phases are therefore uniform and the bulk phases constitute zero resistance to mass transfer. 5 . All the resistance to mass transfer, therefore occurs within the films.

62

1 Basic Concepts

Fick's Law states that the flux j (mol/s m2) for molecular diffusion, for any given component is given by

where D is the molecular diffusion coefficient (m2/s), and dC/dZ is the steadystate concentration gradient (mol/m3 m). Thus applying this concept to mass transfer across the two films

where DG and DL are the effective diffusivities of each film, and ZG and ZL are the respective thicknesses of the two films. The above equations can be expressed as

where, k c and kl, (m2/s) are the mass transfer coefficients for the G-phase and L-phase films, respectively. The total rate of mass transfer, Q (mol/s), is given by Q = j A = j (a V)

where "A" is the total interfacial area for mass transfer; "a" is defined as the specific area for mass transfer or interfacial area per volume (m2/m3). and V is the volume (m3). Thus

or in terms of a and V

Since the mass transfer coefficient, k, and the specific interfacial area, a, vary in a similar manner, dependent upon the hydrodynamic conditions and system physical properties, they are frequently combined and referred to as a "ka value" or more properly as a mass transfer coefficient. In the above theory, the interfacial concentrations CGi and CLi are not measurable directly and are therefore of relatively little immediate use. In order to overcome this apparent difficulty, overall mass transfer rate equations are defined by analogy to the film equations, These are based on overall

63

1.4 Mass Transfer Theory ..

coefficients of mass transfer, KG and KL, and overall concentration driving forces, where

Q = KG A (CG - CG*) = KL A (CL* - CL) and CG* and CL* are the respective equilibrium concentrations, corresponding to the bulk phase concentrations, CL and CG, respectively, as shown in Fig. 1.30.

CL

c,^

Figure 1.30. The bulk phase concentrations determine the equilibrium concentrations.

Simple algebra, based on a combination of the film and overall mass transfer rate equations, lead to the following equations, relating the respective overall mass transfer coefficients and the coefficients for the two films

and

where m is the local slope of the equilibrium line

For a non-linear equilibrium relationship, in which the slope of the equilibrium curve varies with concentration, the magnitudes of the overall mass transfer coefficients will also vary with concentration, even when the film coefficients themselves remain constant. The use of overall mass transfer coefficients in mass transfer rate equations should therefore be limited to the case of linear equilibria or to situations in which the mass transfer coefficient is known to be

64

1 Basic Concepts

relatively insensitive to concentration changes. Design equations based on the use of film mass transfer coefficients and film concentration driving forces make use of the identity that:

and that the interfacial concentrations CGi and CLi are in local equilibrium. Examples with mass transfer are OXIDAT, KLADYN and all examples in Sec. 5.8.

Chemical Engineering Dynami :Modelling

with PC Simulation

John Inph;m. Iruine J. Dunn. Elmar Heinzlc &JiiiE. Picnoul

copyright OVCH VerlagsgesellschaftmbH, 1994

2

Process Dynamics Fundamentals

2.1

Signal and Process Dynamics

2.1.1

Measurement and Process Response

The aim of dynamic simulation is to be able to relate the dynamic output response of a system to the form of the input disturbance, in such a way that an improved knowledge and understanding of the dynamic characteristics of the system are obtained. Fig. 2.1 depicts the relation of a process input disturbance to a process output response. Input disturbance

output

model or

Figure 2.1. Relation of process input to process output.

In testing process systems, standard input disturbances such as the unit-step change, unit pulse, unit impulse, unit ramp, sinusoidal, and various randomised changes can be employed. All the above changes are easily implementable in dynamic simulations, using ISIM and other digital simulation languages. The forms of response obtained differ in form, depending upon the system characteristics and can be demonstrated in the various ISIM simulation examples. The response characteristics of real systems are, however, more complex. In order to be able to explain such phenomena, it is necessary to first examine the responses of simple systems, using the concept of the simple, step-change disturbance.

66

2. Process Dynamics Fundamentals ~~~

._ -

~

~-

2 . 1 . 1 . 1 First-Order Response to an Input Step-Change Disturbance The simplest response of a linear system, is described mathematically by the following standard form of first-order differential equation

dy dt -

YO-J! 'I:

where y is the measured or process response value, t is time, and 'I: is the equation time constant. In its second form, the equation is often described as a first-order lag equation, in that the response, y, lags behind the input value yo, imposed on the system at time t = 0. For the step change condition, shown in Fig. 2.2, the initial conditions are given by y = 0, when t = 0. The solution to the differential equation, with the above boundary conditions, is given by

and is shown in Fig. 2.2

y=yo

- - - - - - - - -- - -

= o.63 yo

------------

,/---

Y y=o I I I I

t=O

I I I I

t = T

t

Figure 2.2. First-order exponential response to an imposed step-change disturbance.

67

2.1 Signal and Process Dynamics

Substituting the value t = z,gives y~ = yo(1 - e - l )

= 0.632 yo

Hence the value of the equation time constant, z,is simply determined as the time at which the response achieves sixty three per cent of its eventual steadystate value, when following a step change disturbance to the system.

2 . 1 . 1 . 2 Case A.

Concentration Response of a ContinuousFlow, Stirred Tank

Liquid flows through a tank of constant volume V, with volumetric flow rate F and feed concentration Co as shown in Fig. 2.3.

F

Figure 2.3. A continuous stirred-tank reactor.

Assuming well-mixed conditions, the component balance equation is given by

This can be expressed as dC1 _ _ _ _ _C_o_- C l dt -

z

where T is V/F. Note that the equation has the general form

68

2. Process Dynamics Fundamentals _.___.

Rate of change of

~.

~~

___

Input variable - Output Variable Processtime constant

( output variable ) = (

Thus the time constant for the process is equal to the mean residence or holdup time in the tank and has units of time, (volumeholumetric flow rate). Integrating with C1 = 0 when t = 0 gives

The response to a step change in feed concentration is thus given by = ~

( 1 e-t'T) = ~ o ( 1 -e - F t / V ) and follows the same form as that shown in Fig. 2.4. ~1

CO 0.98 Co

0.63 Co

I

I I

I I I

I I

0

I

I

t

47

Figure 2.4. Step response of a first-order system.

Note that when time, t, tends to infinity, the value C1 approaches Co, and when time, t, is numerically equal to T

Thus the response is 63.2% complete when the time passed is equal to the time constant, T. Further, for time t equal to four times the value of time constant 2, C1 = Co (1 - ec4) = 0.98 Co

then the response is 98% complete.

69

2.1 Signal and Process Dynamics

____

2.1.1.3

Case B. Concentration Response in a Continuous Stirred Tank with Chemical Reaction

Assuming a chemical reaction i n the tank, in which the rate of reaction is proportional to concentration, the component balance equation now becomes I V dC __ = F C O - F C l - k C 1 V dt

where k is the chemical rate coefficient ( U s ) . This can be rewritten as

and the system time constant now has the value

with high values of k acting to reduce the magnitude of The above equation becomes now

z.

Where C1, is the final steady-state value at t = 00, and is given by

Fig. 2.5 illustrates the effect on the process response of increasing values of k. This shows that the response time of the system is decreased by increasing values of k and that the final effluent concentration leaving the tank is reduced in magnitude. Increasing k has, however, very little influence on the initial rate of response. Note that for k = 0, with zero reaction, the final steady-state response is given by Co, and the response is identical to that of Case A.

70

2. Process Dynamics Fundamentals

c1= co

.-----I

I

I

I

C1 I I

I

I I

I

I

0

I

I

I

I

I

I I I I

ki > kq > k3 > k4 k4 = 0 I 1

2

3

T

4

Figure 2.5. Step response of a stirred tank with first-order chemical reaction ( V = l , z =1, k4=O; k3=0.2; k 2 d . 5 ; k l = l ) .

2 . 1 . 1 . 4 Case C. Response of a Temperature Measuring Element Instrument measurement response can often be important in the overall system response. The thermal response of a simple thermometer bulb, immersed in fluid, as shown in Fig. 2.6, is the result of a simple heat balance in which Rate of accumulation

( of heat by the bulb ) = (to

Rate of heat transfer the bulb from the fluid

dT M CP dt = U A (Ts - T) where, M is the mass of the thermometer bulb contents, cp is the specific heat capacity of the bulb contents, U is the film heat transfer coefficient between the fluid and bulb wall, A is the heat transfer surface area and Ts is the temperature of the surrounding fluid. Note that the stirred-tank or lumped-parameter approach has again been assumed in the modelling approach and that the temperature of the fluid within the bulb is assumed to be uniform. The simple balance equation can be reformulated as

71

2.1 Signal and Process Dynamics

Heat transfer from surroundings

Figure 2.6. Temperature measuring element.

showing that the measurement time constant is

z = Mc, UA Often an instrument response measurement can be fitted empirically to a firstorder lag model, especially if the pure instrument response to a step change disturbance has the general shape of a first-order exponential. As shown in Sec. 2.1.1. I, the time constant for the instrument is then given as the time at which 63 percent of the final response is achieved and the instrument response may be described by the simple relationship

where Tsystem, the actual temperature, and Tmeas, the measured temperature, are related by the measurement dynamics, as shown in Fig. 2.7, and ‘I: is the experimentally obtained instrument time constant.

Measurement dynamics

___)

Figure 2.7. Influence of instrument response on the measured temperature.

72

2. Process Dynamics Fundamentals

The ratio of the time constants, Zmeas and Zsystem, determines whether or not the system value is significantly different from the measurement value when conditions are changing, since the measured value will tend to lag behind that of the system. For a thermometer to react rapidly to changes in the surrounding temperature, the magnitude of the time constant should be small. This involves a high surface area to liquid mass ratio, a high heat transfer coefficient and a low specific heat capacity for the bulb liquid. With a large time constant, the instrument will respond slowly and may result in a dynamic measurement error.

2.1.1.5

Case D. Measurement Lag for Concentration in a Batch Reactor

The meaurement lag for concentration in a reactor is depicted in Fig. 2.8.

Measurement instrument

Batch reactor I

I

b I

I

Figure 2.8. Concentration measurement lag.

The actual reactant concentration, in the reactor at any time t is given by C r , but owing to the slow response of the measuring instrument, the measured concentration, shown by the instrument, C m , lags behind C,,as indicated in Fig. 2.9.

C

t

Figure 2.9. Measured (C,) and actual (C,) concentration responses.

2.1 Signal and Process Dynamics

73

The dynamic error existing between C, and C, depends on the relative magnitudes of the respective time constants. For the reactor, assuming a firstorder, constant volume reaction

giving a time constant for the reaction

Assuming that the instrument response is first order, then as shown in Sec. 2.1.1.1, the instrument time constant 7, is then given by the value of time at the 63% point (response to a step-change disturbance), where

which for this case equals (k ),z will The ratio of the time constants, T&, determine whether Cr is significantly different from Cm. When this ratio is less than 1.0 the measurement lag will be important. If Zr/Zrn > 10, then C, = C, and the measurement dynamics become unimportant. The effects of measurement dynamics are demonstrated in the simulation examples KLADYN, TEMPCONT and CONTUN.

2.1.2

Higher Order Responses

Actual response curves often follow a sigmoidal curve as shown in Fig. 2.10. This is characteristic of systems having a series of multiple lags and hence of systems which are characterised by several time constants.

2. Process Dynamics Fundarnentalc

74

__

~-

t=O

Figure 2.10. Higher order step response.

Examples of higher order response curves are shown by the following case studies.

2 . 1 . 2 . 1 Case A. Multiple Tanks in Series Consider the case of three, constant-volume tanks in series, as represented in Fig. 2.1 1, in which the tanks have differing volumes V1, V2, V3, respectively.. Assuming well-mixed tanks, the component balance equations are for tank 1

for tank 2

V z d g = FC1-FC2 for tank 3 V3d:j

= FC2-FC3

75

2.1 Signal and Process Dynamics

CO

Figure 2.11. Stirred tanks in series.

The above balance equations may be expressed as

and 73

dC3 + c3 = c2 x

or as

In this case, three time constants in series, 71, z2 and 23, determine the form of the final outlet response C3. As the number of tanks is increased, the response curve increasingly approximates the original, step-change, input signal, as shown in Fig. 2.12. The response curves for three stirred tanks in series, combined with chemical reaction are shown in the simulation example CSTR.

76

2. Process Dynamics Fundamentals ~~

c=co

-------

C

c=o

I

I I

0

I I

I I

1

2

I

I

I

I

3

t

4

Figure 2.12. Step response of n stirred tanks in series (n=l; 2; 5; 10; 20).

2 . 1 . 2 . 2 Case B. Response of a Second-Order Temperature Measuring Element The temperature response of the measurement element shown in Fig. 2.13 is strictly determined by four time constants, describing a) the response of the bulk liquid, b) the response of the thermometer pocket, c) the response of the heat conducting liquid between the wall of the bulb and the wall of the pocket and d) the response of the wall material of the actual thermometer bulb. The time constants c) and d) are usually very small and can be neglected. A realistic model should, however, take into account the thermal capacity of the pocket, which can sometimes be significant. Assuming the pocket to have a uniform temperature Tp, the heat balance for the bulb is now of accumulation (Rate of heat by the bulb ) = (to

Rate of heat transfer the bulb from the pocket

dT M cP dt = U1 A1 (Tp - T ) where, U1 is the heat transfer coefficient from the pocket to the bulb, and A1 is the heat transfer surface between the fluid and the bulb.

2.1 Signal and Process Dynamics

Figure 2.13. Thermometer enclosed within a pocket

Since the pocket temperature, Tp, is now a variable in the system, an additional heat balance equation is now required for the pocket. This is of the same form as the bulb, except that heat is now transferred both to the pocket from the surrounding and from the pocket to the bulb. Thus

[

Rate of accumulation of heat by the pocket

]

=

[

Rate of heat transfer to the pocket from the fluid

Rate of heat transfer from the pocket to the bulb

where U2 is the heat transfer coefficient between the fluid and the pocket, A2 is the heat transfer surface between the fluid and the pocket, Mp is the mass of the pocket, and cpp is the specific heat of the pocket. The overall instrument response is thus now determined by the relative magnitudes of the two major time constants, where for the liquid in the bulk,

78

2. Process Dynamics Fundamentals

for the pocket,

For accurate dynamic measurement of process temperature, both z2 and T I should be small compared with the time constant of the actual process.

2.1.3

Pure Time Delay

In contrast to the prior lag type response signals, time delays give no immediate response until the elapse of a given period of dead time or delay. In flow processes, this is equivalent to the time required for the system to pass through the signal in an otherwise unchanged state. An example would be the time taken to pump a sample from the process to a measuring instrument. In this case the magnitude of the time delay would simply be the time taken for the sample to travel along the pipe or the volume of the sample pipe divided by the sample flow rate, and thus equal to the mean residence time of the sample system,. As shown in Fig. 2.14, the input signal from the process is transmitted through the sample pipe until it arrives at the measuring instrument at a delay time t[,. In all other respects, however, the signal arriving at the measurement point is identical to the response of the actual system. Many simulation languages include a standard time delay function, which is pre-programmed into the language structure. This facility, however, is not available in ISIM, but a time delay can be programmed by defining the total delay in terms of a given multiple of the communication interval, CINT. The techniques of programming time delay functions are well described by Franks (1972), as shown in Fig. 2.15. An example of the use of a time delay function within an ISIM program is given in the simulation example DELAY.

79

2.1 Signal and Process Dynamics

~

_

_

.-

Process

device

-' I I

I

TimedelaytL

;I

I

I

t=O

t

t = tL

Figure 2.14. Schematic drawing of a process with immediate (dashed line) and time-delayed responses to a step change of an input signal.

In Fig. 2.15, the pointers are moved one element around the array at every communication interval CINT and thus with N elements separating the two pointers, a total time delay of N * CINT is achieved. Provision must be made to zero the array initially and also to reset the pointers back to 1 as each pointer passes through the final element of the array M.

Figure 2.15. Circular time delay function, Franks (1972).

80

2. Process Dynamics Fundamentals

~

~-

~~

~-

Alternatively as shown in Sec. 2.1.2. I , as more and more tanks are connected in series, the obtained response approximates more and more to that of plug flow. Hence a very simple but approximate presentation of a time delay is that of a cascade of tanks in series, as shown in Fig. 2.16.

1

n

2

N

Figure 2.16. Tanks-in-series, time-delay approximation.

In this procedure, it is necessary to write the balance equations for each tank, where for tank n

and the lag time for each tank,

z is given by z=

time delay N

and N is the equivalent number of tanks. As the number of tanks is increased, the better, of course, is the approximation obtained. A third possibility is to use Pad6 approximations. These are based on transfer function representations which are discussed in Sec. 2.1.4.

2.1.4

Transfer Function Representation

Complex systems can often be represented by linear time-dependent differential equations. These can conveniently be converted to algebraic form using Laplace transformation and have found use in the analysis of dynamic systems (e.g., Coughanowr and Koppel, 1965, Stephanopolous, 1984 and Luyben, 1990). Thus, as shown in Fig. 2.17, the input-output transfer-function relationship, G(s), is algebraic, whereas the time domain is governed by the differential equation.

81

2.1 Signal and Process Dynamics I

I

Process model (differential equation) I

I

Transfer function G(s) (algebraic equation)

Figure 2.17. Time and Laplace domain representations.

Complex models are often slow in execution owing to the large number of equations involved and the large range of time constants. Under these circumstances it is often useful to approximate the transient behaviour of the full model by a simpler model representation which is faster to compute. Such simplifications are commonly achieved by a combination of first-order lags and time delays and are often represented in Laplace transform form, especially when the sub-model is to be used as part of a control engineering application.

2 . 1 . 4 . 1 First-Order Lag Model Well-mixed tank systems (Fig. 2.18) are characterised by a first-order lag response.

Figure 2.18. Simple continuous flow tank system.

For the simple tank problem of Sec. 2.1.1.2 V

and

%

= FCo - FC1

82

2. Process Dynamics Fundamentals

-_______

where T is VIF. Taking transforms where the variables C1 and Co now represent deviations from initial steady state, gives (T s + I ) C , ( s ) = CO(S)

where s is the Laplace operator. Hence the transfer function for the above firstorder lag system is represented by

Inversion of the Laplace transform for a step change in Co gives the analytical solution derived previously. Thus, a general first-order lag element

is equivalent to T-dY + y =

dt

KX

or

Thus as shown previously in Sec. 2.1.1.1, if the step response curve has the general shape of an exponential, the response can be fitted to the above firstorder lag model by determining T at the 63% point. The response can now be used as part of a dynamical model, either in the time domain or in Laplace transfer form.

2 . 1 . 4 . 2 Pure Time Delay The transfer function representation of a time delay discussed in Sec. 2.1.3 is given by

83

2.1 Signal and Process Dynamics

This is identical to the first order Pad6 power series and gives a crude time delay approximation, when transformed back into differential equation form. Alternatively, writing

(e-tD S/2) e-tDS (etD Sl2)

and expanding in a Taylor series gives

As shown by Ord-Smith and Stephenson (1975) however, a better approximation is achieved using a fourth-order power series. As noted by them, the Pad6 approach works quite well with slowly varying inputs.

2 . 1 . 4 . 3 Higher Order Response Curves Many process response curves have an S-shaped form, as given in Fig. 2.19.

Time

Figure 2.19. Higher order response to a step change disturbance.

A first-order lag will not fit the response. However, a combination of first-order lags in series can be used, as shown in Fig. 2.20.

84

CO

--

2. Process Dynamics Fundamentals

71

C1

*

72

c2

T3

c3

As shown in Sec. 2.1.2.1, the response is modelled by the sequence For tank 1

tank 2

tank 3 _ dC3_ _(C2-C3) dt 23

In the Laplace domain, the system is represented as shown in Fig. 2.21.

Figure 2.21. Combination of first-order lags in series; Laplace Transform representation.

where

85

2.1 Signal and Process Dynamics

Since

where G(s) is the overall transfer function C3(S) _ __C0(s) - ( T s~

1

+ l ) ( q s + l ) ( q s + 1)

Usually, a 2nd or 3rd order system with equal values of T often provides a sufficient representation of the actual response. Experience has shown, that most chemical processes can often be modelled by a combination of several first-order lags in series and a time delay (Fig. 2.22).

z +

tD

C

Figure 2.22. Sigmoidal form of response, fitted by a combination of lags and time delay.

Laplace transformation of a simple time delay function with delay time tD gives L (f(c)) = e-tD Hence for a single first-order lag with time delay C(s) e-tD S G(s) = _ _ - CO(S) - z s + I Allowing a step input disturbance of magnitude Co, as shown in Fig. 2.23, the constants z and tD are derived by drawing a tangent to the step response curve at the point of inflexion.

86 -

2. Procew Dynamics Fundamental\

__

1 ; I

I

I

Figure 2.23. Fitting of measured response to a first-order lag plus time delay.

For a better fit of the system response, the method of Oldenbourg and Sartorius, as described in Douglas (1972), using a combination of two firstorder lags plus a time delay, can be used. The method is illustrated in Fig. 2.24. and applies for the case C(s> G ( s )= -~ Co(s) -

e-tD/S (71 s + 1) (72 s + 1)

Here values ZA and TB are determined from the measured response curve and are related to the system time constants 71 and 72, by the formulae

Dynamic models expressed in terms of transform functions can be solved by digital simulation by transposing the transfer function into an equivalent set of differential equations, as shown by Ord-Smith and Stephenson (1975) and by Matko et al. (1992). Also some languages include special transfer function subroutines.

87

2.1 Signal and Process Dynamics

Time Figure 2.24. Response curve characteristics according to Oldenbourg and Sartorius.

A complex transfer function may consist of processes in series, as shown in Fig. 2.25. This then gives the form

By block algebra the overall transfer function G(s) relating output Y(s) to input X(s) may be expressed in terms of a series of intermediate variables A(s), B(s), .. ., W(s).

Figure 2.25. Complex transfer function representation.

As shown in Fig. 2.25, the individual inputs and outputs are related by

88

2. Process Dynamics Fundamentals

Each individual transfer unit can be solved in turn to give the time dependent values of the variables a, b, c etc., eventually leading to the time variation in the output y for a given forcing value of input x, as shown in Fig. 2.26.

'

YO)

+

The values of the intermediate variables a, b, c, etc. have no fundamental significance, of course, and are used simply to link experimental process output x to the corresponding input y. Here the individual transfer functions can be 1. First-order lags as discussed in Sec. 2. I .4.1 2. A time delay as discussed in Sec. 2.1.3 and 2.1.4.2 3. First-order lead-lag elements of the form

Y(s) s +1) -- K ( T ~ - ( 3 s ) = X(s) - T 2 S + 1 ~~

The latter case is also easily solved, since, as shown by Ord-Smith and Stephenson ( I 975)

Y(s)

( ~ s2

+ 1)

= K X(S) (TI s + 1)

or

and

Simulation example TRANSIM is based on the solution of a complex transfer function.

89

2.2 Time Constants

2.2

Time Constants

As shown in the preceding sections, the magnitude of various process time constants can be used to characterise the rate of response of a process resulting from an input disturbance. A fast process is characterised by a small value of the time constant and a slow process by large time constants. Time constants can therefore be used to compare rates of change and thus also to compare the relative importance of differing rate processes. The term time constant is more or less equivalent to process time, characteristic time and relaxation time. Relaxation time is often used in physics, but is applied only to first-order processes and refers to the time for a process to reach a certain fraction of completion. This fraction is given by (1l/e) = 0.63, which for a first-order process, as shown previously, is reached at a time t = T. Time constants also may be used to describe higher order processes and also non-linear processes. In these cases the time constant is defined as the time in which the process proceeds to a specified fraction of the resultant steady state. Higher order processes are often more elegantly described by a series of time constants. A knowledge of the relative magnitude of the time constants involved in dynamic processes is often very useful in the analysis of a given problem, since this can be used to - Discover whether a change of regime occurs during scale up. - Reduce the complexity of mathematical models. - Determine whether the overall rate of a process is limited by a particular rate process, e.g., kinetic limitation or by diffusion, mixing, etc. - Check the controllability of a process. - Check the difficulty of numerical solution due to equation stiffness.

If the differing time constants for a chemical process are plotted as a function of the system variables, it can often be seen which rate process may be limiting. Many dimensionless groups can be considered as a ratio of the time constants for differing processes, and can give a clearer view on the physical meaning of the group. Such factors are discussed in much greater detail in other texts (Sweere et al., 1987), but here the intention is simply to draw attention to the importance of process time constants in the general field of dynamic simulation. As shown previously, the general form of equation serving to define time constants is as follows The rate of change of the variable

(

Final value - Instantaneous value Time constant

) = (--

However, a more general way to define time constants is

90

2. Process Dynamics Fundamentals

Time constant =

Capacity Rate

In this definition both "capacity" and "rate" have to be used in a rather general way. Some examples are presented in Table 2.1.

Table 2.1. Time constants defined by capacity and flow. Capacity Symbol

Rate Dimension Symbol

Dimension

L

m

V

m/S

V

m3

F

m31s

vc vc

Residence time T

kmol

V kLa C

kmolls

Mass transfer time

kmol

v rc

kmolls

Reaction time T,

Vpc,dT

J

U A dT

Jls

Heat transfer time

V p cp dT

J kmol

v rh

Jls

Heat production time Thp

A D C/L

kmolls

Diffusion time Td

vc

Time Constant Travelling time T Tmt

Zht

The symbols are defined in the Nomenclature. The choice of capacity is sometimes a problem, and may change according to the particular circumstance. Sometimes using a definition of time constant, based on the above equations, is not very helpful and other means must be employed. For example, mixing time is a very important time constant relating to liquid mixing, and this is best obtained directly from empirical correlations of experimental data.

2.2.1

Common Time Constants

2.2.1.1

Flow Phenomena

Some common time constants, relating to particular chemical engineering flow applications, are T =

Capacity Length - Travelling time - Velocity Rate

91

2.2 Time Constants

Capacity Volume - Residence time Rate - Volumetric flow -

T=---

Tcirc =

Circulation time

~ Various empirical equations are available for the circulation time constant, ~ in stirred vessels, columns, etc. Usually the value of the time constant, however, will represent a mean value, owing to the stochastic nature of flow. Mixing time constants, Tmix, are also available based on an empirical correlation and are usually closely related to the value of Tcirc (Joshi et al., 1982). A value of ~~i~ = 4 Tcirc is often used for stirred vessels and a value of Tmix = 2 to 4 Tcirc for columns. The exact value strongly depends on the degree of mixing obtained.

2 . 2 . 1 . 2 Diffusion-Dispersion Diffusion and dispersion processes can be characterised by a time constant for the process, given by TD =

~2 Capacity - Rate - D

where L is the characteristic diffusion or dispersion length and D is the diffusion or dispersion coefficient.

2 . 2 . 1 . 3 Chemical Reaction Chemical reaction rate processes can be described by time constants. In general Tr =

Capacity _ _V_C- _ C Rate - V r - r

where C is concentration and r is the reaction rate. Hence for a zero-order process, r = k Zr = C/k for a first-order process, r = kC Tr = l/k for a second-order process, r = kC2 zr = l/k C

i

~

92

2. Process Dynamics Fundamentals

2 . 2 . 1 . 4 Mass Transfer Transfer rate processes can also be characterized by time constants formulated as

tZ ,

=

Capacity Rate ~

VC I W a C = Ka

where Ka is the mass transfer capacity coefficient, with units (l/s). For a first-order process the time constant can be found from the defining differential equation as shown in Sec. 2.1.1.1. For the case of the aeration of a liquid, using a stirred tank, the following component balance equation applies

where CL is the concentration of oxygen in the liquid phase (kg/m3), t is time (s), kLa is the mass transfer coefficient (Us), VL is the liquid volume (m3), and CL* is the equilibrium dissolved oxygen concentration corresponding to the gas phase concentration, CG. By definition, the time constant for the process is thus l/kLa and the dissolved oxygen response to a step change in gas concentration is given by CL = CL"

( 1 - e-kLa t)

One has to be careful, however, in defining time constants. The first important step is to set up the correct equations appropriately. If the prime interest is not the accumulation of oxygen in the liquid as defined previously, but the depletion of oxygen from the gas bubbles, then the appropriate balance equation becomes

Note that in this case, the gas phase concentration, CG, relates to the total mixed gas phase volume VG, whereas the mass transfer capacity coefficient term is more conveniently related to the liquid volume, VL. If we consider the case where CL<
H VG The time constant is now given by -VL kLa'

93

2.2 Time Constants

Thus for the accumulation of oxygen in the liquid phase

and for the depletion of oxygen from the gas phase

thus representing a substantial order of difference in magnitude for the two time constants.

2 . 2 . 1 . 5 Heat Transfer Heat tranfer processs time constants are formulated as zht =

Capacity - M cl, T Mc, Rate - U A T = U A

U is the heat transfer coefficient, M the mass, cp the heat capacity and A the heat transfer area. A knowledge and understanding of the appropriate time constants is important in interpreting many of the simulation examples.

2.2.2

Application of Time Constants

Fig. 2.27 gives an example of a bubble column reactor with growing microorganisms which consume oxygen. Here the individual process time constants are plotted versus the operating variable, the superficial gas velocity. It can be seen that at low rates of oxygen consumption the time constants for mixing (z,i,) and the oxygen mass transfer rate (zmt) are lower in magnitude than z,, (the oxygen reaction rate time constant), if the superficial gas velocity (vs) in the column is sufficiently high. For higher reaction rates (e.g., 1-02 = kg/m3 s) and reasonable values of vs it is impossible to obtain a value of zxi, less than z, and therefore, mixing and mass transfer processes can become limiting at higher reaction rates.

94

2. Process Dynamics Fundamentals

Figure 2.27. Mixing, mass transfer and oxygen consumption in a bubble column bioreactor (Oosterhuis, 1984). 71. - reaction time constant, Zmt - mass transfer time constant, Tmix mixing time constant. r 0 2 - oxygen consumption rate, vs - superficial gas velocity.

2.3

Fundamentals of Automatic Control

Automatic process control involves the maintenance of a desired value of a measured or estimated quantity (controlled variable) within prescribed limits (deviations, errors), without the direct action of an operator. Generally, this involves three steps:

1 . Measuring the present value of the controlled variable. 2. Comparing the measurement with the desired value (set point). 3 . Adjusting some other variable (manipulated variable), which has influence on the controlled variable, until the set point is reached. The most important reasons for applying process control are as follows: Safety for personnel and equipment. Uniform and high quality products. - Increase of productivity. - Minimisation of environmental hazards. - Optimisation and decrease of labour costs.

-

-

95

2.3 Fundamentals of Automatic Control

Successful design of a process control system requires the following steps:

1. Selection of the control variables which are the most sensitive and easily measurable. 2. Formulation of the control objective; for example, the minimisation of some cost function. 3 . Analysis of the process dynamics. 4. Selection of the optimal control strategy. Process control is highly dynamic in nature, and its modelling leads usually to sets of differential equations which can be conveniently solved by digital simulation. A short introduction to the basic principles of process control, as employed in the simulation examples of Sec. 5.7, is presented.

2.3.1

Basic Feedback Control

The concept of an automatic control system is illustrated in Fig. 2.28, based on a temperature-controlled chemical reactor. Desired value

-in

or steam

+out

T

Products

Figure 2.28. Simple feedback temperature control system. M-motor with stirrer, PVpneumatic valve, TT - temperature measurement, TC - temperature controller.

96

2. Process Dynamics Fundamentals

The components of the basic feedback control loop, combining the process and the controller can be best understood using a generalised block diagram (Fig. 2.29). The information on the measured variable, temperature, taken from the system is used to manipulate the flow rate of the cooling water in order to keep the temperature at the desired constant value, or setpoint. This is illustrated by the simulation example TEMPCONT, Sec. 5.7.1. Manipulated variable

F

4

2.3.2

Process

Measured variable

Controller

Types of Controller Action

In the basic conventional feedback control strategy the value of the measured variable is compared with that for the desired value of that variable and if a difference exists, a controller output is generated to eliminate the error.

2.3.2.1

On/Off Control

The simplest and, despite its several drawbacks, the most widely used type of control is the on/off control system. An example is a contact thermometer, which closes or opens a heater circuit. The designation on/off means that the controller output, or the manipulated variable (electric current) is either fully on or completely off. To avoid oscillations around the setpoint, the real on/off controller has built into it, a small interval on either side of the setpoint, within which the controller does not respond, and which is called the differential gap or deadzone. When the controlled variable moves outside the deadzone, the manipulated variable is set either on or off. This is illustrated in Fig. 2.30. Such shifts from the set point are known as offset.

97

2.3 Fundamentals of Automatic Control

100

Manipulated variable 0

Control variable

Set point

Time

Figure 2.30. On/off controller with differential gap or dead zone.

The oscillatory nature of the action and the offset make the resulting control rather imperfect, but the use of on/off control can be justified by its simplicity and low price, and the reasonable control obtained, especially for systems which respond slowly.

2.3.2.2

Proportional-Integral-Derivative

(PID) Control

Three principal functional control modes are proportional (P), integral (I) and derivative (D) control. These are performed by the ideal three-mode controller (PID), described by the equation

Controller modes:

P

I

D

where,

- Po is the controller output for zero error. - K, is the proportional gain. - &(t) is the error or deviation of actual from desired - TI is the integral time or reset time constant. - TD is the derivative time constant.

value.

98

2. Process Dynamics Fundamentals

The response of a controller to an error depends on its mode. In the proportional mode (P), the output signal is proportional to the detected error, E . Systems with proportional control often exhibit pronounced oscillations, and for sustained changes in load, the controlled variable attains a new equilibrium or steady-state position. The difference between this point and the set point is the offset. Proportional control always results in either an oscillatory behaviour or retains a constant offset error. Integral mode controller (I) output is proportional to the sum of the error over the time. It can be seen that the corrections or adjustments are proportional to the integral of the error and not to the instantaneous value of the error. Moreover, the corrections continue until the error is brought to zero. However, the response of integral mode is slow and therefore is usually used in combination with other modes. Derivative mode (D) output is proportional to the rate of change of the input error, as can be seen from the three-mode equation. In industrial practice it is common to combine all three modes. The action is proportional to the error (P) and its change (D) and it continues if residual error is present (I). This combination gives the best control using conventional feedback equipment. It retains the specific advantages of all three modes: proportional correction (P), offset elimination (I) and stabilising, quick-acting character, especially suitable to overcome lag presence (D). Simple control strategies form an integral part of many of the simulation examples, including RELUY, COOL, DEACT, REFRIG, RUN, and COLCON and the special control examples in Sec. 5.7, TEMPCONT, TWOTANK and CONTUN. Fig. 2.31 depicts the responses of the various control modes and their combinations to step and ramp inputs.

99

2.3 Fundamentals of Automatic Control

t Error signal

Step change

I

/ Positive ramp

Controller Response

Figure 2.31. Response of the most common controller modes for step change and ramp function of the error signal.

The performance of different feedback control modes can be seen in Fig. 2.32. The selection of the best mode of control depends largely on the process characteristics. Further information can be found in the recommended texts listed in the reference section. Simulation methods are often used for testing control methods. The basic PI controller equations are easily programmed using a simulation language, as shown in the example programs. In the simulation examples, the general PTD equation is simplified and only the P or P and I terms are used. Note that the I term can be set very low by using a high value for TI.

100

2. Process Dynamics Fundamentals -~

.-

------

,,-' Uncontrolled ,'

response

I' I

t Figure 2.32. Response of controlled variable to a step change in error using different control modes.

If desired, the differential term, d&/dt,can be programmed as follows Since, E = Y - Yset then

de dy d t = dt

and this derivative can be obtained directly from the model equations.

2 . 3 . 2 . 3 Case A. Operation of a Proportional Temperature Controller Liquid flows continuously through a tank of volume, V, provided with an electric heater. A controller regulates the rate of heating directly in accordance with the difference between a required set point temperature, TSet,and the actual temperature, T I , as shown in Fig. 2.33. The heat balance equation for the tank is similar to that of Case A of Sec. 1.2.5.1, i.e.,

101

2.3 Fundamentals of Automatic Control

1

u

'U

I

1

II

F1 TI I

I I I I I

Figure 2.33. Temperature control, TC, of a continuously operated, stirred tank with an electric heater (Q).

but where Q is the rate of electric heating and is expressed by a proportional control equation (omitting the I and D terms)

2.3.3

Controller Tuning

The purpose of controller tuning is to choose the correct controller constants to obtain the desired performance characteristics. This usually means that the control variables should be restored in an optimal way to acceptable values, following either a change in the set point or the appearance of an input disturbance. Simulation examples TEMPCONT and CONTUN, provide exercises for controller tuning using the methods explained below.

2.3.3.1

Trial and Error Method

Controllers can be adjusted by changing the values of gain Kp, reset time TI and derivative time ZD. The controller can be set by trial and error by experimenting, either on the real system or by simulation. Each time a disturbance is made the response is noted. The following procedure may be used to test the control with small set point or load changes:

I02

2. Process Dynamics Fundamentals ~~

I. 2. 3. 4.

5.

-__

~~

Starting with a small value, K, can be increased until the response is unstable and oscillatory. This value is called the ultimate gain K,o. K, is then reduced by about one half. Integral action is brought in with high TI values. These are reduced by factors of 2 until the response is oscillatory, and TI is set at 2 times this value. Include derivative action and increase ZD until noise develops. Set TD at 1/2 this value. Increase K, in small steps to achieve the best results.

2.3.3.2

Ziegler-Nichols Method

This method is an empirical open-loop tuning technique, obtained by uncoupling the controller. It is based on the characteristic curve of the process response to a step change in manipulated variable of magnitude A. The response, of magnitude B, is called the process reaction curve. The two parameters important for this method are given by the slope through the inflection point normalised by A so that S=Slope/A, and by its intersection with the time axis (lag time TL), as determined graphically in Fig. 2.34. The actual tuning relations, based on empirical criteria for the "best" closed-loop response are given in Table 2.2. B

0 4-TL-b

Time

Figure 2.34. Process reaction curve for the Ziegler-Nichols Method.

103

2.3 Fundamentals of Automatic Control

2.3.3.3

Cohen-Coon Controller Settings

Cohen and Coon observed that the response of most uncontrolled (controller disconnected) processes to a step change in the manipulated variable is a sigmoidally shaped curve. This can be modelled approximately by a firstorder system with time lag TL, as given by the intersection of the tangent through the inflection point with the time axis (Fig. 2.34). The theoretical values of the controller settings obtained by the analysis of this system are summarised in Table 2.2. The model parameters for a step change A to be used with this table are calculated as follows

K = B/A

z = B/S

where B is the extent of response, S is the slope at the inflection point, and TL is the lag time as determined in Fig. 2.34.

2.3.3.4

Ultimate Gain Method

The previous transient-response tuning methods are sensitive to disturbances because they rely on open-loop experiments. Several closed loop methods have been developed to eliminate this drawback. One of these is the empirical tuning method, ultimate gain or continuous cycling method. The ultimate gain, K,o, is the gain which brings the system with sole proportional control only to sustained oscillation (stability limits) with frequency fpo, where l/fpo is called the ultimate period. This is determined experimentally by increasing K, from low values in small increments until continuous cycling begins. The controller settings are then calculated from K,o and fpo according the tuning rules given in Table 2.2. While this method is very simple it can be quite time consuming in terms of number of trials required and especially when the process dynamics are slow. In addition, it may be hazardous to experimentally force the system into unstable operation.

104

2. Process Dynamics Fundamentals -

-

Table 2.2. Controller settings based on process responses. Ziegler-Nichols --

~

Controller

KP

_.

-

~~

P PI PID Cohen-Coon

_

_

~

-

Controller

71

ZD

3.33T~ 2TL

T~l2

-

KP

-

21

2D

TI

ZD

P

PI PID Ultimate Gain -

Controller

P PI PID

2.3.3.5

.~

KP 0.5 Kpo

0.45 K,o 0.6 K,o

fo/1.2 fo12

Time Integral Criteria

Several criteria may be used to estimate the quality of control (Stephanopoulos, 1984). One of these is the integral of the time-weighted absolute error (ITAEj, where

105

2.3 Fundamentals of Automatic Control

t

ITAE =

I It E(t)l dt

0

Integral error criteria are ideally suited to simulation applications since only one additional program statement is required for the simulation. The optimal control parameters K,, TI and TD can be then found at minimal ITAE. For this, it is useful to be able to apply the available optimisation tools implemented in such programs as MATLAB, SIMUSOLV and ESL.

2.3.4

Advanced Control Strategies

2.3.4.1

Cascade Control

In control situations with more then one measured variable but only one manipulated variable, it is advantageous to use control loops for each measured variable in a master-slave relationship. In this, the output of the primary controller is usually used as a set point for the slave or secondary loop. An example of cascade control could be based on the simulation example DEACT and this is shown in Fig. 2.35. The problem involves a loop reactor with a deactivating catalyst, and a control strategy is needed to keep the product concentration Cp constant. This could be done by manipulating the feed rate into the system to control the product concentration at a desired level, Cset. In this cascade control, the first controller establishes the setpoint for flow rate. The second controller uses a measurement of flow rate to establish the valve position. This control procedure would then counteract the influence of decreasing catalyst activity.

2 . 3 . 4 . 2 Feed-Forward Control Feedback control can never be perfect as it only reacts to the disturbances which are already measured in the system output. The feed-forward method tries to eliminate this drawback by an alternative approach. Instead of using the process output, the measured variable is taken as the measured inlet disturbances and its effect on the process is anticipated via the use of a model. The action is taken on the manipulated variable using the model to relate the measured variable at the inlet, the manipulated variable and the process output. The success of this control strategy depends largely on the accuracy of the model prediction, which is often imperfect as models can rarely predict the

106

2. Process Dynamics Fundamentals

A-P

Figure 2.35. Cascade control to maintain product concentration by manipulating the reactant concentration in the feed.

effects of all possible process disturbances. For this reason, an additional feedback loop is often used as a backup or to trim the main feed-forward action, as shown in Fig. 2.36. Many of the continuous process simulation examples may be altered to simulate feed-forward control situations.

4 h- A

Disturbance

Controller

Measured variable

Process

b

b

Setpoint Controller

Figure 2.36. Feed-forward control with additional feedback loop.

107

2.3 Fundamentals of Automatic Control

2 . 3 . 4 . 3 Adaptive Control An adaptive control system can automatically modify its behaviour according to the changes in the system dynamics and disturbances. They are applied especially to systems with non-linear and unsteady characteristics. There are a number of actual adaptive control systems. Programmed or scheduled adaptive control uses an auxiliary measured variable to identify different process phases for which the control parameters can be either programmed or scheduled. The "best" values of these parameters for each process state must be known a priori. Sometimes adaptive controllers are used to optimise two or more process outputs, by measuring the outputs and fitting the data with empirical functions.

2.3.4.4

Sampled Data or Discrete Control Systems

When discontinuous measurements are involved, the control system is referred to as a sampled data or discrete controller. Concentration measurements by chromatography would represent such a case. Here a special consideration must be given to the sampling interval AT (Fig. 2.37). In general, the sampling time will be short enough if the sampling frequency is 2 times the highest frequency of interest or if AT is 0.5 times the minimum period of oscillation. If the sampling time satisfies this criterion, the system will behave as if it were continuous. Details of this and other advanced topics are given in specialised process control textbooks, some of which are listed in the references.

AT+-

Figure 2.37.

Sampled control strategy.

time

108

2. Process Dynamics Fundamentals

2.4

Numerical Aspects of Dynamic Behaviour

2.4.1

Optimisation

Optimisation may be used, for example, to minimise the cost of reactor operation or to maximise conversion. Having set up a mathematical model of a reactor system, it is only necessary to define a cost or profit functionOptimisation and then to minimise or maximise this by variation of the operational parameters, such as temperature, feed flow rate or coolant flow rate. The extremum can then be found either manually by trial and error or by the use of a numerical optimisation algorithms. The first method is easily applied with ISIM, or with any other simulation software, if only one operational parameter is allowed to vary at any one time. If two or more parameters are to be optimised this method however becomes extremely cumbersome. Basically two search procedures for non-linear parameter estimation applications apply. (Nash and Walker-Smith, 1987). The first of these is derived from Newton's gradient method and numerous improvements on this method have been developed. The second method uses direct search techniques, one of which, the Nelder-Mead search algorithm, is derived from a simplex-like approach. Many of these methods are part of important mathematical computer-based program packages (e.g., IMSL, BMDP, MATLAB) or are available through other important mathematical program packages (c.g., IMSL).

2 . 4 . 1 . 1 Case A. Optimal Cooling for a Reactor with an Exothermic Reversible Reaction A reversible exothermic reaction A 2 B is carried out in a stirred-tank reactor with cooling. The details of the model are given in the simulation example REVTEMP. Specific heats are functions of temperature. The temperature dependancy of the reaction rate constants are given by the Arrhenius equation, that of the equilibrium constant by the van't Hoff equation. Adiabatic operation restricts conversion because of an unfavourable equilibrium at high temperature. Early cooling favours the equilibrium conversion but reduces reaction rates, according to the Arrhenius equation. It is assumed that the cooling water temperature is constant, and that the cooling water flow, FC, may be either on or off. At time TIMEON the cooling water flow is set to FCON. A profit function is defined as

109

2.4 Numerical Aspects of Dynamic Behaviour

SPTYB = CtB2 This reflects the desire to have high conversion in a short time period. SPTYB always passes through a maximum during a batch run. The problem is defined as finding the optimal times to switch on the cooling water flow (TIMEON) and to harvest the tank contents (TFIN). The model equations are given below in the notation used by SIMUSOLV. The model is identical with the simulation example REVTEMP. The case study here, also provides an opportunity to see how easily ISIM programs may be converted to other simulation software. PROGRAM

' '

' ' '

' '

REVTEMP

CONTINUOUS STIRRED TANK REACTOR ' EQUILIBRIUM REACTION AND JACKET COOLING A <=> B' HEAT AND TEMPERATURE EFFECTS WITH' CP = F(T) ' REACTION ENTHALPY = F (T)' EQUILIBRIUM = F(T)'

initial CONSTANT TST = 298 'Heats of formation' CONSTANT HFAST = -160 T=298 K' CONSTANT HFBST = -240 'Reaction enthalpy' DHRST = HFBST-HFAST 'Equilibrium constant: CONSTANT KEST = 1000 'Heat capacities: cp CONSTANT AA = 0.100 CONSTANT AB = 0.080 CONSTANT BA = 0.0006 CONSTANT BB = 0.0004 DA = AB-AA DB = BB-BA

=

'

$'K' $'kJ MOL-1 Standard: $'kJ MOL-1' Standard a

+

b*T'

'Activation energy' CONSTANT EA = 120 CONSTANT R = 8.3143-3 CONSTANT KST = 0.001 'K1 = KA1 * EXP[-EA/(R*T1)1'

$'kJ MOL-1,' 298 K'

$'kJ $'kJ $'kJ $'kJ

MOL-1 MOL-1 MOL-1 MOL-1

K-1' K-1' K-2' K-2'

$'kJ MOL-1' $'kJ MOL-1 K-1' $'S-1'

110 KA1

2. Process Dynamics Fundamentals __

=

__

~~

KST/EXP(-EA/R/TST)

'Reactor' CONSTANT F = 0.0 CONSTANT V = 2 CONSTANT CAO = 25e3 CONSTANT CBO = 0.0 CONSTANT TI0 = 290 CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT

CPC FCON FCO VC TCO RHOC UA TIMEON TFIN

'Initial CONSTANT CONSTANT CONSTANT CONSTANT end

conditions' CAinit = 20e3 CBinit = 0 Tlinit = 305 TCinit = 305

= 4 = 0.1

= 0.0 = 0.1 = 290 = 1000 = 2000 = 1000 = 50

$'m3 sec-1'

$ 'm3 '

$'mol m3' $'rnol m-3'

$1,'

$'kJ $Im3 $'m3 $ Im3 $IK' $'kg $'kJ

kg-1 K-1'

8-11

8-11 m-3' K-1 8-1'

$'mol m-3' $IKI $ I of initial I

DYNAMIC DERIVATIVE K1 = KA~*EXP(-EA/(R*T~)) DHDA = DHRST - DA*TST - DB/2*TST**2 LNKE = ALOG(KEST) - DHDA/R*(l/Tl-I/TST) + DA/R*ALOG(Tl/TST) + DB/R*(Tl-TST) KE = EXP(LNKE) RA = -Kl*(CA-CB/KE) 'Mass balance' dCA = F/V*(CAO-CA) + RA dCB = F/V*(CBO-CB) - RA CA = integ(dCA,CAinit) CB = integ(dCB,CBinit) 'Heat CPA = CPB = SUMCP FEEDCP

-

~~

balance reactor' AA + BA*T1 AB + BB*T1 = V*(CA*CPA + CB*CPB) = F*CAO*(AA*(Tl-Tlo)

+ BA/2*(T1**2

...

-

T10**2))

111

2.4 Numerical Aspects of Dynamic Behaviour

DHRTl = DHRST + DA*(Tl-TST) DT1 = (-FEEDCP + V*RA*DHRTl T1 = INTEG(DT1,TlINIT) 'Heat dTC = TC = FC = THELP

balance

+

+

DB/2*(T1**2 - TST**2) UA*(TC-Tl))/SUMCP

cooler-jacket'

FC/VC*(TCO-TC)-UA*(TC-T1)/V/RHOC/CPC

INTEG(dTC,TCinit) RSW(T.GE.TIMEON,FCON,O.O) = RSW(T.GT.lE-20,T,lE20) $ 'prevent division by SPTYB = RSW(THELP.LT.lElO,CB**2/THELP,O.o) $'BATCH CASE WEIGHTED SPTYC = F*CB**2/V $'CONTINUOUS CASE' END $'DERIVATIVE' terrnt(t.ge.tfin) END $'of dynamic' END $ ' of program'

zero

I

YIELD'

The optimization in SIMUSOLV requires three self-explanatory statements >VARY TIMEON TFIN >MAXIMIZE SPTYB >OPTIMIZE The Nelder-Mead search technique was used to find the maximum, which was found after 70 complete integrations. The computer time needed was 21 s on a Vax Workstation 4100. The maximal value of SPTYB was 9.23 lo6 and was found at TIMEON = 8.97 and TFIN = 32.72. The time course of the various variables can be created by running the simulation example REVTEMP using the above value of TIMEON and a slightly larger value of TFIN, so that the maximum in SPTYB can be seen. In non-linear systems one can usually not predict a priori whether the optimum found is global or only local. A good judgement on the behaviour of the model can be seen in the contour and threedimensional plots of Fig. 2.38.

112

2. Process Dynamics Fundamentals Contour P l u t S P T Y B S C A L E 0 B Y 10

S u r f a c e Plot

8

TIMEUN

ID

I2

14

Figure 2.38a. Three-dimensional plot o f

Figure 2.38b. Contour plot of the results

the REVTEMP optimisation.

in Fig. 2.38a.

2.4.2

Parameter Estimation

Having set up a model to describe the dynamics of the system, a very important first step is to compare the numerical solution of the model with any experimental results or observations. In the first stages, this comparison might be simply a check on the qualitative behaviour of a reactor model as compared to experiment. Such questions might be answered as: Does the model confirm the experimentally found observations that product selectivity increases with temperature and that increasing flow rate decreases the reaction conversion? Following the first preliminary comparison, a next step could be to find a set of parameters, that give the best or optimal fit to the experimental data. This can be done by a manual, trial-and-error procedure or by using a more sophisticated mathematical technique which is aimed at finding those values for the system parameters that minimise the difference between values given by the model and those obtained by experiment. Such techniques are general, but are illustrated here with special reference to the dynamic behaviour of chemical reactors. Table 2.3 is used to classify the differing systems of equations, encountered in chemical reactor applications and the normal method of parameter identification. As shown, the optimal values of the system parameters can be estimated using a suitable error criterion, such as the methods of least squares, maximum likelihood or probability density function.

113

2.4 Numerical Aspects of Dynamic Behaviour

Table 2.3. Classification of systems of reactor equations with a set of parameters and time-dependent variables. ~~

Examples of linear systems

Examples of non-linear systems

Algebraic equations

Steady state of CSTR with first-order kinetics. Algebraic solution and optimisation (least squares, Draper and Smith, 1981).

Steady state of CSTR with complex kinetics. Numerical solution and optimisation (least squares or likelihood function).

Differential equations

Batch reactor with first-order kinetics. Analytical or numerical solution with analytical or numerical parameter optimisation (least squares or likelihood).

Batch reactor with complex kinetics. Numerical integration and parameter optimisation (least squares or likelihood).

2.4.2.1

Non-Linear Systems Parameter Estimation

The methods concerned with differential equation parameter estimation are, of course, the ones of most concern in this book. Generally reactor models are non-linear in their parameters and therefore we are concerned mostly with nonlinear systems. Given a model in the form of a set of differential equations, = f(k1 ... kn, y)

A model described by this differential equation is linear in the parameters

kl ... kn, if

but is non-linear, if for at least one of the parameters kj

114

2. Process Dynamics Fundamentals

The application of optimisation techniques for parameter estimation requires a useful statistical criterion (e.g., least-squares). A very important criterion in non-linear parameter estimation is the likelihood or probability density function. This can be combined with an error model which allows the errors to be a function of the measured value. A simple but flexible and useful error model is used in SIMUSOLV (Steiner et al., 1986; Burt, 1989). If basic assumptions concerning the error structure are incorrect (e.g., nonGaussian distribution) or cannot be specified, more robust estimation techniques may be necessary. In addition to the above considerations, it is often important to introduce constraints on the estimated parameters (e.g., the parameters can only be positive). Such constraints are included in the simulation and parameter estimation package SIMUSOLV. Because of numerical inaccuracy, scaling of parameters and data may be necessary if the numerical values are of greatly differing order. Plots of the residuals, difference between model and measurement value, are very useful in identifying systematic or model errors. Non-linear parameter estimation is far from a trivial task, even though it is greatly simplified by the availability of user-friendly program packages such as a) SIMUSOLV (Steiner et al., 1986), b) ESL, c) a set of BASIC programs (supplied with the book of Nash and Walker-Smith, 1987) or d) by mathematical software (MATLAB). ISIM itself does not supply these advanced features, but ISIM programs can easily be translated into other more powerful languages.

2 . 4 . 2 . 2 Sensitivity Analysis Sensitivity analysis is a very important tool in analysing the relative importance of the model parameters and in the design of experiments for their optimal determination. In many cases, it is be found that a model may be rather insensitive to a particular parameter value in the region of main interest, and then the parameter obviously does not need to be determined very accurately. Model parameters are usually determined from experimental data. In doing this, sensitivity analysis is valuable in identifying the experimental conditions that are best for the estimation of a particular model parameter. In advanced software packages for parameter estimation, such as SIMUSOLV, sensitivity analysis is provided. The resulting iterative procedure for determining model parameter values is shown in Fig. 2.39.

115

2.4 Numerical Aspects of Dynamic Behaviour

mathematical model

r

r

Compare qualitative behaviour of the model with experiments

Revise model

1

t

No

Perform new experiments

Estimate

Calculate

Enough data

sensitivity

Figure 2.39. Iterative procedure for parameter estimation, sensitivity analysis and experimentation.

2. Process Dynamics Fundamentals

2.4.3

Case B. Estimation of Rate and Equilibrium Constants in a Reversible Esterification Reaction Using ESL and SIMUSOLV

The objective is to demonstrate the power of modern simulation packages in the estimation of model parameters. Here the parameters are estimated using SIMUSOLV running on VAX systems and on a PC using the ESL simulation package, the main features of which can be found on the last page of this book. Ethanol and acetic acid react reversibly to ethyl acetate, using a catalyst, ethyl hydrogensulfate, which is prepared by reaction between sulfuric acid and ethanol. Acetic acid

+

Ethanol A + B

2 kl

2

Ethyl acetate

+

water

C + D

k2 The rate of batch reaction for reactant A (acetic acid) is modelled as

The progress of the reaction is followed by taking samples at regular time intervals and titrating the remaining free acid with alkali (mL). Originally the parameter estimation was done using SIMUSOLV, running on a VAX or workstation. ESL has a somewhat simpler optimisation routine, but has the advantage that both PC's and larger computers can be used. Both programs are shown below and show the great similarity in the model formulation. Although some details are different, the structure is much the same as ISIM. In ESL, the initial conditions for concentrations A, B, C, and D are followed by the table of measured data, time (min.) versus titrated volume (mL). In the DYNAMIC section, the program statements are identical to those of the model equations. Here, a function generator, FGEN1, is used to describe the tabular data. The objective function to be minimised is defined as follows Objective function =

(model solution

This least-squares method is a standard technique.

-

actual data)2

117

2.4 Numerical Aspects of Dynamic Behaviour

At the end of the program is a section for EXPERIMENT, where initial guesses of the parameters are given. The next to last line of the program gives the instructions to OPTIMIZE the values of k l and k2. ESL

Esterification

Equilibrium

Optimization

Program

study - - ESTERIFICATION include "integ"; include "fgenl"; PACKAGE statistics; integer: count/O/; end statistics; model ester(rea1: object-function : = real: kl,k2); use statistics; constant real: a0/5.636/, - - mol/ 1 b0/5.636/, - - rnol/l c o / o .o/, - - mol/l d0/0.364/, - - mol/l m10/16.25/; - - initial ml 1 N NaOH for titr. real: a/aO/, - - mol/l b, c, d,ml, afakt, actual-solution, Table (2,201 / 0 .o, 16.25, 2.0, 15.1, 4.0, 14.25, 6.0, 13.8, 8.0, 12.9, 10.0, 11.7, --measurement error 15.0, 12.1, 20.0, 11.05, 25.0, 10.6, 30.0, 10.45, 35.0, 10.2, 40.0, 9.5, 45.0, 9.5, 55.0, 8.9, 70.0, 8.5, 103.0, 7.5, 143.0, 7.1, 1000.0, 6.5, 1043.0, 6.5,

I18

2 . Process Dynamics Fundamentals

~___-

initial dynamic

___-

-~

~ ~

1100.0,

___

6.5/;

afakt:=aO/mlO;

a':= -kl*a*b + k2*c*d; ml:= a/afakt; b:= a; c:= aO-a; d:= dO+aO-a; actual-solution:=fgenl(Table,t);

object-function:=integ(O.O, (ml-actual-solution)**2);

communication prepare t ,ml; plot t ,ml,0,tf in,5 ,17.5; terminal count:=count+l; print it era t iont1 ,count , ,K 1 = ,kl , ,K 2 = ,k2, ,A= ,a, OF=",object-function; end ester; II

'I,

It

It

'I

II

I'

'I,

- - experiment real: of,k1/0.003/,k2/0.001/; integer:i; tfin:=1200.0; cint:= tf in/50; op-stp: =o. 5; - - Optimize simplex size parameter OPTIMIZE ester(of:=kl,kZ); end-s tudy The optimisation converged after 50 iterations, and gave K l = 0.00328, K 2 = 0.00149. The graph of Fig. 2.40 shows the optimised simulation result and measured data.

119

2.4 Numerical Aspects of Dynamic Behaviour ESL

11.50

35 00

12.50

10.00

7.50

5.00

T

Figure 2.40. Parameter estimation example using ESL to fit kinetic data to a model. The smooth curve with squares is the fitted model and the irregular curve is the data.

The SIMUSOLV listing is again very similar to HIM, and starts with the INITIAL section for definition of constants followed by the DYNAMIC region containing the model equations. PROGRAMM ESTERIFICATION IN1TIAL CONSTANT AO=5.636 $'mol / 1' CONSTANT BO=5.636 $'mol / 1' CONSTANT co=o.0 $'mol / 1' CONSTANT DO=O. 364 $'mol / 1' CONSTANT m10=6 $'initial ml 1 N NaOH f o r titration' K1=0.003 $ ' ~ / m o lmin' CONSTANT K2=0.001 $ ' ~ / m o l min' CONSTANT CONSTANT TSTOP=1200. $ 'min ' VARIABLE TIME=O .O CINT=TSTOP/100. AFAKT=AO/mlO END DYNAMIC DERIVATIVE DADT = -KI*A*B + K2*C*D C = AO-A B = A

120

2. Process Dynamics Fundamentals -______.

D = DO+AO-A ml=A/AFAKT A=INTEG(DADT,AO) END TERMT (TIME.GE.TSTOP) CONV=100.*(AO-A)/AO

END TERM INAL K=Kl/KZ END END

The experimental data are in a separate data file together with the PREPARE statement and the initial titration value. Here the data point at T=10 was assumed to be erroneous because of mishandling and was not included into the parameter estimation. PREPARE TIME,A,B,C,D,ML,CONV set m10=16.25 'measured data: ml 1N NaOH in DATA TIME ML 0.0 16.25 2 15.1 4 14.25 6 13.8 8 12.9 10 11.7* 15 12.1 20 11.05 25 10.6 30 10.45 40 9.5 45 9.5 55 8.9 70 8.5 103 7.7 143 7.1 1000 6.5 1043 6.5 1100 6.5 END

titration'

121

2.4 Numerical Aspects of Dynamic Behaviour

The parameter estimation was started with the three commands

>VARY K1, K2 >FIT ML >OPTIMIZE With the Nelder-Mead search, 58 full integrations and 0.6 s computer time on a VAX 6000 computer were needed to get the following result with initial and final values, standard deviation, residual plot and statistical information. DESCRIPTION

---------__

LOG LlXELIHOOD FUNCTION K1 K2 TIME

0.000 2.000 4.000 6.000 8.000 10.00 15.00 20.00 25.00 30.00 40.00 45.00 55.00 70.00 103.0 143.0 1000. 1043. 1100.

ML OBSERVED

16.25 15.10 14.25 13.80 12.90 11.70 12.10 11.05 10.60 10.45 9.500 9.500 8.900 8.500 7.700 7.100 6.500 6.500 6.500

PARAMETER ESTIMATES

STANDARD DEVIAT T ON

---__-_______---___

INITIAL FINAL -17.137 -11.709 3.00000E-03 3.48640E-03 1.798E-04 1.00000E-03 1.55786E-03 1.580E-04

ML PREDlCTED

% ERROR

STANDARDIZED RESIDUAL

RESIDUAL, PLOT

16.25 15.64 15.07 14.54 14.06 13.60 12.60 11.76 11.05 10.45 9.505 9.130 8.531 7.906 7.194 6.876 6.715 6.715 6 .'I15

0.00 -3.56 -5.75 -5.38 -8.95 -16.26 -4.15 -6.45 -4.29 -0.04 -0.05 3.89 4.15 6.98 6.57 3.16 -3.31 -3.31 -3.31

0.000E+00 -0.758 -1.20 -1.13 -1.81 -3.09 -0.879 -1.34 -0.908 -7.936E-03 -1.135E-02 0.859 0.915 1.54 1.45 0.697 -0.707 -0.707 -0,707

I **I ****I *** 1

****** 1 I

*** 1 **** 1 *** 1 *I

*I

1 ***

I *** I ***** I**** I ** **I **I **I

STATISTICAL SUMMARY

ML

MAX IM IZED LOG LIKELIHOOD FUNCTION --- -- - -11.71

CORRELATION MATRIX

K1 K2

WT RESID WEIGHTED STANDARD SUM OF RESIDUAL, ERROR OF SQUARES SUM ESTIMATE - - - - --- ----- ---------3.6963-02 -1.165E-02 0.451

K1 1.000 0.7030

K2 1.000

PERCENTAGE VARIATION WEIGHTING EXPLAINED PARAMETER ----__-__ _________ 97.168 2.00

122

2. Process Dynamics Fundamentals ~

VARTANCE-COVARLANCE MATRTX K1 K1 3.2345E-08 Ki 1.9982F 08

___________

~

K2 2 . 4 9 7 7 E 08

The parameter values found by the two methods differ slightly owing to the different criteria used which were the least squares method for ESL and the maximum-likelihood method for SIMUSOLV and because the T=10 data point was included with the ESL run. The output curve is very similar and the parameters agree within the expected standard deviation. The quality of parameter estimation can also be judged from a contour plot as given in Fig. 2.41. Conlour P l o t O b j e c t ,w e F u n c i & o n

0.on22 0.0020

O.OOl8 0.0016 0.OOlrl 0.0012 O.UOI0

o.oooa 0.0U1

0.UO2

0.0113

0.004

0.005

0 IflG

K I

Figure 2.41. Objective function contour plot using SIMUSOLV.

2.4.4

Numerical Integration

Only a very short introduction to numerical integration is given here, simply to demonstrate the basic principles and possible sources of error. In the great majority of simulation studies, the numerical integration will not be found to create problems and a detailed knowledge of the differing numerical integration methods is generally unnecessary. For more complex problems, where numerical difficulties may occur, the reader is referred to more specialist texts, e.g., Press et al. (1989), Walas (1991), Noye (1984). In the solution of mathematical models by digital simulation, the numerical integration routine is usually required to achieve the solution of sets of simultaneous, first-order differential equations in the form

2.4 Numerical Aspects of Dynamic Behaviour

123

The differential equations are often highly non-linear and the equation variables are often highly interrelated. In the above formulation, yi represents any one of the dependent system variables and, fi is the general function relationship, relating the derivative, dyi/dt, with the other related dependent variables. The system independent variable, t, will usually correspond to time, but may also represent distance, for example, in the simulation of steady-state models of tubular and column devices. In order to solve the differential equations, it is first necessary to initialise the integration routine. In the case of initial value problems, this is done by specifying the conditions of all the dependent variables, yi, at initial time t = 0. If, however, only some of the initial values can be specified and other constant values apply at further values of the independent variable, the problem then becomes one of a split-boundary type. Split-boundary problems are inherently more difficult than the initial value problems, and although most of the examples in the book are of the initial value type, some split-boundary problems also occur. In both types of problem, solution is usually achieved by means of a step-bystep integration method. The basic idea of this is illustrated in the information flow sheet, which was considered previously for the introductory ISIM complex reaction model example (Fig. 1.4). Referring to Fig. 1.4, the solution begins with the initial concentration conditions Ao, Bo, Coand Do, defined at time t = 0. Knowing the magnitudes of the kinetic rate constants kl, k2, k3 and k4, thus enables the initial rates of change dCA/dt, dCB/dt, dCc/dt and dCD/dt, to be determined. Extrapolating these rates over a short period of time At, from the initial conditions, Ao, Bo, Co and Do, enables new values for A, B, C and D to be estimated at the new time, t = t + At. If the incremental time step At is sufficiently small, it is assumed that the error in the new estimated values of the concentration, A, B, C and D, will also be small. This procedure is then repeated for further small increments of time until the entire concentration versus time curves have been determined. In this approach, the true solution is approximated as a series of discrete points along the axis of the independent variable, t. The solution then proceeds step-by-step from one discrete time step to the next. In the simplest case, the time steps will be spaced at uniform intervals, but the spacing can also be varied during the course of solution. Where there are several dependent variables, involved, all the variables must be updated to their new value, by projecting all the respective rates of change or concentration-time gradients over the identical time increment or integration step length, h. In order to do this, the integration method has to carry out a number of separate evaluations of the gradient terms. These evaluations are known as the "stages" of the computation. Thus taking the single rate equation dy/dt = f (y, t) and knowing the solution at any point Pn(yn, tn), the value of the function at the next point can be

124

2. Process Dynamics Fundamentals ~~~

predicted, knowing the local rate of change dyn/dtn. In the simplest case, this can be approximated by a simple difference approximation

from which Y n + l = Yn + hn

f(Yn, tn)

This procedure is illustrated in Fig. 2.42.

Figure 2.42. The difference approximation for rate of change or slope.

Much effort has been devoted to producing fast and efficient numerical integration techniques, and there is a very wide variety of methods now available. The efficiency of an integration routine depends on the number of function evaluations, required to achieve a given degree of accuracy. The number of evaluations depends both on the complexity of the computation and on the number of integration step lengths. The number of steps depends on both the nature and complexity of the problem and the degree of accuracy required in the solution. In practice, an over complex integration routine will require excessive computing time, owing to the many additional function evaluations that are required, and the use of an inappropriate integration algorithm can lead to an inaccurate solution, excessive computing time and, sometimes, a complete inability to solve the problem. It is thus very important that the output of any simulation is checked, using other integration methods. Most simulation languages allow a choice of integration routine which can be made best on the basis of experience. It is

125

2.4 Numerical Aspects of Dynamic Behaviour

___

~~

important to remember that all methods generate only approximate solutions, but these must be consistent with a given error criterion. As the models themselves, however, are also approximate, errors in the numerical solution must be seen in the general context of the problem as a whole. Numerical errors occur in the approximation of the original function and also are due to limits in the numerical precision of the computation. From experience, it can be shown that most cases of "strange" behaviour, in the results of a simulation, can be attributed largely to errors in the model and inadequate model parameter selection, rather than to numerical inaccuracies. ISIM uses four different integration methods. The user can select a particular method by assigning the variable ALGO a corresponding value. Other variables associated with the integration method are also explained below. Integration Methods Used by I S M 1. 2. 3. 4.

Variable step, 5th-order, Runge-Kutta explicit method (ALGO = 0). Fixed step, 4th-order, Runge-Kutta, explicit method (ALGO = 1 ) . Fixed step, 2nd-order, Runge-Kutta, explicit method (ALGO = 2). Fixed step, 2nd-order, implicit method for stiff systems (ALGO = 3).

Important Integration Parumeters ALGO CINT NOCI TFIN AERR RERR

Choice of integration method as above. Calculation interval or integration step length in fixed step methods. Used to specify the communication interval as NOCI * CINT. Value of the independent variable with which the run is terminated. Permitted absolute error per unit time. Permitted relative error per integration step length.

Typical Default Values ALGO = 0, CINT = 1, NOCI = 1, AERR = 0.01 and RERR = 0.1 The most common numerical problem, as shown by some of the simulation examples, is that of equation stiffness. This is manifested by the need to use shorter and shorter integration step lengths, with the result that the solution proceeds more and more slowly and may come to a complete halt. Such behaviour is shown by systems having combinations of very fast processes and very slow processes. Stiff systems, can also be thought of as consisting of differential equations, having time constants of widely differing magnitudes. Sometimes, the stiffness is the result of bad modelling practice, and can be removed by assuming the very fast processes to be virtually instantaneous, as compared to perhaps the slower overall rate determining processes. In this way, the differential equations, representing the troublesome very fast processes are

126 ~~-

__

2. Process Dynamics Fundamentals ~

~~-

~.

-~

replaced by steady-state algebraic equations, in which the rate of change is, in effect, taken to be zero. Unfortunately, this technique is not always possible and many systems are stiff in their own right and need therefore special means of solution.

2.4.5

System Stability

System instability can also be a problem, in dynamic simulation, and this can origin either from the integration routine or from the model itself. Instability in the integration routine, can arise owing to the approximation of the real functions by finite-differenced approximations, which can have their own parasitical exponential solutions. When the unwanted exponentials decrease with respect to time, the numerical solution will be stable, but if the exponential is positive then this can increase very rapidly and either swamp or corrupt the solution, sometimes in a manner that may be difficult to detect. Many integration algorithms show a dependence of the stability of the solution on the integration step length. There can thus be a critical integration step length for the integration, which if exceeded can lead to instability. This type of instability can be seen in many of the simulation examples, where an injudicious choice of CINT can cause numerical overflow and loss of the program. Reducing the integration step size makes the solution run more slowly, and rounding off errors caused by the limited accuracy of the digital representation may then become important. Practically, one can try to solve such problems by changing the integration routine or by adjusting the error criteria AERR and RERR. Model instability is demonstrated by many of the simulation examples and leads to very interesting phenomena, such as multiple steady states, naturally occurring oscillations, and chaotic behaviour. In the case of a model which is inherently unstable, nothing can be done except to completely reformulate the model into a more stable form In general, the form of the solution to the dynamic model equations will be in the form Yi(t) = steady-state solution

+ transient solution

where the transient part of the solution can be represented by a series of exponential functions

Ytrans = A1 e

+ A2 e h2t + ..........+ A,

e hnt

In the above relationship, the coefficients A1 to A n depend on the initial conditions of the problem and the exponential values, hi, are determined by the

2.4 Numerical Aspects of Dynamic Behaviour

127

~

parameters of the system and in fact represent the eigenvalues or roots of the characteristic solution of the system. In a stable system, the above transient terms will decay to zero, to give the steady-state solution. This applies when all the roots, are either simple negative exponentials, and giving a simple, direct approach to steady state. The system is also stable with all the roots occuring as negative real parts of complex roots, and causing a decaying, oscillatory approach to the eventual steady-state condition. If any of the roots are positive real numbers or complex numbers with real positive parts, the corresponding transient terms in the solution will grow in magnitude, thus directing the solution away from the unstable steadystate condition. Where the roots of the transient solution are pure imaginary numbers, the result is an oscillation of constant amplitude and frequency. The dynamic stability of such systems is often shown most conveniently on a phaseplane diagram, as shown in several of the simulation examples. Model instability is discussed further in Sec. 3.2.7, with regard to the stability of continuous stirred-tank reactors. For a fuller treatment of dynamic stability problems, the reader is referred to Walas (1991), Seborg et al. (1989), Habermann (1976), Perlmutter (1972) and to the simulation examples THERM, THERMPLOT, COOL, STABIL, REFRIGI and 2, OSCIL, LORENZ, HOPFBIF and CHAOS.

Chemical Engineering Dynami :Modelling

with PC Simulation

John Inph;m. Iruine J. Dunn. Elmar Heinzlc &JiiiE. Picnoul

copyright OVCH VerlagsgesellschaftmbH, 1994

3

Modelling of Stagewise Processes

3*1

Introduction

The principle of the perfectly-mixed stirred tank has been discussed previously in Sec. 1.2.2, and this provides essential building block for modelling applications. In this section, the concept is applied to tank type reactor systems and stagewise mass transfer applications, such that the resulting model equations often appear in the form of linked sets of first-order difference differential equations. Solution by digital simulation works well for small problems, in which the number of equations are relatively small and where the problem is not compounded by stiffness or by the need for iterative procedures. For these reasons, the dynamic modelling of the continuous distillation columns in this section is intended only as a demonstration of method, rather than as a realistic attempt at solution. For the solution of complex distillation problems, the reader is referred to commercial dynamic simulation packages.

3.2

Stirred-Tank Reactors

3.2.1

Reactor Configurations

This section is concerned with batch, semi-batch, continuous stirred tanks and continuous stirred-tank-reactor cascades, as represented in Fig. 3.1 Tubular chemical reactor systems are discussed in Chapter 4. Three modes of reactor operation may be distinguished, batch, semi-batch and continuous. In a batch system all reactants are added to the tank at the given starting time. During the course of reaction, the reactant concentrations decrease continuously with time, and products are formed. On completion of the reaction, the reactor is emptied, cleaned and is made ready for another batch.

130 -~

~-____

~-

3 Modelling of Stagewise Processes ~ . _ _ _ _ _

A, 6 , c

A, €3

A B C

A B

c

C

t

I

v A

____

t -B

A C

CQB

A C

Stirred-tank reactors in series Figure 3.1. Stirred-tank reactor configurations.

This ype of operation provides great flexibility with very simple equipment and allows differing reactions to be carried out in the same reactor. The disadvantages are the downtime needed for loading and cleaning and possibly the changing reaction conditions. Batch operation is often ideal for small scale flexible production and high value, low output product production, where an exact knowledge of the reaction chemistry and reaction kinetics is unnecessary. In many reactions pure batch operation is not possible, mainly due to safety or selectivity reasons. In semi-batch operation, one reactant may be charged to the vessel at the start of the batch, and then the others fed to the reactor at perhaps varying rates and over differing time periods. When the vessel is full, feeding is stopped and the contents allowed to discharge. Semi-batch operation allows one to vary the reactant concentration to a desired level in a very flexible way, and thus to control the reaction rate and the reactor temperature. It is, however, necessary to develop an appropriate feeding strategy. Sometimes it is necessary to adjust the feeding rates using feed-back control. The flexibility of operation is generally similar to that of a batch reactor system. Continuous operation provides high rates of production with more constant product quality. There are no downtimes during normal operation. Reactant preparation and product treatment also have to run continuously. This requires

131

3.2 Stirred-Tank Reactors

careful flow control. Continuous operation can involve a single stirred tank, a series of stirred tanks or a tubular-type of reactor. The latter two instances give concentration profiles similar to those of batch operation, whereas in a single stirred tank, the reaction conditions are at the lowest reactant concentration, corresponding to effluent conditions.

3.2.2

Generalised Model Description

The energy and mass balance equations for reacting systems follow the same principles, as described previously in Secs. 1.2.3 to 1.2.5.

3 . 2 . 2 . 1 Total Mass Balance Equation It becomes necessary to incorporate a total mass balance equation into the reactor model, whenever the total quantity of material in the reactor varies, as in the cases of semi-continuous or semi-batch operation or where volume changes occur, owing to density changes in flow systems. Otherwise the total mass balance equation can generally be neglected. The dynamic total mass balance equation is represented by accumulation ( ofRatetotalof mass in system)

=

Mass flow rate ( into the system)

-

flow rate ( outMassof the system

3 . 2 . 2 . 2 Component Balance Equation

I

The general component balance for a well-mixed tank reactor or reaction region can be written as Rate of

system

Rate of

the system

Rate of flow of component from the system

+

Rate of production of component by reaction

For batch reactors, there is no flow into or out of the system, and those terms in the component balance equation are therefore zero. For semi-batch reactors, there is inflow but no outflow from the reactor and the outflow term in the above balance equation is therefore zero.

132

3 Modelling of Stagewise Processes

_ _ _ _ _ _ _ _ _ _ _ _ _ _ ~ - ~ _ _ _ _ _ _ _ _

For steady-state operation of a continuous stirred-tank reactor or continuous stirred-tank reactor cascade, there is no change in conditions with respect to time, and therefore the accumulation term is zero. Under transient conditions, the full form of the equation, involving all four terms, must be employed.

3 . 2 . 2 . 3 Energy Balance Equation For reactions involving heat effects, the total and component mass balance equations must be coupled with a reactor energy balance equation. Neglecting work done by the system on the surroundings, the energy balance is expressed by Rate of accumulation

Flow of energy into

Flow of energy from

Rate of heat transfer

where each term has units of kJ/s. For steady-state operation the accumulation term is zero and can be neglected. A detailed explanation of energy balancing is found in Sec. 1.2.5, together with a case study for a reactor. Both flow terms are zero for the case of batch reactor operation, and the outflow term is zero for semi-continuous or semi-batch operation. The information flow diagram, for a non-isothermal, continuous-flow reactor, in Fig. 1.19, shown previously in Sec. 1.2.5, illustrates the close interlinking and highly interactive nature of the total mass balance, component mass balance, energy balance, rate equation, Arrhenius equation and flow effects F. This close interrelationship often brings about highly complex dynamic behaviour in chemical reactors.

3 . 2 . 2 . 4 Heat Transfer to and from Reactors Heat transfer is usually effected by coils or jackets, but can also be achieved by the use of external loop heat exchangers and, in certain cases, by the vaporisation of volatile material from the reactor. The treatment, here mainly concerns jackets and coils. Other instances of heat transfer are illustrated in the simulation examples of Chapter 5. Fig. 3.2 shows the case of a jacketed, stirred-tank reactor, in which either heating by steam or cooling medium can be applied to the jacket. Here V is volume, cp is specific heat capacity, p is density, Q is the rate of heat transfer, U is the overall heat transfer coefficient, A is the area for heat transfer, T is temperature, H is enthalpy of vapour, h is liquid enthalpy, F is volumetric flow

133

3.2 Stirred-Tank Reactors

rate, and W is mass flow rate. The subscripts are j for the jacket, s for steam and c for condensate. Fj Ti

f1 U

A

Ti 4

Figure 3.2. Model representation of a stirred-tank reactor with heat transfer to or from the jacket.

The rate of heat transfer is most conveniently expressed in terms of an overall heat transfer coefficient, the effective area for heat transfer and an overall temperature difference, or driving force, where Q = UA (T - Tj) ~ J / s

Jacket or Coil Cooling In simple cases the jacket or cooling temperature, Tj, may be assumed to be constant. In more complex dynamic problems, however, it may be necessary to allow for the dynamics of the cooling jacket, in which case T, becomes a system variable. The model representation of this is shown in Fig. 3.3. Under conditions, where the reactor and the jacket are well insulated and heat loss to the surroundings and mechanical work effects may be neglected

134

3 Modelling of Stagewise Processes

Fj Ti

t

Figure 3.3. Dynamic model representation of the cooling jacket.

i Rateof 1 accumulation of energy in the jacket

-

Rate of energy flow to the jacket by convection

-

[

en$;iiw from the jacket by convection

1

+

[

Rate of heat transfer into the jacket

or Rate of

Energy

in the jacket

from Tjin to T j

Rate of

Assuming the liquid in the jacket is well-mixed, the heat balance equation for the jacket becomes V j P j Cpj

dT, d := Fj

P j Cpj (Tjin - Tj)

+ UA(T - Tj)

Here F j is the volumetric flow of coolant to the jacket, Tjin is the inlet coolant temperature, and T j is the jacket temperature. Under well-mixed conditions, T j is identical to the temperature of the outlet flow. Alternatively neglecting the jacket dynamics and assuming that the coolant in the jacket is at some mean temperature, Tjavg, as shown in Fig. 3.4, a steadystate energy balance can be formulated as

135

3.2 Stirred-Tank Reactors

Fjt Tout

t

Figure 3.4. Steady-state model representation of the cooling jacket.

energy gain (Rateby ofcoolant flow )

=

Rate of energy transfer

( from the reactor to the jacket

F j Pj cpj (Tjout - Tjin) =

UA (T - Tjavg 1 = Q

Assuming an arithmetic mean jacket temperature,

Tjavg =

Tjin + Tjout

2

Substituting for Tjout into the steady-state jacket energy balance, solving for and substituting Tjavg into the steady-state balance, gives the result that

Tjavg

Q = U A K1 (T - Tjin)

where K1 is given by 2 F, p i c P j K 1 = UA + 2 F j p j c p j

As shown in several of the simulation examples, the fact that Q is now a function of the flow rate, Fj, provides a convenient basis for the modelling of cooling effects, and control of the temperature of the reactor by regulation of the flow of coolant.

I36

3 Modelling of Stagewise Processes .

3.2.2.5

_ _ _.

~

Steam Heating in Jackets

The dynamics of the jacket are more complex for the case of steam heating. The model representation of the jacket steam heating process is shown in Fig. 3.5.

Figure 3.5. Model representation of steam heating in the jacket.

A mass balance on the steam in the jacket is represented by

Rate of

Mass flow

Mass flow of

in the jacket

jacket

jacket

d . V j - 8 = F s p s - Wc The enthalpy balance on the jacket is given by

where Hj is the enthalpy of the steam in the jacket. The saturated steam density, Pj, depends on the jacket temperature, Tj, in the first approximation in accordance with the Ideal Gas Law and hence

137

3.2 Stirred-Tank Reactors

18 P. pj = 2 R Tj where R is the Ideal Gas Constant. The jacket steam pressure, Pj, is itself a function of the jacket steam temperature, Tj, as listed in steam tables or as correlated by the Antoine equation for vapour pressure, where

Pj = exp (T-

A J

-

abs

+ B)

Here, A and B are the Antoine steam constants and Tj abs is the absolute steam temperature. The determination of the steam density, Pj, therefore requires the simultaneous solution of two algebraic equations. This represents an IMPLICIT algebraic loop and cannot be solved within a simulation program without the incorporation of a trial and error convergence procedure. The implicit loop nature of the calculation is illustrated, by the information flow diagram, shown in Fig. 3.6. In this, Tj is needed in order to calculate Pj,

138 ._

~

__

3 Modelling of Stagewise Processes

Figure 3.6. Implicit loop calculation of the steam jacket dynamics.

but Pj in turn is needed to calculate Tj, so that the normal solution procedure of TSIM and other simulation languages will fail, unless some special IMPLICIT loop function is provided in the software. In practice the implicit loop calculation can sometimes be easily avoided. This can be done, for example, by formulating an empirical polynomial expression for the saturated steam temperature as a function of the saturated steam density

the data of which can be taken directly from steam tables data. Alternatively this can be done by utilising a function generator, if available in the digital simulation language and use this to interpolate a value of the temperature, Tj,

3.2 Stirred-Tank Reactors

139

for a given value of the steam density, Pj, based on tabulated data, for the variation of steam density with temperature. Since the steam in the jacket is saturated, the temperature-density relationship, of course, follows a unique form. This direct method of calculation is illustrated by the information flow diagram of Fig. 3.7.

Figure 3.7. Direct calculation of steam temperature and pressure.

For saturated steam P j = f(Tj), where Tj = f(Pj). Thus the knowledge of density permits the determination of temperature and from this the pressure.

3.2.2.6

Dynamics of the Metal Jacket Wall

In some cases, where the wall of the reactor has an appreciable thermal capacity, the dynamics of the wall can be of importance (Luyben, 1973). The simplest approach is to assume the whole wall material has a uniform temperature and therefore can be treated as a single lumped parameter system or, in effect, as a single, well-stirred tank.

140 ~

-.

3 Modelling of Stagewise Processes -

~

~

~

_

_

_

_

The heat flow through the jacket wall is represented in Fig. 3.8. React&

backet

v

V,

I I I

Tj I

I

Figure 3.8. Model representation of heat flow through the reactor wall

The nomenclature is as follows: Qrn is the rate of heat transfer from the reactor to the reactor wall, Qj is the rate of heat transfer from the reactor wall to the jacket, and

Qj = Uj Aj (T, -Tj) Here, U, is the film heat transfer coefficient between the reactor and the reactor wall, Uj is the film heat transfer coefficient between the reactor wall and the jacket, Am is the area for heat transfer between the reactor and the wall and Aj is the area for heat transfer between the wall and the jacket. The heat balance for the wall gives

and the balance for the jacket becomes dT. d: = Fj Pj Cpj (Tjin - Tj) + Qj In some cases, it may be of interest to model the temperature distribution through the wall. This might be done, by considering the metal wall and perhaps also the jacket as consisting of a series of separate regions of uniform temperature, as shown in Fig. 3.9. The balances for each region of metal wall and cooling water volume, then become for any region, n. vj pj Cpj

3.2 Stirred-Tank Reactors

141

Figure 3.9. Model representation of temperature distribution in the wall and jacket, showing wall and jacket with four lumped parameters.

where V,, and V j n are the respective volumes of the wall and coolant in element n. Am, and A j n are the heat transfer areas for transfer from the reactor to the wall and from the wall to the jacket. Hence:

I42

3 Modelling of Stagewke Processes -

~~

3.2.3

-~

~

._

The Batch Reactor

It is assumed that all the tank-type reactors, covered in this and the immediately following sections, are at all times perfectly mixed, such that concentration and temperature conditions are uniform throughout the tanks contents. Fig. 3.10 shows a batch reactor with a cooling jacket. Since there are no flows into the reactor or from the reactor, the total mass balance tells us that the total mass remains constant.

