Process Plant Simulation

Process Plant Simulation B.V. Babu Birla Institute of Technology and Science, Pilani, Rajasthan

OXFORD UNIVERSITY PRESS

OXFORD UNIVERSITY PRESS

YMCA Library Building, Jai Singh Road, New Delhi 110001

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The moral rights of the authorls have been asserted. Database right Oxford University Press (maker) First published 2004 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, without the prior permission in writing of Oxford University Press, or as expressly permitted by law, or under terms agreed with the appropriate reprographics rights organization. Enquiries concerning reproduction outside the scope of the above should be sent to the Rights Department, Oxford University Press, at the address above. You must not circulate this book in any other binding or cover and you must impose this same condition on any acquirer. ISBN 0-19-566805-7

Qpeset in Times by Le Studio Graphique, Gurgaon 122001 Printed in India by Rani Book Binding House, New Delhi 110020 and published by Manzar Khan. Oxford University Press YMCA Library Building, Jai Singh Road, New Delhi 110001

It was the discovery of crude oil in 1859 by Drake that led to the thought that there was the need to have a separate discipline of chemical engineering. Until then, chemists and mechanical engineers also served as chemical engineers. In the early stages of the evolution of chemical engineering as a separate discipline, it was thought to be a combination of unit operations and unit processes. Since then, chemical engineering has been dynamic and has undergone many changes. The beauty of this discipline lies in its versatility and ability to adapt itself to various new and interdisciplinaryfields of engineering and science. Environmental engineering,biochemical engineering, energy engineering, optimization, and process control and instrumentationcan all come under the single umbrella of chemical engineering. Taking these dynamic changes and the versatility of this discipline into account, it has been redefined as comprising the concepts of process synthesis, analysis, and optimization. Process synthesis deals with the selection of the best process alternative out of the millions of alternate process flowsheets available for manufacturingany product. A systematic methodology has been developed based on the conceptual design principles for this process synthesis activity. It includes making process design decisions at various levels of hierarchy, such as batch versus continuous, input-output structure, recycle structure, general structure, and energy integration. The complexity of the decisions increases as the level goes up from batch versus continuous to energy integration. Process analysis deals with the detailed simulation of process plants for the best process alternative selected from the process synthesis activity. Optimization, as the name suggests, deals with finding the optimum design parameters for maximizing the profit or minimizing the total investment on a process plant (trade-off between capital and operating costs). Process plant simulation, as a subject, deals with concepts on process analysis and optimization. It includes the concepts of modelling, optimization, decomposition of networks, modular and equation-solving approaches, data regression, convergence promotion, specific-purpose simulation and dynamic simulation, etc. I have been teaching the modelling part of this course for the last 17 years of my teaching career. Earlier, modelling was part of a course I was teaching at Gujarat University during 1985-1996 as chemical systems modelling, and now at BITS, Pilani, it is part of process plant simulation, which I have been teaching for the last five years.

About the Book There is no consolidated literature available which covers all these aspects of process plant simulation. The idea of writing a book on process plant simulation originated in my mind two years ago. I thought it appropriate and felt the necessity to bring all the concepts and aspects of this subject together in the form of a single book. The purpose and objectives of this book are fourfold: to provide a textbook for the course on process plant simulation, bringing all related concepts together, offered basically in undergraduate and postgraduate programmes at various engineering colleges and universities across the country and abroad

vi

Preface

to introduce a generalized approach to modelling various engineering systems, so that these principles can be used to model any new situation or system in various engineering, science and other applied disciplines I to introduce non-traditional optimization techniques, which can serve as a research monograph and guide to the methodology for various evolutionary optimization techniques (population-based search algorithms) I to familiarize students with professional software packages such as HYSYS and FLUENT I