Figure 3.10. The batch reactor with heat transfer.

Component Balance Equation Rate of accumulation of reactant A

( Thus.

or for constant volume

Rate of production of A by chemical reaction

143

3.2 Stirred-Tank Reactors

Energy Balance Equation

[

Rate of accumulation of heat in he reactor

]

=

[

Rate of heat generation by reaction

) [ -

Rate of heat transfer to the surroundings

I

Here V is the volume of the reactor, p is the density, cp is the mean specific heat of the reactor contents (kJ/kg K) and rQ is the rate of generation of heat by reaction (kJ/s m3).

3 .2.3 .1

Case A.

Constant-Volume Batch Reactor

A constant volume batch reactor is used to convert reactant, A, to product, B, via an endothermic reaction, with simple stoichiometry, A -+ B. The reaction kinetics are second-order with respect to A, thus

From the reaction stoichiometry, product B is formed at exactly the same rate as that at which reactant A, is decomposed. The rate equation, with respect to A is

giving the component balance equation as

For the second-order reaction, the term representing the rate of heat production by reaction, simplifies to

144

3 Modelling of Stagewise Processes

This gives the resultant heat balance equations as V P cp

dT

= kC

A ( 1 ~- X~A ) (-AH) ~ V

+

UA(Tj - T)

where it is assumed that heat transfer to the reactor occurs via a coil or jacket heater. The component mass balance, when coupled with the heat balance equation and temperature dependence of the kinetic rate coefficient, via the Arrhenius relation, provide the dynamic model for the system. Batch reactor simulation examples are provided by BATCHD, COMPREAC, BATCOM, CASTOR, HYDROL and RELUY.

3.2.4

The Semi-Batch Reactor

A semi-batch reactor with one feed stream and heat transfer to a cooling jacket is shown in Fig. 3.11.

Figure 3.11.

The semi-batch reactor.

Total Mass Balance A total mass balance is necessary, owing to the feed input to the reactor, where

Rate of accumulation of mass in the reactor

)

=

($ewM:)::

145

3.2 Stirred-Tank Reactors

Here, po, is the feed density. The density in the reactor, p, may be a function of the concentration and temperature conditions within the reactor. Assuming constant density conditions dV

dt

= F

Component Balance Equation All important components require a component balance. For a given reactant A Rate of (accumulation of A

)

- (Rate of flow) of A in

+

(Rate of production of A by reaction

where NAOis the molar feeding rate of A per unit time. In terms of concentration. this becomes

where F is the volumetric feed rate and CAOis the feed concentration. Note that both the volumetric flow and the feed concentration can vary with time, depending on the particular reactor feeding strategy.

Energy Balance Equation Whenever changes in temperature are to be calculated, an energy balance is needed. With the assumption of constant cp and constant p, as derived in Sec. 1.2.5, the balance becomes dT

PCPVdt =

Fop cP (To - T)

+ rQ V +

Q

Note that the available heat transfer area may also change as a function of time, and may therefore may also form an additional variable in the solution. Note

3 Modelling of Stagewise Processes

I46

also that although constant p and cp have been assumed here, this is not a restrictive condition and that equations showing the variations of these properties are very easily included in any simulation model.

3 . 2 . 4 . 1 Case B. Semi-Batch Reactor A semi-batch reactor is used to convert reactant, A, to product, B, by the reaction A + 2B. The reaction is carried out adiabatically. The reaction kinetics are as before

and the stoichiometry gives

The balances, for the two components, A, and B, with flow of A, into the reactor, are now

and the enthalpy balance equation is V p cP

dT x= F p cP (To

- T)

+

k CA$ (1

-

X,4)2 (-AHA)V

since, for adiabatic operation, the rate of heat input into the system, Q, is zero. With initial conditions for the initial molar quantities of A and B, (V CA, V CB), the initial temperature, T, and the initial volume of the contents, V, specified, the resulting system of equations can be solved to obtain the time varying quantities, v(t), VCA(t), vcg(t), T(t) and hence also concentrations, CA, and c g , as a functions of time. Examples of semi-batch operations are given in the simulation examples HMT, SEMIPAR, SEMISEQ, RUN and SEMIEX.

147

3.2 Stirred-Tank Reactors

3.2.5

The Continuous Stirred-Tank Reactor

Although continuous stirred-tank reactors (Fig. 3.12) normally operate at steady-state conditions, a derivation of the full dynamic equation for the system, is necessary to cover the instances of plant start up, shut down and the application of reactor control. =A07 FOy TO

1

Figure 3.12. Continuous stirred-tank reactor with heat transfer.

Total Mass Balance The dynamic total mass balance equation is given by Rate of accumulation (of mass in the reactor)

=

flow) (Mass rate in

-

= Fopo-Fp

Under constant volume and constant density conditions dV = F o - F = O dt

-

and therefore

Fo = F

(Mass flow) rate out

148

3 Modelling of Stagewise Processes

Component Mass Balance Equation The component mass balance equation is given by (accumulation] Rate of = of component in the reactor

i

Rateinofof

component

-

j

Rate of flow out of component

+

Rate of iproduction of) component by reaction

For a given reactant, A

where, n A is the moles of A in the reactor, NAOis the molar feeding rate of A to the reactor, and NA is the molar flow rate of A from the reactor. Under constant density and constant volume conditions, this may be expressed as dCA

"dt or

where T (= V/F) is the average holdup time, or residence time of the reactor. For steady-state conditions to be maintained, the volumetric flow rate F, and inlet concentration, CAO,must remain constant and

hence at steady state

and

or

149

3.2 Stirred-Tank Reactors

Energy Balance Equation This, like the other dynamic balances for the CSTR, follows the full generalised form, of Sec. 1.2.5, giving Rate of

Energy needed

the reactor

temperature

Rate of

Rate of

surroundings

or assuming constant cp

3 . 2 . 5 . 1 Case C. Constant-Volume Continuous Stirred-Tank Reactor The chemical reaction data are the same as in the preceding example. The reaction kinetics are

with and the component balances for both A and B are given by

Assuming both constant density and constant specific heat, the heat balance equation becomes dT V P cp dt = F P cp (To - T)

+

k C A V~ (-AH)

- UA

(T - Tj)

Here cooling of an exothermic chemical reaction, via a cooling coil or jacket, is included.

I50

3.2.6

3 Modelling of Stagewise Processes

Stirred-Tank Reactor Cascade

For any continuous stirred-tank reactor, n, in a cascade of reactors (Fig. 3.13) the reactor n receives the discharge from the preceding reactor, n-I, as its feed and discharges its effluent into reactor n + l , as feed to that reactor. F

F

F

1 n-1 n Figure 3.13. Cascade of continuous stirred-tank reactors.

F

F

n+l

Thus the balance equations for reactor, n, simply become

where, for example

Thus the respective rate expressions depend upon the particular concentration and temperature levels, that exist within reactor, n. The rate of production of heat by reaction, rQ, was defined in Sec. 1.2.5 and includes all occurring reactions. Simulation examples pertaining to stirred tanks in series are CSTR, CASCSEQ and COOL.

151

3.2 Stirred-Tank Reactors

3.2.7

Reactor Stability

Consider a simple first-order exothermic reaction, A + B, carried out in a single, constant-volume, continuous stirred-tank reactor (Fig. 3.12), with constant jacket coolant temperature, where rA = - k CA. The model equations are then given by

dT V p cPdt = F p cP (To - T) + V k CA (-AH)

- UA

(T - Tj)

At steady state, the temperature and concentration in the reactor are constant with respect to time and dCA_ - d-T_ dt - dt Hence from the component mass balance

the steady-state concentration is given by

From the steady-state heat balance the heat losses can be equated to the heat gained by reaction, giving -F

i.e.,

P cP (To - T) + U A (T - Tj) = V k CA (-AH) = Net rate of heat loss from the reactor

( or

1

=

(

V k CAO(-AH) 1 k V/F

Rate of heat generation by chemical reaction in the reactor

+

1

I52

3 Modelling of Stagewise Processes

The above equation then represents the balanced conditions for steady-state reactor operation. The rate of heat loss, HL, and the rate of heat gain, HG, terms may be calculated as functions of the reactor temperature. The rate of heat loss, H L , plots as a linear function of temperature and the rate of heat gain, HG, owing to the exponential dependence of the rate coefficient on temperature, plots as a sigmoidal curve, as shown in Fig. 3.14. The points of intersection of the rate of heat lost and the rate of heat gain curves thus represent potential steady-state operating conditions that satisfy the above steady-state heat balance criterion.

"G

Heat quantity

Temperature

Figure 3.14. Heat loss, HL, and heat gain, HG, i n a steady-state, continuous stirred-tank rcactor.

Fig. 3.14 shows the heat gain curve, HG, for one particular set of system parameters, and a set of three possible heat loss, HL, curves. Possible curve intersection points, A1 and C2, represent singular stable steady-state operating curves for the reactor, with cooling conditions as given by cooling curves, I and 111, respectively. The cooling conditions given by curve 11, however, indicate three potential steady-state solutions at the curve intersections, A, B and C. By considering the effect of small temperature variations, about the three steady-state conditions, it can be shown that points A and C represent stable, steady-state operating

153

3.2 Stirred-Tank Reactors

conditions, whereas the curve intersection point B is unstable. On start up, the reaction conditions will proceed to an eventual steady state, at either point A or at point C. Since point A represents a low temperature, and therefore a low conversion operating state, it may be desirable that the initial transient conditions in the reactor should eventually lead to C, rather than to A. However if point C is at a temperature which is too high and might possibly lead to further decomposition reactions, then A would be the desired operating point. The basis of the argument for intersection point, B, being unstable is as follows and is illustrated in Fig. 3.15.

Heat quantity HG or

HL

-AT

+AT

Temperature

Figure 3.15. Characteristics of unstable point B of Fig. 3.14.

Consider a small positive temperature deviation, moving to the right of point B. The condition of the reactor is now such that the HG value is greater than that for HL. This will cause the reactor to heat up and the temperature to increase further, until the stable steady-state solution at point C is attained. For a small temperature decrease to the left of B, the situation is reversed, and the reactor will continue to cool, until the stable steady-state solution at point A is attained. Similar arguments show that points A and C are stable steady states. System stability can also be analysed in terms of the linearised differential model equations. In this, new perturbation variables for concentration C' and temperature T' are defined. These are defined in terms of small deviations in

154

3 Modelling of Stagewise Processes

the actual reactor conditions away from the steady-state concentration and temperature, Css and T,, respectively. Thus C' = C - C,,

and T' = T - T,,

The linearisation of the non-linear component and energy balance equations, based on the use of Taylor's expansion theorem, leads to two, simultaneous, first-order, linear differential equations with constant coefficients of the form dC'

dt =

a l C ' + blT'

dT' dt = a2C'

~

+ b2T'

The coefficients of the above equations are the partial differentials of the two dynamic balance equations and are given by

where F(C, T) represents the dynamic component balance equation and G(C, T) represents the dynamic heat balance equation. The two linearised model equations have the general solution of the form

The dynamic behaviour of the system is thus determined by the values of the exponential coefficients, hl and k2, which are the roots of the characteristic equation or eigenvalues of the system and which are also functions of the system parameters.

155

3.2 Stirred-Tank Reactors

If hl and h2 are real numbers and both have negative values, the values of the exponential terms and hence the magnitudes of the perturbations away from the steady-state conditions, c', and T', will reduce to zero, with increasing time. The system response will therefore decay back to its original steady-state value, which is therefore a stable steady-state solution or stable node. If hl and h2 are real numbers and both or one of the roots are positive, the system response will diverge with time and the steady-state solution will therefore be unstable, corresponding to an unstable node . If the roots are, however, complex numbers, with one or two positive real parts, the system response will diverge with time in an oscillatory manner, since the analytical solution is then one involving sine and cosine terms. If both roots, however, have negative real parts, the sine and cosine terms still cause an oscillatory response, but the oscillation will decay with time, back to the original steady-state value, which, therefore remains a stable steady state. If the roots are pure imaginary numbers, the form of the response is purely oscillatory, and the magnitude will neither increase nor decay. The response, thus, remains in the neighbourhood of the steady-state solution and forms stable oscillations or limit cycles. The types of system behaviour predicted, by the above analysis are depicted in Figs. 3.16 and 3.17. The phase-plane plots of Fig. 3.17 give the relation of the dependant variables C and T. Detailled explanation of phase-plane plots is given i n control textbooks (e.g., Stephanopoulos, 1984). Linearisation of the reactor model equations is used in the simulation example, HOMPOLY.

I

Response

System response with real roots

fi Time

Response

Time

Figure 3.16. Systems response with real and complex roots.

156

3 Modelling of Stagewise Processe\ __

-~~~~ .

~

I

I I

I

a) Stable node

b) Unstable node

c) Limit cycle

Figure 3.17. Phase-plane representations of reactor btability. In the above diagrams the point + represents a possible steady-state solution, which (a) may be stable, (b) may be unstable or (c) about which the reactor produces sustained oscillations in temperature and concentration.

Thus it is possible for continuous stirred-tank reactor systems to be stable, or unstable, and also to form continuous oscillations in output, depending upon the system, constant and parameter, values. This analysis is limited, since it is based on a steady-state criterion. The linearisation approach, outlined above, also fails in that its analysis is restricted to variations, which are very close to the steady state. While this provides excellent information on the dynamic stability, it cannot predict the actual trajectory of the reaction, once this departs from the near steady state. A full dynamic analysis is, therefore, best considered in terms of the full dynamic model equations and this is easily effected, using digital simulation. The above case of the single CSTR, with a single exothermic reaction, is covered by the simulation examples, THERMPLOT and THERM. Other simulation examples, covering aspects of stirred-tank reactor stability are COOL, OSCIL, REFRTG and STABIL.

3.2.8

Reactor Control

Two simple forms of a batch reactor temperature control are possible, in which the reactor is either heated by a controlled supply of steam to the heating jacket, or cooled by a controlled flow of coolant (Fig. 3.18) Other control schemes would be to regulate the reactor flow rate or feed concentration, in order to maintain a given reaction rate (see simulation example SEMTEX).

-

3.2 Stirred-Tank Reactors

157

Temperature

4 Controller

Figure 3.18. Reactor with control of temperature by manipulating the flow of cooling water.

Fig. 3.19 represents an information flow diagram for the above control scheme.

I

Component balance for reactor

Control equation

Heat balance for reactor

Jacket heat balance

I

T

Figure 3.19. Information flow diagram for the above reactor control implementation.

Assuming the jacket is well-mixed, a heat balance on the jacket gives

Consider the case of a proportional controller, which is required to maintain a desired reactor temperature, by regulating the flow of coolant. Neglecting

158

3 Modelling of Stagewise Proce\se\ ~~

~~~

dynamic jacket effects, the reactor heat balance, can then be modified to include the effect of the varying, coolant flow rate, Fj, in the model equation, as:

where the mean temperature of the jacket is accounted for by the term, K1, as shown in Sec. 3.2.2.4 and given by

For a proportional controller

where K, is the proportional gain of the controller and the temperature difference term, (T - TSet),represents the error between the reactor temperature T and controller set point TSet.Note that in this conventional negative feedback system, when the reactor temperature, T, is below the setpoint temperature, TSet, the coolant flow is decreased, in order to reduce the rate of heat loss from the reactor to the jacket. Note also that the incorporation of the controller equation and parameter value, K,, into the dynamic model also alters the stability parameters for the system and thus can also change the resultant system stability characteristics. A stable system may be made to oscillate by the use of high values of K, or by the use of a positive feedback action, obtained with the use of the controller equation with K, negative. Thus the reactor acts to provide greater degrees of cooling when, T < Tset, and conversely reduced degrees of cooling when, T > Tiet. This phenomenon is shown- particularly i n simulation example OSCIL. This example shows that the reactor may oscillate, either naturally according to the system parameters, or by applied controller action. Owing to the highly non-linear behaviour of the system, it is sometimes found that the net yield from the reactor may be higher under oscillatory conditions than at steady state (see simulation examples OSCIL and COOL). It should be noted also that under controlled conditions, Tset need not necessarily be set equal to the steady-state value, T, and TSet,and that the control action may be used to force the reactor to a more favourable yield condition than that simply determined by steady-state balance considerations. The proportional and integral controller equation

3.2 Stirred-Tank Reactors

159

where TI is the integral time constant and E is the error = (T - Tset) or (CA C A ~ can ~ ~similarly ) be incorporated into the reactor simulation model. Luyben (1973) (see simulation example RELUY) also demonstrates a reactor simulation including the separate effects of the measuring element, measurement transmitter, pneumatic controller and valve characteristics which may in some circumstances be preferable to the use of an overall controller gain term.

3.2.9

Non-Ideal Flow

The previous analysis of stirred-tank reactors has all been expressed in terms of the idealised concept of perfect mixing. In actual reactors, the mixing may be far from perfect and for continuous flow reactors may lie somewhere, between the two idealised instances of perfect mixing and perfect plug flow, and may even include dead zones or short-cut flow. The concept of ideal plug flow is usually considered in terms of continuous tubular or continuous column type devices and the application of this to continuous flow reactors is discussed in Sec. 4.3.1. In ideal plug-flow reactors, all elements of fluid, spend an identical period of time within the reactor, thus giving a zero distribution of residence times. All elements of fluid thus under go an equal extent of reaction, and temperatures, concentrations and fluid velocities are completely uniform across any flow cross section. For a well-mixed, continuous stirred-tank reactor, however, there will be elements of fluid with, theoretically, a whole range of residence times varying from zero to infinity. Practical reactors will normally exhibit a distribution of residence times, which lie somewhere between these two extreme conditions and which effectively determine the performance of the reactor. The form of the residence time distribution curve can therefore be used to characterise the nature of the flow in the reactor and models of the mixing behaviour can be important in simulations of reactor performance. Often different combinations of tanks in series or in parallel can be used to represent combinations of differing mixed flow regions and provide a powerful tool in the analysis of actual reactor behaviour.

E-curve Characterisation One method of characterising the residence time distribution is by means of the E-curve or external-age distribution function. This defines the fraction of material in the reactor exit which has spent time between t and t + dt in the reactor. The response to a pulse input of tracer in the inlet flow to the reactor gives rise to an outlet response in the form of an E-curve. This is shown below in Fig. 3.20.

160

3 Modelling of Stagewise Processes -

- .-

~~

-r Total area = 1.O 0

01

Figure 3.20. E-curve response to pulse input of tracer.

where O = t/z and C0 is the concentration in the tank at t = 0; C0 is calculated from the quantity of tracer added by the pulse to the reactor. Important relationships, relevant to the E-curve are: m

E dO = 1.0 0

where E dO is the fraction of material in the exit stream with age between O and dO and since all tracer eventually leaves the reactor, this integral is equal to unity.

where F is the fraction of material younger than 01.

F-curve Characterisation This is the response, in the outlet flow, which is obtained from a step change of tracer concentration in the reactor inlet and is shown below in Fig. 3.21.

161

3.2 Stirred-Tank Reactors

1.o

0

Figure 3.21. F-curve response to a step change in tracer concentration.

Here 0 = t/z and Co is the tracer inlet feed concentration. The corresponding relationships for the F-curve response are:

This means that the F-curve can be obtained from the E-curve as its slope. Referring to Fig. 3.21 0 F(@) =

dEd@

Tracer Diagnosis The forms of actual tracer response curves may be used to formulate models of the actual mixing processes in the reactor. One has, however, to be careful since the tracer response curve does not give a unique solution. It does, for example, not allow one to distinguish between early and late mixing, which may be important when used in the estimation of conversion in a particular reactorreaction system. In a real stirred tank with bypassing or short-cut flow (Fig. 3.22), highly concentrated tracer comes out early, and the residence time distribution depends on the fraction a of the flow in the bypass (Fig. 3.23). The tailing of the response curve is caused by the perfect mixing in the main part of the tank.

I62

3 Modelling of Stagewise Processes

~ _ _

-

F

1

I

v

Figure 3.22. Representation of stirred-tank reactor with bypassing.

Figure 3.23. Residence time distribution in a tank with bypassing.

Another frequently observed phenomenon in stirred tanks are dead zones. The experimental E-curve response for a system with dead zones may appear as in Fig. 3.24. In this, the response is initially too rapid and then has a long tail. The tailing is a result of the tracer being retained in the dead zones.

163

3.2 Stirred-Tank Reactors

1.o

0

Figure 3.24. Residence time distribution in a tank with a dead zone in comparison with an ideal tank.

A possible model for this phenomenon is shown in Fig. 3.25 (see simulation example NOSTR).

' I

I

Figure 3.25. Stirred-tank reactor with dead zones.

In special cases, the departure from ideal single stirred-tank mixing behaviour is intended, e.g., when a series of stirred tanks are used to approach that of plug-flow, as shown in Fig. 3.26.

164

3 Modelling of Stagewise Processes

1.o E = ClCo

1.o

0

Figure 3.26. Tanks-in-series response to a pulse tracer input from different number of tanks.

More complex situations can be imagined, for example, the combination of stirred-tanks, bypassing and stagnant zones. Care must be taken to establish the model parameters carefully however and not to merely fit the tracer response data. Experimental E-curve data is given in Fig. 3.27 and a possible model for this is shown in Fig. 3.28.

Figure 3.27. Tracer pulse response of a tanks-in-series system with and without dead zones.

3.2 Stirred-Tank Reactors

Active zones

165

’

Figure 3.28. Combination of tanks and stagnant zones.

Here the pulse response curve has a multiple tanks-in-series character but with significant tailing. This is often observed in packed columns where the stagnant zones are a result of the packing structure. Sometimes bypassing occurs as a result of channelling through the packing. In principle, any type of residence time distribution can be described by combinations of tanks in series or in parallel. This type of modelling of residence time distribution can very easily be implemented in simulation programs by adding the various tanks and adjunct flow streams. Other model representations of flow mixing cases in chemical reactors are described by Levenspiel (1972), Fogler (1992) and Szekely and Themelis (197 1). Simulation tank examples demonstrating non-ideal mixing phenomena are CSTR, NOSTR, TUBMIX, MIXFLO, GASLIQ and SPBEDRTD.

Using Tracer Information in Reactor Design Unfortunately RTD studies cannot distinguish between early mixing and late mixing sequences of different types. It can be shown that mixing history does not influence a first-order reaction but other reaction types are affected; the more complex the reaction kinetics the more the reaction selectivity or product distribution generally will be influenced. Thus for certain kinetic cases a detailed knowledge of the mixing can be important to the reactor performance. In practice this information is difficult to obtain, since normal tracer studies are not sufficient to provide this type of information. In principle, tracer injection and sampling at different points in the reactor can supply the needed information. In practice, the usual procedure is to develop a model based on RTD experiments and intuition. A comparison of the actual reactor performance with the model predictions for known reaction kinetics reveals whether the model assumptions were correct. The procedure is outlined below in Fig. 3.29.

I66 __

-_______

3 Modelling of Stagewise Processes ~___ ~

.

~

____ _ _

Physical model for mixing

Mathematical model

Parameter estimation

&

Reaction kinetlcs model combined with mixing model

Dynamic tracer experiment

Kinetic parameters

Simulation Steady-state experiment

I

*

Final design

I

Application for control and operation Figure 3.29. Reactor design procedure with reactors having residence time distributions deviating from those of ideal reactors.

3.3

Stagewise Mass Transfer

3.3.1

Liquid-Liquid Extraction

Liquid-liquid extraction is an important chemical engineering separation process and a knowledge of the process dynamics is important since many solvent extraction operations are still carried out batchwise. In addition, although most continuous solvent extraction plants are still designed on a steady-state basis, there is an increasing awareness of the need to assess possible safety and environmental risks at the earliest possible design stage. For this, a knowledge of the probable dynamic behaviour of the process becomes

167

3.3 Stagewise Mass Transfer

increasingly important. This applies, especially, in the fields of nuclear reprocessing and heavy metal extraction. The modelling of solvent extraction is also of interest, since the dynamic behaviour of both liquid phases can be important, and because of the wide range of equipment types that can be employed and the wide range of dynamic behaviour that results. The equipment is typified by mixer-settlers at one extreme, often representing high capacity, stagewise contacting devices, in which near equilibrium conditions are achieved with slow, stable, but long lasting dynamic characteristics and, on the other hand, differential column devices, representing high throughput, low volume non-equilibrium differential contacting devices, with fast acting dynamic behaviour; a limited range of permissible operating conditions and often an inherent lack of stability, especially when plant conditions are changing rapidly. Truly differential column devices are considered in Chapter 4, but some types of extraction columns can be regarded basically as stagewise in character, since the modelling of the dynamic characteristics of this type of device, leads quite naturally from the equilibrium stagewise approach. The treatment is confined to the use of two completely immiscible liquid phases, the feed or aqueous phase and the solvent or organic phase. No attempt is made to apply the modelling methodology to the case of partially miscible systems. Although one of the phases, the dispersed phase, will be in the form of droplets, dispersed in a continuum of the other, this is neglected and each liquid phase is assumed to consist of separate, well-mixed stage volumes. The modelling approach, shown in this chapter, follows the general modelling methodology, in that it starts with the simplest case of a single component, batch extraction and then builds in further complexities. Finally a complex model of a non-ideal flow in a multistage, multicomponent, extraction cascade, which includes a consideration of both hydrodynamic effects and control, is achieved.

3 . 3 . 1 . 1 Single Batch Extraction Volumes VL and VG of the two immiscible liquid phases, are added to the extraction vessel and a single solute distributes itself between the phases as concentrations X and Y, respectively, at a rate, Q, as shown in Fig. 3.30. For batch extraction, with no feed into the system, the component balances on each phase are given by: Rate of accumulation of solute in the given phase

Effective rate of mass transfer to or from the phase

Neglecting the effects of concentration changes on solvent density, the phase

168

3 Modelling of Stagewise Processes ~.

~.

Figure 3.30. Single-solute batch extraction between immiscible liquid phases.

volumes will remain constant. Thus for the liquid phase with volume VL

v dXq

= -

Q

and for liquid phase, with volume VG

where Q is the rate of solute transfer with units (mol/s) or (kg/s)

Q

= K L ~ ( X- X*)V

KL is the mass transfer coefficient for the L phase ( d s ) , a is the interfacial area per unit volume (m2/m3), referred to the total liquid volume of the extractor, V is the total holdup of the tank, and is equal to (VL+VG). X* is the equilibrium concentration, corresponding to concentration Y, given by X" = feq(Y) The information flow diagram, Fig. 3.31, for this system shows the two component mass balance relations to be linked by the equilibrium and transfer rate relationships. Note that the transfer rate equation is based on an overall concentration driving force, (X-X*) and overall mass transfer coefficient, KL. The two-film theory for interfacial mass transfer shows that the overall mass transfer coefficient, KL, based on the L-phase is related to the individual film coefficients for the L and G-phase films, k L and kG, respectively by the relationship 1

1

KL - kL

m + kG -

169

3.3 Stagewise Mass Transfer

4

~.

Q

Q=-KL~(X-X*)V

0 x * = feq(Y) V . 5

= +Q

b

Figure 3.31. Information flow diagram for simple batch extraction.

where ni is the local slope of the equilibrium curve, and where

For a linear equilibrium curve with constant film coefficients, kL and kG, the overall coefficient, KL, will also be constant, but for the case of a non-linear equilibrium relationship, the value of m, which is the local slope of the equilibrium curve, will vary with solute concentration. The result is that the overall coefficient, KL, will also vary with concentration, and therefore in modelling the case of a non-linear equilibrium extraction, further functional relationships relating the mass transfer coefficient to concentration will be required, such that

KL = f ( X )

3 . 3 . 1 . 2 Multisolute Batch Extraction Two solutes distribute themselves between the two phases as concentrations XA and YA, and XB and YB and with rates QA and QB, respectively as shown in Fig. 3.32. The corresponding equilibrium concentrations XA * and XB * are functions of both the interacting solute concentrations, YA and YB,and can be expressed by functional relationships of the form

170

3 Modelling of Stagewise Processes

Figure 3.32. Two-solute batch extraction.

xA* = fAey (YAI YB) xB* = fBeq (YA, YB) Typical representations of the way that the two differing equilibrium relationships can interact are shown in Fig. 3.33, and it is assumed that the equilibria can be correlated by appropriate, explicit equation forms. increasing XB

,I

Increasing XA

XA

XB

Figure 3.33. Interacting solute equilibria for the two solutes A and B.

For multi-component systems, it is necessary to write the dynamic equation for each phase and for each solute, in turn. Thus for phase volume VL, the balances for solute A and for solute B are

For phase volume VG the balances for both solutes are

171

3.3 Stagewise Mass Transfer

The overall mass transfer coefficients are also likely to vary with concentration, owing to the complex multisolute equilibria, such that:

Again, these functional relationships should ideally be available in an explicit form in order to ease the numerical method of solution. Two-solute batch extraction is covered in the simulation example TWOEX.

3 . 3 . 1 . 3 Continuous Equilibrium Stage Extraction Here the extraction is carried out continuously in a single, perfectly mixed, extraction stage as shown in Fig. 3.34. It is assumed that the outlet flow concentrations, XI and Y 1, achieve equilibrium and that density variations are negligible.

Figure 3.34. Continuous equilibrium stage extraction.

Following an initial transient condition, the extractor will achieve a steady state operating condition, in which the outlet concentrations then remain constant with respect to time. At steady state, the quantity of solute entering the extractor is equal to the quantity of solute leaving. A steady-state balance for the combined two-phase system gives LXo

+

GYo = LX1

+

GY1

172

3 Modelling of Stagewise Processes

where for an equilibrium stage extraction

Here, L and G are the volumetric flow rates of the heavy and the light phases, respectively, Xo and Yo are the respective inlet solute concentrations of the two phases, X I and Y1 are the respective outlet solute concentrations. For a linear equilibrium relationship,

A simple substitution of the value Y1 in the balance equation enables the steady-state concentration X 1 to be determined, where

The steady-state approach, however, provides no information on the initial transient conditions, whereby the extractor achieves eventual steady state or on its dynamic response to disturbances. For this it is necessary to derive the dynamic balance equations for the system. For unsteady-state operation the component balance equations, for each phase are now of the form Rate of

Rate of

Rate of

Rate of

The sign of the transfer term will depend on the direction of mass transfer. Assuming solute transfer again to proceed in the direction from volume VL to volume Vc;, the component mass balance equations become for volume VL V LdX1 T - LXo

-

LX1

-

Q

and for volume VG

where

Q = KL a(X1

-

Xi*) V

Adding the above two component balance equations gives the dynamic equation for the complete stage as

173

3.3 Stagewise Mass Transfer

LX1

+

GYo

-

Note that this could also be obtained by defining the mass balance system to include both phases and that the mass transfer rate expression, Q, now no longer appears in the balance equation. If equilibrium between the two phases is assumed, the above balance equation can be solved in conjunction with the equilibrium relationship, as shown by Franks (1967) and as indicated in the information flow diagram, Fig. 3.35.

Figure 3.35. Information flow diagram for the continuous equilibrium extraction stage.

Note that the total component balance equation is linked here only to the equilibrium relationship. Franks has suggested that the solution procedure should be based on the concentration value, Y 1 or X I , whichever having the greatest magnitude. Thus for a linear equilibrium in which Yl = m X I , the equilibrium equation can be differentiated to give

from which the balance equation now becomes

This resulting equation dX1 ~dt

-

LXo

+

G Y o - X1 (L VL + m VG

-

m G)

can then be solved to obtain the value X I and hence Y 1 , for any value of time t. With complex forms of equilibrium relationships, the above procedure can cause difficulties, since the equilibrium relationship must be able to be

174

3 Modelling of Stagewise Processes

__.

~

~~~

differentiated to obtain the derivative quantity dXl/dt or dY I/dt, and this may not always be mathematically convenient or possible. This will especially be the case of a non-ideal multicomponent extraction, where the equilibrium is nonlinear and where the different solutes exhibit mutually interacting effects. For example, if the equilibrium relationship is of the form

differentiation gives

such that the resulting balance equation V1

+

V2(A2

+

2 A 3 X l ) T = LXo

+

-

LXI

G (A1

+

+

GYo+

A2X1

+

A3 Xi2)

is again soluble by direct numerical integration. However with multicomponent equilibria of the form

solution is not so straightforward (Wilkinson and Ingham, 1983). An alternative approach to the solution of the system dynamic equations, is by the natural cause and effect mass transfer process as formulated, within the individual phase balance equations. This follows the general approach, favoured by Franks (1967), since the extractor is now no longer constrained to operate at equilibrium conditions, but achieves this eventual state as a natural consequence of the relative effects of solute accumulation, solute flow in, solute flow out and mass transfer dynamics. The individual component balance equations for each phase are VL~;]

where

= LXo - L X I - Q

175

3 3 Stagewise Mass Transfer ~

~-

Q = KL ~ ( X -I XI*) V

and X1* = feq(Y1) The method of solution is to employ an arbitrarily large value for the mass transfer coefficient KL, in the mass transfer rate equation, as suggested by Franks (1963). This acts effectively as a high gain factor in the rate equation, with the effect of forcing conditions towards equilibrium. Thus, for a finite rate of transfer Q, a high value of KL will force the driving force term (Xn-Xn*) to be very low and hence the value of Xn* to approach very closely to X,. Thus the outlet concentration conditions will approach very closely to equilibrium. This closely follows the natural cause and effect sequence, since for equilibrium to be obtained, high rates of mass transfer are required and high rates of mass transfer are favoured by high values of the mass transfer coefficient. While offering a more inherently realistic method of solution, however, the technique may cause some additional problems in the numerical solution, since high values of KL can lead to increased stiffness in the differential equations. Thus in using this technique, a compromise between the approach to equilibrium and the speed of numerical solution may have to be adopted. Continuous single-stage extraction is treated in the simulation example EQEX. Reaction with integrated extraction is demonstrated in simulation example REXT.

3 . 3 . 1 . 4 Multistage Countercurrent Extraction Cascade For a high degree of extraction efficiency, i t is usual to connect several continuous flow stages to form a countercurrent flow extraction cascade. as indicated in Fig. 3.36.

Figure 3.36. Multistage countercurrent extraction cascade.

In each stage, it is assumed that the two phases occupy well-mixed, constant volumes VL and V c . The phase volumes VL and VG can, however, vary from stage to stage along the cascade. This effect is easily included into any

176 -

3 Modelling of Stagewise Processes ~~

-~

simulation program. Additional complexity, in the formulation of the model, is now provided by the requirement of having to write balance equations for each of the stages of the cascade. The total number of equations to be solved is thus increased, but the modelling procedure remains straightforward. For any given stage, n, the component material balance equations for each phase are thus defined by

where and

as shown in Fig. 3.37.

Figure 3.37. Flow and composition inputs to stage n of the cascade.

Note that the rate of transfer is defined by the local concentrations X, and Xn" appropriate to the particular stage, n. It is straightforward in the formulation of the model to allow for variations of the parameter values, VL, VG, KL and a from stage to stage and for both KL and a to vary with respect to the local concentration. In order to do this, it is necessary to define new constant values for VL and VG for each stage and to have functional relationships, relating the mass transfer capacity coefficient to stage concentration. In order to model equilibrium stage behaviour, actual values of the mass transfer capacity product term, ( K L an), ~ would be again replaced by an arbitrary high value of the gain coefficient, KL", which is used to force actual stage concentrations close to the

3.3 Stagewise Mass Transfer

equilibrium. EQMULTI.

177

Multistage extraction is treated in the simulation example

3 . 3 . 1 . 5 Countercurrent Extraction Cascade with Backmixing The extension of the modelling approach to allow for backmixing between stages, cascades with side streams, or multiple feeds is also accomplished, relatively easily, by an appropriate modification of the inflow and outflow terms in the component balance equations. Backmixing reduces the efficiency of countercurrent mass transfer cascades, owing to its effect on the concentration profiles within the cascade and in decreasing the effective concentration driving forces. The effects of backmi Xing are especially severe in the case of solvent extraction columns. The stagewise model with backmixing is a well-known model representation, but the analytical solution is normally mathematically very complex and analytical solutions, both for steady-state and unsteady-state operating conditions, only apply for single-solute extraction where parameter values remain constant and furthermore where a linear equilibrium relationship applies. Compared to this, the solution of the dynamic model equations by digital simulation is far more general, since this has the ability to encompass varying parameter values, non-linear equilibria and multisolute systems. A multistage extraction cascade with backmixing is shown in Fig. 3.38. Here the backmixing flow rates LB and GB act in the reverse direction to the main phase flows, between the stages and along the cascade. One important factor in the modelling process is to realise that, as a consequence of the backmixing flows, that since phase volumes remain constant, then the interstage flow rates along the cascade, in the forward direction, must also be increased by the magnitude of the appropriate backmixing flow contribution. With a backmixing flow LB in the aqueous phase, the resultant forward flow along the cascade must now be (L + LB), since the backmixing does not appear exterior to the column. Similarly with a backmixed flow GB, the forward flow for the organic phase is also increased to (G + GB). Taking into account the changed flow rates, however, the derivation of the component balance equations follows normal procedures.

178 ~~

3 Modelling of Stagewise Processes

I +

-

~-

.

~~

Forward flow L + Lp Backmixing flow Lg

Forward flow G + GB

Figure 3.38. Multistage extraction cascade with backmixing of both phase\.

The relative inflow and outflow contributions for each phase of any stage n, of the cascade is shown in Fig. 3.39.

Figure 3.39. Stage n of a multistage extraction cascade with backmixing.

Allowing for the additional backmixing flow contributions, the component balance equation for the two phases in stage n of the cascade are now

or

179

3 3 Stagewise Mass Transfer -~

~

.

~

~

~

_

_

_

~

_

_

.

~~

Although more complex in form, the resulting model equations provide no major additional difficulty, and solution is also easily obtained. Multistage extraction with backmixing is covered in the simulation example EQBACK.

3.3.1.6

Countercurrent Extraction Cascade with Slow Chemical Reaction

A countercurrent extraction cascade with reaction A + B -+ C was considered by Ingham and Dunn (1974), as shown in Fig. 3.40. The reaction takes place between a solute A in the L-phase, which is transferred to the G-phase by the process of mass transfer, where it then reacts with a second component, B, also in the G-phase, to form an inert product, C. +1 +1 +1

Figure 3.40. Solute A is transferrcd from the L-phase to the G-phase, where it reacts with a component B, to form C.

The general component balance form of equation gives Rate of accumulation

Inflow

in phase i

in phase i

outflow reaction

180

3 Modelling of Stagewise Processes

where thus for phase L

and for phase G

Components B and C are both immiscible in phase L and remain in phase G. Therefore

where the transfer rate is

and the reaction rate is

Fig. 3.41 shows the graphical output in the G-phase concentrations of component A with respect to time, starting the cascade at time t = 0 with initially zero concentrations throughout. The maximum in the YAl profile for stage 1 is, due to the delay of reactant B in reaching A. This is because B is fed into the opposite stage N of the cascade, and requires a certain time of passage through the extractor in order to be able to react with A in stage 1.

3 . 3 . 1 . 7 Multicomponent Systems The previous treatments have been confined to the case of single-solute extraction between immiscible liquid phases. Assuming the liquid phases

181

3.3 Stagewise Mass Transfer

1.6 yAi 1.2 Concentration 0.8

0.4 0

y/

1

0

I

I

2

I

I

4

I

I

6

I

Time

Figure 3.41. Concentrations in the solvent phases of stages 1, 2 and 3 in a countercurrent extraction column with slow chemical reaction.

remain immiscible, the modelling approach for multicomponent systems remains the same, except that it is now necessary to write additional component balance equations for each of the additional solutes, as shown previously, in Sec. 3.2.2. Thus for a three component mixture with components A, B and C, the component balance equations can be written as follows:

For component A

For component B

3 Modelling of Stagewise Procescs ~

~~

For component C

where LB and GB are again backmixing flow rates. The respective phase transfer fluxes are

and where the equilibrium concentration Xi'k,for each solute will probably be a complex function of all three solute concentrations, i.e.,

The increased number of differential model equations obviously represents a considerable increase in computational effort but apart from this, no major additional difficulty should occur, except from that possibly arising from the equilibrium equations, which most favourably should be in an explicit form for easy solution.

3.3.1 . 8 Control of Extraction Cascades A typical control problem might be the maintenance of a required raffinate outlet concentration, YN, with the controller action required to compensate the effect of variations in the feed concentration Yo, as indicated in Fig 3.42. The proportional-integral control equation, as given in Sec. 2.3.2.2 is:

where L is the manipulated variable, i.e., the solvent flow rate, Lo is the base value for L, K, is the proportional gain, TI is the integral action timc and E is the error between the actual concentration, YN, and the desired value YSct. These relationships can be directly incorporated into a digital simulation program as

3.3 Stagewise Mass Transfer ____

Figure 3.42. rate.

183

~.

Maintenance of raffinate outlet concentration, by regulation of solvent flow

shown in example COLCON. If required, it is also possible to include the dynamic effects of the measuring elements, measurement transmitters, and the control valve characteristics into the simulation program as shown by Franks (1972) and by Luyben (1973 and 1990).

3 . 3 . 1 . 9 Mixer-Settler Extraction Cascades The archetypal, stagewise extraction device is the mixer-settler. This consists essentially of a well-mixed agitated vessel, in which the two liquid phases are mixed and brought into intimate contact to form a two phase dispersion, which then flows into the settler for the mechanical separation of the two liquid phases by continuous decantation. The settler, in its most basic form, consists of a large empty tank, provided with weirs to allow the separated phases to discharge. The dispersion entering the settler from the mixer forms an emulsion band, from which the dispersed phase droplets coalesce into the two separate liquid phases. The mixer must adequately disperse the two phases, and the hydrodynamic conditions within the mixer are usually such that a close approach to equilibrium is obtained within the mixer. The settler therefore contributes little mass transfer function to the overall extraction device. Ignoring the quite distinct functions and hydrodynamic conditions which exist in the actual mixer and settler items of the combined mixer-settler unit, it is possible, in principle, to treat the combined unit simply as a well-mixed equilibrium stage. This is done in exactly the way, as considered previously in Secs. 3.2.1 to 3.2.6. A schematic representation of an actual mixer-settler

I84

3 Modelling of Stagewise Processes ~

~

~

~

device is shown in Fig. 3.43 and an even more simplified representation of the equivalent simple well-mixed stage is given in Fig. 3.44.

Figure 3.43. Schematic representation of a mixer-settler unit.

Figure 3.44. The well-mixed stagc representation of a mixer-settler unit.

A realistic description of the dynamic behaviour of an actual mixer-settler plant item should however also involve some consideration of the hydrodynamic characteristics of the separate mixer and settler compartments and the possible flow interactions between mixer and settler along the cascade. The notation for separate mixer-settler units is shown in Fig. 3.45, for stage n of the cascade.

185

3.3 Stagewise Mass Transfer

Figure 3.45. cascade.

The separate mixer and settler compartments for stage n of a mixer-settler

In this representation, the heavy phase with a flow rate, Ln-1, enters the mixer from the preceding stage n-1, together with solvent flow, G n + l , from stage n+l. The corresponding phase flow rates, in the dispersion, leaving the mixer and entering the settler are shown as Lmn and Gmn and with concentrations X m n and Ymn respectively. This is to allow for possible changes in the volumetric holdup of the mixer following changes in flow rate. The modelling of the separate mixer and settler compartments follows that of Wilkinson and Ingham (1983).

Mixer Dynamics Owing to the intensive agitation conditions and intimate phase dispersion, obtained within the mixing compartment, the mixer can usually be modelled as a single, perfectly mixed stage in which the rate of mass transfer is sufficient to attain equilibrium. As derived previously in Sec. 3.3.1.3, the component balance equations for the mixer, based on the two combined liquid phases, is thus given by

where subscript m refers specifically to the conditions, within the mixer and hence to the effluent flow, leaving the mixer and entering the settler. The total mass balance for the mixer is expressed by Rate of change o Mass (mass in the mixer? = (flow in)

-

Mass (flow out)

Neglecting the effects of any density changes, the total mass balance then provides the relationship for the change of total volume in the mixer with respect to time.

186

3 Modelling of Stagewise Processes

Under well-mixed flow conditions, it is reasonable to assume that the mixer holdup volumes, VLmn and VGmn will vary in direct proportion to the appropriate phase flow rate, and that the total liquid holdup in the mixer will vary as a function of the total flow rate to the mixer. The total flow rate, (Lmn + Gmn), leaving the mixer will be related to the total ~ a~ hydrostatic equation, which will depend on phase volumes VLm and V G by thc net difference in the head of liquid, between the levels in the mixer and in the settler. The actual form of this relationship might need to be determined experimentally, but could, for example, follow a simple, square-root relationship of the form, in which flow rate is proportional to the square root of the difference in liquid head, or indeed to the total volume of liquid in the mixer, e.g.,

The further assumptions are that the respective phase volumes are in direct proportion to the phase flow rate, i.e.,

and the concentrations leaving the mixer are in equilibrium, according to

These equations coinplete a preliminary model for the mixer. Note that it is also possible, in principle, to incorporate changing density effects into the total inass balance equation, provided additional data, relating liquid density to concentration are available.

Settler Dynamics The simplest settler model is that in which it is assumed that each phase flows through the settler in uniform plug flow, with no mixing and constant velocity, with the effect that the concentrations leaving the settler, Xn and Y,, are simply and Y,,,,. In this the time delayed values of the exit mixer concentrations, the magnitude of the time delay is thus siinply the time required for the phase to pass through the appropriate settler volume. The outlet phase flow rates, L, and G,,, may again be related to the inlet settler phase flow rates, L,,,, and G,,,,,, and settler phase volumes by hydraulic

187

3.3 Stagewise Mass Transfer -~

~~~

~-

~~

considerations, using similar formulations to those proposed for the mixer, as required. In practice, some mixing will, however, occur in each phase of the settler, and various models involving either an arbitrary number of perfect mixing stages or various flow combinations, with and without recycle effects, can be postulated. Some of these are indicated in Fig 3.46, where Fig 3.46(A) represents settler mixing, given by a series of stirred tanks, Fig 3.46(B) a series of well-mixed tanks interconnected to stagnant regions and Fig 3.46(C) a series of two tanks with recycle. The actual representation adopted for a given situation, would, of course, have to depend very much on the actual mechanical arrangement and flow characteristics of the particular settler design, together with actual observations of the flow behaviour.

Figure 3.46. Alternative settler-flow representations for the separate phases.

Fig. 3.47 shows one possible representation, in which a proportion of each phase, passes through the settler in plug flow, while the remaining proportion is well mixed. The resultant outlet concentration from the settler is then given by the combined plug-flow and well-mixed flow streams.

188

3 Modelling of Stagewise Processes .

~

~~

Ymn

I, -

, 'yl

I I

,

I

Y-phase

I I

~-

-~

I I I

I

I

G n , Yn

~

Plug - flow

I

I

I

I

I

I

Ihase

..

I

I I I

I I I

I

I

Figure 3.47. Combined plug-flow and well-mixed settler-flow rcprcscntation.

The notation for the above flow model, in respect of the aqueous-phase settler volume, is shown in Fig. 3.48. For each phase

I I

I I

I I

i

Lmn f @

Plug flow time delay tDn

Xspn

I I I

I

Lmn xmn

I I I

I I

Mixed flow

I

Figure 3.48. Combined plug-flow and well-mixed flow representation for the heavy phase settler flow.

3.3 Stagewise Mass Transfer

189

If Lmn is the volumetric flow rate of the heavy phase entering the settler from the mixer, and f is the fraction of flow passing through the plug-flow region with time delay tDn, then f Lmn is the volumetric flow passing through the plugflow region. The concentration at the plug-flow region outlet, Xspn, is the inlet concentration at time t - tDn and is given by

The fractional flow rate, (1 - f) Lmn, is then also the volumetric flow passing through the well-mixed region of settler phase volume, Vmix. The flows leaving the plug-flow and well-mixed regions, Xspn and Xsmn, respectively, then combine to give the actual exit concentration from the settler, Xn. The model equations for the heavy phase settler region then become for the well-mixed region

and for the combined outlet phase flow

Mixer-Settler

Cascade

The individual mixer and settler model representations can then be combined into an actual countercurrent-low, multistage, extraction scheme representation as shown in Fig. 3.49 as simulated by Ingham and Dunn (1974). This includes an allowance for backmixing, between the stages of the cascade caused by inefficient phase disengagement in the settlers, such that a fraction, fL or fG, of the appropriate phase flow leaving the settler is entrained and actually carried back, in the reverse direction along the cascade, by entrainment in the other phase. It is assumed, for simplicity, that the total flows of each phase, L and G, remain constant throughout all the stages of the cascade. Entrainment fractions, fL and fG, are also assumed constant for all settlers. The above conditions, however, are not restrictive in terms of the capacity of the solution by digital simulation.

I90

3 Modelling of' Stagewise Proccsscs __ -~

1

n

N

Figure 3.49. Multistage mixer-settler cascade with entrainment hackmixing. M and ni refers 10 mixer a n d S and s to scttlcr; stagers are 1 to N.

The conditions in the mixer of any stage n are represented in Fig. 3.50. Allowing for the additional flow contributions due to the entrainment backmixing, the component balance equations, for any mixer, n, along the cascade, are now expressed by

191

3.3 Stagewise Mass Transfer

Figure 3.50. One mixer-settler stage, n

and X m n " = feq(Yrnn)

The settler equations are as shown previously, but must, of course, be applied to both phases. Fig. 3.51 shows the output obtained from a full solution of the mixer-settler model equations, including entrainment backmixing, and combined plug flow and mixing in the settler. The simulation started from an initial steady-state condition and followed the resultant change in the aqueous phase stage concentrations, following a step change in feed concentration. The effect of the time delay in the settlers, as the disturbance, as propagated through the system from stage to stage, is very evident. The above model of settler flow behaviour, combined with entrainment backmixing was used by Aly (1972) to model the unsteady-state extraction of copper from aqueous solution, using Alamine 336 solvent. An identification procedure for the relevant flow parameters showed an excellent fit to the experimental data with very realistic entrainment backmixing factors, f L = f G = 3.5 percent, the fraction of well-mixed flow in the settlers, a x = a y = 5 percent and an overall mass transfer capacity coefficient, Ka = 25 s-'.

192

3 Modelling of Stagewise Processes -

y2 y3 y4 y5

4.0

-

~~

-\.

xs1

\--

3.0

.

.

L

0

5

Concentrations

I

I

10

-

xs2

I

I

15

20

I

25

I

30

35

Time

Figure 3.51.

Computer simulation output for a five stage mixer-settler

cascade with

entrainment.

3.3.1.10 Staged Extraction Columns A wide variety of extraction column forms are used in solvent extraction

applications and many of these, such as rotary-disc contactors (RDC), OldshueRushton columns, and sieve-plate column extractors, have rather distinct compartments and a geometry, which lends itself to an analysis of column performance in terms of a stagewise model. As the compositions of the phases do not come to equilibrium at any stage, however, the behaviour of the column is therefore basically differential in nature. At the prevailing high levels of dispersion normally encountered in such types of extraction columns, the behaviour of these essentially differential type contactors, however, can be represented by the use of a non-equilibrium stagewise model. The modelling approach to multistage countercurrent equilibrium extraction cascades, based on a mass transfer rate term as shown in Sec. 1.4, can therefore usefully be applied to such types of extractor column. The magnitude of the

193

3.3 Stagewise Mass Transfer

mass transfer capacity coefficient term, now used in the model equations, must however be a realistic value, corresponding to the hydrodynamic conditions, actually existing within the column and, of course, will be substantially less than that leading to an equilibrium condition. In Fig. 3.52, the column contactor is represented by a series of N nonequilibrium stages, each of which is of height H and volume V. The effective column height, Z, is thus given by Z = N H.

I

NoEndstageo mass transfer

1

Ho

I

1

Xn Xn+1

Yn Yn-1

I

Model representation Figure 3.52. Model representation of a non-equilibrium staged extraction column.

The stagewise model with backmixing is an essential component of any model representation of a stagewise extraction column. As shown in Sec. 3.3.1.5 the non-ideal flow behaviour is represented by the presence of the N stages in

194 ~

~~

~

~

_

_

_

3 Modelling of Stagewise Processes ~-~

series and the constant backflow contributions, LB and GB, as indicated in Fig. 3.52 by the dashed lines, appropriate for each phase. Special attention has to be given to the end compartments of an extraction column, since the phase inlet and outlet points are usually located at different points of the column and are complicated by the presence of phase distributors and at one end by the coalescence zone for the dispersed phase droplets. In Fig 3.52, the end sections are represented quite simply as well-mixed zones, in which some limited degree of mass transfer may be present, but at which the mass transfer rate is much lower than in the main body of the column. The standard equations for a stagewise extraction cascade with backmixing as developed in Sec. 3.3.1.5 are

Here, an is the interfacial area per unit volume. In extraction column design, the model equations are normally expressed in terms of superficial phase velocities, L' and G', based on unit cross-sectional area. The volume of any stage in the column is then A H, where A is the crosssectional area of the column. Thus the volume occupied by the total dispersed phase is h A H, where h is the fractional holdup of dispersed phase, i.e., the droplet volume in the stage, divided by the total volume of the stage and thc volume occupied by the continuous phase, in the stage, is (I-h) A H. Taking the phase flow rate, G', to represent the dispersed phase, the component balance equations now become, for any stage n

In thc above equations, KL is the overall mass transfer coefficient (based on phase L), a is the specific interfacial area for mass transfer related to unit column volume, X and Y are the phase solute concentrations, X" is the

195

3.3 Stagewise Mass Transfer ~~

~~~~~

~~

equilibrium concentration corresponding to concentration Y and subscript n refers to stage n of the extractor. Normally the backmixing flow rates LB and GB are defined in terms of constant backmixing factors, a~ = LB/L and CXG= GB/G. The mass balance equations then appear in the form

H,h

dY dt

= G'(1

+

a ~ ) Y , + j - G'(1

+

2 a ~ ) Y n+ a~ G'Y,-j

-

Qn

Considering the end regions of the column as well-mixed stages, with small but finite rates of mass transfer, component balance equations can be derived for end stage 0

and for end stage N

The correct modelling of the end sections is obviously of great importance, and depending on the geometrical arrangement, it is possible to consider the column end sections as combinations of well-mixed tanks, exterior to the actual column.

3 . 3 . 1 . 1 1 Column Hydrodynamics Under changing flow conditions, it can be important to include some consideration of the hydrodynamic changes within the column (Fig. 3.53), as manifested by changes in the fractional dispersed phase holdup, h,, and the phase flow rates, L, and G,. which, under dynamic conditions, can vary from stage to stage. Such variations can have a considerable effect on the overall dynamic characteristics of an extraction column, since variations in h, also

196

_

_

3 Modelling of Stagewise Processes

~

-

affect the solute transfer rate terms Q n , by virtue of the corresponding variation in the specific interfacial area, an.

N

%+I

Ln

t'

LN

Figure 3.53. Hydrodynamic model representation of an extraction column.

A dynamic balance for the dispersed phase holdup in stage n gives

Since liquid phases are incompressible

and for the overall column

L

+

G = LN

+

GI

The fractional dispersed phase holdup, h, is normally correlated on the basis of a characteristic velocity equation, which is based on the concept of a slip velocity for the drops, v,lip, which then can be related to the free rise velocity of single drops, using some correctional functional dependence on holdup, f(h).

197

3.3 Stagewise Mass Transfer

The normal method of correlating dispersed phase holdup is normally of the form

where Vchar is the characteristic velocity for the dispersed phase droplets. Knowing the value of Vchar, the value of h can be determined for any values of L' and G', using an iterative procedure. Under normal circumstances, the use of a characteristic velocity equation of the type shown above can cause difficulties in computation, owing to the existence of an implicit algebraic loop, which must be solved, at every integration step length. In this the appropriate value of Ln or Gn satisfying the value of hn generated in the differential mass balance equation, must be found as shown in the information flow diagram of Fig. 3.54.

t I

z

? I

t 1 Ln

t 1 2 1 1 1 t

dhn vn dt

Gn Gn+l-Gn

zz $

Gn

1

s1

z z t

1

2 1

- _L' - -

G'

- "char f(h)

Gn

1

Figure 3.54. Implicit loop calculation of dispersed-phase holdup.

An alternative approach is to use explicit relationships to correlate the experimental data, which are then immediately solvable by the computer. This usually means some prior manipulation of the data. For example, the dispersed

198

3 Modelling of Stagewise Processes -~

~

- -

-~

phase holdup data could be expressed as direct polynomial expressions for the variation of h with G' and L' as in Fig. 3.55.

h

G' Figure 3.55. Correlation of dispersed phase fractional holdup values with aqueous ( L ) rind solvent (G') flow rates.

The resulting calculation procedure is then straightforward and the variation of holdup in the column may be combined with the component balance equations to give a model for a countercurrent extraction column, incorporating both concentration and flow rate changes, together with the effects of flow on column dispersed phase holdup. It is interesting to note that the modelling approach here differs from the purely scientific approach in which the use of a characteristic velocity equation in an attempt to relate holdup to a fundamental factor such as a single drop free rise velocity, whereas the modelling approach is to get the job done by the simplest method possible. Thc above modelling approach to the column hydrodynamics is illustrated by the simulation example HOLDUP.

3.3.2

Stagewise Absorption

Gas-liquid contacting systems can be modelled in a manner similar to liquidliquid contactors. There are however some modelling features which are peculiar to gas-liquid systems. The single well-mixed contacting stage is shown in Fig 3.56. In this, G and L are the volumetric flow rates of the two phases, and X and Y are the concentrations of any component in each phase. Q is the transfer rate of the component.

199

3.3 Stagewise Mass Transfer

GO3YO

Figure 3.56. Well-mixed gas-liquid contacting stage.

For single-solute, gas-liquid mass transfer, the component balances are as before

+ Q

where VG is the volume of the well-mixed gas phase, and VL is the volume of the well-mixed liquid phase. In the preceding solvent extraction models, it was assumed that the phase flow rates L and G remained constant, which is consistent with a low degree of solute transfer relative to the total phase flow rate. For the case of gas absorption, normally the liquid flow is fairly constant and Lo is approximately equal to L1, but often the gas flow can change quite substantially, such that Go no longer equals G I . For highly concentrated gas phase systems, it is therefore often preferable to define flow rates, L and G, on a solute-free mass basis and to express concentrations X and Y as mass ratio concentrations. This system of concentration units is used in the simulation example AMMONAB. The transfer term Q is written as

where a is the transfer surface per volume of liquid, KL is the overall mass transfer coefficient for phase L, VL is the liquid phase volume and Xi * is given by the equilibrium relation Xi" = feq 0'1)

200 ~-

~

3 Modelling of Stagewise Processes

_ _ ~

As shown for the case of extraction, a high value of KLa will result in X I approaching very close to the value Xi*, and therefore the outlet concentrations of the two phases will be close to equilibrium. Owing to the substantially greater density of liquids, as compared to gases, the volumetric flow rate of the gas is usually much greater than that of the liquid, or C >> L. Also as a general consequence

meaning that ZL

>>

TG

The significance of the large difference in the relative magnitudes of the time constants, for the two phases, is that the gas concentrations will reach steady state much faster than the liquid phase. In the component balance equations, dY 1 /dt will therefore be zero, whereas dXl/dt may still be quite large. This can obviously cause considerable difficulties in the integration procedure, owing to equation stiffness. For gas absorption, this problem can often be circumvented by the assumption of a quasi-steady-state condition for the gas phase. In this, the dynamics of the gas phase are effectively neglected and the steady state, rather than the dynamic form of component balance is used to describe the variation in gas phase concentration. The gas phase balance then becomes for the above situation

0 = Go Yo

-

G1 Y1

-

Q

Hence

Thus Y1 is obtained not as the result of the numerical integration of a differential equation, but as the solution of an algebraic equation, which now requires an iterative procedure to determine the equilibrium value, Xi *. The solution of algebraic balance equations in combination with an equilibrium relation has again resulted in an implicit algebraic loop. Simplification of such problems, however, is always possible, when XI* is simply related to Y 1, as for example

Combining the two equations then gives an explicit solution for concentration Y 1 and hence also X I

3.3 Stagewise Mass Transfer

20 1

and the implicit algebraic loop is eliminated from the solution procedure. Assuming equilibrium conditions and a linear equilibrium relationship, where Y1 = m X1, and a quasi-steady-state conditions in the gas with dYl/dt = 0 to be achieved, a component balance for the entire two phase system of Fig. 3.56, gives

which can be expressed as

This equation has the form

where the time constant for the system, 2, is thus shown to be dependent on the value of the equilibrium constant, m. Since the value of m depends on the nature of the particular solute concerned, this has the consequence that in multicomponent applications the value of the time constant will vary according to the system component. This can cause problems of equation stiffness in the solution of the often quite large sets of simultaneous, multicomponent balance equations. The importance of eliminating unnecessary stiffness, by careful consideration of the relative magnitudes of the various system time constants, thus becomes very apparent.

3.3.3

Stagewise Distillation

3 . 3 . 3 . 1 Simple Overhead Distillation A simple overhead topping distillation process, without fractionation, is illustrated in Fig. 3.57.

202 -~

3 Modelling of Stagewise Processes ~~

-

~

~~

Figure 3.57. Model representation of a simple overhead distillation.

The total mass balance is given by Rate of accumulation

( of mass in the still ) = (Ratetoofthemassstillinput giving

where M is the total moles of liquid in the still and V is the vapour removal rate in molesltime. For a simple binary distillation, the component balance equation becomes

d(M X A ) dt

= - vYA

where X A and YA are the liquid and vapour mole fraction of component A o f the liquid and vapour phases, respectively, where A is the more volatile component. The relative volatility, a , is usually related to the compound having the higher boiling point, which in this case is B and hence

Assuming that the liquid and vapour compositions in the still are in equilibrium, i.e., that the still acts as a theoretical stage

203

3.3 Stagewise Mass Transfer

or in terms of relative volatility, a

The combination of the two mass balance equations, together with an explicit form of equilibrium relationship gives a system that is very easily solvable by direct numerical integration, as demonstrated in the simulation example BSTILL. Extending the method to a multicomponent mixture, the total mass balance remains the same, but separate component balance equations must now be written for each individual component i, i.e.,

xi = dM ~dt

-

vyj

and where now the equilibrium condition is given by

Again solution is straightforward, as illustrated in the simulation examples MUBATCH, DIFDIST and MCSTILL.

3 . 3 . 3 . 2 Binary Batch Distillation A batch distillation represents a complete dynamic process since everything, apart from the geometry of the column and the nature of the equilibrium relationship, varies with time. Owing to the removal of a distillate containing more of the volatile component, the compositions of the vapour and the liquid on all plates of the column vary with time. The total quantity of liquid in the still decreases with time, and its composition becomes successively depleted in the more volatile component. This makes the separation more and more difficult, requiring the use of higher reflux ratios to maintain a high distillate composition. The increased reflux increases the liquid flow down the column, and hence the liquid holdup on each plate. As a result of the increasing concentration of less volatile component in the still, the still temperature increases during distillation, thus reducing the rate of heat transfer to the still by reducing the temperature driving force in the reboiler and hence reducing the vapour boil-up rate. Despite this, conventional textbooks persist in analysing batch distillation in terms of quasi-steady-state graphical techniques applied at different concentration levels during the distillation process. These are also

204

3 Modelling of Stagewise Processes

. _

__

.___

based on rather idealised and unrealistic conditions of operating a batch distillation process, i.e.: 1. Distillation at constant reflux ratio but varying top product composition. 2. Distillation at constant top product composition but varying reflux ratio.

Compared to this a solution approach based on digital simulation is much more realistic. Consider the binary batch distillation column, represented in Fig. 3.58, and based on that of Luyben (1973, 1990). The still contains MB moles with liquid mole fraction composition X B . The liquid holdup on each plate n of the column is M, with liquid composition x, and a corresponding vapour phase composition y,. The liquid flow from plate to plate L, varies along the column with consequent variations in M,. Overhead vapours are condensed in a total condenser and the condensate collected in a reflux drum with a liquid holdup volume MD and liquid composition XD. From here part of the condensate is returned to the top plate of the column as reflux at the rate Lo and composition X D . Product is removed from the reflux drum at a composition XD and rate D which is controlled by a simple proportional controller acting on the reflux drum level and is proportional to MD. For simplicity the following assumptions are made, although a more general model could easily be derived, in which these assumptions could be relaxed. 1. The system is ideal, with equilibrium described by a constant relative volatility, the liquid components have equal molar latent heats of evaporation and there are no heat losses or heat of mixing effects on the plates. Hence the concept of constant molar overflow (excluding dynamic effects) and the use of mole fraction compositions are allowable. 2. The liquid volumes in the still, reflux drum and on the column plates are well-mixed regions of uniform composition. 3. The dynamics of the overhead pipework and condenser are negligible. 4. The dynamics of the vapour phase in the column are much faster than that of the liquid phase and are neglected. 5. The still provides a constant vapour boil-up rate, which remains constant with respect to time. 6. The column plates have 100 percent plate efficiency and act as theoretical plates.

Since the vapour phase dynamics are negligible, the vapour flow rate through the column is constant from plate to plate, at the rate of V, k mol/s. The liquid flow rates L,, and the liquid holdup on the plate, however, will vary, under changing hydrodynamic conditions in the column. The corresponding notation, for any plate n in the column, is as indicated in Fig. 3.58.

3.3 Stagewise Mass Transfer

205

Figure 3.58. Model representation of a batch distillation column and typical plate n, as per Luyben (1973).

Total mass and component mass balance equations are written for all the plates of the column, for the still and for the top reflux drum. Here L, G and D are the molar flow rates and x and y are mol fraction compositions For plate n the total mass balance is given by _ dMn _dt - Ln-1 + Vn+l - Ln - Vn Since vapour phase dynamics are neglected and "Constant Molal Overflow" conditions also apply, Vn+l = V, = V, and

206 ~-~~~ -.

3 Modelling of Stagewise Procecxes -

-~

~~

The component balance equation is given by

The corresponding equations for the boiler are

and

where N refers to conditions on the bottom plate of the column. For the reflux drum

and

where Lo/D = R is the reflux ratio for the column. Assuming theoretical plate behaviour, i.e., equilibrium between the gas and liquid phases, for plate n a x n -~ YN = i+(a - 1) x, where a is the relative volatility. The above equation also applies to the liquid and vapour compositions of the still, where equilibrium plate behaviour is again assumed. The maintenance of constant liquid level in thc reflux drum can be expressed by the following proportional control equation

D = Kp (MD

-

MD(set))

where Kp is the controller gain and MD(,,~) is the level controller set point. The hydraulic condition of the plates is represented in Fig 3.59.

3.3 Stagewise Mass Transfer

207

Figure 3.59. Model representation of the plate hydraulics

A mass balance for the liquid on plate n is given by

In this simplified model, it is assumed that liquid may leave the plate, either by flow over the weir Ln(we,r)or by weepage Ln(weep). Both these effects can be described by simple hydraulic relations, in which the flow is proportional to the square root of the available hydrostatic liquid head. The weir flow depends on the liquid head above the weir and hence

Ln(weir)is zero for the condition Mn< Mns. M, is the mass of liquid on plate n, and Mns is the mass of liquid on the plate corresponding to the weir height or static liquid holdup on the plate. The rate of loss of liquid from the plate by weepage, however, will depend on the total mass of liquid on the plate

The total flow of liquid from the plate, is therefore given by

where K1 is an effective weir discharge constant for the plate, K2 is the weepage discharge constant and Mns is the static holdup on the plate.

208

3 Modelling of Stagewise Processes ~.

3 . 3 . 3 . 3 Continuous Binary Distillation The continuous binary distillation column of Fig. 3.60 follows the same general representation as that used previously in Fig. 3.58. The modelling approach again follows closely that of Luyben (1973, 1990).

v, Y l

4

water

1 1 11

bDXD

vB, YB 4 f

I

4

Steam

\ I

Figure 3.60. Model representation of a continuous binary distillation column. PC is the cooling water controller, LC the reflux controller.

3.3 Stagewise Mass Transfer

209

The relationships for the section of column above the feed plate, i.e., the enriching section of the column, are exactly the same as those derived previously for the case of the batch distillation column. The mass balance relationships for the feed plate, the plates in the stripping section, of the column and for the reboiler must, however, be modified, owing to the continuous feed to the column and the continuous withdrawal of bottom product from the reboiler. The feed is defined by its mass flow rate, F, its composition XF and the thermal quality or q-factor, q. The column bottom product is defined by its mass flow rate, W, and composition, xw and is controlled to maintain constant liquid level in the reboiler. The liquid and vapour molar flow rate in the enriching section, are denoted by L and V, as previously and in the stripping section as L' and V'. The relationship between L, V, L' and V' is determined by the feed rate F and the thermal quality of the feed "q". The situation on the column feed plate, plate f, is shown in Fig. 3.61. Neglecting plate hydraulic dynamics allows all the liquid and vapour flow rates Li and Vi with i < f, in the top section of the column, to be set to L and V and with all i L f, in the bottom section of the column, to be set to L' and V'.

Figure 3.61. Column feedplate flow rates and compositions.

Assuming constant liquid holdup on the plate, which is equivalent to neglecting the plate hydraulic variations, a component balance for the feed plate is given by

210

3 Modelling of Stagewise Processes ~~

-

where

--.

L' = L

+

qF

and where for a saturated liquid feed, with q = 1

L ' = L + F

and

v

= V'

For any plate, n, above the feed, as shown previously dx M --ndt = L ( ~ ~ --1 x,)

+

V (yn+l

-

y,)

and for any plate m, in the stripping section, below the feed

For the reboiler

M BdXB T ~= L'XN - W XB

-

V'YB

The controller equations

are also required to complete the model. The relationships around the top part of the column and control of reflux drum level remain the same as those for the batch situation, described in Sec. 3.3.3.2.

3 . 3 . 3 . 4 Multicomponent Separations As discussed previously in Sec. 3.3.1.7, each additional component of the feed mixture must be expressed by a separate component mass balance equation and by its own equilibrium relationship. Thus for component i of a system of j components, the component balance equation, on the nth plate, becomes

21 1

3.3 Stagewise Mass Transfer

where i = 1 to j. Assuming the equilibrium to be expressed in terms of relative volatilities CXi and theoretical plate behaviour, the relation between the vapour and liquid mole fraction compositions leaving the plate is given by Yin =

a in

Xin

J

C a i n xin 1

or where the equilibrium can be based in terms of K values, the relationship becomes Yin = Kin Xin

Using the above form of equilibrium relationship, the component balance equations now becomes

As explained by Franks (1972), this again shows that the component balance equations, for the different components of the mixture, will thus have different time constants, which depend on the relative magnitudes of the equilibrium constants Ki and which again can lead to possible problems of numerical stiffness. Again one way of dealing with this is to replace those differential rate equations, having low time constants (i.e., high K values) and fast rates of response, by quasi-steady-state algebraic equations, obtained by setting M -dxi = 0 dt and effectively neglecting the dynamics in the case of those components, having very fast rates of response.

212

3 Modelling of Stagewise Processes

3 . 3 . 3 . 5 Plate Efficiency The situation for any plate n, with liquid composition x n corresponding to an equilibrium vapour composition yn*, but with actual vapour composition yn, is represented on a small section of the McCabe-Thiele diagram in Fig. 3.62.

Figure 3.62. Actual and theoretical plate compositions for plate n

The actual plate efficiency can be defined as

q =

actual change of composition maximum possible change of composition

_-

where

Hence by simple algebra

Additional equations, as above, can thus be used to correct the values of Yn*, obtained from the equilibrium data to give actual plate values, Yn.

213

3.3 Stagewise Mass Transfer

3.3.3.6

Complex Column Simulations

More complex situations where ideal behaviour can no longer be assumed require the incorporation of activity coefficient terms in the calculation of the equilibrium vapour compositions. Assuming ideal behaviour in the gas phase, the equilibrium relation for component i is

y , - Yi xi Pi I P where P is the total pressure in the column. Since 'the saturated vapour pressure of the pure compound i, Pi, is a function of temperature, the calculation of the equilibrium vapour composition requires that a plate temperature must be determined such that the condition Zy, = 1 is obtained. An example of this technique is illustrated, in Sec. 3.3.4 based on steam distillation and in the corresponding simulation example STEAM and also in the simulation example BUBBLE. Furthermore heat effects on the plates may also have to be accounted for, by means of a dynamic heat balance for each plate, including allowances for the enthalpies of the liquid and vapour streams, entering and leaving the plate, heat of mixing, etc. This thus represents a much more complicated and timeconsuming computational procedure, than has been considered so far. In such cases, it obviously becomes much more meaningful to employ larger simulation packages, with their sophisticated physical property data bases and estimation procedures. The general principles of the modelling procedure, however, remain very much the same.

3.3.3.7

Multicomponent Equilibria

The condition for boiling, of a liquid mixture, is that the sum of the partial pressures of the components of the liquid phase mixture is equal to the total pressure of the system cpi = P where P is the total pressure. The partial pressures, pi, are related to liquid phase mole fraction composition, Xi, and vapour pressure of the pure compound Pi, and activity coefficient, yi by

pi = xi yi Pi where the activity coefficient is a function of composition and vapour pressure is a function of temperature

214 ~

3 Modelling of Stagewiw Processes ~.

~

and

Pi = f(T)

As discussed by Franks (1972), in order to solve this system of equations, a value of temperature T must be found to satisfy the condition that the difference term 6 = P - Epi is very small, i.e., that the equilibrium condition is satisfied. This is known as a bubble point calculation. The above system of defining equations, however represent, an implicit algebraic loop and the trial and error solution procedure can be very time consuming, especially when incorporated into a dynamic simulation program. An information flow diagram showing the iteration for 6 is shown in Fig. 3.63. Examples of multicomponent equilibria with activity coefficients are given in the simulation examples BUBBLE and STEAM.

I

t

I

L

I

Revised T

Figure 3.63. Information flow diagram for multicomponent equilibrium determination.

3.3.4

Multicomponent Steam Distillation

Steam distillation is a process whereby organic liquids may be separated at temperatures sufficiently low to prevent their thermal decomposition or whereby azeotropes may be broken. Fats or perfume production are examples of applications of this technique. The vapour-liquid equilibria of the threephase system is simplified by the usual assumption of complete immiscibility of the liquid phases and the validity of the Raoult and Dalton laws. Systems containing more than one volatile component are characterised by complex dynamics (e.g., boiling point is not constant). Steam distillation is normally carried out as a semi-batch process whereby the organic mixture is charged into the still and steam is bubbled through continuously, as depicted in Fig. 3.64.

215

3.3 Stagewise Mass Transfer

--b

TE

Q

Figure 3.64. Schematic drawing of the apparatus for steam distillation.

As discussed, modelled and simulated by Prenosil (1976), the dynamics of the process bring in the question of steam consumption, steam flow rate, starting time of the distillation, and shut-down time when the desired degree of separation has been reached. The modelling of steam distillation often involves the following assumptions.

1) 2) 3) 4)

Ideal behaviour of all components in pure state or mixture. Complete immiscibility of the water and the organic phases. Zero temperature gradients in the bulk phases (ideal mixing in the boiler). Equilibrium between the organic vapour and its liquid at all times.

The mathematical model is divided into two time periods: The heating period until boiling point is reached. (i) (ii) The distillation period after boiling has started.

Heating Period To describe the dynamic behaviour of this semi-batch process, unsteady-state mass and energy balances are needed. Their interrelationships are depicted in Fig. 3.65.

216

3 Modelling of Stagewise Processes

Heat transfer

Enthalpy

T

balance

Mass balance

Molar heat

Figure 3.65. Information flow diagram for the heating period.

For the water phase,

Here, it is assumed that all the steam condenses in the distillation vessel. In this period, the organic phase component masses remain constant. The rate of heat accumulation is balanced by the heat of condensation and the heat losses. An energy balance therefore gives

The enthalpy changes are calculated from molar heat capacities given by the usual functions of temperature, according to

from which it follows that

217

3.3 Stagewise Mass Transfer

The heat transfer Q to the surroundings is calculated from the simple relation Q = UA(TE-T)

The solution of the above model gives the temperature of the mixture at any time during the heating period. Distillation Period The distillation starts when the boiling point is reached. Then a vapour stream at flow rate V is obtained, which condenses as a distillate. The mass balances can be written as follows: For water

and for the total organic phase

The energy balance is now

The vapour enthalpies are calculated from the molar heat capacity functions for the vapour components and the latent heats of vaporisation at standard temperature. The vapour overflow, V, is then obtained from the energy balance as

218

3 Modelling of Stagewise Processes

Phase Equilibria Assuming ideal liquid behavior, the total partial pressure of the organic phase is given by the sum of the partial pressures of its components according to Raoult's law.

where Pi is the vapour pressure of pure component i. For non-idealities, this must be modified with appropriate activity expressions (Prenosil (1976). For water, the vapour pressure varies only with temperature so that Pw = p w

Boiling will commence when the sum of the organic partial pressures and the water vapour pressure is equal to the total pressure, or in terms of the mole fractions

cyi+y w = 1 As boiling proceeds, the loss of the lightest organic vapours will cause the boiling point to increase with time. The vapour pressures Pi and Pw of the pure components can be calculated using the Antoine equation lOgP = A

-

B C+T

~

The highly interactive nature of the balance and equilibria equations for the distillation period are depicted in Fig. 3.66. An implicit, iterative algebraic loop is involved in the calculation of the boiling point temperature at each time interval. This involves guessing the temperature and calculating the sum of the partial pressures, or mole fractions. The condition required is that Ey, + y, = 1. The iterative loop for the bubble point calculation is represented by the five interconnected blocks in the lower right hand corner of Fig. 3.66. The model of Prenosil (1976) also included an efficiency term E for the steam heating, dependent on liquid depth L and bubble diameter D.

219

3.3 Stagewise Mass Transfer -.

___

-

CTS

T

Water enthalpy

enthalpy

r w

HLw

/-----\ Heat transfer

balance

enthalpies

T

enthalpies gas and liquid

Total mass balance

'"

organic material

balance

IYi D

Steam

I

J

(=el fraction in vapour

Figure 3.66. Information flow diagram for the distillation period.

Chemical Engineering Dynami :Modelling

with PC Simulation

John Inph;m. Iruine J. Dunn. Elmar Heinzlc &JiiiE. Picnoul

copyright OVCH VerlagsgesellschaftmbH, 1994

Differential Flow and Reaction Applications

Introduction 4.1.1

Dynamic Simulation

The main process variables in differential contacting devices vary continuously with respect to distance. Dynamic simulations therefore involve variations with respect to both time and position. Thus two independent variables, time and position, are now involved. Although the basic principles remain the same, the mathematical formulation, for the dynamic system, now results in the form of partial differential equations. As most digital simulation languages permit the use of only one independent variable, the second independent variable, either time or distance is normally eliminated by the use of a finite-differencing procedure. In this chapter, the approach is based very largely on that of Franks (1967), and the distance coordinate is treated by finite differencing. In this procedure, the length coordinate of the system is divided into N finitedifference elements or segments, each of length AZ, where N times AZ is equal to the total length or distance. It is assumed that within each element, any variation, with respect to distance, is relatively small. The conditions at the midpoint of the element can therefore be taken to represent the conditions of the element as a whole. This is shown, in Fig 4.1, where the average concentration of any element n, is identified by the midpoint concentration C,. The actual continuous variation in concentration with respect to length is therefore approximated by a series of discontinuous variations. The dynamic behaviour of element n is affected by the conditions in its neighbouring elements n- 1 and n+ 1 and each original partial differential equation is approximated by a system of N simultaneous difference differential equations. In practice, the length of each element AZ may be kept constant or may be varied from segment to segment. A greater number of elements usually improves the approximation of the profile, but the computational effort required is also greater. The approach is demonstrated in the simulation examples DISCRET, AXDISP, MEMSEP, HEATEX, DRY, ENZDYN and BEAD. Note that owing to the large number of equations, some of the above

4 Differential Flow and Reaction Applications

222

__

examples are rather slow in computation and when carrying out multiple studies on a particular example it may be preferable to start from some prior determined steady state rather than from initially zero conditions for each simulation run.

-

L

'

Figure 4.1. Finite-differencing a tubular reactor with the stepwise approximation of the continuous concentration profile.

4.1.2

Steady-State Simulation

Under steady-state conditions, variations with respect to time are eliminated and the steady-state model can now be formulated in terms of the one remaining independent variable, length or distance, In many cases, the model equations now result as simultaneous first-order differential equations, for which solution is straightforward. Simulation examples of this type are the steady-state tubular reactor models TUBE and TUBED, TUBTANK, ANHYD, BENZHYD and NITRO. Some situations, however result in the form of second-order diffferential equations, which often give rise to problems of the split boundary type. In order to solve this type of problem, an iterative method of solution is required, in which an unknown condition at the starting point is guessed, the differential equation integrated twice and the resulting solution compared with a known boundary condition, obtained at the end point of the calculation. Any error between the known value and the calculated value can then be used to revise the initial starting guess for the next iteration. This procedure is then repeated until

223

4.1 Introduction

convergence is achieved. Examples of the steady-state split-boundary type of solution are shown by the simulation examples ROD and ENZSPLIT. In order to overcome the problem of split boundaries, it is sometimes preferable to formulate the model dynamically, and to obtain the steady-state solution, as a consequence of the dynamic solution, leading to the eventual steady state. This procedure is demonstrated in examples ENZPLIT and ENZDYN.

4.2

Diffusion and Heat Conduction

This section deals with problems involving diffusion and heat conduction. Both diffusion and heat conduction are described by similar forms of equation. Fick's law for diffusion has already been met in Sec. 1.2.2 and the similarity of this to Fourier's law for heat conduction is apparent. With Ficks Law

and Fourier's Law

dT q = - k z

Here j A is the diffusional flux of component A (kmol/m2 s), D is the diffusion coefficient (m2/s), CA is the concentration of component A (kmol/m3), q is the heat transfer flux (kJ/m2 s), k is the thermal conductivity (kJ/m s K), T is the temperature (K) and Z is the distance (m). In diffusional mass transfer, the transfer is always in the direction of decreasing concentration and is proportional to the magnitude of the concentration gradient; the constant of proportionality being the diffusion coefficient for the system. In conductive heat transfer, the transfer is always in the direction of decreasing temperature and is proportional to the magnitude of the temperature gradient; the constant of proportionality being the thermal conductivity of the system. The analogy also extends to Newton's equation for momentum transport, where

4 Differential Flow and Reaction Applications

where, for Newtonian liquids: p is the viscosity, z is the shear stress, v is velocity and Z is again distance.

4.2.1

Unsteady-State Diffusion Through a Porous Solid

This problem illustrates the solution approach to a one-dimensional, nonsteady-state, diffusional problem, as demonstrated in the simulation examples, DRY and ENZDYN. The system is represented in Fig. 4.2. Water diffuses through a porous solid, to the surface, where it evaporates into the atmosphere. It is required to determine the water concentration profile in the solid, under drying conditions. The quantity of water is limited and, therefore, the solid will eventually dry out and the drying rate will reduce to zero.

Figure 4.2. Unsteady-state diffusion through a porous solid.

The movement of water through a solid, such as wood, in the absence of chemical reaction, is described by the following time-dependent diffusional equation.

where at steady state, dC/& = 0, and

0 = D-

d2C dZ2

225

4.2 Diffusion and Heat Conduction

Integrating

dC

= constant

Thus at steady state the concentration gradient is constant. Note that since there are two independent variables of both length and time, the defining equation is written in terms of the partial differentials, aCldt and aCfdZ, whereas at steady state only one independent variable, length, is involved and the ordinary derivative function is used. In reality the above diffusion equation results from a combination of an unsteady-state mass balance, based on a small differential element of solid length dZ, combined with Fick’s Law of diffusion. To set up the problem for simulation involves discretising one of the independent variables, in this case, length, and solving the time-dependent equations, obtained for each element, by means of a simulation language. By finite-differencing the length coordinate of the solid, as shown in Fig. 4.3, the drying process is approximated to that of a series of finite-differenced solid segments. Evaporation Water

Figure 4.3. Finite-differenced equivalent of the depth of solid.

The diffusional fluxes from segment to segment are indicated in Fig. 4.4.

Figure 4.4. Diffusional fluxes from segment to segment.

4 Differential Flow and Reaction Applications

226

Note that the segments are assumed to be so small, that any variation in concentration within the segment, with respect to length, can effectively be ignored. The effective concentration of the segment can therefore be taken as that at the midpoint. A component mass balance is written for each segment, where

(

Rate of accumulation in the segment

) ( Diffusional ) ) = ( Diffusional flowin flow out -

Here jn is the mass flux leaving segment n (kg/m2 s), Cn is the concentration of segment n (kg/m3), A is the cross-sectional area (m2), t is time (t) and AVn is the volume of segment n (m3). By Fick’s Law dC j = - D z with dimensions

M - L2 M L ~ T T L ~ L The concentration gradient terms, dC/dZ, both in and out of segment n, can be approximated by means of their finite-differenced equivalents. Substituting these into the component balance equation, gives

where AZ is the length of the segment and AVn = A AZ. Thus

The above procedure is applied to all the finite-difference segments in turn. The end segments (n=l and n=N), however, often require special attention according to particular boundary conditions: For example, at Z=L the solid is in contact with pure water and CN+1=Ceq, where the equilibrium concentration C,,, would be determined by prior experiment.

227

4.2 Diffusion and Heat Conduction

At the air-solid surface, Z=O, the drying rate is determined by the convective heat and mass transfer drying conditions and the surrounding atmosphere of the drier. Assuming that the drying rate is known, the component balance equation for segment 1 becomes Rate of accumulation in segment I

Rate of input from segment 2

-

Rate of drying from surface

1

Alternatively, the concentration in segment 1 C 1, may be taken simply as that in equilibrium with the surrounding air. The unsteady model, originally formulated in terms of a partial differential equation, is thus transformed into N difference differential equations. As a result of the finite-differencing, a solution can be obtained for the variation with respect to time of the water concentration, for every segment, throughout the bed. The simulation example DRY is based directly on the above treatment, whereas ENZDYN models the case of unsteady-state diffusion, when combined with chemical reaction. Unsteady-state heat conduction can be treated in an exactly analogous manner, though for cases of complex geometry, with multiple heat sources and sinks, the reader is referred to specialist texts, such as Carslaw and Jaeger (1959).

4.2.2

Unsteady-State Heat Conduction and Diffusion in Spherical and Cylindrical Coordinates

Although the foregoing example in Sec. 4.2.1 is based on a linear coordinate system, the methods apply equally to other systems, represented by cylindrical and spherical coordinates. An example of diffusion in a spherical coordinate system is provided by simulation example BEAD. Here the only additional complication in the basic modelling approach is the need to describe the geometry of the system, in terms of the changing area for diffusional flow through the bead.