Content and Structure The book is divided into five parts. A brief introduction to process synthesis, process analysis, and optimization is followed by the presentation of process plant simulation as a subject in Chapter 1. Subsequently, Part I gives an overview of modelling aspects in Chapter 2, and the classification of mathematical modelling in Chapter 3. Part II discusses chemical systems modelling in four chapters: modelling mass-transfer systems in Chapter 4,heat-transfer systems in Chapter 5 , and fluid mechanics and reaction engineering systems in Chapters 6 and 7, respectively.Apart from focusing on modelling various chemical engineering systems, various mathematical techniques to be used for solutions of the models proposed (such as algebraic equations, ordinary differential equations, partial differential equations, finite difference equations, Laplace transformations, solution by series, etc.) have also been dealt with. Part 111deals with the treatment of experimental results, wherein error propagation and data-regression techniques (Chapter 8) are discussed in detail. Y focuses on traditional optimizationtechniques (analyticalmethods, Lagrangian Part J multiplier method for constraint optimization, gradient methods such as steepest descent and sequential simplex methods, etc.) in Chapter 9, and non-traditional optimization techniques (such as simulated annealing, genetic algorithms, differential evolution, evolutionary strategies, etc.) in Chapter 10. Finally, Part V, consisting of five chapters, gives a detailed overview .of all aspects related to simulation. Chapter 11 covers the modular and equation-solving approaches. Chapter 12 deals with the decomposition of networks and the various associated tearing algorithms available in literature. Chapter 13 covers convergence promotion and the physical and thermodynamic properties. Chapter 14 discusses case studies on specificpurpose simulation and dynamic simulation. Chapter 15 gives an overview of two most widely used professional software packages, namely, HYSIS of HyproTech and FLUENT, and demonstrates the step-by-step procedure of solving problems using these software packages.

Key Features Offers exhaustive coverage of all topics related to plant simulation I Covers traditional as well as non-traditional optimization techniques I Discusses case studies on specific-purpose and dynamic simulation I

Preface

vii

Includes an overview of professional softwarepackages used in plant simulation, such as HYSIS and FLUENT rn Includes a CD-ROM containing program codes and related useful information Comes with a Solutions Manual which provides the solutions to selected problems

Acknowledgements Many people have contributed directly or indirectly in bringing this book to its present form since its conception in the year 2000. To begin with, I would like to accord my deep sense of gratitude and thanks to Prof. S. Venkateswaran, Vice Chancellor, BITS, Pilani, for encouraging this course development work. I must acknowledge the active guidance and encouragement I received from Prof. L.K. Maheshwari, Director, BITS Campus, Pilani; Prof. K.E. Raman, Deputy Director (Administration),BITS, Pilani, who constantly inspired me in time management and encouraged me to complete this work; and Prof. V.S. Rao, Deputy Director (Off-campus Programmes), BITS, Pilani, for being a fatherly figure for me. I must thank Prof. R.K. Patnaik, Prof. G.P. Srivastava,Prof. A.K. Sarkar, Dean (InstructionDivision), BITS, Pilani, and Prof. R.N. Saha, Dean (Educational Development Division), BITS, Pilani, who made the necessary arrangements for the manuscript to be ready in time. I am deeply grateful to my colleagues Mr Rakesh Angira, Mr Ashish Chaurasia, Mr V. Ramakrishna, Mr Nitin Maheshwari, Mr Sheth Pratik, and Mr Suresh Gupta, who are doing their PhD under my guidance, for helping me in reviewing portions of the manuscript and providing valuable comments apart from making the figures and proofreading. I am indebted to all my students at BITS, Pilani (especially Mubeen, Pallavi, Ravindra, Gautam, Shobhana, Vishnu, Simli, Mandar, Angarasu, Amol, and Subha Jyothi), for their tireless effort and help in bringing out the preliminary draft of the book. I cannot find words to describe the debt I owe to Dr Rainer Storn of ICSI, Berkeley, Prof. David E. Goldberg of the University of Illinois at UrbanaChampaign, and Prof. Kalyanmoy Deb of IIT, Kanpur, whose inputs have helped me to understand non-traditional optimization techniques better. Last but not the least, this work would not have been completed without the understanding and support I got from my wife Shailaja and my children Shruti and Abhinav.