4.2.3

Steady-State Diffusion with Homogeneous Chemical Reaction

The following example, taken from Welty et al. ( 1976), illustrates the solution approach to a steady-state, one-dimensional, diffusional or heat conduction problem.

228

4 Differential Flow and Reaction Applications ~-

As shown in Fig. 4.5, an inert gas containing a soluble component, S, stands above the quiescent surface of a liquid, in which the component, S is both soluble and in which it reacts chemically to form an inert product. Assuming the concentration of S at the gas-liquid surface to be constant, it is desired to determine the rate of solution of component S and the subsequent steady-state concentration profile within the liquid. Inert gas and solute S

Quiescentliquid

z = 0, cs = CSO

W Z

AZ

z k)Z+AZ

Z=L

0

L

Figure 4.5. Steady-state diffusion with chemical reaction.

Under quiescent conditions, the rate of solution of S within the liquid, is determined by molecular diffusion and is described by Fick's law, where

At steady-state conditions, the rate of supply of S by diffusion is balanced by the rate of consumption by chemical reaction, where assuming a first-order chemical reaction

229

4.2 Diffusion and Heat Conduction

with typical units of kmol/m3 s. Thus considering a small differential element of liquid volume, dV, and depth, dZ, the balance equation becomes js As = rAdV where As is the cross-sectional area of the element and dV = As dz. Hence

k Cs As dZ or -

Dd2Cs p dZ2

+ k CS

= 0

where each term has the dimensions masshime or units (kmol/s). The above second-order differential equation can be solved by integration. At the liquid surface, where Z=O, the bulk gas concentration, Cso, is known, but the concentration gradient dCs/dZ is unknown. Conversely at the full liquid depth, the concentration Cso is not known, but the concentration gradient is known and is equal to zero. Since there can be no diffusion of component S from the bottom surface of the liquid, i.e., js at Z=L is 0 and hence from Fick's Law dCs/dZ at Z=L must also be zero. The problem is thus one of a split boundary type, but one which can be solved by an iterative procedure based on an assumed value for one of the unknown boundary conditions. Assuming a value for dCs/dZ at the initial condition Z=O, the equation can be integrated twice to produce values of dCs/dZ and Cs at the terminal condition, Z=L. If the correct value has been taken, the integration will lead to the correct boundary condition that dCs/dZ=O at Z=L and hence the correct value of Cs. The value of the concentration gradient dCs/dZ is also obtained for all values of Z, throughout the depth of liquid. Further applications of this approach are given in the simulation examples, ENZSPLIT and ROD.

4.3

Tubular Chemical Reactors

Mathematical models of tubular chemical reactor behaviour can be used to predict the dynamic variations in concentration, temperature and flow rate at various locations within the reactor. A complete tubular reactor model would however be extremely complex, involving variations in both radial and axial

230

4 Differential Flow and Reaction Applications

positions, as well as perhaps spatial variations within individual catalyst pellets. Models of such complexity are however beyond the scope of this text and variations only with respect to both time and axial position are treated here. Allowance for axial dispersion is however included, owing to its very large influence on reactor performance, and the fact that the modelling procedure using digital simulation is relatively straightforward.

The Plug-Flow Tubular Reactor

4.3.1

Consider a small element of volume, AV, of an ideal plug-flow tubular reactor, as shown in Fig. 4.6.

cAO

b

L

NAO

Ffi

CAf

NAf

Figure 4.6. Component balancing for a tubular plug-flow reactor.

Component Balance Equation A component balance equation can be derived for the element AV, based on the generalised component balance expression, where for any reactant, A

[

[

Rate of Mass a c c u r t t i o n ) = n);;o:

[

-u J;;o:

Mass

+

[

Rate of formation of A by reaction

23 1

4 . 3 Tubular Chemical Reactors

The rate of accumulation of component A in element AV, is (AV dCA/dt), where, dCA/dt is the rate of change of concentration. The mass rate of flow of A into element AV is F CA, and the rate of flow of A from element AV is F CA + A(F CA), where F is the volumetric flow rate. The rate of formation of A by reaction is rAAV, where rA is the rate per unit volume. Substituting these quantities, gives the resulting component balance equation as

The above equation may also be expressed in terms of length, since

where Ac , is the cross-sectional area of the reactor. Allowing AV to become very small, the above balance equation is transformed into the following partial differential equation, where

For constant volumetric flow rate, F, throughout the reactor

ac

-

2

where F/Ac, is the superficial inear fluid velocity v, through the reactor. Under steady-state conditions

and hence, at steady state

232

4 Differential Flow and Reaction Applications

This equation can be integrated to determine the resulting steady-state variation of CA with respect to Z, knowing the reaction kinetics, rA = f(CA) and the initial conditions, CA at Z = 0. Cases with more complex multicomponent kinetics will require similar balance equations for all the components of interest. The component balance equation can also be written in terms of fractional conversion, XA, where for constant volumetric flow conditions

and CAOis the inlet reactor feed concentration. Thus

The mass balance, in terms of XA, is thus given by

Reactant A is consumed, so rA is negative, and the fractional conversion will increase with Z. In terms of molar flow rates

and

where NAOis the molar flow of reactant A entering to the reactor.

Energy Balance Equation The energy balance, for element AV of the reactor again follows the generalised form, derived in Sec. 1.2.5. Thus ’

Rate of accumulation of energy

(

)

=

Rate of energy

to heat incoming stream from T to T+AT

Rate of energy generated at T+AT

Rate of transfer

233

4 . 3 Tubular Chemical Reactors

Cin

AZ Figure 4.7. Energy balancing for the tubular plug-flow reactor.

Referring to Fig. 4.7, the general energy balance for segment n with S components and R reactions is

When in some more complex cases, cp and AH are functions of temperature, then the substitutions would be as shown in Case C of Sec. 1.2.5.3. This general form, which may give rise to very complex expressions, may be important for gas phase reactions (see also Sec. 4.3.3). Simplifying by assuming only one reaction (R=l) and constant cpi gives

S

Assuming the total heat capacity to be constant, FC(Ci Cpi) = F p cp, gives i= 1

234

4 Differential Flow and Reaction Applications

dT AV p CPdt = - F p cp AT,

+

AV%

Vi

AH^)

- Q,,

The heat loss through the wall to the jacket is

The term AT can be approximated by (dT/dZ)AZ (see also Sec. 4.3.5). Noting that AV = A, AZ, that the linear velocity v = F/Ac and letting AZ approach zero, gives the partial differential equation

For steady-state conditions, the above equation reduces to

This equation can be integrated together with the component balances and the reaction kinetic expressions where the kinetics could, for example, be of the form

thus including variation of the rate constant k with respect to temperature. The component mass balance equation, combined with the reactor energy balance equation and the kinetic rate equation, provide the basic model for the ideal plug-flow tubular reactor.

4.3.2

Liquid-Phase Tubular Reactors

Assuming the case of a first-order chemical reaction, (rA = - k CA), and a noncompressible liquid system, the generalised mass and energy balance equations reduce to

and

235

4 . 3 Tubular Chemical Reactors

where v, p and cp are assumed constant and k is given by the Arrhenius equation, k = ko e - E/RT. The general solution approach, to this type of problem, is illustrated by the information flow diagram, shown in Fig. 4.8. The integration thus starts with the initial values at Z = 0, and proceeds with the calculation of rA, along the length of the reactor, using the computer updated values of T and CA, which are also produced as outputs.

z=o

+

cA = cAO

Figure 4.8.

Component balance for A

cA CA

Flow diagram for the simultaneous integration of the balances.

The simultaneous integration of the two continuity equations, combined with the chemical kinetic relationships, thus gives the steady-state values of both, CA and T, as functions of reactor length. The simulation examples BENZHYD, ANHYD and NITRO illustrate the above method of solution.

4.3.3

Gas-Phase Tubular Reactors

In gas-phase reactors, the volume and volumetric flow rate frequently vary, owing to the molar changes caused by reaction and the effects of temperature and pressure on gas phase volume. These influences must be taken into account when formulating the mass and energy balance equations. The Ideal Gas Law can be applied both to the total moles of gas, n, or to the moles of a given component of the gas mixture ni, where PV = nRT

236

4 Differential Flow and Reaction Applications

Here P is the total pressure of the system, pi is the partial pressure of component, i, V is the volume of the system, T is temperature and R is the Ideal Gas Constant. Using Dalton’s Law

a relationship for the concentration Ci, in terms of mole fraction yi, and total pressure is obtained as

This can also be expressed in terms of the molar flow rate Ni, and the volumetric flow rate G, where,

Assuming first-order kinetics, for the reaction, A

+m B

and by stoichiometry rg = + m k C ~ =m kRYTA P

AV

Figure 4.9. Mass balancing for a gas-phase, tubular reactor.

The steady-state mass balance, for a volume element, AV, as shown in Fig. 4.9, for reactant A, is given by

237

4 . 3 Tubular Chemical Reactors

which, since A V =Ac A Z, then gives

Similarly for component B,

where A, is again the cross-sectional flow area of the reactor. The variation in molar flow can be written as dNA ~dZ

-

d(yA P G/R T) dZ

which for constant temperature and pressure conditions becomes

Substituting this and the reaction kinetics rA, into the component balance equation for reactant A gives

I Similarly for B

d(ydBZG)

=

+ m k YA A,

I1

The volumetric flow rate depends on the total molar flow of the gas, and the temperature and pressure of reaction, where

where the molar flow rates of A and B are given by IV and

238

4 Differential Flow and Reaction Applications

V

The model equations, 1 to V above, provide the basis for solution, for this case of constant temperature and pressure with a molar change owing to chemical reaction. This is illustrated by the information flow diagram, Fig. 4.10. The step-by-step calculation procedure is as follows:

I . The initial molar tlow rates of each Component at the reactor inlet, (YA G)o and (YB G)0, are known. 2. The component balance equations I and I1 are integrated with respect to distance to give the volumetric flow rate of each component, (YA G)z and (YB G)z, at any position Z, along the length of the reactor.

3 . The total molar flow rate of each component, ( N A ) ~and ( N B ) ~can then be calculated at position Z, from equalions IV and V. 4. The total volumetric flow rate G, is calculated at each position, using equation Ill and hence:

5 . The composition of the gas mixture at position Z is obtained by dividing the individual molar flow rate by the volumetric flow rate.

I Ninert

v-

=

YB YB Figure 4.10. Information flow diagram for a gas-phase tubular reactor with molar change.

239

4 . 3 Tubular Chemical Reactors

The results of the calculation are thus the mole fraction compositions, YA and YB, together with the total volumetric flow rate G, as steady-state functions of reactor length. The step-by-step evaluation is, of course, effected automatically by the computer, as shown in the ISIM simulation example VARMOL. To obtain the fractional conversion at any position along the reactor the appropriate equation is

In the case of non-isothermal situations with significant pressure drop through the reactor, the term d(yi G P/R T) dZ must be retained in the model equations. The variation of reactor pressure with reactor length, (P)z, can be obtained by the use of available pressure drop-flow correlations, appropriate to the reactor geometry and flow conditions. The variation in temperature, with respect to length, (T)z must be obtained via a steady-state energy balance equation, as described in Sec. 4.3.1.

4.3.4

Batch Reactor Analogy

The ideal plug-flow reactor is characterised by the concept that the flow of liquid or gas moves with uniform velocity similar to that of a plug moving through the tube. This means that radial variations of concentration, temperature and flow velocity are neglected and that axial mixing is negligible. Each element of fluid flows through the reactor with the same velocity and therefore remains in the reactor for the same length of time, which is given by the flow volume of the reactor divided by the volumetric flow rate. The residence time of fluid in the ideal tubular reactor is thus analogous to the reaction time in a batch reactor. With respect to reaction rates, an element of fluid will behave in the ideal tubular reactor, in the same way, as it does in a well-mixed batch reactor. The similarity between the ideal tubular and batch reactors can be understood by comparing the model equations. For a batch reactor, under constant volume conditions, the component mass balance equation can be represented by

240 -

4 Differential Flow and Reaction Applications ~

-

For a plug-flow tubular reactor, the flow velocity v, through the reactor can be related to the distance travelled along the reactor or tube Z, and to the time of passage t, where

Equating the time of passage through the tubular reactor to that of the time required for the batch reaction, gives the equivalent ideal-l-low tubular reactor design equation as

4.3.5

Dynamic Simulation of the Plug-Flow Tubular Reactor

The coupling of the component and energy balance equations in the modelling of non-isothermal tubular reactors can often lead to numerical difficulties, especially in solutions of steady-state behaviour. In these cases, a dynamic digital simulation approach can often be advantageous as a method of determining the steady-state variations in concentration and temperature, with respect to reactor length. The full form of the dynamic model equations are used in this approach, and these are solved up to the final steady-state condition. at which condition

The procedure is to transform the defining model partial differential equation system into sets of difference-differential equations, by dividing the length or volume of the reactor into disc-shaped segments. The concentrations and temperatures at the boundaries of each segment are approximated by a midpoint average. Each segment can therefore be thought of as behaving in a similar manner to that of a well-stirred tank. This is shown below in Fig. 4.11, where segment n, with volume AV, is identified by its midpoint concentration CA"and midpoint temperature T,. The procedure again follows very closely that of Franks (1967).

24 1

4 . 3 Tubular Chemical Reactors

I I

CAn-l

I

Tn-1

c

I

I 1

CAn

I

Tn

L

I

I

CAn+l

I

Tn+l

L

AV

I

I

L

I

I

Figure 4.11. Finite-differencing for a dynamic tubular reactor model.

The concentration of reactant entering segment n, from segment n-1, is approximated by the average of the concentrations in the two segments and is given by CAn-1

2

CAn

Similarly, the concentration of reactant leaving segment n, and entering segment n+l, is approximated by CAn -!- C A n + l

2

This averaging procedure has the effect of improving the approximation. Alternatively, each segment could be treated as a well-mixed tank with the outflow variables equal to the tank values. This simpler approach would require a greater number of segments for the same accuracy. Applying the generalised component balance equation to each segment n gives

and results in

242

4 Differential Flow and Reaction Applications

where AV = A AZ. The application of the energy balance equation to segment n, results similarly in the relationship

-

U AAt (Tn - Tj)

where, the outer surface area for heat transfer in segment n is given by

At is the total surface for heat transfer and for a single tube is given by

where D is the tube diameter and A 2 is the length of segment. The resulting forms of the component and energy balance equations are thus

The above equations are linked by the reaction rate term rA, which depends on concentration and temperature. In the above equation, z, is the mean residence time in segment n and is equal to the volume of the segment divided by the volumetric flow rate

The modelling of the end sections, however, needs to be handled separately, according to the appropriate boundary conditions. The concentration and temperature conditions at the inlet to the first segment, n = 1, are, of course with no axial dispersion, identical to those of the feed, CAOand To, which gives rise to a slightly different form of the balance equations for segment, 1 . The outlet stream conditions may be taken as CAN and TN or, more accurately, as suggested by Franks (1967), one may assume an extrapolated value of C A N + I and T N + I through segments N-1, N and N+1. The averaging for the outlet could then be done as before. This would be done by assuming a constant gradient for CA and T and is treated in greater detail in the following section.

243

4.3 Tubular Chemical Reactors

4.3.6

Dynamics of an Isothermal Tubular Reactor with Axial Dispersion

Axial and radial dispersion or non-ideal flow in tubular reactors is usually characterised by analogy to molecular diffusion, in which the molecular diffusivity is replaced by eddy dispersion coefficients, characterising both radial and longitudinal dispersion effects. In this text, however, the discussion will be limited to that of tubular reactors with axial dispersion only. Otherwise the model equations become too complicated and beyond the capability of a simple digital simulation language. Longitudinal diffusion can be analysed using the unsteady-state diffusion equation ~

ac at

= -

=(DE) a

based on Fick’s Law. In the above case, D is an eddy dispersion coefficient and Z is the axial distance along the reactor length. When combined with an axial convective flow contribution, and considering D as constant, the equation takes the form

where v is the linear flow velocity. Written in dimensionless form, this equation is seen to depend on a dimensionless group v L / D, which is known as the Peclet number. Its inverse is called the dispersion number. Both terms represent a measure of the degree of dispersion or axial mixing in the reactor. Thus low values of the Peclet number correspond to high dispersion coefficients, low velocities and short lengths of tube and thus characterise conditions approximating to those of perfect mixing. For high values of the Peclet number, the converse conditions apply and thus characterise conditions approximating to perfect plug flow. In the extreme, the Peclet number corresponds to the following conditions

P, P,

-+ 0 +

03

perfect mixing prevails plug flow prevails

4 Differential Flow and Reaction Applications

244

4.3.6.1

Dynamic Difference Equation for the Component Balance Dispersion Model

The development of the equations for the dynamic dispersion model starts by considering an element of tube length AZ, with a cross-sectional area of A,, a superficial flow velocity of v and an axial dispersion coefficient, or diffusivity D. Convective and diffusive flows of component A enter and leave the element, as shown by the solid and dashed arrows respectively, in Fig. 4.12.

- iAn -*

-i An-1 CAn-l

A2 Figure 4.12. Fluxes for the axial dispersion model.

For each element, the mass balance is R a t e of

Convec-

Convec-

ulation of reactant A

flow of A in

flow of A out

Diffusive

Diffusive

A in

A out

Rate of reaction

As before, the concentrations are taken as the average in each segment, and the diffusion fluxes are related to the concentration gradients at the segment boundaries. The concentrations of reactant entering and leaving section n are CAin = and CAout

=

CAn-1

CAn

f

2

CAn

C An+ 1 2

The concentration gradients at the inlet and outlet of the section are

245

4 . 3 Tubular Chemical Reactors

and

1 (%)out = CAn A CAn+ Z -

The convective mass flows in and out are obtained by multiplying the respective concentrations by the volumetric flow rate, which is equal to A, v. The diffusive mass flows are calculated from the inlet and outlet concentration gradients using the multiplying factor of A, D. Dropping the A subscript for concentration, the component balance for reactant A, in section n, becomes

+AcD(

Cn-1 - Cn AZ

-

Cn

i?+')

-

kn Cn Ac A Z

where here, the reaction rate is taken as first order, rA = k CA. Dividing by A, A Z gives the defining component mass balance equation for segment n, as

A dimensionless form of the balance equation, can be obtained by substituting the following dimensionless variables

The model for section n, now in dimensionless form, yields

where D/Lv = UPe, the inverse Peclet number. The boundary conditions determine the form of balance equation for the inlet and outlet sections. These require special consideration as to whether diffusion fluxes can cross the boundaries in any particular physical situation. The physical situation of closed ends is considered here. This would be the case if a smaller pipe were used to transport the fluid in and out of the reactor, as shown in Figs. 4.13 and 4.14.

246

4 Differential Flow and Reaction Applications

Figure 4.13. Inlet section for the tubular reactor.

Since no diffusive flux enters the closed entrance of the tube, the component balance for the first section, becomes

Dividing by A, AZ gives

Similarly, the outlet of the reactor is closed for diffusion as shown in Fig. 4.14.

I

I Cout

Figure 4.14. Outlet section of the tubular reactor

An extrapolation of the concentration profile over the last half of element N is used to calculate the outlet concentration Gout, giving

4 . 3 Tubular Chemical Reactors

247 -~

with the balance for section N becoming

A similar finite-differenced equivalent for the energy balance equation (including axial dispersion effects) may be derived. The simulation example DISRET involves the axial dispersion of both mass and energy and is based on the work of Ramirez (1976). A related model without reaction is used in the simulation example FILTWASH.

4.3.7

Steady-State Tubular Reactor Dispersion Model

Letting the element distance AZ approach zero in the finite-difference form of the dispersion model, gives

Defining the following dimensionless variables

where z = L/v. The dimensionless form for an nth-order reaction is

At steady state, acAhT can be set to zero, and the equation becomes an ordinary second-order differential equation, which can be solved using ISIM. Again the entrance and exit boundary conditions must be considered. Thus the two boundary conditions at Z = 0 and Z = L are used for solution, as shown in Fig. 4.15. Note, that these boundary conditions refer to the inner side of the tubular reactor. A discontinuity in concentration at Z = 0 is apparent in Fig. 4.16.

248

4 Differential Flow and Reaction Applications

I Z=O

Z=L

Figure 4.15. Convective and diffusive fluxes at the entrance (Z=O) and exit (Z=L) of the tubular reactor.

Balancing the mass flows at the inlet gives a relation for the boundary conditions at Z=O

The zero flux condition at the closed outlet requires a zero gradient, thus

According to the boundary conditions, the concentration profile for A must change with a discontinuity at the reactor entrance, as shown in Fig. 4.16. -

In dimensionless form, the boundary condition at Z = 0 is represented by

and at Z = 1 by dZ

249

4 . 3 Tubular Chemical Reactors

0

L

Z

Figure 4.16. Concentration profiles in the tubular reactor for extreme and intermediate values of the dispersion number.

The solution requires two integrations as shown in Fig. 4.17.

d2

EA

dT2

L

Integration

Integration

Figure 4.17. Flow diagram for solving the second-order differential equation from the axial disperion model.

Referring to Fig. 4.15, it is seen that the concentration and the concentration gradient are unknown at Z=O. The above boundary condition relation indicates that if one is known, the other can be calculated. The condition of zero gradient at the outlet (Z=L) does not help to start the integration at Z=O, because, as Fig.

250

4 Differential Flow and Reaction Applications

4.17 shows, two initial conditions are necessary. The procedure to solve this split- boundary value problem is therefore as follows:

I . Guess (CA) z=o and calculate

dC, (-=-)

\dZ)Z=O

from the boundary condition

relation. 2. Integrate to

? = 1 and check whether

3. Vary the guess and iterate between obtained.

4.4

("")

dZ

Z=1

equals zero.

2 = 0 and 2 =

1 until convergence is

Differential Mass Transfer

This section concerns the modelling of countercurrent flow, differential mass transfer applications, for both steady-state and non-steady-state design or simulation purposes. For simplicity, the treatment is restricted to the case of a single solute, transferring between two inert phases, as in the standard treatments of liquid-liquid extraction or gas absorption column design.

4.4.1

Steady-State Gas Absorption with Heat Effects

Fig. 4. I8 represents a countercurrent-flow, packed gas absorption column, in which the absorption of solute is accompanied by the evolution of heat. In order to treat the case of concentrated gas and liquid streams, in which the total flow rates of both gas and liquid vary throughout the column, the solute concentrations in the gas and liquid are defined in terms of mole ratio units and related to the molar flow rates of solute free gas and liquid respectively, as discussed previously i n Sec. 3.3.2. By convention, the mass transfer rate equation is however expressed in terms of mole fraction units. In Fig. 4.18, G, is the molar flow of solute free gas (kmol/m2 s), L, is the molar flow of solute free liquid (kmol/m2 s), where both L, and G, remain constant throughout the column. Y is the mole ratio of solute in the gas phase (kmol of solute/kmol of solute free gas), X is the mole ratio of solute in the liquid phase (kmol of

25 1

4.4 Differential Mass Transfer

solute/kmol of solute free liquid), y is the mole fraction of solute in the gas phase and x is the mole fraction of solute in the liquid phase and T is temperature (K).

I

Y+dY X+dX TL + dTL

I

Figure 4.18. Steady-state gas absorption with heat effects.

Mole ratio and mole fraction contents are related by

Y Y = - 1-Y

X

a n d X = - -1-x

Subscripts L and G refer to the liquid and gas phases respectively, and subscripts 'in' and 'out' refer to the inlet and outlet streams.

4 . 4 . 1 . 1 Steady-State Design In the steady-state design application, the flow rates L, and G,, and concentrations Yin, Xin, Yo,t and Xout will either be specified or established by an overall, steady-state solute balance, where

252

4 Differential Flow and Reaction Applications

Temperatures TL in and TG in will also be known. The problem then consists of determining the height of packing required to obtain the above separation.

Component Mass Balance Equations For a small element of column volume dV

-

GI, A, dY = L, A, dX = K L a~ ( x* - x) dV

Here K L a~ (kmol/m3 s) is the overall mass transfer coefficient for the liquid phase, based on mole fraction in the L-phase, x* is the equilibrium liquid phase mole fraction, and A, is the cross-sectional area of the column (m2). Hence dY

K L a~ ( x* GI,

a=and

dX

dz =

K L a~ ( x* Lm

- x)

- x)

Energy Balance It is assumed that there are no heat losses from the column and that there is zero heat exchange between the gas and liquid phases. Consequently the gas phase temperature will remain constant throughout the column. A liquid phase heat balance, for element of volume dV is given by Rate of gain of heat by the liquid

( or

Rate of generation of heat by absorption

L A, cP dTI> = K L a~ ( X*

-

X)

dV AHabs

where L is the total mass flow rate of liquid(kg/m2 s), cp is the specific heat capacity of the liquid (kJ/kg K) and AHabs is the exothermic heat of absorption (kJ/kmol solute transferred). Hence

253

4 . 4 Differential Mass Transfer

The temperature variation throughout the column is important, since this affects the equilibrium concentration x*, where x* = feq

( Y 3

TL)

Solution of the required column height is achieved by integrating the two component balance equations and the heat balance equation, down the column from the known conditions xjn, yout and TLin, until the condition that either Y is greater than Yi, or X is greater than XOutis achieved. In this solution approach, variations in the overall mass transfer capacity coefficient both with respect to temperature and to concentration, if known, can also be included in the model as required. The solution procedure is illustrated by the simulation example AMMONAB. Using the digital simulation approach to steady-state design, the design calculation is shown to proceed naturally from the defining component balance and energy balance equations, giving a considerable simplification to conventional text book approaches.

4 . 4 . 1 . 2 Steady-State Simulation In this case, the flow rates L, and G,, concentrations Yi, and Xi,,, temperatures TGin and TLin, are known and in addition the height of packing Z is also known. It is now, however, required to establish the effective column performance by determining the resulting steady-state concentration values, Yout and Xoutr and also temperature TLout. The problem is now of a splitboundary type and must be solved by assuming a value for YOut, integrating down the column for a column distance Z and comparing the calculated value for Yi, with the known inlet gas concentration condition. A revised guess for the starting value You[ can then be taken and the procedure repeated until convergence is achieved.

4.4.2

Dynamic Modelling of Plug-Flow Contactors: Liquid-Liquid Extraction Column Dynamics

A plug-flow, liquid-liquid, extraction column is represented in Fig. 4.19. For convenience, it is assumed that the column operates under low concentration conditions, such that the aqueous and organic flow rates, L and G, respectively

254

4 Differential Flow and Reaction Applications -~

are constant. At low concentration, mole fraction x and y are identical to mole ratios X and Y, which are retained here in the notation for convenience. This however leads to a more complex formulation than when concentration quantities are used, as in the example AXDISP.

I

x

Y

I X+(dWdZ)AZ

Figure 4.19. Liquid-liquid extraction column dynamic representation.

Consider a differential element of column volume, A V , height AZ and crosssectional area, A,, such that AV=Ac AZ. Component mass balance equations can be written for each of the liquid phases, where Rate of (accumulation) = of solute

(

Convective flow of solute in

-

(

Convective flow of solute out

) +(

Rate of solute transfer

)

The actual volume of each phase in element AV is that of the total volume of the element, multiplied by the respective fractional phase holdup. Hence considering the direction of solute transfer to occur from the aqueous or feed phase into the organic or solvent phase, the mass balance equations become: For the aqueous phase

255

4 . 4 Differential Mass Transfer

ax

ax

P L ~ L A = V -~L A ,

zAZ

- KLXa(X-X*)AV

for the organic phase

where each term in the equations has units of kmol solute/s and where the symbols are as follows: a is the specific interfacial area related to the total volume (m2/m3). is the column cross-sectional area (m2) A, G is the molar flow rate of the light, organic phase per unit area (kmol / m2 s). is the volumetric holdup fraction of the light, organic phase (-). hG is the volumetric holdup fraction of the heavy, aqueous phase (-). hL K ~ x a is the overall mass transfer capacity coefficient based on the aqueous phase mole ratio X (kmol/m3 s) L is the molar flow rate of the heavy, aqueous phase per unit area (kmol / m2 s). X is the aqueous phase mole ratio (kmol solute/kmol water). X* is the equilibrium mole ratio in the heavy phase, corresponding to light phase mole ratio Y (kmol solute/kmol water). Y is the organic phase mole ratio (kmol solute/kmol organic ). Z is the height of the packing (m). AV is the total volume of one column segment with length AZ (m3). is the density of the solute-free light phase (kmol/m3). PG is the density of the solute-free heavy phase (kmol/m3). PL The above balance equations simplify to

ax

ax

P L ~ L= L~

-

KLx a (X - X*)

with the equilibrium relationship represented by

X* = feq (Y). Thus the system is defined by two coupled partial differential equations, which can be solved by finite-differencing. Consider a finite-difference element of length AZ, as shown in Fig. 4.20.

25 6

4 Differential Flow and Reaction Applications

I I

A2

t I

Figure 4.20. Finite-difference element for the two-phase transfer system.

The concentrations entering and leaving each section can be approximated by the following averages

-

(Xn-1 + Xn)

-

(Xn + Xn+1) 2

Bulk flow concentration of Y entering element n

-

(Yn+1 + Yn)

Bulk flow concentration of Y leaving element n

-

_( Y_n _ + Yn_ _ I~)

Bulk flow concentration of X entering element n Bulk flow concentration of X leaving element n

2

2

2

Hence the component balance equations for each phase in stage n , become

and

where Q n (kmol soluteh) is rate of solute transfer in element n given by Qn =

Hence

KLX a (X - X*) AV

4.4Differential Mass Transfer

257

The boundary conditions are formulated with the help of Figs. 4.21 and 4.22, and in accordance with the methodology of Franks (1967).

Figure 4.21. Aqueous and organic phase streams at the inlet.

xN-l

i I

1 1

*

xN

Figure 4.22. Aqueous and organic phase streams at the outlet.

At the aqueous phase inlet, the concentration of the aqueous phase entering section 1 is Xo. The concentration of the aqueous phase leaving section 1 is (XI + X2)/2. Formulating the balance for the aqueous phase gives

p~ hL A, AZTdX 1 = L A, (Xo -

"

x2)

-

KLx a A, AZ (XI - X I * )

Similarly for the organic phase, the concentration of the organic phase entering section 1 equals (Y2 + Y1)/2, and by extrapolation the concentration of the organic phase leaving section 1, Yo equals Y 1 + (Y 1 - Y2)/2. The balance then gives pchcA,AZT

= GA, ( y 2 ~ y 1 - Y ~ ) + KLXaA,AZ(Xl - X I * )

25 8

4 Differential Flow and Reaction Applications

Simplifying the balance for each phase

The balance for section N can be formulated similarly. Thus extrapolating to approximate the concentration of the aqueous phase at the outlet gives

The balances for end stage, N, thus become

The representation of the boundary conditions for both the top and bottom of the column are really more mathematical than practical in nature and fail to take into account the- actual geometry and construction of the upper and lower parts of the column and the relative positioning of the inlet and outlet connections. They may therefore require special modelling appropriate to the particular form of construction of the column, as discussed previously in Sec. 3.3.1 . l o .

4.4.3

Dynamic Modelling of a Liquid-Liquid Extractor with Axial Mixing in Both Phases

Axial mixing is known to have a very significant effect on the performance of agitated liquid-liquid extraction columns and any realistic description of column performance must take this in to account. Fig. 4.23 represents a small differential element of column volume AV and height AZ. Here the convective flow rates, as in Fig. 4.19, are shown by the solid arrows, and the additional dispersion contributions, representing axial mixing, are shown by dashed arrows. It is assumed that axial mixing in both phases can be described by analogy to Ficks Law , but using an effective eddy dispersion coefficient appropriate to the respective liquid phase.

259

4 . 4 Differential Mass Transfer

4

L

G

I

I

I QnL I

hln

kl

I

t hGn

1.z

n

I

4 I I

I

L

7

4 G

I

I

t

n+l

Writing unsteady-state component balances for each liquid phase results in the following pair of partial differential equations which are linked by the mass transfer rate and equilibrium relationships

Here the nomenclature is the same as in Sec. 4.4.2 and in addition, DG is the effective eddy dispersion coefficient for the organic or extract phase (m2 / s) and DL is the effective eddy dispersion coefficient for the aqueous or feed phase (m2 / s). The above equations are difficult to solve analytically (Lo et al., 1983) but are solved with ease, using digital simulation. Referring to Fig. 4.23, the extractor is again divided into N finite-difference elements or segments of length AZ. The convective terms are formulated for each segment using average concentrations entering and leaving the segment, as shown in Sec. 4.4.2. The backmixing terms j, are written in terms of dispersion coefficients times driving-force, mole-ratio gradients. The resulting equations, for any segment n, are then for the aqueous feed phase with each term in kmol solute/s.

260 ~

4 Differential Flow and Reaction Applications ~~

-~

~~

~-

Rearranging

Similarly for the light solvent-extract phase

Rearranging

Note that the above formulation includes allowance for the fractional phase holdup volumes, h L and hG, the phase flow rates, L and G, the diffusion coefficients DL and DG, and the overall mass transfer capacity coefficient KLX a, all to vary with position along the extractor.

Boundary Conditions The column end sections require special treatment to allow for the fact that zero diffusive flux enters through the end wall of the column. The equations for the end section are derived by setting the diffusion flux leaving the column to zero. In addition, the liquid phase outlet concentrations leaving the respective end sections of the column are approximated by an extrapolation of the concentration gradient from the preceding section. The resulting model equations give the concentrations of each segment in both phases as well as the

26 1

4 . 4 Differential Mass Transfer

outlet concentrations, as a function of time. The resulting model formulation is shown in the simulation example AXDISP.

4.5

Heat Transfer Applications

4.5.1

Steady-State Tubular Flow with Heat Loss

Here a steady-state formulation of heat transfer is considered (Pollard, 1978). A hot fluid flows with linear velocity v, through a tube of length L, and diameter D, such that heat is lost via the tube wall to the surrounding atmosphere. It is required to find the steady-state temperature profile along the tube length. Consider an element of tube of length AZ, distance Z from the tube inlet as shown in Fig. 4.24. If the temperature at the inlet to the tube element is T, then the temperature at the outlet of the element can be written as T + (dT/dZ) AZ.

....;+ Z

Q TW

Figure 4.24. Tubular flow with heat loss.

[

1

The energy balance for the element of tube length can be stated as accumulation] Rate of = [inlet Heating stream] of of enthalpy in the element

to element temperature

-

(

heat loss Rate of

to the wall

As shown in Sec. 1.2.5, the heat balance equation, assuming constant fluid properties, becomes

dT M c p x = WcP

-

UA(T-Tw)

262

4 Differential Flow and Reaction Applications

~~

~~

where A D M T, T, U v W

is the heat transfer surface area for the element = 2 71: D AZ (m2). is the tube diameter (m). is mass of fluid in the element = (n: D2/4) p AZ (kg). is the wall temperature (K). is the temperature of the wall (K). is the heat transfer coefficient beween the fluid and the wall (kJ/m2 s). is the linear velocity of the fluid ( m / s ) . is the mass flow rate along the tube = (n: D2/4) p v (kg/s).

Simplifying the above equation gives dT dt

Under steady-state conditions, along the tube is given by

dT

4 u

= -v d Z - p c p D (T - Tw)

dT dt =

0, and the resulting temperature profile

dT 4U v -++-d Z p c p D (T-T,)

= 0

Assuming constant coefficients, both the dynamic and steady-state equations describing this system can be solved analytically, but the case of varying coefficients requires solution by digital simulation.

4.5.2

Single-Pass, Shell-and-Tube, CountercurrentFlow Heat Exchanger

4 . 5 . 2 . 1 Steady-State Applications Fig. 4.25 represents a steady-state, single-pass, shell-and-tube heat exchanger. For this problem W is the mass flow rate (kg/s), T is the temperature (K), cp is the specific heat capacity (kJ/m2 s), A (= n: D Z) is the heat transfer surface area (m2), and U is the overall heat transfer coefficient (kJ/m2 s K). Subscripts c and h refer to the cold and hot fluids, respectively.

263

4 . 5 Heat Transfer Applications

Thin

-

A

Thout

b

Figure 4.25. Steady-state, countercurrent flow heat exchange.

Heat balances on a small differential element of heat transfer surface area, AA, give

[

][

Rate of accumulation of enthalpy = in the element

Rate of heat transfer to the element

)

Thus for the hot fluid (W

Cp)h

ATh =

-U

(Th - Tc) AA

-U

(Th - TC) AA

and for the countercurrent cold fluid (W cp)c ATC =

In the limit, the defining model equations for countercurrent flow become

and

where for cocurrent flow the sign in cold-fluid equation would be positive. For design purposes the two equations can be integrated directly starting for the known temperature conditions at one end of the exchanger and integrating towards the known conditions at the other end, hence enabling the required heat exchange surface to be determined. This procedure is very similar to that

264

4 Differential Flow and Reaction Applications

of the steady-state mass transfer column calculation of Sec. 4.4.1.1. The design approach for a steady-state two pass exchanger is illustrated by simulation example SSHEATEX. However, the simulation of the performance for a heat exchanger with a known heat transfer surface area will demand an iterative split boundary solution approach, based on a guessed value of the temperature of one of the exit streams, as a starting point for the integration.

4 . 5 . 2 . 2 Heat Exchanger Dynamics The modelling procedure is again based on that of Franks (1967). A simple, single-pass, countercurrent flow, heat exchanger is considered. Heat losses and heat conduction along the metal wall are assumed to be negligible, but the dynamics of the wall (thick-walled metal tube) are significant. Fig. 4.26 shows the temperature changes over a small differential element of exchanger length AZ.

Tsout

t

I I

I

ws

I

I

I

I

I

I

I

I

I--+

I

I I

I

I

I I

1

I

A2

Z

Ts in

I

ws

Figure 4.26. Shell-and-tube heat exchanger: differential model.

In this problem, W is the mass flow rate (kg/s), T is temperature (K), cp is the specific heat capacity (kJ/kg K), D is the diameter (m), U is the heat transfer coefficient (kJ/m2 s K), Q is the rate of heat transfer (kJ/s), V is the volume (m3), p is the density (kg/m3), and A is the heat transfer area, (m2). The

265

4 . 5 Heat Transfer Applications

subscripts are as follows: t refers to tube conditions, s to shellside conditions, and m to the metal wall. Heat balance equations on the element of heat exchanger length AZ, according to enthalpy balance relationship,

[

Rate of accumulation of enthalpy in the element

][

Flow rate of enthalpy = into the element

1- [

Flow rate of enthalpy out of the element

]+ (

Rate of into heat thetransfer element

lead to three coupled first-order partial differential equations, which can be converted into difference equations for simulation language solution using the formulae of Sec. 4.6. Alternatively, the difference-equation model form can be derived directly by dividing the length of the heat exchanger into N finite-difference elements or segments, each of length AZ, as shown in Fig. 4.27. I

+ 1

I

Tsn-1

+

Shellside fluid

1

Tsn 4Qmn

Tsn+1

1

1 Tube wall

Ttn-1

+ I

Ttn

1-

Ttn+l

Qtn

I

I 1

I I

1 1

1

I

I

L

Tubeside fluid

1

Figure 4.27.

Finite-differencing of heat exchanger length.

The heat balance equation can now be applied to segment n, of the heat exchanger. The heat transfer rate equations are given by the following terms Rate of heat transfer from tube contents to the metal wall

where Ut is the tubeside film heat transfer coefficient, and A A t is the incremental tubeside area ,

266

4 Differential Flow and Reaction Applications -~

~-

Rate of heat transfer from the metal wall to the shellside contents

where Um is the film heat transfer coefficient from the wall to the shell and AAm is the incremental metal wall outside area

The inlet and outlet temperatures for the segment, are given by the following approximations Temperature of tubeside fluid entering element n = Temperature of tubeside fluid leaving element n =

1

+ Ttn)

(T,n+1

+ Tsn)

z1 (Ttn + Ttn+i) 1

Temperature of shellside fluid entering element n = Temperature of shellside fluid leaving element n =

(Ttn-1

1

2 (Tsn + Tsn-1)

Assuming constant heat capacity, the following balance equations apply For the tubeside fluid

For the metal wall

For the shellside fluid

where

AVs = AZ K (Ds2 - Dm2>

AVt

=

AZ 7c Dt2

AV,,, = AZ n: ( D m 2 - D t 2 )

267

4 . 5 Heat Transfer Applications

Simplification gives the resulting finite-differenced model equations as

Bo u nda ry Conditions The consideration of the boundary conditions again follows Franks (1967). The position at the tube inlet and shell outlet section, segment number 1 is shown in Fig. 4.28.

Ts1

TtO

___)

Ttl

f

*

Ts2

Tt2

I

Figure 4.28. Tube inlet and shell outlet segment.

Again the heat balance equation can be applied, but in this case, the entering shellside temperature is (Ts2 + Ts1)/2, and the leaving shellside temperature is TSO.

The outlet shellside fluid temperature can also be approximated, either as TsO = Ts1

268

4 Differential Flow and Reaction Applications

or

TsO = Ts1+

Ts1 - Ts2 2

The heat balance equations for end segment 1 , thus become cps ps AVs

dTs 1 7

= Ws cps

( Ts2

Tsl - Tso)

+ Qmi

and cPt pt AVt

dTt 1

= Wt cPt (Tto -

Similar reasoning for the tube outlet and shell inlet segment, number N, give For the tubeside fluid

and for the shellside fluid,

Note that the outlet approximations must be consistent with a final steady-state heat balance. Note also that is easy to allow in the simulation for variations in the heat transfer coefficient, density and specific heats as a function of temperature. The modelling methods demonstrated in this section are applied in the simulation example HEATEX.

4.6

Difference Formulae for Partial Differential Equations

As shown in this chapter, in the simulation of systems described by partial differential equations, the differential terms involving variations with respect to length are replaced by their finite-differenced equivalents. These finitedifferenced forms of the model equations are shown to evolve as a natural consequence of the balance equations, according to the manner of Franks (1967). The approximation of the gradients involved may be improved, if necessary, by using higher order approximations. Forward and end sections can

269

4.6 Difference Formulae for Partial Differential Equations

be better approximated by the forward and backward differences given below. The various forms of approximation based on the use of central, forward and backward differences have been listed by Chu (1969). a) First-Order Approximations Central difference as extensively used in this chapter

Forward difference

Backward difference

(gl =

(Un - Un-1) AX

b) Second-Order Central Difference Approximations

(-Un+2

+ 16Un+1- 30Un + 16Un-1 1 2(AX)2

-

Un-2)

270

4.7

References

References Cited in Chapters 1 to 4

Aly, G. Ottertun, H. (1973) J. Appl. Chem. Biotechnol. 23, 643. Aris, R. (1989) Elementary Chemical Reactor Analysis, Butterworth Publ., Stoneham. Bird, R. B., Stewart, W. E. and Lightfoot, E. N. (1960) Transport Phenomena, Wiley. Brodkey, R. S. and Hershey, H. C. (1988) Transport Phenomena, McGraw-Hill. Burt, C. J. (1989) SIMUSOLV, A New Code for Modeling and Optimization, CACHE News, 29, 13-16. Cambridge University Press. Carslaw, H. S. and Jaeger, J. C. (1959) Conduction of Heat in Solids, Clarendon Press. Chu, Y. (1969) Digital Simulation of Continuous Systems, McGraw-Hill. Coughanowr, D. R. and Koppel, L. B. (1965) Process Systems Analysis and Control. McGraw-Hill. Douglas, J. M. (1972) Process Dynamics and Control, Vol. 1 and Vol. 2, Prentice Hall. Draper, N R. and Smith, H. (198 1) Applied Regression Analysis, Wiley, Fogler, H. S. (1 992) Elements of Chemical Reactor Engineering, Prentice-Hall. Franks, R. G. E. (1 967) Mathematical Modeling in Chemical Engineering, Wiley. Franks, R. G. E. ( I 972) Modelling and Simulation in Chemical Engineering, W iley-Interscience. Geankoplis, C. J. (1983) Transport Processes and Unit Operations, 2nd edition, Allyn and Baker. Habermann, R. ( 1 976) Mathematical Models: Mechanical Vibration, Population Dynamics and Traffic Flow, Prentice-Hall.

4.7 References Cited in Chapters 1 to 4

27 1

Heinzle, E., and Saner, U. (1991) Methodology for Process Control in Research and Development. In: Bioprocess Monitoring and Control, Ed. Pons, M.N., Hanser, Munich, p. 223-304. Ingham, J. and Dunn, I. J. (1974) Digital Simulation of Stagewise Processes with Backmixing. The Chem. Eng., June, 354-365. Joshi, J. B., Pandit, A. B. and Sharma, M. M. (1982) Mechanically Agitated Gas-Liquid Reactors. Chem. Eng. Sci. 37, 813. Kapur, J. N. (1988) Mathematical Modelling, Wiley. Levenspiel, 0. ( 1972) Chemical Reaction Engineering, Wiley. Lo, T. C., Baird, M. H. I., and Hanson, C. (1983) Handbook of Solvent Extraction, Wiley. Luyben, W. L. (1973) Process Modelling, Simulation, and Control for Chemical Engineers, McGraw-Hill. Luyben, W. L. (1990) Process Modelling, Simulation, and Control for Chemical Engineers, 2nd edition, McGraw-Hill. Matko, D., Karbar, R. and Zupancic, B. (1992) Simulation and Modeling of Continuous Systems, Prentice-Hall. Noye, J. Ed. ( 1984) Computational Techniques for Differential Equations, North-Holland. Oosterhuis, N. M. G. (1984) PhD Thesis, Delft University, The Netherlands. Ord-Smith, R. J. and Stephenson, J. (1975) Computer Simulation of Continuous Systems, Cambridge Univ. Press. Perlmutter, D. D. ( 1 972) Stability of Chemical Reactors, Prentice-Hall. Pollard, A. (1978) Process Control, Heinemann Educational Books. Prenosil, J. E. (1976) Multicomponent Steam Distillation: A Comparison between Digital Simulation and Experiment. Chem. Eng. J., 12, 59-68. Press, W. H. (1989) Numerical Recipes: The Art of Scientific Computing. Cambridge Univ. Press.

272

References

Ramirez, W. F. (1 976) Process Simulation, Lexington Books. Russell, T. W. F. and Denn, M. M. (1972). Introduction to Chemical Engineering Analysis. Wiley. Seborg, D. E., Edgar, T. F., Mellichamp, D. A. (1989) Process Dynamics and Control, Wiley . Smith, C. L., Pike, R. W. and Murrill, P. W. (1970) Formulation and Optimisation of Mathematical Models, Intext. Smith, J. M. (1987) Mathematical Modelling and Digital Simulation for Engineers and Scientists, 2nd edition, Wiley-Interscience. Steiner, E. C., Blau, G. E. and Agin, G. L. (1986) SIMUSOLV - Modeling and Simulation Software, Mitchell and Gauthier Ass., Concord, MA, USA. Stephanopoulos, G. (1 984) Chemical Process Control, Prentice-Hall. Sweere, A. P. J., Luyben, K. Ch. A. M. and Kossen, N. W. F. (1987) Regime Analysis and Scale-Down: Tools to Investigate the Performance of Bioreactors. Enzyme Microb. Technol., 9, 386-398. Szekely, J. and Themelis, N. J. (197 1) Rate Phenomena in Process Metallurgy, Wiley. Walas, S. M. (1991) Modelling with Differential Equations in Chemical Engineering, Butterworth-Heinemann Series in Chemical Engineering. Welty, J. R., Wicks, C. E. and Wilson, R. E. (1976) Fundamentals of Momentum, Heat and Mass Transfer, Wiley. Wilkinson, W. L. and Ingham, J. ( I 983) In: Handbook of Solvent Extraction, Eds. Lo, T. C., Baird, M. H. I., and Hanson, C., Wiley, pp. 853-886. Wozny, G. and Lutz, J. (1991) Dynamische Prozess Simulation in der industriellen Praxis. Chemie Ing. Technik 63, 3 13-324.

4.8 Additional Books Recommended

4.8

27 3

Additional Books Recommended

Adams, J. A. and Rogers, D. F. (1973) Computer-Aided Heat Transfer Analysis. McGraw-Hill. Al-Khafaji, A. W. and Tooley, J. R. (1986) Numerical Methods in Engineering Practice, Holt-Rinehart-Winston. Andrews, J. G. and McLone, R. R. (1976) Mathematical Modelling, Butterworths. Aris, R. (1 978) Mathematical Modelling Techniques, Pitman. Aris, R. and Amundson, N. R. (1973) First Order Partial Differential Equations with Applications, Prentice-Hall. Aris, R. and Varma A. (1980) The Mathematical Understanding of Chemical Engineering Systems: Collected Papers of Neal R. Amundson, Pergamon. Belter, P. A., Cussler, E. L. and Hu, W. S. (1988) Bioseparations: Downstream Processing for Biotechnology, Wiley. Beltrami, E. (1987) Mathematics for Dynamic Modelling, Academic Press. Bender, C. and Orszag, S. A. (1978) Advanced Mathematical Methods for Scientists and Engineers, McGraw-Hill. Bender, E. A. (1978) An Introduction to Mathematical Modelling, WileyInterscience. Bronson, R. (1989) 2500 Solved Problems in Differential Equations, McGrawHill. Burghes, D. N. and Borris, M. S. (1981) Modelling with Differential Equations. Ellis Horwood. Carberry, J. J. (1976) Chemical and Catalytic Reaction Engineering, McGrawHill. Carberry, J. J. and Varma, A. (Eds.) (1987) Chemical Reaction and Reactor Engineering, Dekker. Champion, E. R. and Ensminger, J. M. (1988) Finite Element Analysis with Personal Computers, Dekker.

274

References ~~

.~

Chapra, S. C. and Canale, R. P. (1988) Numerical Methods for Engineers, McGraw-Hill. Churchill, S. W. (1974) The Interpretation and Use of Rate Data, McGraw-Hill. Constantinides, A. ( 1987) Applied Numerical Methods with Personal Computers, McGraw-Hill. Coulson, J. M. and Richardson, J. F. (1979) Chemical Reactor Design, Biochemical Reaction Engineering Including Computation, Pergamon Press. Cross, M. and Mascardini, A. 0. (1985) Learning the Art of Mathematical Modelling, Ellis Horwood. Davis, M. E. (1984) Numerical Methods and Modeling for Chemical Engineers, Wiley. Dunn, I. J. , Heinzle, E. Ingham, J. and Prenosil, J. E. (1992) Biological Reaction Engineering: Principles, Applications and Modelling with PC Simulation. VCH. Denn, M. M. (1987) Process Modelling, Longman Scientific and Technical Publishers. Finlayson, B. A. (1980) Nonlinear Analysis in Chemical Engineering, McGrawHill. Fishwick, P. A. and Luker, P. A. (Eds.) (1991) Qualitative Simulation, Modelling and Analysis, Springer. Friedly, J. C. (1972) Dynamic Behaviour of Processes, Prentice Hall, . Froment, G. F. and Bischoff, K. B. (1990) Chemial Reactor Analysis and Design, Wiley. Gordon, G. (1978) System Simulation, 2nd edition, Prentice Hall. Guenther, R. B. and Lee, J. W. (1988) Partial Differential Equations of Mathematical Physics and Integral Equations, Prentice-Hall. Himmelblau, D. M. (1985) Mathematical Modelling. In: Bisio, E. and Kabel, R. L. (Eds.) Scaleup of Chemical Processes, Wiley. Hines, A. L., and Maddox R. N. (1985) Mass Transfer, Prentice-Hall.

4.8 Additional Books Recommended

275

Holland, C. D. and Anthony, R. G. (1989) Fundamentals of Chemical Reaction Engineering, 2nd edition, Prentice-Hall. Holland, C. D. and Liapis, A. I. (1983) Computer Methods for Solving Dynamic Separation Problems, McGraw-Hill. Huntley, I. D. and Johnson, R. M. (1983) Linear and Nonlinear Differential Equations, Ellis Horwood. Hussam, A. (1986) Chemical Process Simulation, HalsteaWiley Jenson, V. G. and Jeffreys, G. V. (1977) Mathematical Methods in Chemical Engineering, Academic Press. Kaddour, N. Ed. (1993) Process Modeling and Control in Chemical Engineering, Marcel Dekker. Kocak, H. (1989) Differential and Difference Equations Through Computer Experiments, With Diskette, Springer. Korn, G. A. and Wait, J. V. (1978) Digital Continuous System Simulation, Prentice Hall. Lapidus, L. and Pinder, G. F. (1982) Numerical Solution of Partial Differential Equations in Science and Engineering, Wiley-Interscience. Leesley, M. E. Ed. (1982) Computer Aided Process Plant Design, Gulf Publishing Company. Nauman, E. B. (1987) Chemical Reactor Design, Wiley. Nauman, E. B. (1991) Introductory Systems Analysis for Process Engineers, Butterworth-Heinemann. Neelamkavil, F. (1 987) Computer Simulation and Modelling, Wiley. Rademaker, O., Rijnsdarp, J. E. and Maarleveld, A. (1975) Dynamics and Control of Continuous Distillation Units, Elsevier Scientific. Ramirez, W. F. (1 989) Computational Methods for Process Simulation, Butterworth-Heinemann. Rase, H. F. (1977) Chemical Reactor Design for Process Plants. Vol. 11. Case Studies and Design Data, Wiley.

276

References

Reilly, M. (Ed.) (1972) Kinetics, Computer Programs for Chemical Engineering Education, Vol. 2, CACHE, A.1.Ch.E. Riggs, J. B. (1988) An Introduction to Numerical Methods for Chemical Engineers, Texas Techn. Univ. Press. Robinson, E. R. (1975) Time-Dependent Chemical Processes, Applied Science. Satterfield, C. N. (1970) Mass Transfer in Heterogeneous Catalysis, MIT Press. Shah, M. J. (1976) Engineering Simulation Using Small Scientific Computers, Prentice Hall. Sherwood, T. K., Pigford, R. L. and Wilke, C. R. (1975) Mass Transfer, McGraw-Hill. Smith, C. A. and Corripio, A. B. (1985) Principles and Practice of Automatic Process Control, Wiley. Smith, J. M. (1 970) Chemical Engineering Kinetics, 2nd edition. McGraw-Hill. Smith, J. M. (1987) Mathematical Modelling and Digital Simulation for Engineers and Scientists 2nd edition, Wiley-Interscience. Stewart, W. E., Ray, W. H. and Conley, C. C. (1980) Dynamics and Modeling of Reactive Systems, Academic Press. Tarhan, M. 0. (1 983) Catalytic Reactor Design, McGraw-Hill. Vemuri, V. and Karplus, W. J. (1981) Digital Computer Treatment of Partial Differential Equations, Prentice-Hall. Walas, S. M. (1989) Reaction Kinetics for Chemical Engineers, McGraw-Hill. Reprinted with corrections by Butterworths. Walas, S. M. (1991) Modelling with Differential Equations in Chemical Engineering, Butterworth-Heinemann. Wankat, P. C. (1990) Rate Controlled Separations, Elsevier Applied Science. Weber, T. W. (1978) An Introduction to Process Dynamics and Control, WileyInterscience. Wen, C. Y. and Fan, L. T. (1975) Models for Flow Systems and Chemical Reactors, Dekker.

4.8 Additional Books Recommended

277

Westerberg, A. Ed. (1972) Control. Computer Programs for Chemical Engineering Education Vol. 111, CACHE Committee/A.I.Ch.E. Whitwell, J. C. and Toner R. K. (1969) Conservation of Mass and Energy, Blaisdell. Zwillinger, D. (1 989) Handbook of Differential Equations, Academic Press.

Chemical Engineering Dynami :Modelling

with PC Simulation

John Inph;m. Iruine J. Dunn. Elmar Heinzlc &JiiiE. Picnoul

copyright OVCH VerlagsgesellschaftmbH, 1994

5

Simulation Examples of Chemical Engineering Processes

In this chapter the simulation examples are described. As seen from the Table of Contents, the examples are organised according to twelve application areas: Batch Reactors, Continuous Tank Reactors, Tubular Reactors, Semi-Continuous Reactors, Mixing Models, Tank Flow Examples, Process Control, Mass Transfer Processes, Distillation Processes, Heat Transfer, and Dynamic Numerical Examples. There are aspects of some examples which relate them to more than one application area, which is usually apparent from the titles of the examples. Within each section, the examples are listed in order of their degree of difficulty. Each simulation example is identified by a file name and title, and each comprises the qualitative physical description with drawing, the model equation development, the nomenclature, the ISIM program, suggested exercises, sample graphical results and literature references. The diskette in the pocket at the back of the book contains the programs and the ISIM software. As explained in detail in the Appendix, to run the examples, ISIM must be called and the desired file name called with the command READ (R). The program will run with START (S) and will continue until TFIN is reached. Any run can be interrupted by pressing any key. The command HOLD (H) then allows graphical output via the command GRAPH. Parameters can then be changed with VAL, and the program can be continued from the interrupt point with GO. Most of the programs include an INTERACT; RESET statement, which allows parameters to be changed after TFIN is reached. The command GO creates a new run from T=O using the new parameters. In this way, many runs with parametric changes can be made, and the results can be plotted for comparison with GRAPH. The Appendix should be consulted for the installation procedure and for further details on the use of KIM. The objective of the exercises with each example is, of course, to aid in understanding the example models. Usually changes in parameters are suggested, which can either be accomplished by comparing complete runs or by changing parameters within a run, as described above. Some exercises involve making minor changes in a program statement, some require adding a few lines to the program and some even require writing a new program. The reader-simulator should try to relate the simulation results to the parameters used and to the model equations. For this understanding, it is necessary to refer back to the model equations and to the program. Most will find it useful to make notes of the parameters investigated and to sketch the approximate graphical results.

280

5 Simulation Examples of Chemical Engineering Processes

5.1

Batch Reactor Examples

5.1.1

BATSEQ

-

Complex Batch Reaction Sequence

System The following complex reaction sequence can be used to study various reaction kinetics of interest simply by varying the rate constants.

Figure 5.1. A complex reaction sequence.

Model For batch kinetics the model equations become

For this batch reaction case, the sum of all molar species must equal the initial moles changed.

28 1

5.1 Batch Reactor Examples

Program :Example :

BATSEQ

COMPLEX

CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT 1

SIM;

BATCH

REACTION

SEQUENCE

K1= 0.01 K2= 0.01 K3= 0.05 K4= 0.02 CINT=10 TFIN=500

RESET;

INTERACT;

GOT0

1

INITIAL A=1 B=O

c=o D=O

DYNAMIC A'=-Kl*A B1=K1*A-K2*B+K3*C-K4*B C8=K2*B-K3*C D'=K4*B MOLE=A+B+C+D OUTPUT T,A,B,C,D,MOLE PREPARE T,A,B,C,D,MOLE PLOT T,A,O,TFIN,O,l

Nomenclature

Symbols C k t

Component concentration First-order reaction rate constant Time

l/s kmol/m3 S

5 Simulation Examples of Chemical Engineering Processes

282 -~

.

Indices 1, 2, 3 , 4 A, B, c,D

Refer to reactions Refer to components

Exercises 1.

Study the effect of varying rate coefficients k l , k2, k3, k4 in the range (0.001 to 0.1) on the time-dependent concentrations of A, B, C and D. Note that CB and Cc pass through maximum values.

2.

Set k3 =

= 0, to simulate a simple consecutive reaction sequence.

kl

k2

d

A - B

C

3.

Set k2 = k3 = k4 = 0, to simulate a simple first-order reaction.

4.

Set CAO= CCO= CDO= 0, CBO= 1 and kl = first-order reversible reaction. B

-

k2

c

= 0, to simulate a simple

1

k3

5.

Note that with CAO= Cco = CDO= 0, CBO= 1, k l = k3 = 0 the program models a simple first-order parallel reaction sequence. B - c

6.

k2

Write the model and program in terms of mole fractions.

283

5.1 Batch Reactor Examples

Results 4

Figure 5.2. Profiles of concentrations versus time as obtained from the rate constants in the program.

ISIU

Figure 5.3. Changing K4 from 0.02 to 0.002 resulted in these profiles, otherwise the same constants as Fig. 5.2.

Reference Dunn, I.J., Ingham, J. (1972) Verfahrenstechnik. 399, 6, Nr. 11.

5.1.2

BATCHD - Dimensionless Kinetics in a Batch Reactor

System An nth-order homogeneous liquid phase reaction is carried out in a batch tank reactor.

284

5 Simulation Examples of Chemical Engineering Processes

k

A

d

B

Figure 5.4. Batch reactor with simple nth-order reaction.

The model is written in both dimensional and dimensionless forms. example provides experience in the use of dimensionless equations.

Model The batch balance for the reactant A with nth-order kinetics is dCA V - -dt= - k C

n A

v

Dimensionless variables are defined as

where CAOis the initial concentration. The dimensionless balance equation is then d ~

d t

= - -cn,

Defined in this way, there is no governing dimensionless parameter.

This

285

5.1 Batch Reactor Examples

Program :EXAMPLE BATCHD : Batch reactor dimensionless,kinetics nth-order : Dimin. time defined a s t*k*CAO** (n-1) Constant CINT=O.1, TFIN=10, NOCI=l,ALGO=O Constant n = 2 Constant k = 3 Constant CAO = 1 1 S1M;INTERACT;RESET;GOTOl INITIAL CA=CAO/CAO DYNAMIC CA'=-CA**n Time=T/(k*(CAO**(n-l))) ConcA=CAO*CA PLOT T,CA,O,10,0,1 OUTPUT T,CA PREPARE T,CA,Time,ConcA

Nomenclature Symbols CA

Concentration of A

CA

CA0 k n t

Dimensionless concentration of A Initial concentration of A Reaction rate constant Reaction order Time

t V

Dimensionless time Volume

-

-

mol/L

-

L

Exercises 1.

For a first-order reaction n=l, and the dimensionless time is given by k t. Make a series of runs with different initial concentrations and compare the results, plotting the variables in both dimensional and dimensionless terms.

286

5 Simulation Examples of Chemical Engineering Processes

2.

For n=2 the dimensionless time is k t CAO. Make the runs and study the results as in Exercise 1. Verify that a change in k and a change in CAO have the same influence as long as the product term k CAOis maintained constant.

3.

The sensitivity of the reaction is given by the order of reaction, n. Experiment with this, in a series of runs for n=O, n=0.5, n=l, n=1.5 and n=2.

Results Dimensionless and normal outpur variables are plotted below. Ism

RCR

Figure 5.5. Profile of dimensionless concentration of A versus dimensionless time for n = 2 and n = 1 (light curve). Here CAO

Ism

R

Figure 5.6. Profile of concentration of A versus time for the same parameters as in Fig. 5.5.

= 15.

5.1.3

COMPREAC - Complex Reaction

System The complex batch reaction between formaldehyde, A, and sodium p-phenol sulphonate, B, proceeds in accordance with the following complex reaction scheme. All the reactions follow second-order kinetics. Components C, D and F are intermediates, and E is the final product.

287

5.1 Batch Reactor Examples

Reaction

Rate Coefficient

A+B+C A+C+D

C+D+E B+D+F C+C+F C+B-+G A+G+F A+F+E

Model For a constant volume batch reaction, the balance equations for each component lead to

Program Example COMPREAC COMPLEX REACTION SCHEME BETWEEN FORMALDEHYDE : SODIUM PARA PHENOL SULPHONATE CONSTANT K1=0.16, K2=0.05, K3=0.15, K4=0.14 :

:

AND

288

5 Simulation Examples of Chemical Engineering Processes

CONSTANT K5=0.03, K6=0.058, K7=0.05, K8=0.05 CONSTANT AO=0.15, BO=0.1 CONSTANT TFIN=100, CINT=10 1 SIM; INTERACT; RESET; G O T O 1 INITIAL A=AO; B=BO DYNAMIC Rl=Kl*A*B R2=K2*A*C R3=K3*C*D R4=K4*B*D R5 =K5*C*C R6=K6*C*B R7=K7*A*G R8=K8*A*F A1=-R1-R2-R7-R8 B1=-R1-R4-R6 C1=R1-R2-R3-R5-R6 D1=R2-R3-R4 E'=R3+R8 F1=R4+R5+R7-R8 G'=R6-R7 P L O T T,A,O,TFIN,0,0.15 PREPARE T,A, B, C, D, E, F, G, R 1 , R 2 , R3, R4, R5, R6, R7 O U T P U T T,A,B,C,D $ VAL TFIN = 1000.0

Nomenclature

Symbols C k R

Concentration Rate coefficient Rate

Indices A,B, ..., G 1,2, ..., 7

Refer to components Refer to reactions

mol/L Wmol min m o l b min

5.1 Batch Reactor Examples

289

Exercises 1.

Vary the initial concentrations of the reactants A and B and observe their effect on the relative rates of the individual reactions and on the resulting product distribution. Note how the distribution varies with time,

2.

Write the model in dimensionless form, program it and compare the results with those from Exercise 1 .

Results ISIW

c c

D D 8.18

Figure 5.7. Concentration-time profiles using the parameters as given in the program.

Figure 5.8. Phase-plane plot showing variation of R2 and R6 versus R1.

References Ampaya, J.P. and Rinker, R.G. (1974) Kinetics: Computer Programs for Chemical Engineeering Education, Ed. M.Reilly, Aztec Publ.Co. C.A.Ch.E. Comm. AIChE. Stults, F.C., Moulton, R.W. and McCarthy, J.L. (1952) Chem. Eng. Prog. Symp. Series, Vol4, 38.

5 Simulation Examples of Chemical Engineering Processes

290

BATCOM - Batch Reactor with Complex Reaction Sequence

5.1.4

System Product B is manufactured by the liquid phase batch reaction, whose complexity gives a range of byproducts.

Figure 5.9. This complex reaction can be altered to fit various situations.

The batch reactor can be run in the temperature range 190 to 250°C. Reactant X is in large excess and C, D, and E are undesired byproducts. Reactions follow first-order relationships, where, r = -kC

Model For the batch reactor with these complex kinetics, the model is as follows _ dCA __-

dt

-

- (kl

+ k2) CA

dCB dt~-~= kl CA - k3 CB

29 1

5.1 Batch Reactor Examples

Program The program can be used to generate repeated runs for differing values of reactor temperature TR, using the ISIM interactive facility. The range of interest is 190 to 250°C. :Example

BATCOM

BATCH REACTION OF KANTYKA TEMPERATURE OPTIMIZATION OF BATCH : CONSECUTIVE AND PARALLEL REACTION CONSTANT T R = 1 9 0 CONSTANT R=1.9869 CONSTANT CINT=1.0 CONSTANT TFIN=150 1 S1M;RESET; INTERACT; GOT0 1 : :

INITIAL A=l B=O

c=o

D=O Tl=R* (TR+273) K1=3.46831E+12*EXP(-29777/Tl) K2=2.3539OE+14*EXP(-35919/Tl) K3=985521*EXP(-18619/Tl) K4=3.75840E+lO*EXP(-26638/Tl) K5=1.30000E-02 DYNAMIC A'=-(Kl+K2)*A B'=Kl*A-K3*B C1=K2*A-K4*C D'=K3*B-K5*D PLOT T,B,O,TFIN,O,l OUTPUT T, A, B, C, D PREPARE T, A , B, C, D

REACTOR SEQUENCE

292

~

_

_

5 Simulation Examples of Chemical Engineering Processes _

-~

Nomenclature Symbols Concentration Activation energy Reaction rate constant Gas constant Time Reactor temperature Reactor temperature Arrhenius constant

C Ea

k R t T1

TR Z

mol/L cal/mol 1 /min cal/mol K min K "C Umin

Indices A, B, C, D, E 1, 2, 3 , 4 , 5

Refer to components Refer to reactions

Exercises 1.

Determine the optimum batch time and batch temperature giving maximum yield of B.

2.

Under optimum conditions, determine the concentrations of A, B, C, D and E obtained.

3.

Check your results with the following analytical expressions

Note: These equations can be easily programmed using ISIM.

4.

Introduce a new variable rg describing the net formation rate of B. Determine the optimum batch time and batch temperature.

293

5.1 Batch Reactor Examples

Results Ism

R B

1

e. im 3

Figure 5.10. Profiles of B versus time for a range of reactor temperatures, TR, from 180 to 240 K, curves A to E. The optimum corresponds to TR = 220 (curve C).

Figure 5.11. Profiles of all concentrations for TR = 220.

Reference Binns, D.T., Kantyka, K.A. and Welland, R.C. (1969) Trans. Inst. Chem. Engrs. 47, 53.

CASTOR - Batch Decomposition of Acetylated Castor Oil

5.1.5

System The batch decomposition of acetylated castor oil to drying oil proceeds via the following reaction: Acetylated Castor Oil

Heat

Drying Oil

+ Acetic Acid Vapour

294

5 Simulation Examples of Chemical Engineering Processes

Acetic acid vapor

Figure 5.12. The reactor for CASTOR, showing acetic acid vapour escaping.

The reaction is endothermic, and heat transfer to the reactor is required in order to accomplish the decomposition of the acetylated oil, to liberate acetic acid vapour. The example has been considered previously by Perona (1972), Smith (1972), Cooper and Jeffreys (1971) and Froment and Bischoff (1990), although their solution procedures differ from that here. Data values are based on those used by Froment and Bischoff.

Model The solution here treats the process conventionally as occurring at constant reactor mass, although actually some mass is lost via the acetic acid vapour. The rate of production of acetic acid (kg acidkg charge sec) is rp = k C A where

1 k = 60 exp ~~

(35.2-

E $ 3 )

The acetyl component balance for constant mass conditions is given by

and hence

295

5.1 Batch Reactor Examples

dX - = k(1-X) dt

The energy balance equation is given by dT MCpx

= -kCAMAH+Q

= -kMCAO(l-X)AH+Q = -kMl(l-X)AH+Q

where M1 = M CAOis the equivalent acetic acid content of the castor oil. The acetyl balance is

where

and the fraction conversion is

The heat balance is given by dMT

'P - d T

= -kCAM(AH)+Q

Program An IF statement could be used to set the value of the heating rate equal to zero, once a fractional conversion of 90 per cent is achieved. :Example CASTOR BATCH DECOMPOSITION OF ACETYLATED ACETIC ACID : CONSTANT MASS MODEL :Total batch charge CONSTANT M = 2 2 7 CONSTANT CAO=O.l56:Acetic acid equivalent of charge CONSTANT CP=2.51 :Specific heat :

296

5 Simulation Examples of Chemical Engineering Processes

CONSTANT H=1050 :Endothermic heat of reaction CONSTANT TO=631 :Initial batch temperature CONSTANT Q = O :Heat input CONSTANT TFIN=2000, CINT=20 1 SIM; INTERACT; RESET; GOT01 INITIAL TR=TO : Equivalent mass of acetic acid in charge Ml=M*CAO DYNAMIC K=(1/60)*EXP(35.2-2245O/TR) : Component balance X' =x* ( 1 - X ) : Energy balance TR'=(-Ml*H*K*(l-X)+Q)/(M*CP) IF ( X GE 0.9) Q = O PLOT T, X, 0, TFIN, 0 , 1 OUTPUT T, X I TR PREPARE T IX I TR, Q ,K

Nomenclature

Symbols M M1

Total mass of castor oil charge Acetyl content of reactant Equivalent acetic acid content Initial equivalent acetic acid content Kinetic rate constant Heat of reaction with vaporisation Reactor temperature Rate of production of acetic acid Time Specific heat Heat input

CA CAO

k AH TR 'P

t

cP

Q

Exercises 1.

Study the effect of differing rates of heat input on the fractional conversion X and reactor temperature TR, for the range Q = 0 to 200 kW, and note particularly the effect on the reactor time required to obtain a given conversion efficiency.

297

5.1 Batch Reactor Examples

2.

Repeat Exercise 1 for differing batch starting temperatures.

3.

In your simulations, plot temperature and conversion versus time and also plot temperature versus fractional conversion.

4.

Change the program to consider the adiabatic case. Calculate the temperature drop for adiabatic operation from the equation.

Verify this by simulation.

Results

Figure 5.13. Temperature profiles obtained by setting Q=O, 30, 60, and 90 (A, B, C , and D)

Figure 5.14. Conversion correspond-ing to Fig. 5.13.

profiles

References Smith ( 1 972) Chemical Engineering Kinetics., McGraw-Hill, Vol. 2. Cooper and Jeffreys (1971) Chemical Kinetics and Reactor Design, Chem Engr. Texts. Perona, J. (1 972) Computer Programs for Chemical Engineering Education: Kinetics, Ed. M. Reilly, AIChE.

298

5 Simulation Examples of Chemical Engineering Processes

Frornent, G.F. and Bischoff, D.B. (1990) Chemical Reactor Analysis and Design. 2nd edition, Wiley.

5.1.6

HYDROL - Batch Reactor Hydrolysis of Acetic Anhydride

System The liquid phase hydrolysis reaction of acetic anhydride to form acetic acid is carried out in a constant volume, adiabatic batch reactor. The reaction is exothermic with the following stoichiometry (CH3C)20 + H20

+

2 CH3 COOH + heat

Figure 5.15. The adiabatic batch rcactor with variables of concentration and temperature.

Model The acetic anhydride balance can be formulated as

(

Rate of accumulation in reactor

hence, for the first-order reaction

of loss due to > = ( Rate chemical reaction

299

5.1 Batch Reactor Examples

and for constant volume conditions

The fractional conversion for the batch is

The temperature influence on the rate is given by

k = Z e-E/RTR The heat balance under adiabatic conditions is as follows: of accumulation ( Rate of heat in the reactor ) = (

Rate of heat generation by chemical reaction

Program :Example

HYDROL

BATCH HYDROLYSIS OF ACETIC ANHYDRIDE EXAMPLE OF REACTION WITH HEAT EFFECTS : UNDER ADIABATIC CONDITIONS

:

:

CONSTANT R = 1.987 CONSTANT CINT = 0.5 CONSTANT TFIN = 50 : PHYSICAL PROPERTIES CONSTANT RO = 65.5 CONSTANT CP = 0.9 CONSTANT H = -90000 :EXOTHERMIC : KINETIC DATA CONSTANT 2 = 6.915003+07 CONSTANT E = 21265

HEAT

OF

REACTION

300

5 Simulation Examples of Chemical Engineering Processes

: INITIAL CHARGE CONDITIONS CONSTANT M = 500 CONSTANT A0 = 1.350003-02 CONSTANT TRO = 600

1

SIM;

INTERACT;

RESET;

GOT0

1

INITIAL A = A0 TR = TRO V = M/RO DYNAMIC K = Z*EXP ( - E /(R*TR)) A' = -K*A TR' = (-K*A*V*H)/(M*CP) XA = (AO-A)/AO PLOT T,XA, O,TFIN, 0,l OUTPUT T I A, TR I XA PREPARE T IA, TR, XA

Nomenclature Symbols CA

C A0

AH k M R TR V X P

Concentration of A Initial concentration of A Specific heat Activation energy Heat of reaction Rate constant Mass in tank Gas constant Temperature in reactor Volume of charge Fractional conversion Solution density

lb mol/ft3 lb mol/ft3 BTU/lb O F BTU/lb mol BTU/lb mol l/min lb BTU/lb mol O O R

ft3

-

lb/ft3

R

301

5.1 Batch Reactor Examples

Exercises 1.

Study the effect of differing initial batch charge temperatures TRO (500 to 600"R) on the fractional conversion XA, the reactor temperature TR, and the required batch time.

2.

What initial temperature is required to give 99% conversion if the batch temperature must not exceed 600"R? What is the corresponding batch time required?

3.

What is the batch time requirement for the reactor operating isothermally at 60°F. How must the heat removal requirement vary during the batch in order to maintain the batch temperature constant?

Results

AM

e m

+-/

Figure 5.16. Temperature variations with time for a range of starting temperature TRO, as shown in Fig. 5.17.

Figure 5.17. Variation of fraction conversion during the runs in Fig. 5.16.

Reference Cooper, A.R. and Jeffreys, G.V. (1971) Chemical Kinetics and Reactor Design, Chem. Engng. Texts, 133.

302

5.1.7

5 Simulation Examples of Chemical Engineering Processes

OXIDAT - Oxidation Reaction in an Aerated Tank

System The influence of gassing rate and stirrer speed on an oxidation reaction, in an aerated batch reactor, are to be investigated. The outlet gas is assumed to be essentially air, which eliminates the need for a gas balance for the well-mixed gas phase.

~1

00 0

O

000

0

t

2A

+ 02 +P

0 Air

I

Figure 5.18. Schematic of the batch enzymatic oxidation reactor.

Model The reaction kinetics are described by a second-order relation in reactant A and oxygen

The stoichiometry gives

rp =

1 -2 rA

303

5.1 Batch Reactor Examples

The batch mass balances are

v dA dt

---

= rA

v

V dCL dt = KLa (Co* - Co) V ~

+ ro V

V-dCP dt = r p V

KLa varies with stirring speed (N) and aeration rate (G) according to KLa = kt N3

where kt = 4.78 x

with N in l/min, G in L/h and KLa in l/h.

Program :

Example

OXIDAT

BATCH OXIDATION IN AGITATED TANK KLA VARIES WITH STIRRING RATE N AND GASSING RATE G : UNITS ARE IN MOLS,L,H CONSTANT N=500, G=60000, KT=0.9E-9 CONSTANT KR=2000 CONSTANT COS=0.25E-3,CAO=0.2 CONSTANT CINT=0.0001, TFIN=O.1 1 SIM; INTERACT; RESET; GOT0 1 INITIAL CA=CAO :IN MOL/L co=o :IN MOL/L CP=O :IN MOL/L DYNAMIC KLA=KT*(N**3)*(G**O.5) :INFLUENCE OF N(RPM) AND G(L/H) ON KLA(l/H) RA=-KR*CA*CO :KINETICS RO=O. 5*RA RP=-0.5*RA :MASS BALANCES CA' = R A CO'=KLA*(COS-CO)+RO : :

304

5 Simulation Examples of Chemical Engineering Processes

CP ' =RP PLOT T, CO, 0, TFIN, 0,O. 253-3 PREPARE T, CA, CO, CP, RA, RO, K L A OUTPUT T, CA, CO, CP, KLA, RA

Nomenclature Symbols Dissolved oxygen concentration Saturation oxygen concentration Aeration rate Transfer coefficient Second-order rate constant Constant in KLa correlation Stirring rate Product concentration Reaction rate

mol/L molL m3/h l/h L/mol h min3/ho.s Lo.5 l/h mol/L mol/L h

Indices 0 A 0 P

Refers Refers Refers Refers

to feed to reactant to oxygen to product

Exercises 1.

Vary stirrer speed N and aeration rate G and observe the response in dissolved oxygen Co and product concentration cp.

2.

Vary operating parameters in the very low Co range (2*1OU6mol/L) and note the influence on the rates.

3.

Change the program so that the oxygen transfer rate is calculated and can be plotted. Verify that the rate of reaction is limited by oxygen transfer.

4.

Calculate the value of KLa required to give Co = C0*/2 when CA = 0.2. Verify by simulation.

305

5.1 Batch Reactor Examples

5.

Reduce the aeration rate until the reaction becomes oxygen-limited. This can be done interactively. Maintain this and then reduce the catalyst amount by 1/2 (reduce v, by one-half). Does the initial rate decrease by 1/2? Is the reaction still oxygen-limited?

6.

Run the process oxygen-limited and observe the increase of Co at the end of the reaction.

7.

Increase N by 25% and notice the change of KLa and Co.

8.

Vary N and G under non-oxygen-limiting conditions. What is the effect on the reaction rate?

Results The results in Figs. 5.19 and 5.20 show the influence of stirrer speed and aeration rate.

ISM

r m

k--

RCO BCP

Bcp

8.

ISIN

-E 4.

m

8.858

E.888

E z

8.’M

8.489

8.688

8.888

“E -1

Figure 5.19. Response of dissolved oxygen and product concentration for a change in stirrer speed from N = 500 to 1000 to 100 rpm.

8.888 8.888

8.288

8.488

8.6m

8.888

-f -1

Figure 5.20. Response of Co (in the 1% saturation range) and Cp to changes in G (60000 to 80000 L/h) and N (100 to 200 rpm). Note that the rate of P formation is proportional to Co levels.

306

5.1.8

5 Simulation Examples of Chemical Engineering Processes

RELUY - Batch Reactor of Luyben

System The first-order consecutive exothermic reaction sequence, A -+ B + C, is carried out in a thick-walled, jacketed batch reactor, provided with both jacketed heating and cooling, as shown below.

Figure 5.21. Batch reactor with temperature control, adding steam and water via a split-range valve to the reactor jacket.

The reactants are charged and the contents heated to the starting batch temperature via the split range valve V,, using steam to the jacket which is supplied until the reactor temperature, T I , reaches the temperature Tmax. Then the steam supply is stopped, and cooling water is supplied to the jacket via the split range valve Vw at a controlled rate, set in accordance with a prescribed temperature-time profile, by a controller cam operating on the set point. Tma, is held for a period, tihold, then a linear ramp is followed. B is the desired product of the reaction, and the aim is to find the optimum batch time and temperature to maximise the selectivity for B. Saturated steam density data are taken from steam tables and fitted to a polynomial. The model and data for this example are taken from Luyben (1973). A full comparison of actual experimental results and computer simulation study is given by Marroquin and Luyben (1973).

307

5.1 Batch Reactor Examples

Model The balances for the reactor liquid are as follows Total mass balance, assuming constant density

Thus V = constant. Component balances,

where

kl = Z1 e -

The reactor energy balance is dT p V cP dt =

-V

kl CA (-AH])

+ V k2 CB (-AH2)

-

U1 A (T - Tm)

The energy balance for the metal wall at constant cp is

For both heating and cooling, the energy balance remains the same except that differing values of the overall heat transfer coefficient now apply. Thus for the coolant Us is replaced by U,. A mass balance for the jacket is used to calculate the temperature of the saturated steam.

From steam table relations

308

5 Simulation Examples of Chemical Engineering Processes

During the cooling period, the energy balance for the jacket is

The steam valve is initially opened slowly by increasing X, from 0 to 1 and then kept open until the reactor temperature, T I , reaches Tmax. Then it is closed for time tihold. Then the cooling ramp and the temperature is controlled at T,, is started. The control equation for the water valve is as follows

x,

=

(9 - PC) 6

where X, is bound to 0 5 X, 5 1. Additional equations are required for the flow rates of the steam and the cooling water to the jacket. Mass flow rates for steam

F, = ~ , , ~ , . \ j p , o - p j Volumetric flow rate for water

The temperature transmitter has a range of 50 to 250°F and an output 3 to 15 psig. Hence,

Ptt = 3 + (T - 50) 0.06 For the proportional feedback controller

Set point for the water cooling is given by

309

5.1 Batch Reactor Examples

where tSt is the time when the reactor temperature first reaches a temperature, and Tmax and tihold is the time period to keep Tmax.

Program The ISIM program makes use of conditional statements to create the necessary heating and cooling periods: This is done by changing a logical variable PERIOD from an initial value of 1 , during the heating period, to a value of 2 for the cooling period. Using these values in GOT0 statements, the calculation is moved from one part of the program to the other. Further details for each period are as follows: PERIOD = 1 is the heating period using a steam mass flow rate of Fs. The mass balance on the steam jacket determines the steam density ps, which together with a steam table function determines the temperature of the jacket Tj. When the reactor temperature (TI) becomes greater than Tmax (=240 F), PERIOD = 2, and the program turns the cooling water on with flow rate Fw. This flow is controlled with a proportional controller using control constant Kc, whose set point (Pset) is varied according to the time ramp function with setting kR and whose output to the valve is Pc. This ramp is horizontal until time period Tihold has passed. Then the setpoint is decreased linearly. The temperature is sensed using a pressure transmitter with output Ptt. In running the program, the initial steam heating period is unfortunately rather slow in execution, owing to equation stiffness. To reduce stiffness at the beginning, an appropriate initial value of the steam density is calculated in the FORTRAN subroutine START, which uses the halfinterval method for the non-linear algebraic equation. Note that the execution may be very slow because of equation stiffness. Increasing the value of CINT during the initial heating period may terminate ISIM execution. :Example

RELUY

:Control of a Batch :Problem of Luyben CONSTANT CONSTANT CONSTANT CONSTANT

Reactor

TFIN=4 0 0 ALGO= 0 CINT=O. 0 5 NOCI= 10

CONSTANT CAO=O . 2 const ant CBO=O constant T10=184.07 const ant RHOsO=O. 3

:initial conc

[mol ft-31

:initial reactor temp [Fl :init. steam density [lb ft-31

310

5 Simulation Examples of Chemical Engineering Processes - -~

constant ACCUR=O.l t e m p e r a t u r e [Fl CONSTANT NMAX=30

:accuracy

CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT

:CONTROLLER

KC=O.l PO=9 KXs=1 CvS=lO kR=0.03 Tmax=240 Tihold=100 Ps0=35 V=42.5 Vm=9.42 Vj=18.83 Cvw=lO

:HEAT TRANSFER DATA CONSTANT U1= 5 CONSTANT U ~ = 1 6 . 6 7 CONSTANT U ~ = 6 . 6 7 CONSTANT A =56.5

for

:steam valve

:RAMP

initial

constant

:temp. where cooling started :time period t o hold Tmax :psig :reactor volume :metal jacket volume :jacket volume [ft31 :WATER VALVE CONSTANT :btu min-1 ft-2 F-1 :ft2

: REACTION DATA CONSTANT 21=729.5 CONSTANT E1=15000 :K1 = Zl*exp(-El/RT) [min-11 CONSTANT Z2=6567.6 CONSTANT E2=20000 :K2 = Z2*exp(-E2/RT) [min-1] CONSTANT H1=40000 :[btu m01-1] CONSTANT H2=50000 CONSTANT R=1.987 : PHYSICAL PROPERTY DATA CONSTANT C p = l :specific heat vessel contents [btu lb-1 F-11 CONSTANT RH01=50 :dens. vessel contents [lb ft-31 CONSTANT Cpm=.l2 :spec. heat wall [btu lb-1 ft-31 CONSTANT RHOm=512 :density wall [lb ft-31 : STEAM DATA CONSTANT KA=-8744.4 CONSTANT KB=15.7 CONSTANT Lambda=940 :latent heat of steam [btu lb-1] CONSTANT TO=70 :temp of cooling water [Fl CONSTANT P j = 2 5 : Temperature from steam density CONSTANT STDO=145.01

5.1 Batch Reactor Examples

311

CONSTANT STD1=2287.4 CONSTANT STD2=-1.4657E4 CONSTANT STD3=4.3092E5 :WATER DATA CONSTANT CpW= 1 CONSTANT RHOw=62.3 INITIAL T1=T10 CA=CAO CB=CBO Tm=T1 Tj = T 1 T j O=Tmax :only valid when cooling ramp started CALL START(RHOsO,RHOs,Tj,ACCUR,STDO,STDl,STD2,STD3,NMAX) PERIOD=l : initial valve settings xs=o xw=o Ptt = 3+(T1-50)*0.06 : pressure temperature transmitter PsetO = 3+(Tmax-50)*0.06 : pressure setpoint at tmax Tjabs = Tj + 460 : absolute jacket temperature (Rankine) Pj = EXP(KA/(Tjabs)+KB) : Pjacket [regression from steam tables1 FsA = PsO-Pj IF(FsA.LT.O.0) FSA=O.O FS = Xs*Cvs*SQRT(FaA) :Fs is steam Flow Qj = Us*A*(Tj-Tm) SEL=O kRA=O

DYNAMIC : REACTION KINETICS Tabs=T1+460 :absolute temp. of vessel [Rankin] Kl=Zl*EXP(=-El/(R*Tabs)) K2=22*EXP(=-Ez/(R*Tabs)) CA'=-Kl*CA CB'=Kl*CA-K2*CB : HEAT BALANCE ON REACTOR Ql=Ul*A*(Tm-Tl) T~'=(K~*CA*H~+K~*CB*H~)/(RHO~*C!~)+Q~/(V*RHO~*C~)

312

5 Simulation Examples of Chemical Engineering Processes

: JACKET CONDITIONS Fcw = Qj/Lambda : TRANSMITTER, CONTROLLER, VALVE, STEAM and Water FLOW : Tj temperature of jacket content : Tj from steam density IF(PERIOD.EQ.1) T~=STDO+STD~*RHO~+STD~*RHOS**~+STD~*RHOS**~ : Tj from heat balance Tj '=Fw* ( T O - T j) / V j - Q j / ( V j*RHOw*cpw) I F (PERIOD.EQ. 2) : HEAT BALANCE ON WALL IF(PERIOD.EQ.2) Qj = Uw*A*(Tj-Tm) IF(PERIOD.EQ.1) Qj = Us*A*(Tj-Tm) Tm'=(Qj-Ql)/(Vm*RHOm*cpm)

Ptt = 3 + (Tl-50)* O . 06 : pressure temperature transmitter Tjabs = Tj + 460 : absolute jacket temperature (Rankine) Pj = EXP(KA/(Tjabs)+KB) : Pjacket [regression from steam tables1 IF(PERIOD.EQ.2) Pset = PsetO-kRA*(T-(TST+Tihold)) IF (PERIOD.EQ. 1) Pset = 12.6 Pc = PO+KC*(Pset-Ptt) IF(PERIOD.EQ.1)Xs = KXs*T IF(Xs.GT.1.0) Xs=l.O :Xs position of steam valve IF(PERIOD.EQ.2) XS=O.O XW = ( 9 - P ~ ) / 6 IF(Xw.LT.O.0) XW=O.O IF(Xw.GT.1.0) XW=l.O :Xw position of water valve FSA = PSO-Pj IF(FsA.LT.O.0) FsA=O.O Fs = Xs*Cvs*SQRT(FsA) :Fa is steam Flow RHOS' = (Fs-Fcw)/Vj IF(PERIOD.EQ.~)FS=O.O FW = 37.25*Cvw*XW :Fw flow of cooling water IF(PERIOD.EQ.1) TST = T :start time for cooling if(T.GE.(TST+Tihold))kRA=kR :start cooling ramp IF ((PERIOD.EQ.l).AND.(T1.LT.Tmax)) GO TO 500 PERIOD = 2 5 0 0 CONTINUE IF(CAO.GT.CA)SEL=(CB-CBO)/(CAO-CA) :PLOT T,Tl,O,TFIN,O,250 OUTPUT T, TI, TJ, FW, CA

313

5.1 Batch Reactor Examples

PREPARE

T,CA,CBlT1,TJ,TM,FW,RHOS,PSET,SET,SEL

SUBROUTINE START(RHO~~,RHO~,Tj,ACCURlSTDO,STDllSTD2,STD3,NM) RHOS=RHOsO RLOW= 0 RHIGH=2*RHOs 100 CONTINUE Tjl=STDO+STDl*RHOS+STD2*RHOS**2+STD3*RHOS**3 IF (Tjl.LT.Tj) GOTO 110 RHIGH=RHOs RHOs=RLOW+(RHIGH-RLOW)/2 GOTO 130 110 CONTINUE RLOW=RHOs RHOs=RLOW+(RHIGH-RLOW)/2 13 0 NUM=NUM+1 IF (NUM.GT.NM) GOTO 150 IF (ABS(Tj-Tjl).GT.ACCUR) GOTO 100 150 CONTINUE

Nomenclature Symbols Heat transfer area of jacket Concentration Specific heat of vessel contents Specific heat of metal wall Specific heat of cooling water Valve constant for steam Valve constant for water Activation energy of reaction Steam flow rate Cooling water rate Condensed water flow rate Reaction rate constant Constant in steam pressure relation Constant in steam pressure relation Controller constant Ramping constant Controller pressure Controller pressure output Steam vapour pressure

ft2

mol/ft3 BTUAb O F BTUAb O F BTUAb OF lb/min ( ~ s i ) O . ~ ft3/min (psi)O.5 BTU/mol lb/min ft3/min ft3/min

m1n-1 OR

-

psi/min Psig Psig psig

3 14

5 Simulation Examples of Chemical Engineering Processes

Steam supply pressure Set point pressure Pressure of temperature transmitter Water supply pressure Gas constant Temperature in reactor Absolute reactor temperature Temperature of jacket contents Steam supply temperature Temperature of metal wall Water supply temperature Temperature for batch start is first reached Time when T,, Time period to hold Tmax Heat transfer coeff. contents to wall Heat transfer coeff. wall to coolant Heat transfer coeff. wall to steam Vessel volume Jacket volume Wall volume Steam valve position Water valve position Arrhenius rate constant Reaction enthalpy Latent heat of steam Density of vessel contents Density of steam in jacket Steam supply density Density of wall Density of cooling water Refers to component A Refers to component B Refers to inlet Refers to reaction A --> B Refers to reaction B -> C

Psig PSk

Psig Psig BTU/mol "R O F

"R "F "F "F "F "F min min BTUh ft2 "F BTUh ft2 "F BTUlh ft2 "F ft3

ft3 ft3

-

min-1 BTU/mol BTUAb Ib/ft3 lb/ft3 Ib/ft3 lb/ft3 lb/ft3

5.1 Batch Reactor Examples -

315

-

Exercises 1.

2. 3.

Vary the temperature of the heating period limit, Tma,, and note the influence on the concentration profiles. Study the effects of the parameters of the cooling water ramp function (Tihold and kR) on the selectivity, SEL. Investigate the influence of the wall mass and the heat transfer coefficients on the maximum yield of B.

Results IS11

Figure 5.22. The temperature of the reactor rises to TI = 240 during the heating period. TI is hold for the period Tihold and then the flow FC is controlled to let T1 follow a ramp decrease.

ISI!

Figure 5.23. Concentration profiles.

References Luyben, W.L. (1973) Process Modelling, Simulation and Control, McGraw-Hill. Marroquin and Luyben, W.L. (1973) Chem. Eng. Sci. 28, 993.

5 Simulation Examples of Chemical Engineering Processes

316

5.2

Continuous Tank Reactor Examples

5.2.1

CSTRCOM - Isothermal Reactor with Complex Reaction

System Complex reactions can be conveniently investigated by simulation techniques. In this example from Russell and Denn (1972), the characteristics of the following complex reaction can be investigated. k

A + B A X

B+X&Y B + Yk a Z The reaction is sequential (A to B.

Fg

+ X -+

Y

+ Z), as well as parallel with respect

CAOi cBO

7 V

03

CY Ct

Figure 5.24. Stirred tank with series-parallel reaction.

Model For each component

3 17

5.2 Continuous Tank Reactor Examples

[accumulation) Rate of = in reactor

(“p) (:$) -

+ [formation) Rate of by reaction

All three reactions are assumed to follow simple second-order kinetics, and thus the mass balances become

Dividing by V, the equations can be written in terms of the mean residence time of the reactor, ‘t = F/V.

Program :Example :Single Constant Constant Constant Constant 1

SIM:

INITIAL A=AO B=BO

x=o Y=O

CSTRCOM Isothermal

CSTR

with

complex

AO=0.4, BO=0.6, XO=O, YO=O, K1=5E-4, K2=5E-2, K3=2E-2 TAU=5000 CINT=500, TFIN=25000 INTERACT:

RESET:

GOT0

1

reaction

ZO=O

318

_ _ _ ~ ._ -

5 Simulation Examples of Chemical Engineering Processes - - -~

~~

-

z=o

DYNAMIC A'=(AO-A)/TAU-Kl*A*B B'=(BO-B)/TAU-Kl*A*B-KZ*X*B-K3*Y*B X'=(XO-X)/TAU+Kl*A*B-K2*X*B Y'=(YO-Y)/TAU+K2*B*X-K3*B*Y Z1=(ZO-Z)/TAU+K3*B*Y PLOT T i A i O , T F I N I O i A O OUTPUT Ti A, Bi X i Y i Z PREPARE T,A,B,X,Y,Z

Nomenclature Symbols C F k V z

Concentration Flow rate Rate constant Volume Mean residence time

kmol/m3 m3/s m3krnol s m3 S

Indices A,B,X,Y,Z 1, 2, 3

Refer to components Refer to the reaction steps

Exercises 1. 2. 3.

Compare the steady-state values of A, B, X, Y, Z with calculated values obtained by solving the steady-state component balance equations for this problem. Study the effect of varying residence time z, feed concentration, and rate constants on reactor performance. Note that when kl = k2 = k3 = 0, CAO= 1, CBO= 0, the program solves the case of a step input of tracer solution, which can be used to generate the typical F-diagram for a single perfectly mixed tank. Compare this result with the analytical solution.

319

5.2 Continuous Tank Reactor Examples

Results The concentration levels are very sensitive to the kinetic parameters. Using the parameters in the program the following results were obtained

151N

B,lak+T

8.8888.888

8.875

8.158

8.225

8.m

8 . b

*E 5

Figure 5.25. After steady state was reached, the value of T was increased form 5000 to 10000 s.

8.888 8.888

8.875

8.158

8.225

*E

8.388 5

Figure 5.26. The components X and Y are present at very low concentration. Note that the scales are different from Fig. 5.25.

Reference Russell, T.W.F. and Denn, M.M. (1972) Introduction to Chemical Engineering Analysis, McGraw-Hill.

5.2.2

DEACT

- Deactivating

Catalyst in a CSTR

System Solid catalysts can be conveniently studied in loop reactors, which allow measuring the rates by difference measurement across the catalyst bed. When operated continuously, they usually can be modelled as well-stirred tanks. Here the case of catalyst deactivation is studied.

3 20 -

5 Simulation Examples of Chemical Engineering Processes ___

-__~___-_____-__-__-__

catalyst

A

-

P

Figure 5.27 A recycle-loop reactor for catalyst studies.

Model The kinetics are

and T h e mass balance for component A is

T h e mass balance for product P is

The batch balance for the catalyst with first-order deactivation is

__

5.2 Continuous Tank Reactor Examples

32 1

Program :Example

DEACT

Deactivating catalytic reaction : Langmuir-Hinshelwood kinetics : in a completely mixed reactor. :

with

Constant V=10 :Reactor volume [m3I Constant F=0.1 :Flow rate [m3/hl Constant Cc0=20 : Initial catalyst conc. [kg/m3] Constant Ca0=500 :Initial reactant conc. [mol/m3] :Reaction constant Constant k=O. 9 Constant Ki=O. 2 :Inhibition constant Constant kd=0.05 :Deactivation constant Constant CINT=l, TFIN=200, NOCI=10 1 SIM; INTERACT; RESET; GOT0 1 INITIAL Ca=CaO cc=cco CPIO TAU=V/F :Mean residence time [hl DYNAMIC rp=k*Cc*Ca/(l+Ki*Ca) Ca'=(CaO-Ca)/TAU-rp Cc'=-kd*Cc Cp'=rp-Cg/TAU PLOT t Cp, 0, TFIN, 0, CaO PREPARE T, Ca, CC, Cp, rp

Nomenclature Symbols C CC F k

Concentration Catalyst concentration Flow rate Reaction constant

mo11m3 kg/m3 m31h m3kg h

5 Simulation Examples of Chemical Engineering Processes

3 22

m3/m01 1/h mol/m3 h h m3

Kinetic constant Deactivation constant Reaction rate Residence time in loop Reactor volume

KA

KD r z V

Indices A, C, P 0

Refer to reactant, catalyst and product Refers to initial value

Exercises FI

Observe the influence of the catalyst decay constant, KD, on the outlet product concentration, which should slowly decrease. Make runs for a range of KD values.

2.

Attempt to maintain the product concentration constant, either by decreasing the flow rate (increasing TAU) or increasing the inlet reactant concentration.

3.

Assuming P can be measured, devise and program a control system to keep the product level constant.

Results

0.198

Figure 5.28. As product concentration, CP, dropped, the catalyst concentration, CC, was set to its original value at T = 75 and 165.

Figure 5.29. The influence of CC changes on the product rate, RP.

5.2 Continuous Tank Reactor Examples

5.2.3

323

TANK and TANKD - Single Tank with Nth-Order Reaction

System An nth-order reaction is run in a continuous stirred-tank reactor. The model and program are written in both dimensional and dimensionless forms. This example provides experience in the use of dimensionless equations.

Figure 5.30. A continuous-tank reactor with nth-order reaction.

Model The balance for the reactant A is

and the kinetics are

Dimensionless variables are defined as

3 24

5 Simulation Examples of Chemical Engineering Processes

where 7,the residence time, is V/F. The dimensionless balance equation is then

and the rate equation becomes

-

rA =

-k

n -

n

CAo CAI

Thus the governing dimensionless parameter is ( k z C.,-d ).

Program Program TANK solves the normal dimensional model equation for the problem, whereas TANKD is formulated in terms of the dimensionless model equations. :Example TANK Outlet concentration of a CSTR : nth-order kinetics Constant V=lOOO : [dm31 Volume Constant F=10 : [dm3/minl Feed Constant CAO=100 :[g/dm31 Feed concentration Constant n = 2 : [ - I Reaction order constant k=O. 1 :[various1 Reaction constant Constant CAI= 0 : [g/dm31 Initial conc. Constant CINT=O.1, TFIN=5, NOCI=10 1 SIM; INTERACT; RESET; GOT0 1 INITIAL CAl=CAI DYNAMIC TAU=V/F Ra=-k*CAl**n :Reaction rate CAI’=l/TAU*(CAO-CAl)+Ra PLOT T,CAl,O,TFIN,O,CAO OUTPUT T, CA1, Ra PREPARE T, CAI, Ra :

325

5.2 Continuous Tank Reactor Examples

:Example TANKD :Continuous stirred-tank reactor. Dimensionless form :Compare with TANK :nth-order kinetics Constant V=lOOO : [dm31 Volume Constant F=10 : [dm3/minl Feed Constant CAO=100 :[g/dm31 Feed concentration Constant n = 2 :[ - I Reaction order Constant k=0.1 : [various1 Reaction constant Constant CINT=O.Ol, TFIN=l, NOCI=10 1 SIM; INTERACT; RESET; GOT0 1 INITIAL CAlD=O DYNAMIC TAU=V/F TIME=TAU*T CAl=CAO*CAlD PAR=k*TAU*(CAO**(n-1)) CAlD'=(l-CAlD)-PAR*CAlD**n PLOT T, CAlD 0,l 0,O. 4 OUTPUT T , C A I D I P A R , C A 1 PREPARE T, TIME, CAID, CA1, PAR I

I

Nomenclature Symbols CAO CA 1 F k n T

t V

Feed concentration Concentration in reactor Feed Reaction constant Reaction order Residence time Time Volume

Exercises Starting with program TANK:

min h dm3

326

__

5 Simulation Examples of Chemical Engineering Processes _____~

~~

1.

Run the simulation for a range of residence times. When are the steady states reached?

2.

Change the initial concentration and rerun the simulations.

3.

Turn off the flow by setting the value of TAU initially very high, and start the reactor as a batch system. Change over to continuous operation using the interactive facility to define a value of TAU.

4.

Run for several values of n and plot rA versus CAI for each case. Calculate the steady-state concentrations for n = 0, n = 1 and n = 2.

5.

Compare the results from TANK and TANKD for a range of parameters.

6.

Calculate the steady state for n = 1 by hand. Run the simulation and compare. Change CAOfor this case and run. Compare the effects on and CAI.

7.

Investigate the time to reach steady state for various parametric values.

8.

With zero initial conditions, run for n = 0, n = 1 , n = 1.5 and n = 2 by using VAL n = "value" and GO. Plot rA versus CAI and compare.

Results A CAI

m.B

Figure 5.31. The program T A N K is started at very high flow rate (F = 100000) which is reduced at 2 min. (F = 1000).

R Cdl

38.8

ISIN

Figure 5.32 A change in program TANKD was made at TIME = 5 , such that the dimensionless parameter PAR varied from 9 to 5, and the results were plotted as concentration versus time. The dimensionless plot looks quite different.

321

5.2 Continuous Tank Reactor Examples

5.2.4

CSTR

- Continuous Stirred-Tank Cascade

System A system of three continuous stirred-tank reactors is used to carry out the firstorder isothermal reaction A

Figure 5.33.

Products

Tanks-in-series with first-order reaction

Model

Tank 3

or dividing by V

328

5 Simulation Examples of Chemical Engineering Processes

where the mean residence time

z in tank n is given by

Program The program is formulated with all three tanks having an equal volume, V. :Example :

CSTR

CONTINUOUS

STIRRED

TANK

REACTOR

CONSTANT CO = 1 CONSTANT TAU = 3 CONSTANT K = 0.1 CONSTANT CINT = 0.2 CONSTANT TFIN = 20 1

SIM;

INTERACT;

RESET;

GOT01

INITIAL c1 = 0 C2 = 0 c3 = 0 DYNAMIC C1' = (CO-C1)/TAU-K*Cl C2' = (Cl-C2)/TAU-K*C2 C3' = (C2-C3)/TAU-K*C3 x1 = (CO-Cl)/CO x2 = (CO-C2)/CO x3 = (CO-C3)/CO P L O T T IX 1 0 TFIN, 0 Co OUTPUT T,CllC2,C3,X1,X2,X3 PREPARE T,Cl,C2,C3,X1,X2,X3 I

I

CASCADE

(CSTR)

329

5.2 Continuous Tank Reactor Examples

Nomenclature Symbols C F k V z

Concentration Flow rate Rate constant Volumes of each tank Residence time in each tank

mol/L L/min l/min L min

Indices 0, 1,2, 3, n

Refer to tank number

Exercises 1.

Study the effect of varying the residence time, z, and the rate constant values.

2.

Setting k = 0, simulate the tracer response (F-curves) for 3 perfectlystirred tanks in series.

3.

Compare the steady-state values obtained with the solution of the steadystate algebraic equations.

4. Modify the program to enable you to study the response of the system, using tanks of differing volumes and differing types of reaction kinetics.

5.

Derive the dimensionless form of the balances to show kz to be the governing parameter. Run simulations to confirm this.

6.

Change the program to compare the conversion from various combinations of tank volumes such that the total volume is constant. For a total volume of 1 0 L this could be a tank of 2.5 L followed by 2.5 L and 5 L, or this sequence could be reversed.

Results Residence time is an important operating parameter, as these results demonstrate.

330

5 Simulation Examples of Chemical Engineering Processes

R x1 B x2

c

A C1

B cz

c

x3

c3

A 0.a888.8

58

T

18.8

15.0

28.0

Figure 5.34. A residence time of z = 3 h produces these conversion-time profiles.

Figure 5.35. An interactive decrease of residence time (z from 3 to 0.5) at hour 3 produced this dynamic response of c o nce n tr a t i o ns .

Reference Dunn, I.J. and Ingham, J. (1972) Verfahrenstechnik. 6, No. 11, 399.

5.2.5

CASCSEQ - Cascade of Three Reactors with Sequential Reactions

System A cascade of 3 tanks in series is used to optimise the selectivity of a complex sequential-parallel reaction. Depending on the kinetics, distributing the feed of one reactant among the tanks may lead to improved selectivity.

33 1

5.2 Continuous Tank Reactor Examples

FB2 cB20

FB1 cB1 0

F, cAO

FB3 cB30 cA33 cB3 +

cA23

kl

A+B

P

t

k2

P+B

Q

Figure 5.36 Stirred tanks in series with feeding to accomplish the best performance with this complex reaction.

Model The kinetics are rA = - kl CnACnB1

A

rp = + k l C n A C n B ' -

A

B

B

kz Cp "P

rQ = + k 2 CnpCnB2 P B The component balances for tank 2 are

c"B2

332

5 Simulation Examples of Chemical Engineering Processes

v dCP2 dt

= F1 Cpl - F2 Cp2 + rp V

The other tanks have analogous balances. Note that the flow rates vary from tank to tank F1 = F + F g l F2 = F + FBI + F B ~ F3 = F + F s l + F B +~ F B ~

Program :Example

CASCSEQ

:Cascade of Three Tank Reactors :Sequential-Parallel Reactions Constant CINT=l, TFIN=200, NOCI=l,ALGO=O Constant nA=l :React ion order Constant nP=l :React ion order Constant nBl=l.O :Reaction order Constant nB2=2.0 :Reaction order Constant F=O.O5 :Flow rate of A into tank (L/min) Constant Fbl=0.05,Fb2=0.05,Fb3=0.05 :Flow rates of B (L/min) Constant Vl=l :Initial tank volume (L) Constant V2=1 :Initial tank volume (L) Constant V3=1 :Initial tank volume (L) Constant AO=1 :Conc of A in feed (mol/L) Constant B10=1 :Conc of B in feed to tank l(mol/L) Constant B20=1 :Conc of B in feed to tank 2(mol/L) Constant B30=1 :Conc of B in feed to tank 3(mol/L)

5.2 Continuous Tank Reactor Examples

Constant Ain=O :Initial conc of A (rnol/L) Constant Bin=0 :Initial conc of B (mol/L) Constant Pin=O :Initial conc of P (rnol/L) Constant Qin=O :Initial conc of Q (mol/L) Constant kl=l :Rate constant reaction 1 Constant k2=0.1 :Rate constant reaction 2 Constant END = 0.85 :Graph-Range of conc. [P31 1 SIM; INTERACT; RESET; GOT0 1 INITIAL Al=Ain;Bl=Bin;Pl=Pin;Ql=Qin A2=Ain;B2=Bin;P2=Pin;Q2=Qin A3=Ain;B3=Bin;P3=Pin;Q3=Qin DYNAMIC :Flow rates leaving tanks 1,2 and 3. Fl=F+Fbl F2=Fl+Fb2 F3=F2+Fb3 :Tank 1 Al'=(F*AO-Fl*Al)/Vl-rll B1~=(Fbl*B1O-F1*B1)/Vl-(rll+r21) P11=rll-r21-F1*P1/V1 Q11=r21-F1*Q1/V1 rll=kl*(Al**nA)*(Bl**nBl) r21=k2*(Pl**nP)*(Bl**nB2) :Tank 2 A2'=(Fl*Al-F2*A2)/V2-r12 B2'=(Fl*Bl+Fb2*B2O-F2*B2)/V2-(r12+r22) P2'=r12-r22+(Fl*Pl-F2*P2)/V2 Q2I=r22+(Fl*Ql-F2*Q2)/V2 rl2=kl*(A2**nA)*(B2**nBl) r22=k2*(P2**nP)*(B2**nB2) :Tank 3 A3'=(F2*A2-F3*A3)/V3-r13 B3l=(F2*B2+Fb3*B3O-F3*B3)/V3-(r13+r23) P3'=r13-r23+(F2*P2-F3*P3)/V3 Q3'=r23+(F2*Q2-F3*Q3)/V3 rl3=kl*(A3**nA)*(B3**nBl) r23=k2*(P3**nP)*(B3**nB2) IF (A3.LT.O) A3=0 IF (B3.LT.0) B 3 = 0 PLOT T IP3,0, Tf in, 0, END OUTPUT T,A3,B3,P3,Q3 PREPARE T,Al,A2,A3,Bl,B2,B3,Pl,P2,P3,Ql,Q2,Q3,Prod,Sel TERMINAL

333

5 Sitnulation Examples of Chemical Engineering Processes

334

Sel=P3/(Q3+P3) Prod=P3*F3/(Vl+V2+V3) PRINT Prod, "Sel=", S el

Nomenclature Symbols Initial tank volume Concentration of A in feed Concentration of B in feed to tank 1 Concentration of B in feed to tank 2 Concentration of B in feed to tank 3 Initial concentration of A Initial concentration of B Initial concentration of P Initial concentration of Q Rate constant reaction 1 Rate constant reaction 2

L mol/L mol/L niol/L mol/L mol/L moHL mol/L mol/L

Exercises 1.

Set nB1 = ng2, and feed B at the same molar rate as A to tank 1 only. Vary the volumetric flows F and FBI to obtain a maximum in Cp3.

2.

Set n g l < ng2 and repeat Exercise 1.

3.

Feed B into each tank separately, keeping the total molar flow rate as in Exercise 1. Run for the case ngl < ng2 and optimise the selectivity.

4.

Repeat Exercises 1, 2 and 3 for n g l > ng2.

Results Startup and steady states are shown for the kinetic case nA = 1, np = 1 , ng 1 = 1 and n ~ =2 2.

335

5.2 Continuous Tank Reactor Examples

n BI

8. m

8.888

B BZ

882

r

Figure 5.37. This result was obtained when all of reactant B was fed into tank 1 with flow rate FBI = 0.15. The resultant selectivity was 0.74.

5.2.6

Figure 5.38. Distributing reactant B with feed rates FBI = FB2= FB3 = 0.05 gave lower concentrations of B in all tanks and consequently a higher selectivity of 0.83. The productivity was also slightly higher (0.13 compared to 0.12).

REXT - Reaction with Integrated Extraction of Inhibitory Product

System Consider the necessity to extract a product which impedes a reaction in a continuous tank system. This can be accomplished with an integrated extraction unit, here a liquid-liquid extraction system.

336

5 Simulation Examples of Chemical Engineering Processes _ .

3

Fa I

I

1

CPl

I

I ~

1 4

Figure 5.39.

F1 9 cA29 cP2

Integrated reaction and liquid-liquid extraction.

Model The kinetics exhibit inhibition by the product,

The balances for A and P in the reactors are for A

and for P V

R

T - ~1 cP2- (FO

+ ~ 1cPl ) + VR

rp

The single-stage extraction is modelled by balances for P in both phases X and Y. ("'

dt cp2' = (Fo + F1) Cpl - F1 Cp2 - Fo Cp2 - Ka VE (Cp3* - Cp3)

5.2 Continuous Tank Reactor Examples

337

(" dt cp31 = F3 Cp30 - F3 Cp3

+ Ka VE (Cp3* - Cp3)

where the equilibrium is given by

The reactant is not removed by the extraction. Thus,

The recycle flow, F1, can be written in terms of fraction recycled as

Thus,

Fo+ F1 = (1 + R ) Fo

Program :Example :Reaction

REXT with

Extraction

and

Recycle

Constant FO=100 : [dm3/h] Feed of A Constant ca0=0.9 :[kg/dm3] Feed concentration Constant F3=20 : [dm3/h] Extraction solvent Constant cp30-0.05 :[kg/dm3] Prod. conc. in solvent Constant R=O.5 :[-I Reflux ratio Constant Vr=1000 : [dm31 Reactor volume Constant Vx=200 : [dm31 Ext. volume of product Constant Vy=lOO : [dm31 Ext. volume of solvent Constant kl=0.7 : f l / h l Reaction constant Constant ki=0.3 :[kg/dm3] Inhibition constant Constant m = 2 :[-I Saturation constant Constant K a = l :[m2/m3minl Transfer coeff. Constant CINT=O.1, TFIN=50, NOCI=10 1 SIM; INTERACT; RESET; GOT0 1 INITIAL cal=O cp3=cp30

338

______

5 Simulation Examples of Chemical Engineering Processes

__ ____ _____

_____

_____

___~

__

ca2=0 cp2=0 DYNAMIC :Reactor ra=-kl*cal/(l+cpl/ki) rp=-ra cal'=(FO*caO+R*FO*ca2-(1+R)*FO*cal)/Vr+ra c p l ' = ( R * F O * c p 2 - ( 1 + R ) * F O * c p l )/ V r + r p :Extractor Ve=Vx+Vy cp3s=m*cp2 C ~ ~ ~ = ( ( ~ + R ) * F O * ~ ~ ~ - F O * C ~ ~ - R * F O * C ~ ~ ) / V ~ cp2~=((1+R)*FO*cpl-R*FO*cp2-FO*cp2-Ka*Ve*(cp3sCP3))/VX cp3'=(F3*cp3O-F3*cp3+Ka*Ve*(cp3s-~p3))/Vy PL O T T,cpl, O,TFIN, 0,caO OUTPUT T,cal,cpl,ca2,cp2,cp3 PREPARE T,cal,cpl,ca2,cp2,cp3

Nomenclature Symbols Feed rate of A Concentration Extraction solvent flow Product concentration in solvent Recycle ratio Reactor volume Solvent volume Extract volume of solvent Reaction constant Inhibition constant Saturation constant Specific transfer coefficient Total extractor volume

dm31ti kgldm3 dm31h kgldm3 dm3 dm3 dm3 llh kg/dm3 -

1lmin dm3

339

5.2 Continuous Tank Reactor Examples .

._

~~

Indices Refer to reactant and product Refer to steams Refer to extractor phases

A, p 0, 1, 2, 3 y

x,

Exercises 1.

Vary the mass transfer coefficient Ka, to see its influence.

2.

Vary the recycle ratio, R, and note its influence on the concentration levels.

3.

Change the operating variables and note their influence on the reaction rate for given values of the inhibition coefficient, KI.

4.

Investigate the influence of the relative reactor and extraction volumes.

Results R La1 8 GP1

R

0.mn.n

25.0

T

50.0

75.0

188.8

Figure 5.40. Startup with recycle (R = 0.5) and turning off recycle at T = 55. The concentrations in the reactor are shown.

Figure 5.41. The product concentrations in the extractor are shown for the case of Fig. 5.40.

340

5 Simulation Examples of Chemical Engineering Processes

5.2.7

THERM and THERMPLOT - Thermal Stability of a CSTR

System A first-order, exothermic reaction occurs within a continuous stirred-tank reactor, equipped with jacket cooling, where the kinetics and reactor schemes are A

+

products

+ heat

Figure 5.42. Continuous tank with heat transfer to jacket.

Model The dynamic model involves a component mass balance, an energy balance, the kinetics and the Arrhenius relationship. Hence V -dCA -dt = F C A O - F C A - ~ V C A

dTR = F p cp (To - TR) - k V CA (-AH) V p cp dt

-

UA(TR - Tj)

34 1

5.2 Continuous Tank Reactor Examples

The steady-state equations allow evaluation of the possible steady operating points. For the component balance Giving

CAO

= I + ( k V/F)

Separating the energy balance into heat loss and heat generation terms Rate of heat loss (HL) = Rate of heat generation (HG)

which with the mass balance gives F cP (TR - To) + UA (TR - Tj) = k V CAOAH / (1 + (k V / F))

Programs Program THERMPLOT models the steady-state variation in HL and HG as a function of temperature. For the conditions of this reaction, there may be three possible steady-state operating conditions for which HG = HL. Only two of these points, however, are stable. Program THERM solves the dynamic model equations. The initial values of concentration and temperature in the reactor can be changed after each run using the ISIM interactive commands. The plot statement causes a composite phase-plane graph of concentration versus temperature to be drawn. Note that for comparison both programs should be used with the same parameter values. :Example

THERM

REACTOR STABILITY CSTR WITH EXOTHERMIC REACTION AND JACKET : Dynamic solution f o r phase-plane plots : Located steady-states with THERMPLO : and use same parameters. CONSTANT C P = 1 CONSTANT R H O = 1 CONSTANT F = O 0 1 CONSTANT V = 2 CONSTANT UAZ1.356E-03 : :

.

COOLING

342

5 Simulation Examples of Chemical Engineering Processes

CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT 1

SIM;

HR=-10 TO=300 TJ=305 AO=5 Z=8.03000E+12 R=1.987 E=22500 CINT=20 TFIN=2000 A=5 TR=325

RESET;

INTERACT;

GOT0

1

DYNAMIC KO=Z*EXP(-E/(R*273))

K=Ko*EXP(-(E/R)*(l/TR-1/273))

:This form avoids math overflow A'=F*(AO-A)/V-K*A :Dynamic mass balance XA=(AO-A)/AO HL=F*(TR-TO)/V+UA*(TR-TJ)/(V*RHO*CP) HG=-K*A*HR/(CP*RHO) TR' =HG-HL :Dynamic energy balance PLOT TR,A,300,350,0,5 OUTPUT T,A,TR,HL,HG,XA PREPARE T IA, TR, HL I HG, XA :Example

THERMPLOT

REACTOR STABILITY CSTR WITH FIRST-ORDER EXO. REACTION AND JACKET COOLING : PLOTS THE STEADY-STATE HEAT GAINED VERSUS HEAT LOST : To be use with example THERM CONSTANT C P = 1 CONSTANT R H O = 1 CONSTANT F=0.01 CONSTANT V = 2 CONSTANT UAz1.356E-03 CONSTANT H R = - 1 0 CONSTANT T O = 3 0 0 CONSTANT TJ=305 CONSTANT A O = 5 : :

343

5.2 Continuous Tank Reactor Examples

CONSTANT CONSTANT CONSTANT CONSTANT 1

Z=8.0300OE+12 R=1.987 E=22500 TFIN=150

S1M;INTERACT;RESET;GOTO

1

INITIAL A=O DYNAMIC T R = T +2 5 0 : TEMP IS CHANGED USING T AS A VARIABLE KO=Z*EXP(-E/(R*273)) K = K O * E X P ( - ( E / R ) * ( l / T R - 1 / 2 7 3 ) ) :AVOIDS MATH OVERFLOW A=AO/ (l+K*V/F) :STEADY-STATE MASS BALANCE :HEAT LOSS HL=F*(TR-TO)/V+UA*(TR-TJ)/(V*RHO*CP) HG=-K*A*HR/(CP*RHO) :HEAT GAIN TR, HG, 2 5 0,4 O O , O , 0.5 PLOT OUTPUT TR,A,HL,HG,TR PREPARE T ITR, A, HG ,H L

Nomenclature Symbols cP

P

F V UA AH T TR

C Z E R

Specific heat Density Volumetric feed rate Reactor volume Heat transfer capacity coefficient Exothermic heat of reaction Feed temperature Reactor temperature Concentration Collision frequency Activation energy Gas constant

Indices 0 A J

Refers to feed condition Refers to reactant Refers to jacket

calJg C g/cm cm3/s 0

3

cavc s cal/mol K

K

mole/cm us caVmol cal/mol K

344

5 Simulation Examples of Chemical Engineering Processes

Exercises 1.

Using THERMPLO, locate the steady states. With the same parameters, verify the steady states using THERM. To do this, change the initial conditions (A and TR) using VAL and GO. Plot as a phase-plane and also as concentration and temperature versus time.

2.

Make small changes in the operating and design values (V, F, AO, TO, TJ, UA) separately and note the effect on the steady states, using THERMPLO. Verify each with THERM.

3.

Choose a multiple steady-state case and try to upset the reactor by changing AO, F, TO or TJ interactively during a run. Only very small changes are required to cause the reactor to move to the other steady state. Plot as time and phase-plane graphs.

Results

4.58

3,m 1.9

E.W

Figure 5.43. Program T H E R M P L O provides the graphical steady-state solution. The parameters in the program give two steady states and one unstable state.

Figure 5.44. This plot was produced by changing the initial conditions TR and A at the end of each run of THERM, using V A L and GO. The steady states for concentration and temperature lie at (4.1, 305.7) and (1.5, 331.0).

Reference Bilous, O., and Amundson, N. R. (1956) Chem. Eng. Sci., 5 , 81.

345

5.2 Continuous Tank Reactor Examples

5.2.8

COOL - Three-Stage Reactor Cascade with Countercurrent Cooling

System A cascade of three continuous stirred-tank reactors arranged in series, is used to carry out an exothermic, first-order chemical reaction. The reactors are jacketed for cooling water, and the flow of water through the cooling jackets is countercurrent to that of the reaction. A variety of control schemes can be employed and are of great importance, since the reactor scheme shows a multiplicity of possible stable operating points. This example is taken from the paper of Mukesh and Rao (1977).

coolant Figure 5.45.

A three-stage reactor system with countercurrent controlled cooling.

Model The kinetics are r = kC

346

...

..

5 Simulation Examples of Chemical Engineering Processes ~

The mass and heat balances in any reactor i are dCi V dt = F (Cj-1 ~

where

k.1 = p Cp V

dT,

=

- Ci) - V

ki Ci

z -E/RTi

F P cp (Ti-1 - Ti) - V k i Ci (-AH)

-

UA (Ti - Tcj)

For the cooling jackets of any reactor i the energy balance is dTci

PCcpc VC dt = FC P c cpc Uci-1 - Tci) + UA ( T i - Tci) Program For the parameters used in the program, the system is characterised by multiple steady-state operating conditions. For example, four of them are as follows Case

CAI TI

1

5.76 x 0-4 308

5.41 316

10-4

4.76 x 0-4 332

2

5.75 x 0-4 308

7.79 44 3

10-5

9.96 x 0-5 446

3

8.13 443

10-5

1.01 447

10-5

1.74 x 432

4

3.45 482

10-5

2.51 x 47 1

2.73 45 3

10-7

Obviously the aim in operating the reactor cascade is to ensure operation at the most favourable conditions, and for this both the startup policy and control strategy are important.

347

5.2 Continuous Tank Reactor Examples

:Example

COOL

:CONTROL AND STARTUP OF A 3-STAGE :MULTIPLE OPERATING STATES AND :COUNTERCURRENT COOLING CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT

1

CSTR

CASCADE

F=2500 V=800000 CO=6E-4 TO=302 CP=1.35E-4 RHO=l HR=-35 R=1.98E-3 Z=2000 E=10.1 A = 90 0 U=4E-6 VC=100000 FC=2000 RHOC=1 CPC=lE-3 CINT=100 TFIN=1E4 TCO=298 T10=300,T20=310,T30=340 :INITIAL TEMPS C10=6.OE-04,C20=6.OE-04,C30=6.0E-04

S1M;RESET;INTERACT;GOTO

1

INITIAL Cl=ClO c2=c20 C3=C30 Tl=T10 T2=T20 T3=T30 TCl=TCO TC2=TCO TC3=TCO DYNAMIC Kl=Z*EXP(-E/(R*Tl)) Cl'=F*(Co-Cl)/V-Kl*Cl Hl=F*(TO-T1)/V-(K1*CI*HR)/(RHO*CP)

348

5 Simulation Examples of Chemical Engineering Processes

Tla=H1-U*A*(T1-TC3)/(RHO*CP*V) TC3'=FC*(TC2-TC3)/VC+U*A*(Tl-TC3)/(RHOC*CPC*VC) K2=Z*EXP(-E/(R*T2)) C2'=F*(Cl-C2)/V-K2*C2 H2=F*(Tl-T2)/V-(K2*C2*HR)/(RHO*CP) T2'=H2-U*A*(T2-TC2)/(RHO*CP*V) TC2'=FC*(TCl-TC2)/VC+U*A*(T2-TC2)/(RHOC*CPC*VC) K3=Z*EXP(-E/(R*T3)) C3'=F*(C2-C3)/V-K3*C3 H3=F*(T2-T3)/V-(K3*C3*HR)/(RHO*CP) T3'=H3-U*A*(T3-TCl)/(RHO*CP*V) TCl'=FC*(TCO-TCl)/VC+U*A*(T3-TC1)/(RHOC*CPC*VC)

xl=(co-cl)/co x2=(c0-c2)/c0 x3=(co-c3)/co P L O T T, X3,0, TFIN, 0 , 1 OUTPUT T,Xl,X2,X3,Tl,T2,T3,TCl,TCTC3 PREPARE T,Xl,X2,X3,Tl, T2,T3,TCl,TC2, T C 3

Nomenclature Symbols

VC

z

AH

Heat transfer area Inlet concentration of A Specific heat Coolant specific heat Activation energy Volumetric flow rate to reactor Coolant flow Gas constant Density Coolant density Inlet temperature Coolant inlet temperature Heat transfer coefficient Reactor volume Jacket volume Frequency factor Heat of reaction

cm2 mo~cm3 kcal/g "C k caYg "C kcal/mol

cm3/5 cm3/5

kcal/mol K g/cm3 g/cm3 K K kcal/cm2 s K cm3 cm3 I /s kcal/mol

349

5.2 Continuous Tank Reactor Examples

Exercises 1.

Study the normal start up procedure with the reactors empty of reactant and cold (CAI = CA2 = CA3 = 0, TI = T2 = T3 = 0) and confirm that the system proceeds to the low yield condition of Case 1 .

2.

By modifying the initial concentration and/or temperature profile along the cascade show that other stable operating conditions are realisable.

3.

Investigate alternative start up policies to force the cascade to a more favourable, stable yield condition, as given below.

3a. Introduction of a feed-preheater and the effect of a short-burst preheat condition can be studied by starting with an increased value of the feed temperature, To, which is returned to normal once the reaction is underway. 3b. Temperature control of feed flow rate can be described by F

=

FO

-

KC

*

(TSET

-

T3)

Proportional control can be based on the temperature of the third stage. Here FO is the base flow rate, KC is the proportional controller gain, and TSET is the temperature set point. Note that in order to guard against the unrealistic condition of negative flow, a limiter condition on F should be inserted into the DYNAMIC region. This can be accomplished with ISIM by the following statement IF

(F.LT.O.0)

F

=

0

3c. Temperature control of coolant flow can be described by FC

=

FCO

-

KC

*

(TSET - T3)

A limiter on FC should be used as in 3b. Control values (FCO, KC, TSET) in the range (2000, 10, 400) can be tried. Remember to revise the PREPARE statement.

4.

The response to step changes in feed flow rate, feed concentration or temperature may also be simulated.

350

5 Simulation Examples of Chemical Engineering Processes

- -~

~.

~

. _ _ _ _ _ ~

~

Results A change in the feed temperature will move the system to another steady state, as these results show.

"'7r'

Ac n f2

"."1

Ism

e. SBB

Figure 5.46. A burst of warm feed at 310 K for the first 1400 seconds causes the system to shift into a higher temperature and higher conversion steady state. The fractional conversion curves A and C are for normal startup and B and D are for the startup with a warm feed period.

Figure 5.47. Temperature profiles corresponding to the conditions for Fig. 5.46. The curve labels are the same.

Reference Mukesh, D. and Prasada Rao, C.D. (1977) Ind. Eng. Chem. Proc. Des. Dev. 16, 186.

5.2.9

OSCIL

-

Oscillating Tank Reactor Behaviour

System A water-cooled, continuous stirred-tank reactor is used to carry out the following exothermic parallel reactions

2A

3

B +heat(-AHI)

35 1

5.2 Continuous Tank Reactor Examples

A

9 C + heat (-AH2)

I

Tjo

TR

Figure 5.48.

Continuous tank reactor with coolant flow control.

The parameters used in the program give a steady-state solution, representing, however, a non-stable operating point at which the reactor tends to produce natural, sustained oscillations in both reactor temperature and concentration. Proportional feedback control of the reactor temperature to regulate the coolant flow can, however, be used to stabilise the reactor. With positive feedback control, the controller action reinforces the natural oscillations and can cause complete instability of operation.

Model The reaction kinetics are given by

and

rg = kl C A ~

The component mass balances become

352

5 Simulation Examples of Chemical Engineering Processes

- = F (CBO- CB) + kl V CA* V dCB dt

where kl = Z I e-E1/RTR and k2 = Z2 e-E2/RTR. The heat balance equation is given by

Note that this includes terms for the rates of heat generation of both reactions. As shown in Sec. 1.2.5, the heat balance equation is equivalent to

where and

U' = UA K F, 1+KF,

Both the effects of coolant flow rate and coolant inlet temperature are thus incorporated in the model. The effective controller action is represented by its effect on U' and hence on F,, where

U' = Uo'

+ K,

(TR - TseJ

Here K, = 0 represents open loop conditions, K, < 0 represents positive feed back conditions, and K, > 0 represents conventional negative feedback control.

Program The constant parameters for the program give a steady-state solution for the reactor, with the following conditions of

T = 333.4 K, CA = 0.00314, CB = 0.000575 mol/cm3

353

5.2 Continuous Tank Reactor Examples

but the system at this steady-state condition is non-stable, and the simple open loop system of the reactor with no imposed controller action is such that the reactor itself generates sustained oscillations which eventually form a limit cycle. The program can be used to generate time dependent displays of A, B and TR for input values of the effective proportional gain KC and set point temperature TSET, for initial conditions A0 = BO = 0, T = TO at t = 0. Dimensionless variables are defined in the program, where TR CB CA C A = CA [steady-state] ’ cB = CB [steady-state] and TR = TR [steady-state]

as these are useful in the following phase-plane analysis. :Example : :

OSCILLATIONS IN A CSTR WITH PARALLEL REACTIONS BOTH OPEN LOOP AND CLOSED LOOP CONTROL

CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT 1

OSCIL

TO=319.1 AO=0.01 BO=O F=10 V = 10 0 CP=1 RHO=1 Zl=l.OE+19 Z2=9.493+12 E1=28000 E2=21000 H1=-27000 H2=-20000 TJ=319.1 KC=0.5 TSET=350 U=88.5 R=1.987 CINT=1.0 TFIN=100

S1M;RESET;INTERACT;GOTO

INITIAL A=O B=O

1

354

5 Sirnulation Examples of Chemical Engineering Processes

TR=TO :UNSTABLE STEADY-STATE ASTEADz3.14E-03 BSTEADz5.75E-04 TSTEAD=333.4

VALUES,ASTEAD,BSTEAD,TSTEAD

DYNAMIC Kl=Zl*EXP(-El/(R*TR)) K2=22*EXP(-E2/(R*TR)) At=F*(AO-A)/V-K1*A*A-K2*A B'=F*(BO-B)/V+Kl*A*A RH=(-l)*(Hl*Kl*A*A+H2*K2*A)/(CP*RHO) :REACTION HEAT,RH TR'=F*(TO-TR)/V-Ul*(TR-TJ)/(V*CP*RHO)+RH :ACCUM.HEAT=CONVECT.+ TRANSFER + REACTION Ul=U+KC*(TR-TSET) :CONTROLLER ADIM=A/ASTEAD BDIM=B/BSTEAD TDIM=TR/TSTEAD :DIMENSIONLESS VARIABLES PLOT T , T D I M , O , T F I N , O I 2 OUTPUT T,A,B,RH,TR,ADIM,BDIM,TDIM PREPARE T, A, B, R H ITR, ADIM, BDIM, TDIM $ VAL TSET=325.00

HEAT

Nomenclature

Symbols A C cP

AH

E F k

Kc

P T U

Heat transfer area Concentration Specific heat Exothermic heat of reaction Activation energy Volumetric flow rate Rate coefficient Controller constant Density Temperature Heat transfer coefficient

cm* mol/cm3 caVg "C caVmol calimol K cm3/s l/s cal/cm2 K2 g/cm K cal/cm2 K

355

5.2 Continuous Tank Reactor Examples

Heat transfer conductance factor Volume Frequency factor

U' V Z

calls "C cm3 1Is

Indices A and B 1 and 2 0 j in 0

R

W

Refer to components Refer to reactions Refers to inlet Refers to jacket Refers to inlet Refers to outlet Refers to reactor Refers to coolant water

Exercises Study the simple, open-loop (KC = 0) a d clos d-loop responses (KC = -1 to 5, TSET = TDIM, and 300 to 350 K) and the resulting yields of B. Confirm the oscillatory behaviour and find appropriate values of KC and TSET to give maximum stable and maximum oscillatory yield. For the open-loop response, show that the stability of operation of the CSTR is dependent on the operating variables by carrying out a series of simulations with varying To in the range 300 to 350 K. Modify the program to generate a dimensionless, phase-plane display of BDIM versus ADIM, repeating the studies of Exercise 1. Note that the oscillatory behaviour tends to form stable limit cycles in which the average yield of B can be increased over steady-state operation.

356

5 Simulation Examples of Chemical Engineering Processes ~-

-~

Results

A RDM

1,13

1,28

1.85

0.88

0.98

0.48

0.93

.E

0.m

.E

Figure 5.49. Dynamic response o f dimensionless temperature versus time for two setpoints; TSET = 325 (curve A) and 375 K (curve B).

Figure 5.50. Response of dimensionless concentration of A for the conditions of Fig. 5.49.

Figure 5.51. A phase-plane pIot of dimensionless A versus dimensionless temperature for the conditions of Fig. 5.49.

Figure 5.52. A phase-plane plot of dimensionless concentrations of A and B at TSET = 375 K.

Reference Dorawala, T.G.and Douglas, J.M. (1971) A.1.Ch.E.J. 17, 974.

5.2 Continuous Tank Reactor Examples

357

5.2.10 REFRIGl and REFRIG2 - Auto-Refrigerated Reactor System In the auto-refrigerated reactor shown below, an exothermic reaction A + B is carried out using a low boiling solvent C. The heat of reaction is removed from the reactor by vapourising the solvent, condensing the vapour in the reflux condenser and returning the condensate as saturated liquid to the reactor. The total holdup of liquid in the reactor is maintained constant, but the temperature of the reactor is controlled by regulating the mass flow of vapour to the condenser. The example is taken from the paper of Luyben (1960).

CONDENSER

Figure 5.53. rate.

A schematic drawing of' the example REFRIG, showing control of the vapour

5 Simulation Examples of Chemical Engineering Processes

358

Model The balance equations are as follows:

Energy balance equation

a- w cP (To

M cp d TI< =

-

TR)- M k CA (-AHR)

Arrhenius equation k = Ze-”/RTR

Control equation S = S o - K p (TSet-TR)

At steady-state conditions

Rate of ky;ra:;n]

where

HG =

-M

=

[

Rate of heat removal

CA k (-AHK)

where

and

HL = W cP (TR - To) + S AHv

-S

AHv

359

5.2 Continuous Tank Reactor Examples

Program The program REFRlGl calculates the steady-state heat generation and heat loss quantities, HG and HL, as functions of the reactor temperature, TR, over the range 320 to 410 K. It is shown that, according to the van Heerden steady-state stability criterion that the simple loop control response, KP=O is unstable. Program REFRIG2 calculates the dynamic behaviour and generates a phaseplane plot for a range of reactor concentrations and temperatures. :

Example

REFRIGl

AUTOREFRIGERATED REACTOR O F LUYBEN STEADY-STATE STABILITY ANALYSIS CONSTANT W=900, M = 1 1 0 0 C O N S T A N T CAO=0.5, TEMPO=330 CONSTANT HR=-1.4E3, HVz637, C P = 3.14 CONSTANT Z=7.08E10r E=76.2E3,R=8.314 CONSTANT SO= 330, SMAX=1000, TSET=360, = 90, CINT=.5 CONSTANT TFIN 1 SIM; INTERACT; RESET; G O T 0 1 DYNAMIC T R = T + 3 20 K=Z*EXP(-E/(R*TR)) CA=CAO/(l+M*K/W) HG=-M*CA*K*HR HL=W*CP*(TR-TEMPO)+S*HV S=SO-KP*(TSET-TR) S=O IF (S.LE.0) IF ( S GT. SMAX) S = S M A X P L O T TR, CA, 3 2 0,4 10,0, C A O P R E P A R E T,CA,TR,S O U T P U T T,CA,S :

KP=30

.

:

Example

REFRIG2

: :

AUTOREFRIGERATED REACTOR OF DYNAMIC STABILITY ANALYSIS

LUYBEN

CONSTANT W=900, M=1100 C O N S T A N T CAO=0.5, T E M P O = 3 3 0 CONSTANT HR=-1.4E3, HV=637, CP= 3.14 C O N S T A N T Z=7.08E10, E=76.2E3,R=8.314 CONSTANT S O = 330, SMAX=1000, TSET=360, CONSTANT TFIN = 10, CINT=.5

KP=30

360

5 Simulation Examples of Chemical Engineering Processes

1 SIM; INTERACT; RESET; GOT0 1 IN ITIAL TR=TEMPO DYNAMIC CAI= (W*(CAO-CA)-K*CA*M)/M TR'=(W*CP*(TEMPO-TR)-M*K*CA*(HR)-S*HV)/(M*cP) K=Z*EXP(-E/(R*TR)) S=SO-KP*(TSET-TR) IF (S.LE.0) S = o IF (S.GT.SMAX) S=SMAX PLOT T,CA, O,TFIN,O, 1 PREPARE T, CA, TR, S OUTPUT T, CA, S

Nomenclature Symbols Mass of reactor contents Reactant concentration Time Mass flow rate Kinetic rate constant Specific heat Feed temperature Reactor temperature Exothermic heat of reaction Vapour reflux rate Latent heat of vapourisation Frequency factor Activation energy Ideal Gas Constant Base level for vapour rate Controller gain Set point temperature Net rate of heat generation Net rate of heat loss

M CA t W k cP

TO TR AHR S

kglh l/h kJkg K K K kJkg kglh kJkg l/h Jlmol J/mol K kglh kglh K K kJ/h kJ/h

Exercises 1.

Using program REFRIG1, find the magnitude of the steady-state vapourisation rate, So, required to give stable operation.

36 1

5.2 Continuous Tank Reactor Examples

2.

Using the value of So = 330 kg/h, use program REFRIG2 to find the value of the proportional gain constant kp required to give stability.

Results ISIU

I

4.8488.328

8.360

rR

8.488

8.488

9.448

/----+=--b

8.M -E3

Figure 5.54. Using REFRIG 1, the plot of heat-loss and heat-gain terms versus temperature exhibits the stable state for KP=30.

Figure 5.55. Plot of the concentrationtemperature phase-plane, corresponding to the run in Fig. 5.54 using REFRIG2.

Reference Luyben, W.L. (1960) A.I.Ch.E.J, 12, 663.

5.2.11 STABIL - Stability of Chemical Reactors with Disturbances System The stability of a first-order exothermic reaction A+B, in a single CSTR with jacket cooling has been studied by Seborg (1971), and the usefulness of simulations for this type of investigation has been emphasised by Luus (1972). The influence of sinusoidal, feed-temperature variations is corrected by simple

362

5 Sirnulation Exaniples of Chemical Engineering Processes

feedback control of reactor temperature based on a regulation of the coolant temperature.

Model Component balance

dC dt = F ( C o - C ) - V k C

Energy balance dT V P Cp dl = F p Cp (To - T) - V k C (-AH)

-

UA, (T - Tc)

Feed temperature variation

To = Tavg+ A sin 03 t Controller equation

TC = TCO+ KC V s e t

-

T)

Substituting the data values of' Seborg (197 I), as listed in the Nomenclature, the model equations become the same as used by Luus (1972)

dC = 2 (0.2704 - C) - k C dt ~~~

dT dt = 2 ( 5 3 0 - T ) + 2 0 0 k C - 0 . 1 ( T P 5 3 0 ) + 2 v + 0 . l u Here the feed temperature (To) and the base-level, coolant temperature in the jacket (Tco) are both taken as 530 R. The feed temperature perturbation is v = A sin

and the controller action is

03

t

363

5.2 Continuous Tank Reactor Examples

which has the effect of varying the cooling water temperature, above and below the value of 530 R.

Program Since the reduced form of the model is programmed here, the parametric values cannot be easily changed. Data values are this time included in the equations within the DYNAMIC region. This is not really good practice as it removes from the generality of any given program. The feed temperature To and base cooling water temperature Tavgare both 530 R. :

Example

STABIL

:STABILITY OF REACTOR WITH SUSTAINED Constant KC=O, TSET=530, A=O, W = l Constant CO=O, TEMPO=550 Constant CINT=0.5,TFIN=50 1 SIM; INTERACT; RESET; GOT01 INITIAL

c=co

DISTURBANCES

TEMP=TEMPO DYNAMIC K=le8*EXP(-lOOOO/TEMP) C'=2*(0.2704-C)-K*C TEMP'=2*(530-TEMP)+2OO*K*C-O.1* (TEMP-530)+2*VAR+o.l*U : Feed temp. variation VAR=A*SIN(=W*T) : Controller action U=KC*(TSET-TEMP) PLOT T, C, 0, TFIN, 0,O. 2 5 OUTPUT T,C,TEMP PREPARE T,C,TEMP,VAR,U

Nomenclature The numerical values are given in brackets.

3 64

5 Simulation Examples of Chemical Engineering Processes -

Symbols Amplitude of sine wave disturbances Heat transfer area ( 100) Feed concentration (0.2704) Specific heat (1 .O) Exothermic heat of reaction (1 * lo4) Activation energy factor (1 * lo4) Volumetric flow rate (200) Rate coefficient factor (1* lo8) Controller constant Density (50) Feed temperature (530) Cooling water temperature (530) Controller action on cooling water Heat transfer coefficient (5.0) Volume (100) Perturbation on feed temperature Angular velocity of sine disturbances

A Ar

co cP

AH

m F

ko KC

P TO Tco U

U V V

03

"R ft2 Ib mole/ft3 BTU/lb "F BTU/lb mol BTU/lb mol "F ft3/h l/h BTU/ h ft2 OF2 wft3

"R OR R/h BTU/ h ft2 "F ft3

"R/h radiandh

Indices 0 avg C set

Refers to inlet Refers to average Refers to coolant Refers to setpoint

Exercises Show that in the absence of control and feed disturbances (u = v = O), the system has a singular, stable, steady-state solution of C = 0.1654 and T = 550. This can best be done by carrying out runs with different initial conditions (CO and TEMPO) and plotting the results as a phase-plane, TEMP versus C. Setting the reactor initial conditions, (say Co = 0, TEMP0 = 550) study the effect of feed disturbances for A between 0 and 48 and o between 128 and 0.008. Compare the results with those reported by Luus (1972).

3 65

5.2 Continuous Tank Reactor Examples

3.

Study the influence of control, both by varying the controller gain, Kc, and the set point temperature, Tset. Compare the results with the conclusions stated by Luus (1972). Note whether the amplitude of the reactor oscillations is greater or smaller than those of the disturbances.

4.

Operate the controlled reactor with temperature disturbances to see if the oscillations can be removed or reduced by control. Operate with a setpoint of 645 R. What is the steady-state value of temperature compared with the cooling water?

5. 6.

Reprogram this example in terms of the actual reactor parameters, as given in the Nomenclature and the model equations. Investigate the influence of these parameters individually.

Results

8.8888.538

8.548

8.558

TEMP

8.568

8.578

*E 3

Figure 5.56. A phase-plane plot of oxcillations of C and TEMP without control for amplitude of feed temperature disturbance, A = I .

I 0.538 8 . 8

15.8

Figure 5.57.

T

38.8

45.8

60.0

Starting with no disturbances (A=O), A is set to 1.0 at T=10, and at T=29 the control is turned on (KC=lOO).

References Luus, R. (1972) Can. J. Chem. Eng. 50, 412. Seborg, D. E. (1971) Can. J. Chem. Eng. 49, 535.

366

5 Simulation Examples of Chemical Engineering Processes

5.2.12 HOMPOLY - Homogeneous Free-Radical Polymerisat ion System A certain free-radical polymerisation reaction is described by the following sequence of initiation, addition and termination

$ 2.R

Initiation

1

Addition

*R+M

+ *Mi

2 *M2 * M x + M 9 *Mx+l 'MI + M

Termination

M x + My kt +, Mx+y

The reaction is carried out in a non-cooled, continuous stirred-tank reactor (Fig. 5.58), and it is required to find the effect of changes in the reactor inlet conditions on the degree of polymerisation obtained. The model is that of Kenat, Kermode and Rosen (1967).

Figure 5.58. The reactor for example HOMPOLY is a constant-volume stirred tank.

Model Only the propagation reactions are significantly exothermic and the energy balance is given by

3 67

5.2 Continuous Tank Reactor Examples

the monomer balance is represented by

v

F dM (Mo-M)-rp dt =

__

and the initiator balance by dI

=

vF

(10 - I)

-

ri

The reaction kinetics are as follows 1)

Initiation or initiation decomposition reaction

where f is the fraction of free-radicals, OR,initiating chains.

2)

Termination

3)

Propagation

rt = 2 kt Mo2 rp = k, M Mo

Assuming equal rates of initiation and termination

The kinetic rate coefficients k d , kt are described by Arrhenius temperature dependencies. Average chain length

The model equations are expressed in dimensionless form and defining new variables in terms of the fractional deviation from steady state. Dimensionless pertubation variables are defined as

368

5 Simulation Examples of Chemical Engineering Processes

~~~~~

where the bar superscripts relate to steady-state conditions, and the dash superscript represents the perturbation from steady state. The model equations can then be recast into dimensionless form as

(l'+l)OV5(M'+l) exp (Ez

(T'+1)

x~ =

dM'

[

-

MO

(Mo'+l) I-(M'+l) M

+ d1'

50

[

1-

-

=

A)

=- (I'+l)

(lo'+l) I0

I

3

[$I, [$I, [+I, -

In the above model equations, the terms

(11+1)O-~(M'+I) exp ( E I

now represent dimensionless constants, where

-

-

El and E2

A)

369

5.2 Continuous Tank Reactor Examples

10

I

= 1

Ad + vy

F

exp (-El)

For a full derivation of the above model equations, the reactor is referred to the original paper (Kenat et al., 1967).

Program The dimensionless model equations are programmed into the ISIM simulation program HOMPOLY, where the variables, M, I, X and TEMP are zero. The values of the dimensionless constant terms in the program are realistic values chosen for this type of polymerisation reaction. The program starts off at steady state, but can then be subjected to fractional changes in the reactor inlet conditions, Mo, 10, To and F of between 2 and 5 per cent, using the ISIM interactive facility. The value of T in the program, of course, refers to dimensionless time. : : :

Example HOMPOLY DYNAMICS OF A HOMOGENEOUS FREE-RADICAL POLYMERISATION REACTION WITH HEAT EFFECTS

: TYPICAL POLYMERISATION PARAMETERS CONSTANT MBAR=4, IBAR=4, E1=35, E2=25, FEED PERTURBATION VARIABLES CONSTANT F=O,MO=O,IO=O,TO=O CONSTANT TFIN=5, CINT=O.1 1 SIM; INTERACT; RESET; GOT0 1 INITIAL TEMP=TO M=MO I=IO

x=o

TEMPBAR=.9

DYNAMIC :ENERGY BALANCE TEMP'=((TO+l)*TEMPBAR-(TEMP+1))*0+(1TEMPBAR)*FACTORl FACTOR1=EXP(=(E2*TEMP/(1+TEMP)))*SQRT(I+l)*(M+l) :MONOMER BALANCE M'=((MO+l)*MBAR-(M+1))*(F+l)+(1-MBAR)*FACTORl :INITIATOR BALANCE

370

5 Simulation Examples of Chemical Engineering Processes

I'=((IO+l)*IBAR-(I+1))*(F+1)+(1-IBAR)*o+l)*FACTOR2

FACTOR2=EXP(El*TEMP/(I+TEMP)) :MEAN CHAIN LENGTH X=((M+l)/SQRT(I+l))*EXP((E2-E1)*TEMP/(l+TEMP))-l PLOT T,M, O,TFIN, -1,l PREPARE T ,M ,I ,X I TEMP OUTPUT T, M, I, X, TEMP

Nomenclature Symbols A cP

E

F

1

M 'M Mx 'MY *R T

v

X

f

k r t 0

r AHP

Indices d 1

P t

Arrhenius constant Specific heat Energy of activation Flow rate Initiator's concentration Monomer's concentration Chain radical concentration Dead polymer chain of x units Growing polymer chain of y units Initiator free radical Absolute temperature Reactor volume Average degree of polymerisation or Chain length Reactivity fraction for free radicals Rate constant Rate of reaction Time Dimensionless time Density Heat of reaction per monomer unit Refers Refers Refers Refers

to decomposition of initiator to initiation process to propagation reaction to termination reaction

l/s or moYL s cal/g K cal/mol us mol/L mol/L molk mol/L mol/L mol/L K L

l/s or mol/L s mol/L s S

-

g k

cal/mol

5.2 Continuous Tank Reactor Examples

37 1

Exercises 1.

Use the program to study the effects of both positive and negative fractional changes in feed temperature, monomer and initiator concentration and feed flow rate. Compare your observations with the results and explanations of Kenat et al.

2.

One of the main conclusions of Kenat et al., was that the largest changes in polymer, mean-chain length, occur from the effect of inlet temperature changes and that, therefore, controlling inlet temperature, rather than reactor temperature, is beneficial to reactor performance.

Results

Figure 5.59. An increase in the dimensionless feed temperature from 0 to 0.02 results in increased reactor temperature as shown.

Figure 5.60. The temperature change in Fig. 5.59 causes pronounced decreases in the dimensionless concentrations of monomer, initiator, and chain length.

Reference Kenat, T.A., Kermode, R.I. and Rosen, S.L. (1967) Ind. Eng. Chem. Proc. Des. Devel, 6, 363.

372

5 Simulation Examples of Chemical Engineering Processes _ _

~

5.2.13 REVTEMP - Reversible Reaction with Variable Heat Capacities System A reversible reaction A

ki + B is carried out in a well-mixed, tank reactor, k2

which can be run in either batch or continuous mode. A high temperature at the beginning of the batch will increase the rate of approach to equilibrium, while a low temperature at the end of the run will give a favourable final equilibrium condition. After starting the reactor under adiabatic conditions, there will be therefore an optimal time to turn on the cooling water, A heat exchanger is used for temperature control. This simulation example is also used in Sec. 2.4.1.1, Case A, to demonstrate the use of optimization with dynamic models.

I

lQ

Figure 5.61. Continuous stirred-tank reactor with cooling.

Model The standard heat of reaction can be calculated from the heats of formation as AHRSt = HFBSt

-

HFASt

The specific heats are dependent on the temperature, according to C ~ A=

aA

+

bAT

313

5.2 Continuous Tank Reactor Examples

The equilibrium constant at standard conditions is

Its temperature dependency is given by the van't Hoff equation as

where AHR at any temperature is calculated by

and therefore

this reduces to

where Aa = ag

- aA

and Ab = bB

- bA.

Integration of the van't Hoff equation gives then In KE = In K E S ~-

_____

(&

-

&) + R Aa

In

The balances for the continuous reactor are as follows The total mass balance, assuming constant density is

T Tst

___

+

Ab R

- (T - Tst)

374

5 Simulation Examples of Chemical Engineering Processes

Thus V is a constant. The component balances are

dCB V--dt = - F C g

+

rgV

The kinetics are each assumed to be first order. At equilibrium ki CAE = k2 CBE Since

then

allowing the kinetics to be written as rA = -rg =

- kl

(CA -

it)

The rate constant is a function of temperature according to

The reactor energy balance as derived in Sec. 1.2.5 is

The individual terms are now

5.2 Continuous Tank Reactor Examples

375

The energy balance for the reactor becomes

The energy balance for the jacket is

An objective function, SPTYB, is defined for batch operation SPTYB = CB2 T and the equivalent for continuous operation

These functions represent space-time yields with special weight on high conversion.

Program The reactor is run as a batch reactor (F = 0) under adiabatic conditions (Fco = 0). At time TIMEON the cooling flow rate is set to FCON.

5 Simulation Examples of Chemical Engineering Processes

376

’ TIMEON

T (time)

Figure 5.62. Cooling flow rate profile. :

Example

REVTEMP

: CONTINUOUS STIRRED TANK REACTOR : REVERSIBLE REACTION AND JACKET COOLING : HEAT AND TEMPERATURE EFFECTS WITH : CP = F(T) : REACTION ENTHALPY = F(T) : EQUILIBRIUM = F(T)

CONSTANT

TST =

298

:Activation energy CONSTANT EA = 120.0 CONSTANT R = 8.3143-3 CONSTANT KST = 0.001 :Reactor CONSTANT F = 0.0 CONSTANT V = 2.0 CONSTANT CAO = 25e3

A

<=>

B

:K

:Heats of formation CONSTANT HFAST = -160 CONSTANT HFBST = -240 :Equilibrium constant CONSTANT KEST = 1000 :Heat capacities cp CONSTANT AA = 0.100 CONSTANT AB = 0.080 CONSTANT BA = 0.0006 CONSTANT BB = 0.0004

:

:kJ MOL-1; Standard :kJ MOL-1 Standard 298 K =

a

+

b*T :kJ MOL-1 :kJ MOL-1 :kJ MOL-1 :kJ MOL-1

K-1 K-1 K-2 K-2

:kJ MOL-1 :kJ MOL-1 K-1 :s-1 :m3 sec-1

:m3

:mol m3

T=298 K

377

5.2 Continuous Tank Reactor Examples

CONSTANT CBO = 0.0 CONSTANT TI0 = 290

:mol m- 3 :K

4.0 1.0 0.0 0.1 290 1000 2000 = 1000

CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT

CPC = FCON = FCO = VC = TCO = RHOC = UA = TIMEON

:Initial CONSTANT CONSTANT CONSTANT CONSTANT

conditions CAinit = 20e3 CBinit = 0 Tlinit = 305 TCinit = 305

:kJ :m3 :m3 :m3 :K :kg :kJ :s;

kg-1 K-1 8-1 8-1 m-3 K-1 8-1 time to

start

cooling

:mol m-3 :mol m-3 :K :K

CONSTANT TFIN = 35 CONSTANT CINT=O.Ol CONSTANT NOCI=5 CONSTANT ALGO=3 1 SIM; INTERACT; RESET; 30TO 1 INITIAL :Reaction enthalpy DHRST = HFBST-HFAST :kJ MOL-1 DA = AB-AA DB = BB-BA KA1= KST/EXP(-EA/R/TST) FC = FCO CA = CAinit CB = CBinit T1 = Tlinit TC = TCinit SPTYB=O.0 DYNAMIC K1 = KAl*EXP(-EA/(R*Tl)) DHRSTl = DHRST - DA*TST - DB/2*TST**2 LNKE 1 = DA/R*LOG(Tl/TST)+DB/R*(Tl-TST) LNKE = LOG(KEST)-DHRSTl/R*(l/Tl-l/TST) +LNKE1 KE = EXP(LNKE) RA = -Kl*(CA-CB/KE) :Mass

balance

378 dCA dCB CA' CB'

5 Simulation Examples of Chemical Engineering Processes

F/V*(CAO-CA) F/V*(CBO-CB) = dCA = dCB = =

+ RA - RA

:Heat balance reactor CPA = AA + BA*T1 CPB = AB + BB*T1 SUMCP = V*(CA*CPA + CB*CPB) FEEDCP = F*CAO*(AA*(TlO-Tl) + BA/2*(T10**2 - T1**2)) DHRTl = DHRST + DA*(Tl-TST) + DB/2*(T1**2 - TST**2) DT 1 = (FEEDCP + V*RA*DHRTl + UA*(TC-Tl))/SUMCP T1' = DT1 :Heat balance cooler-jacket = FC/VC*(TCO-TC)-UA*(TC-Tl)/V/RHOC/CPC dTC TC' = dTC IF(T.GE.TIMEON) FC=FCON IF(T.GE.O.1) SPTYB=CB**2/T SPTYC=F*CB**2/V OUTPUT t,ca,cb,Tl, t c P L O T T,CA,O,TFIN,O,CAINIT PREPARE T,CA,CB,T1,TC,CPA,CPB,DHRTllKl,SPTYBISPTYC

Nomenclature Symbols A C cP cPC

Ei

F Fc

kA 1 KE

ki N or n R SPTYB

T1

Heat transfer area of jacket Concentration Specific heat of vessel contents Specific heat of cooling water Activation energy of reaction Flow rate to the reactor Coolant flow rate Arrhenius rate constant Equilibrium constant Reaction rate constant Number of moles Gas constant Objective function for batch operation Temperature in reactor

m2

m011m3 kJkg K kJkg K kJlmol kgls m31s

11s

-

11s mol kJ1mol K moPIrn6 s K

379

5.2 Continuous Tank Reactor Examples

Temperature of jacket contents Coolant supply temperature Heat transfer coefficient Vessel volume Jacket volume Reaction enthalpy Density of vessel contents Density of cooling liquid

TC TcO U V VC AHR P Pc

Indices A B 1 2 E 0 St

K K kJ/s m2 m3 m3 kJ/mol kg/m3 kg/m3

Indicates component A Indicates component B Refers to reaction A -+ B Refers to reaction B + A Refers to equilibrium Refers to reactor inlet Refers to standard conditions (298 K)

Exercises 1.

Compare the batch reactor performance at constant cp with that for variable cp. Do this by setting both cp values constant at a temperature TI and calculating cP1= a + b TI for particular values of aA = C ~ I Aag, = C ~ I B , bA = 0 and bB = 0.

2.

Optimise the reactor performance (space-time yield SPTB) by modification of TIMEON. Compare results obtained with those in Sec. 2.4.1.1.

3.

Modify FO and FCON to improve SPTYB.

4. Run the reactor continuously. Observe the space-time yield, SPTYC. Set TIMEON = 0, and set FCON to the desired cooling flow rate.

380

5 Simulation Examples of Chemical Engineering Processes

~

Results

R Ti

BK

8,388

R

ml 8,368

BCPB

-

0.328

-

8,348 -

0,288

-I

-

8.248-

8,368

0,328

&

Figure 5.63. Temperature profiles for the batch reactor, T I , and for the jacket TC, which lags behind.

Ism

8.368 0.328

P

n

I

I '

Figure 5.64. Variation of the cp values as a result of the temperature profiles, corresponding to Fig. 5.63.

rsm

4

Figure 5.65. Temperature profiles caused by turning on the cooling water flow by setting TIMEON to 10.

Figure 5.66. Space-time yields, SPTYB, profiles in the batch reactor as a result of varying TIMEON (A - 1000, B - 8, C - 9, D 15).

38 1

5.3 Tubular Reactor Examples

5.3

Tubular Reactor Examples

5.3. 1

TUBE and TUBED - Tubular Reactor Model for the Steady State

System The tubular reactor, steady-state design equation is of interest here. The dimensional and dimensionless forms are compared for an nth-order reaction.

Figure 5.67. Isothermal tubular reactor.

Model The component mass balance for steady state is

where, v = HA. The kinetics here are rA = - k C ”A Using the variables

CA =

CA/CAO and

Note that L/v is the residence time.

2=

Z/L, the dimensionless model.is

382

5 Simulation Examples of Chemical Engineering Processes

Program Programs for the separate dimensional and dimensionless cases are given here. :Example TUBE : Concentrations in a steady-state tubular reactor. : nth-order kinetics : To be compared with TUBED,the dimensionless version. Constant A = 2 0 :[dm21 Cross-sectional area : [dm3/sl Flow rate Constant F=10 Constant CAO=0.8 :[kg/dm31 Feed concentration Constant L=20 : [ml Length Constant n=1.7 : [-I Reaction order Constant k=0.1 : [various] Reaction constant Constant CINT=O.1, NOCI=20 1 SIM; INTERACT; RESET; GOT0 1 INITIAL CA=CAO TFIN=L :Sets TFIN DYNAMIC v=F/A :velocity of fluid Z = t :Transform to length variable CA’=rA/v rA=-k*CA**n PAR=k*Z/v*CAO**(n-1) PLOT Z, CA, 0, L, 0, CAO OUTPUT Z, CA, PAR PREPARE Z,CA

:Example

TUBED

:Axial concentration profile in a tubular reactor. :Dimensionless form,n-th- kinetics :TO be compared to TUBE :The length should be set and TFIN always set to 1 .o Constant A = 2 0 :[dm21 Cross-sectional area Constant F=10 : [dm3/s] Flow rate Constant L=20 :[am] Length Constant CAO=0.8 :[kg/dm31 Feed concentration Constant n=1.7 : [-I Reaction order Constant k=0.1 : [various] Reaction constant Constant CINT=O.Ol,TFIN=l, NOCI=10

383

5.3 Tubular Reactor Examples

1 SIM; INTERACT; RESET; GOT0 1 INITIAL CAd=1 DYNAMIC v=F/A :velocity Zd=t :Dimensionless length PAR=k*L/v*CAO**(n-1) :With TFIN=l, Z = L at CAd'=-PAR*CAd**n Length=Zd*L CA=CAO*CAd PLOT Zd, CAd, 0, 1, 0'1 OUTPUT Zd, CAd, length, C A PREPARE Zd, CAd, PAR

Zd=l

Nomenclature Symbols A

Area Concentration Flow rate Rate constant Reaction order Reaction rate Time Flow velocity Length

CA

F k n rA t V

Z

dm2 k/dm3 dm3/s various

-

kg/dm3 s S

dds dm

Exercises 1.

2.

3.

Run TUBE for various k, v and CAOand compare with corresponding runs from TUBED. kL Make plots of 5, versus Z for n = 1 and various values of [ 7 Ci: ] Repeat Exercise 2 for n = 2.

Results Using the same set of parameters, the results from both programs will be the same.

5 Simulation Examples of Chemical Engineering Processes

384 -

~~

e.me.e

5. 8

I

.~-

18.8

15.8

28.8

Figure 5.68. With F = 30 and n = 1.7 this steedy-state profile is obtained from TUBE.

n.mn.m

e.258

~~

ZD

8.588

8.758

1.888

Figure 5.69. This profile was obtained from TUBED.

Reference Levenspiel, 0. ( 1972) Chemical Reaction Engineering, Wiley.

5.3.2

TUBTANK - Comparison of Tubular and Tank Reactors

System Steady-state conversions for both continuous tank and tubular reactors are compared for nth-order reaction kinetics.

385

5.3 Tubular Reactor Examples

Figure 5.70. Comparison of tank and tubular reactors.

The comparison is made by calculating the residence times required for a particular value of fraction conversion. This allows a comparison of the reactor volumes for a given flow rate. Levenspiel (1972) presents graphical plots similar to those generated by this program.

Model For a tank reactor with nth-order reaction at steady state n

O = FCAO-FCA-kCAV For n = 1

~ the mean residence time for the tank. where T T is For a plug-flow tubular reactor with nth-order reaction at steady state,

For n = 1 CA = CAO eckTTu , equal to Z/v. where the tubular reactor residence time, Z T ~ is

386

5 Simulation Examples of Chemical Engineering Processes

For a given fraction conversion, XA, the residence time for n=l can be calculated from TT, =

1

1 lni-x,

For nth-order reaction in terms of fraction conversion, XA for a tank

and for a plug-flow tubular reactor (nzl) the integration gives

Program For this program, n must not be set exactly to 1.0. :T U B T A N K

:Comparison of tank and tubular reactors steadystate. :Residence time versus fraction conversion Constant Ca0=800 : [kg/m31 Feed concentration of A Constant k=0.8 : [l/sl Reaction rate Constant n=1.5 :[-I Reaction order Constant F=0.04 :[m3/sl Flow rate Constant CINT=0.005, TFIN=0.9, NOCI=20 1 SIM; INTERACT; RESET; GOT0 1 INITIAL if (n.eq.1) ns1.001 T=0.0001 DYNAMIC Xa=T if (Xa.ge.1) Xa=0.9999 Cata=CaO-Xa*CaO Catu=Cata :Tank TAUta=Xa/(k*CaO**(n-l)*(l-Xa)**n) Vta=TAUta*F :Volume of tank :Tube TAUtu=((l-Xa)**(l-n)-l)/(k*CaO**(n-l)*(n-l))

387

5.3 Tubular Reactor Examples

Vtu=TAUtU*F :Volume of t u b e Vtuta=Vtu/Vta :Volume ratio, t u b e t o t a n k if (Xa.gt.1) X a = l PLOT Xa,Vtuta,O,TFIN,O,l OUTPUT Xa,TAUta,TAUtu,Vta,Vtu Xa,TAUta,TAUtu,Vta,Vtu,Vtuta,Cata,Catu PREPARE

Nomenclature Symbols Concentration of A Flow rate Reaction rate constant Reaction order Flow velocity Volume Fraction conversion Length Residence time

CA F k n V

V XA

Z z

d

S

m3

m h

Indices Refers to tube Refers to tank Refers to inlet conditions

Tu Ta 0

Exercises 1.

Set the volumetric flow rate and feed concentration for the tank and tubular reactors to desired values. Set also the order of reaction to n = 1.01. Run for a range of fraction conversions from 0 to 0.99. Compare the required volumes for the two reactor types.

2.

Rerun Exercise 1 for n = 2 and compare the ratio of volumes, Vtuta. Answer the question in Exercises 1, regarding the required volumes. Suppose a conversion of 90% is desired, and the flow rate to the tank reactor is to be one-half that of the tubular reactor. What would be the ratio of volumes?

388 3.

5 Simulation Examples of Chemical Engineering Processes

Repeat Exercise 2 with a 2-fold increase in CAO. How do the required volumes compare. Repeat for n = 1.01. Explain the results.

Results Sample results for TUBTANK are given below.

A

:I

““‘“11 ISIU

R TAulA B TAUIII

1.28

,

1.88

-

ISIU

1.28-

0.88-

8.48-

\

n

Figure 5.71. The ratio of required volumes is plotted versus conversion for n = 1.5.

/B

Figure 5.72. The plot of the required residence times shows that the differences become important above XA = 0.5.

Reference Levenspiel, 0. (1972) Chemical Reaction Engineering, Wiley

5.3.3

BENZHYD

-

Dehydrogenation of Benzene

System The gas-phase dehydrogenation of benzene to diphenyl (D) and further to triphenyl (T) is conducted in an ideal isothermal tubular reactor. The aim is to maximize the production of D and to minimize the formation of T. Two parallel, gas-phase reactions occur at atmospheric pressure

389

5.3 Tubular Reactor Examples

Figure 5.73. The partial-pressure variables for the tubular reactor are shown.

Model The kinetics are given in terms of Arrhenius relationships and partial-pressure relationships as

At steady-state conditions, the mass balance design equations for the ideal tubular reactor apply. These equations may be expressed as

V

F

=

I-,-

dX2

- 7

where X I is the fractional conversion of benzene by reaction 1, X2 is the fractional conversion of benzene by reaction 2 and z is the reactor space time. In differential form 1 dX _ _ d7 -- - r l

390

5 Simulation Examples of Chemical Engineering Processes _ _ _ ~ ~-

~

~ ~ _ - _ _ _ ~ _ _ _ _ ~

Expressing PB, PD, PH and PT in terms of XI and X2.

PB = (1 - X I -x2)

Substituting into the mass balance,

where

Program The ISIM program for this example with typical output is shown. :Example

BENZHYD

: Isothermal tubular react ions : dehydrogenation of

constant constant constant constant constant

tr=1033 keql=0.312 keq2=0.480 cint=0.1 t€in=l

reactor benzene

with

two

consecutive

39 1

5.3 Tubular Reactor Examples

1

S1M;INTERACT;RESET;GOTO

initial stfin=tfin kl=1.496e7*exp(-1520O/tr) k2=8.670e6*exp(-1520O/tr) x1=0 x2=0

1 :Final space time

dynamic PB=l-xl-x2 PD=x1/2-x2 PH=x1/2+x2 PT=x2 rl=kl*(PB*PB-PD*PH/keql) r2=k2*(PB*PD-PH*PT/keq2) xl =rl x2 = r 2 ST=T :ST is space time plot plot ST,xl,O,tfin,O,l output ST,xl, x2, rl, r2 prepare ST,X~,X~,PB,PD,PH,PT,~~,~~

Nomenclature Symbols Equilibrium constants Partial pressure of benzene Partial pressure of diphenyl Partial pressure of hydrogen Partial pressure of triphenyl Reaction rates Temperature Fractional conversion Space time

-

atm atm atm atm lb/h ft3 R

-

h

Exercises 1.

Study the effect of varying space time on the fractional conversions X1 and X2 and evaluate the compositions of benzene, hydrogen, diphenyl and triphenyl.

392

5 Simulation Examples of Chemical Engineering Processes .

~~

-

-~

2.

Show that the diphenyl composition passes through a maximum.

3.

Study the effect of varying pressure on the reaction by modifying the partial-pressure relationships.

4. Plot the rates rl and r2 as functions of space time. 5.

Study the effects of temperature.

Results Two outputs for BENZHYD are shown.

Ism

Ism

Figure 5.74. The maximum in diphenyl (PD) composition is shown.

the

Figure 5.75. The reaction rate of the second reaction step exhibits a maximum, as a function of reactor space time.

References Crum, E.H. (1972) in "Computer Programs for Chemical Engineering" Education: Kinetics, Ed. M. Reilly. AIChE. Smith, J.M. (1970) Chemical Engineering Kinetics, 2nd edition, p. 158, McGraw-Hill.

393

5.3 Tubular Reactor Examples

5.3.4

ANHYD - Oxidation of 0-Xylene to Phthalic Anhydride

System Ortho-xylene (A) is oxidised to phthalic anhydride (B) in an ideal, continuous flow tubular reactor. The reaction proceeds via the complex consecutive parallel reaction sequence, shown below. The aim of the reaction is to produce the maximum yield of phthalic anhydride and the minimum production of waste gaseous products (C), which are C02 and CO.

A+Op

kl

B

k2,

C

Figure 5.76. The fixed-bed reactor and the reaction scheme. Symbols: A = o-xylene, B = phthalic anhydride, C = waste gaseous products (C02 and CO).

The reaction is carried out with a large excess of air, and the total gas mass flow rate can be taken as constant. Detailed modelling of the fixed bed production of phthalic anhydride from o-xylene is discussed by Froment and Bischoff (1990), involving both axial and radial temperature profile effects.

Model Kinetics The reaction kinetics are as follows

3 94

5 Simulation Examples of Chemical Engineering Processes

It is assumed that the mole fraction of oxygen does not change, owing to the large excess of air. The temperature dependencies are, according to Arrhenius, In kl

RT

+ 19.837

31400 RT

+ 20.86

28600 RT

+ 18.97

= --27000

In

k2 =

--

In

k3 =

--

Component Mass Balance Equations Considering Fig. 5.77, the steady-state balance around any segment AV, for component A, is given by

where nAF is the molar flow rate of A in kmol/h, and Fm is the mass flow rate in kg/h ( = G Ac).

--& "AF

Fm cpT "AF

+

Fm cpT +AFm cpT

A2

Figure 5.77.

The balance region AV

Thus the changes in the molar flow rates (kmol/h m) at steady state can be set equal to the rates of production of each component.

395

5.3 Tubular Reactor Examples ~-

~dnAF -

A,prA

dZ

_ d _ n -~ ~ d Z - A,prB _ dnCF _dZ - ACprc The cross-sectional area of the reactor is given by dt2 A, = TC- 4 The total molar flow (mostly air) does not change. Thus, ntF =

M -G

m

The mole fractions can be calculated by

A,

nAF YA = ___ ntF YB = ~ B F ntF Y c = nCF ntF ~

The fractional conversions of A to B, and C are given by

xc

=

nFC . nAFO - n A F

Energy Balance At steady state the energy balance for element AV is 0 = - A (F, cPT)

+

p A V C r AH

-

U At (T-Tj)

396

5 Simulation Examples of Chemical Engineering Processes

where the heat transfer area, At, is n dt AZ, and ihe element AV is A, AZ Thus,

The above equations are solved with the initial conditions at the reactor entrance, given by

Program The particle diameter, not used in the program, can be used to normalize the length Z. The factors 1000 on k,, k2 and k, convert gmol to kmol. :Example ANHYD : NON-ISOTHERMAL FIXED BED REACTOR : OXIDATION OF 0-XYLENE TO PHTHALIC ANHYDRIDE : STEADY-STATE AXIAL TEMPERATURE AND CONVERSION CONSTANT G=4684 :superf. mass velocity [kg/m2 hl CONSTANT MM=0.02948 :mean molecular weight CONSTANT NAO=9.27E-3:inlet mole fraction o-xylene CONSTANT NO=0.208 :mole fraction oxygen (constant) CONSTANT CP =0.25 :specific heat [kcal/kg K1 CONSTANT H1 =-307 :heat of react. A->B[kcal/molel CONSTANT H3 =-I090 :heat of react. A->C[kcal/molel CONSTANT RHOB=1300 :catalyst bulk density [kg/m31 CONSTANT DP = 3 E - 3 :catalyst particle diameter [ml CONSTANT U =82.7 :heat trans. coeff.[kcal/m2 h Kl CONSTANT DT = O .025 :tube diameter [ml CONSTANT R = 1 . 8 9 7 :gas constant [cal/mole Kl CONSTANT TJ = 6 5 0 :cooling jacket temperature [K] CONSTANT TEMPO=650 :inlet temperature [Kl CONSTANT XBO=O :initial fract. conv. of A to B CONSTANT XCO=O :initial fract. conv. of A to C CONSTANT TFIN=8 CONSTANT CINT = 0 . 1 CONSTANT NOCI=2

397

5.3 Tubular Reactor Examples :

Ki

: Z : Ri : XA : Ni

1

kinetic const. o f reaction i [mole/kg(kat) hl distance along bed length t m l total reaction rate of component i total conversion of o-xylene mole fraction of component i

S1M;INTERACT;RESET;GOTO

1

INITIAL XB=XBO

xc=xco

TEMP=TEMPO B 1 =RHOB*MM/ (G*NAO) B 2 =RHOB*(-Hl)/(G*CP) B 3 =RHOB*(-H3)/(G*CP) B 4 =4*U/ (G*CP*DT) DYNAMIC Z=T TEMP'=B2*RB + B3*RC - B4*(TEMP-TJ) XB' =B1*RB XC' =Bl*RC X A =XB + XC N A =NAO* (1-XA) N B =NAO*XB K 1 = l o 0 0 * EXP(-27000/(R*TEMP) + 19.837) K 2 =lo00 * EXP(-31400/(R*TEMP) + 20.86) K3 = l o 0 0 * EXP(-28600/(R*TEMP) + 18.97) R A = - ( K l + K 3 )* N A * N O R B =KI*NA*NO - K2*NB*NO R C =K3*NA*NO + K2*NB*NO PREPARE Z,TEMP,XA,XB,XC,RA,RB,RC,TEMP' O U T P U T Z ,TEMP, XA, XB, X C PLOT Z,TEMP,O,TFIN,600,700

Nomenclature Symbols Area cross-section Specific heat Heat of reaction A Heat of reaction A Tube diameter

-+ B +C

m2 kcal/kg K kcal/kmol kcal/kmol m

398

5 Simulation Examples of Chemical Engineering Processes ~

~~

Mass flow rate Superficial mass velocity Rate constant Mean molecular weight Molar flow rate Universal gas constant Catalyst bulk density Reaction rate Temperature Heat transfer coefficient Volume Mole fraction Distance along tube Refers to total

Fm

G k Mnl n R P r T U V Y Z t

~~

kg/h kg/m2 h kmol/(kg cat.) h g/11101 kmol/h cal/mole K kg/m3 kmol/(kg cat.) h K kcal/m2 h K m3

m

Indices A, B, c,0 2 1, 2, 3 F j

Refer to components Refer to reactions Refers to flow Refers to jacket

Exercises I.

With an inlet temperature, Tempo, o 650 K, study the effect of tube diameter (d, = 0.025, 0.03, 0.04, 0.05).

2.

With dt = 0.025 m study the effect of varying inlet temperature (To = 600, 640, 660 K, with constant jacket temperature. Note the hot spot effect in the reactor temperature profile.

3.

Study the effect of jacket temperature on the temperature profile, by keeping To constant and varying Tj.

4.

Repeat 1) and 2) setting k3 = 0. Observe that the reaction rate A + C is relatively small but has a large influence on the temperature of the reactor.

Results Some important temperature effects are seen below. Selection of high TJ and TEMPO causes arithmetic error.

399

5.3 Tubular Reactor Examples

Figure 5.78. The influence of jacket temperature (TJ = 600, 620, 640, 660) is seen i n these steady-state conversion profiles.

ISIB

Figure 5.79. The temperature profiles are strongly influenced by the jackct temperature, corresponding to the runs in Fig. 5.78

Am

rE3

0.128

Ism

0,705

e. 698 0.675

e, 668

Figure 5.80. The influence of jacket temperature (TJ = 650, 655, 657, and 658) is seen i n these steady-state temperature profiles, for TEMPO=700.

Figure 5.81. The temperature profiles are strongly influenced by the inlet temperature, (TEMPO = 670, 680,690, and 695) with TJ = 657. Higher inlet temperatures lead to a math overflow.

References Froment, G.E., and Bischoff, K.B. (1990) Chemical Reactor Analysis and Design, 2nd edition, Wiley. Rase, H.F. (1978) Chemical Reactor Design for Process Plant, Vol. I, Wiley. Satterfield, C. N. and Jones, R.L. (1972) Computer Programs for Chemical Engineering Education. Kinetics, p. 180, Ed. M. Reilly, AIChE.

400

5 Simulation Examples of Chemical Engineering Processes

_ _

~

___

~

_

_

~

~

_

NITRO - Conversion of Nitrobenzene to Aniline

5.3.5

System Nitrobenzene and hydrogen are fed at the rate of ntF (mol/h) to a tubular reactor operating at atmospheric pressure and 450 K and containing catalyst with a voidage E = 0.424.

TO* NtFO, NAFO

4 . - I Ti

"

At

T, CA

b

Figure 5.82. Tubular reactor with heat transfer.

The heat transfer coefficient, U, to the reactor wall (calfh cm2 K) and the coolant temperature is Tj. The nitrobenzene feed rate is NAFO (mol/h), with hydrogen in very large excess. The heat of reaction is AH (cal/mol). The heat capacity of hydrogen is cp (cal/mol K). The reaction rate is given by

where rA has the units of (moles of nitrobenzene reacting) / (cm3 of void space h), and CA is the concentration of nitrobenzene (mol) / (cm3). This problem is considered in many chemical engineering texts, but here the solution by digital simulation is shown to be very convenient.

Model Mass and Energy Balances Consider the balances around the segment AV

_

40 1

5.3 Tubular Reactor Examples

where NtF is the total molar flow rate (mol / h), which can be assumed constant owing to the large excess of hydrogen in the feed, and NAF is the molar flow rate of A (mol / h).

NAF 4 NtF cpT

--b

Figure 5.83. Balance region for a segment.

The changes in the molar flow rates with distance Z at steady-state conditions are

where

The fractional conversion of A is

and, The balance becomes

Energy Balance At steady-state

0 =

-A

(NtF cpT)

+

A V rA AH

-

U At (T-Tj)

where the transfer area At is n dt A Z, and the volume element AV is A, AZ.

402

5 Simulation Examples of Chemical Engineering Processes ~-

Thus,

Kinetics The reaction rate is rA =

-

5.79 x 104

cA0.578 e - 2958 / T

The above equations are solved with the initial conditions at the reactor entrance X A = 0 and T = Ti,

Program :Example

NITRO

:NONISOTHERMAL HYDRATION OF NITROBENZENE :IN A PACKED BED TUBULAR REACTOR

TO

ANILINE

CONSTANT PTOT=l. 0 : Pressure [atml CONSTANT TJ=400 : Coolant Temperature [Kl CONSTANT TEMPO=450 : Feed Temperature [KI CONSTANT F0=65.9 : Feed Rate [mole/h] CONSTANT FAO=1.0 : Feed Rate Nitrobenzene [ m o l e/ h I CONSTANT X O = O : Initial Conversion [-I CONSTANT H ~ = - 1 5 2 1 0 0 : Heat of Reaction [cal/molel CONSTANT EAR=2958 : Activation Energy/Gas Constant [Kl CONSTANT KA=5.79E4 : Kinetic constant CONSTANT U=8.67 : Heat Transfer Coefficient [cal/( h cm2 K) ] CONSTANT D=3.0 : Diameter of Reactor [cml CONSTANT VE=0.424 : Voidage of Catalyst

403

5.3 Tubular Reactor Examples

CONSTANT CP=6.62 : Heat Capacity [cal/(mole K ) I CONSTANT R=82.06 : Gas Constant K)1 CONSTANT PI=3.14159 CONSTANT T F I N = 2 0 CONSTANT CINT=O.1 CONSTANT NOCI=5 : CA=Concentration of Nitrobenzene : X=Conversion of Nitrobenzene : RA=Reaction Rate : Z=Reactor Bed length [cml 1

S1M;INTERACT;RESET;GOTO

of

Feed

[cm3

Gas

atm/ (mole

1

INITIAL TEMP=TEMPO

x=xo

DYNAMIC X'=- PI*D*D*VE/(4*FAO) * RA - PI*D*U*(TEMP-TJ))/(Fo*CP) TEMP'=(FAO*(-HR)*X' CA=(l-X)*(FAO/FO)*PTOT/(R*TEMP) IF (CA .LT. 0) CA=O RA=-KA*CA**0.578*EXP(-EAR/TEMP) Z=T PLOT Z,X,O,TFIN,O,l PREPARE Z,X,TEMP,CA,RA,X',TEMP' OUTPUT Z I X, TEMP, CA, RA

Nomenclature Symbols A, At CA CP

AH dt E

NAF

Cross-sectional area for flow Transfer area of tube Concentration of nitrobenzene Heat capacity of hydrogen Heat of reaction Diameter of tube Catalyst with voidage Nitrobenzene molar flow rate

cm* cmL mol/cm3 cal/mol K cal/mol cm -

mol/h

404

5 Simulation Examples of Chemical Engineering Processes - ...

Total molar flow rate Total pressure Gas constant Reaction rate Reactor temperature Coolant temperature Heat transfer coefficient Fraction conversion Reactor length

NtF Ptot

R rA T

Ti U X

z

molh atm cm3 atm/mol K mol/cm3 of void h K K cal/h cm2 K -

cm

Indices 0

Refers to entering conditions

Exercises 1.

Run the program for varying inlet temperatures in the range 350 to 500 K.

2.

Study also the effect of coolant temperature.

3.

Comment on the monotonously decreasing temperature profiles in the last part of the reactor.

4.

Modify the program such that it models adiabatic behaviour and study the influence of varying feed temperature.

405

5.3 Tubular Reactor Examples

Results The sensitivity of this example to operating conditions is demontrated by these results

rE3

8.688

RUL

1

Ism

0.525 0.450 0.315

Figure 5.84. The inlet temperatures were set at 350, 450 and 500 for these axial concentration profiles.

Figure 5.85. The reactor for the conditions of Fig. 5.84 exhibit a temperature maximum for TEMPO = 450 and 500.

References Fahidy, T.Z. (1972) in Computer Programs for Chemical Engineering Education. Kinetics, Ed. M. Reilly, AIChE. Smith, J.M. (1972) Chemical Engineering Kinetics, 2nd edition, McGraw-Hill

5.3.6

TUBDYN - Dynamic Tubular Reactor

System Tanks-in-series reactor configurations provide a means of approaching the conversion of a tubular reactor. In modelling, they are employed for describing axial mixing in non-ideal tubular reactors. Residence time distributions, as measured by tracers, can be used to characterise reactors, to establish models and to calculate conversions for first-order reactions.

406

5 Simulation Examples of Chemical Engineering Processes

~~~~~~

Figure 5.86.

Tanks-in-series approximation of a tubular reactor.

Model For the nth tank the balance for a single component is

If applied to a tracer, rAn is zero. Otherwise it is taken as nth-order

The residence time distributions can be measured by applying tracer pulses and step changes as explained in Sec. 3.2.9. The response curves are best normalised such that the dimensionless time is @ = -

t

‘I:

where T, the residence time, applies to the entire N-tank system and is given for N tanks of volume V by N V/F. The dimensionless concentration for a pulse of tracer is

where Mt is the mass of tracer injected at t = 0 into the first tank. For a step change in the feed tracer from 0 to CAO and giving an F-curve response,

407

5.3 Tubular Reactor Examples

The pulse tracer response, giving the E-curve response, can be used directly to calculate the steady-state conversion for a first-order reaction according to the relationship w

_ _ -- jE(t) ,-kt dt CAs-s CAO

Program This program is designed to simulate tracer experiments for residence time distributions based on a cascade of 1 to 8 tanks-in-series. An nth-order reaction can be run, and the steady-state conversion can be obtained. The important parameters to change are as follows for the tracer experiments: k, CAINIT, and CAO ( = 0 for E curve, = 1 for F curve). For reaction studies, the parameters to change are n, k, CAO, and CAINIT.

Program :Example

TUBDYN

:Eight tanks in series with nth-order reactions. Constant V=lOO : [dm31 Volume of tanks Constant F=10 : [dm3/minl Flow rate Constant CAO=500 : tg/dm31 Feed concentration Constant n = 2 :[-I Reaction order Constant k=0.0005 : [ l / s ] Rate constant Constant CAinit=O : [ g / d m 3I Initial tank conc. Constant CINT=O.1, TFIN=50, NOCI=10 1 SIM; INTERACT; RESET; GOT0 1 INITIAL CAl=CAinit CA2=CAinit CA3=CAinit CA4=CAinit CAS=CAinit CA6=CAinit CA7=CAinit CA8=CAinit DYNAMIC TAU=V/F Ral=-k*CAl**n :Reaction rate

408

5 Simulation Examples of Chemical Engineering Processes

CAl'=l/TAU*(CAO-CAl)+Ral Ra2=-k*CA2**n CA2'=1/TAU*(CAl-CA2)+Ra2 Ra3=-k*CA3**n CA3'=1/TAU*(CA2-CA3)+Ra3 Ra4=-k*CA4**n CA4'=1/TAU*(CA3-CA4)+Ra4 RaS=-k*CAS**n CA5'=1/TAU*(CA4-CAS)+RaS Ra6=-k*CA6**n CA6'=l/TAU*(CAS-CA6)+Ra6 Ra7=-k*CA7**n CA7'=1/TAU*(CA6-CA7)+Ra7 ~a8=-k*CAB**n CAB'=l/TAU*(CA7-CAB)+Ra8 PLOT T , C A 1 , O I T F I N I O , C A O OUTPUT T,CAl,CA2,CA3,CA5,CA7,CA8 PREPARE T,CA1,CA2,CA3,CA4,CA5,CA6,CA6,CA71CA8

Nomenclature Symbols

cA0

CAinit CAN F k

n N V

Feed concentration Initial tank concentration Concentration in tank N Flow rate Reaction constant Reaction order Number of tanks Volume of tanks

Indices n t

Refers to nth tank Refers to time

g/dm3 g/dm3 g/dm3 dm3/min depends on n -

dm3

409

5.3 Tubular Reactor Examples

Exercises 1.

Set k = 0 and simulate tracer experiments.

2.

Use the E-curve to calculate the conversion for n = 1 and compare with simulation. This requires changing the program.

3.

Vary the operating variables of feed flow rate and CAOto see the influence with n = 1 and n = 2.

4.

Calculate the steady-state conversion for 1 , 3 and 8 tanks using the same value of total residence time.

Results ISIH

Figure 5.87. The feed rate was changed at T = 20 minutes from 10 to 5 for this reaction with n = 2.

Figure 5.88. A pulse of tracer was simulated by setting k = 0 and setting CAO= 0 for a duration of AT = 1. The residence time TAU in each tank was 20, or 160 for the entire 8-tank system.

410

5.3.7

5 Simulation Examples of Chemical Engineering Processes

DISRE - Isothermal Reactor with Axial Dispersion

System This example models the dynamic behaviour of an non-ideal isothermal tubular reactor in order to predict the variation of concentration, with respect to both axial distance along the reactor and flow time. Non-ideal flow in the reactor is represented by the axial dispersion flow model. The analysis is based on a simple, isothermal first-order reaction.

n Figure 5.89

Tubular reactor with finite-difference element n.

Model The reaction involves the conversion: A

+

products, with the kinetics

As discussed in Sec. 4.3.6, the axial dispersion flow model is given by

where Z is the distance along the reactor, D is the eddy diffusivity and u is the superficial flow velocity. Thus the problem involves the two independent variables, time t and length Z. The distance variable can be eliminated by finite-differencing the reactor length into N equal-sized segments of length AZ such that N AZ equals L, where L is the total reactor length. Thus for a segment n

5.3 Tubular Reactor Examples

41 1

or defining Cn' = Cn / Co, Az = A Z / L, 0 = t u / L in dimensionless terms and omitting the primes for simplicity

The term Lu / D is known as the Peclet number, Pe, and its inverse as the dispersion number. The magnitude of the Peclet number defines the degree of axial mixing in the reactor. Thus,

The boundary conditions are satisfied by Co' = 1 for a step change in feed concentration at the inlet, and by the condition that at the outlet C'N+1 = C'N, which sets the concentration gradient to zero. The reactor is divided into 8 equal-sized segments.

Program :Example

DISRE

:UNSTEADY-STATE :REACTOR LENGTH

FINITE DIFFERENCED MODEL DIVIDED INTO 8 EQUAL ELEMENTS

CONSTANT L=10, U=l, CO=1 CONSTANT P=O.5, K=0.01 :PECLET NO. AND KINETIC CONST. CONSTANT CIO=O :INITIAL CONDITIONS CONSTANT T F I N = 3 0 J C I N T = 1 1 S1M;INTERACT;RESET;GOTOl INITIAL L1=L/8 L2=Ll*Ll*P Kl=K*L/U Cl=CIO C2=CIO C3=CIO C4=CIO C5=CIO C6=CIO

412

5 Simulation Examples of Chemical Engineering Processes

C7=CIO C8=CIO DYNAMIC C1'=(CO-C1)/L1+(CO-2*Cl+C2)/L2-Kl*Cl C2'=(Cl-C2)/Ll+(Cl-2*C2+C3)/L2-Kl*C2 C3'=(C2-C3)/Ll+(C2-2*C3+C4)/L2-Kl*C3 C4'=(C3-C4)/Ll+(C3-2*C4+CS)/L2-Kl*C4 C5'=(C4-CS)/Ll+(C4-2*C5+C6)/L2-Kl*C5 C6'=(CS-C6)/Ll+(C5-2*C6+C7)/L2-Kl*C6 C7'=(C6-C7)/Ll+(C6-2*C7+C8)/L2-Kl*C7 C8'=(C7-C8)/Ll+(C7-2*C8+C9)/L2-Kl*CE C9=C8 OUTPUT T,Cl,C2,C4,C6,C8,C9 PREPARE T , C I ~ C ~ , C ~ I C ~ ~ C ~ ~ C ~ , C ~ , C B PLOT T, C8,0, TFIN, 0,l

Nomenclature Symbols C D k L Pe t U

Z z 0

Concentration Effective axial diffusivity Rate constant Total length Peclet number (L u I D) Time Velocity Length variable Dimensionless length Dimensionless time

Indices A n !

Refers to reactant Refers to increment number Refers to a dimensionless variable

mol/m3 and m2/h 1/h m -

h m/h m

413

5.3 Tubular Reactor Examples

Exercises 1.

Obtain the tracer response curve to a step input disturbance of tracer solution by setting k = 0.

2.

Study the effect of varying Peclet number, Pe, on the resulting tracer response.

3.

Note that as Pe becomes large, the conditions approach plug flow, and as Pe approaches zero conditions approach perfect mixing.

4.

Study the effect of varying Pe on the performance of the reactor, and compare the resulting performance with perfect plug flow.

5.

Use the tracer response curve to calculate the first-order conversion.

Results

A C1

B

c

1.888

-

cz

c3 DCA E C5

F

C6 G Cl

H C8 8.298

9.BBBB.B

1.5

T

15.8

22.5

38.8

Figure 5.90. Dynamic concentration profiles at each axial position in the reactor.

n.ean.8

15.8

T

38.8

45.8

68.8

Figure 5.91. Response of reactor to changes in Co (from 1.0 to 2.0 at T = 30). The reactor flow rate was initially changed from U = 1 to 0.1.

5 Simulation Examples of Chemical Engineering Processes

414

5.3.8

DISRET - Non-Isothermal Tubular Reactor with Axial Dispersion

System The dispersion model of example DISRE is extended for non-isothermal reactions to include the dispersion of heat from a first-order reaction. A

+ B +

heat (-AH)

A2 Figure 5.92.

Tubular reactor showing the finite element.

Model From the axial dispersion flow model the component balance equation is

with the energy balance equation given by

where h is the eddy diffusivity for heat transfer. The reaction kinetics are rA = kCA

Defining Z = z / L, t = T / To, C = CA / CAO,0 = D t / L2, the model equations can be recast into dimensionless terms to give the following dimensionless

model equations

415

5.3 Tubular Reactor Examples -

Mass Balance

Energy Balance

where

B2 = (-AH)and &=-

CAO P c p To

E R To

A full treatment of this problem is given by Ramirez (1976, 1989) and by Clough and Ramirez (1971), Finite-differencing the dimensionless length of the reactor (Z = 1) into N equal-sized segments of length AZ such that N AZ = L, gives for segment n,

and

The boundary conditions used here are Co = to = 1 for a step change of inlet concentration or temperature, and at the outlet by fN+1 = tN and CN+1 = CN.

416

5 Simulation Examples of Chemical Engineering Processes

Program :

Example

: :

Non-Isothermal Unsteady-state

CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT

DISRET Tubular Model

Reactor,Axial

Dispersion

L=lOO : reactor length [cm] CAO=0.5 : inlet conc. [lb mole/ft31 T0=637 : inlet temperature [Kl U=O. 3 : flow velocity [cm/sl RHO=60 : density [lb/ft3] D=l : mass transfer eddy diff. [cm2/sl LAMBDA=48:heat conductiv. coeff.[BTU/Kcm s] HR=-36000: heat of reaction [BTU/lb mole] CP=O. 8 : specific heat [BTU/lb K] EA=40000 :activation energy [BTU/lb molel KO=1.2E4 : frequency factor [l/sl R=3.577 :gas constant [BTU/lb mole Kl TFIN=0.3 CINT=0.002

SIM IN ITIAL DUMMY = 0 cc0=1;cc1=0;cc2=0;cc3=0;cc4=0;cc5=0 CC6=0;CC7=0;CC8=0;CC9=0 TTO=l;TTl=l;TT2=1;TT3=1;TT4=1;TT5=1 TT6=1;TT7=1;TT8=1;TT9=1 DYNAMIC DZ=1/8 ALPHA=LAMBDA/(RHO*CP) Bl=Z*L*L/D B2=HR*CAO/(RHO*CP*TO) Pl=U*L/D P2=U*L/ALPHA E=EA/ (R*TO) Rl=Bl*CCl*EXP(-E/TTl) R2=Bl*CC2*EXP(-E/TT2) R3=Bl*CC3*EXP(-E/TT3) R4=Bl*CC4*EXP(-E/TT4) RS=Bl*CC5*EXP(-E/TTS) R6=Bl*CC6*EXP(-E/TT6)

417

5.3 Tubular Reactor Examples

R7=Bl*CC7*EXP(-E/TT7) R8=Bl*CC8*EXP(-E/TT8) CCl'=(CCO-2*CCl+CC2)/(DZ*DZ) + Pl*(CCO-CCl)/DZ CC2'=(CC1-2*CC2+CC3)/(DZ*DZ) + Pl*(CCl-CC2)/DZ CC3'=(CC2-2*CC3+CC4)/(DZ*DZ) + Pl*(CC2-CC3)/DZ CC4'=(CC3-2*CC4+CC5)/(DZ*DZ) + Pl*(CC3-CC4)/DZ CC5'=(CC4-2*CC5+CC6)/(DZ*DZ) + Pl*(CC4-CC5)/DZ CC6'=(CC5-2*CC6+CC7)/(DZ*DZ) + Pl*(CC5-CC6)/DZ CC7'=(CC6-2*CC7+CC8)/(DZ*DZ) + Pl*(CC6-CC7)/DZ CC8'=(CC7-2*CC8+CC9)/(DZ*DZ) + Pl*(CC7-CC8)/DZ CC9=CC8 TTl'=ALPHA/D*(TTO-2*TTl+TT2)/(DZ*DZ)+P2*(TTOTT1) /DZ-B2*R1 TT2'=ALPHA/D*(TT1-2*TT2+TT3)/(DZ*DZ)+P2*(TTlTT2) /DZ-B2*R2 TT3'=ALPHA/D*(TT2-2*TT3+TT4)/(DZ*DZ)+P2*(TT2TT3) /DZ-B2*R3 TT4'=ALPHA/D*(TT3-2*TT4+TT5)/(DZ*DZ)+P2*(TT3TT4) /DZ-B2*R4 TT5'=ALPHA/D*(TT4-2*TT5+TT6)/(DZ*DZ)+P2*(TT4TT5) /DZ-B2*R5 TT6'=ALPHA/D*(TT5-2*TT6+TT7)/(DZ*DZ)+P2*(TT5TT6) /DZ-B2*R6 TT7'=ALPHA/D*(TT6-2*TT7+TT8)/(DZ*DZ)+P2*(TT6TT7) /DZ-B2*R7 TT8'=ALPHA/D*(TT7-2*TT8+TTg)/(DZ*DZ)+P2*(TT7TT8) /DZ-B2*R8 TT9=TT8 PREPARE T,CCl,CC2,CC3,CC4,CC5,CC6,CC7,CC8, TTl,TT2,TT3,TT4,TTS,TT6,TTTT8 OUTPUT T,CCl,CC2,CC4,CC8,DUMMY,TTl,TT2,TT4,TT8 PLOT T,CC8,0,TFINI0,1

-

-

Nomenclature Symbols CA CA0 cP

D DZ E (-AH)

Concentration Inlet concentration Specific heat Mass transfer eddy diffusiviy Length increment Activation energy Heat of reaction

lb mol/ft3 lb mol/ft3 BTUAb K cm% cm BTU/lb mol BTU/lb mol

R1 R2 R3 R4 R5 R6 R7 R8

418

5 Simulation Examples of Chemical Engineering Processes

L

Reactor length Heat transfer eddy diffusivity Universal gas constant Reaction rate Density Time Inlet temperature Flow velocity Frequency factor Diffusivity of heat

h

R Rn P t TO U

Z

a

cm BTU/ft s K BTU/lb mol K

-

lb/ft3 S

K cds 11s cm2/s

Dimensionless Factors Bl B2

Dimensionless group Dimensionless group Activation energy group Peclet number for mass transfer Peclet number for heat transfer Dimensionless length Dimensionless temperature Dimensionless concentration Dimensionless time

&

P1 p2 Z t

cc 0

Exercises 1.

Setting Z and hence k = 0, models the reactor response curve to a step input disturbance CAO= 1.

2.

Vary reactor inlet temperature up to 700 K.

3.

Study the effect of varying mass transfer and heat transfer diffusivities (D and h, respectively) and hence Peclet numbers (PI and P2) on the resulting dimensionless concentration and temperature reactor profiles.

4.

Based on a linearisation approach as applied to the model equations, Clough and Ramirez (197 1) predict multiple, steady-state solutions for the reactor. Is this confirmed and if not, why not?

5.

Verify that the unusual mixed units used here actually give the dimensionless groups as stated and that the dimensionless balances are consistent.

419

5.3 Tubular Reactor Examples

Results Noting that a hot region was developing at T = 0.14 in the last half of the reactor, the inlet concentration was reduced at T = 0.14 to 0.3, then raised to 0.4 and finally to 0.45. The graphical results are shown below.

,225

8.388

Figure 5.94. The above temperature profiles were obtained for this case.

Figure 5.93. The reactor responded to the changes according to these concentration profiles.

References Clough, D.E. and Ramirez, W.F. (1971) Simulation, 207. Ramirez, W.F. (1976) Process Simulation, Lexington.

5.3.9

VARMOL - Gas-Phase Reaction with Molar Change

System A tubular reactor is considered in which tLLegas-phase reaction leads to a change in the molar flow rate and thus in the linear gas velocity. The reaction stoichiometry is represented by A

-+

mB

420

5 Simulation Examples of Chemical Engineering Processes ~~

Figure 5.95.

Tubular reactor with variable molar flow.

The model is described in Sec. 4.3.3. The steady-state balances are written in terms of moles. The Ideal Gas Law is used to calculate the volumetric flow rate from the molar flow at each point in the reactor. This gives also the possibility of considering the influence and temperature or pressure profiles along the tube.

Program :Example VARMOL :Gas Phase Tubular Reactor :Variable volumetric flow rate :Molar change due to reaction (A - - > m*B) CONSTANT R=8.3144 :Gas constant [J/mol K] CONSTANT Temp=350 :Temperature [K] CONSTANT P=le5 :Pressure [N/m21 CONSTANT k=15 :Reaction rate [l/sl CONSTANT m=2.0 :Stoichiometric constant[-] CONSTANT NAO= 5 :Molar feed flow rate A [mol/sl CONSTANT NBO=O :Molar feed flow rate B [mol/sl CONSTANT Ninert=l :Molar inert flow rate [mol/sl CONSTANT Area=O.O12 :Cross-sectional area [m2] CONSTANT CINT=0.5, TFIN=20 1 SIM; INTERACT; RESET; GOT0 1 INITIAL NA=NAO NB=NBO G=(NA+NB+Ninert)*R*Temp/P yA=NA/(NA+NB+Ninert) yB=NB/(NA+NB+Ninert)

42 1

5.3 Tubular Reactor Examples

y A G =y A * G yBG=yB*G DYNAMIC Z=T G=(NA+NB+Ninert)*R*Temp/P yAG'=-k*yA*Area yBG'=m*k*yA*Area NA=yAG*P/(R*Temp) NB=yBG*P/(R*Temp) yA=yAG/G yB=yBG/G XA=(NAO-NA)/NAO OUTPUT Z,NA,NB,G Z, NA, NB, YAr Y B r XA, G PREPARE PLOT Z, G, 0, TFIN, 0,l

Nomenclature Symbols A G k m N NAO NBO Ninert P R Temp XA

Y

Z

Cross-sectional area Volumetric flow rate Reaction rate Stoichiometric constant Molar flow rate Molar feed flow rate A Molar feed flow rate B Molar inert flow rate Pressure Gas constant Temperature Fraction conversion Mole fraction Length

Indices 0 inert A B

Refers to inlet Refers to inerts Refers to component A Refers to component A

m2 m31s 11s molls molls molls molls N/m2 Jlmol K K -

m

422

5 Simulation Examples of Chemical Engineering Processes

Exercises 1.

Vary the stoichiometry to see the influence of m. Note that if m=l, G must be constant through the reactor.

2.

Set the molar feed rate of inerts to a high value, and note that G does not change much with position.

3.

Change the model and program to account for a linear temperature profile.

4.

Change the model and program to account for a linear pressure profile, allowing for pressure drop through the reactor.

Results

Figure 5.96. Variation of volumetric flow rates with position for three operating temperatures (200, 300 and 400 K).

Figure 5.97. Fraction conversion as a function of position for the three temperatures in Fig. 5.96.

5.4 Semi-continuous Reactor Examples

5.4

Semi-Continuous Reactor Examples

5.4.1

SEMIPAR - Parallel Reactions in a Semi-Continuous Reactor

423

System A semi-continuous reactor is used to carry out the parallel reaction shown below. It is of interest to investigate how the amount of desired product, P, depends on the differing orders of the two reactions and on the feeding rate.

Figure 5.98. Parallel reactions in a semi-continuous reactor.

Model The total mass balance is

The mass balances for components A, B, P and Q are

424

5 Simulation Examples of Chemical Engineering Processes

The kinetics for the parallel reactions are

r2 = k2 CnA2 A cnB2 B

Program :Example SEMIPAR :Semicontinuous parallel

reaction

Constant CINT=O.l, TFIN=15, NOCI=l,ALGO=O Constant nAl=l :Reaction order Constant nA2=1 :Reaction order Constant nB1=1.5 :Reaction order Constant nB2=2.5 :Reaction order Constant F=0.05 :Flow rate into tank (L/min) Constant Vin=0.5 :Initial tank volume (L) Constant Vtank=l :Volume of tank (L) Constant Bfeed=l :Conc. of B in feed (moles/L) Constant VAin=l :Initial mass of A (mol) Constant VBin=O :Initial mass of B (mol) Constant kl=l :Rate constant reaction 1 Constant k2=2 :Rate constant reaction 2 1 SIM; INTERACT; RESET; GOT0 1 INITIAL VA=VAin VB=VBin VP=O vq=o V=Vin DYNAMIC VA'=-(rl+r2)*V

425

5.4 Semi-continuous Reactor Examples

VBt=F*Bfeed-(rl+r2)*V VP' =rl*V VQr=r2*V rl=kl*(A**nAl)*(B**nBl) r2=k2*(A**nA2)*(B**nB2) V'=F A=VA/V B=VB/V P=VP/V Q=VQ/V SEL=VP/(VQ+O.OOl): Avoids division by IF (V.GE.Vtank) F=O IF (VA.LT.0) A=O IF (VB.LT.0) B=O PLOT T,P,O,Tfin,O,l OUTPUT T,VA,VB,VP,VQ T I VA t VB I V P VQ I At B I P I Q t v I S E L PREPARE

zero

Nomenclature Symbols Concentration Volume Flow rate Reaction rate

C V F r

mol/L L L/min mol/L min

Indices A, B, P, Q F n 1, 2

Refer to components Refers to inlet feed Refers to reaction orders Refer to reactions

Exercises 1.

Set nA1 = n ~ and 2 n g l < ng2. Run the reactor in two ways: semicontinuously with slow feeding of B and as a batch reactor with all of B initially in the reactor. Compare and explain the differing amounts of desired product P, obtained.

426

5 Simulation Examples of Chemical Engineering Processes

2.

Repeat Exercise 1 but with ngl > ng2.

3.

Study the semi-batch case ngl < ng2 for various constant feed rates.

Results

Figure 5.99. T h e concentrations of reactant A and the two products, P and Q, are shown. Reactant B is maintained at very low concentration by the slow feeding.

Figure 5.100. The volume reaches a maximum at approximately t = 10. The selectivity, ratio of desired to total products, remains constant over much of‘ the reaction time. It is strongly influenced by feeding rate of B for this case, for which nB2 > nB ,.

5.4.2 SEMISEQ - Sequential Reactions in a Semi-Continuous Reactor System A complex reaction is run in a semi-batch reactor with the purpose of improving the selectivity for the desired product, P. The kinetics are sequential with respect to components A, P and Q but parallel with respect to B. The relative orders of the reactions for the reactions determine the feeding policy.

427

5.4 Semi-Continuous Reactor Examples

kl

A+B

P

k2 b

P+B

Q

Figure 5.101. Complex sequential reactions in a semi-continuous reactor.

Model The total mass balance is

and the component

- 1-2

V

V

The kinetics are r l = kl

"A C"B1

B1

r2 = k2 CnpCnB2 P

B2

The orders, ngl and ng2, will determine whether CB should be maintained high or low in the reactor. The total reaction time must be adjusted such that P is

5 Simulation Examples of Chemical Engineering Processes

428

obtained at its maximum concentration, and this will also influence how B is fed.

Program :Example

SEMISEQ

:Semicontinuous

sequential

reaction

Constant CINT=O.l, TFIN=50, NOCI=l,ALGO=O Constant n A = l :Reaction order Constant n P = l :Reaction order Constant nBl=l.O :Reaction order Constant nB2=1.5 :Reaction order Constant F=0.05 :Flow rate into tank (L/min) Constant Vin=0.5 :Initial tank volume ( L ) Constant Vtank=lO :Volume of tank (L) Constant Bfeed=l :Conc of B in feed (mol/l) Constant VAin=l :Initial mass of A (mol) Constant VBin=O :Initial mass of B (mol) Constant VPin=O :Initial mass of P (moll Constant VQin=O :Initial mass of Q (mol) Constant k l = l :Rate constant reaction 1 Constant k 2 = 2 :Rate constant reaction 2 1 SIM; INTERACT; RESET; GOT0 1 INITIAL VA=VAin VB=VBin VP=VPin VQ=VQin V=Vin DYNAMIC VA'=-rl*V VB1=F*Bfeed-(r1+r2)*V VP'=(rl-r2)*V VQV=r2*V A=VA/V B=VB/V P=VP/V Q=VQ/V rl=kl*(A**nA)*(B**nBl) r2=k2*(P**nP)*(B**nB2) V'=F

429

5.4 Semi-continuous Reactor Examples

SEL=VP/(VQ+0.001) :Avoids division by IF (V.GE.Vtank) F=O F=O IF (P.le.O.and.T.gt.10) IF (A.LT.0) A=O IF (B.LT.0) B=O PLOT T, P, 0, Tf in, 0,1 OUTPUT T,A,B,P,Q,V,SEL T,A,B,P,Q,V,VA,VB,VP,VQ,SEL PR E P A R E

zero

Nomenclature Symbols Concentration Volume Flow rate Reaction rate

C V F r

mol/L L Wmin mol/L min

Indices A, B, P, Q F n 1, 2

Refer to components Refers to inlet feed Refers to reaction orders Refer to reactions

Exercises 1.

Determine the optimal time to stop the reaction for a range of flow rates. How do the values of k l and k2 influence this?

2.

Investigate the influence of flow on the maximum concentration of P. Study this for ngl greater and less than ng2.

3.

For ng2 > ngl, operate the reactor with interactive variation in F to keep CB low. How is the selectivity influenced by this?

430

5 Simulation Examples of Chemical Engineering Processes

Results IS18

Figure 5.103. Plotting the component masses versus time, it is seen that the maximum for VP comes much later, at about 22 minutes.

Figure 5.102.

The concentration profiles show P passing through a maximum around time 15 minutes.

5.4.3

HMT - Semi-Batch Manufacture of Hexamethylenetriamine

System Aqueous ammonia is added continuously to an initial batch charge of formaldehyde solution. The reaction is instantaneous and highly exothermic, and is given by

4 NH3 + 6 HCHO

+

N4(CH2)6 + 6 H20

43 1

5.4 Semi-Continuous Reactor Examples

Figure 5.104

Variables for the HMT example.

Model For the semi-batch operation the total mass balance is

For the instantaneous reaction, the rate is equal to the reactant feed rate. The energy balance then becomes dT M cP dt = W cP (To - T)

+

W XAO(-AH)

-

UA (T - T,)

where XAOis the mass fraction of ammonia in the feed stream.

Program :Example

HMT

:SEMI-BATCH

PRODUCTION

OF

HEXAMETHYLENETRIAMINE

CONSTANT MO=lOOO,XFO=0.42,TEMPI=5O,TC=25 CONSTANT W = 1 4 0 , X A O = 0 . 2 5 , T E M P O = 2 5 , CP=l,TEMPMAX=lOO CONSTANT HR=1100,U=415,A=1 CONSTANT T F I N = 9 0 , C I N T = 5 1 SIM; INTERACT; RESET; GO TO 1 INITIAL

432

5 Simulation Examples of Chemical Engineering Processes

M F = M O *XFO : INITIAL FORMALDEHYDE : INITIAL MASS M=MO : INITIAL TEMPERATURE TEMP=TEMPI : INITIAL HEAT QUANTITY QH=M*CP*TEMP DYNAMIC MI = W Q=U*A*(TEMP-TC) TEMP'=(W*CP*(TEMPO-TEMP)+W*XAO*HR-Q)/(M*CP) P L O T T,TEMP,O,TFIN,O,TEMPMAX P R E P A R E T,TEMP,M

Nomenclature Symbols Heat exchange area Specific heat Mass in reactor Temperature Overall heat transfer coefficient Mass feeding rate Ammonia feed mass fraction Formaldehyde charge fraction Exothermic heat of reaction

A cP

M T U

w X A0 XFO

-AH

m2 kcallkg "C kg "C kcal/m2 "C kglh kg/k kg/kg kcallkg

Indices Refers to cooling Refers to feed values

C

0

Exercises 1.

Find the area of cooling area required such that the reactor temperature just rises to 100 "C, by the end of the feeding at TFIN of 90 min.

2.

Carry out simulations with differing inlet temperatures, initial charges and feeding rates. Modify the program in order to ensure that reaction does not continue, as shown by the simulation, once the formaldehyde charge has been exhausted.

433

5.4 Semi-Continuous Reactor Examples

Results The results indicate that the mass flow rate has a strong influence on the temperature.

l.q

-E 4A H 1.000 A H

11.298

I

0 . W 8.8

ism

I

I

25.8

r

58.8

75.0

im.0

Figure 5.105. The mass flow rate, W was decreased at T=45, which caused a lower temperature in the reactor.

8.88$ 8 . 8

25.8

T

58.8

75.8

180.8

Figure 5.106. The decrease in the mass flow rate caused the mass in the reactor, M to increase more slowly.

Reference Smith, J. M. (1970) Chemical Engineering Kinetics, 2nd edition, McGraw Hill.

5.4.4

RUN - Relief of a Runaway Polymerisation Reaction

System The manufacture of a polyol lubricant by the condensation of an alcohol with an alkylene oxide in a semi-batch reactor proceeds according to the following reaction C4H90H

+ (n + 1) C 3 H 6 O -+

C4H9 (OC3H6),0CH2CHOHCH3

+ heat

434

5 Simulation Examples of Chemical Engineering Processes

The reaction is highly exothermic and the reactor contains large quantities of volatile oxide. Careful control of temperature is therefore required to avoid a runaway reaction and excessive pressure generation.

Oxide feed

#i

Bursting disc

- -lnltial - - -level ------- I

w -1

Figure 5.107.

Recycle

The polymerisation reactor with cooling and bursting disc.

The catalysed alcohol is charged initially to the reactor and the oxide then fed to the reactor at constant rate. Heat generated by reaction is removed via an external heat transfer loop. In this, the batch contents are recycled through an external cooler with sufficient capacity to maintain a constant return temperature to the reactor. With high rates of cooling, the reactor runs stably but produces a product of low molecular weight. As the cooling is decreased, higher molecular weight product is obtained, but at the risk of a runaway reaction and reactor failure. In the event of this or the case of a cooling failure, a bursting disc in the reactor will rupture. This will discharge oxide vapour to the atmosphere, and the feed of oxide to the reactor will be stopped.

435

5.4 Semi-Continuous Reactor Examples

Model The total mass balance is

dM_ _ dt - F - V

where M is the reactor mass at time t (kg) and V is the vapour venting rate (kg/min) and F is the feed rate (kg/min). The oxide component mass balance is

where C is the oxide concentration (kg/kg) and R1 is the oxide reaction rate (kg/min) . The mass of oxide reacted is given by dX dt = + R 1

-

where X is the mass of oxide reacted at time t. The energy balance is dT M cp dt = F cp (To - T) - V h - R1 (-AH) - Q where T is the temperature of reaction ("C), Q is the rate of cooling (kJ/min) and h is the heat of vapourisation of the oxide. The heat removed via the recirculation through the external cooler is given by

where Fc is the recycle mass flow rate (kg/min). The reaction kinetics are given by R1 = k C M

(kg oxide/min)

where N is the kmol of initial alcohol charge.

436

5 Simulation Examples of Chemical Engineering Processes

Assuming instantaneous equilibrium, the reactor pressure and oxide vapour pressure is given by:

P =

[exp + !;(

11.7)

~

+ 1.45 *

MI] C

The vapour discharge rate is given by the following. (1) For normal conditions and up to disc rupture

v=o (2) For sonic discharge to atmospheric pressure

k, 0.85 P =

4T-T-m

kg/min for P > 1.9 bar

(3) For sub-sonic discharge for 1 < P < 1.9 bar V =

.$T+273

d

m

kglmin

where k, is the valve discharge coefficient.

Program The ISIM program RUN with KV = 0 represents operation of the reactor under zero venting (V = 0) conditions, i.e., the bursting disc remains closed. With KV > 0 ( 100 - 1000) the program simulates the performance under emergency venting conditions. The program runs stably until the user interacts with HOLD and causes the coolant failure to occur with HOLD, VAL FC = 0 and GO. The program ensures that the bursting disc remains integral up to the bursting pressure, that the vapour discharge is sonic for reactor pressures greater than critical, that the discharge is subsonic for pressures less than critical and that the discharge stops when the pressure is atmospheric. For convenience, the exit temperature from the cooler, Tc, is assumed equal to the feed temperature, To, i.e., the heat transfer capacity of the cooler is in large excess

5.4 Semi-continuous Reactor Examples :Example :

RUN

RELIEF

OF

CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT 1

SIM;

437

A

RUNAWAY

POLYMERIZATION

REACTION

MO=4400 MOL=74 FO=100 FC=5000 TO=80 CP=3.5 LAMBDA=670 E=21000 Z=9.00000E+09 R=1.987 HR=-1660 PBURST=8 KV=O CINT=2 TFIN=2000

RESET;

INTERACT;

GOT0

1

INITIAL M=MO MC=O MTR=MO*TO

x=o

N=MO /MOL F=FO :OPEN=O :OPEN=l

Bursting disk OK disk burst OPEN=O

DYNAMIC C=MC/M :Conc. from mass K=Z*EXP(-E/(R*(TR+273))) :Vapor

pressure

:Arrhenius

P=(EXP(-3430/(TR+273)+11.7)+1.45E-3*MWEI)*C

IF (P.LT.l) P = l IF(P.EQ.l) V=O

:Bursting disc IF(P.GE.PBURST)

control OPEN=l

43 8

5 Simulation Examples of Chemical Engineering Processes

:Venting velocity control IF (OPEN.EQ.l) F=O IF(OPEN.EQ.l.AND.P.GE.1.9) V=O.SS*KV*P/SQRT(=(TR+273)) HILF=OPEN.EQ.l.AND.P.LT.l.g.AND.P.GT.1 IF ( H I L F ) V=KV*P/SQRT((TR+273))*SQRT((l+l/P/P)) MWEI=(MO + X)/N :Molecular weight RR=K*MC :Kinetics :DIFFERENTIAL B A L A N C E S M' =F-V MC'=F-V-RR TR'=(F*TO-TR)-V*LAMBDA/CP-RR*HR/CP-FC*(TR-To))/M X' =RR :Oxide reacted P L O T T,P,O,TFIN,O,lO O U T P U T T,M,C,TR,P,MWEI P R E P A R E T, M, C, TR, P, MWEI, V

Nomenclature Symbols C cP

E F

FC

h

M M1 Mol N P P1 p2 R R1 T (TR) t TO V X Z -AH

Oxide concentration Specific heat Activation energy Oxide mass feed rate Recycle rate Heat of vapourisation of oxide Charge mass Molecular weight Molecular weight of charge Initial alcohol charge Reactor pressure Bursting disc relief pressure Maximum safe working pressure Gas constant Reaction rate Reaction temperature Batch time Charge, feed and recycle temperature Venting rate Mass of oxide Frequency factor Exothermic heat of reaction

kg/kg kJkg "C caltmol kglmin kglmin kJkg kg kglmol kglkmol kmol bar bar bar callmol K kglmin "C min "C kglmin kg llmin

5.4 Semi-Continuous Reactor Examples

439

Indices Refers to initial and inlet values Refers to cooler

0 C

Exercises The simulation emphasises that when reactions start to run away they do so extremely fast. In this example the rate of pressure generation is compounded by the double exponential terms in the Arrhenius and vapour pressure equations. 1.

With the bursting disc remaining closed (KV = 0 and V = 0), establish the minimum external recycle cooling flow (Fc) needed to ensure stable operating conditions and satisfactory polymer product (Mol wt = 2500). Determine the minimum cooling capacity required of the cooler. Study the effects of "runaway" following coolant failure by interactive computation. By carefully decreasing F, from an initially "safe" value, the absolute limit of safe operation can be established. In this the reaction starts to run away, but then the reaction corrects itself by having consumed all the oxide and then continues with further stable operation. Using the value of Fc found in Exercise 1, study the effect of differing valve discharge capacities (KV) following cooling failure at different stages of reaction time (FC = 0). As an alternative to venting vapour to atmosphere, the vapour may be condensed and returned to the reactor. Modify the program accordingly and study the resultant behaviour. Alter the program to include a complete model for the heat exchanger. This will allow a more realistic calculation of the heat removed.

440

5 Simulation Examples of Chemical Engineering Processes __

Results

8'ml , I $ , 0.m

,

8.8888.888

Figure 5.108. A cooling water failure caused the disc to burst, as shown by the pressure profile.

?

8.875

.

8.158

8.225

8.388

*E 3

Figure 5.109. The venting velocity for KV = 1000 is shown, as caused by the disc rupture in Fig. 5.108.

References Kneale, M. and Forster, G.M. (1968) 1.Chem.E.Symp. Series No. 25, 98. Kneale, M. and Binns, J.S. (1977) 1.Chem.E.Symp. Series No. 49.

5.5

Mixing-Model Examples

5.5.1

NOSTR

-

Non-Ideal Stirred-Tank Reactor

System In this model of non-ideal reactor mixing, a fraction, f l , of the volumetric feed rate, F, completely by-passes the mixing in the reactor. In addition, a fraction, f2, of the reactor volume, V, exists as dead space. F3 is the volumetric rate of exchange between the perfectly mixed volume V l and the dead zone volume V2 of the reactor.

44 1

5.5 Mixing-Model Examples

I

v1

c1

I

Figure 5.110. The non-ideal flow reactor model.

Model The reaction is assumed to be first order, where r = kC The component mass balances are as follows.

For the well-mixed volume

where V1 = (1

-

f2)V and F1 = (1 - fl)F

For the dead space volume

442

5 Simulation Examples of Chemical Engineering Processes

where

v2 = f2

and

v

F~=cxF

Note that the dead space volume is treated as having a uniform well-mixed composition. For the reactor outlet F C 3 = F1 C1 + F 2 Co where

F2 = fl F

For a perfectly mixed tank with outlet concentration C4 and total volume V V dC4x = F ( C o - C 4 ) - V

kC4

Program In this program FRACl is fl and FRAC2 is f2. Note also that while FRACl can be set to zero, FRAC2 cannot be set to zero without modifying the program, since the derivative value for C2 is divided by V2, which is then zero. :

Example

:

NON-IDEAL

NOSTR CSTR

WITH

BYPASSING

CONSTANT CO = 1 CONSTANT V = 10 CONSTANT F = 1 CONSTANT K = 0.1 CONSTANT FRACl = 0.5 CONSTANT FRAC2 = 0.25 CONSTANT ALPHA = 5 CONSTANT CINT = 1.0 CONSTANT TFIN = 50 1

S1M;INTERACT;RESET;

GOT0

1

AND

DEAD

SPACE

443

5.5 Mixing-Model Examples

INITIAL F 1 = F*(l-FRACl) F2 = F*FRAC1 V 1 = V*(l-FRAC2) V2 = V*FRAC2 F3 = ALPHA*F c1 = 0 c2 = 0 c4 = 0 DYNAMIC C1' = (F1*(CO-Cl)+F3*(C2-Cl)-K*Vl*Cl)/V1 C2' = ( F 3 * ( C l - C 2 ) - K * V * C 2 ) / V 2 C 3 = ( F l * C l + F 2 * C O )/ F C4' = (F*(CO-C4)-K*V*C4)/V x3 = (CO-C3)/CO x 4 = (CO-C4)/CO PLOT T,X3,0,TFINfO, 1 OUTPUT TfC1,C2,C3,C4,X3,X4 P R E P A R E TfC1.C2,C3,C4,X3,X4

Nomenclature Symbols CO C1

c2 c3 c4

F

fl

F2 f2

F3 R X

a

Feed concentration Active zone concentration Deadzone concentration Effluent concentration Ideal tank concentration Total flow rate Fractional by-pass flow By-pass flow rate Dead volume fraction Deadzone exchange flow Rate constant Fractional conversion Fractional deadzone flow

Indices 0

Refers to feed

mol/L mol/L molL molL mol/L L/min Wmin L/min 1/min

444

5 Simulation Examples of Chemical Engineering Proce5ses ~~~

1 2 3

Refers to interchange flow and outlet concentration Refers to dead volume region Refers to deadzone flow

Exercises 1.

Study the effects of fractional by-passing and fractional dead space volume on the steady-state conversion, X3, compared to that of the ideal tank. X4.

2.

Note if K = 0, the program generates an F-curve for the non-ideal reactor.

3.

Study the effect of variations in the rate of exchange F3 between the two reactor volumes, by varying ALPHA.

4.

Modify the program to allow for zero dead space volume and study the effect of fractional by-passing only. Compare the results to that of a perfectly stirred tank.

5.

Modify the system to that shown below by Froment and Bischoff ( I 990), in which the volumetric flow, F, having passed through a partial wellmixed volume, V1, is then diverted such that a fraction, f, then passes through the residual volume, V2, before rejoining the main flow stream. Formulate the model equations and record the F-curve response for the system for varying values of V1 and V2 keeping the total volume (V = V1 +V2 ) constant and varying values of f.

Figure 5.111. Mixing model of Froment and Bischoff (1990).

445

5.5 Mixing-Model Examples

Results

A x3

EX4

r Figure 5.112. The fraction conversion for the system, X3, is lower than the ideallymixed situation, curve Xq.

B

Figure 5.113. These tracer response curves were obtained by setting k = 0.

References Dunn, I.J., Ingham, J. (1972) Verfahrenstechnik. 399, 6, Nr 11. Froment, G.F., Bischoff, K.B. (1990) Chemical Reactor Analysis and Design, 2nd edition, Wiley Interscience.

5.5.2

TUBEMIX - Non-Ideal Tube-Tank Mixing Model

System Non-ideal mixing conditions in a reactor can often be modelled by combinations of tanks and tubes. Here, three, stirred tanks are used to simulate the tubular, by-passing condition.

446

5 Simulation Examples of Chemical Engineering Processes

Figure 5.114. A combined tank-tube mixing model.

Model The model involves mass balances according to the scheme of Fig. 5.114. For the tubular by-pass the balances, for reactant A, have the form v1

:FA"

= F1 (CAn-1 - CAn)

4- rAn

v

for sections n = 1, 2 and 3. The main reactor region has the balance

:FA'

v2

+

= F2 CAO F 3 c

~ - 5F2 c.44- F 3 c A 4

The dead-zone region is balanced by v3

iFA5=

F3 c A 4 - F3 c A 5

+r

~ v53

The final outlet concentration is given by the mixing point FO c A 6 = F1 c A 3 + F 2 c A 4

+ rA4 v 2

447

5.5 Mixing-Model Examples

The flow rates are given by the flow fractions as

and

F3 = F2 R2

The volumes are given by the volume fractions as

and v3 =

The reaction rates are

V2

R4

rAn = kcAnn

where n refers to the reaction and the reaction order.

Program :Example

TUBEMIX

:Tank reactor, with a bypass and a dead-zone. :The bypass is modelled by a series of tanks. :The react ion is nth-order Constant Vtub=1000 : [dm3lVolume of tube reactor Constant R3=0.2 : [-]Volume fraction tube/tankl Constant R4=0.6 : [-]Volume fraction tankl/tank2 Constant FO=100 : [dm3/minl Feed to reactor Constant R1=0.8 : [-I Feed fraction tube/tank Constant R2=0.5 :[-I Feed fraction tankl/tank2 Constant CAO=500 :[g/dm3] Feed concentration Constant n=2 :[-I Reaction order Constant k=0.0005 :[depends on nl Rate constant Constant CAinit=O :[g/dm3] Initial concentrations Constant CINT=l, TFIN=50, NOCI=10 1 SIM; INTERACT; RESET; GOT0 1 IN ITIAL

.

448

5 Simulation Examples of Chemical Engineering Processes

CAl=CAinit CA2=CAinit CA3=CAinit CAQ=CAinit CA5=CAinit CA6=CAinit CAIM=CAinit DYNAMIC Fl=FO*Rl F2=FO-F1 F3=F2*R2 Vl=Vtub/3 V2=Vtub/R3 V3=V2/R4 Vt0t=Vtub+VZ+V3 Ral=-k*CAl**n :Reaction velocity CAl'=Fl/Vl*(CAO-CAl)+Ral Ra2=-k*CA2**n CA2'=Fl/Vl*(CAl-CA2)+Ra2 Ra3=-k*CA3**n CA3'=Fl/Vl*(CA2-CA3)+Ra3 Ra4=-k*CA4**n CA4'=1/V2*(F2*(CAO-CA4)+F3*(CA5-CA4))+Ra4 Ra5=-k*CA5**n CA5'=1/V3*F3*(CA4-CA5)+Ra5 CA6=(F2*CA4+Fl*CA3)/FO CAIM'=(FO*(CAO-CAIM))/Vtot-K*CAIM**n :Ideally-mixed model PLOT T,CA6,0,TFIN,OICAO OUTPUT T,CAI,CA2iCA3,CA4,CA51CA6tCAIM PREPARE T,CA1,CA2,CA3,CA4,CA5,CA6,CA6,CAIM

Nomenclature Symbols CA0 CAinit FO F1 F2 F3 k

Feed concentration Initial concentrations Feed to reactor system Flow rate to tube Flow rate to tank Interchange flow rate Reaction constant

g/dm3 g/dm3 dm3/min dm3/min dm3/min dm3/min various

449

5.5 Mixing-Model Examples

Reaction order Feed fraction tubehank Flow fraction F I F O Feed fraction tankl/tank2 Flow fraction F3/F2 Volume fraction tubehank1 Volume fraction tankUtank2 Volume of tank sections in tube Volume of main reactor section Volume of deadzone section Volume of tube reactor

;:=/I: :

Results

Sample results for TUBEMIX are given below ISIU

ISM

A CA6

/-

L

8.815

8.888 8 . 8

1.5

I

15.8

22.5

38.8

Figure 5.115. Here for second-order reaction the ideally-mixed tank has a larger conversion.

Figure 5.116. For a first-order reaction the rate is much slower and the response time is greater. Note the different scales.

450

5.5.3

5 Simulation Examples of Chemical Engineering Processes

MIXFLOl and MIXFL02 - Mixed-Flow Residence Time Distribution Studies

System The following systems represent differing combinations of ideal plug-flow, mixing, dead space, flow recycle and flow by-pass.

c3 "D FR Figure 5.117.

Mixing model for residence time distribution Case 1.

1 VD

Figure 5.118.

Mixing model for residence time distribution Case 2.

c4

5 . 5 Mixing-Model Examples

45 1

It is desired to compare the residence time distributions (RTD) for these cases.

Model 1. Residence Time Distributions Ideal plug-flow regions VT are represented by time delay functions, which are programmed into the ISIM examples. The mixing junction is given by the steady-state balance. Case 1

Fo Co + FR C4 = (Fo + FR) Ci Thus the concentration leaving the junction is

The plug-flow region represents a time delay; thus C2 is equal to C1 delayed in time by

The well-mixed region is given by V mdC3 T - (Fo + FR - FB) (C2 - C3)

The final outlet concentration is given by a steady-state junction relation as,

Case 2

The relations used here are analogous to Case 1.

For the junction

452

5 Simulation Examples of Chemical Engineering Processes -

For the mixing tank

For the by-pass junction

For the plug-flow time delay C4 is equal to C3 delayed in time by VT - VD FO FR

Program ISIM program MIXFLOl models the situation of Case 1, program MIXFLO2 models that of the Case 2. The plug-flow section of the plant is modelled as a circular time delay function, as described in Sec. 2.1.3, using the program subroutine TD. This is first initialised in the INITIAL section of the program and accessed during the DYNAMIC execution with the use of the reserved ISIM logical control variable LCOM. The size of the delay is restricted to a maximum of 50 integer multiples of CINT, based on the use of the integer function IFIX. :Example MIXFLOl :MIXED FLOW RESIDENCE TIMES : CASE1 CONSTANT FO=l,FR=.5,FB=.2 CONSTANT VT=lO,VD=l,VM=Z CONSTANTCO=l CONSTANT TFIN=501CINT=.5 1 S1M;INTERACT;RESET;GOTO 1 INITIAL :TIME DELAY IN PLUG FLOW REGION DELAY=(VT-VD)/(FO+FR) :TIME DELAY ELEMENTS (INTEGER VALUE) N=IFIX(DELAY/CINT) :MAX TIME DELAY ELEMENTS M=N+Z :INITIAL POINTER POSITIONS J =1 I=J+N-1

45 3

5.5 Mixing-Model Examples

:INITIALISE TIME DELAY ELEMENTS CALL TD(T,M,N,YIN,YOUT, I, J,LCOM) c1=0;c2=0;c3=0;c4=0 DYNAMIC C1=(FO*CO+FR*C4)/(FO+FR) YIN=Cl C2=YOUT C3'=(FO+FR-FB)*(C2-C3) C4=(FB*C2+(FO+FR-FB)*C3)/(FO+FR) OUTPUT T,ClfC2,C3,C4 PLOT T ,C4 ,0 ,TFIN, 0,1 PREPARE T f C 1 , C 2 , C 3 , C 4 :TIME DELAY SUBROUTINE IS EFFECTED :EVERY COMMUNICATION INTERVAL USING :THE RESERVED LOGICAL CONTROL VARIABLE IF(LC0M)

CALL

LCOM

TD(T,M,N,YIN,YOUT, I, J,LCOM)

SUBROUTINE TD(T,M,N,YIN,YOUT, I, J,LCOM) DIMENSION Y ( 5 0 ) IF(T.GT.0) GOT0 10 :SET TIME DELAY ELEMENTS TO ZERO DO 5 COUNT=1,50 5 Y(COUNT)=O YOUT=O RETURN :INPUT NEWEST VALUE 10 Y(1) = YIN :EXTRACT OLDEST VALUE YOUT=Y (J) :RESET POINTERS I=I+1 J=J+1 IF(1.GT.M) 1=1 IF ( J GT M ) J = 1 RETURN

.

:Example MIXFL02 :MIXED FLOW RESIDENCE TIMES : CASE 2 CONSTANT FO=l,FR=.5,FB=.2 CONSTANT VT=lO,VD=l, V M = 2 CONSTANT CO = 1 CONSTANT T F I N = 5 0 f C I N T = . 5

454

5 Simulation Examples of Chemical Engineering Processes

1 SIM; 1NTERACT;RESET;GOTOl INITIAL :TIME DELAY IN PLUG FLOW REGION DELAY=(VT-VD)/(FO+FR) :TIME DELAY ELEMENTS (INTEGER VALUE) N=IFIX (DELAY/CINT) :MAX TIME DELAY ELEMENTS M=N+2 :INITIAL POINTER POSITIONS J=l I=J+N-l :INITIALISE TIME DELAY ELEMENTS CALL T D (T,M, N, YIN, YOUT, I, J, L C O M ) c1=0;c2=0;c3=0;c4=0 DYNAMIC Cl=(FO*CO+FR*C4)/(FO+FR) C2'=(Fo+FR-FB)*(Cl-C2)/VM C3=(FB*Cl+(FO+FR-FB)*C2)/(FO+FR) YIN=C3 C4=YOUT OUTPUT T,Cl,C2,C3,C4 PLOT T IC4,0, TFIN, 0,1 PREPARE T,Cl,C2,C3,C4 :TIME DELAY SUBROUTINE IS EFFECTED :EVERY COMMUNICATION INTERVAL USING :THE RESERVED LOGICAL CONTROL VARIABLE IF (LCOM)

CALL

LCOM

T D ( T IM, N, YIN, YOUT, I, J, LCOM)

SUBROUTINE TD(T,M,N,YIN,YOUT, I, J , L C O M ) DIMENSION Y(50) IF(T.GT.0) GOT0 10 :SET TIME DELAY ELEMENTS TO ZERO DO 5 COUNT=1,50 5 Y(COUNT)=O YOUT=O RETURN :INPUT NEWEST VALUE 10 Y(1) = YIN :EXTRACT OLDEST VALUE YOUT=Y (J) :RESET POINTERS I=I+l J=J+1

455

5.5 Mixing-Model Examples

IF(1.GT.M) IF(J.GT.M) RETURN

I=l J=l

Nomenclature Symbols V F C

Volume Volumetric flow Tracer concentration

m3 m3/s "units"/m3

Indices T

M

D B R 192,374

Refers to plug flow Refers to mixing Refers to dead space Refers to by-pass Refers to recycle Refer to positions according to figure

Exercises 1.

Vary the relative system parameters VT, VD and VM for the two differing flow cases, keeping the total volume of the system constant. Compare the resulting residence time distributions.

2.

Keeping the total volume of the system constant, vary the relative flowrates, FO, FB and FR and note the influence on the RTD's.

3.

Revise the program to include nth-order reaction. Evaluate the conversion for first-order reaction from the dynamic model. Compare this with the conversion calculated from the E curve. Show by simulation that the two are equal only if the reaction is first order.

456

5 Simulation Examples of Chemical Engineering Processes

Results

B.BBB0.0

5.8

I

r

1B.B

1t.0

2E:E

Figure 5.119. Residence time distributions from program MIXFLOl with varying recycle flow rate, FR = 0.5, 2 and 10 (curves A, B and C).

Figure 5.120. Concentrations at location 2 and 4,Cz and Cq, using MIXFLOl with FR = 0.5.

Residence time Figure 5.121. distributions from program MIXFL02 with varying recycle flow rate, FR = 0.5, 2 and 10 (curves A, B and C).

Figure 5.122. Concentrations at location 3 and 4, C2 and Cq, using MIXFL02.

Reference Szekely, J., Themelis, N.J. ( I 97 1) Rate Phenomena in Process Metallurgy, Wiley Interscience.

457

5.5 Mixing-Model Examples

5.5.4

GASLIQl and GASLIQ2 - Gas-Liquid Mixing and Mass Transfer in a Stirred Tank

System The gas-liquid flow characteristics of stirred vessels depend both on the level of agitation and the rate of gas flow and can vary from the case of bubble column type operation to that of a full circulating tank, as shown in Fig. 5.123. The mixing characteristics and gas distribution obtained, obviously exert a considerable influence on the rate of mass transfer obtained (Harnby et al.. 1985).

Air

b) Figure 5.123 a and b. Gas-liquid mixing modes: a) bubble column behaviour ( flow, low rotor speed); b) circulating tank behaviour (high rotor speed, reduced gas flow).

In this example it is required to model the behaviour of the mixing vessel according to the conceptual models indicated below.

Liquid Mixing The basic liquid mixing pattern is assumed to remain constant, as shown in Fig. 5.124.

45 8

5 Simulation Examples of Chemical Engineering Processes

.-

FL1

FL1

FL2

FL2

Figure 5.124. Model for the liquid mixing pattern in a stirred tank.

In this, the liquid acts as three well-stirred regions, the smaller one representing the region close to the impeller. The two larger ones represent the volumes of liquid above and below the impeller, with the flow from the impeller discharging equally into the two main tank regions.

Gas Mixing The gas phase mixing is represented by three equally-sized tanks in series. In the case of the full bubble column, the bottom gas phase mixed region is coincident with the impeller, and no significant gas mixing occurs below the impeller, in the lower region of the tank, as shown in Fig. 5.125. The dashed arrows in Fig. 5.125 and Fig. 5.126 represent mass transfer interchange between the gas space and liquid volumes of the tank. The liquid circulation, which is not shown, is identical to Fig. 5.124.

459

5.5 Mixing-Model Examples

vG3

vG2

t

c

/

/

/

Q2

cG1

"G1

FGy cGO

Figure 5.125. Gas phase mixing pattern and mass transfer for full bubble column.

Figure 5.126. Gas phase mixing and mass transfer for full circulation case.

460

5 Simulation Examples of Chemical Engineering Processes

-___

With full circulation, substantial gas phase mixing occurs in all tank regions, as shown in Fig. 5.126.

Model Bubble Column Operation Liquid phase balances VL1

dCLl dt =

FL2 (CL2 - C L I )

Pump discharge rate FD = FL1 -k FL2

Gas phase balances

46 1

5.5 Mixing-Model Examples

The equilibrium relationships are represented by means of Henry's Law where

Full Circulation Liquid phase balances

Rates of mass transfer Qi =

KLa ( CL1* - CLi) V G ~

Q2 = K L (~C L ~ " C L ~V)

G ~

Q3 = K L (~C L ~ " C L ~V) G ~

The equilibrium relationships are represented by Henry's Law as before.

462

5 Simulation Examples of Chemical Engineering Processes

Program Program GASLIQI simulates the case of the full bubble column and program GASLIQ2 that of the full circulation case. A tank of total liquid volume, V = 1 m3, is assumed, giving VL1 = 0.3 m3, V L =~ 0.1 L and V L = ~ 0.6 L. An impeller discharge rate of 0.22 m3/s is assumed to apply, consistent with conditions of Fig. 5.123a and 0.44 m3/s consistent with the conditions of operation in Fig. 5.123b. In the case of the full bubble column, the volumes of the gas phase in the two tanks in the upper part of the vessel are three times that located at the impeller, giving V C ; ~= 0.007 m3 and v G 2 = v G 3 = 0.021 m3, consistent with a total gas holdup fraction of 0.0476. With full circulation, the holdup volume of the gas phase in the upper region is assumed to be twice that in the lower region and six times that in the middle region. A total gas holdup fraction of twice that for the full bubble column, gives V G ~= 0. 03m3, VG2 = 0.01m3 and v G 3 = 0.06 m3. Gas circulation rates of five percent of the corresponding liquid discharge rates are assumed. A total gassing rate of 1 m3/m3 s is assumed. For simplicity a constant value of overall mass transfer capacity coefficient, KLa, is assumed to apply for all the liquid regions, but the value obtained in the case of full circulation is taken as five times that for the bubble column. :Example GASLIQl : GAS-LIQUID MIXING AND MASS TRANSFER : IN BUBBLE COLUMN-TYPE MIXING TANK OPERATION : EXAMPLE FOR OXYGEN CONSTANT VL1-300, VL2=100,VL3=600 :L CONSTANT VG1=7,VG2=21,VG3=21 :L CONSTANT FG=16.7,CGO=.0086,FL1=110,FL2=110 :L/S,MOL/L CONSTANT KLA = 0.09 :1/SEC CONSTANT R=.082,H=.84E3,TEMP=293:LATM/MOLK,ATM L/MOL CONSTANT CL11=.00025,CL21=0.00025,CL31=0.00025:M0L/L CONSTANT CGlI=O, CG2I=O, CG3I=O CONSTANT TFIN = S O , CINT-0 1 1 SIM; INTERACT; RESET; GO TO 1 INITIAL CLl=CLlI; CL21CL21; CL3=CL31 CGl=CGlI: CG2=CG2I; CG3=CG3I DYNAMIC : LIQUID PHASE MASS BALANCES CL1' = (FL2*(CL2-CLl))/VLl CL2' = (FL2*(CLl-CL2)+FLl*(CL3-CL2) +Ql)/VL2 CL3' = (FLl*(CL2-CL3) + 4 2 + Q3)/VL3 : GAS PHASE MASS BALANCES

.

5.5 Mixing-Model Examples

463

CGl'= (FG *(CGO-CGl) - Ql)/VGl CG2'= (FG *(CGl-CG2) - Q2)/VG2 CG3'= (FG *(CG2-CG3) - Q3)/VG3 : MASS TRANSFER RATES Q1 = KLA *(CLlEQUIL-CL2)*VLl 4 2 = KLA * (CL2EQUIL-CL3)*VL2 43 = KLA *(CL3EQUIL-CL3)*VL3 : EQUILIBRIA CLlEQUIL = R*TEMP*CGl/H CL2EQUIL = R*TEMP*CG2/H CL3EQUIL = R*TEMP*CG3/H PLOT T, CL1, O,TFIN,0,0.00025 PREPARE T, CL1,CL2,CL3,CG1,CG2,CG3,QlfQ2,Q3 OUTPUT T , CL1 ,CL2 ,CL3 :Example GASLIQ2 GAS-LIQUID MIXING AND MASS TRANSFER WITH FULL CIRCULATION : MIXING TANK OPERATION CONSTANT VL1=300, V L 2 = 1 0 O f V L 3 = 6 O 0 :L CONSTANT VG1=7,VG2=21,VG3=21 :L CONSTANT FG=16.7, CGO = 0.0086, FL1=110, FL2=110 :L/S, MOL/L CONSTANT FGR1=2,FGR2=2 :L/S CONSTANT KLA = 0.09 :1/SEC CONSTANT R=0.082, H=0.84E3, TEMP=293 :L ATM/MOL K, ATM L/MOL, CONSTANT CLI=O, CGI=O.O CONSTANT TFIN =50, CINT=O.Ol 1 SIM; INTERACT; RESET; GO TO 1 INITIAL CLl=CLI; CL2=CLI; CL3=CLI CGl=CGI;CG2=CGI;CG3=CGI DYNAMIC : LIQUID PHASE MASS BALANCES CL1' = (FL2*(CL2-CLl)+Ql)/VLl CL2' = (FL2*(CLl-CL2) + FLl*(CL3-CL2) + Q2)/VL2 CL3' = (FLl*(CL2-CL3) + Q3)/VL3 : GAS PHASE MASS BALANCES CG1'= (FGRl*(CG2-CGl)- Ql)/VGl CG2'= (FG*CGO+FGRl*(CGI-CG2)+FGR2*CG3-(FGR2+FG)*CG2Q2)/VG2 CG3'= ((FGR2+FG)*(CG2-CG3) -Q3)/VG3 : MASS TRANSFER RATES :

464

5 Simulation Examples of Chemical Engineering Processes

KLA * ( CLlEQUIL-CL2 ) *VL1 KLA * ( CLZEQUIL-CL3 ) *VL2 KLA * ( CL3EQUIL-CL3)*VL3 : EQUILIBRIA CLlEQUIL= R*TEMP*CGl/H CL2EQUIL= R*TEMP*CG2/H CL3EQUIL= R*TEMP*CG3/H PLOT T, CL1, O,TFIN,0,0.00025 PREPARE T, CLl,CL2,CL3,CGl,CG2,CG3,Ql,Q,Q3 OUTPUT T, CL1, CL2, CL3

Q1 42 Q3

= = =

Nomenclature Symbols V F C

Volume Volumetric flow rate Oxygen concentration Rates of oxygen transfer Mass transfer capacity coefficient Ideal Gas Constant Temperature Henry's Law constant

Q

KLa R T H

L US molL us L atdmol K K L atdmol

Indices 0, 1, 293 L G

*

Refer to streams and zones (see figures) Refers to liquid Refers to gas Refers to equilibrium

Exercises 1.

Simulated tracer experiments can be performed on either program by setting KLa = 0 and starting with suitable initial conditions and inlet concentrations. For example: a) In G A S L I Q l set CL2I = 1, C L l I and CL31 = 0. Observe the distribution of tracer for different values of FL1 and FL2.

465

5.5 Mixing-Model Examples

b) In GASLIQ2 set CGI = 0 and the inflow CGO = 1. Observe the response of the outflow CG3 for various values of the circulation flow FGRl and FGR2.

2.

Set KLa to a suitable value and experiment with the influence of the mixing parameters FLl and FL2 in GASLIQl and FGRl and FGR2 in GASLIQ2.

Results Ism

Ism

R 8.225

em 8.815

r Figure 5.127. Tracer experiments using GASLIQ1 according to Exercise 1 for two values of F L and ~ F L (Curves ~ C and A: FLI and FL2 = 110; curves D and B: F L and ~ F,2 = 50.

r Figure 5.128. Oxygen transfer experiment with GASLIQZ showing response of CL1 to changes in liquid and gas circulation (Curve A: parameters as in program; Curve B: F L and ~ F L =~ 50; Curve C: FL1 and F L =~5, FGRl = F G R=~ 1.0.

References Campbell, S. (1988) NELIBradford University Report, East Kilbride. Harnby, N., Edwards, M.F. and Nienow, A.W. (Eds.) (1985) Mixing in the Process Industries, Butterworths.

466

5.5.5

5 Simulation Examples of Chemical Engineering Processes

SPBEDRTD Model

- Spouted

Bed Reactor Mixing

System A stirred-tank model has been proposed, (Daly, 1980), to model the mixing behavior of an air-solid, spouted, fluidised-bed reactor. The central spout is modelled as two tanks in series, the top fountain as a further tank and the down flowing annular region of the bed as 6 equal tanks in series. It is assumed that a constant fraction of the total solids returns from each stage of the annular region into the central two tank region, as depicted below.

Fountain

t

Annulus

7

t

'$

I I Spout(

I

\ r /

I

Air

Figure 5.129.

The spouted bed and its mixing model.

,

73c6

467

5.5 Mixing-Model Examples

Experimental residence time studies are to be carried out in which the solid in the bottom half of the bed is initially mixed with tracer, the bed started and timed samples taken from various locations in the bed annular region. It is desired to model the resulting tracer distribution in the bed in order to find the fractional slippage rate between the annular tanks and the central bed regions.

Model The initial expansion of the bed consequent to the start of the air flow is neglected from the analysis. Component balance equations for each region of the reactor are given. For the bed central flow region

v 3 d Cx3

For the annular flow region

-

F2C2-F3 C3

468

where

5 Simulation Examples of Chemical Engineering Processes

F1 = F o + 3 F F2 = F1 + 3 F F3 = F4 = F5 = F6 =

and

F2 F3 - F

F4-F F5 - F F7 = Fg-F F8 = F7-F F9 = F8 - F

co

= c9.

The initial distribution of tracer within the bed is represented by

Program Example SPBEDRTD SPOUTED BED MIXING MODEL CONSTANT V1=0.4, V2=0.3 CONSTANT V3=0.05 CONSTANT FO=l,F=0.05 CONSTANT Vtot=1.8 CONSTANT CSTART=1 CONSTANT TFIN=5,CINT=O.1 1 SIM; INTERACT; RESET; GOT0 INITIAL V=(Vtot-Vl-V2-V3)/6 C 0 = CS T A R T Cl=CSTART c2=0 C3=0 C4=0 : :

C5=0

C6=0 C7-CSTART C8=CSTART C9=CSTART DYNAMIC

Volume central spout volume fountain Flow rates Total reactor volume

: : : :

1

469

5.5 Mixing-Model Examples

Cl'=(FO*CO+F*(C7+C8+C9)-(F0+3*F)*Cl)/Vl C2'=(Fl*Cl+F*(C4+C5+C6)-(F1+3*F)*C2)/V3 C3'=F2*(C2-C3)/V C4'=F3*(C3-C4)/V C5'=F4*(C4-C5)/V C6'=F5*(C5-C6)/V C7'=F6*(C6-C7)/V C8'=F7*(C7-C8)/V C9'=F8*(C8-C9)/V co=c9 Fl=F0+3*F F2=F1+3*F F3=F2 F4=F3-F F5=F4-F F6=F5-F F7=F6-F F8=F7-F PLOT T,Cg,O,TFIN,O,CSTART PREPARE T,Cl,C2,C3,C4,C5,C6,C7,C8,C9 OUTPUT T,CO,Cl,C2,C3

Nomenclature Symbols V F C

Volume Volumetric flow rate Tracer concentration

m3 m3/h part s/m3

Indices 0, 1, ...) 9

Refer to designated regions and flow rates (see Fig. 5.129).

Exercises 1.

Determine the time required for the tracer to be nearly completely mixed throughout the vessel. Which criterion did you use for uniform mixing?

470

5 Simulation Examples of Chemical Engineering Processes

2.

Change the initial conditions to place the tracer at a different region of the vessel. Does this effect the time for complete mixing?

3.

Double the flow rates in the vessel and note the influence on the time for mixing.

Results

RC4 B c9

1.0W

-

A

0.7580.5880.258-

J

0.BBB0.88

1,b

I

3388

4.Q

RCI BC3

2,00

1.58

1.88

6.b

Figure 5.130. Distribution of a tracer in the annular region, starting with an initial concentration, CSTART= I , in the lower sections.

8-88

Figure 5.131. Response of Sections 1 and 3, corresponding to Fig. 5.130, with an additional tracer addition at T=1.8 to Section 3.

Reference Daly, W.F. (1980) University of Bradford.

5.5.6

BATSEG, SEMISEG, COMPSEG - Mixing and Segregation in Chemical Reactors

System Chemical reactions will take place only when the reactant molecules are in intimate contact. In some cases, especially with very fast reactions or viscous liquids, segregation of the reactants can exist, which make the reaction rates and

5.5 Mixing-Model Examples

47 1

~

selectivities dependent on the mixing intensity. In chemical reactor engineering, the assumption is usually made that only mean concentrations need be considered. In reality, concentration values fluctuate about a mean, and in some cases these fluctuations must be considered in detail. This field is very complex and is still the subject of much research. This example serves only to introduce these concepts and to show how simulations can be made for certain simple situations. The concept of segregation and its meaning to chemical reactors was first described by Danckwerts, 1953. The intensity or degree of segregation is given the symbol I, which varies between one and zero. Shown in Fig. 5.132 is a tank with two components, A and B, which are separated into volume fractions, q A and 1 - q A ; this condition represents complete initial segregation (I=l). Stirring or mixing this tank would cause the segregation to decay until complete homogeneity would be achieved (I=O). In this situation, the rate of any reaction between the reactants could obviously be influenced by the rate of mixing, as measured by the change in I.

Figure 5.132.

Complete segregation (I = I ) of reactants A and B in two volume fractions.

Segregation may also be important if the reactants are fed to a reactor in an unmixed condition. This could be the case in any continuously fed reactor, either tubular (Fig. 5.133) or tank (Fig. 5.134).

Figure 5.133. Feeding of unmixed reactants to a tubular reactor.

472

,

5 Simulation Examples of Chemical Engineering Processes

FA~ C A O

I

1

Figure 5.134.

FgxAt

Unmixed reactants fed into a continuous stirred tank.

The intensity of segregation, I, can be described in terms of concentration fluctuations, as shown in Fig. 5.135. Here the time-varying fluctuations above and below an average value are shown.

Time Figure 5.135.

Time-variant turbulent fluctuations of concentration about a mean value showing the fluctuation value CAI.

G,

The unmixedness can be characterised by the mean of the square of the fluctuations or concentration variance, according to

473

5.5 Mixing-Model Examples

To compare the “state of unmixedness”, Danckwerts introduced the intensity of segregation, which is calculated in terms of the mean square of the fluctuations, as

where the subscript 0 denotes the initial or feed value. In the case shown in Fig. 5.132, the initial mean square fluctuations would be

For unpremixed feed streams, the mean volume fraction is defined in terms of the mean entering flow rates as

-

-

qA = -

FA

FA

+ FB

Segregation or unmixedness of non-reactive systems, as measured by I or

[P

I , can be balanced using the general macroscopic population balance

equation as follows

Rate of accumulation of unmixedness in the system

=

-

[ [

[

Flow of

Flow of u n m ~ ~ ~ $ e s s ]unmixedness out of the system the system

Dissipation rate of by unmixedness mixing in the system

]

+

)

-

Rate of [off~$~i%k] in the system

The dissipation rate term, rdm, can be taken as a first-order decay of concentration variance whose rate constant is the inverse mixing time

The mixing time can be related to power per unit volume and the geometry. Turbulent dispersion can produce unmixedness within the system, if there are gradients of mean concentration. This is not considered here. A useful discussion of these equations was given by Brodkey (1975) and Baldyga (1989).

474

5 Simulation Examples of Chemical Engineering Processes

Model In what follows, the above balance for unmixedness is applied to individual reactor cases. The relation for reaction rate in terms of I is then considered, and finally this is applied for simple and complex reactions.

Intensities of Segregation in a Batch Reactor For a batch reactor system only the accumulation and dissipation terms are important. Thus

With the initial value of

[ v ] o , integration gives

This equation predicts the intensity of segregation to decay with time in the batch reactor. It is also applicable to a steady-state plug flow system, where t is the residence time.

Intensities of Segregation in a Continuous Reactor For a steady-state continuous tank reactor only the flow terms and the dissipation terms are important. Thus

With the mean residence time z = V/F, this becomes

475

5.5 Mixing-Model Examples

Intensities of Segregation in a Semi-Continuous Reactor In a semi-batch system the volume varies from zero to Vmax with total flow rate F as dV dt = FA

+

-

or for constant F

FB = F

V = Ft

During filling, the accumulation, inflow and dissipation terms are important. Thus the balance gives dI

dt=

1 -_1 _- -I _ t

Tm

where feed and the initial tank contents are completely segregated, i.e., I(0) = 1. Introducing dimensionless time, 0 = t/z, and dimensionless mixing time, 0, = T&, during the filling period (0 c 1) gives dI 1 - 1 - I a = 7 Om __

After the reactor is full and overflowing V = Vmax, and therefore T = Vm,x/F. Then 0 2 1 and the outflow term is also important. The balance becomes dI I a = l - I - -

Om

Reaction Rate in Terms of Intensity of Segregation The rate of a simple A + B + C reaction, which is given by the instantaneous concentrations is expressed in terms of the mean concentrations and the extra variable CA' CB', becoming

The derivation of this rate expression depends on the theory developed by Toor (1969) and assumes that the covariance of the reactant fluctuations is the same with and without reaction, as given by

476

5 Simulation Examples of Chemical Engineering Processes

Mixing effects are particularly important for complex reactions, since selectivity is changed. The following reaction is considered A

+

B

kl, R

R

+

B

k2,

(desired reaction)

S (undesired reaction)

The kinetics in terms of the intensities of segregation and the mean concentrations can be shown to be

This result was obtained by Toor and Bourne (1977) by neglecting the influence of the covariance, CB' CR' and by assuming that CA' CB' could be estimated by Toor's hypothesis, as explained above. This simplification is restricted to a relatively slow second reaction and to a volume ratio qA/qB near one.

Batch Reactor The batch reactor balance for a simple A

+ B -+C reaction becomes

_ -

Dimensionless variablescan -be defined as follows: 0 = t/Tm, T\ = (&&A(), = cBO qB/(qA + 9s) and CAO = Da = k CAOT, and y = CA/CAO. Here CTO CAOqA/(qA + Q). The dimensionless balance becomes

a dY = - D a [ y ( y + q - 1 )

where the initial value y(0) = I .

-IT\] = -Da(y(y+q-l)-e-@q)

477

5.5 Mixing-Model Examples

For finite y, when Da

4

00

this gives

[Y(Y+r\-l)

-

Irll’O

and thus ym is given by the quadratic formula

In this case only the + sign has a physical meaning. The conversion X in the batch tank is given by X = 1 dX a = Da[(l-X)(q-X)

-

-

y, which becomes

Iq] = Da[(l-X)(q-X)

-e-@q]

The above equations also apply to a plug flow reactor, where 0 is the dimensionless residence time, which varies with distance.

Continuous Stirred-Tank Reactor For a simple A + B 4 C reaction in a continuous stirred-tank reactor, as shown in Fig. 5.134, in terms of fraction conversion of the reactant A, the balance becomes X = Da [ ( l - X ) ( q - X )

-

Iq]

with the fractional conversion defined as

where C?O= CAOFA/(FA+ FB) If X is to be finite when Da -+00, the term [ (1 - X) (q - X) and then as before

x,

=

(1 + q ) * d ( l+ q ) 2 - 4 q (1 - I ) 2

since 0 < X, < 1, only the minus sign has meaning here.

-

I q ] + 0,

5 Simulation Examples of Chemical Engineering Processes

478

Generally

Semi-Batch Reactor For a simple A + B

+ C reaction the semi-batch reactor balance is

using the same dimensionless quantities, this becomes

with y(0) = 1. Here the constant A = 0 during the filling period, and A = 1 after the filling period, when 0 > 1. The requirement to have y finite when Da + 00 gives,

The component balances for the complex reaction A

+

B

kl, R

(desired reaction)

R

+

B

k2, S

(undesired reaction)

become as follows: A-balance

B-balance

479

5.5 Mixing-Model Examples

R-Balance

During the filling period VCVmax. The dimensionless form of these equations when combined with the total balance with constant F, apply both to the filling and full periods. These are

--

where -the dimensionless terms are 0 = tj2, q = C B ~ C A ODa , = k2 CAOz, y = CiEAO and p = kl/k2. Here z = V/F, and the feed concentrations are YAO = 1, YBO = 11, YRO = 0 and yso = 0. During the filling period A = 0, and A = 1 when 0 > 1. For the desired selectivity, p = kl/k2 > I , and q c 1. Selectivity at any time 0 can be defined as:

xs

=

2 YS YR + 2YS

Here a value of Xs = 1 would mean that no desired product was obtained.

480

5 Simulation Examples of Chemical Engineering Processes

Program :

Example

BATSEG

:Simple Reaction with :in Batch Reactor

Segregation

Constant Da = 5, Eta = 0.6 Constant Cint = 0.01, Tfin =

5

1 S1M;INTERACT;RESET;GOTOl INITIAL y=l 1=1 DYNAMIC Theta=T I=Exp(-Theta) y'= -Da*(y*(y+Eta-1)-I*Eta) x ' = Da*((l-x)*(Eta-x)-I*Eta) PLOT Theta,y,O,Tfin,O,l OUTPUT Theta,y,x,I PREPARE T h e t a , y D X DI

:

Example

SEMISEG

:Simple Reaction with Segregation :in Semi-Batch Reactor Constant Da=5, Constant Cint =

Eta=0.8,Thetamix=0.3 0.01, Tfin = 5

1 S1M;INTERACT;RESET;GOTOl INITIAL y=l 1=1

v=o

T=O. 01

DYNAMIC Theta=T If(Theta.lt.l.O)A=Theta V'=l

48 1

5.5 Mixing-Model Examples

I'=(l-I)/Theta-I/Thetamix y'= 1 / A *(1-y)-Da*(y*(y+Eta-1)-I*Eta) If (Theta.gt.l.O)A=l PLOT Theta, y, 0, Tfin, 0,l OUTPUT Theta,y, I,V PREPARE Theta,y, I,V

:

Example

COMPSEG

:Complex Reaction with Semi-Batch Reactor :A+B--->R :R+B--->S

:in

Segregation

Constant Da=5, Eta=0.3,Thetamix=0.6 Constant Beta=2000 Constant Cint = 0.01, Tfin = 5 1 S1M;INTERACT;RESET;GOTOl INITIAL yA=1 yB=Eta yR=O. 01 ys-0 1=1 v=0 T=O.01 DYNAMIC Theta=T If(Theta.lt.l.O)A=Theta V'ml I'=(l-I)/Theta-I/Thetamix yA'= (1-yA)/A-Da*Beta*(yA*yB-I*Eta) yB'= ( E t a - y B ) / A - B e t a * D a * ( y A * YB - I * E t a ) - D a * y A * y R yR'= -yR/A+Beta*Da*(Ya*yB-I*Eta)-Da*yB*yR yS'= -Ys/A+Da*yB*yR Sel= 2 * y y ( y R + 2 * y S ) If (Theta.gt.l.O)A=l PLOT Theta, Sel, 0, Tf in, 0 , l OUTPUT Theta,yA,yB,yR,yS,Sel,I,V PREPARE Theta,yA,yB,yR,yS, Sel, I,V

482

5 Simulation Examples of Chemical Engineering Processes -

Nomenclature Symbols C

Concentration Concentration fluctuation variance Damkohler Number (= k C ~ Oand

Da

kg/m3 kg2/m6

k2 G O 7) Volumetric flow rate Degree of segregation Fractional flow rate Time Volume Fractional conversion Dimensionless concentration Ratio of reaction rate constants (= klJk2) -Initial concentration ratio (= CBO/CAO) Dimensionless time (= t/z) or (=z/z,) Time constant or mean residence time

F I 9 t

V X Y

P

rl

0

z

Indices , 0 dm m max T

Refers to fluctuations around mean value Refers to initial or input value Refers to dissipation of unmixedness Refers to mixing Refers to maximum volume Refers to tank Refers to Da = infinity Refers to mean value

M

Exercises 1.

Very high values of Da correspond to instantaneous reactions. Verify that simulations with programs BATSEG and SEMISEG give results corresponding to the quadratic formulae presented.

483

5.5 Mixing-Model Examples

2.

Show by simulation that the corresponding conversion of B for instantaneous reaction is given by solving the equation (1-X)(q-X)

-

Iq=O

3.

Investigate the influence of concentration ratio, ETA, from 0.05 to 0.8 on the selectivity and concentration of desired product using COMPSEG.

4.

With all three programs, vary the Damkohler number, Da over a range from 1 to 500 and explain the results. Note that when Da no longer has any influence on the rate then mixing must be controlling

References Baldyga, J. (1989) Chem. Eng. Sci. 44, 1175. Bourne, J. R. and Toor, H. L (1977) AIChE J. 23, 602. Brodkey, R. S . (1975) Ch. 2 in "Turbulence in Mixing Operations", R. S . Brodkey ed. Academic Press. Danckwerts, P. V (1953) Appl. Sci. Res., A3, 209. Toor, H. L. (1969) Ind. Eng. Chem. Fundam. 8, 655. This example was developed by J. Baldyga., Dept. of Chem. and Proc. Eng., Warsaw University of Technology, Poland.

484

5 Simulation Examples of Chemical Engineering Processes

Results

A Y D I

1,m 8,758 8. SBB 8.258 Y

e.45e:w

Figure 5.136. Program BATSEG: Dimensionless concentration y versus dimensionless time 0 for Da = 0.1, 5.0, and 100 (curves B, A and C). Shown also is the decay of segregation I with time (curve DEF).

8,'15

1:58

WR

2.'25

3 h

Figure 5.137. P r o g r a m SEMISEG: Varying THETAMIX (0.03, 0.1, 0.3 and 0.6, curves D, C, A, B) gave these results for the semi-batch reactor. Note that when THETA > 1 the reactor becomes continuous.

Ism

Ism 8.158

0.858

0.m

Figure 5.138. Program COMPSEG: Dimensionless concentration of desired product Y R versus dimensionless time THETA obtained by varying THETAMIX (0.03, 0.1,0.3, and 0.6, curves D, B, A and C). Parameter values used: ETA=0.3, BETA=2000. DA=1.0.

WR

Figure 5.139. Selectivity SEL profiles corresponding to the simulations i n Fig. 5.134. Low values of SEL are desired. Mixing is seen to greatly influence the selectivity, even though the second reaction is much slower.

485

5.6 Tank Flow Examples

5.6

Tank Flow Examples

5.6.1

CONFLO1, 2 and 3

- Continuous Flow

Tank

System Liquid flows from an upstream source, at pressure PI, via a fixed position valve into the tank, at pressure P2. The liquid in the tank discharges via a second fixed position valve to a downstream pressure P3. The tank can be open to atmosphere, closed off to the atmosphere and can work either under isothermal or adiabatic temperature conditions. The detailed derivation of this problem is discussed both by Franks (1972) and Ramirez (1976, 1989). Data values in the problem are similar to those of Ramirez, who also made simulations of this problem.

Figure 5.140. Schematic of the tank flow problem.

Model A mass balance for the liquid gives

486

5 Simulation Examples of Chemical Engineering Processes

For constant density, p dV = A--dz - F1-F3 dt dt -

-

where V = A z. The flow through each fixed position valve is given by

where Kvl and K v are ~ the effective valve constants. Tank pressure and temperature a) For the tank open to atmosphere, the outlet pressure from the tank, P2, is given by

b) For a closed tank under isothermal conditions, the air space above the liquid obeys the Ideal Gas Law PV = nRT

so that

where PGOand TGOare the defined initial conditions. c) Under well insulated conditions, the air in the tank will behave adiabatically and its temperature will change isotropically, where

where y is the adiabatic expansion coefficient for air, equal to 1.4. Both the upstream pressure P, and final downstream pressure P3 are fixed, together with the atmospheric pressure Po. The above model equations are also presented in dimensionless terms by Rarnirez (1989).

487

5.6 Tank Flow Examples

Program The programs for the open tank (CONFLOl), for the closed isothermal tank (CONFL02) and for the closed adiabatic tank (CONFL03) are given below. :Example

CONFLOl

:CONTINUOUS

OPEN

TANK

FLOW

PROBLEM

OF

RAMIREZ

CONSTANT A=l,G=9.81,20=3.0 P1=150,P0=101.325,P3=101.35 CONSTANT CONSTANT KV1=.001,KV2=.001,RHO=~~~~ CONSTANT TFIN =2000, CINT=10 1 SIM; INTERACT; RESET; GO TO 1 INITIAL

z=zo

DYNAMIC : Convert time to minutes TMIN=T/60 :Total mass balance Z' = (Fl-F2)/A : Units of pressure are kN/m2 P2=PO+RHO*G*Z/1000 : Flow rates, avoid negative arguments IF(P2.GT.P1)PZ=P1 Fl=KVl*SQRT(Pl-P2) F2=KV2*SQRT(P2-P3) PLOT T,Z,O,TFIN,O,ZO T,TMIN, Z, F1, F2, PI, P2, P3 PREPARE OUTPUT TMIN,Z,FI,FZ,PZ :Example

CONFLO2

:CONTINUOUS CLOSED ISOTHERMAL :PROBLEM OF RAMIREZ

FLOW

TANK

CONSTANT A=lrG=9.81,ZO=3.0,ZMAX=4 CONSTANT P1=150,PO=101.325,P3=101.~5 CONSTANT KV1=.001,KV2=.001,RH0=1000 CONSTANT TFIN=2000, CINT=10 1 SIM; INTERACT; RESET; GO TO 1 INITIAL

z=zo

: Initial PGO=PO

gas

phase

conditions

in

SQRT

488

5 Simulation Examples of Chemical Engineering Processes

PG=PGO VGO=(ZMAX-Z)*A VG=VGO DYNAMIC : Convert time to minutes TMIN=T/60 2' =(Fl-F2) /A : Units of pressure are kN/m2 P2=PG+RHO*G*Z/1000 : Flowrates, avoid negative argument in IF(PZ.GT.P1)P2=Pl Fl=KVl*SQRT(Pl-P2) F2=KV2*SQRT(P2-P3) : Gas space volume VG=(ZMAX-Z)*A PG=PGO*VGO/VG PLOT T, Z, O,TFIN, 0, ZO T, TMIN, Z, F 1 ,F 2 ,PI, P 2 ,P3,VG, P G PREPARE OUTPUT TMIN,Z,F1,FZfP2 :Example CONFL03 :CONTINUOUS CLOSED ADIABATIC :PROBLEM OF RAMIREZ

FLOW

TANK

CONSTANT A=1,G=9.81,Z0=3.OfZMAX=4 CONSTANT P1=150,P0=101.325,P3=101.35 CONSTANT KV1=.001,KV2=.001,RHO=~~OO CONSTANT GAMMA= 1.41fTGO=330 CONSTANT TFIN=2000, CINT=10 1 SIM; INTERACT; RESET; GO TO 1 INITIAL :Initial liquid depth

z=zo

: Initial gas phase conditions PGO=PO PG=PO VGO=(ZMAX-Z)*A VG=VGO TG=TGO DYNAMIC : Convert time to minutes TMIN=T/60 Z' =(Fl-F2)/A : Units of pressure are kN/m2 P2=PG+RHO*G*Z/1000

SQRT

489

5.6 Tank Flow Examples : Valve constants, avoid negative arg. in SQRT IF(PZ.GT.Pl)PZ=P1 Fl=KVl*SQRT(Pl-P2) F2=KV2*SQRT(P2-P3) : Gas space volume VG=(ZMAX-Z)*A PG=PGO*VGO*TG/(VG*TGO) TG=TGO*(VGO/VG)**(GAMMA-1) PLOT T, Z ,0, TFIN, 0, Z 0 PREPARE T,TMIN,Z,F1,F2,P1,P2,P3,VGlPGlTG OUTPUT TMIN, Z ,F1, F2, P 2

Nomenclature Symbols Area Valve proportionality factor Acceleration of gravity Mass Moles Pressure Ideal gas constant Density Temperature Volume Depth of liquid Adiabiatic expansion coeff., cp/cv

A KV

g M n P R

P T V Z

Y

m2 m2/N1/2 s d S 2

kg mol N/m2 cal/gmol K kg/m3 K m3 m

Indices 0, 1, 2, 3

Refer to positions (see figures)

Exercises 1.

For the open valve case (CONFLOl), make a step change in P1 and observe the transients in flow rate and liquid level. Try this for a sinusoidally varying inlet pressure. Experiment with a sudden change in the outlet valve setting.

490

5 Simulation Examples of Chemical Engineering Processes

2.

For the closed valve case (CONFL02), repeat the changes of Exercise 1 , observing also the variation in the gas pressure and volume. Change the gas volume, and note the influence of this variable.

3.

Using CONFL03, determine the importance of including the adiabatic work effects in the model, over a range of flow rate, pressure and valve conditions by following the changes in the gas space temperature and pressure.

4.

Study the dimensionless forms of the model equations given by Ramirez (1989); program these and compare the results.

Results xsxn 0.758

1

I

0,388 a:w

a.ka

?

dl00

a.'isB

a,h

a4

Figure 5.141. The results for CONFLOl were obtained by making a step increase i n P I from 150 to 200 N h 2 . Steady state is not completely reached here.

Figure 5.142. Same conditions as Fig. 5.141 CONFLOI, showing changes of height.

5.6 Tank Flow Examples

49 1

Figure 5.143. For CONFL02, at t = 3.5 min. the pressure PI was increased from 150 to 300 N/m2, at t = 8.5 it was decreased to 200 and at t = 19 further to 150. This resulted i n an increase in Z and a compression of the gas phase volume VG and opposite changes later in the simulation.

Figure 5.144. The flow rates F I and F, for the conditions of Fig. 5.141 become equal at steady-state.

Figure 5.145. For this run of CONFL03, the same conditions as those for CONFL02 were kept. The curves look very similar but with important differences: The gas volume is higher, and Z reaches a lower value.

Figure 5.146. The reason for the increased volume in Fig. 5.145 is the temperature rise for this adiabatic compression case.

492

5 Simulation Examples of Chemical Engineering Processes

References Franks, R. G. E. (1972) Modelling and Simulation in Chemical Engineering, Wiley -1nterscience. Ramirez, W. F. (1976) Process Simulation, Lexington Books. Ramirez, W. F. (1 989) Computational Method for Process Simulation, Butterworth.

5.6.2

TANKBLD - Liquid Stream Blending

System Two aqueous streams containing salt at differing concentrations flow continuously into a well-mixed tank and a mixed-product stream is removed. In this case, the densities of the streams are not assumed constant but vary as a function of concentration. All three streams can have time varying flow rates, and hence the concentration and volume of liquid in the tank will also vary with time.

F1 p l c l l

Figure 5.147.

;

,

Tank with density variation

5.6 Tank Flow Examples

493

Model Total mass balance

Component mass balance

The density of salt solution varies linearly with concentration, according to

where b is a constant of proportionality. The above system is treated in detail by Russell and Denn (1972), and an analytical solution is obtained.

Program :Example : :

LIQUID Liquid

TANKBLD BLENDING PROBLEM OF RUSSELL AND DENN densities depend o n concentration

F1=2,F2=2,C1=5,C2=15,F3=3 CONSTANT CONSTANT RHO = 62.4, B = 0.6 CONSTANT VMIN =0.05, VMAX=20,VO=2,C30=0 CONSTANT TFIN=10, CINT=.Ol 1 SIM; INTERACT; RESET; GO TO 1 INITIAL

v=vo

C3=C30 RH03=RHO+B*C3 : Initial mass of liquid VRH03=VO*RH03 : Initial mass of salt vc3=vo*c3o

494

5 Simulation Examples of Chemical Engineering Processes .~

~

~~

DYNAMIC :Liquid densities RHOl=RHO+B*Cl RH02=RHO+B*C2 RH03=RHO+B*C3 : Total mass balance (Fl*RHOl+F2*RH02-F3*RH03) VRH03'= : Component salt balance VC3'=(Fl*Cl+F2*C2-F3*C3) : Salt concentration c3=vc3/v :Tank volume V=VRH03/RH03 IF(V.LE.VMIN) V=VMIN STOP IF ( V GE .20 ) P L O T T,C3,0,TFIN,0,20 T,C1,C2,C3,VIF1,F2,F3,VC3,VRH03,RHO3 PREPARE O U T P U T T,C1,C2,C3,V

.

Nomenclature Symbols Volume Density Concentration

V P C

ft3

Ib/ft3 lb/ft3

Indices 1,233 0

Refer to stream values Refers to water

Exercises 1.

Carry out simulations for differing tank volumes, flow rates and feed concentrations, in which the inlet and outlet flow rates are set equal, (F1 + F2 = F3) and observe the approach to steady state. Relate the time taken to approach steady state to the mean tank residence time (z = V/F3).

5.6 Tank Flow Examples

495

2.

Study the case of tank washout, with the tank initially full, at high salt concentration and the liquid input streams free of salt, CI = C2 = 0. Show that the tank concentration, C3, decreases exponentially with respect to time.

3.

Russell and Denn solved the model equations analytically, assuming constant liquid density, and showed that

where

Here C30 is the starting tank concentration at time t = 0, and Vo is the starting tank volume at t = 0. Compare the results of the simulation to the analytical solution which can also be calculated with ISIM. For the constant density assumption, simply set b = 0. The analytical solution shows that the approach to steady state is very rapid when Vo is small and that the concentration in the tank is always constant, when starting with a relatively empty tank. Analytical solution also shows that the rate of change of volume in the tank is equal to the net volumetric flow rate, but only for a linear density concentration relationship. Check the above analytical conclusions numerically and test the case of a non-linear density-concentration relationship. 4.

Modify the program to enable the outlet flow to vary as a function of liquid depth and density, e.g.,

and test the system with time-varying inlet flows F1 and F2.

496

5 Simulation Examples of Chemical Engineering Processes

____-

_.

Results The output from two runs are shown.

A W

B V

.I

s.aeYm

2 . i

T

4.88

6 .'ME

n

S . i

Figure 5.148. The volume increases linearly and C3 increases from an initial zero value. At T = 3 the inlet concentrations C1 and C2 were set to zero.

Figure 5.149. These results confirm the analytical solution that C3 remains constant if V is initially zero. V was set very low (VMIN=0.001, VO=O.OOl) and CINT was set initially to 0.001. The numerical errors are caused by increasing CINT to 0.1.

Reference Russell, T.W.F, and Denn, M.M. (1972) Introduction to Chemical Engineering Analysis, Wiley

5.6.3

TANKDIS

-

Ladle Discharge Problem

System The simple tank discharge problem of Sec. 1.2.3.1 has been extended by Szekely and Themelis (1971) to cover the discharge of molten steel from a ladle into a vacuum degassing chamber; the transfer being effected via a discharge nozzle, in which the effects of both friction and wear can be important.

497

5.6 Tank Flow Examples

t

hl

- H

p2

h2

Figure 5.150.

Schematic drawing of the tank discharge problem.

Model Total mass balance

where for the nozzle mass flow

For the cylindrical tank geometry the mass as a function of head is

x D2

M = A H p = AQ(h-L)

where A = __ and gives the result that

M h=-+L AP

498

5 Simulation Examples of Chemical Engineering Processes

____.

.~

Neglecting acceleration effects in the metal, the energy balance equation is given by

where AHf represents the losses due to friction and 2fLvi AHf = DTg The friction factor, f, depends on the relative surface roughness of the nozzle (E) and the Reynolds number (Re) where

A constant value of the friction factor f = 0.009 is assumed, for fully developed turbulent flow and a relative pipe roughness E = 0.01. The assumed constancy of f, however, depends upon the magnitude of the discharge Reynolds number which is checked during the program. The program also uses the data values given by Szekely and Themelis (1971 ), but converted to SI. Since the velocity of the metal at the ladle surface is very low (vl = O), the metal discharge velocity is then given by

Note that a full treatment of this problem would also involve an energy and a momentum balance together with relations for possible physical property changes, during discharge and cooling. Acceleration effects are ignored in this analysis, which is solved analytically by Szekely and Themelis.

Program : :

Example LADLE

CONSTANT

TANKDIS DISCHARGE

PROBLEM

OF

SZEKELY

AND

THEMELIS

D=2.5,DT=0.05,L=0.05,E=0.01,F=O.009

499

5.6 Tank Flow Examples

CONSTANT RHO=7800,MHU=1.5E-4,PI=3.14,M1=4.5E4 CONSTANT Pl=lE4,P2=O,G=9.81 CONSTANT TFIN=1500,CINT=20 1 SIM; INTERACT; RESET; GO TO 1 INITIAL M = MI DYNAMIC M' = -W IF(M.LT.O)M=O ATANK = PI*D*D/4 H = M/(ATANK*RHO)+L ANOZZLE = PI*DT*DT/4 W = ANOZZLE*RHO*V = SQRT((G*H+(Pl-P2)/RHO)/(0.5+2*F*L/DT)) V RE = RHO*V*DT/MHU PLOT T, H I 0, TFIN, 0,2 PREPARE T,H,V,M,W,RE OUTPUT T,H,M,W,V,RE

Nomenclature Symbols D DT f g H h L M P P Re t V

w P

AHf

Tank diameter Nozzle diameter Friction factor Acceleration of gravity Tank height Effective head Nozzle length Mass Pressure Metal density Reynolds number time Metal velocity Mass flow rate Metal viscosity Pressure head lost by friction

m m m/S2

m m m kg kglm s2 kg/m3 S m/S

kgls kg m s

m

5 00

5 Simulation Examples of Chemical Engineering Processes

Indices 1

Metal surface Tank outlet

2

Exercises 1.

Compare the rates of discharge with and without allowance for friction.

2.

Modify the program to allow for nozzle wear.

3.

How would you expect cooling to affect the discharge rate? How could you make allowance for this in your model?

Results

A

n.48 8.88

"

"1

Ism

6.00

7

n.m

8.848

8.888

n.1a

*E 9.168 4

Figure 5.151. Steel level in ladle, H, as a function of time.

I

1.988.88

8.1

H

8.88

1.a

1.b

Figure 5.152. Discharge velocity as a function of steel level in ladle.

Reference Szekely, T. and Themelis, N.T. (1971) Rate Phenomena in Process Metallurgy, Wiley Interscience.

50 1

5.6 Tank Flow Examples

5.6.4

TANKHYD - Interacting Tank Reservoirs

System Two open reservoirs with cross-sectional areas Al, A2 and liquid depths hl and h2 respectively are connected by a pipeline of length L and diameter D. The resulting liquid levels form a decaying oscillatory response to final steady state. Ramirez (1976) provides a detailed model derivation for this problem and shows how the problem is solved by analogue computation. The parameters are the same as those used by Ramirez.

S D Figure 5.153.

Dynamics of interacting reservoir levels.

Model The mass balance for tank 1 at constant liquid density is d z ( A 1 hl) = - F and for tank 2

d ;Ti(Azh2) = F

where F is the volumetric exchange flow rate between tank 1 and tank 2.

502

5 Simulation Examples of Chemical Engineering Processes

Momentum balance for the pipeline:

(

accumulation) The rate of = ( m Rate o y y of um)

-

(-omenturn) Rate of

of momentum

(

+ Sum forces of

out

]

acting

Since the pipeline is of uniform diameter, the rate of momentum at the inlet is equal to that at the outlet and of accumulation ) = ( Sum of the forces (Rateofmomentum acting on the pipeline

where PI and P2 are the pressure forces acting on the system

and F, is the frictional force, given by the Fanning frictional equation as

Fx =

8 F2

n D3

__

fpL

where f is the Fanning friction factor, assumed constant in this analysis.

Program Note that the frictional force must always act in the opposite direction to that of the direction of flow and that this must be allowed for in the program. :Example

TANKHYD

:HYDRAULICS OF INTERACTING :PROBLEM OF RAMIREZ CONSTANT CONSTANT CONSTANT

A1=1,A2=0.5 L = 10 0, D = 0.3 F=0.005

TANK

RESERVOIRS

503

5.6 Tank Flow Examples

CONSTANT H10=1.5,H20=1.2 CONSTANT T FI N = 2 0 0 , C I N T = 1 CONSTANT G=9.81,PI=3.142 1 S1M;RESET;INTERACT;GOTOl INITIAL Hl=HlO;H2=H20 DYNAMIC Hl'=-FLOW/Al H2'=FLOW/A2 TERMI=PI*D*D*G/(d*L) TERM2=F*8/(PI*D*D*D) FLOW2=FLOW*ABS(FLOW) FLOW'=TERMl*(Hl-H2)-TERM2*FLOW2 PLOT T,Hl, 0, TFIN, 1,1.5 PREPARE T, H1, H2, FLOW

Nomenclature Symbols A D F f FX g h L

P t P

Cross-sectional area Diameter of pipe Flow rate Overall loss factor Frictional force Acceleration of gravity Liquid depth Length of pipe Hydrostatic pressure force Time Liquid density

Indices 1, 2

Refer to tanks 1 and 2

m2 m m3/s kilopond dS2

m m kilopond/m2 S

kg/m3

5 04

5 Simulation Examples of Chemical Engineering Processes ____.

.~

Exercises I.

Observe the oscillatory approach to steady-state for different initial liquid depths. Plot the depths and the flow rate versus time.

2.

Investigate the influence of pipe length and diameter on the time to achieve steady state.

3.

Suppose tank 1 represents a body of tidal water with tidal fluctuations and tank 2 is a tide meter used to measure the tide. The connection is made by 100 m of pipeline and tank 2 is 1 m in diameter. Assume that the tide varies in a sine wave fashion with an amplitude of 2 m with a wave length of 12 h. Change the program to investigate the minimum pipe diameter required to ensure less than a 1 cm difference between the actual tide level and the level in the tide meter.

Results

1.65

1

A F U U

.E -1, 0.158

8.075

-

0.m -

-8.875 -

1

-8.158 1.38

Figure 5.154. Oscillating heights of the liquid in the tanks.

1.35

ni

1.48

1.45

1.58

Figure 5.155. Phase-plane of liquid height in tank 1 versus flow rate. Note the reversal in the flow direction.

Reference Ramirez, W.F. (1976) Process Computation, Lexington Books.

505

5.7 Process Control Examples

5.7

Process Control Examples

5.7.1

TEMPCONT - Control of Temperature in a Water Heater

System The simple feedback control system below consists of a continuous-flow stirred tank, a temperature measurement device, a controller and a heater.

Figure 5.156. Feedback control of a simple continuous water heater.

Model The energy balance for the tank is

V p cp

2

= F p cp (To - TR) + Q

where Q is the heat input from the heater, delayed with a lag.

506

5 Simulation Examples of Chemical Engineering Processes ~

~

A proportional-integral feedback controller can be modelled by Q = Q o + K p & + KP --&dt TI

where the control error is given by &

= (TRset - TR)

Program The program makes use of the random number generator in ISIM, RAND, to cause random fluctuations in the feed temperature, To. Eliminating RAND will shorten the run time. :Example : :

TEMPCONT

FEEDBACK CONTROL OF A CONTINUOUS UNITS DEG C,KCAL,KG,L,HR

WATER

HEATER

CONSTANT TO=20, V=100, F=10,RHO=l,CP=l CONSTANT Q0=6, KP=10, TIr5, TS=80, TAUQ=O.1 CONSTANT CINT=O.1, TFIN=75.0 1 SIM; INTERACT; RESET; GO TO 1 INITIAL TR=20 SEED=l DYNAMIC :HEAT BALANCE TRI=(F/V)*(TO-TR)+Q/(V*RHO*CP) :CONTROL EQUATIONS QC=QO+KP*E+(KP/TI)*INTG(E,O) Q'=(QC-Q)/TAUQ :HEATER DELAY IF (QC.LT.O.0) Q=O E= (TS-TR) :VARIABLE INLET TEMPERATURE (15-25 C) TO=TO+(RAND(=SEED)-O.5) IF (T0.gt.25) TO=25.0 IF (TO.lt.15) TO=15.0 P L O T T, TR, 0, TFIN, 0,100 OUTPUT T, TR, TO PREPARE T,TR,TO,Q,E,QC

5 07

5.7 Process Control Examples

Nomenclature

Symbols Specific heat Frequency of oscillations Flow rate Proportional control constant Differential control constant Heor volume Error Density Inteat input Temperature Reactgral control constant Time constant for measurement Time constant for heater

cP

f F

KP KD

Q E

P

TI

T V

Trn TQ

kcal/(kg "C) lfh m3/h kcaV"C (kcal h)/"C m3 "C kg/m3 kcalk "C h h h

Indices Refers to measured value Refers to reactor Refers to setpoint Refers to inlet

m R set 0

Exercises 1.

Disconnect the controller (set Kp = 0), allow the temperature to reach steady state, and measure the temperature response to changes in water flow.

2.

Repeat Exercise 1 for changes in inlet temperature.

3.

Measure the controlled response to a step change in F with proportional control only (set T I very high). Notice the offset error E and its sensitivity to Kp. Change TI to a low value. Does the offset disappear?

4.

Obtain the process reaction curve for the process with disconnected controller, as explained in Sec. 2.3.3. Analyse this curve to obtain the parameters for the Ziegler-Nichols Method. Use Table 2.2 to obtain the best controller settings for P and PI control. Try these out in a simulation.

508

5 Simulation Examples of Chemical Engineering Processes

5.

Proceed as in Example 4, but use the Cohen-Coon settings from Table 2.2.

6.

As explained in Sec. 2.3.3.1, try to establish the controller settings by trial and error.

7.

With proportional control only, increase Kp to Kpo, until oscillations in the response occur. Use this frequency, fo, to set the controller according to the Ultimate Gain Method ( K p = 0.45 Kpo, TI = f0/1.2), where fo is the frequency of the oscillations at Kp = K p o (see Sec. 2.3.3). Measure the response to a change in To.

8.

Adjust the controller by trial and error to obtain the best results to a change in set point.

9.

Include a loss term in the energy balance.

10. Change the controller to give PID-control by adding a differential term,

11. Include a first-order time lag in the temperature measurement, using the equations

and

Results The results in Fig. 5.157 show the approach to the steady-state for a setpoint of 80°C. The influence of control parameter changes is shown in Fig. 5.158.

509

5.7 Process Control Examples

d

Figure 5.157. A change in setpoint from 80 to 50 resulted in these dynamics for an inlet random disturbances.

5.7.2

Figure 5.158. A sequence of changes in K P ( 10 to 50 to 20 to 30 to 40 to 45) gave these results.

TWOTANK - Two Tank Level Control

System Liquid flows through a system of two tanks arranged in series, as shown below. The level control of tank 2 is based on the regulation of the inlet flow to the tank 1. This tank represents a considerable lag in the system. The aim of the controller is to maintain a constant level in tank 2, despite disturbances which occur in the flow F3.

5 Simulation Examples of Chemical Engineering Processes

5 10

h1

f

Figure 5.159. Level control in a two-tank system.

Model Mass balance equations with constant density yields for tank 1

and for tank 2

A2 dh2 d t = F1

+ F3

-

F2

Flow equations are given by F1 = Ki

6

1

F2 = K 2 6 2 The disturbance is generated with a sine wave F3 = F30 + F3AMP sin

3t

511

5.7 Process Control Examples

and the proportional control equation is

FO = FO+ KC (h2 - b e t )

Program :

Example

TWOTANK

Level control in second tank by flow control to first tank Constant A1=4 : Area tank 1 Constant A2=3 : Area tank 2 Constant K1=0.5 : Outflow constant tank 1 Constant K2=1.2 : Outflow constant tank 2 Constant FOin=l : Initial flow to tank 1 Constant F30=1 : Average flow of disturbance Constant F3AMP=0.5 : Amplitude of disturbance Constant H10=0 : Initial level tank 1 Constant H20=0 : Initial level tank 2 Constant H2set=2 : Set-point level for tank 2 : Controller constant Constant Kc=4 Constant CINT=2,TFIN=80 1 SIM; INTERACT; RESET; GOT01 INITIAL H1=H10 H2=H20 FO=FOin DYNAMIC H I ' = ( F O - F 1 )/ A 1 H2'=(Fl+F3-F2)/A2 i f (Hl.le.0) H1=0 if (H2.le.0) H2=0 Fl=Kl*SQRT(=Hl) F2=K2*SQRT(=H2) F3=F30+SIN(=T/3)*F3AMP : Controller FO=FOin+Kc*(H2set-H2) if (FO.le.0) FO=O PLOT T, H2,0, TFIN, 0,4 OUTPUT T,H1,H2 PREPARE T, HI, H2, FO, F1, F2, F3 : :

512

5 Simulation Examples of Chemical Engineering Processes

Nomenclature Symbols A F h K

Tank area Flow rate Liquid depth Valve constant Controller constant Volume

KC

V

m2 m3/h m m31h m0.5 m2/h m3

Indices 0 1 2 3 amp set

Refers to inlet Refers to tank 1 Refers to tank 2 Refers to feed of tank 2 Refers to amplitude Refers to setpoint

Exercises 1.

With no control Kc = 0 and constant flow of F3, study the response of the levels hl and h2 and final steady-state values for differing flow rates Fo and F3.

2.

Apply sinusoidal variations in the flow rate F3, F3AMP = 0.05, and again study the responses in h2 to changes in F3.

3.

Study the influence of the ratio of A2/A1 on the response in h2.

4.

Set F3AMP = 0, for constant flow F3 and study the system response to h2set. Repeat the simulations with control added.

5.

Repeat Exercise 3 but with a sinusoidal variation in F3, as in Exercise 2.

6.

Change the valve constants and note the influence on the dynamics.

7.

The following diagrams indicate other common combinations of filling and emptying tanks, which can be easily programmed to form the basis of simple simulation examples.

5.7 Process Control Examples

8

Figure 5.160. Tank flow configurations for exercises.

513

5 14

5 Simulation Examples of Chemical Engineering Processes

Results

7.91

5.m 2.9

4

0.m0.m

8.058

T

9.1m

9.158

P.is$

-E3

Figure 5.161. At T=58 the value of the controller constant was increased (KC from 4 to 20).

8.WH.W

0.858

T

0.1m

8.158

u.m

-E 3

Figure 5.162. The increase in KC at T=58 gives very high peaks in FO (curve A). The sine wave disturbance of F3 is seen i n curve D.

References Smith, C.L., Pike, R.W. and Murril, P.W. (1970) Formulation and Optimisation of Mathematical Models, Intext.

5.7.3

CONTUN - Controller Tuning Problem

System The temperature of a continuous flow of material through a steam-heated stirred tank is controlled by regulating the flow of steam. The tank temperature is measured by a thermocouple set inside a thermowell, giving a delayed temperature measurement response. This example is based on that of Robinson ( I 975).

5 15

5.7 Process Control Examples

Figure 5.163. Tank reactor with control.

Model Tank heat balance dTR - W cp (To - TR) + Ws h

M C P d t

where h is the latent heat of the steam entering at rate Ws. The temperature of both thermocouple and thermowell are each described by first-order lag equations, so that: for the thermocouple

_ dTc_ - T w - T c

dt

-

and for the thermowell

where zc and TW are the lag constants.

ZC

516

5 Simulation Examples of Chemical Engineering Processes

Controller Equation For proportional and integral control

Program :

Example

CONTUN

:Controller tuning problem in a stirred water heater Constant M = 5 0 : Mass in tank Constant Cp=l : Specific heat Constant WO-250 : Feed Constant TO=20 : Feed temperature Constant Lambda=537 : Latent heat of vaporization : Average steam mass flow Constant WsO=O.2 Constant Kc=1.5 : Proportional gain Constant TRset=40 : Set temperature of tank Constant TAUc=O.l : Time constant of thermocouple Constant TAUw=0.3 : Time constant of thermowell Constant TAUi=5:Time constant of integral control Constant CINT=O.l,TFIN=3O,NOCI=3 1 SIM; INTERACT; RESET; GOT01 INITIAL TR=TO

w=wo

EPSO=TRset-TC TC=TO TW=TO DYNAMIC :STEP CHANGES IN WATER FLOW i f (T.ge.10) W=6*WO i f (T.ge.20) W=WO TR'=W/M*(TO-TR)+Ws*Lambda/(M*Cp) TC'=(TW-TC)/TAUC TW'=(TR-TW)/TAUw EPS=TRset-TC Ws=WsO+KC*EPS+(Kc/TAUi)*INTG(=EPS,EPSO) if (Ws.le.0) Ws=O

517

5.7 Process Control Examples

i f (Ws.ge.30) Ws=30 PLOT T,TR,O,TFIN,0,100 OUTPUT T,TR,W,WS PREPARE T,TR.TC,TW,W,WS

Nomenclature Symbols Specific heat Proportional gain Integral time constant Mass Temperature of tank Time Mass flow rate Latent heat of vapourisation of steam Time constant

cP KC KI

M

TR t

W

h T

kcalkg K kg/min K kg/min2 kg kg K min kglmin kcalkg min

Indices Thermocouple Steam Thermowell Feed condition Integral Setpoint

C S W 0 I set

Exercises 1.

Allow the tank to achieve steady-state operation in the absence of control (Kc = 0). Use the resulting process reaction curve to estimate combined proportional and integral control parameters. Then use the obtained steady-state values as the initial values for a following sequence of runs.

2.

Using proportional control only (71 very high), vary the proportional gain constant, in conjunction with a temperature set point change of 10 OC and determine the critical value of K c giving continuous oscillation. Tune the controller for both proportional and integral actions using the ZieglerNichols-criteria, given in Sec. 2.3.3.2.

518 3.

5 Simulation Examples of Chemical Engineering Processes

Experiment with the influence of different values of the measurement time constants, zc and zw,Plot all the temperatures to compare the results.

Results ISIH

IsIn

66.8

22,5

45.8

15,8

38.0

1.5

T

8

Figure 5.164. Tank temperature versus time for two values of KC (1.5 and 2.0), with TI = 10000. The changes at T=10 and T=20 are programmed step changes in the inlet water flow rate. Oscillations and offset are caused by suboptimal controller tuning.

8.8

?

Figure 5.165. Variations in steam flow corresponding to the run in Fig. 5.164.

References Robinson, E.R. ( 1975) Time Dependent Processes, Applied Science Publishers.

5.7.4

SEMIEX - Temperature Control for Semi-Batch Reactor

System An exothermic reaction involving two reactants is run in a semi-continuous reactor. The heat evolution can be controlled by varying the feed rate of one component. This is done with feedback control with reactor temperature

519

5.7 Process Control Examples

measurement to manipulate the feed rate. The reactor is cooled by a water jacket, whose heat transfer area varies with volume. Additional control could involve the manipulation of the cooling-water flow rate. -_____----_------_-----

-I

I

iC Figure 5.166. Exothermic semi-batch reactor with feed control.

Model The component balances with second-order kinetics for the variable volume reactor are as follows

The volume is given by the total mass balance

The reaction rate constant varies with absolute temperature as

k = ko e - E l

RTA

5 20

5 Simulation Examples of Chemical Engineering Processes ~-

____.

The heat balance for the reactor contents is V p cP%

= FB p cP (Tf-TR)

- u A (TR-Tc)

+ ( - A H ) (- k CA CB v)

which simplifies, for constant properties, to

The cooling jacket is assumed to be well-mixed and is modelled by

simplifying to

The level in the tank, h, varies with volume and tank diameter, D, according to

h =

V .n (D I 2)2

The heat transfer area changes with liquid level as

A = nDh The reactor temperature is controlled by manipulating the feed rate according to

A similar equation could be applied to the manipulation of the cooling water flow.

5.7 Process Control Examples

521

Program :Example

SEMIEX

:Semi-batch reactor with heat production :Feedback control of feed stream :Jacket cooling Constant VO= 6 :[m3] Initial tank volume Constant Ca0=10 :[kmol/m3] Init. conc. of A in tank Constant CbO=O : [kmol/m3] Init. conc. of B in tank Constant Cbf=lO :[kmol/m31 Conc. of B in feed Constant CcO= 0 : [kmol/m3JInitial conc.of C in tank Constant Tf =18 : [deg Cl Feed temperature Constant FbO= 1 :[m3/h] Base feed rate Constant Kp=20 : [m3/h K] Proportional constant Constant TI=1000 : [h] Integral time constant Constant TRO=20 :[deg Cl Initial temp. in tank Constant Cpr=4.14: [kJ/kg K] Cp of reactants Const ant Fc=lO : [m3/h] Cooling water stream Constant Cpc=4.18: [kJ/kg Kl Cp of cooling water Constant Vc=O.5 :[m31 Volume of cooling jacket Constant Tcin=15 :[deg C] Cooling water temp. Constant U=lOOOO :[kW/m2K]Heattransfer coefficient Constant AO=3 :[m21 Heat transfer area if empty Constant C=1.6 : [mlCircumference of reactor Constant Rho=1000 : [kg/m3]Densities Constant k0=8.OE+17 :[m3/kmol h1Reaction constant Constant H=30000 : [kJ/kmoll Reaction heat : [KlE/R in Arrhenius rel. Constant ER=11000 Constant Fbmax=2 :[m3/h]Maximum feed flow rate Constant Fbmin=O.Ol : [m3/hl Minimum feed flow rate Constant Tset=25 : [C]Setpoint temp. Constant CINT=O.Ol, TFIN=6 1 SIM; INTERACT; RESET; GOT0 1 INITIAL TC=TCin TR=TRO Ca=CAO Cb=CbO cc=cco VCa=CaO*VO VCb=CbO*VO vcc=cco*vo

v=vo

5 22

_____________

5 Simulation Examples of Chemical Engineering Processes __.__

DYNAMIC TRabs=TR+273 V' =Fb :TOTAL BALANCE A=AO+(V/C) :VARIABLE TRANSFER AREA k=kO*EXP(-ER/TRabs) :ARRHENIUS EQ. :MASS BALANCES VCa'=-k*Ca*Cb*V VCbI=Fb*Cbf-k*V*Ca*Cb VCc'=k*Ca*Cb*V Ca=VCa/V Cb=VCb/V cc=vcc/v :ENERGY BALANCES Q = U*A*(TR-TC)/(Rho*Cpr) TR'=(Fb*(Tf-TR)-Q+H*k*Ca*Cb*V/(Rho*Cpr))/V TC1=(Fc/Vc)*(TCin-TC)+U*A*(TR-TC)/(Rho*Cpc*Vc) :FEED FLOW RATE CONTROL Err=TR-Tset Fb=FbO-Kp*Err-(Kp/TI)*intg(Err,O) if (Ca.le.0) Ca=O if (Cb.le.0) Cb=O if (Cc.le.0 ) Cc=O if (Fb.le.0) Fb=Fbmin if (Fb.gt.Fbmax) Fb=Fbmax PLOT T,Ca,T,TFIN,O,CaO OUTPUT T,CA,CB,FB,TR PREPARE T,Ca,Cb,Cc, Fb,TR,TC,V

Nomenclature Symbols A0 CA

CBf cP cPC

D ER

FBO

Fc H h

Initial heat transfer area Concentration of A in tank Concentration of B fed into tank Heat capacity of reactor and feed Heat capacity of cooling water Diameter Activation energy / gas constant Feed rate basis Cooling water flow Heat of reaction Level of reactor liquid

m2 kmol/m3 kmol/m3 kJkg K kJ/kg K m K m3/h m3/h kJ/kmol

m

523

5.7 Process Control Examples

Reaction rate constant Frequency factor Proportional control constant value of n: Density of all streams Initial temp. in tank Absolute reactor temperature Cooling water temp. Feed temperature Integral control constant Reactor temperature Heat transfer coefficient Initial tank volume Volume of cooling jacket

m3/kmol h m3/mol h m3/h K

-

kg/m3 "C K "C "C h "C kW/m2 K m3 m3

Indices Refers to feed values Refers to initial values Refers to cooling water Refers to cooling jacket Refers to cooling water inlet Refers to absolute temperature

f I C out in ab

Exercises 1.

Operate the reactor without control using a constant flow of reactant B. Note the reactor temperature.

2.

Choose a suitable temperature setpoint and simulate the reactor with control, first with proportional control only and then including integral control. Adjust the controller constants to obtain adequate control.

3.

Using the controller constant found in Exercise 2, make a step decrease in the cooling water flow rate. Did the controller decrease the reactant flowrate to compensate for this change?

4.

Modify the program and change the controller to manipulate the cooling water flow rate. Experiment with this program.

5 24

5 Simulation Examples of Chemical Engineering Processes ~~

~~

Results

Figure 5.167. The feed rate of B, FB, was manipulated by the control equation to control the reactor temperature. At the end of the batch, when the reaction rate decreased due to decreasing CA, FB increased gradually to its maximum.

5.7.5

Figure 5.168. During the run of Fig. 5.167, the temperature, TR, at first increased to its setpoint and then decreased at the end of the run. Liquid volume, V , increased throughout the run.

TRANSIM - Transfer Function Simulation

System This is based on the example of Matko, Korba and Zupancic (l), who describe methods of simulating process transform functions, based on partitioned and nested forms of solution. Here the process transfer function is given by

The partitioned method of solution as described in Sec. 2.1.4 is used as the basis of solution.

525

5.7 Process Control Examples

Model The overall transfer unit G(s) may be written as

(3s) = Gl(S) G2(s) G3(S)

.

where the transfer functions relate the outlet response Y(s) to the inlet , below. disturbance U(s), via the dummy variables Wl(s) and W ~ ( S )shown

U l s ) t / F /

q-q t/--q

W+)

W,(S)

Y(S)

Figure 5.169. Transfer function consisting of three transfer functions in series.

Here

G3(s) =

Hence

Y(S>

1

w2(s) - 1 + 10s

s W I = 0.25 U - 0.25 W1 w2 = w1- 4 s w 1 SY

= Of.1 w2- 0.1 Y

Note that the above formulation avoids the need to know the rate of change of the forcing disturbance, U. Defining the variables Y, W1 and W2 as deviations from an initial steady state, the original transfer function can now be expressed as ~dwl

dt

- 0.25 U - 0.25 W1

5 Simulation Examples of Chemical Engineering Processes

526

ddty = 0.1 w2 - 0.1 y

An example of the use of an ISIM sub-model representing a complex pole transfer function is also given in the ISIM manual (obtainable separately).

Program Note that the derivative term in the expression for W2 can be programmed directly into the ISIM program as a variable. : :

Example TRANSIM Transfer Function

constant 1

Sim;

tfin=50, Reset;

Simulation

cint=0.2,

Interact;

GOT0

noci=lO 1

Dynamic : sinusoidal disturbance u=sin(=t) : dummy variables wl'=0.25 * u-0.25 * wl w2=wl-4*wl' : function output y'=O.l * w2-0.1 * y plot t,y,O,tfin,-0.5,0.5 output t,u,y prepare t ,u, y

Exercises 1.

Note the initial negative deviation in response to a positive forcing disturbance, caused by the right half plane zero in the transfer function.

2.

Vary the magnitude of the frequency in the forcing disturbance and study its effect on the phase angle and amplitude ratio of the system. Use the information to construct the Bode diagram for the system, i.e., plots of phase angle and amplitude ratio versus frequency.

3.

Vary the numerical parameters and the form of the transfer function.

5.7 Process Control Examples

5 27

Results

R U B Y

Figure 5.170. Transfer function input and output

References 1. Matko, D., Korba, R. and Zupancic, B. (1992)."Simulation and Modelling of Continuous Systems: A Case Study Approach", Prentice-Hall. 2. ISIM Simulation Manual, KIM International Simulation Ltd.

5.8

Mass Transfer Process Examples

5.8.1

BATEX - Single Solute Batch Extraction

System Consider a batch two-phase extraction system, with a single solute.

528

5 Simulation Examples of Chemical Engineering Processes -

Figure 5.171.

~

~~

~

~~

Batch liquid-liquid extraction

Model As described in Sec. 3.3.1, separate component balances on each phase are expressed by

of accumulation of Effective rate of ( Ratesolute in one phase ) = ( mass transfer Thus

where the constant K contains the mass transfer coefficient, the specific interfacial area, and the total liquid volume. In this case, the equilibrium is expressed by a simple linear relation

Program :BATEX CONSTANT CONSTANT

SIMPLE

BATCH

EXTRACTION

CINT=0.5, TFIN=10 Vl=lO,V2=40 :BATCH

VOLUMES

529

5.8 Mass Transfer Process Examples

.

CONSTANT K=2.5 :MASS TRANSFER COEFF CONSTANT M=O .8 :EQUILIBRIUM C O N S T A N T 1 S1M;INTERACT;RESET;GOTO 1 INITIAL : INITIAL CONCENTRATIONS Xl=l; xo=1; Yl=O; YO=O DYNAMIC : COMPONENT MASS BALANCES Xl'=-K*(Xl-Yl/M)/Vl Yl'=K*(Xl-Yl/M)/V2 : FRACTIONAL EXTRACTION z1=(xo-x1)/xo P L O T T, X1,0, TFIN, 0,1 O U T P U T T IX1, Y 1, Z 1 P R E P A R E T, X1, Y1,Zl

Nomenclature Symbols K M V X1 Y1

Rate constant Equilibrium constant Volume Concentration in X-phase Concentration in Y-phase Fractional extraction

Z

m3/h

-

m3 kg/m3 kg/m3

Indices

*

Refers to equilibrium Refers to X and Y phases

132

Exercises 1.

Observe the dynamic approach to equilibrium, and note that the driving force for transfer changes.

2.

Change K over a range of values and observe how this changes the time to reach equilibrium. Is there a relationship between K and the system time constant Z ?

530

5 Simulation Examples of Chemical Engineering Processes

3.

Vary the equilibrium constant M and study its effect on the resulting dynamic behaviour.

4.

Verify that the total mass of solute is conserved.

5.

Experiment with the influence of V 1, V2 and M.

6.

Simulate the effect of a process operator adding an additional charge of solvent part way through the extraction. Note that you now need to allow for both total mass balance and component mass balance changes. Use the interactive facility of ISIM, to effect the changes.

Results ISIM

ISIH

8.2501

_---

Figure 5.172. The addition of solute is simulated by changing XI suddenly to 0.75.

5.8.2

Figure 5.173. Solvent was added suddenly by doubling V2 and reducing Y l by one-half.

TWOEX - Two-Solute Batch Extraction with Interacting Equilibria

System This example considers the interactions involved in multicomponent extraction as shown in Sec. 3.3.1.2 and takes the particular case of a single batch extractor with two interacting solutes.

531

5.8 Mass Transfer Process Examples

v1, X I , w1

Q1 Q2

Model Two solutes distribute between the phases as concentrations X I and Y1, and W1 and Z1, respectively. The respective equilibrium concentrations El and E2 are each functions of Y 1 and Z I E l = 1.25 Y1 E2 = Z1

-

0.2 Z1

+ 0.5 Yl2

It is necessary to write the dynamic equations for each solute in each phase. For phase X

For phase Y

As in example BATEX, the constants Kl and K2 contain the mass transfer coefficient, the specific interfacial area and the total liquid volume.

Program :Example TWOEX COMPLEX TWO-SOLUTE BATCH EXTRACTION WITH INTERACTING EQUILIBRIA

: :

532

5 Simulation Examples of Chemical Engineering Processes

: BATCH VOLUMES AND SOLUTE MASS TRANSFER COEFFS CONSTANT V1=10, V2=40, K1=2.5, K2=2 CONSTANT TFIN=15, CINT=0.5 1 SIM; INTERACT; RESET; GOT01 INITIAL : INITIAL CONCENTRATIONS Xl=l; xo=1; Y1=0; YO=O w1=.75; wo=.75: z1=.1; zo=.1 DYNAMIC : COMPONENT MASS BALANCES Xl'=-Kl*(Xl-El)/Vl Yl8=Kl*(X1-E1)/V2 Wl'=-K2*(Wl-E2)/Vl Zl1=K2"(W1-E2)/V2 : EQUILIBRIUM RELATIONS : El IS THE EQUILIBRIUM CONCOF X1 (X1*) : E2 IS THE EQUILIBRIUM CONC OF W1 (W1*) E1=1.25*Y1-0.2*Zl E2=Z1+.5*Yl*Y1 : FRACTIONAL SOLUTE EXTRACTION x = ( X O - X 1 )/xo w=(wo-wl)/wo P L O T T IX1,0, TFIN, 0,l OUTPUT T, XI, Y1, W1,Zl PREPARE T IXI, Y 1,W1, Z 1, X IW

Nomenclature Symbols

Y1 Z1

E2

Volume of solute phase Volume of solvent phase Concentration of solute 1 in the solute phase Concentration of solute 2 in the solute phase Concentration of solute 1 in the solvent phase Concentration of solute 2 in the solvent phase Equilibrium concentration corresponding to solute 1 Equilibrium concentration corresponding to solute 2

m3

m3

kg/m3 kg/m3 kg/m3

kg1m3 kg1m3 kgfm3

533

5.8 Mass Transfer Process Examples

K1 K2

Mass transfer coefficient for solute 1 Mass transfer coefficient for solute 2

m3/h m3/h

Exercises 1.

Observe the influence of the magnitudes of the relative mass transfer coefficients K1 and K2 on the dynamics of the approach to equilibrium.

2.

Confirm that the total balances for both solutes are correct at equilibrium.

3.

Overall mass transfer coefficients are only constant when both liquid film coefficients are constant and also when the slope of the equilibrium line is constant. Thus, for a non-linear equilibrium relationship, the overall mass transfer coefficient will vary with concentration. How would you implement this effect into the program?

Results The results give system response to parametric and concentration changes.

A x1

A x1

D Y1

8.1981

n. 988

B c

1.888

‘h

ISIM

Y1 w1 0 I1

\

i\

0.298

D . 8.8888.8

’

4.0

T

0 .0

12.8

16.8

Figure 5.175. Changing K1 from 0.25 (dark curve) to 10.0 (light curve) resulted in faster attainment of equilibrium.

8.8888.8

4.8

r

8.0

lz.n

16.0

Figure 5.176. A sudden increase in concentration of the second solute concentration in phase 2 (Z, from 0.2 to 0.5) caused a shift in equilibrium.

534

5 Simulation Examples of' Chemical Engineering Processe5 ~-

5.8.3

-

KLADYN

~

~

~

- Dynamic

~

.~

~~

-~

Method for KLa

System A simple and effective means of measuring the oxygen transfer coefficient (KLa) in an air water contacting system involves degassing the well-mixed water system with nitrogen, starting the air flow at time t = 0, and measuring the increasing dissolved oxygen concentration by means of an oxygen electrode.

Figure 5.177. Aerated tank with oxygen electrode.

As shown below, the influence of three quite distinct dynamic processes play a role in the overall measured oxygen concentration response curve. These are the processes of the dilution of nitrogen gas with air, the gas-liquid transfer, and the electrode response characteristic, respectively. Whether all of these processes need to be taken into account when calculating KLa can be seen by examining the mathematical model and making simulations.

Gas phase

Liquid phase

I I

Electrode diffusion film

Electrode

' - - - - - - - _ - - - - - - - - - J

Figure 5.178. Representation of the process dynamics

1

535

5.8 Mass Transfer Process Examples

Model Model relationships include the mass balance equations for the gas and liquid phases and equations representing the measurement dynamics.

Oxygen Balances The oxygen balance for the well-mixed flowing gas phase is described by

where VG/V = ZG , and KLa is based on the liquid volume. The oxygen balance for the well-mixed batch liquid phase, is

The equilibrium oxygen concentration CL* is given by the combination of Henry's law and the Ideal Gas Law equation where

and CL* is the oxygen concentration in equilibrium with the gas concentration, CG. Defining dimensionless variables as

The component balance equations then become CG'

dt' =

(1 - CG') - KLa ZG

VL RT -- (CG' - CL') VG H

and

Initial conditions corresponding to the experimental method are, t' = 0 ;

CLI = CG' = 0

536

5 Simulation Examples of Chemical Engineering Processes

Oxygen Electrode Dynamic Model A semi-empirical, second-order response lag is used. This consists of a firstorder lag equation representing the diffusion of oxygen through the liquid film on the surface of the electrode membrane

and a first-order electrode response relation

TF and TG are the time constants for the film and electrode lags, respectively. In dimensionless form, the equations are

and

where CF' is the dimensionless diffusion film concentration

and CE is the dimensionless electrode output

As shown by Dang et al. (1977), solving the model equations by Laplace transformation gives

where a is the area above the CE' versus t response curve, as shown in Fig. 5.179.

537

5.8 Mass Transfer Process Examples

Time (s) Figure 5.179. Determination of a from the CE' versus time curve.

Program For simplicity the prime (') is dropped in the dimensionless equations. :

Example

: : :

DYNAMIC KLA-MEASUREMENT GAS PHASE, LIQUID PHASE IN DIMENSIONLESS FORM

KLADYN AND

ELECTRODE

CONSTANT KLA=500, G=3000, VL=45, V G = 5 CONSTANT RTH=0.03, TE=2.78E-3, TF=1.1E-3 CONSTANT CINT=O.1, TFIN=20.0 1 SIM; INTERACT; RESET; GOT0 1 INITIAL CG=O CL=O CE=O CF=O TH=T*TG TG=VG/G DYNAMIC :MASS BALANCES CG'=(l-CG)-KLA*TG*(VL/VG)*RTH*(CG-CL) CL'=XLA*TG*(CG-CL) :ELECTRODE DYNAMICS

DYNAMICS

538

5 Simulation Examples of Chemical Engineering Processes

CF'=(CL-CF)/(TF/TG) CE'=(CF-CE)/(TE/TG) :AREA UNDER 1-CE VS.TIME CURVE ALPHA=TG*INTG(l-CE,O) :ALPHA IS COMPARED TO VALUE OF AREA(SEE AREA=(l/KLA)+(RTH*VL/VG+l)*TG+TE+TF PLOT T, CE, 0, TFIN, 0 , l PREPARE T, CG, CL, CF, C E OUTPUT T,CG,CL,CF,CE

TEXT)

Nomenclature Symbols C G H KLa RT/H

Oxygen concentration Gas flow rate Henry coefficient Oxygen transfer coefficient Gas constant - absolute temperature/Henry coefficient time Reactor volume Area above CEl-time (h) curve Time constant

t V

a

z

g/m3 m3/h Pa m3h01 l/h -

h m3 h h

Indices E F

Refers to electrode Refers to film Refers to gas phase Refers to liquid Prime denotes dimensionless variables

G L ,

Exercises 1.

Compare the values of the time constants (zc, l/KLa, TF and ZE) and the dimensionless groups. For the parameter values used, can you estimate whether all processes will be influential? How would the situation be changed for G = 30 m3/h and T E = ZF = Is? Check your reasoning with simulations, observing the plots of CG, CL, CF and CE versus time.

539

5.8 Mass Transfer Process Examples

2. Verify by simulation that KLa can be estimated from a. 3.

The textbook method for calculating KLa proposes plotting In( l/l-CL') versus t to obtain KLa from the slope. What is the reasoning behind this, and when would it yield accurate KLa results? Change the program to give this plot, and investigate its validity for a range of parameters.

4.

Vary the time constants for transfer (l/KLa) and measurement (ZE + ZF) to demonstrate the response with electrodes that are very fast and too slow.

6.

Using the simplification of Exercise 5 , estimate the time constants and show how their sum can be estimated from the CG' = CL' = CE' = 0.63 position on the curves?

7.

Design, model and simulate a steady-state experiment for KLa determination.

8.

Alter the program to investigate the response of the electrode to a step change in CL.

Results

7

16.8

Figure 5.180. The profiles display all the oxygen concentration variables; gas, liquid, film and electrode from left to right.

Figure 5.181. The profiles are shown for a lower gas rate (G = lOOO), a lower transfer rate (KLa = 200), a small film constant (ZF = 0.0001) and a faster electrode (zE = 0.001).

540

5 Simulation Examples of Chemical Engineering Processes

Reference Dang, N.D.P., Karrer, D.A. and Dunn, I.J. (1977). Oxygen Transfer Coefficients by Dynamic Model Moment Analysis, Biotechnol. Bioeng. 19, 853.

5.8.4

EQEX - Simple Equilibrium Stage Extractor

System The extraction is represented by a single perfectly mixed constant volume, continuous flow equilibrium stage. This may actually consist of separate mixer-settler units.

Figure 5.182. Single continuous extraction stage.

Model The separate phase balances form the dynamic model.

(L XO- L X I - Q) dX1 ~dt Vl _dY1 _ - _ _ _ _(C _ _ YO-

dt -

where

G Y1 v2

+ Q)

Q = K (XI - X I * )

54 1

5.8 Mass Transfer Process Examples

and

X i * = Y1/M

As shown in Sec. 3.3.1.3, the equilibrium condition is satisfied by the use of an arbitrarily high value for K to force conditions towards equilibrium.

Program : :

EXAMPLE EQEX SINGLE CONTINUOUS

:

SIMPLE

STIRRED

EXTRACTION

TANK

STAGE

MODEL

CONSTANT V1=10, V2=10 :PHASE V O L U M E S CONSTANT XO=l,YO=O,L=l,G=3 :INLET CONC. A N D FLOWS CONSTANT M = . 8 :EQUILIBRIUM CONSTANT CONSTANT K = 5 0 :MASS TRANSFER COEFF SET HIGH FOR EQUIL. CONSTANT TFIN=30 SIM INITIAL : INITIAL CONCENTRATIONS x1=0; Y 1 = 0 DYNAMIC Ql=K*(Xl-Yl/M) : MASS TRANSFER RATE E Q N : COMPONENT BALANCE EQUATIONS Xl'=(L*(XO-Xl)-Ql)/Vl Yl'=(G*(YO-Yl)+Ql)/VZ : FRACTIONAL EXTRACTION x = ( X O - X 1 )/xo P L O T T, XI, 0, TFIN, 0, XO O U T P U T T, XI, Y1, Q1, X P R E P A R E T IXI, Y 1, X, Q 1

Nomenclature

Symbols G K

Flow rate Rate constant

m3/h m3/h

542

5 Simulation Examples of Chemical Engineering Processes

L M

Flow rate Equilibrium constant Transfer rate Volume Fractional extraction Concentration in X-phase Concentration in Y-phase

Q

V X X1 YI

m3/h

-

kglh m3

-

kglm’ kg/m3

Indices

*

Refers to equilibrium Refers to inlet Refer to X and Y phases

0

1.2

Exercises 1.

Observe how the approach to equilibrium is dependent on the magnitude of K for given values of flow rates. Check the concentrations for equ ili bri urn.

2.

Compare the residence time for the X phase, V,/L, with the time constant for transfer, VlIK. Maintain a constant ratio of time constants by varying the individual parameters. How is the approach to equilibrium influenced?

543

5.8 Mass Transfer Process Examples

Results

5 '

6.BBBB.B

I.5

T

1s.n

n.888n.n 22.5

3.8

Figure 5.183. Approach to steady state, showing how equilibrium is maintained.

5.8.5

1.5

T

22.5

Figure 5.184. Response of the equilibrium stage to a change in flow (L from 1.0 to 1.5).

COLCON - Extraction Cascade with Backmixing and Control

System A five-stage extraction column with control on the outlet raffinate phase is to be simulated. The solute balance equations for each phase are formulated according to Sec. 3.3.1.5.

544

5 Simulation Examples of Chemical Engineering Processes

L, x5

y6 I I

---

I I

@--*

Figure 5.185. Five stage extraction column with control.

Model Allowing for the backmixing flow contributions in the inflow and outflow terms, the continuity equations for each phase, in stage n, are given by

The transfer rates, Qn, are given by the equation

Proportional plus integral control manipulates the solvent flow rate based on the exit solute concentration in the raffinate. Thus G is a function of GO, KC,

545

5.8 Mass Transfer Process Examples

TI, XSET, and X5 according to the controller equation, described in Sec. 2.3.2.2.

Program The simulation starts with the extractor operating at steady-state conditions, but with a relatively high outlet raffinate concentration. Control is implemented in order to reduce the raffinate concentration in accordance with a lower controller set point. :EXAMPLE COLCON :NON-EQUIL. EXTR. : :

COLUMN

WITH

BACKMIXING

AND

CONTROL

THIS PROGRAM STARTS FROM STEADY-STATE CONDITIONS CONTROL CAN BE STARTED BY SPECIFYING KC (KC=5)

CONSTANT V1=2, V2=2 :PHASE VOLUMES IN EACH STAGE CONSTANT L=l, G=3 :VOLUMETRIC FLOW RATES CONSTANT XO=l,Y6=0 :INLET CONCENTRATIONS CONSTANT M=O. 8, K=O. 8 :EQUIL AND MASS TRANSFER CONSTS CONSTANT E1=0.5,E2=0.5 :BACKMIXING FACTORS CONSTANT KC=O :PROPORTIONAL CONTROL CONSTANT CONSTANT TI=.2 :INTEGRAL TIME CONSTANT CONSTANT GO=3 :BASE SOLVENT FLOW RATE CONSTANT XSET=.10 :RAFFINATE SET POINT CONSTANT CINT=.5, TFIN=10 SIM INITIAL :INITIAL STEADY-STATE CONCENTRATION PROFILES X1=.6642;X2=.5009;X3=.3711;X4=.2513;X5=.1892 Yl=.2773;Y2=.2208;Y3=.162O;Y4=.1125;Y5=.0582 : CONSTANT FACTORS IN DYNAMIC EQUATIONS LB=El*L L2=L+LB DYNAMIC :NON-CONSTANT FLOW FACTORS GB=G*E2 G2=G+GB : COMPONENT BALANCES FOR STAGE 1 Xl'=(L*XO+LB*X2-L2*Xl-Ql)/Vl Y11=(G2*(Y2-Y1)+Q1)/V2

546

5 Simulation Examples of Chemical Engineering Processes _.

~~

: BALANCES FOR STAGES 2 T O 4 X2'=(L2*(Xl-X2)+LB*(X3-X2)-Q2)/Vl X3'=(L2*(X2-X3)+LB*(X4-X3)-Q3)/Vl X4'=(L2*(X3-X4)+LB*(X5-X3)-Q4)/Vl Y2'=(G2*(Y3-Y2)+GB*(Yl-Y2)+Q2)/V2 Y3'=(G2*(Y4-Y3)+GB*(Y2-Y3)+Q3)/V2 Y4'=(G2*(Y5-Y4)+GB*(Y3-Y5)+Q4)/V2 : COMPONENT BALANCES FOR STAGE 5 X5'=(L2*(X4-X5)-Q5)/Vl Y5'=(G*Y6+GB*Y4-G2*Y5+Q5)/V2 : MASS TRANSFER RATE EQUATIONS Ql=K*(XI-Yl/M) Q2=K*(X2-Y2/M) Q3=K*(X3-Y3/M) Q4=K*(X4-Y4/M) Q5=K* (X5-Y5/M) : CONTROLLER EQUATIONS E=X5-XSET :CONTROLLER ERROR G=G~+KC*(E+EINT/TI) EINT' =E :I N T E G R A L ERROR : FRACTIONAL EXTRACTION x=(xo-x5)/xo O U T P U T T, XI, X 2 X3, X 4 X 5 PLOT TfX5,0,TFINfOfO.S P R E P A R E T IX 1 X 2 X 3 X4, X5, G

Nomenclature

Symbols G K L M

Q

V X X1 Y1

Flow rate Rate constant Flow rate Equilibrium constant Transfer rate Volume Fractional extraction Concentration in X-phase Concentration in Y-phase

m3/h m3/h m3/h -

kglh m3

kg/m3 kg/m3

547

5.8 Mass Transfer Process Examples

Indices

*

Refers to equilibrium Refers to inlet Refer to X and Y phases Refers to backmixing Refers to stage

0 132

B n

Exercises 1.

Experiment with the response of the column to changes in the operating variables in the absence of control (KC = 0). Note the response times to reach steady state and the fulfilment of the steady-state balance for the solute.

2.

Add control (KC positive) and experiment first with proportional control (TI very large) to measure the offset. Add integral control and see the effect.

3.

Try tuning the controller to obtain the best response to a change in feed concentration and feed rate.

Results

ISIH

-_--

I B . m ; d / / r -

8.48

I

i

8.W8.W

2.50

I

5.m

1.50

I.88

Figure 5.186. The response of the raffinate concentrations to an increase in L from 1 to 2 are shown.

n

8.08.88

2.50 '

1.'50

1B.W

Figure 5.187. Turning the controller on (KC = 5 ) causes G to be changed until XI reaches the setpoint.

548

5.8.6

5 Simulation Examples of Chemical Engineering Processes

EQMULTI - Continuous Equilibrium Multistage Extraction

System A countercurrent multistage extraction system is shown below, which is to be modelled as a cascade of equilibrium stages.

Figure 5.188. Countercurrent multistage equilibrium extraction unit.

Model As explained in Sec. 3.3.1.4, the unsteady-state solute balance equations for each phase are defined for any stage n by

where Qn = K (Xn - Xn")

and Xn" = Yn/M

As in example EQEX, the constant K is set sufficiently large to obtain equilibrium, for the compositions, X, and Y n , of the streams leaving each stage.

549

5.8 Mass Transfer Process Examples

Program : : :

EXAMPLE EQMULTI F I V E STAGE COUNTERCURRENT EQUILIBRIUM STAGE MODEL

EXTRACTION

CASCADE

CONSTANT V1=2, V 2 = 2 : PHASE VOLUMES I N EACH STAGE CONSTANT L=l, G=3 : VOLUMETRIC FLOW RATES CONSTANT XO=l,Y6=0 : INLET CONCENTRATIONS CONSTANT M=0.8, K = 5 0 : EQUIL A N D MASS TRANSFER CONSTS C O N S T A N T CINT=.5, TFIN=10 SIM INITIAL : INITIAL CONCS x1=0;x2=0;x3=0;x4=0;x5=0 Y1=0;Y2=0;Y3=0;Y4=0;Y5=0 DYNAMIC : COMPONENT BALANCES FOR X- PHASE Xl'=(L*(XO-Xl)-Ql)/Vl X2'=(L*(Xl-X2)-Q2)/Vl X3'=(L*(XZ-X3)-Q3)/Vl X4'=(L*(X3-X4)-Q4)/Vl X5'=(L*(X4-X5)-QS)/Vl : COMPONENT BALANCES F O R Y-PHASE Yl'=(G*(Y2-Yl)+Ql)/V2 Y2'=(G*(Y3-Y2)+Q2)/V2 Y3'=(G*(Y4-Y3)+Q3)/V2 Y4'=(G*(Y5-Y4)+Q4)/V2 YS'=(G*(YS-Y5)+Q5)/V2 : MASS TRANSFER RATE EQUATIONS Ql=K*(XI-Yl/M) Q2=K* (X2-Y2/M) Q3=K* (X3-Y3/M) Q4=K*(X4-Y4/M) QS=K*(XS-YS/M) : FRACTIONAL EXTRACTION X=(XO-XS)/XO O U T P U T T, X1, X 2 I X3, X4 , X 5 P L O T T,Xl,O,TFIN, 0,XO PREPARE T,X1,X2,X3,X4,X5,Y1,Y21Y31Y4~Y51X

550

5 Simulation Examples of Chemical Engineering Processes

Nomenclature Symbols G K L M

Flow rate Rate constant Flow rate Equilibrium constant Transfer rate Volume Fractional extraction Concentration in X-phase Concentration in Y-phase

Q

V X XI YI

m3/h m3/h m31h kglh m3

kg/m3 kg/m3

Indices

*

Refers to equilibrium Refers to inlet Refer to X and Y phases Refers to stage

0 I ,2 n

Exercises 1.

Run the model and ascertain if equilibrium is attained for each stage. Reduce K by a factor of 10 and check again. Reduce L and G by the same factor. Explain the effects in terms of time constants.

2.

Confirm that the overall solute balance is satisfied at steady state.

3.

Modify the above program to model the case of a discontinuous five stage extraction cascade, in which a continuous flow of aqueous phase L is passed through the cascade. The solvent is continuously recycled through the cascade and also through a solvent holding tank, of volume Vs, as shown below. This problem of a stagewise discontinuous extraction process has been solved anlytically by Lelli (1966).

55 1

5.8 Mass Transfer Process Examples

Solvent tank VS

A L, XO

G, Y1

4.

Try to predict the form of the X and Y phase responses before carrying out the simulations.

5.

Vary L and G and observe the influence on the dynamics. Explain the results in terms of the system time constants.

Results Shown below are responses of the extraction system to changes in flow rates.

552

5 Simulation Examples of Chemical Engineering Processes

A X1 0 X?

0 xi

I: Y B

F X5

1

18.88

Figure 5.190. Increasing L from 1 to 10 causes a rapid response, but the short residence times lead to lower degrees of extraction.

Figure 5.191. Increasing both flow rates by a factor of 10 results in much lower fractional extraction values for these equilibrium stages.

References Lelli, U. (1966) Annali di Chimica. 56, 113

5.8.7

EQBACK - Multistage Extractor with Backmixing

System This model is described in Sec. 3.3.1.5 and consists of a countercurrent stagewise extractor cascade with backmixing in both phases.

553

5.8 Mass Transfer Process Examples .-

1

-

2

3

-

-

5

-

Backmixing Flow GR

1

Forward Flow G + G B

Figure 5.192. Multistage extractor with backmixing.

Model Allowing for the backmixing flow contributions in the inflow and outflow terms, the continuity equations for each phase in stage n are given by

where

Qn = K (Xn - Xn*) and Xn* = Y n I M

Special balances for the end sections must be written to account for the absence of backflow.

Program :

Example

: : :

Five stage countercurrent cascade with backmixing Equilibrium stage model

EQBACK

extraction

554

5 Simulation Examples of Chemical Engineering Processes

CONSTANT VL=2 :Volume of extr. phase in one stage CONSTANT VG=2 :Volume of solvent phase in one step CONSTANT L=l :Flow rate of extr. phase CONSTANT G=3 :Flow rate of solvent phase CONSTANT EL=5 :Backmixing factor of extr. phase CONSTANT EG=5 :Backmixing factor of solvent phase CONSTANT XO=1 :Inlet conc. of extraction phase CONSTANT Y6=0 :Inlet conc. of solvent phase CONSTANT K=50 :Mass transfer capacity coeff. : (arbitrarily high for equilibrium stage) CONSTANT M = 0.8 :Distribution coefficient M=Y/X CONSTANT CINT=0.2 CONSTANT TFIN=10 :

: :

Xi=Extraction phase concentration at stage i Yi=Concentration in solvent at stage i Qi=Mass flow from extraction phase to solvent

SIM INITIAL

x1=0;x2=0;x3=0;x4=0;x5=0 Y1=0;Y2=0;Y3=0;Y4=0;Y5=0

DYNAMIC Xl'=(L*(XO-Xl + X2'=(L*(Xl-X2 + X3'=(L*(X2-X3 + X4'=(L*(X3-X4 + X5'=(L*(X4-X5 + Y11=(G*(Y2-Y1 + Y2'=(G*(Y3-Y2 + Y3'=(G*(Y4-Y3 + Y4 ' = (G* (Y5-Y4 + Y5'=(G*(Y6-Y5 + Ql=K*(Xl-Yl/M) Q2=K*(X2-Y2/M) Q3=K*(X3-Y3/M) Q4=K*(X4-Y4/M) Q5=K*(X5-Y5/M)

EL*(X2-X1)) - Ql)/VL EL*(Xl+X3-2*X2)) - Q2)/VL EL*(X2+X4-2*X3)) - Q3)/VL EL*(X3+X5-2*X4)) - Q4)/VL EL*(X4-X5)) - Q5)/VL EG*(Y2-Y1)) + Ql)/VG EG*(Yl+Y3-2*Y2)) + Q2)/VG EG*(Y2+Y4-2*Y3)) + Q3)/VG EG* (Y3+Y5-2*Y4)) + 44)/VG EG*(Y4-Y5)) + Q5)/VG

555

5.8 Mass Transfer Process Examples

PLOT T,Yl,O,TFIN, 0 , 1 O U T PUT T,Xl,X2,X3,X4,X5,Y1,Y2,Y3,Y4 PREPARE

T,X1,X2,X3,X4,X5,Y1,Y2,Y3,Y4,Y5,Q1,Q2,Q2Q3,Q4,Q5

Nomenclature Symbols Flow rate Rate constant Flow rate Equilibrium constant Transfer rate Volume Fractional extraction Concentration in X-phase Concentration in Y-phase

G K L M

Q

V X XI Y1

m3/h m3/h m3/h -

kglh m3 kg/m3 kg/m3

Indices

*

Refers to equilibrium Refers to inlet Refer to X and Y phases Refers to backmixing Refers to stage

0 132 B n

Exercises 1.

Vary the backmixing flows for each phase separately and notice the effect of changes in conditions between plug flow and full backmixing on the performance of the extractor.

2.

For which conditions of backmixing does the extraction process operate best?

3.

Check the steady-state overall solute balances.

556

8.19

5 Simulation Examples of Chemical Engineering Processes ~

Results

A X1 B x2 c x3 0 x4 8.225 E X5

8.875

0.8888.88

Figure 5.193. The response of the cascade for a ten-fold increase in the backmixing flows EL and EG.

5.8.8

2.88

1

4.88

6.88

7

8.88

Figure 5.194. A change to plug-flow (EL = 0) of the extract phase gives more complete extraction.

HOLDUP - Transient Holdup Profiles in an Agitated Extractor

System As explained in Sec. 3. 3. 1.11, the fractional holdup of the dispersed phase in agitated extraction columns will vary with changing phase flow rate. This dynamic variation in the holdup along the column will cause a dynamic velocity profile in both phases.

557

5.8 Mass Transfer Process Examples

Figure 5.195. Flows of a single stage agitated extraction column.

Model A dynamic balance for the dispersed phase holdup in stage n gives

The above equation can be divided by cross-sectional area to give the balance in terms of superficial velocities. Since the liquid phases are incompressible Ln-1

+ Gn+l

= Ln

+ Gn

and for the overall column Lo+G8 = L ~ + G I

As explained in Sec. 3.3.1.11, in order to avoid difficulties in the IMPLICIT LOOP solution of the characteristic velocity equation for dispersed phase

558

5 Simulation Examples of Chemical Engineering Processes -. .

-

..

~~

holdup, the holdup relationship is assumed to be given by a direct, explicit, empirical function of the continuous and dispersed phase flow rates. For a given range of L, it is assumed that this relationship takes the form

where n and m are empirical constants.

Program In the program the flow rates are taken as the superficial velocities. :Example HOLDUP :TRANSIENT HOLDUP

IN

AN

AGITATED

COLUMN

EXTRACTOR

CONSTANT L0=13.05 : INLET AQUEOUS FLOW RATE : INLET ORGANIC FLOW RATE CONSTANT GO=10.5 CONSTANT Z0=20, Z8=20 : HEIGHTS OF END SECTIONS CONSTANT Z=8.5 : HEIGHT OF COMPARTMENT CONSTANT M=.7605 : CONSTANT IN HOLDUP RELATION CONSTANT N=2.805E-2 : CONSTANT IN HOLDUP RELATION CONSTANT H=0.2 : INITIAL FRACTIONAL HOLDUP CONSTANT TFIN=10, CINT=0.25 SIM INITIAL :INITIAL HOLDUP AND FLOW RATE PROFILES Hl=H;H2=H;H3=H;H4=H;H5=H;H6=H;H7=H L1=LO;L2=LO;L3=LO;L4=LO;L5=LO;L6=O;L7=LO G1=GO;G2=GO;G3=GO;G4=GO;G5=GO;G6=GO;G7=GO DYNAMIC :HOLDUP RELATIONS Gl=Hl**M/N G2=H2**M/N G3=H3**M/N G4=H4**M/N G5=H5**M/N G6=H6**M/N G7=H7**M/N G8=GO :LIQUID PHASE BALANCES Ll=LO+GZ-Gl L2=Ll+G3-G2

559

5.8 Mass Transfer Process Examples

L3=L2+G4-G3 L4=L3+G5-G4 L5=L4+G6-G5 L6=L5+G7-G6 L7=L6+G8-G7 :HOLDUP BALANCES Hl'=(GZ-Gl)/ZO H2'=(G3-G2)/Z H3'=(G4-G3) / Z H4'=(G5-G4)/Z H5'=(G6-G5)/Z H 6 ' = ( G 7 - G 6 )/ Z H 7 ' = ( G 8 - G 7) / Z 8 PLOT T, H4,0, TFIN, 0,O. 5 T, H1, H2,H3, H4, H5, H6, H7 PREPARE OUTPUT T, H1, H3, H5, H7

Nomenclature Symbols V h t G L

Volume Fractional holdup of dispersed phase Time Organic, dispersed phase flow rate Aqueous, continuous phase flow rate

cm3

-

h cm3/h cm3/h

Index n

Stage number

Exercise The simulation starts with a uniform holdup distribution through the column and with the column operating at steady-state. The response to changing inlet organic phase flow rate Go can be studied by use of the ISIM interactive facility.

5 60

5 Simulation Examples of Chemical Engineering Processes

Results The results show a) The fractional holdup response along the column and b) The aqueous (continuous) phase flow rate along the column resulting from a step change in the organic (dispersed) phase flow rate.

0.255

8.218

1

4

t-

R. 195 a

nn

ISIH

2.58

r

5.m

I . 50

1B.M

Figure 5.196. Response of the holdup on stages 1, 3, 5 and 7 to an increase of organic phase flow, GO, from 10.5 to 12.0 crn3/h.

5.8.9

-

ie. a 8. m

2.'5e

5.88

7.50

1s 88

Figure 5.197. Response of the organic phase rates to a change in organic flow inlet rate. The time lags through the column can be related to the residence times on the stages.

AXDISP - Differential Extraction Column with Axial Dispersion

System In Sec. 4.4.3 the modelling of differential extraction columns with axial dispersion is discussed. The extractor is considered to be approximated by eight finite-difference elements as shown below.

561

5.8 Mass Transfer Process Examples

Section 1

AZ = Zd8

Section n

Section 8

Figure 5.198. Axial dispersion model for an extractor.

Model The derivation method of the model equations for extraction columns with backmixing are explained in Sec. 4.4.3. Here the balances are easier to formulate because concentrations are used. The form of the component balance equations in terms of concentrations are as follows.

562

5 Simulation Examples of Chemical Engineering Processes

Here the factors 7 1 and ~ 7 segment n

1 represent ~

the residence times of the liquid in the

and the factors E2x and E2y are the inverse dispersion times for the phases 72L =

Z2 HL

where HL and HG are the fractional holdups of the X and Y phases. The end sections require special treatment to account for the fact that the no diffusive flux enters through the end wall of the column and are derived by omitting the diffusion term from the column end. In addition, the outlet concentrations X9 and Yo are extrapolated from the previous section. The component balances for section 8 can be derived with the aid of Figs. 5.199 and 5.200.

Ai!

Figure 5.199. The end section 8 with the designation of the streams. X

8

I

1

x6 (x6 -+ x7)/2 (X7+ X8)/2

x*

-

(X7

x7 "

- X8)/2

X8

k--Azai

2

Figure 5.200. Approximation of concentration profiles with end section 9.

9

563

5.8 Mass Transfer Process Examples

Formulating the balance for section 8 in the L-phase

Where Xy is obtained by extrapolation as

Rearrangement gives

In terms of the program nomenclature

and 72L =

AZ2A~

^DzA

=

AZ2H~

The G-phase balance becomes

The balances for the other end section n =I are analogous, as seen in the program. The resulting solution gives the concentrations of each segment in both phases, and the outlet concentrations, as a function of time.

Program The simulation starts with the initial axial concentration profiles set to zero. The program is rather slow to run and therefore after the first run, the program should perhaps be modified so that the steady-state values from the first run are used as starting values for subsequent simulations.

5 64

5 Simulation Examples of Chemical Engineering Processes

:Example AXDISP :2-PHASE DIFFERENTIAL DISPERSION :8

FINITE

DIFFERENCE

EXTRACTION

COLUMN

WITH

AXIAL

ELEMENTS

CONSTANT L=13.05,6=10.5 :PHASE FLOWRATES CONSTANT DL=100,DG=100 :DISPERSION COEFFICIENTS CONSTANT 20-160 :EXTRACTOR LENGTH CONSTANT K=4 :MASS TRANSFER COEFF CONSTANT M-0.6 :EQUILIBRIUM CONSTANT CONSTANT HL= 0.3 :DISPERSED PHASE HOLDUP CONSTANT XO = 1 :INLET FEED CONCENTRATION CONSTANT Y 9 = 0 :INLET SOLVENT CONC CONSTANT T F I N = 2 0 , C I N T = l 1 S1M;INTERACT; RESET; GOT0 1 INITIAL :INITIAL AQUEOUS PHASE CONCENTRATIONS X1=0;X2=0;X3=0;X4=0;X5=0;X6=0;X7=0;X8=0 :INITIAL SOLVENT PHASE CONCENTRATIONS Y1=0;Y2=0;Y3=0;Y4=0;Y5=0;Y6=0;Y7=0;Y8=0 :FACTORS IN MODEL EQUATIONS dZ=Z0/8 :LENGTH OF FINITE ELEMENT HG=l-HL :HOLDUP OF AQUEOUS PHASE E3=L/(dZ*HL*2) E4=DL/(dZ*dZ*HL) E5=G/(dZ*HG*2) E6=DG/(dZ*dZ*HG) DYNAMIC :ELEMENT 1 Xl'=E3*(2*XO-Xl-X2)+E4*(X2-Xl)-Ql/HL Yl'=E5*(Y2+Y1-2*YO)+E6*(Y2-Yl)+Ql/HG Ql=K* (Xl-M*Yl) YO=Yl+O.S*(Yl-Y2) :ELEMENTS 2 TO 6 :MASS TRANSFER RATES Q2=K* (X2-M*Y2) Q3=Kf(X3-M*Y3) Q4=K*(X4-M*Y4) QS=K*(X5-M*Y5) Q6=K*(X6-M*Y6) Q7=K*(X7-M*Y7) :AQUEOUS PHASE COMPONENT BALANCES X2'=-E3*(X3-Xl)+E4*(X1-2*X2+X3)-Q2/HL X3'=-E3*(X4-X2)+E4*(X2-2*X3+X4)-Q3/HL

.

.

565

5.8 Mass Transfer Process Examples

X4'=-E3*(X5-X3)+E4*(X3-2*X4+X5)-Q4/HL X5'=-E3*(X6-X4)+E4*(X4-2*X5+X6)-Q5/HL X6'=-E3*(X7-X5)+E4*(X5-2*X6+X7)-Q6/HL X7'=-E3*(X8-X6)+E4*(X6-2*X7+X8)-Q7/HL :SOLVENT PHASE COMPONENT BALANCES Y2'=E5*(Y4-Y2)+E6*(Y1-2*Y2+Y3)+Q2/HG Y3'=E5*(Y5-Y3)+E6*(Y2-2*Y3+Y4)+Q3/HG Y4'=E5*(Y6-Y4)+E6*(Y3-2*Y4+Y5)+Q4/HG Y5'=E5*(Y7-Y5)+E6*(Y4-2*Y5+Y6)+Q5/HG Y6'=E5*(Y8-Y6)+E6*(Y5-2*Y6+Y7)+Q6/HG Y7'=E5*(Y9-Y7)+E6*(Y6-2*Y7+Y8)+Q7/HG :ELEMENT 8 X9=X8-0.5*(X7-X8) XB1=E3*(X7+X8-2*X9)+E4*(X7-X8)-Q6/HL Y8'=E5*(2*Y9-Y7-Y8)+E6*(Y7-X8)+Q8/HG Q8=K* (XB-M*Y8) O U T P U T T, X 2 ,X 4 ,X6, X 8 PLOT T,X4,0,TFIN,0r1 T, X2, X4, X6, X8, Y1, Y2, Y4, Y6, Y 8 PREPARE

Nomenclature Symbols D ElX E1Y E2X E2Y H Ka L and G

Q

X and Y Z

Eddy dispersion coefficient Inverse residence time for X phase Inverse residence time for Y phase Inverse dispersion time for X phase Inverse dispersion time for Y phase Fractional phase holdup Mass transfer capacity coefficient Superficial phase velocity Mass transfer rate Solute concentration Column height

Indices

*

n L and G

Equilibrium value Stage n Phases for X and Y

cm2/5 115 115 115 115

-

115

cds gIcm3 s

g1cm3 cm

566

5 Simulation Examples of Chemical Engineering Processes

Exercises 1.

The steady-state outlet values should be checked for the overall balance.

2.

The extent of extraction can be investigated as a function of solvent flow rate.

3.

The mass transfer coefficient can be increased to obtain an approach to equilibrium conditions.

4.

The dispersion coefficients can be increased to observe an approach to ideal mixing, and setting them to zero should give the best performance.

5.

The dynamic responses for changes in inlet conditions can be investigated.

Results

R x2 8x4 C X6

0.688’

ISIN

D X8 0,458-

R Y2 B Y4 C YE

Lml

Ism

D YB 8.158-

@‘a 8.268-

a,ma.e Figure 5.201. Aqueous phase concentration profiles moving to steady state (D1 and D2 = 100).

5.b

-

1b.B

Figure 5.202. Solvent phase profiles corresponding to Fig. 5.201.

5 67

5.8 Mass Transfer Process Examples

5.8.10 AMMONAB - Steady-State Absorption Column Design System Ammonia is recovered from an air-ammonia gas mixture by absorption into water, using a countercurrent packed column. The absorption of ammonia in water is accompanied by the evolution of heat which causes a rise in the liquid water temperature and hence a change in the equilibrium relationship. The problem and data values are taken from Backhurst and Harker (1 990).

l-4 Y+dY

L+dX

Xout Xout

Figure 5.203.

Steady-state ammonia absorption column.

Model The gas is at high concentration and therefore the column component balance equations are based on mole ratio concentration units. The form of the balance equations follow those of Sec. 4.4.1.

568

5 Simulation Examples of Chemical Engineering Processes

For the gas

For the liquid

The energy balance for the liquid is given by

The temperature in the gas phase is assumed constant with no heat losses, and zero heat exchange between gas and liquid. The equilibrium relationship is represented by

where the equilibrium constant, K, is a function of temperature. The inlet gas flow is 1.25 m3/m2 s measured at NTP containing 5 per cent by volume NH3. Hence Gm = 1.25 * 0.95 / 22.4 = 0.053 k mol/m2 s For y1 equal 0.05 Y1 = 0.05 / 0.95 = 0.0526 kmol NH3kmol air For 95% ammonia recovery y 2=

0'0526 * 0'05

r+0.0526 + 0.05

= 0.00263 kmol NH3/kmol air

The inlet water flow is 1.95 kg/m2 s, hence Lm = 195 / 18 = 0.108 k mol/m2 s An overall component balance for the column gives G m (Yin - Youd = Lm G o u t - Xin)

569

5.8 Mass Transfer Process Examples

Calculating the concentration in the outlet liquid Xout =

0'053

* (0*0526 0.108 - 0'00263)

= 0.0245 kmol NH3/kmol water

Backhurst and Harker present the equilibrium data as straight line relationships for the temperatures 293 K, 298 K and 303 K. This data was curve fitted to the form

where K = a + b T + c T2, and the constants are given by a = 0.0059, b = 0.1227 and c = 3.01.

Program The program starts from the known concentration and temperature conditions at the top of the column and integrates down the column until the condition Yin (calculated) is equal to the known inlet value Yin. :Example

AMMONAB

:Steady-state Effects

Gas

Absorption

Column

with

Constant GM=0.053,LM=0.108 Constant YIN=0.052,YOUT=0.00263 Constant KGA=0.113 :Constants in equilibrium relationships A=O.0059,B=-0.1227,C=3.01 Constant Constant CP=4.18 Constant HABS=36200 Constant ATMOS=101.33 Constant TG=23,TLIN=20 Constant CINT=O.Ol, TFIN=2 1 SIM; INTERACT; RESET; GOT0 1 INITIAL :Conditions at column top Y=YOUT; X = O ; TEMP=TLIN DYNAMIC Z=T

Heat

570

5 Simulation Examples of Chemical Engineering Processes

:Solvent phase balance Y'=DQ/GM :Rate of mass transfer DQ=KGA*(YMF-YMFEQ) :Aqueous phase mas8 balance X ' =DQ/LM :Energy balance TEMP'=DQ*HABS/(LM*18*CP) DQ=KGA*(YMF-YMFEQ) :Convert t o mole fractions YMF=Y/(l+Y) XMF=X/(l+X) :Equilibrium relationship YMFEQ=K*XMF/ATMOS K=A*TEMP*TEMP+B*TEMP+C :Check inlet gas concentration condition TERMINATE Y.GE.YIN PLOT Z,Y,O,TFIN,0,0.5 OUTPUT Z,X,Y,TEMP PREPARE 2 , X, Y, TEMP, XMF, YMF, YMFEQ

Nomenclature Symbols cP

Gm K KGa

Lm PT

P

T X X

Y

Y Z

AH

Specific heat Molar flow of inert air Equilibrium constant Overall mass transfer capacity coefficient base on the gas phase Molar flow of solute-free water Pressure Density Liquid temperature Mole ratio c in the liquid phase Mole fraction in the liquid phase Mole ratio in the gas phase Mole fraction in the gas phase Height of packing Exothermic heat of adsorption

kJ/kmol "C kmol/m2 s kN/m2 kmol/m3 s Ay kmol/m2 s kN/m2 kg/m3

"C

kmol NH3/kmol water -

kmol NH3/kmol air -

m kJ/kmol

57 1

5.8 Mass Transfer Process Examples

Indices in, out

Refer to inlet and outlet conditions

Exercises 1.

Stud the effects of varying liquid rate, inlet water temperature, inlet gas concentration and pressure on the absorption.

2.

Modify the program so that you are able to graph the driving force for mass transfer (y - ye) and the resulting "pseudo" equilibrium curve, i.e., ( y e versus x), allowing for the changing temperature. Note however that the equilibrium relationship is only valid for the range 293 to 303 K.

Results A X

1.m

Figure 5.204. Concentrations of the gas and liquid phases from top (Z=O) to bottom (Z=1.45) of the column.

A

m

36.0

Figure 5.205. Temperature profiles in the column, corresponding to Fig. 5.204.

References Backhurst, J.R., and Harker, J.H. (1990). Chemical Engineering, Vol. 5 , Pergamon Press.

572 ~

~

5 Simulation Examples of Chemical Engineering Processes ~~~~

5.8.11 MEMSEP - Gas Separation by Membrane Permeation System An internally-staged, gas-permeation module is used for the oxygen enrichment of air, using the flow arrangement shown in Fig. 5.206. Enrichment depends on differing membrane permeabilities for the oxygen and nitrogen to be separated. The permeation rates are proportional to the differences in component partial pressures.

Membrane 2 ate Membrane 1 Feed

Section 1

P

Permeate

Figure 5.206. Configuration of the gas-permeation module

Model The module system consists of three ideally mixed sections, separated by two membranes. Overall steady-state balances can be used to check the solution and are as follows: Total mass Nf = N1 + N 2 + N 3 Oxygen Nitrogen

Nf XOF = N1 xO1 + N 2 x02 + N 3 x03

5.8 Mass Transfer Process Examples

573

Steady-state total mass balances can also be written for Sections 1, 2 and 3 , based on the flow rates in each section and the permeation rates between sections. They are as follows

Here the rates are converted from mol/s to m3/s using the ideal gas law. The component balances in terms of concentrations are as follows:

Section 1 Oxygen

Nitrogen

Section 2 Oxygen

Nitrogen

Section 3 Oxygen

Nitrogen

574

5 Simulation Examples of Chemical Engineering Processes

The permeation rates are written in terms of partial pressures, permeabilities, membrane areas and thicknesses as follows: 1. From Section 1 to Section 2:

For 0 2 ROl =

k

(P1 xo1- p2 x02) A1

2. From Section 2 to Section 3:

For 0 2 R02 =

k

(p2 X02 - p3 X03) A2

For N2

Program :Example :

Gas

MEMSEP

Separation

by

Membrane

Constant Nf=50 Constant d=3.81E-5 Constant kg0=0.163E-12 Constant Constant Constant Constant Constant Constant Constant Constant Constant

KpN=0.0516E-12 A1=5 A2=5 V1=0.01 V2=0.01 V3=0.01 P1=26E5 P2=5.6E5 P3=1E5

Permeation.

:Feed rate (mol/s) :Membrane thickness (m) :Permeability of oxygen (mol/m 8 Pa) :Permeability of nitrogen (mol/m s Pa) :Membrane area (m2) :Membrane area (ma) :Volume stage 1 (m3) :Volume stage 2 (m3) :Volume stage 3 (m3) :Pressure in stage 1 (Pa) :Pressure in stage 2 (Pa) :Pressure in stage 3 (Pa)

5.8 Mass Transfer Process Examples

575

Constant R=8.314e3 :Gas constant (m3 Pa/mol K ) Constant T e m p 2 9 8 :Absolute temperature ( K ) Constant CINT=O.l, TFIN=5, NOCI=l,ALGO=O 1 SIM; INTERACT; RESET; GOT0 1 INITIAL Gf = Nf*R*Temp/P1/1000 :Val. flow (m3/s) XO1=0.209 x02=0.209 X03=0.209 XN1=0.791 XN2=0.791 XN3=0.791 COf=Pl*XOl/(R*Temp) C N f = P l * X N l /( R * T e m p ) COl=COf CNl=CNf C02=P2*X02/(R*Temp) CN2=P2*XN2/(R*Temp) C03=P3*X03/(R*Temp) CN3=P3*XN3/(R*Temp) DYNAMIC :Transfer rates [moles/sl R0l=(kpO/d)*(Pl*~Ol-P2*~02)*Al RNl=(kpN/d)*(Pl*XNl-P2*xN2)*Al R02=(kpO/d)*(P2*~02-P3*~03)*A2 RN~=(~PN/~)*(P~*XN~-P~*XN~)*A~ :Flow rates [m3/sl Gl=Gf-(ROl+RNl)*(R*Temp/Pl) G2=(Rol+RNl-R02-RN2)*(R*Temp/P2) G3=(R02+RN2)*(R*Temp/P3) IF (Gl.lt.0) Gl=O IF (G2.lt.0) G2=0 IF (G3.lt.0) G3=0 :Accumulation [mOleS/m3/s] COl'=(Gf*COf-G1*C01-ROl)/V1 CNl'=(Gf*CNf-G1*CNl-RNl)/Vl C02'=(ROl-RO2-G2*C02)/V2 CN2'=(RNl-RN2-G2*CN2)/V2 C03'=(R02-G3*C03)/V3 CN3'=(RN2-G3*CN3)/V3 IF (COl.lt.0) C01=0 IF (C02.lt.O) C02=0 IF (C03.lt.0) C03=0 :Mole fractions XOl=COl/(COl+CNl)

576

5 Simulation Examples of Chemical Engineering Processes

X02=C02/(C02+CN2) X03=C03/(C03+CN3) X N 1 =1-XO1 XN2=1-X02 X N 3 = 1-X03 P L O T T, XO3,O ,TFIN, 0,l OUTPUT T,X01fX02,X03 T,X01,X02,X03,G1,G2,G3,R01,R02,RN1,RN2 PREPARE

Nomenclature Symbols Membrane area Concentration Membrane thickness Flow rate Cost constant Permeability coefficient Molar flow rate Pressure Gas constant Permeation rate Section volume Mole fraction of oxygen

A C d G KC

kP N

P R

Rin

v X

m2 m0i/m3 m m3/s 1/Pa2

mol/m s Pa moVs k Pa m3 Pa/mol K moVs m3 -

Indices 1,2, 3 0 N f

Sections 1, 2, 3 and membrane 1, 2 Oxygen Nitrogen Feed

Exercises 1.

Study the influence of P1 and P3 on the separation performance.

2.

Introduce new membranes with improved permeabilities to give better separation.

577

5.8 Mass Transfer Process Examples

3.

Define a cost function assuming that the cost increases with P12 and Gf and that the enrichment of stream 3 gives profits according to COST = Pi2 Gf

-

Kc ~ 0 G3 3

Use a reasonable value for K c and then try to find the optimum by varying the pressure in the three stages.

Results

Figure 5.207. Mole fractions in the three stages at varying feed flow rate (Gf = 0.05, 0.2 and 2 m3 / s).

Figure 5.208. Influence of P2 on oxygen mole fractions (P2 = 0.056, 0.8 and 0.2 l o 6 Pa).

Reference Li, K., Acharya, R. and Hughes, R. (1991) Simulation of Gas Separation in an Internally Staged Permeator, Trans IChemE, 69, Part A, 35-42.

578

5 Simulation Examples of Chemical Engineering Processes

5.8.12 FILTWASH

- Filter

Washing

System The washing of filter cake is carried out to remove liquid impurities from valuable solid product or to increase recovery of valuable filtrates from the cake. Wakeman (1990) has shown that the axial dispersion flow model, as developed in Sec. 4.3.6, provides a fundamental description of cake washing. It takes into account such situations as non-uniformities in the liquid flow pattern, non-uniform porosity distributions, the initial spread of washing liquid onto the topmost surface of the filter cake and the desorption of solute from the solid surfaces.

Model As shown by Wakeman, the solute material balance for the flowing liquid phase, allowing for axial dispersion and desorption of solute is given by the following defining partial differential equation

The concentration of the solute adsorbed on to the particles is related to the concentration of the solute in the liquid by an equilibrium relationship of the form

Expressing the model equations in dimensionless form gives

where

c* W is the wash ratio given by

C -

CO

and x* =

X

L

579

5.8 Mass Transfer Process Examples

and P is the Peclet number

For solution by digital simulation, the depth of filter cake is divided into nine segments, each of which has an equivalent dimensionless thickness Ax. For any element n, the form of the resulting difference differential equation is given by

where the term Z is given by

In these equations the * designation for dimensionless concentration c*, has been dropped. Note that in the above equation, the finite differencing of the convection term has been done over two neighbouring segments. Again special relationships apply to the end segments, owing to the absence of axial dispersion, exterior to the cake.

Program Program FILTWASH models the dimensionless filtration wash curves for the above case of a filter cake with constant porosity, axial dispersion in the liquid flow and desorption of solute from the solid particles of the filter bed (Boyd, 1993). The IF-statements in the listings prevent the solute concentrations from going below 0.0001 (a value near to zero). This is because these concentrations exist in the denominators of some of the expressions used in the program and therefore cannot be allowed to become zero. :EXAMPLE FILTWASH :WASHING WITH DESORPTION Constant U=4E-4 Constant CO=O Constant CI=1 Constant L=4.3E-2 Constant DL=8.5E-8 Constant E=0.85

FROM

PARTICLE

SURFACES

:Pore velocity [m/sl :Solute conc. in cake prior washing [kg/m31 :Initial sol. conc. [kg/m31 :Cake thickness [ml [ma / s 1 :Axial dispersion coef :Cake porosity [ - I

.

580

5 Simulation Examples of Chemical Engineering Processes

Constant N=l :Reciprocal order of ads. Constant K = l :Adsorption constant CONSTANT ALGO=O CONSTANT TFIN=3,CINT=O.O1,NOCI=20

1 SIM

;

RESET

INITIAL Cl=CI C2=CI C3=CI C4=CI C5=CI C6=CI C7=CI C8=CI C9=CI X=1/8 P=DL/(U*L) DN=l/P G= (1-E)/E M = ( 1 - N )/ N H=K*(CI**M)/N

;

INTERACT

;

[-I

GOT0 1

:Initial concentrations :of solute in elements

:Peclet number :Constant factors

DYNAMIC W=T :Prevents the concentrations IF (Cl.LT.0.0001) C1=0.0001 IF (C2.LT.0.0001) C2=0.0001 IF (C3.LT.0.0001) C3=0.0001 IF (C4.LT.0.0001) C4=0.0001 IF (C5.LT.0.0001) C5=0.0001 IF (C6.LT.0.0001) C6=0.0001 IF (C7.LT.0.0001) C7=0.0001 IF (C8.LT.0.0001) C8=0.0001 IF (C9.LT.0.0001) C9=0.0001 Z1 = l+H*G*(Cl**M) 22 = 1+H*G*(C2**M) 23 = 1+H*G*(C3**M) 24 = 1+H*G*(C4**M) Z5 = 1+H*G*(C5**M) 26 = 1+H*G*(C6**M) 27 = l+H*G*(C7**M) 28 = l+H*G*(C8**M)

from

going

<0.0001

:Adsorption terms

5.8 Mass Transfer Process Examples

58 1

c1'=(P*(c2-c1)/(x**2)+(2*co-cl-c2)/(2*x)~/ c 2 ' = ( P * ( c 1 - 2 * c 2 + c 3 ) / ( x * * 2 ) - ~ c 3 - c 1 ) / ( 2 * x ) ~/ c3'=(P*(c2-2*c3+c4)/(x**2)-(c4-c2)/(2*x)~/ C4'=(P*(C3-2*C4+C5)/(X**2)-(C5-C3)/(2*X)) / C5'=(P*(C4-2*C5+C6)/(X**2)-(C6-C4)/(2*X)) 1 C6' = ( P * ( C 5 - 2* C 6 + C 7 ) / ( X * * 2 ) - ( C 7 - C 5 ) / ( 2 * X ) ) / C7'=(P*(C6-2*C7+C8)/(X**2)-(C8-C6)/(2*X)) / C8'=(P*(C7-C8) /(X**2)+(C7-C8)/ X) / C9= C8 + (CS-C7) / 2 PREPARE OUTPUT PLOT

z1 22 23 24

25 26 27 28

W,C1,C2,C3,C4,C5,C6,C7,C8,C9

WfC1,C2,C3,C4,C5,C6,C7,C8,C9 W, C9,0, TFIN, 0, CI

Nomenclature

Symbols

DL KO

Solute concentration in liquid Dimensionless solute concentration Solute concentration in cake prior to washing Axial dispersion coefficient Constant in equilibrium isotherm

Kc

Adsorption equilibrium constant

L N P S

Cake thickness Reciprocal order of the adsorption Peclet number per unit volume of voids Fractional saturation, volume of liquid Time Pore velocity Wash ratio Distance measured from filter cloth Dimensionless distance Cake porosity Solute concentration on solid

C C*

CO

t

U

w

X X* E:

rl

kg/m3 m2/s kg/m3

(3$) m -

-1/N

582

5 Simulation Examples of Chemical Engineering Processes _

.-

_

_

_

Exercises 1.

Use the program to assess the effects of differing degrees of axial dispersion, for values of Peclet number ranging from 0.005 to 0.25. Modify the program to account for zero axial dispersion.

2.

Wakeman found that the effect of increasing the equilibrium constant in the description isotherm equation is to increase the wash-ratio required. Is this confirmed by simulation and what is the explanation of this effect.

3.

Increasing the value of N (reciprocal power term) reduces the effect of the desorbed solute and agrees ultimately with the case of zero solute desorption.

4.

In compressible filter cakes, the porosity varies from a minimum next to the filter septum to a maximum at the cake surface. How could you include such a variation within the context of a simulation program? Remember that as the porosity changes, the local fluid velocity will also change as a result of

A necessary assumption would also be that the initial porosity distribution remains constant and independent of flow rate.

5.

The presence of a liquid layer on the surface of the filter cake will cause solute to diffuse from the top layer of cake into the liquid. Also if disturbed the layer of liquid will mix with the surface layer of filter cake. This effect can be incorporated into the digital simulation by assuming a given initial depth of liquid as an additional segment of the bed which mixes at time t=O with the top cake segment. The initial concentrations in the liquid layer and top cake segment are then found by an initial mass balance.

Cases 4 and 5 are solved also by Wakeman (1990).

~

583

5.8 Mass Transfer Process Examples

Results ISIH

ISIH

Figure 5.209. Washing curve with no dispersion.

Figure 5.210. Variation of wash curve with dispersion coefficient, DL = 8.5 lo-*, and

1,284

c91'm1

ISM

ISIN 1.281

1.28

8.88

0.48

0.48

W

Figure 5.211. Effect of varying the adsorption equilibrium constant, K = I , 2 and 5 (curves A, B and C).

e.m

Y

Figure 5.212. Variation of wash curve with reciprocal order of adsorption isotherm. N = 1, 2 and 3 (curves A, B and C for C3 and D, E, and F for C9).

References 1. Wakeman, R.J. (1990) Chem. Eng. Res. Des. 68, 161-171.

2. Wakeman, R.J. (1986) Chem. Eng. Res. Des. 64, 308. 3. Wakeman, R.J. (1981) in "Solid-Liquid Separation", Ed. L. Svarovsky, Butterworths, London.

4. Boyd, J. (1993) M. Eng. Research Project, University of Bradford.

584

5 Simulation Examples of Chemical Engineering Processes .~

5.9

Distillation Process Examples

5.9.1

BSTILL - Binary Batch Distillation Column

System This distillation equipment is as described in Sec. 3.3.3.2. For convenience a column containing seven theoretical plates and reboiler is assumed, together with constant volume conditions in the reflux drum. The example is based on that of Luyben (1973, 1990).

I

-+ - '-44-+ - -4 -

-T 3-i-4 :-+ -4 -+-;-

--f

-

Steam

t

-

MB

*8

I c

Figure 5.213. The seven-stage bar.ch distillation column.

585

5.9 Distillation Process Examples ~~

Model The component balance for the more volatile component on any plate n yields

M dXtl x = LXn-1-LXn+V(Yn+1-Yn) For the still

where 7 refers to the bottom plate and 8 to the still and column base. For the reflux drum and condenser M OdXo T - VYl-(L+D)Xo where the reflux ratio, R, is given by Lo/D and

L = V-D Assuming theoretical plate behaviour in the column Yn =

a Xn 1 + ( a - 1) Xn

where a is the relative volatility.

Program It may be noted that the liquid holdup on the plates is set rather large. This is necessary to avoid very slow rates of computation.

586

5 Simulation Examples of Chemical Engineering Processes

_ _ ~ __

:Example

-

BSTILL

:CONSTANT HOLDUP BINARY BATCH DISTILLATION : EQUILIBRIUM PLATE BEHAVIOUR : : : :

COLUMN

ASSUMPTIONS CONSTANT BOIL-UP RATE CONSTANT MOLAR OVERFLOW AND CONSTANT PLATE CONSTANT HOLDUP IN REBOILER AND SURGE DRUM

:For new

charge,change MB,X8

and

set

HOLDUP

DIST=O

CONSTANT A=3.5 :RELATIVE VOLATILITY CONSTANT MO=100 :MOLAR HOLDUP IN SURGE DRUM CONSTANT MB=2500 :MOLAR HOLDUP IN STILL CONSTANT M=20 :MOLAR HOLDUP ON PLATES CONSTANT X8 = .8 :STILL CHARGE CONCENTRATION CONSTANT V=10 :VAPOUR BOIL-UP RATE CONSTANT R=lE10 :START UP AT TOTAL REFLUX CONSTANT CINT=1, TFIN=500 SIM INITIAL : INITIAL CONCENTRATIONS x1=0;x2=0;x3=0;x4=0;x5=0;x5=0;x7=0 CHARGE =MB DYNAMIC :CONDENSER AND REFLUX DRUM XO'=(V*Yl-(L+D)*XO)/MO D = V / (R+1) L=V-D : PLATES 1 TO 7 Xl'=(L*(XO-Xl)+V*(Y2-Yl))/M Yl=A*Xl/(l+(A-l)*Xl) XZ'=(L*(Xl-XZ)+V*(Y3-YZ))/M Y2=A*XZ/(l+(A-l)*X2) X3'=(L*(X2-X3)+V*(Y4-Y3))/M Y3=A*X3/(l+(A-l)*X3) X4'=(L*(X3-X4)+V*(Y5-Y4))/M Y4=A*X4/(1+(A-l)*X4)

X5'=(L*(X4-X5)+V*(Y6-Y5))/M

Y5=A*X5/(1+(A-l)*X5)

X6'=(L*(X5-X6)+V*(Y7-Y6))/M

Y6=A*X6/(l+(A-l)*X6)

x7'=(L*(X6-X7)+V*(YE-Y7))/M

:

587

5.9 Distillation Process Examples

Y7=A*X7/(l+(A-l)*X7) :STILL AND COLUMN BASE MB'=L-V X8'=(L*X7-V*Y8)/MB Y8=A*X8/(l+(A-l)*X8) : TOTAL DISTILLATE REMOVED DIST ' = D : FRACTION OF CHARGE DISTILLED FRAC=DIST/CHARGE PLOT T, XO, 0, TFIN, 0,1 OUTPUT T IXO ,X2, X4, X6, X 8 T,XO,Xl,X2,X3,X4,XS,X6,X7,X8,R,FRAC,MB PREPARE

Nomenclature Symbols Molar holdup Liquid flow rate Vapour flow rate Distillate rate Liquid phase mole fraction Vapour phase mole fraction Reflux ratio Relative volatility

M L V D X Y R

a

kmol kmol/h kmol/h kmol/h

-

Indices 1 to 8 B 0

Plate number Reboiler Reflux drum

Exercises 1.

Investigate the response of the column to changes in the boilup rate, V.

2.

The program starts up the column at total reflux (R very high). After steady state is reached on all plates, vary the reflux ratio interactively and attempt to carry out the distillation in minimum time, while attempting to

588

5 Simulation Examples of Chemical Engineering Processes ~

maintain a distillate composition with X0 > 0.9. Note the response of the distillate composition, XO. 3.

Recharge the column by changing X8 and MB and note the transients on each plate.

4.

Devise a control scheme for the column to maintain the distillate composition xu > 0.9.

5.

Modify the program to allow for continuous feeding of fresh feed into the still during the process of distillation. The feed rate should be adjusted to equal the rate of withdrawal of distillate. What is the effect of feeding on the distillation?

Results Outputs are shown below. Ism

RXB B x2

c

x4

0 X6

E XB

F F W

Figure 5.214. The batch column was started with total reflux; the reflux was then reduced to 0.25 at t = 75 and then increased (from 1.0 to 5.0 to 50.0) whenever the distillate tank composition X0 fell below 0.9. The response of the plate compositions and the fraction distilled, FRAC, are shown.

Figure 5.215. The contents of the still decreased throughout the run of Fig. 5.214.

References 1. Luyben, W.L. (1973, 1990). Process Modelling, Simulation and Control for Chemical Engineers McGraw-Hill.

589

5.9 Distillation Process Examples

5.9.2

DIFDIST - Multicomponent Differential Distillation

System A mixture of hydrocarbons A, B, C and D is charged to a still pot and distilled over in the absence of reflux.

Distillate tank

Figure 5.216.

Single multicomponent batch distillation.

Model Total mass balance for still

where V is the molar vapourisation rate. Component mass balance

Equilibrium is given by

y' = 1

(Xi X i ___

C

xi

5 90

5 Simulation Examples of Chemical Engineering Processes --__. .

~~

based on constant relative volatilities. Mass balances for distillate in the receiver tank,

where D is the total distillate in tank and Di is the mass of component i in tank.

Program :Example

DIFDIST

: Four component differential distillation Constant S0=50 : Initial moles in still Constant V=5 : Vaporisation rate Constant XA=0.2 : Molar fraction of A Constant XB=0.3 : Molar fraction of B Constant XC=O.l : Molar fraction of C Constant XD=O.Q : Molar fraction of D Constant ALA=10 : Relative volatility of A Constant ALB=5 : Relative volatility of B Constant ALC=1 : Relative volatility of C Constant ALD=0.8 : Relative volatility of Constant CINT=O.l,TFIN=lO 1 SIM; INTERACT; RESET; GOT01 INITIAL

s= s o

D

: Molar quantities in still MA=SO*XA MB=SO*XB MC=SO *XC MD=SO*XD :Moles in distillate tank DA=O;DB=O;DC=O;DD=O;D=O

DYNAMIC :Total balance S'=-V :

Equilibrium vapor

composition

SUM=ALA*XA+ALB*XB+ALC*XC+ALD*XD

YA=ALA*XA/SUM

59 1

5.9 Distillation Process Examples

YB=ALB*XB/SUM YC=ALC*XC/SUM YD=ALD*XD/SUM : Component balances MA'=-V*YA MB'=-V*YB MC' =-V*YC MD'=-V*YD : Liquid mole fractions XA=MA/S XB=MB/S XC=MC/S XD=MD/S :Component balances for distillate tank DA' = V * Y A DB'=V*YB DC'=V*YC DD'=V*YD :Total balance for distillate tank D'=V : Average composition in distillate tank if (D.eq.O)D=(D+O.OOOl):Avoids zero division A v A =D A / D AvB =DB/ D AvC=DC/D AvD=DD/D IF(XA.LE.O)XA=O IF(XB.LE.O)XB=O IF(XC.LE.O)XC=O IF(XD.LE.O)XD=O PLOT T,XA, O,TFIN, 0,1 T, XA, XB, XC, XD, AvA, AvB, AvC, A v D OUTPUT PREPARE T,XA,XB,XC,XD,S,D,AVA,AvB,AvC,AvD

Nomenclature Symbols S V Xi

Yi ai

Liquid in the still Distillate removal rate moles Mole fraction in the liquid Mole fraction in the vapour Relative volatility of i

k mol k mol/h

592

5 Simulation Examples of Chemical Engineering Processes

Indices Component i Initial condition Refers to average values in distillate tank

i 0 avg

Exercises 1.

Vary the relative volatilities of the mixture and study the effect on the distillation.

2.

Vary the initial proportions of the four components.

3.

Compare the computer predictions with the Rayleigh equation prediction, where:

4.

Check the overall material balance in all the above exercises.

5.

Using the interact feature of ISIM, make an additional charge to the still at a point when the light component A has been completely distilled. This requires changing the mass quantities (S, MA, MB, MC and MD) in the still, which can be done by using VAL to obtain the current values and to set (new mass) = (current mass) + (additional mass charged).

593

5.9 Distillation Process Examples

Results 8.888

ISIH

1

8.498

0.388

8.198

8.EMB.W

2.98

I

5.00

1.98

18.88

Figure 5.217. Variation in still compositions with time. At T=7.2, a new charge of 4 moles (MA=2, MB=3, MC=4 and MD=5) was added to the still.

5.9.3

MUBATCH Distillation

9.8888.W

2.98

r

5.1

1.98

i8.m

Figure 5.218. Change of distillate tank composition with time, corresponding to the run in Fig. 5.217.

- Multicomponent

Batch

System A batch still corresponding to a total separation capacity equivalent to eight theoretical plates (seven plates plus the still) is used to separate a hydrocarbon charge containing four (A, B, C, D) simple-hydrocarbon components. Both the liquid and vapour dynamics of the column plates are neglected. Equilibrium data for the system is represented by constant relative volatility values. Constant molar overflow conditions again apply, as in BSTILL. The problem was originally formulated by Robinson ( I 975).

5 94

5 Simulation Examples of Chemical Engineering Processes

I

Lr-l

Figure 5.219. The multicomponent, batch distillation column.

Model For the still, the total mass balance is given by

Component mass balances for the four liquid components in the still, are given by

where i = A to D.

5.9 Distillation Process Examples

595

For any plate n, neglecting the plate dynamics, the compositions of the passing liquid and vapour streams are represented by the standard operating line equation, where

or as

where R = L/D and V = (R+1)D. Assuming a well-mixed, constant holdup reflux drum, the balance equation for each component i, in the drum, is

Assuming theoretical plate behaviour for both the column plates and the still, the corresponding vapour compositions are given by

Program Program MUBATCH is rather stiff and tends to run rather slowly using the integration routine ALGO = 0. With the other integration routines, the calculation tends to be unstable. Execution of the program may also stop owing to the PREPARE statement becoming full. For this reason, the program uses a single SIM statement for single runs rather than that for repeated runs i.e., 1 SIM; INTERACT; RESET; GOT0 1. Example MUBATCH MULTICOMPONENT BATCH DISTILLATION : ASSUMPTIONS ARE THE SAME AS IN BSTILL : EXCEPT THAT THE : PLATE DYNAMICS ARE NOW EXCLUDED FROM THE ANALYSIS CONSTANT MB=1000 :Molar holdup in still CONSTANT MO=100 :Molar holdup in reflux drum CONSTANT V=10 :Vapor boil up rate CONSTANT R=10 :Reflux rate CONSTANT AA=2.0 :Relative volatilities : :

596

5 Simulation Examples of Chemical Engineering Processes

C O N S T A N T AB=1.5 C O N S T A N T AC=1.0 C O N S T A N T AD=0.5 CONSTANT TFIN=500, CINT=2, N O C I = 1 0 SIM INITIAL XA8=0.25 : Initial Still Concentrations XB8=0.25 XC8=0.25 XD8=0.25 XDO=l;XD1=1;XD2=1;XD3=1;XD4=1;XD5=1;XD6=1;XD7=1 : ALL OTHER CONCENTRATIONS A R E IMPLICITLY : SET T O ZERO, USING THE ISIM DEFAULT VALUES DYNAMIC : LIQUID A N D VAPOUR FLOWRATES D=V/(R+1) L=V-D : REFLUX DRUM DYNAMICS XAO’=(V*YAl-(L+D)*XAO)/MO XBO’=(V*YBl-(L+D)*XBO)/MO XCO’=(V*YCl-(L+D)*XCO)/MO XDO=l-XAO-XBO-XCO : LIQUID COMPOSITIONS ON PLATES 1 T O 7 : Liquid Plate Dynamics neglected Rl=(R+1) XAl=(Rl*YA2-XAO)/R XBl=(Rl*YB2-XBO)/R XCl=(Rl*YC2-XCO)/R XDl=l-XAl-XBl-XCl XA2=(Rl*YA3-XAO)/R XB2=(RI*YB3-XBO)/R XC2=(Rl*YC3-XCo)/R XD2=1-XA2-XB2-XC2 XA3=(Rl*YA4-XAO)/R XB3=(Rl*YB4-XBO)/R XC3=(Rl*YC4-XCO)/R XD3=1-XA3-XB3-XC3 XA4=(Rl*YAS-XAO)/R XB4=(Rl*YBS-XBO)/R XC4=(Rl*YC5-XCO)/R XD4=1-XA4-XB4-XC4 XA5=(Rl*YA6-XAO)/R XB5=(Rl*YB6-XBO)/R XCS=(Rl*YC6-XCO)/R XD5=1-XA5-XB5-XC5

5 97

5.9 Distillation Process Examples

XA6=(Rl*YA7-XAO)/R XB6=(Rl*YB7-XBO)/R XC6=(Rl*YC7-XCO)/R XD6=1-XA6-XB6-XC6 XA7=(Rl*YAE-XAO)/R XB7=(Rl*YBE-XBO)/R XC7=(Rl*YCE-XCO)/R XD7=1-XA7-XB7-XC7 : VAPOUR COMPOSITIONS ON PLATES 1 T O SUMAl=AA*XAl+AB*XBl+AC*XC1+AD*XDI. SUMA2=AA*XA2+AB*XB2+AC*XC2+AD*XD2 SUMA3=AA*XA3+AB*XB3+AC*XC3+AD*XD3 SUMA4=AA*XA4+AB*XB4+AC*XC4+AD*XD4 SUMA5=AA*XA5+AB*XB5+AC*XC5+AD*XD5 SUMA6=AA*XA6+AB*XB6+AC*XC6+AD*XD6 SUMA7=AA*XA7+AB*XB7+AC*XC7+AD*XD7 YAl=AA*XAl/SUMAl YBl=AB*XBl/SUMAl YCl=AC*XCl/SUMAl YDl=YAl-YBl-YCl YA2=AA*XA2/SUMA2 YB2=AB*XB2/SUMA2 YC2=AC*XC2/SUMA2 YD2=YA2-YB2-YC2 YA3=AA*XA3/SUMA3 YB3=AB*XB3/SUMA3 YC3=AC*XC3/SUMA3 YD3=1-YA3-YB3-YC3 YA4=AA*XA4/SUMA4 YB4=AB*XB4/SUMA4 YC4=AC*XC4/SUMA4 YD4zl-YA4-YB4-YC4 YAS=AA*XAS/SUMA5 YBS=AB*XB5/SUMAS YC5=AC*XC5/SUMA5 YDS=l-YAS-YBS-YCS YA6=AA*XA6/SUMA6 YB6=AB*XB6/SUMA6 YC6=AC*XC6/SUMA6 YD6=1-YA6-YB6-YC6 YA7=AA*XA7/SUMA7 YB7=AB*XB7/SUMA7 YC7=AC*XC7/SUMA7 YD7=1-YA7-YB7-YC7 STILL DYNAMICS

7

598

5 Simulation Examples of Chemical Engineering Processes

MB'zL-V XA~'=(IJ*XA~-V*YA~)/MB XBe1=(L*XB7-V*YB8)/MB XC8'=(L*XC7-V*YC8)/MB XD8=1-XA8-XB8-XC8 SUMAB=AA*XA8+AB*XB8+AC*XC8+AD*XD8 YA8=AA*XA8/SUMA8 YBB=AB*XB8/SUMA8 YC8=AC*XC8/SUMA8 YD8=1-YA8-YA8-YA8 :TOTAL D I S T I L L A T E DIST' = D P L O T T, XAO, 0, TFIN, 0,1 PREPARE T,XA0,XBOIXC0,XD0,XA8,XB8,XC8,XD8 : P R E P A R E T , Y A 1 , YB1,YC1,YD1,YA8,YBa,YCa,YD8 :PREPARE T,XA2,XA4,XA6,RlDIST OUTPUT T,XAO,XBO,XCO,XDO,XA8,XB8,XC8,XD8

Nomenclature Symbols M L

v

D X Y R

a

Molar holdup Liquid flow rate Vapour flow rate Distillate rate Liquid phase mole fraction Vapour phase mole fraction Reflux ratio Relative volatility

Indices i 1 to 8

B 0

Component Plate number Reboiler Reflux drum

kmol kmol/h kmol/h kmollh -

599

5.9 Distillation Process Examples

Exercises 1.

Study the effect of varying reflux ratio on the separation.

2. Study the effect of varying relative volatility on the separation.

Results

Figure 5.220. Liquid phase mole fractions of components A, B, C and D in the reflux drum.

Figure 5.221. Liquid phase mole fractions in the bottom plate in the same run as Fig. 5.220.

References Robinson, E.R. (1975) Time Dependent Process, Applied Science Publishers.

5.9.4

CONSTILL Column

-

Continuous Binary Distillation

System For simplicity, a column containing a total of eight theoretical plates plus a reboiler is assumed, with the feed entering on plate 5 , numbered from the column top, as described in Sec. 3.3.3.3.

600

5 Simulation Examples of Chemical Engineering Processes

Figure 5.222. An eight-plate continuous distillation column.

Model The following model can be written for this system: For the feed plate (plate 5 )

d(Xf> = L Xf-1- L' Xf + V' Y f + l - V Yf + F Xf M7

60 1

5.9 Distillation Process Examples

Feed plate 5

Figure 5.223. Streams in and out of the feed plate.

For the plates above the feed plate

For the plates below the feed

where, assuming constant molar overflows

L'= L + q F and

V = V'-(1 - 9 ) F

For a saturated liquid feed with q = 1, L' = L + F and V = V'.

For the reboiler

where N = 8 is the bottom plate of the column.

602

5 Simulation Examples of Chemical Engineering Processes

Assuming theoretical plate behaviour in the column, the equilibrium is calculated by

where a is the relative volatility.

Program :Example :8

PLATE

CONSTILL CONTINUOUS

DISTILLATION

COLUMN

CONSTANT F=1000 :COLUMN FEEDRATE CONSTANT Q=1 :SATURATED LIQUID FEED CONSTANT X=0.5 :FEED COMPOSITION CONSTANT R=2.25 :REFLUX RATIO CONSTANT A=2.4 6 :RELATIVE VOLATILITY CONSTANT MD=200 :HOLDUP IN SURGE DRUM CONSTANT MR=400 :HOLDUP IN REBOILER CONSTANT M=50 :HOLDUP ON PLATES CONSTANT V1=1575 :VAPOUR BOILUP RATE CONSTANT TFIN=10, CINT=0.5 SIM INITIAL :START UP WITH ZERO LIQUID CONCENTRATION PROFILE XO=O;X1=O;X2=O;X3=O;X4=O;X5=O;X6=O;X7=O;X8=O;X9=O DYNAMIC :FLOW RATES IN COLUMN V = V 1 + ( 1 - Q )* F D=V/(R+l) L=R*D Ll=F+Q*F W=Ll-Vl :CONDENSER AND REFLUX DRUM XO'=(V*Yl-(L+D)*Xo)/MD :LIQUID PHASE COMPONENT BALANCES FOR PLATES 1 TO 8 Xl'=(L*(XO-Xl)+V*(Y2-Yl))/M X2'=(L*(Xl-X2)+V*(Y3-Y2))/M X3'=(L*(X2-X3)+V*(Y4-Y3))/M X4'=(L*(X3-X4)+V*(Y5-Y4))/M :FEED PLATE

603

5.9 Distillation Process Examples

X5'=(F*X+L*X4-Ll*X5+Vl*Y6-V*Y5)/M X6'=(Ll*(X5-X6)+Vl*(Y7-Y6))/M X7'=(Ll*(X6-X7)+Vl*(Y8-Y7))/M X8'=(Ll*(X7-X8)+Vl*(Y9-Y8))/M :VAPOUR P H A S E EQUILIBRIA ~ 1 = ~ * ~ 1 / ( 1 + ( ~ - 1 ) * ~ 1 ) Y2=A*X2/(1+(A-l)*X2) Y3=A*X3/(1+(A-I)*X3) Y4=A*X4/(l+(A-l)*X4) Y5=A*X5/(1+(A-l)*X5) Y6=A*X6/(1+(A-l)*X6) Y7=A*X7/(1+(A-l)*X7) Y8=A*X8/(1+(A-l)*X8) :REBOILER A N D COLUMN BASE X9'=(Ll*X8-Vl*Y9-W*Xg)/MR Yg=A*Xg/(l+(A-l)*Xg) O U T P U T T,XO,X3,X6,X9 P L O T T, XO, 0, TFIN, 0,1 PREPARE T,X0,X1,X2.X3,X4,X5.X6,X7,X8fX9

Nomenclature Symbols M L V D X Y R

a

Molar holdup Liquid flow rate Vapour flow rate Distillate rate Liquid phase mole fraction Vapour phase mole fraction Reflux ratio Relative volatility

Indices 1 to 8 9 (in program)

B 0

Plate number Refers to flows below the feed plate Reboiler Reflux drum

kmol kmol/h kmol/h kmol/h

-

604

5 Simulation Examples of Chemical Engineering Processes ~

-~

Exercises 1.

Study the response of the column to changes in the operating variables feed rate, feed composition and reflux ratio.

2.

Verify that the overall steady-state balance is attained.

3.

How long does it take for the steady state to be reached?

4.

What influence does the reflux ratio have on the separation?

5.

In the program, the column starts up assuming an initial zero concentration profile of the more volatile component. Use the results of 1 ) to 4) to establish an initial steady-state concentration profile through the column and using this as the starting point, then carry out further simulations to study the effects of step changes in R, F, q and XF.

Results

Figure 5.224. Response caused by decrease of reflux ratio (R from 2.5 to 1.0) after startup.

Figure 5.225. Response of column to change in feed rate (F from 1000 to 1200).

605

5.9 Distillation Process Examples

MCSTILL - Continuous Multicomponent Distillation Column

5.9.5

System This problem is similar to CONSTILL except that three components benzene, xylene and toluene are considered. Thus, as explained in Sec. 3.3.3.4, each component of the mixture may be expressed by a separate component mass balance. Using mole fractions one balance can be omitted, and replaced by the condition that the sum of the mole fractions must be equal to unity. 'J, Y i l

Cooling water

0

L Xi

I I I

F, Xi

I

Benzene Toluene

xi

Xylene

1

al Y Y

L Steam

4I

Figure 5.226. The three-component, eight-plate, continuous distillation column.

606

5 Simulation Examples of Chemical Engineering Processes

Feed plate 5

'-1 Xi5

"1 yi6

Figure 5.227. The feed plate showing the streams and components i.

Model For component i on any nth plate above the feed plate

where i = 1 to j. Expressing the equilibria in terms of relative volatility values ai,

Program The program is rather slow in execution and therefore the model is limited to an eight-plate column, which is rather unrealistic for this multicomponent separation. The program is therefore given only for example purposes and a real simulation should involve rather more plates. As in BSTILL, the speed of calculation is also very sensitive to the magnitude of the liquid holdup on the plates, which again are large compared to normal practice. :Example

MCSTILL

:8-PLATE CONTINUOUS DISTILLATION COLUMN :BENZENE,XYLENE,TOLUENE SEPARATION

5.9 Distillation Process Examples

607

CONSTANT F=40 :COLUMN FEEDRATE CONSTANT Q=l :SATURATED LIQUID FEED CONSTANT BF=0.6,TF=0.25 :FEED COMPOSITIONS CONSTANT R=5 :REFLUX RATIO CONSTANT A1=2.75,A2=1,A3=.4 :RELATIVE VOLATILITIES CONSTANT MO=75,M9=150fM=30 :LIQUID HOLDUPS CONSTANT V1=150 :VAPOUR BOILUP RATE CONSTANT TFIN=50, CINT=2 SIM INITIAL :BENZENE STEADY-STATE COMPOSITIONS BO=.967;Bl=.914;B2=.813;B3=.65l;B4=.457;B5=.289 B6=.137;B7=.56E-l;B8=.2OE-l;B9=.60E-2 :TOLUENE STEADY-STATE COMPOSITIONS TO=.325E-l;T1=.849El;T2=.185;T3=.343;T4=.522 T 5 = . 6 4 9 ; T 6 = . 7 8 1 ; T 7 = . 8 1 7 ; T 8 = . 7 55 ; T 9 = . 5 9 4 :FLOW RATES IN COLUMN V=V1; D=V/(R+l); L=V-D; Ll=L+F; W=Ll-Vl DYNAMIC :CONDENSER AND REFLUX DRUM BO'=(V*VBl-(L+D)*BO)/MO TO'=(V*VTl-(L+D)*TO)/MO XO=l-BO-TO :LIQUID PHASE COMPONENT BALANCES FOR PLATES 1 TO 8 Bl'=(L*(BO-Bl)+V*(VB2-VBl))/M B2'=(L*(Bl-B2)+V*(VB3-VB2))/M B3'=(L*(B2-B3)+V*(VB4-VB3))/M B4'=(L*(B3-B4)+V*(VB5-VB4))/M :FEED P L A T E B5'=(F*BF+L*B4-Ll*BS+Vl*VB6-v*VB5)/M B6'=(Ll*(B5-B6)+Vl*(VB7-VB6))/M B7'=(Ll*(B6-B7)+Vl*(VB8-VB7))/M B8'=(Ll*(B7-B8)+Vl*(VBg-VB8))/M Tl'=(L*(TO-Tl)+V*(VT2-VTl))/M T 2 ' = ( L * ( T l - T 2 )+ V * ( V T 3 - V T 2 ) ) /M T3'=(L*(T2-T3)+V*(VT4-VT3))/M T4'=(L*(T3-T4)+V*(VT5-VT4))/M T5'=(F*TF+L*T4-Ll*T5+Vl*VT6-V*VT!j)/M :FEED PLATE T6'=(Ll*(T5-T6)+Vl*(VT7-VT6))/M T7'=(Ll*(T6-T7)+Vl*(VT8-VT7))/M T8'=(Ll*(T7-T8)+Vl*(VT9-VT8))/M X1=1-Bl-T1 X2=1-B2-T2 X3=1-B3-T3 X4=1-B4-T4 X5=1-B5-T5

608

5 Simulation Examples of Chemical Engineering Processes _-

..

X6=1-B6-T6 X 7 = 1 - B 7 -T7 X8=1-B8-T8 :REBOILER AND COLUMN BASE B9'=(Ll*B8-VI*VB9-W*Bg)/Mg Tg'=(Ll*T8-Vl*VT9-W*Tg)/Mg X9=1-B9-T9 :FACTORS I N VAPOUR-LIQUID EQUILIBRIA Fl=Al*BI+A2*Tl+A3*X1 F2=AI*B2+A2*T2+A3*X2 F3=AI*B3+A2*T3+A3*X3 F4=AI*B4+A2*T4+A3*X4 F5=AI*B5+A2*T5+A3*X5 F6=Al*B6+A2*T6+A3*X6 F7=AI*B7+A2*T7+A3*XI F8=AI*Ba+A2*T8+A3*X8 F9=AI*B9+A2*Tg+A3*Xg :VAPOUR COMPOSITIONS VBI=AI*Bl/F1; VTl=A2*T1/Fl; VXl=l-VBl-VTl VB2=AI*B2/F2; VT2=A2*T2/F2; VX2=I-VB2-VT2 VB3=AI*B3/F3; VT3=A2*T3/F3; V X ~ P I - V B ~ - V T ~ VB4=AI*B4/F4; VT4=A2*T4/F4; VX4-1-VB4-VT4 VB5=AI*B5/F5; VT5=A2*T5/F5; VX5=I-VB5-VT5 VB6=Al*B6/F6; VT6=A2*T6/F6; VX6=I-VB6-VT6 VB7=AI*B7/FI; VT7=A2*T7/F7; VX7=1-VB7-VT7 VB8=AI*Ba/Fa; V T ~ = A ~ * T ~ / F vxa=i-vBa-vTa ~ ; VB9=AI*B9/F9; VT9=A2*T9/F9; VXg=l-VBg-VTg OUTPUT T, BO, B3, B6, B9 PLOT T, BO, 0 , TFIN, 0,l T, BO,Bl,B2, B3, B4, B5, B6,B7,B8, B9,T9, X 9 PREPARE

Nomenclature Symbols M L V D X Y

R a

Molar holdup Liquid flow rate Vapour flow rate Distillate rate Liquid phase mole fraction Vapour phase mole fraction Reflux ratio Relative volatility

kmol kmollh kmollh kmollh

609

5.9 Distillation Process Examples

Indices I to 8 9 (in program) B 0

Plate number Refers to flows below the feed plate Reboiler Reflux drum

Exercises 1.

Study the response of the column to changes in the operating variables feed rate, feed composition and reflux ratio.

2.

Verify that the overall steady-state balance is attained.

3.

How long does it take for the steady state to be reached?

4. What influence does the reflux ratio have on the separation? 5.

Modify the program to increase the number of plates in the column.

Results

I e.mmn.8

15.8

I

3.0

45.8

60.0

Figure 5.228. Column response to a change in feed flow rate from F = 40 to F = 42.

I

8.845 8 . 8

15.8

I

3E.8

45.8

m.8

Figure 5.229. Response of reboiler benzene fraction to a change in feed flow rate from F = 40 to F = 42 to F=38.

5 Simulation Examples of Chemical Engineering Processes

610

BUBBLE - Bubble Point Calculation for a Batch Distillation Column

5.9.6

System As in Example BSTILL, a column containing four theoretical plates and reboiler is assumed, together with constant volume conditions in the reflux drum. The liquid behaviour is, however, non-ideal for this water-methanol system. The objective of this example is to show the need for iterative calculations required for bubble point calculations in non-ideal distillation systems, and how this can be achieved with the use of simulation languages.

L. lt:

x4

Steam

cI

Q f

t I

Figure 5.230. The four-stage batch distillation column.

Model Component balances for the more volatile component on any plate n give

61 1

5.9 Distillation Process Examples

for the still

where 4 refers to the bottom plate and B to the still and column base. For the reflux drum and condenser

where the reflux ratio, R, is given by Lo/D and D=-

V ( R + 1)

Throughout the column the total mass balance gives L = V-D Theoretical plate behaviour in the column is assumed, where the liquid-vapour equilibrium is described by Pi = 'Yi

Xi

Ppi

Here ppi is the vapour pressure of the pure fluid i, xi is its mole fraction in the liquid phase, "/i is its activity coefficient and pi is the actual vapour partial pressure. Vapour pressures ppi are given by

For the estimation of activity coefficients the Van Laar equation is used.

612

5 Simulation Examples of Chemical Engineering Processes

These equations represent an implicit loop to satisfy the condition

Since P is constant, the temperature will adjust at each plate such that the above equation is satisfied.

Program For convenience, only four stages were used in this model. An iterative solution is required for the bubble point calculations and this is based on the half-interval method. A FORTRAN subroutine EQUIL, incorporated in the ISIM program, estimates the equilibrium conditions for each plate. The iteration routine was taken from Luyben and Wenzel (1988). The program runs very slowly. :

Example

: :

Bubble Binary

BUBBLE Point Batch

Calculation Distillation

Column

CONSTANT BOIL-UP RATE CONSTANT MOLAR OVERFLOW AND CONSTANT PLATE : CONSTANT HOLDUP IN REBOILER AND SURGE DRUM CONSTANT M 0 = 1 0 0 :MOLAR HOLDUP IN SURGE DRUM CONSTANT MB=2500 :MOLAR HOLDUP IN STILL CONSTANT M = 2 0 :MOLAR HOLDUP ON PLATES CONSTANT XB=.8 :STILL CHARGE CONCENTRATION CONSTANT V=10 :VAPOUR BOIL-UP RATE CONSTANT R = l e l O :STARTUP A T TOTAL REFLUX CONSTANT Pbar=l :TOTAL PRESSURE CONSTANT Accur=le-2 :ACCURACY FOR SUBROUTINE CONSTANT CINT=0.2, NOCI=5, TFIN=300 1 SIM; INTERACT; RESET; GOT0 1 INITIAL : INITIAL CONCENTRATIONS x1=0;x2=0;x3=0;x4=0 MBXB=MB*XB CHARGE=MB : :

HOLDUP

5.9 Distillation Process Examples

613

DYNAMIC :CONDENSER AND REFLUX DRUM XO'=(V*Yl-(L+D)*XO)/MO D=V/ (R+1) L=V-D : PLATES 1 TO 4 Xl8=(L*(X0-X1)+V*(Y2-Yl))/M CALL EQUIL (Xl,T1, Y1, Pbar,ACCUR) X2'=(L*(Xl-X2)+V*(Y3-Y2))/M CALL EQUIL(X2,T2,Y2,Pbar,ACCUR) X3'=(L*(X2-X3)+V*(Y4-Y3))/M CALL EQUIL(X3,T3,Y3,Pbar,ACCUR) X4'=(L*(X3-X4)+V*(YB-Y4))/M CALL EQUIL(X4,T4,Y4,Pbar,ACCUR) :STILL AND COLUMN BASE WITH VARIABLE MB MB'=L-V MBXB1=(L*X4-V*YB) XB=MBXB/MB CALL EQUIL(XB,TB,YB,Pbar,ACCUR) : TOTAL DISTILLATE REMOVED DIST ' = D : FRACTION OF CHARGE DISTILLED FRAC=DIST/CHARGE : FRACTION OF STILL CHARGE DISTILLED P L O T T, XO, 0, TFIN, 0,1 OUTPUT T,X2,X4,XB,Y2,Y4,YB PREPARE T,XO,X1,X2,X3,X4,XB,R,Yl,Y2,Y3,Y4,YBlTl,T2,T3,T4,TB SUBROUTINE EQUIL (X,TEMP,Y,P,ACCUR) : BUBBLE POINT CALCULATIONS FOR METHANOL-WATER : USING INTERVAL-HALVING. DM=-4617.8 CM=13.676 DW=-5042.6 CW=13.519 A=0.85 B = O .48 XM=X XW=l-X GAMM=EXP(=A*XW*XW/((XW+A*XM/B)**2)) GAMW=EXP(=B*XM*XM/((XM+B*XW/A)**2)) 3 DT=2 LOOP=O FLAGM=l FLAGP=l 10 LOOP=LOOP+1

614 -

5 Simulation Examples of Chemical Engineering Processes ~~

IF (LOOP.GT.50) GOTO 100 PPM=EXP(=(CM+DM/(TEMP+273))) PPW=EXP(=(CW+DW/(TEMP+273))) PCALC=XW*PPW*GAMW+XM*PPM*GAMM IF (ABS(P-PCALC).LT.(P*Accur)) GOTO 100 IF ((P-PCALC).GT.O) GOTO 30 IF (FLAGM.GT.0) GOTO 22 DT=DT/2 22 TEMP=TEMP-DT FLAGP=-l GOTO 10 30 IF (FLAGP.GT.0) GOTO 32 DT=DT/2 32 TEMP=TEMP+DT FLAGM=- 1 GOTO 10 100 Y=XM*PPM*GAMM/P

Nomenclature Symbols

D Di

L M R V X

Y

a

Y

Coefficient in Van Laar equation Constant in the vapour pressure relation Distillate rate Constants in the vapour pressure relation Liquid flow rate Molar holdup Reflux ratio Vapour flow rate Liquid phase mole fraction Vapour phase mole fraction Relative volatility Activity coefficient

Indices 0 1 to 4 B

Reflux drum Plate number Reboiler

kmol/h kmol/h kmol kmol/h -

615

5.9 Distillation Process Examples

M

Methanol Water

w Exercises 1.

Investigate the response of the column to changes in the boilup rate, V

2.

The program starts up the column at total reflux (R very high). After steady state is reached on all plates, vary the reflux ratio interactively and attempt to carry out the distillation in minimum time, while attempting to maintain a distillate composition, so far as is possible, that xo > 0.9 and note the response of the distillate composition, xo.

3.

Recharge the column by changing XB and MB and note the transients on each plate.

4.

Devise a control scheme, such that the column maintains a distillate composition at xo > 0.9.

Results Outputs are shown below.

ISIU

; 1.a

ISIU

A ri B 12 c 13

D T4 E l l

78.8

m.ss.'sse

Figure 5.231. The column was started with total reflux; the reflux was then reduced to 10 at T = 20.

8.'158

0:w

8.458

8.

rE3

i

Figure 5.232. The temperature profiles here are from the run of Fig. 5.231. The highest temperature is in the still (TB) and the lowest is on the top plate (TI).

616

5 Simulation Examples of Chemical Engineering Processes

References Luyben, W.L. & Wenzel, L.A. (1988) Chemical Process Analysis, Prentice-Hall.

5.9.7

STEAM - Multicomponent, Semi-Batch Steam Distillation

System This example is based on the model description of Sec. 3.3.4, and involves a multicomponent, semi-batch system, with both heating and boiling periods. The compositions and boiling point temperatures will change with time. The water phase will accumulate in the boiler. The system simulated is based on a mixture of n-octane and n-decane, which for simplicity will be assumed to be ideal but which has been simulated using detailed activity coefficient relations by Prenosil (1976).

I)

v, Jil J,

-1

0 U

Figure 5.233. Schematic of the semi-batch steam distillation.

Steam MS "S

617

5.9 Distillation Process Examples

The model involves first a heating period and, when the boiling temperature is reached, and then a distillation period. All the steam is assumed to condense during the heating period, therefore the steam efficiency, E, equals 1. The standard state for enthalpy calculations is considered to be that of the pure liquid components at 0°C and 1 atm.

Program The heating period begins with FLAG set initially to zero. When cy > 1 then FLAG becomes 1, and the distillation period begins at statement 10. At each time interval the subroutine TCALC is used to make the iterative bubble point calculation. The component mass balance determines the removal of volatiles in the vapour, where the total molar flow rate, V, is determined from the energy balance. : : : : : :

Example STEAM Model of water steam distillation. Steam distillation: n-octane (comp.1) and n-decane (comp2). Ideal behavior of all components Bubble point calculation

______-------------....................

Simulation parameters

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

CONSTANT CINT = 0.1, NOCI = 10 CONSTANT TFIN = 40 :terminate at CONSTANT MFIN = 0.01

MFIN

[moll

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

Physical constants and model parameter

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

CONSTANT K = 273.1 :abs. Temp. at 0-C [Kl :gas constant [ccm*atm/ (mol*K)I CONSTANT R = 82.06 CONSTANT U = 15:Heat transf. coeff. [cal/(grad*min)] :Bottom area of vessel CONSTANT AREA = 10 :Initial water amount CONSTANT Mwo = 0 :Ambient pressure [Torrl CONSTANT P = 738 7.96681 :Constants Antoine-equation CONSTANT AW = CONSTANT BW = 1668.21 :water ( W ) and organic compounds CONSTANT CW = 228.0 :octane (1) und decane (2) CONSTANT A1 = 6.92374

61 8

5 Simulation Examples of Chemical Engineering Processes

CONSTANT

B1

=

1355.13

Bi CONSTANT C1= 209.517: loglO(Pi0) = Ai - - - - - - - - CONSTANT A2 = 6.95367 : Ci + Temp CONSTANT B2 = 1501.27 CONSTANT C2 = 194.48 :Pi0 [Torrl, Temp [Grad C] CONSTANT CPLW = 18.0 :Cp constants of liquid phase CONSTANT CPLl = 66.025:CPLi [cal/(mol*Grad K)] CONSTANT CPL2 = 83.67 CONSTANT AVW = 7.256 :Cp constants of vapor phase CONSTANT BVW = 0.002898 CONSTANT CVW = 2 . 8 3 3 - 0 7 :cpvi=AVi+BVi*Temp+CVi*Temp2 CONSTANT AV1 = 8.163 CONSTANT BV1 = 0.140217 :CpVi[Cal/(mOl K)] ,T[ K] CONSTANT CV1 = -4.41273-05 CONSTANT AV2 = 10.357 CONSTANT BV2 = 0.173551 CONSTANT CV2 = -5.48033-05 CONSTANT LW=10751.4:Lat.vapor.heat at 0 C[cal/moll CONSTANT L1 = 10274.1 CONSTANT L2 = 12917.5 CONSTANT G2 = 1.008 CONSTANT V1 = 178.647 :molar volume [ml/mol] CONSTANT V2 = 214.10 CONSTANT MS = 2.31 :Steam stream [mol/min] CONSTANT MO = 15.160 :Initial amount [moll CONSTANT TEMPS = 99.2 :Steam temp. [Grad Cl CONSTANT TEMPE = 25 :Ambient temperatur [Grad Cl CONSTANT TEMPO = 25 :Initial temp. [Grad Cl CONSTANT X01=0.725 :Init. mole fraction component 1 CONSTANT x02=0.275 :Init. mole fraction component 2 2 SIM; INTERACT; RESET; GOT02 INITIAL x1 = xo1 x2 = x02

- - - - - - - - - - - - - - - - --

:

Vapor

enthalpy

- -- -- -- -- -- -- -- -- -- -- -- -- --------

K2 = K*K K3 = K2*K TKS = K+TEMPS HS=LW+AVW*(TKS-K)+BVW*(TKS**2-K2)/2+CVW*(TKS**3K3)/3 TEMP = TEMPO

5.9 Distillation Process Examples

619

620

5 Simulation Examples of Chemical Engineering Processes

CALL

TCALC(Xl,X2,YW,Yl,Y2,Y,TEMP) CPL MW*CPLW+M*Xl*CPLl+M*X2*CPL2 :Component balances -MX1' -v*y 1 MX2 ' - -v*y2 MW' = MS - V*YW M = MX1 + MX2 x1 = MXl/M x2 = MX2/M IF (X1.LE.O) X1=0 IF (M.LE.0) M =O

- -___________--___- _ ----____ - -_- -- -_-_- - - - -

:

Enthalpy

and

mass

balance

. ._. ._. .- ._. -. .- .-.-.-._ - - - - -. .-.-.-.-.-.-. -. .- .-._. _

HLW = CPLW * TEMP :Enthalpy of water HL1 = CPLl * TEMP HL2 = CPLl * TEMP HL = X1 * HL1 + X2 * HL2 Q=MS*(HS-HLW) + U*(TEMPE-TEMP) :Heat input and loss TK = K+TEMP HVW=LW+AVW*(TK-K)+BVW*(TK**2-K2)/2+CVW*(TK**3-K3)/3 HVl=Ll+AVl*(TK-K)+BVl*(TK**2-K2)/2+CVl*(TK**3-K3)/3 HV2=L2+AV2*(TK-K)+BV2*(TK**2-K2)/2+CV2*(TK**3-K3)/3 HV=YW*HVW + Yl*HVl + Y2*HV2 :Enthalpy of vapor :Vapor rate from energy balance V=Q/(HV-HLW*YW-HLl*Yl-HL2*Y2) MlDIST = MO*XO1 - MX1 M2DIST = MO*X02 - MX2 MWDIST = MS*T-MW MDIST= MlDIST + M2DIST:Distilled 'organic phase [moll XlDIST = M1DIST/(MDIST+0.00001) X2DIST = M2DIST/(MDIST+0.00001) GOT0 20 20 CONTINUE TERMINATE M.LE.MFIN PLOT T,TEMP,O,TFIN,0,100 OUTPUT T ITEMP, Y, XI, X2, M PREPARE T ITEMP, Y 1 Y2, YW, Y, XI, X2, M, MW, VOL V SUBROUTINE TCALC(X1,X2,YW,Y1,Y2,YlTEMP) I

I

The program continues with a subroutine for the bubble point calculation, which is given on the diskette.

62 1

5.9 Distillation Process Examples

Nomenclature Symbols cP

H MS m or M m, or M, P P" PO

Q

T

TE

U V X

Y

Molar heat capacity Enthalpy Flow rate of steam Mass of organic phase in the still Mass of water in the still Total pressure Partial pressure Vapour pressure Heat transfer rate Absolute temperature Absolute ambient temperature Overall heat transfer coefficient Vapour flow rate Mole fraction in the liquid Mole fraction in the vapour

cal/mol K cal/mol mol/min mol mol Torr Torr Torr cal/min K K cal/K min mo1/min -

Indices Refers to components Refers to liquid Refers to steam Refers to vapour Refers to water Refers to gaseous Refers to liquid

1

L S V

w G L

Exercises 1.

Investigate the response of the system to changes in steam rate Ms.

2.

Change the initial composition by varying the mole fractions XOl and x02.

622

5 Simulation Examples of Chemical Engineering Processes -

Results

I

Figure 5.234. The dynamic changes of the organic mole fractions in the gas and liquid phases are seen here.

Figure 5.235. Temperature and water mass variations during the heating and distillation periods.

References I . Prenosil, J.E. (1976) Multicomponent Steam Distillation: A Comparison between Digital Simulation and Experiment, Chem. Eng. J. 12, 59-68, Elsevier Sequoia S.A., Lausanne.

5.10

Heat Transfer Examples

5.10.1 HEATEX - Dynamics of a Shell-and-Tube Heat Exchanger System A shell-and-tube heat exchanger is to be investigated for its dynamic and steady-state behaviour.

623

5.10 Heat Transfer Examples

1

I

I

I

2

3

4

5

6

I

7

8

Figure 5.236. Finite differencing the heat exchanger for dynamic modelling.

Figure 5.237. Detail of section n showing heat flows, volumes and transfer areas.

Model The model equations follow from Sec. 4.5.2.2. Here, except for the end sections, the average temperatures of the sections are used to calculate the convection terms. For section n, the rate of heat transfer from the tube contents to the metal wall is given by

The rate of heat transfer from metal wall to shellside content is

For the tubeside fluid

624

5 Simulation Examples of Chemical Engineering Processes

dTTn cpT PT V T T = FT Cp’T ( TTn-1

- TTn

- Q T ~

For the metal wall dTMn

CpM P M V M d t -

= QTn

-

Q M ~

For the shellside fluid

The respective volumes and area are calculated by

VT = A L ~ D T AT ~ , =ALTcDT V M = A L T c ( D M ~ - D T ~ AM ), = A L ~ D M As = AL n Ds

VS = AL TC (Ds2 - D & ,

where DT, DM, Ds are the respective diameters ,and AT, AM, As are the areas. The tube inlet and outlet end sections are modelled by averaging the temperatures entering the section. For example for the shell side section 1 Entering shellside temperature (averaged) = (Ts2 + T S l ) Leaving shellside temperature = So Thus,

Program :

Example

:DYNAMIC CONSTANT

SHELL MODEL

AND WITH

TUBE

8

HTM=20,HMS=20

HEAT

FINITE :

EXCHANGER DIFFERENCE

HEAT

EXCHANGE

ELEMENTS CONSTANTS

5.10 Heat Transfer Examples

CONSTANT DT=5E2,DM=7.5E2,DS=O.I:TUBE AND SHELL CONSTANT L = 2 : FINITE ELEMENT LENGTH CONSTANT C P T = Z , C P M = l , C P S = 2 : SPECIFIC HEATS CONSTANT FT=7.5E-3,FS=1.5E-2 : FLOW RATES CONSTANT TTO=375 : TUBE INLET TEMPERATURE CONSTANT TS9=293 : SHELL INLET TEMPERATURE CONSTANT R T = 1 5 0 0 , R M = 5 0 0 0 , R S = 1 0 0 0 : DENSITIES CONSTANT PI=3.1427 CONSTANT T F I N = 5 0 , C I N T = 2 1 SIM; INTERACT; RESET; GOT01 I N 1T IA L : INITIAL TUBE SIDE TEMPERATURES TT1=293;TT2=293;TT3=293;TT4=293 TT5=293;TT6=293;TT7=293;TT8=293 : INITIAL SHELL SIDE TEMPERATURES TS1=293;TS2=293;TS3=293;TS4=293 TS5=293;TS6=293;TS7=293;TS8=293 : INITIAL TEMPERATURES OF METAL WALL ELEMENTS TM1=293;TM2=293;TM3=293;TM4=293 TM5=293;TM6=293;TM7=293;TM8=293 : AREAS AND VOLUMES OF ELEMENTS SAT=PI*DT*L AM=PI*DM*L VT=PI*DT*DT*L/4 VM=PI*(DM*DM-DT*DT)*L/4 VS=PI*(DS*DS-DM*DM)*L/4 : COMMON FACTORS IN MODEL EQUATIONS YT=CPT*VT*RT ZT = F T / V T YM=CPM*VM*RS ZS=FS/VS YS=CPS*VS*RS UTM=HTM*AT UMS=HMS*AM DYNAMIC : ELEMENT 1 QTl=UTM*(TTl-TMl) QMl=UMS*(TMl-TSl) TSO=TS1+0.5*(TSl-TS2) TTl'=ZT*(TT0-0.5*(TT1+TT2))-QT1/YT

TMl'=(QTl-QMl)/YM TSl'=ZS*(0.5*(TSl+TS2)-TSO)+QMl/YS : ELEMENTS 2 TO 6 : TUBE HEAT TRANSFER RATES QT2=UTM*(TT2-TM2)

625 DIAM.

626

5 Simulation Examples of Chemical Engineering Processes

QT3=UTM*(TT3-TM3) QT4=UTM*(TT4-TM4) QT5=UTM*(TT5-TM5) QT6=UTM*(TT6-TM6) QT7=UTM*(TT7-TM7) : SHELL SIDE HEAT TRANSFER RATES QM2=UMS*(TM2-TS2) QM3=UMS*(TM3-TS3) QM4=UMS*(TM4-TS4) QM5=UMS*(TM5-TS5) QM6=UMS*(TM6-TS6) QM7=UMS*(TM7-TS7) : TUBE SIDE HEAT BALANCES TT2'=ZT*(TTl-TT2)-QT2/YT TT3'=ZT*(TT2-TT3)-QT3/YT TT4'=ZT*(TT3-TT4)-QT4/YT TTs1=ZT*(TT4-TT5)-QT5/YT TT6'=ZT*(TT5-TT6)-QT6/YT TT7'=ZT*(TT6-TT7)-QT7/YT : METAL WALL HEAT BALANCES TM2'=(QT2-QM2)/YM TM3'=(QT3-QM3)/YM TM4'=(QT4-QM4)/YM TM5'=(QTS-QM5)/YM TM6'=(QT6-QM6)/YM TM7'=(QT7-QM7)/YM : SHELL SIDE HEAT BALANCES TS2'=ZS*(TS3-TS2)+QM2/YS TS3'=ZS*(TS4-TS3)+QM3/YS TS4'=ZS*(TS5-TS4)+QM4/YS TS5'=ZS*(TS6-TSS)+QM5/YS TS6'=ZS*(TS7-TS6)+QM6/YS TS7'=ZS*(TS8-TS7)+QM7/YS : ELEMENT 8 QT8=UTM*(TT8-TM8) QM8=UMS*(TM8-TS8) TT9=TT8-0.5*(TT7-TT8) TT8'=ZT*(O.5*(TT7+TT8)-TTg)-QT8/YT TS8'=ZS*(TS9-0.5*(TS7+TS8))+QM8/YS TM8'=(QT8-QM8)/YM OUTPUT T ITT2 I TT4, TT6, TT8 PLOT T,TT8,O,TFIN,273,400 PREPARE T,TT2,TT4,TT6,TT81TS2,TS4,TS6,TS6,TS8,TM41TM6

627

5.10 Heat Transfer Examples

Nomenclature Symbols A C DM DT F HMS

Transfer area (AA = 71: D2 AL) Specific heat Outside diameter of metal wall Internal diameter of tube wall Mass flow velocity Film heat transfer coefficient from the wall to the shell Film heat transfer coefficient, tubeside Heat transfer rate Metal wall temperature in element n Shellside temperature in element n Temperature in element n, tubeside Volumes of shell side fluid, tubeside fluid and metal wall in length AL Density of the shellside fluid, tubeside fluid and metal wall

P

m2 kJkg K m m

m3/s kJ/m2 K s kJ/m2 K s kJ/S

K K K

kg/m3

Indices S M T 1, 2, ..., n

Refers to shell Refers to metal Refers to tube Refers to section

Exercises 1.

Establish that the overall steady-state balance is satisfied.

2.

Investigate the response of the system to a change in entering temperatures.

3.

Increase the heat transfer coefficients and observe the resulting trend in the temperatures T, M, and S.

4. Increase the flow of shellside fluid and observe the effect on the inlet shellside and outlet tubeside temperatures.

628

5 Simulation Examples of Chemical Engineering Processes

5.

Increase the mass of the wall and observe the influence on the dynamics.

6.

Derive a dimensionless form of the equations and thus obtain the important dimensionless groups governing the dynamic behaviour of the heat exchanger.

Results

R rT4 B TI

cw

I

LI. 2 1 8 . B

15.8

T

3.8

45.a

sB.8

Figure 5.238. Response of tube-side temperature for a step change in the inlet tube-side temperature at T=2 1.

5.10.2 SSHEATEX Exchanger

8.275 8.8

15.8

I

38.8

45.8

6n.B

Figure 5.239. Response of the shell, metal and tube temperature corresponding to Fig. 5.238.

- Steady-State, Two-Pass

Heat

System A purified water stream for a food processing plant is raised to required temperature by heat exchange with warm untreated process water. A conventional two-pass heat exchanger is to be used. This problem is based on the formulation of Walas (1991) and data taken from Backhurst, Harker and Porter ( 1974). In the first pass, both the hot and cold fluids flow in cocurrent flow through the heat exchanger, whereas in the second pass, the cold fluid now flows countercurrent to the hot shellside fluid. Half the heat exchange area is therefore in cocurrent flow and half in countercurrent flow.

629

5.10 Heat Transfer Examples

Thin

-

Th out

L

Tcin

1

I

Figure 5.240. Two-pass heat exchanger schematic, showing two passes for the cold fluid.

Model The derivation of the model equations follows that of Sec. 4.5.2.1. For the hot shellside fluid

For the cold side fluid in forward cocurrent flow

and for the cold side fluid in reverse countercurrent flow

For a given inlet and outlet cold water temperature, the outlet hot water temperature is determined by an overall heat balance where

630

5 Simulation Examples of Chemical Engineering Processes

Program Starting at the given temperature conditions Tc in, Tc out and T h in, the model equations are integrated to the condition that T,f = T,,, which must be true when the fluid reverses directions. It should be noted that a solution exists only for certain values of initial conditions and flow rates. :

Example

SSHEATEX

: Steady-State Heat Exchanger Constant U=1.5:Heat transfer coefficient [kJ/s m2 K] Constant CP=4.18 :Specific heat capacity [kJ/kg K] Constant WH=8.5 :Mass flow rate of hot fluid [kg/sl Constant WC=4.17:Mass flow rate,cold fluid [kg/sl Constant TCIN = 323:Inlet cold water temp.[Kl Constant TCOUT = 343:Outlet cold water temp. [Kl = 353:Inlet hot water temperature [Kl Constant THIN Constant CINT=0.5, TFIN=50 1 SIM; INTERACT; RESET; GOT01 INITIAL TCF=TCIN TCR=TCOUT TH=THIN DYNAMIC L=T :Length variable definition

:Heat balance for hot fluid TH' = -(U/(2*WH*CP))*(TH-TCR)-(U/(2*WH*CP))*(TH-TCF) :Heat balance for cold fluid in forward flow TCF' = (U/(2*WC*CP))*(TH-TCF) :Heat balance for cold fluid in reverse flow TCR' = -(U/(2*WC*CP))*(TH-TCR) :Check heat exchanger end condition if (TCF.ge.TCR) T=TFIN :stops run PLOT L,TCF, O,TFIN, 320,340 OUTPUT L,TH,TCR,TCF PREPARE L,TH,TCF,TCR

63 1

5.10 Heat Transfer Examples

Nomenclature Symbols A

Area Specific heat capacity Length Temperature Time Overall heat transfer coefficient Mass flow rate

cP

L T t U

w

in2

kJ/kg s m K S

kJ/m2 s K kgls

Indices Refer to hot and cold fluids Refer to the cold fluid in forward (cocurrent) and reverse (countercurrent) flow.

h, c f, r

Exercises 1.

Study the effects of varying flow rates of both hot and cold fluids on the heat transfer area required.

2.

Vary the hot fluid temperature in the range 383 to 353 K and study the effect on the area.

3.

The model assumes a constant specific heat for water. Assuming that this varies with temperature, according to cp = a + b T, how could this effect be implemented in the program, and what additional difficulties does it cause?

4. Is any advantage to be gained by changing the hot fluid to the two-pass arrangement?

632

5 Simulation Examples of Chemical Engineering Processes ~

Results

Figure 5.241. The parameters in the program calculate a length of 20 m to satisfy the conditions.

Figure 5.242. Choosing a value of THIN of 340 K does not satisfy the overall energy balance, but a value of 350 K does give a solution with L = 35 m. Here

References 1, Walas, S.M. (1991) "Modeling with Differential Equations in Chemical Engineering", Butterworth. 2. Backhurst, J.R., Harker, J.H. and Porter, J.E. (1974) "Problems in Heat and Mass Transfer", Edward Arnold.

5.10.3 ROD - Radiation from Metal Rod System A metal rod is in contact with a constant temperature source at each end. At steady state the heat conducted towards the center is balanced by the heat loss by radiation. This leads to a symmetrical temperature profile in the rod, as shown.

633

5.10 Heat Transfer Examples

Heat source at TI

.

Heat conduction

Heat source at TI

1

Radiation heat loss T

T1

I

0

u2

x-

L

Figure 5.243. The temperature profile is symmetrical for this heat conduction problem.

Model The model is based on the steady-state energy balance combined with Fourier's law which gives

where

a

= 1.25 x

Boundary conditions are TI = 2000Katx = 0 At the midpoint of the rod,

634

5 Simulation Examples of Chemical Engineering Processes -.

The second midpoint boundary condition is deduced from the symmetrical nature of the rod profile. L = 0.5 m.

Program Knowledge of the derivative at the centre of the rod requires a solution involving repeated estimates of the temperature gradient at the rod end. The integration proceeds from the end to the rod centre, defined by the condition TFIN = L/2 = 0.25, and a check made for the zero gradient. The initial estimate is revised at X = W 2 , accordingly to

The INTERACT; RESET command allows making the next iteration with GO. The value of the first derivative, TEMP ', at the end of each run can be checked with VAL. :Example ROD :TEMPERATURE DISTRIBUTION IN A RADIATING ROD : CONSTANT A=1.25003E-08 :RADIATION CONSTANT CONSTANT K = O . 05 :ITERATION PARAMETER CONSTANT TEMPO=2000 :INITIAL TEMPERATURE CONSTANT I=-15000 :INITIAL GUESS FOR SLOPE A T Z = L CONSTANT TFIN=0.25 :INTEGRATION LENGTH CONSTANT CINT=0.005 CONSTANT EPS=1000 1 SIM; INTERACT; RESET; GOT0 1 INITIAL I=I-K*TEMP' TEMP=TEMPO TEMP'=I DYNAMIC Z=T TEMP"=A*(TEMP**4) IF (TEMP.LE.0) T=TFIN TDIM=TEMP/TEMPO PREPARE Z,TEMP,TEMP',TEMP' ',TDIM T E R M IN A L IF (ABS (TEMP I ) LT. EPS ) PRINT "YOU HAVE THE SOLUTION"

.

635

5. I0 Heat Transfer Examples

Z , TEMP, TEMP ' ,TDIM OUTPUT PLOT Z,TDIM,O,TFIN,O,1

Nomenclature

Symbols T

K cal/min K4 cm2 caVcm K

Temperature Emissivity Conductivity Distance along rod

(3

k

X

cm

Exercises 1.

Experiment with the program by varying EPS and K to refine the final runs.

2.

Change the program to give automatic runs by removing INTERACT.

3.

Reformulate the problem in a dynamic form to be solved with respect to time variations along the rod.

Results

R

Figure 5.244. Runs from A to E represent iterations, each starting with new values of the initial temperature gradient at Z=O.

ISIM

Figure 5.245. The temperature gradient changes are shown here for each run, corresponding to Fig. 5.244.

636

5 Simulation Examples of Chemical Engineering Processes

5.11

Diffusion Process Examples

5.11.1 DRY

- Drying

of a Solid

System A thin slab of solid material dries first by evaporation from the top surface and then by diffusion from the interior of the solid. The water movement is approximated by the diffusion equation

Vapor

-- -I

4 f 4

Porous solid

Diffusionfluxes

-Figure 5.246.

Finite-differencing of the solid.

637

5.1 1 Diffusion Process Examples

Model Surface Drying Water is removed from the top surface where X = 0. Drying rate

Water mass evaporated t

W1 =

0

M1 dt

Depth of water at time t

where the initial water mass in the layer is

w = vop Solid Drying At the instant that the depth of the water layer is reduced to zero, the drying of the solid begins. The solid is considered as 10 finite difference elements, each of length AX. For the drying of element n, applying Ficks law,

v.2

= D( C n 2 i C n ) ~- D(

cn Ax - cn-l)

A

and

where Cn represents the concentration at the midpoint of element n and the gradient dCldX is approximated by ACIAX. The boundary conditions are

5 Simulation Examples of Chemical Engineering Processes

638

At X = 0, Co = constant, At X = L,

dC/dX = 0 corresponding to zero flux

Thus, Cl1 = Clo. The drying rate is obtained by approximating the concentration gradient at the surface

where C O is the water concentration of the outer solid in equilibrium with the air. The total water removed from the solid is

Total water removed in both stages of drying

The water remaining is W4 = W + ( A L C ) - W 3 The percent dryness is

Program Conditional control using the control variables Z and X and the ISIM "COMP" comparater function determines the end of the free surface drying and the start of the diffusional drying. :Example

DRY

:DRYING OF A NON-POROUS SOLID :A WATER LAYER EVAPORATES FIRST

5.1 1 Diffusion Process Examples

639

CONSTANT D=1.000E-03, L=l.O, CO=2.000E-05, C=5.000E03 CONSTANT U=1.000E-02, A=100 CONSTANT T2=20, T1=60 :WET AND DRY BULB TEMP. CONSTANT H=9700, V=10, RHO=5.000E-02 CONSTANT CI=5.0000E-03 CONSTANT TFIN=600, CINT=l INITIAL C1=CI; C2=CI; C3=CI; C4=CI C5=CI; C6=CI; C7=CI C8=CI; C9=CI Dl=CI X=L/10 Y=l DYNAMIC Z=COMP(Y-0.001) :Z IS "TRUE" INITIALLY Ml=Z*U*A*(Tl-T2)/H :Z CONTROLS EVAPORATION W 1 ' = M I :WATER LAYER BALANCE W = V * R H O :MASS OF WATER LAYER Y = ( W - W l )/ R H O * A :DEPTH OF WATER LAYER K = .NOT. Z :ANOTHER CONTROL VARIABLE Kl=K*D/(X*X) :DIFFUSION FACTOR WITH CONT. VAR. M2=K*A*D*(Cl-CO)/X :TRANSFER FROM SURFACE W2'=M2 :MASS OF WATER REMOVED FROM SOLID C11=K1*(C2-2*C1+CO) :INCREMENTED MASS BALANCES FOR SOLID C2'=Kl*(C3-2*C2+Cl) C3'=Kl*(C4-2*C3+C2) C4'=Rl*(C5-2*C4+C3) C5'=Kl*(C6-2*C5+C4) C6'=Klf(C7-2*C6+C5) C7'=Kl*(C8-2*C7+C6) CE'=Kl*(C9-2*CE+C7) C9'=Kl*(D1-2*C9+CE) D11=K1*(D1-2*D1+C9) :WALL BOUNDARY CONDITION W3=Wl+W2 :TOTAL WATER W4=W+A*L*C-W3 :WATER REMAINING P=W3*100/(W+A*L*C) :PERCENT WATER DRIED Rl=Z*Ml+K*M2 :COMPOSITE RATE OF DRYING OUTPUT T , R l r P , M 1 , M 2 PREPARE T,R1, P,Ml,M2,Cl,C3,C5,C7,C9,Dl PLOT T,C9,0,600,O,CI

640 _. .~

~

5 Simulation Examples of Chemical Engineering Processes -

Nomenclature Symbols Area Surface area Water concentration initially Surface equilibrium concentration Water concentration of solid Diffusion coefficient Latent heat of evaporation Depth of finite element Depth of solid Thickness Drying rate of water layer Drying rate of solid Percent dryness Water density Air temperature Water temperature (wet bulb) Heat transfer coefficient Volume of water layer Initial mass of water layer Water mass evaporated Water removed from solid Total water removed Remaining water Length variable Water layer depth

A A C CO C" D AH

AX L L MI M2

P P

Tg Tw

U V W

w1 w2 w3 w4

X Y

cm2 cm2 mole/cm3 mol/cm3 mole/cm3 cm2/min cal/mol cm cm cm mol/min mol/min -

mo~cm3 "C "C cal/cm2 "C min. cm3 mol mol rnol rnol mol cm cm

Exercise 1.

Run the program and graph the water concentrations in the solid versus time. Graph also the rate R1 versus percent dryness P.

2. The drying time for the solid can be characterised by L2/D, a diffusion time. Verify this by simulation by varying D and L separately.

3.

Improve the model by assuming a mass transfer rate at the solid surface, such that M2 = KT A C1. Assume a suitable value for KT.

64 1

5.1 1 Diffusion Process Examples

Results

A C1 B L 3

cu 0 c I E C 3

F

0.450

01

I

8.388

E.l_] I

8.888 8 . M

8.158

r

8.388

8.458

Figure 5.247. Concentration-time profiles in the solid.

0.m rE3

8.888 8.0

25.n

P

58.0

75.8

7

1W.B

Figure 5.248. Drying rate versus percent dryness. The peak is caused by the somewhat unrealistic model for the surface drying rate.

Reference Dunn, I.J. and Ingham, J. (1972) Verfahrenstechnik 399, 6.

5.1 1.2 ENZSPLIT - Diffusion and Reaction: Split Boundary Solution System A rectangular slab of porous solid supports an enzyme. For reaction, reactant S must diffuse through the porous lattice to the reaction site, and, as shown in Fig. 5.249, this occurs simultaneously from both sides of the slab. Owing to the decreasing concentration gradient within the solid, the overall rate of reaction will generally be lower than that at the exterior surface. The magnitude of the concentration gradient determines the effectiveness of the catalyst.

642

5 Simulation Examples of Chemical Engineering Processes

--

~~

Biocatalyst matrix

4-

-I+ Liquid

Substrate diffusion

I

Liquid

L+

+L+

Produc; diffusion

w PO

x=o

X=L

x=o

Figure 5.249. Symmetrical concentration gradients for substrate and product.

Model Under steady-state conditions Rate of diffusion of (reactant into the slab) =

Rate at which reactant

(is consumed by reaction

A quasi homogeneous form for the reaction term is assumed. The boundary conditions are given by AtX=L AtX=O

s

=

so,

P = Po

643

5.1 1 Diffusion Process Examples

The external concentration is known, and the concentration profile throughout the slab is symmetrical, The reaction rate for this enzyme kinetics example is expressed by the Michaelis-Menten equation and with product inhibition. kES R = KM (1 + P/KI) + S where k, KM and KI are kinetic constants and E and P are the enzyme and product concentrations. At steady state, the rate of diffusion of substrate into the slab is balanced by the rate of diffusion of product out of the slab. Assuming the simple stoichiometry S + P,

which on integration gives

P = (So- S ) DS DP where P is assumed zero at the exterior surface. Defining dimensionless variables

gives

where R' =

kES' (K~lSo)(+ l (So P'/Kl))

and P' = (1 - S') Dp/Ds The boundary conditions are then given by X ' = l

s'=1

+ S'

644

5 Simulation Examples of Chemical Engineering Processes

-

-~

X' = 0

dS'/dX' = 0

The catalyst effectiveness may be determined from

where Ro is the reaction rate determined at surface conditions,

Program The dimensionless model equations are used in the program. Since only two boundary conditions are known, i.e., S at X' = L and dS'/dX' at X' = 0, the problem is of a split-boundary type and therefore requires a trial and error method of solution. Since the gradients are symmetrical, as shown in Fig. 5.249, only one-half of the slab must be considered. Thus starting at the midpoint of the slab at X' = 0, where dS'/dX' = 0, an initial value for S' is assumed (SGUESS). After integrating twice, the computed value of S is compared with the known value of SO at X' = 1. A revised guess for S' at X' = I is then made using the half-interval method. This is repeated until convergence is achieved. Here a DO loop is used to cause a new integration with a new value of SGUESS, which is increased or decreased by (SO - SGUESS)/2 . :Example ENZSPLIT : DIFFUSION AND REACTION IN A : STEADY-STATE SPLIT BOUNDARY CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT CONSTANT

POROUS ENZYME SOLUTION

DS=3E-6, DP=6E-6, E=1.4E-3 K = 20, KM=3.2E-2, KI=9.4E-3 PO=O, So-1, L=O .01 CINT=O.O1, NOCI-5 TFIN=l, ENDIT=20 EPS=O.O1, SPLOT=1

:ITERATION ROUTINE,HALF-INTERVAL METHOD SGUESS=S0/2 SOGUESS=O DO I=l,ENDIT SIM (ABS(S0GUESS-SO).LT.EPS) I=ENDIT+l IF

CARRIER

5.1 1 Diffusion Process Examples

645

SGUESS=SGUESS+(SO-SOGUESS)/2 T=O END STOP : DIFFERENTIAL EQUATIONS DESCRIBING THE PROCESS 1 SIM; INTERACT; RESET; GOT0 1 INITIAL S=SGUESS DSDX=O DYNAMIC X=T :CHANGING TO LENGTH VARIABLE P=(l-S)*DP/DS :STEADY-STATE STOICHIOMETRY R = K * E * S / ( K M / S O ) / ( l + ( S O * P / K I ) + S ) :REACTION RATE

:INTEGRATING TWICE DSDX'=L*L*R/DS/SO S ' =DSDX OUTPUT X, S ,P ,DSDX PREPARE X, S ,P, DSDX PLOT X, S, 0, TFIN, 0, SPLOT TERMINAL S OGUESS=S RETURN : END OF MODEL

Nomenclature

Symbols D E K

k L P R S X rl

Diffusion coefficient Enzyme concentration Kinetic constant Reaction rate constant Distance from slab centre to surface Product concentration Reaction rate Substrate concentration Length variable Effectiveness factor

m2/h m01/m3 kmol/m3 l/h m kmol/m3 kmol/(m3 h) kmol/m3 m

-

646

5 Simulation Examples of Chemical Engineering Processes - .~

~

-~

Indices I M P S 1

0 GUESS

Refers to inhibition Refers to Michaelis-Menten Refers to product Refers to substrate Refers to dimensionless variables Refers to bulk concentration Refers to assumed value

Exercises 1.

Vary the parameters L, Ds, So, k and E, but maintaining the parameter [L2 k E/Ds So] at a constant value less than 0.1

2.

Vary the above parameters separately, noting the value of (L2 k E/Ds So) and its influence.

3.

Alter the program to calculate q . The necessary value of the slope at X = 1 is variable at each interaction, before each iteration. It can best be programmed in a TERMINAL section.

4.

Vary KI and KM and note their influence on the gradients and on q.

5.

Alter the program for no reaction by setting E = 0, and note the results. Run with various bulk concentrations.

6.

Write a model and a program for nth-order kinetics.

Results The graphical outputs in Figs. 5.250 and 5.251 exhibit the iterations necessary to satisfy the boundary conditions, involving integration from the centre at X = 0 to the outside at X = 1.

647

5.11 Diffusion Process Examples

A S

1.28

A P

1,28

ism

e. m

e. m

8-48

0.68

0.m

Figure 5.250. Twelve iterations were required to find the solution, using the parameters in the program. Note the interesting excursions in the region of X = 1, owing to values of S > 1. The final values at X = 1 were S = 0.996 and DSDX = 0.882.

Figure 5.251. The P profiles with excursions to negative values for the same case.

5.11.3 ENZDYN - Dynamic Diffusion and Enzymatic Reaction System This example involves the same diffusion-reaction situation as in the previous example, ENZSPLIT, except that here a dynamic solution is obtained, using the method of finite differencing. The substrate concentration profile in the porous biocatalyst is shown in Fig. 5.252.

648

5 Simulation Examples of Chemical Engineering Processes

Biocatalyst matrix Outside

Centre

i i i i i i i 1 1 1 1 I I I I I I I I I I

Liquid

SO

S1

s2

I

I

I

I

I

I 1 I i 1

I 1 I i 1

I 1 I i I

I I I i I

I I I i I

SO s4

I 1 I i 1

I 1 I i 1

X

0

L

Figure 5.252. Finite differencing for ENZDYN.

Model With complex kinetics a steady-state split boundary problem of the type of Example ENZSPLIT may not converge satisfactorily. To overcome this, the problem may be reformulated in the more natural dynamic form. Expressed in dynamic terms, the model relations become,

where at the centre of the slab dSIdX = dP/dX = 0

649

5.1 1 Diffusion Process Examples

The kinetic equation used here is an enzymatic Michaelis-Menten form with product inhibition

This behaves as a zero-order reaction if S>>KM and P << KI/KM. Using finite differencing techniques for any given element n, these relations may be expressed in semi-dimensionless form by

2P'n + P'n+1 AX12

)

R'n +

where

Here So is the external substrate concentration and AX' is the length of the finite difference element. Boundary conditions are given by the external concentrations So and Po and by setting SN+1 = SN and PN+1 = PN, at the slab centre. Catalyst effectiveness may be determined by two different methods, based on:

(a) an estimate of the slope of the reactant concentration, s, at the solid surface q l =

where

D,So

1-S' ( d)

kESo Ro = KM(I + PO/KI) + So

(b) the effectiveness factor based on the ratio of actual rate to maximum rate

q 2 = AX'

R1 + R2 + R3 + R4 + R5 + R6 + R7 + Rg RO

5 Simulation Examples of Chemical Engineering Processes

650

- ~-

The constant values are the same as used as in Example ENZSPLIT.

Program The numerical results of example ENZSPLIT should be essentially the same as the steady-state of ENZDYN. :Example ENZDYN :DYNAMIC SOLUTION FOR AN ENZYMATIC REACTION :WITHIN A POROUS SUPPORT CONSTANT D1=3.OE-06, D2=6.OE-06 :DIFF COEFFS : KINETIC CONSTANTS. CONSTANT K=8.7, K1=9.40E-03, K2~3.20E-02

.

.

CONSTANT E=1.40E-03 :ENZYME CONCENTRATION ON SUPPORT CONSTANT PO=o, S O = 1 :EXTERNAL CONCENTRATIONS CONSTANT L=l. OE-02 :1/2 THICKNESS OF SUPPORT CONSTANT TFIN=60,CINT=O.l CONSTANT PI=O,SI=o :INITIAL CONDITIONS 1 SIM; INTERACT; RESET; GOT0 1 INITIAL S1=SI;S2=SI;S3=SI;S4=SI;S5=SI;S6=SI;S7=SI;S8=SI Pl=PI;P2=PI;P3=PI;P4=PI;P5=PI;P6=PI;P7=PI;P8=PI X=1/8 :INCREMENT DELTA X=X*L Zl=K*E 2 2 =K2 /SO :DIMENSIONLESS KM Z3=SO/K1 :DIMENSIONLESS K I Z4=Dl/(L*L*X*X) :INVERSE DIFFUSION TIMES Z5=D2/(L*L*X*X) DYNAMIC :SUBSTRATE BALANCES FOR EACH ELEMENT THICKNESS X*L Sl'=Z4*(SO-2*Sl+S2)-Rl/SO S2'=Z4*(S1-2*S2+S3)-R2/SO S3'=Z4*(S2-2*S3+S4)-R3/SO S4'=24*(S3-2*S4+S5)-R4/SO S5'=24*(S4-2*S5+S6)-R5/SO S6'-24*(S5-2*S6+S7)-R6/SO S7'=Z4*(S6-2*S7+S8)-R7/SO S8'=Z4*(S7-S8)-R8/SO :SYMMETRY A T CENTER

5.1 1 Diffusion Process Examples

65 1 _

_

_

~

:PRODUCT BALANCES Pl1=Z5*(PO-2*P1+P2)+R1/SO P2'=Z5*(P1-2*P2+P3)+R2/SO P3'=Z5*(P2-2*P3+P4)+R3/SO P4'=Z5*(P3-2*P4+P5)+R4/SO P5'=Z5*(P4-2*P5+P6)+R5/SO P6'=ZS*(P5-2*P6+P7)+R6/SO P7'=Z5*(P6-2*P7+P8)+R7/SO P8'=Z5*(P7-P8)+R8/SO :SYMMETRY AT CENTER :RATE EQUATIONS Rl=Zl*S1/(Z2*(1+Z3*P1)+Sl) R2=Zl*S2/(Z2*(1+Z3*P2)+S2) R3=Z1*S3/(Z2*(1+Z3*P3)+s3) R4=2l*S4/(Z2*(1+Z3*P4)+S4) R5=Z1*S5/(Z2*(1+Z3*P5)+S5) R6=Zl*S6/(Z2*(1+Z3*P6)+S6) R7=Zl*S7/(Z2*(1+Z3*P7)+S7) R8=Zl*S8/(Z2*(1+Z3*P8)+S8) PREPARE T iS1, S3, s5,57, s8, p3, p 5 1 p 7 1 P 8 TERMINAL PRINT "CATALYST EFFECTIVENESS" RO=K*E*SO/(K2*(1+(PO/K1))+SO)

El=X*(Rl+R2+R3+R4+R5+R6+R7+R8)/Ro E2=(Dl*SO)*(l-Sl)/(L*L*RO*X) PRINT "El BASED ON REACTION IN EACH ELEMENT" PRINT "E2 BASED ON DIFFUSION RATE AT SURFACE'' PRINT "VALUES El AND E2 APPLY TO STEADY-STATE Ti S 1 S5 S8 1 P1, P5, P 8 i El r E2 OUTPUT PLOT T iS8,O I TFIN, 0,1

ONLY"

I

Nomenclature The nomenclature is the same as ENZSPLIT but with the additons as follows:

Symbols AX

q2 171 K2

Increment of length Effectiveness factor based on rates Effectiveness factor based on flux Same as KM

m

-

652

5 Simulation Examples of Chemical Engineering Processes

-~

-

Indices n

Refers to segment n

Exercises 1.

Vary K1 and KM and note the inhibition effects. It might be useful to plot the rates of reaction to see where inhibition is important.

2.

Vary E and L and note the effect on the gradients and q.

3.

Experiment with parametric changes, while keeping (Ds So/L2 k E) constant. How is q influenced?

4.

Vary L2/Ds, and note that this influences the time to reach steady state.

5.

Compare q 1 and q 2 after steady-state is definitely reached.

6.

Compare the steady-state results with those of ENZSPLIT.

7.

Verify the form of the dimensionless equations.

8.

Develop a model and program for a simple nth-order reaction.

Results The plots in Fig. 5.253 and 5.254 show the substrate and product concentrations in four elements approaching a steady-state from initial zero values.

653

5.1 1 Diffusion Process Examples ISIU

1.888

A P1

B P3 cF5 D P7 E.875 EP8

8.758

8.588

8.858

8.258

8.825

n.Lmn.8

ISIM

a. 1BB

15.8

I

38.8

45.8

68.8

Figure 5.253. A plot of the substrate profiles indicate substantial gradients, but the substrate concentrations are not low enough to influence the rates.

E.Lm0.8

H

15.8

I

38.8

45.0

Figure 5.254. The product profiles indicate that no inhibition is present, since P is everywhere less than KI/KM.

References Blanch, H.W. and Dunn, I.J. (1973) Modelling and Simulation in Biochemical Engineering. in Advances in Biochemical Engineering Eds. T.K. Ghose, A. Fiechter, N. Blakebrough, Vol. 3, Springer. Goldman, R., Goldstein, L. and Katchalski, C.I. (1971) in Biochemical Aspects of Reactions on Solid Supports, Ed. G.P. Stark, Academic Press.

5.11.4 BEAD - Diffusion and Reaction in a Spherical Bead System Biological flocs and films, which are present in waste treatment systems and in fermentations, are subject to potential oxygen limitation.

654

5 Simulation Examples of Chemical Engineering Processes Sphere

Oxygen

ml

n+l

A

Product

AV

I

Rp

)I

0

Rp

Figure 5.255. The finite differencing of the spherical bead geometry.

Diffusion and reaction takes place within a bead of volume 4/3 .n Rp3 and area 4 .n Rp2. It is of interest to find the penetration distance of oxygen for given specific activities and aggregate diameters. The system is modelled by taking small spherical shell increments of volumes 4/3 n (rn3- rn-l3). The outside area of the nth increment is 4 n rn and its inside area is 4 71: rn-l 2 , The diffusion fluxes are considered as shown below:

Figure 5.256. The diffusion fluxes entering and leaving the spherical shell.

Model Here the single limiting substrate is taken to be oxygen.

655

5.1 1 Diffusion Process Examples ~~

~

The oxygen balance for any element of volume AV is given by

The diffusion fluxes are j,

=

Ds-

Ar

Sn

Substitution gives

The balance for the central increment 1 (solid sphere) is

4

~n (rl

3>

4 -2 jl 4 n r l * + RsnTn (rl

dS

=

3,

Since rl = Ar, this becomes

_dS1_ -_ ~ dt Ar

(S2 - Sl) + R s ~

The reaction rate is expressed by a Monod-type equation

where OUR is the oxygen uptake rate.

Program In the program the segments are numbered from 1 to 6 (outside to centre). :Example

BEAD

:Diffusion and reaction within a spherical Constant D=3.OE-06 :Diffusion coefficient Constant ~ = 0 . 1 :Radius of aggregate ( m )

bead (m2/h)

656

5 Simulation Examples of Chemical Engineering Processes

Constant OURmax=O.Ol :Max. oxygen uptake rate (g/kg h) Constant Ks=l.O :Saturation constant (kg/m3) Const ant X = 1 :Biomass concentration (kg/m3) Constant TFIN=1500,CINT=101NOCI=2 1 SIM; INTERACT; RESET; GOT0 1 INITIAL SO=1O;S1=1;S2=1;S3=1;S4=1;S5=l;S6=1 RS1=O;RS2=O;RS3=O;RS4=O;RS5=O;RS6=0

deltar=R/6 r6=deltar r5=2*deltar r4 = 3 * d e l t a r r3=4*deltar r2 = 5 *deltar rl=R DYNAMIC :SUBSTRATE BALANCES FOR EACH DELTA R ELEMENT S1'=3*D*(((rl**2)*(SO-Sl))-((r2**2)*(Sl-S2))) S11=S11/(deltar*((rl**3)-(r2**3)))+RS1 S2'=3*D*(((r2**2)*(Sl-S2))-((r3**2)*(S2-S3))) S2'=S2'/(deltar*((r2**3)-(r3**3)))+RS2 s31=3*D*(((r3**2)*(S2-S3))-((r4**2)*(S3-S4))) S3'=S3'/(deltar*((r3**3)-(r4**3)))+RS3 s4'=3*D*(((r4**2)*(S3-S4))-((r5**2)*(S4-S5))) S4'=S4'/(deltar*((r4**3)-(r5**3)))+RS4 S5'=3*D*(((r5**2)*(S4-S5))-((r6**2)*(S5-S6))) S5'=S5'/(deltar*((r5**3)-(r6**3)))+RS5 S6'=3*D*(S5-S6)/(deltar**2)+RS6 IF (Sl.le.O.0) S1=0 IF (S2.le.0.0) S 2 = 0 IF (S3.le.0.0) 5 3 = 0 IF (S4.le.0.0) S 4 = 0 IF (S5.1e.0.0) S5=0 IF (S6.le.0.0) S6=0 :RATE EQUATIONS RSl=-OURmax*X*Sl/(Ks+Sl) RS2=-OURmax*X*S2/(Ks+S2) RS3=-OURmax*X*S3/(Ks+S3) RS4=-OURmax*X*S4/(Ks+S4) RS5=-0URmax*X*SS/(Ks+S5) RS6=-OURmax*X*S6/(Ks+S6) PREPARE TIS1,S2,S3,S4,S5,S6 OUTPUT T,Sl,S2,S4,S5,S6 PLOT T, S5,O ,TFIN, 0,s

657

5.1 1 Diffusion Process Examples

Nomenclature Symbols D KS

OUR (max) r RP

RS S X Ar (deltar)

Diffusion coefficient Saturation constant in Monod equation Maximum oxygen uptake rate Radius at any position Outside radius of sphere Reaction rate in the Monod equation Oxygen substrate concentration Biomass concentration Increment length, 1-16

cm2/s mg/L l/s cm cm mg/s L mg/L mgk cm

Indices Refers Refers Refers Refers Refers

1 2 n P S

to segment 1 to segment 2 to segment n to particle to substrate

Exercises 1.

Investigate the response of the floc to changes in the outside to the bulk concentrations. Can you relate the time constant to D/r2?

2.

Note the position of zero oxygen when changing Do2 and r.

3.

Change the program to calculate the effectiveness factor. Hint: this can be done two ways, using diffusion fluxes and reaction rates.

4.

Write the model in dimensionless form. What are the governing parameters for first and zero-order kinetics? Verify by simulation.

5.

Compare the results for a sphere with those for rectangular geometry taking the thickness as Vsphere/Asphere.

658

5 Simulation Examples of Chemical Engineering Processes ~

~

Results

A s1 B 52

c

s3

111.00

1

ISIM

Ds4

E S5 F S6

Figure 5.257. Oxygen concentration time profiles at six different positions in the floc.

0

1

2

3

4

5

6

Dimensionless distance Figure 5.258. Dissolved oxygen profiles at time T = 1500 for three different floc sizes, all other conditions identical. Note how the larger floc becomes oxygen depleted at its centre. These results were replotted from the steady-state results.

659

5.1 2 Dynamic Numerical Examples

5.12

Dynamic Numerical Examples

5.12.1 LORENZ - Random Differential Equation Behaviour System Walas (1991) has shown that the following set of ordinary differential equations gives random chaotic behaviour.

Model XI' = s (- Xl

+ x2)

x2' = r x l - x2 - X I x3 ~ 3 = ' -b ~3

+ X I ~2

Program The values of the parameters b, r and s and the initial values for XI,x2 and x3 in LORENZ are identical to those employed by Walas.

As noted by Walas, the solution exhibits random behaviour and the solution jumps erratically between two critical point regions, as shown very clearly by plotting the variables as a phase-plane plot. :Example L O R E N Z :RANDOM BEHAVIOUR S H O W N BY ORDINARY :PROBLEM OF WALAS CONSTANT B=2.6666667, R=28, S=10 C O N S T A N T TFIN=30, CINT=0.5 1 SIM; INTERACT; RESET; G O T 0 1 INITIAL X1=5;X2=5;X3=5

DIFF.EQNS

660

5 Simulation Examples of Chemical Engineering Processes

DYNAMIC Xl'=S*(-Xl+X2) X2'=R*Xl-X2-X1*X3 X3'=-B*X3+Xl*X2 PLOT T,Xl,O,TFIN,-30,30 P R E P A R E T, XI, X2, X3 O U T P U T T, XI, X2, X3

Nomenclature The variables are

XI,

x2 and x3. No specific nomenclature is necessary.

Exercises I.

Enjoy playing with this system using different parameter values, including CINT, and the differing numerical integration routines. For further study of the phenomenon read the references of Walas (1991) and Denn (1987).

2.

Matko, Karba and Zupancic (1992) quote the Lorentz equations in the form

x' = CT (y - x) y' = ( 1 + h - z ) x - y

where (r = 10, h = 24 and b = 2 and the initial conditions are x(0) = y(0) = z(0) = 0.1. Program the equations in ISIM and study the resulting phase-plane response.

66 1

5.12 Dynamic Numerical Examples

Results

A X1

ISIM

ISIN

28.8 1. n

E.E

-18.8

I

-28.E E. E

I

1.5

1

15.8

22.5

33.1

Figure 5.259. The chaotic behaviour is apparent.

-28.8-38.8

-15.E

x2

8.8

15.8

38.8

Figure 5.260. The run of Fig. 5.259 is plotted as a phase-plane of X1 versus X2.

References 1. Walas, S.M. (1991) Modeling with differential Equations in Chemical Engineering, Butterworth-Heinemann Series in Chemical Engineering. 2. Denn, M.M. (1987) Process Modeling, Longman. 3. Matko, D., Karba, R., and Zupancic, B. (1992) Simulation and Modelling of Continuous Systems: A Case Study Approach, Prentice-Hall.

5.12.2 HOPFBIF - The Hopf Bifurcation System Differential equations are known for their sensitivity to parametric changes. An example of a simple but extremely sensitive model can be investigated here.

662

5 Simulation Examples of Chemical Engineering Processes

Model Walas (1991) quotes the following set of equations and notes its sensitivity to the magnitude of the parameter, A. XI'

= -x2

~ 2 =' -XI

+ X I (A - (xi2 - x**) ) + ~2

(A

-

(xi2 - ~

2) ~

)

Program :Example

HOPFBIF

:THE HOPF BIFURCATION OR THE CHANGING NATURE :EQUILIBRIUM POINTS: PROBLEM OF WALAS CONSTANT A=O CONSTANT TFIN=100, CINT=O.1 1 SIM; INTERACT; RESET; GO TO INITIAL Xl=-l;X2=-1 DYNAMIC Xl'=-X2+Xl*(A-(Xl*Xl+X2*X2)) X2'=Xl+X2*(A-(Xl*Xl+X2*X2)) PLOT Xl,X2,-1,1,-1,1 PREPARE T, X1, X2 OUTPUT T, XI, X2

OF

1

Nomenclature The variables are X1 and X2. No specific nomenclature is necessary.

Exercises 1.

Vary the value of A from -1.0 to +1.0 and examine the changing stability of the system as shown on a phase-plane diagram.

2.

Vary A interactively during the course of a simulation and note how quickly the system stability changes.

663

5.12 Dynamic Numerical Examples

Results

A X Z

1.588

-

ISIN

0.2918.888 -

-8.258/ 1

1.888

Figure 5.261. For A = 0.0 the solution gives the above result.

Figure 5.262. This result was obtained by changing A interactively many times in the range from - 0.5 to + 0.5.

Reference Walas, S.M. (1991) Modeling with Differential Equations in Chemical Engineering, Butterworth.

5.12.3 CHAOS

-

Chaotic Oscillatory Behaviour

System Variations in the values of the parameters will produce attractive oscillations. The equations are d2x dt2

-=

where and b is a constant.

dx - a X + b s

a = a1 t + b l

664

5 Simulation Examples of Chemical Engineering Processes

Program :

Example

: :

Chaotic oscillatory behaviour No physical significance

Constant Constant Constant Constant Constant 1

SIM;

CHAOS

Al=1 B1=0.1 B=-0.1 Xin= 5 CINT=O.l,TFIN=100,NOCI=20 INTERACT;

INITIAL

RESET;

GOT01

X=Xin

DYNAMIC A=Al*T+Bl X"=-A*X+B*X' PLOT T IX, 0, TFIN, - 5 , s OUTPUT T,X PREPARE T,X,X' ,X' '

Nomenclature The variables are X and its first and second derivatives. nomenclature is necessary.

No specific

Exercises I.

Vary A, A l , B, and B1 individually to see the influence of each. Plot the variables versus time and as phase planes.

2.

Experiment with time variations of the parameters, for example, with sine wave variations for B1.

3.

Matko, Karba and Zupancic (1992) also quote the following chaotic system equations

665

5.12 Dynamic Numerical Examples

dx dt

- = y d y x (c - x2) - ay dt =

-.

+ b cos wt

with a = 0.05, b = 7.5, c = 0 and w = 1 and initial conditions x(0) = y(0) = 0.

Program this system using ISIM and study the response.

Results

A X

8.W

1

ISIN

ox'

A X'

om1

ISIH

-

4.98

4.88

8.88

0.m-

-4.W

4.88 -

Figure 5.263. Results from CHAOS may look like this.

Figure 5.264. A phase-plane plot of the run in Fig. 5.263.

Reference 1 . Matko, D., Karba, R., and Zupancic, B. (1992) Simulation and Modelling of Continuous Systems: A Case Study Approach, Prentice-Hall.

Chemical Engineering Dynami :Modelling

with PC Simulation

John Inph;m. Iruine J. Dunn. Elmar Heinzlc &JiiiE. Picnoul

copyright OVCH VerlagsgesellschaftmbH, 1994

Appendix: Using the ISIM Language 1

ISIM Installation Procedure

1. Computer requirements: DOS 2.00 or higher a) b) Hard disk Free RAM required: 0.25 Mbyte c) Supported graphics are VGA, CGA and EGA. d)

2. Before installation: a) Check your computer configuration: 1) This ISIM version requires a hard disk. 2) Do you have a 8087 mathematical coprocessor ? 3) What kind of graphics adapter do you have? Make a copy of the original diskette. b)

3. The installation diskette contains the following 91 files: INS-ISIM .BAT INSTALL .EXE ISIM86 .EXE ISIM87 .EXE ISIMREF .DOC UPDATE .DOC plus all of the 85 simulation examples: FILE.SIM

4.Installing ISIM 2.06 and the simulation examples. This version of ISIM requires the following: a hard disk, with 0.25 Mbyte of free space; approximately 350 Kbyte memory; and a graphics adaptor of the types CGA, EGA or VGA. Place distribution disk in drive A. Type:

A:INSTALL

The installation program will prompt you for the following information:

668

Appendix: Using the ISIM Language

~

~~~

-~

a)

Directory name which is created for the ISIM program, and the library files , e.g., C:\ISIM.PRG

b)

Directory name for the ISIM start file, this can be same as the directory in (a), or you may specify any directory in the "path". Such a directory must exist!

c)

Type of screen; VGA, EGA, CGA.

d)

Whether the computer has a numerical coprocessor.

Following installation, the directory for the ISIM start file must appear in the "path". If necessary edit the C:\AUTOEXEC. BAT file to include the directory specified in (b) above. It is important that only one path command exists.

To call the ISIM program, type:

ISIM If you need to place the distribution diskette in drive B, proceed as follows: Use the DOS "SUBST" command to re-assign the drives so the ISIM disk is in assigned drive A:. That is: (*)

Place ISIM diskette in drive B: then type: SUBST A: B:\ A: INSTALL

Use Ctrl-Alt-Del to re-boot the system (to reset A: and B: drives).

2 Programming with ISIM

2 2.1

669

Programming with ISIM Getting Started

To enter ISIM, simply type ISIM from DOS and press ENTER. After a few seconds, a dollar sign prompt should appear. The user now has two options; to read an existing file from disk, or to create a new model. To exit ISIM and return to DOS type QUIT or the abbreviation Q.

2.2

Reading Files from Disk

To read a file from the current directory into ISIM type READ (can be abbreviated to just R) followed by the filename. A file type specifier of SIM is assumed and need not be entered. Files from directories other than the current one can be read in by specifying the full DOS filename. Examples:

READ SIMULATE READ A:MODELS\SIMULATE R SIMULATE.SIM

The file that is being read is displayed on the screen. The user can then decide whether to run the simulation as it stands or modify the model. Only one model can be operational at any one time.

2.3

Running Simulations

Files that have been previously read into ISIM can be run by typing START after the $ prompt. A graph of one variable against time is plotted on the screen, and numerical values of other variables may also be displayed. In Section 8 it will be shown how to control this output. Runs with complex examples may sometimes fill up a hard disk. The variables in the PREPARE statement should be reduced and the run repeated.

670

2.4

Appendix: Using the ISIM Language

Interacting with ISIM Simulations

An ISIM simulation can be interrupted by hitting any key. The user then has a choice of one of four options. HOLD:

This option allows the value of variables to be be modified or inspected using the VAL command. For example, VAL K I , VAL C3 would give the current values of the variables K l and C3. The command VAL KI=100 would change the value of the variable K1 to 100. The run can be continued by typing GO or can be restarted by typing START. The variable TFIN can also be changed in this way to allow longer or shorter runs. The command XVAL resets all variables to their original value. VAL will list the values of all variables and is useful for debugging.

QUIT: SKIP: COMMAND:

Returns the user to DOS. Current run is aborted; next run commences. Current run is aborted; control is passed to user allowing values to be modified or inspected as with HOLD.

Note: Only the first letter is required (i.e., H, Q, C or S). Any other key causes the simulation to continue.

2.5

Editing ISIM Files

ISIM files can be edited either by exiting to DOS, using any convenient editor, and then returning to ISIM and rereading the file or by using the ISIM line editor. After completion of an edit, the program should be returned to command mode by typing the $ sign followed by return. The important commands of the ISIM line editor are: APPEND:

The current line is added to the end of the program. This is also used to start writing a new program.

2 Programming with ISlM

67 1

-

LIST:

Displays the whole of the program. By specifying a number, or range of numbers before the LIST command, the number of lines displayed can be modified. For example, LIST 25.301

INSERT:

The current line is inserted at the line number specified, e.g., 25INSERT or 251.

CHANGE:

Specify the line that needs to be changed, e.g., 12C. After the question mark prompt, type in the new line. Specify the line that needs to be deleted, e.g., 12D. After the question mark prompt, type in the new line.

DELETE: CLEAR:

The screen is cleared.

SAVE:

The file is saved as the specified filename in the default directory, unless otherwise stated. For example SAVE SIMULATE SAVE C:MODEL\SIMULATE Care must be taken when saving; a changed file will replace the old one if the same name is used. Saving is advisable before running a new example.

KILL:

The current file is removed and a new file can now be read in. It must be used when a new file is read.

QUIT:

Return to DOS.

As an editing example, a simulation file named BATFERM is given below. 1 2 3 4 5 6 7 8 9 10 11 12 13

:FILE, BATFERM : BATCH GROWTH WIIH PRODUCT FORMATION CONSTANT UM=0.3, KS=O.1, K1=0.03, K2=0.08, Yr0.8 CONSTANT CINT=2, TFIN =40 1 SIM; INTERACT; RESET; GOT0 1 INITIAL X=O.Ol S=10 P=O DYNAMIC : BIOMASS BALANCE X'=RX : SUBSTRATE BALANCE S'=RS : PRODUCT BALANCE P'=RP

672 14 15 16 17 18 19 20 21

Appendix: Using the ISIM Language

RX=U*X U=UM*S/(KS+S) RS=-RX/Y RP=(Kl+K2*U)*X IF (S.LT.O.0) S=O.O PLOT T,P,0,40,0,6 OUTPUT T,S,X,P PREPARE T,X,SIP

BIOMASS RATE EQUATION EQUATION : SUBSTRATE RATE EQUATION : PRODUCT RATE EQUATION :

: MONOD

Assuming it is desired to change the kinetics term to include substrate inhibition, then one would proceed as follows after first typing $. $ 1 5 CHANGE (enter) ? U = UM*S/(KS+(S*S/KI)) (enter) $ LIST (This gives the new listing, where it is seen that KI is not defined) (enter) $ 5 INSERT (enter) ? CONSTANT KI=5 (This inserts a new line 5 for the KI value) (enter) $ LlST (The changed program can be inspected) (enter) $ START (enter)

2.6

Writing ISIM Models

General Comments Start the program writing with $APPEND, and the "?" prompt will appear. When writing your own program, it is best to follow an existing program for guidance. ISIM programs usually consist of four sections, namely CONTROL, INITIAL, DYNAMIC and TERMINAL. In addition, ISIM programs can contain blank lines, comment lines and end-of-line sections, which must start with a colon. Any line (except PLOT and OUTPUT statements) may be labelled by a number (1 to 8 digits) before the statement. More than one statement may be written on one line if each statement is separated by a semicolon. ISIM does not distinguish between upper and lower case characters. Derivatives with respect to time are indicated b y an apostrophe, for example X', P', S'.

2 Programming with ISIM

673

a) Always save a new program, using a suitable name, before trying to run it; a failure, such as numerical overflow, could cause the computer to quit ISIM, and your program will be lost unless previously saved. b) Debugging is easiest when all the important variables are in the PREPARE statement. A list of all the calculated values can be obtained at any time by VAL. Any individual value can be obtained by VAL "variable name". A parameter value can be changed with VAL = "parameter name". c) When changing an existing program, save it under a different or related name, otherwise the original program will be replaced. d ) When changing an existing program list the program after every change to see if it is correct. e) Returning to the original constant values is easiest with XVAL. f ) Problems with integration are often caused by errors in the model equations or program. Reducing the value of CINT may sometimes solve problems of numerical overflow at the beginning of an integration. Changing from the default integration method using ALGO will not usually give a faster integration.

g) ISIM is best learned by studying and running the examples.

Program Structure CONTROL section This section must be started by a SIM statement. Commands in this section are executed after the simulation has been completed and include commands for printing messages and resetting values to their original values. INITIAL section In the this section the initial values of variables can be defined or calculated by the INITIAL statement. DYNAMIC section All differential equations and all calculations to be performed at each time interval must be in this section. ISIM does not sort the equations, so place them so that the calculations can proceed from the initial conditions. TERMINAL section End of run calculations and print instructions are placed here.

674

Appendix: Using the ISIM Language

Subroutines can be included in ISIM programs. The simulation examples MIXFLO1, STEAM and BUBBLE should be referred to.

Reserved Variables The variable names listed below are used as ISIM system variables and can not be used for any other purpose. T The independent variable. The value of T at which the run is terminated. TFIN ALGO Selects the integration method. 0 variable step. 1 4th-order fixed step. 2 2nd-order fixed step. 3 implicit (stiff) 2nd-order fixed step. The calculation interval. This is the integration step length CINT for fixed-step methods. It also influences the first step of all methods, and if it is too large an error with crash may occur when running new programs - so save them first. Integration difficulties can often be solved by reducing CINT. The values of the variables displayed on the screen and NOCI defined in the OUTPUT statement are updated at a time interval of CINT * NOCI. NOCI must be a positive integer. The permitted per-unit relative error for variable-step RERR integration.

2.7

ISIM Statements and Functions

Arithmetic Functions Y=ABS(=X) Y=SIN(=X); Y=cos(=X); Y=ATAN(=X) Y =EXP(=X) Y=INTG(=X,Y0)

Absolute value X in radians I, 1,

ex Integrates X with respect to time with Y=YO at time T=O. Must be in the dynamic section

675

2 Programming with ISIM

Y=RAND(=X) Y=LOG(=X) Y=SQRT(=X) .LT. .GT. .EQ. .LE. .GE.

*

I

**

Returns a pseudo-random number in Y in the range 0 to 1. X is the seed for the random number generator Natural logarithm X must be positive

< >

Less than or equal to Greater than or equal to Multiply Divide Exponent

ISIM Statements GOTO

Transfers control to the line indicated by the label, e.g., GOTO 7 ...code ... 7 RESET

IF

Conditional execution of a statement, e.g., IF (OUR.LT.20) AFR=AFR+l

DO

Allows repeated execution of a group of statements, e.g., DO J=O.l, 1.0, 0.2 ...code.. . END The above loop is executed 5 times with values of J= 0.1, 0.3, 0.5, 0.7 and 0.9.

CONSTANT X=?

Variables defined as CONSTANT may be modified interactively via the VAL command.

RESET

The values of all variables are reset to the values defined in the INITIAL section.

676

Appendix: Using the K I M Language

INTERACT

2.8

A simulation can be stopped at any point by the INTERACT command, for instance when a variable exceeds a certain value. The user can then change the values of any variables by the VAL command as described in section 4.

Output

Output to the Screen The PLOT statement usually appears in the DYNAMIC part of the model and controls the output to the screen during a run after a START command. Only two variables may be specified. The format of this command is PLOT x variable, y variable, xmin, xmax, ymin, ymax The GRAPH keyboard command, however, allows any number of variables to be plotted against one other variable. A GRAPH command can be given after a simulation has finished, or after an interrupt. The axes can be defined for only one pair of variables or will be automatically assigned. Only variables that have previously appeared in a PREPARE statement can be used in a GRAPH statement. If the simulation has been run several times, using a DO loop or RESET, the user can select which runs are to be plotted by specifying the number of the run immediately after the GRAPH command. For example, GRAPH 1,5,7,T,CA,CB,O,100,0,25 will plot CA, and CB against T for run numbers 1, 5 and 7 with x axis 0 to 100 and y axis 0 to 25. If no run numbers are specified in the GRAPH command all runs are plotted. For example, GRAPH T,CA,CB Graphing too many variables will often disguise the details of the smallest, due to the automatic scaling. Choosing the range of the scaling (for two variables only) is useful for revealing details. The AGRAPH command is similar to the GRAPH command except the output is produced on the alternative graphics device. The TGRAPH command is to be used to produce an alpha-numeric output if the user does not have a graphics terminal.

2 Programming with ISIM

677

The OUTPUT statement allows the values of the specified variables to be displayed on the screen at intervals of CINT * NOCI during the course of the simulation, e.g., OUTPUT T,CA,CB

An output file can be stored and used for further applications, such as graphing with special graphing software. This done by reading the file as usual and before running the simulation give the command OPEN "File name.data" Run the simulation and close the file with ENDFILE.

Printing ISIM Output The ISIM command MONITOR will cause all subsequent output to the screen to be printed. The command NORMAL terminates this printing procedure. Graphs can be printed on most systems simply by pressing the Print Screen Key. If problems are found with monochrome VGA and EGA when trying to print graphs of multiple runs, then one should change to CGA by reinstalling or editing the ISIM BAT file.

2.9

Useful Sequences of Statements 1 SIM INTERACT RESET GOT0 1

This series of statements in the program allows the user, after the completion of a run, to change the values of any of the variables (using the VAL command) and rerun the simulation (using the GO command). In this way, the influence of the value of one or more variables on the simulation can easily be seen.

67 8

Appendix: Using the ISIM Language

IF (CA.LT.O.0) CA=O

This is a useful program statement to prevent variables, such as concentrations,, becoming negative. Place it just after the differential equation for the variable. PLOT T,CA,O,TFIN,O,CAO OUTPUT T,CA,CB,CC,RATE PREPARE T,CA,CB,CC,RATE This series of program statements will plot CA against T on the screen as the simulation progresses. Values of T, CA, CB, CC and RATE will also appear on the screen. Subsequently, any of the variables T, CA, CB, CC and RATE can be plotted against any other variable by using the GRAPH command. A typical interrupt from the keyboard might consist of the following commands:

Any Key HOLD GRAPH T,CA,CB,CC VAL K1, KLA VAL KLA=O.8

eo

causes an interrupt or simply H plots CA, CB and CC against T, auto-scaling inspect values change value rerun simulation with new value

2 . 1 0 Further Information A summary of ISIM commands is found below. The ISIM manual contains more details on writing models in ISIM and on the numerical methods that are used in the simulations. It can be obtained from: Prof. John L. Hay, ISIM lnternational Simulation Limited, Technology House, Salford University Business Park, Lissadel Street, Salford M6 6AP, England, (Tel: +44-(O)61-745 7444; Fax: +44-(O)61-737 7700).

3 Summary of ISIM Commands

3

679

Summary of ISIM Commands

Unless otherwise indicated, spaces are ignored in ISIM commands and program statements.

3.1

COMMAND and INPUT Modes

When awaiting input from the user, ISIM is always in one of two modes: COMMAND INPUT

mode is indicated by the prompt $ mode is indicated by the prompt ?

Some commands (e.g., APPEND) switch to INPUT mode. To switch to COMMAND mode, type $ or $ command.

3.2

ISIM Commands

Note: Commands may be abbreviated to their initial letter except where otherwise indicated.

Text Editing APPEND

Append the following input lines to the end of the text buffer.

nCHANGE n,mCHANGE

Delete the specified lines (i.e., line n or lines n to m inclusive) and replace them by the following input lines.

nDELETE n, mDELETE

Delete the specified lines.

nINSERT

Insert the following input lines before line n.

680

Appendix: Using the ISIM Language

KILL

MUST BE TYPED IN FULL. Delete the entire text buffer and cancel all interactive commands. This is necessary when waiting to read another program.

nLIST n,mLIST LIST

List the specified lines or the complete text buffer on the screen.

nPRlNT n,mPRINT PRINT

List the specified lines or the complete text buffer on the printer.

Hard Copy Output (to log details of session) MONITOR NORMAL

Direct text output to the printer as well as to the screen. Stop logging output on printer.

File Handling In the following commands the file specification must be preceded by one or more spaces. READ tree-structured-filename Take commands and program text from the specified input file rather than from the keyboard. Input reverts to keyboard when a) an end-of-file is detected. b) an error occurs. c) a USER command is encountered. d) a user interrupt key is pressed. A filename with an extension of .SIM is assumed if an explicit extension is not used. READ

Continue reading from the previously specified input file.

USER

Switch from file input to keyboard input. This command is only effective when it appears within an input file.

SAVE tree-structured-filename

3 Summary of ISIM Commands

68 1

ABBREVIATION IS SA NOT S. Save the contents of the text buffer and all interactive assignments into the specified output file. The contents of the file will be in a format suitable for subsequent use in a READ command. OPEN tree-structured-filename Opens specified file and redirects PRINT and OUTPUT text to the file for post-mortem analysis. ENDFlLE or SAVE will close the file. ENDFILE

Closes an OPENed file and restores normal role of PRINT and OUTPUT.

To Initiate a Simulation START

Complete the compilation process and execute the user's simulation program.

GO

Continue execution of the user's program following an interruption caused by an INTERACT statement or the use of an interrupt key.

To Dismiss ISIM QUIT

Exit from ISIM to the host computer operating system. All files are closed.

Interactive Commands The interactive commands, which cannot be abbreviated, are mainly for use following a simulation run or whenever the INTERACT statement is invoked. VAL VAL A,B,ETC

List values of all calculated program variables. List values of A, B and ETC.

682

Appendix: Using the ISIM Language

VAL A= 1 .B=2

Set values of A and B.

OUTPUT A,B OUTPUT

Tabulate A,B at communication intervals. Tabulate T and all state variables.

PLOT X,Y,XMlN,XMAX,YMIN,YMAX Plot Y (vertical) against X (horizontal) over the ranges Y from YMIN to YMAX and X from XMIN to XMAX. GRAPH 1,3,4,X,Y7Z,XMIN,XMAX,YMIN7YMAX Produce a post-mortem plot using data stored by means of a PREPARE statement. 1, 3, 4 are the run numbers. If the run list is omitted, all runs are plotted. X, Z etc., are the vertical variables. The last four parameters specify axis scaling and are optional, i.e., GRAPH X,Y,Z plots all runs with automatic scaling. AGRAPH ...

As for GRAPH but a graph is produced on the alternative graphics device.

TGRAPH ...

As for GRAPH but a graph is produced on the screen terminal using alpha-numeric characters.

XPLOT XOUT XVAL CLEAR

Cancel the PLOT instruction. Cancel the OUTPUT instruction. Cancel all interactive VAL assignments. Clear display screen.

User Interrupts Pressing a key will halt execution, and pressing a further key provides the following actions:

Q H S C

Quit. Exit to operating system. Hold. Proceed to next communication point pass control to user for further command. Skip. Abort remainder of current run and then continue with subsequent runs. Command. Abort remainder of current run and then pass control to user for interactive commands.

Any other key causes execution to continue.

683

3 Summary of ISIM Commands

3.3

ISIM Program Statements

Numbers Identifiers

Are always regarded as real and treated in a floating-point form. Examples: 21, 36, 123, -62, -0.562, -0.137E-21, 98E2 Always start with a letter which may be followed by further letters or digits up to a maximum of 8 characters. Differential identifiers are identifiers followed by one or Each ' counts as one character. more primes consist of up to eight digits and may precede most executable statements (not PLOT or OUTPUT). The first three variables indicate the state of the integration. (I).

Labels Reserved variables LSTPLPRP LCOM

TRUE when the integration procedure has completed an integration step. TRUE when the integration procedure has completed a communication interval of NOCI*CINT.

Note - User setting: LCOM controls PRINT and OUTPUT; LSTP controls the PLOT; LPRP controls PREPARE. The following variables are given default values if the user does not explicitly define them: T TFIN CINT NOCI

Independent variable, usually time. (default=O) Termination value of T. (default=lO) Calculation interval (integration step). (default=l) Number of steps (CINT) per communication interval. (default=l) RERR Relative integration error. (default=0.00 1) ALGO Integration method. (default=O) 0 = variable-step 5th-order Runge-Kutta 1 = fixed-step 4th-order Runge-Kutta 2 = fixed-step 2nd-order Runge-Kutta 3 = fixed-step 2nd-order implicit (stiff) Comments: All text on a line appearing to the right of a colon (:) is regarded as a comment.

Specification Statements DIMENSION ARRAY(2,3),VEC( 17) - same as FORTRAN

684

Appendix: Using the ISIM Language ~~~

~~~~

~~~

CONSTANT ARRAY=2*4.0,3*8,lO.O,A=5 INTTIAL

introduces INITIAL region

DYNAMIC

introduces DYNAMIC region

TERMINAL

introduces TERMINAL region

SUBROUTINE AVERAG(VEC,N) - same as FORTRAN FUNCTION HYP(X,Y) - same as FORTRAN

Arithmetic Assignments A = (arithmetic expression} conforms to FORTRAN rules.

-

where the arithmetic expression generally

Y = ARRAY(J+K,N)+4*ATAN(L)

The following standard functions are available:

ABS, ATAN, SIN, COS, EXP, LOG (natural), IFIX (truncate), SQRT, RAND, COMP (comparator returns one if argument is greater than or equal to zero, otherwise zero).

Logical Assignments A = {logical expression) B = L.AND.M.0R.NOT.Q C = X.GE.Y.OR.A.LT.3.6 The logical operators are .NOT. .AND. .OR. and the relational operators .GT. .GE. .EQ. .NE. .LE. .LT.

685

3 Summary of ISIM Commands

Control Statements GOT0 210 IF(A.GT.B )A=B IF(LOG1C)GOTO 2 15 DO 220 I=1,5 DO 220 A=0.5,5.0,0.5 DO J=1,3 DO K=I*J,L**2 END :OF K LOOP END :OF J LOOP

220 C0,JTINUE: END OF A AND I OOPS RETURN STOP CALL AVERAG(ARRAY,6) Stop the simulation and pass control to the ISIM monitor. A INTERACT command may then be issued, and the simulation may be resumed by a GO.

Simulation Statements SIM

Calls the simulation model. Following execution of a simulation run, control is returned to the next statement. means dX/dT = -K and will be automatically integrated.

Y = INTG(=-K,YO)

Y= integral( -K dT), with Y=YO initially.

X" = -w**2*x

means d(dX/dT)/dT = -W**2*X and will be automatically integrated.

TERMINATE X.LE .0.3

pass control to TERMINAL region.

RESET

Reset T and the state variables to the values they had at the start of the previous run.

686 RESERVED SAVE A,B

STATE X/DX DERIV

Appendix: Using the ISIM Language

Declaration which makes the reserved variable available in subroutine/function. Also makes routine a sub-model. Declaration to save the values of variables when routine is called more than once. Also makes routine a sub-model. Declares variables as statdderivative pairs. Also makes routine a sub-model. Indicates start of dynamic code in sub-model.

Input/Output Statements PRINT "A IS =n,A,/,"A**2 IS",A**2 For A=2 this statement would output A IS = 2.0000 A**2 IS 4.0000 < and > may be used in place of " PREPARE A,B,T

Store values of A, B and T in a workfile This prepares data for post-mortem plotting of A, B and T. This statement is effective when LPRP is TRUE.

OUTPUT

Tabulate T and all state variables at each communication interval.

OUTPUT X,Y ,B

Tabulate X, Y and B.

PLOT X,Y,XMIN,XMAX,-100,100 Plot Y against X on the specified axis (X axis annotated from XMlN to XMAX, Y axis from -100 to 100) as the simulation proceeds PLOT or OUTPUT statements may not be conditional or labelled.

687

4 ISIM Error Messages

ISIM Error Messages

4

All ISIM error messages are of the form FAIL NO. n where n is a failure number. When an error message occurs immediately after the statement or command which caused it a may be printed under the character at which the error was detected, or the identifier causing the failure may be printed.

Monitor and Syntax Failures (note: the faulty statement or command is ignored) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20-38 39. 40. 41.

Spare. Too many lines. Illegal character in command. Number expected in command. Letter expected in command. Too many characters in identifier or label. Digit expected after - (e.g., -.TRUE. is illegal). Letter other than (E) in a number. Exponent expected in a number. Letter expected in a logical string. Illegal logical string. Illegal character. Illegal use of Line is too long. Missing terminator < >) after text string. Illegal use of GO. Either the INTERACT statement has not been used, or changes have been made which require recompilation (i.e., START). File specification omitted. File error due to (a) illegal file name, or (b) non-existent input file specified, or (c) attempt to open file already open. Number too large or too small. Spare. Wrong number or axis specification parameters in GRAPH or TGRAPH (should be zero or four). A variable named in a VAL command does not appear in the symbol table. A variable named in a VAL=number command does not appear in the symbol table. (I).

(I'

688

Appendix: Using the ISIM Language ~~~

~~

A variable listed in an OUTPUT statement or command does not appear in the symbol table. A variable listed in a PLOT statement or command does not appear in 43. the symbol table of the main program. Maximum and minimum graph range equal. 44. Too many variable redefinitions in VAL command. 45. Variable undefined. 46. GRAPH or TGRAPH issued when there is nothing to plot (e.g., 47. PREPARE omitted from program). GRAPH or TGRAPH specifies a variable not included in PREPARE. 48. GRAPH or TGRAPH specifies a run which was not made. 49. Statement duplicated (DYNAMIC, OUTPUT, PLOT or PREPARE. 50. Wrong number of arguments in PLOT (should be 6) or too many 51. arguments in OUTPUT, SUBROUTINE, FUNCTION or ARRAY. (OUTPUT limited to 10, others to 15). Number expected. 52. Unsubscripted variable expected or positive run number in GRAPH or 53. TGRAPH or syntax error in output or declaration = expected. 54. , expected. 55. Invalid item at end of command (RETURN expected. 56. Invalid item at end of statement (; or RETURN expected). 57. Array name expected (no left parenthesis). 58. Left parenthesis expected. 59. Positive integer expected. 60. Right parenthesis expected. 61 Unsubscripted number or variable expected. 62. Function or subroutine name expected (no left parenthesis). 63 Non-differential unsubscripted variable or array name expected. 64. Non-differential unsubscripted variable expected. 65. Label expected. 66. Subscripted or unsubscripted variable or function expected or 67. unrecognized key word. Illegal syntax in STATE or SAVE. 68. 69-96 Spare. Expression too complex (break it down). 97. Stack 1 overflow - reduce statement complexity. 98. Stack 2 overflow - reduce statement complexity. 99.

42.

Compilation Failures (note: compilation will continue except where stated) 101. 103.

Too many differential equations - program returns to ISIM monitor. Too many user symbols - program returns to ISIM Monitor (try breaking program into more subprograms).

4 ISIM Error Messages

104. 105. 106. 107. 108. 109. 110. 111. 112. 113. 114. 115. 116. 117. 118. 119. 120. 121. 122. 123. 124. 125. 126. 127. 128.

689

Too many object code instructions - program returns to ISIM Monitor. Too many data elements required - program returns to ISIM Monitor. No program text. Array without subscripts used illegally. Wrong number of arguments in subroutine or function or of subscripts in an array or an illegal GOT0 i.e. to a label in another region. Illegal number. Subscripted variable not been declared as an array used to left of =. Duplicated label, subroutine or function. A reserved variable, or a dummy argument of a function, or a variable set by an earlier CONSTANT statement has been redefined as an array. Variable defined twice as an array. A statement is in the wrong region or subprogram: Possibly RESET, INTERACT, PREPARE, INTG, INITIAL, DYNAMIC, TERMINAL, SIM, COMP. No DYNAMIC region although SIM statement has appeared. Spare. Label not set or in the wrong region. Local variable in a subprogram was never given a value. A dummy argument or a function name appears twice in a subprogram definition statement. Declaration in illegal position, i.e., the statement does not appear amongst the first non-executable statements of the main program or subprogram. Subroutine or function is missing. Spare. Too many nested DO loops. Target label for DO not set or misplaced. Subroutine or function dummy argument cannot be set in a CONSTANT statement. Too few numbers in a CONSTANT statement. Too many numbers in a CONSTANT statement. Redeclaration of variable is illegal.

Execution Phase Errors (except where stated control is returned to ISIM monitor; cannot be continued with a GO) 15 1. 152. 153. 154. 155.

Divide by zero or overflow. Stack overflow - too many nested subroutines or functions. Negative number raised to fractional power. Spare. Negative or zero argument in LOG.

690 156.

157. 158. 159. 160. 161. 162. 163.

Appendix: Using the ISIM Language

Subscripted variable with subscripts negative or too large (i.e., addressing location outside available data area). Argument of EXP has absolute value > 85. Spare. PREPARE error. PREPARE file full - non-catastrophic, program continues. Negative argument in SQRT. Spare. Illegal call of submodel (e.g., no DYNAMIC) Dummy in subroutine or function should be a state.

-

ESL THE LANGUAGE OF SIMULATION I

force

IL

I

Over ten years development maturity makes ESL THE language of simulation for simple or advanced applications. Developed to meet the simulation requirements of the European Space Agency: used by such leading companies as British Gas, Lucas Aerospace, BNFL, British Aerospace.

ESL offers a full range of simulation facilities. Whatever the system or process, if it can be modelled, it can be simulated by ESL. Its features include:

Fully integrated submodel capability simplifies the simulation of complex systems. Elegant constructs support simple descriptions and efficient processing of discontinuities. Powerful mouse/menu controlled graphical interface creates system block diagrams, generates error-free simulation models, executes the simulation, and displays graphical results. Real-time distributed (parallel processing) supported on Unix platforms. Post-simulation graphics display package. Interpretive running for testing, or compiled FORTRAN for optimum speed. Eight integration algorithms, including improved Gear/Hindmarsh methods. Hardware supported includes: IBM-PC, SUN, Silicon Graphics, HP, IBM RS/6000, and DEC Unix workstations; VAX workstations, Encore Unix systems. Further information from: ISlM International Simulation Limited Technology House, Salford University Business Park Lissadel Street, SALFORD M6 6AP, England Tel: +44 (0)61 745 7444, Fax: +44 (0)61 737 7700

H.- G. Elias

An Introduction to Plastics 1993. XVIII, 349 pages with 105 figures and 85 tables. Hardcover. DM 78.4 6s 608.-/ sFr. 76.-. ISBN 3-527-28578-4 This book presents a precise, yet non-mathematical introduction to plastics, their raw materials, syntheses, properties, and applications. The manufacture and properties of plastics are discussed as a function of the molecular and supermolecular properties of polymers. Polymer composites and waste disposal are also treated. Clear and comprehensive, this volume is not only an informative guide for practitioners in the field of plastics but also an indispensable aid for plastics engineers, polymer scientists, chemists, physicists, and materials scientists.

W.A. Warr I C. Suhr

Chemical Information Management 1992. XIX, 261 pages with 20 figures and 15 tables. Hardcover. DM 128.40s 998.-/sFr 120.-. ISBN 3-527-28366-8 The authors, capitalizing on their rich experience gained in leading positions in industry, present a comprehensive, up-to-date survey of the management of chemical and related information. Equal importance is attached to scientific and patent literature. Topics considered in this timely book include the needs of the various user communities and methods to monitor, index, store and retrieve structures, chemical reactions and non-structural concepts. Emphasis is placed on data processing, database searching and recent developments in artifical intelligence applications. The book fills a gap and is mandatory reading for everyone involved in using or administering chemical information in academic and industrial research and development, patent matters or corporate management.

INFORMATION TECHNOLOGY

Chemical Engineering Dynami :Modelling

with PC Simulation

John Inph;m. Iruine J. Dunn. Elmar Heinzlc &JiiiE. Picnoul

copyright OVCH VerlagsgesellschaftmbH, 1994

Index

Absorption 567 ammonia 567 Acceleration of gravity 499, 503 Accumulation term - energy balance 39 - mass balance 17, 22 - momentum 502 Activation energy 54 Activity coefficient 213, 611, 614 - relations 616 Adaptive control 107 Adiabatic 486 - semi-continuous reactor 44 -tank 487 Adsorption - adsorption isotherm 583 - equilibrium constant 583 Aerated tank with oxygen electrode 534 Agitated extraction column 556 Ammonia 567 Antoine equation 137 Area - effective, for heat transfer 133 - for heat transfer 35, 140 - heat transfer, varies with volume 519 - interfacial 168 - specific interfacial 62, 528 Arrhenius equation 53, 144 see also the nonisotherrnal reactor examples Auto-refrigerated reactor 357 Automatic control 94 Axial concentration profiles 563 Axial dispersion 560, 578 - coefficient 581 - flow model 578 - model 244, 561 Axial mixing in liquid-liquid extraction columns 258 - of

Backmixing 193, 544, 553, 555 177

- flow rates

Balance region 17, 22 for lumped and distributed parameter systems 19 Balances for the end sections 553 Batch - distillation 203, 584, 589, 610 - distillation column 584, 594, 610 - extraction 167, 527 - extractor 530 - liquid-liquid extraction 528 - operation 130 - reactor 142, 280-315, 476 - semi-batch, and continuous stirred tanks 129 - still 593 Benzene 605 Binary batch distillation 203 - column for 584 Biological flocs and films 653 Books recommended 273 Bottom plate 585, 599 Boundary conditions 247, 257, 260, 633, 637, 642, 643, 646 Bubble point - calculation 214, 610, 612, 620 - calculation for a batch distillation column 610 - iterative loop calculation for 218 Bypassing 164 - and dead space 442 - or short-cut flow 161 -

Cake washing 578 Cascade control 105 Cascade - continuous stirred-tank-reactor 129 - countercurrent extraction with backmixing 177 - countercurrent stagewise extractor cascade 552 - extraction control 182 - extraction with backmixing and control 543

692 extraction with slow chemical reaction 179 mixer-settler 189 - mixer-settler extraction 183 - multistage countercurrent extraction 175 - multistage, multicomponent, extraction 167 - of equilibrium stages 548 - of three reactors 330 - reactor 150, 345 -with side streams 177 Catalyst effectiveness 644, 649 Chaotic behaviour 661 - oscillatory 663 Chemical - equilibrium 56 -kinetics 51 - reaction time constant 91 Cocurrent flow 628 Cohen-Coon controller settings 103, 508 Coils or jackets 132 Column hydrodynamics 195 -batch reaction sequence 280 - distillation 213 - reaction 286 - sequential-parallel reaction 330 -transfer function 87 Component balances 16, 31, 131 Concentration driving force 63, 168 Concentration profiles 58 - for countercurrent differential contactor 59 Condenser 585,611 Constant molal overflow 205 Continuous -and dispersed phase flow rates 558 -binary distillation 208 -binary distillation column 599 -column 159 -distillation column 600, 605 -equilibrium multistage extraction 548 -equilibrium stage extraction 171 -extraction stage 540 -flow 159 -flow equilibrium stage 540 -flow reactor 132, 159 -flow, stirred tank 67, 505 -heating tank 41 - multicomponent distillation column 605 -operation 130 -phase 559 - stirred-tank reactor 17, 47, 147, 477 - stirred-tank-reactor cascade 129 -tubular device 159 -tubular reactor 229,381-422 -

Index Control 94-107 -adaptive 107 -automatic 94 -cascade 105 -differential constant 507 -discrete systems 107 - extraction cascade with backmixing 543 -extraction column with 543 -feed 519 - feed-forward 105 -feedback 95,518 -integral 547 4 -integral constant 507 -level 509 -of extraction cascades 182 -of temperature 135, 505 -on/off 96 - PID 97,505-518 - process examples 505-524 -proportional 547 -proportional and integral 516 -proportional constant 507 -proportional equation 511 -proportional plus integral 544 - proportional-integral equation, 182 - proportional-integral-derivative (PID) 97, 508 -reactor 147, 156 -temperature 95 -temperature for semi-batch reactor 518 -time integral criteria 104 -two tank level 509 Controller 96-107, 509 -adjusting constants, 523 - Cohen-Coon tuning, 103 -equation for 158 -proportional 157 -proportional temperature 100 - proportional-integral feedback 506 -response 99 -trial and error tuning 101 - tuning 101,547 -tuning problem 514 -ultimate gain tuning 103 - Ziegler-Nichols tuning 102 Cooling 345 - countercurrent 263 -extraction cascade with backmixing 177 -flow 628 -jacket 520 -multistage equilibrium extraction 548 -multistage extraction 548 -packed column 567 - stagewise extractor cascade 552

Index Deactivating catalyst 319 Dead zones 159, 162, 163 Degree of segregation 471 Density influences 492 Desorption of solute 578, 579 Difference differential equation 579 Difference formulae for partial differential equations 268 Differential column 167 - control constant 507 - extraction column with axial dispersion 560 - extraction columns 560 - mass transfer 250 Diffusion 24, 233-237, 636-658 - and heat conduction 223 - and reaction 647, 654 - and reaction in a spherical bead 653 - coefficient 645 -drying 638 - dispersion time constant 91 - dynamic and enzymatic reaction 647 - film 534 - fluxes 225, 655 - from the interior of the solid 636 - molecular coefficient 62 - process examples 636 - split boundary solution for 641 - steady-state 227 -time 640 - unsteady-state 224 -with chemical reaction 228 Digital simulation 6 Dimensionless - concentration 579 - equations 323, 486 - form 578, 657 - kinetics 283 -model 381 - model equations 47, 323, 644 - semi-dimensionless form 649 -solute 581 - thickness 579 - variables 535, 643, 646 Discharge nozzle 496 Discharge of molten steel 496 Discrete control systems 107 Dispersed phase 559 -holdup 196, 557 Dispersion model 247-250 Dispersion number 249 Dispersion time 562 Dissolved oxygen concentration 534 -profile 658

693 Distillation 201-214, 584-621 -batch 589 - batch column 584, 594, 610 - binary batch 203 - binary batch column 584 -bubble point calculation for 610 -complex 213 - continuous binary 208 - equipment for 584 - in continuous binary column 599 - in continuous column 600, 605 - in continuous multicomponent column 605 - multicomponent batch 593 - multicomponent differential 589 - multicomponent steam 214 - multicomponent, semi-batch steam 616 - non-ideal 610 -period 617 - process examples 584 - simple overhead 201 - stagewise 201 - steam distillation 616 Distribution of residence times 159-166, 467 Disturbance 101, 106 Driving force 133 Drying 637 - of a solid 636 -rate 641 Dynamic -balance equation 15, 35 - diffusion and enzymatic reaction 647 -method 534 - model formulation 15 - of a shell-and-tube heat exchanger 622 - of the cooling jacket 133 - simulation 1 - tubular reactor 240, 405 - velocity profile 556 Dynamics of metal jacket wall 139 E-curve 159 Early and late mixing 161 Eddy dispersion coefficient 565 Effective area for heat transfer 133 Effectiveness factor 641, 645, 649, 651, 657 Eigenvalues 154 Electrode membrane 536 Electrode response characteristic 534 Empirical approach 4 End section modelling 562 Endothermic reaction 143

694

Index

Energy balance

35, 216, 505, 568

- accumulation term in 30 - and mass balance 131

- equation

for

132, 232, 498

- flow term in 39 - for reactor 307, 374

-heat transfer term in 40 - reaction heat term in 40 - simplified 41 - steady state 395, 633 Enriching section 209 Enthalpy 36 Entrainment fractions 189 Enzyme 641 Equation stiffness 125 Equilibrium 167, 528, 530, 540, 589, 606 - adsorption constant 583 - cascade of stages 548 - continuous flow stage 540 - continuous multistage extraction 548 - continuous stage extraction 171 - countercurrent multistage extraction 548 - interacting in two-solute batch extraction 530 - interacting solute 170 -line 63 - liquid-vapour 611 - multicomponent 174, 213, 606 - rnultisolute 171 - oxygen concentration 535 - phase 60, 218 - relationship 173, 568, 578 -stage 59 - stage behaviour 176 - stage extraction 172 - stage extractor 540 ESL 114 Evaporated 637 Exothermic - reaction 151, 298, 340-365, 372-380 - reaction sequence 306 -reactor 434 - semi-batch reactor 519 Extraction 166-195, 527-566 - agitated columns 556 - axial mixing liquid-liquid 258 -batch 167 -batch extractor 530 - batch liquid-liquid 528 -batch two-phase system 527 - cascade with backmixing and control 543 - cascade with slow chemical reaction 179 - column with control 543 - columns with backmixing 561

-

continuous equilibrium multistage 548 171

- continuous equilibrium stage - continuous stage 540 -

control of cascades

182

- countercurrent cascade with backmixing

I77

548 countercurrent stagewise extractor cascade 552 - differential column with axial dispersion 560 - differential columns 560 - equilibrium stage 172 - extractor 560, 561 - integrated 335 - liquid-liquid 166, 335 - liquid-liquid column 254 - liquid-liquid column dynamics 253 - mixer-settler cascades 183 - multicomponent 530 - multisolute batch 169 - multistage countercurrent cascade 175 - multistage extractor with backmixing 552 - multistage, multicomponent, cascade 167 - non-equilibrium staged column 193 - reaction with integrated 175 - single batch 167 - single solute batch 527 - single stage agitated column 557 - staged columns 192 - stagewise discontinuous 550 - two-solute batch 531 - two-solute batch with interacting equilibria 530 -vessel 167 Extractor 560, 561 - equilibrium stage 540 - multistage 552 - countercurrent multistage -

F-curve 160 Fanning friction factor 502 Feed - composition 609 -control 519 - feed plate 209, 600, 606 - or aqueous phase 167 Feedback control 95, 518 -system 505 Feed-forward control 105 Fermentation 653 Fick’s Law 62, 223, 226, 637 Filling and emptying tanks 512 Film heat transfer coefficient 140, 627 Filter bed 579

695

Index Filter washing 578 Finite-differencing 225, 410, 579, 623, 647, 654 - approximations for 126 - elements 560, 637 - for a heat exchanger 265 - for a tubular reactor 222 First-order lag 66, 81, 515 - equation 536 -response 66 -time lag 508 Fitting of measured response 86 Floc sizes 658 Flow phenomena time constant 90 Flow terms - energy balance 39 -mass balance 24 Fourier’s law 223, 633 Fractional - conversion 232 -holdup 556, 560 - holdup of dispersed 559 -phase holdup 565 Free surface drying 638 Free-radical polymerisation 366 Frequency of the oscillations 508 Friction factor 498 Frictional force 502 Gas absorption

199

- simulation 253 - steady-state design

251 -with heat effects 250 Gas-liquid - contacting systems 198 - interface 61 - mass transfer 50 - mixing and mass transfer in a stirred tank 457 -transfer 534 Gas-permeation 572 Gas-permeation module 572 Gas-phase - reaction with molar change 419 - tubular reactors 235 Gas separation by membrane permeation 572 Half-interval method 644 Heat -balance 515, 629 - capacity 37, 44, 55 - conduction 633 - conductivity 635

emissivity 635 gain curve 152 - IOSS 152 - loss by radiation 632 - of combustion 55 - of formation 55 - of reaction 38, 54 - of vapourisation 517 - reaction term in energy balance 40 - reaction with effects 299 - variable capacities 44, 372 Heat exchanger 132, 261-264, 623 -boundary conditions 267 - differential model 264 - dynamics 264 - dynamics of shell-and-tube 622 - finite-differencing 265 - heat exchange 263 - shell-and-tube 622 - steady-state 262 - steady-state, two-pass 628 - two-pass 628 Heat transfer 132, 261-264, 623 - area varies with volume 519 - coefficient 621, 627, 631 - examples 622 - from metal wall 623 -rate 627 - steady-state tubular flow 261 - term in energy balance 40 -time constant 93 Heater 505 Heating and boiling periods 616 Heating period 617 Heerden steady-state stability criterion 359 Henry coefficient 538 Henry’s law 535 High gain factor 175 Higher order responses 73 Holdup 556, 560, 587, 595, 608 - distribution 559 -time, or residence time 148 Hopf bifurcation 661 Hydrocarbon 593 Hydrostatic equation 185 Hydrostatic pressure force 503 -

Ideal Gas Law 535 Immiscible liquid phases 167, 180 Implicit algebraic loop 200, 557 - calculation 197 -nature 137 Information flow diagram 13, 28, 168, 216, 235, 238

696 Information flow in model 7 Integral action time 182 Integral control 97, 547 Integral control constant 97, 507 Integrated extraction 335 Integration - parameters 125 -routine 124 - step length 126 Interacting solute equilibria 170 Interacting tank reservoirs 501 Interfacial concentrations 62 Interfacial mass transfer 26 Interphase - mass transfer 60 -transport 26 Interstage flow rates 177 Introductory ISIM example 12 ISIM 10, 114 Isothermal - reactor with axial dispersion 410 - reactor with complex reaction 316 -tank 487 - tubular reactor 388 Isothermal or adiabatic temperaturc conditions 485 Iterative - calculations 610 - loop for the bubble point calculation 218 solution 612

~

Jacket cooling 340 Jacketed batch reactor 306 Kinetic energy 498 K,,a 534 Lag in the system 509 Langmuir-Hinshelwood kinetics 321 Laplace transformation 80, 536 Latent heat of vapourisation 517 Least squares 112 Level control 509 Likelihood or probability density 114 Limit cycles 155 Linearisation 154 Liquid - film 536 - holdup 585, 606 - impurities 578 - stream blending 492 Liquid-liquid - axial mixing in extraction columns 258 - batch extraction 528

Index - extraction

166, 335 extraction column 254 - extraction column dynamics 253 Liquid-vapour equilibrium 611 Lorentz equations 659 Lumped parameter 139 -

Manipulation of the cooling water flow 519, 520 Mass balance 4, 15, 22 - accumulation term in 22 and energy balance 131 - flow terms in 24 - procedures 17 - production rate in 27 -total 15, 29 , 131 Mass transfer - coefficient 62, 168 - process examples 527 -theory 58 - time constant 92 MATLAB 114 Maximum likelihood 112 Measured oxygen concentration response curve 534 Measurement - and process response 65 - dynamics 71, 535 - l a g 72 - time constant 71 Membrane permeabilities 572 Metal rod 632 Methanol 615 Michaelis-Menten 643, 646, 649 Micromixing 470 Mixer 183 Mixer dynamics 185 Mixer-settler 167, 540 -cascade 189 - extraction cascades 183 Mixing - and segregation in chemical reactors 470 - behaviour 163 -history 165 - model 440, 445, 450, 466 Mixture of hydrocarbons 589 Modelling - a non-isothermal, chemical reactor 35 -approach 4 - fundamentals 1 - of cooling effects 135 -procedure 6 Modes of reactor 129 ~

697

Index Molar feeding rate 145 - quantities 146 Mole ratios 254 Molecular diffusion coefficient 62 Momentum -balance 46, 502 - transport 223 Monod-type equation 655 Multicomponent 594 -batch distillation 593 - differential distillation 589 - equilibria 174, 213 - extraction 530 -mixture 203 - semi-batch steam distillation 616 - separation 210, 606 - steam distillation 214 - system 180, 616 Multiple -feeds 177 - steady states 126 Multisolute -batch extraction 169, 530 - equilibria 171 Multistage - countercurrent extraction cascade 175, 548 - extractor with backmixing 552, 553 - multicomponent, extraction cascade 167 -

n-Decane 616 n-Octane 616 Natural cause and effect 175 Naturally occurring oscillations 126 Negative feedback 158 Nelder-Mead search algorithm 108 Newton’s gradient method 108 Nitrogen 572 Non-equilibrium - differential contacting 167 - staged extraction column 193 Non-ideal - distillation 610 -flow 159 - liquid behaviour 610 -mixing 165 - stirred-tank reactor 440 Non-isothermal 402 Non-linear parameter estimation 108, 113 Nozzle mass flow 497 Numerical - aspects of dynamic behaviour 108 - integration 122

On/off control 96 Operating line equation 595 Optimisation 108 Order of reaction 53, 323 Oscillating tank reactor 350 Oscillations 507, 518, 663 - attractive oscillations 663 - frequency of 508 - naturally occurring 126 Overall - heat transfer coefficient 133 - mass transfer coefficient 168, 533 Oxidation reaction in an aerated tank Oxygen 572, 654 - electrode 534 - electrode dynamics 536 - enrichment 572 - limitation 653 -transfer coefficient 534 - uptake rate 655 Parameter estimation 112 Partial differential equation 578 Partial differentials 154 Peclet number 243, 579 Penetration distance 654 Perfect mixing 159 Perfectly mixed 142 - stirred tank 129 Permeabilities, membrane 574 Permeability coefficient 576 Permeation rate 572, 576 Perturbation variables 153 Phase - equilibria 60, 218 - flow rate 186 -volume 170, 186 Phase-plane 661 -diagram 127 -plot 659 PID-control 97, 508 Plant start up, shut down 147 Plate - dynamics 595 - efficiency 212 - hydraulics 207 Plug flow 159, 555 - tubular reactor 230-235, 239 Porosity 579, 581 - distributions 578 Porous - biocatalyst 647 - solid supports 641 Potential energy 498

302

698 Principles of mathematical modelling 2 Probability density function 112 Process control examples 505-524 Product inhibition 643, 649 Production rate in mass balance 27 Profit function 108 Proportional - and integral control 97, 516 -control 547 - control constant 507 -control equation 511 - controller 157 -gain 158, 182, 517 - integral control equation 182 - - integral-derivative (PID) control 97 - - integral feedback controller 544, 506 - tcmpcrature controllcr 100 Pulse -input 159 - response curve 165 Pure time delay 78, 85 ~

Quasi-steady-state 200, 211 Radiation from metal rod 632 Raffinate 182 Random -chaotic behaviour 659 - differential equation behaviour 659 - fluctuations 506 Rate of - chemical reaction 51 energy generated by reaction 232 -heat transfer 140 - mass transfer 167, 528 - solute transfer 168, 172 Rayleigh equation prediction 592 Reaction -enthalpy 38 - exothermic 298 - fractional conversion 56 - heat term in energy balance 40 - kinetics 143 -rate 475 - selectivity 57 - with heat effects 299 -with integrated extraction 175 yield 57 Reactor - adiabatic, semi-continuous 44 - auto-refrigerated 3.57 -batch 142, 476 -batch analogy 239 ~

~

Index - cascade 150, 330, 345 - continuous-flow 32 - continuous-flow stirred tank

67 continuous stirred-tank 17, 47, - continuous stirred-tank cascades 129, 327 - control 147, 156 - dynamic tubular 240, 405 - exothermic 434 - exothermic semi-batch 519 - finite-differencing a tubular 222 - gas-phase tubular 235 - isothermal tubular 388 - isothermal with axial dispersion 410 - isothermal with complex reaction 316 -jacketed batch 306 - mixing and segregation in 470 - modelling a non-isothermal 35 -modes of 129 - non-ideal stirred-tank 440 - oscillating tank 350 - plug-flow tubular 230, 239 - response of a stirred tank 70 - segregation in a batch 474 - segregation in a continuous 474 - segregation in a semi-continuous 475 - semi-batch 144, 146, 433, 478 - semi-continuous 423, 426, 518 - spouted bed 466 - stability of chemical 151, 361 - steady-state tubular dispersion 247 - stirred-tank 129 - tanks-in-series approximation of a tubular 406 temperature 308 - temperature control for semi-batch 518 -tubular 18, 234, 393, 402 - tubular and tank 384 - tubular chemical 229 - tubular with axial dispersion 243, 416 - tubular with unmixed reactants 471 - variable volume 519 Reboiler 584, 587, 599, 601, 610 Recovery of valuable filtrates 578 Rectangular slab 641 Reflux 611 - drum 584, 585, 587, 595, 599, 603 -ratio 203, 585, 587, 604, 608, 609, 611, 614 Relative volatility 202, 211, 585, 587, 590, 592, 602, 606, 609 Relaxation time 89 Residence time 467, 560, 562 - distribution 159, 450 -

~

Index Response controller 99 -curve 87 - of a stirred-tank reactor 70 - of measuring element VI 70 - of the measuring instrument 72 Reversible reaction 372 Reynolds number 498 RTD experiments 165 -

Sampled data 107 Second-order -kinetics 519 -measuring element VI 76 - response lag 536 Segregation 471, 475 - in a batch reactor 474 - in a continuous reactor 474 - in a semi-continuous reactor 475 Semi-batch 430, see also semi-continuous - operation 130 - reactor 144, 146, 423-433, 475, 478, 518 -system 616 Semi-continuous reactor 423, 426, 475, 518 Semi-dimensionless form 649 Sensitivity - analysis 114 - t o parametric changes 661 Sequential reactions 330 Settler 183 - dynamics 186 Shell-and-tube heat exchanger 622 Short-cut flow 159 Signal and process dynamics 65 Simple overhead distillation 201 Simplified energy balance 41 Simulation -languages 7 - software 8, 9 - teaching applications 11 -tools 8 SIMUSOLV 114 Single -batch extraction 167 - solute batch extraction 527 - stage agitated extraction column 557 Slip velocity 196 Solid drying 637 Solute 167 - mutual interaction 174 Solvent o r organic phase 167 Specific interfacial area 62, 528 Spherical bead 654 Spherical shell 654

699 Split-boundary problems 123, 644 Spouted bed reactor 466 Stability of chemical reactors 361 Stage - continuous equilibrium extraction 171 - continuous equilibrium multistage extraction 548 - continuous extraction 540 - continuous flow equilibrium 540 - countercurrent multistage equilibrium extraction 548 - countercurrent multistage extraction 548 - countercurrent stagewise extractor cascade 552 - equilibrium behaviour 176 - equilibrium cascade 548 - equilibrium extraction 172 - equilibrium extractor 540 - extraction columns 192 - interstage flow rates 177 - multistage countercurrent extraction cascade 175 - multistage extractor with backmixing 552, 553 - multistage, multicomponent, extraction cascade 167 - non-equilibrium staged extraction column 193 - single stage agitated extraction column 557 Stagewise -absorption 198 - and differential mass transfer 58 - contactor 58 -discontinuous extraction 550 - distillation 201 -extraction columns 192 -mass transfer 166 -model with backmixing 177 -processes 129 Stagnant zones 165 Startup 604 Steady state 504, 566 - absorption column design 567 -balance 547,609 -conditions 545 -criterion 156 -diffusion 227 -energy balance 395,633 -operation 132 - simulation 222 -split boundary problem 648 -tubular reactor dispersion model 247 - two-pass heat exchanger 628 -values 563

700 Steam -density 137 - distillation 616 - -heated stirred tank 514 - tables 137 Step change 539 Stiffness 175 Still 585, 589 Stirred-tank reactors 129 Stoichiometric coefficient 52 Stoichiometry 143 Stripping section 209 Superficial - phase velocities 194 - velocity 557, 558, 565 Surface drying rate 641 System stability 126 Tank drainage 30 Tank mixing 492 Tanks in series 74, 159, 164, 330, 509 - approximation of a tubular reactor 406 Taylor's expansion theorem 154 Temperature -control 95 - control for semi-batch reactor 518 - difference 133 - distribution 140 -measurement 508, 514, 518 - measurement device 505 - profile 615, 632 Theoretical plate 584, 593, 599, 610 - behaviour 585, 595, 602, 611 Thermal - capacity 139 - stability of a CSTR 340 Thin slab 636 Time constants 51, 68, 80-93 - application of 93 - for transfer 539 - of chemical reaction 91 - of diffusion-dispersion 91 - of flow phenomena 90 - of heat transfer 93 - of mass transfer 92 - of measurement 71, 507, 539 Time delay 186 Time lags 560 Toluene 605 Total mass baiance 15, 29, 131 Total reflux 588, 615 Tracer 162 - diagnosis 161 - pulse response 164

Index 164, 407 studies 165 Transfer fluxes 182 Transfer function 80-83 - first-order lag model 81 - higher order response curves 83 - in series 525 - lags in series 84 - pure time delay 82 - simulation 524 Transfer rate 544 - relationships 168 Transient holdup profiles in an agitated extractor 556 Trial and error method controller tuning 101 Tube 623 Tubular and tank reactors 384 Tubular reactor 8, 18, 229-247, 393, 402 - and tank reactors 384 -batch reactor analogy 239 -boundary conditions 245 - component balance dispersion model 244 - dynamics of 240,405 - finite-differencing for 222 - gas-phase 235 - isothermal 388 - model for the steady state of 381 - plug-flow in 230 - steady-state dispersion model for 247 - tanks-in-series approximation of 406 -with axial dispersion 243, 414, 416 -with unmixed reactants 471 Tuning the controller 547 Turbulent fluctuations 472 Two-film theory 61, 168 Two-pass heat exchanger 628 -0-solute batch extraction 531 -with interacting equilibria 530 Two tank level control 509 - response -

Ultimate gain method 508 -controller tuning 103 Unmixed reactants to a tubular reactor 471 Unsteady-state - diffusion 224 -model 416 Van Laar equation 611, 614 Van? Hoff equation 56, 373 Vapour pressure 137 Variable -heat capacities 372 -volume reactor 519 Volumetric feed rate 145

701

Index Wash curve with dispersion coefficient 583 Washing of filter cake 578 Waste -holding tank 32 -treatment 653 Water 615 -heater 505 -jacket 519 - - methanol system 610 Well-mixed region 189

Yield

158

Zero-order - kinetics 657 - reaction 649 Ziegler-Nichols method 507, 517 - controller tuning 102