I heartily welcome any constructivesuggestions,criticism, and appraisal by the readers, which would help me in improving the quality of this book in subsequent editions. B.V. BABU

Contents Preface

V

1. Introduction

1

1.1 1.2 1.3 1.4 1.5

Process Synthesis 3 Process Analysis 5 Optimization 8 Process Plant Simulation 10 Organization of the Text 10 Exercises 12

PARTI MODELLING 2. Modelling Aspects 2.1 Deterministic Versus Stochastic Processes 15

15

2.1.1 Deterministic Process 15 2.1.2 Stochastic Process 16

2.2 Physical Modelling 16 2.3 Mathematical Modelling 17 2.4 Chemical Systems Modelling 20 2.4.1 Model Formulation Principles 21 2.4.2 Fundamental Laws used in Modelling

22

2.5 Cybernetics 31 2.6 Controlled System 3 1 2.7 Pri,iciples of Similarity 32 2.7.1 Geometric Similarity 32 2.7.2 Kinematic Similarity 32 2.7.3 Dynamic Similarity 33

Exercises 38

3. Classification of Mathematical Modelling

39

3.1 Independent and Dependent Variables, and Parameters 39 3.2 Classification based on Variation of Independent Variables 42 3.2.1 Distributed Parameter Models 42 3.2.2 Lumped Parameter Models 42

3.3 Classification Based on the State of the Process 42 3.3.1 Static Model 42 3.3.2 Dynamic Model 43 3.3.3 The Complete Mathematical Model

44

3.4 Classification Based on the Type of the Process 45 3.4.1 Rigid or Deterministic Models 45 3.4.2 Stochastic or Probabilistic Models 45 3.4.3 Comparison Between Rigid and Stochastic Models

3.5 Boundary Conditions 46 3.6 The Black Box Principle 46

45

Contents

3.7 Artificial Neural Networks 47 3.7.1 Network Training 48 3.7.2 Modes of Training 49 3.7.3 Network Architecture 49 3.7.4 Back-propagation Algorithm 3.7.5 ANN Applications 51 3.7.6 Example Problems 51 Exercises 57

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50

PARTII CHEMICAL SYSTEM MODELLING 4. Models in Mass-transfer Operations 4.1 Steady-state Single-stage Solvent Extraction 6 1 4.2 Steady-state Two-stage Solvent Extraction 65 4.3 Steady-state TWO-stage Cross-current Solvent Extraction 69 4.4 Unsteady-state Single-stage Solvent Extraction 73 4.5 Unsteady-state Mass Balance in a Stirred Tank 79 4.6 Unsteady-state Mass Balance in a Mixing Tank 82 4.7 Unsteady-state Mass Transfer (Fick’s Second Law of Diffusion) 85 4.8 Steady-state N-stage Counter-current Solvent Extraction 90 4.9 Multistage Gas Absorption (Kremser-Brown Equation) 94 4.10 Multistage Distillation 98 Exercises 103

61

5. Models in Heat-transfer Operations 5.1 Steady-state Heat Conduction Through a Hollow Cylindrical Pipe (Static Distributed Parameter Rigid Analytical Model) 108 5.2 Unsteady-state Steam Heating of aLiquid (Dynamic Lumped Parameter Rigid Analytical Model) 115 5.3 Unsteady-state Heat Loss Through a Maturing Tank (Dynamic Lumped Parameter Rigid Analytical Model) 117 5.4 Counter-current Cooling of Tanks 124 5.5 Heat Transfer Through Extended Surfaces (Spine Fin) 129 5.6 Temperature Distribution in a Transverse Cooling Fin of Triangular Cross Section 135 5.7 Unsteady-state Heat Transfer in a Tubular Gas Preheater 138 5.8 Heat Loss Through Pipe Flanges 139 5.9 Heat Transfer in aThermometer System 141 5.10 Unsteady-state Heat Transfer by Conduction 145 Exercises 148

107

6. Models in Fluid-flow Operations

152

6.1 The Continuity Equation 152 6.2 Flow Through a Packed Bed Column 155 6.3 Laminar Flow in a Narrow Slit 156

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Contents

6.4 6.5 6.6 6.7 6.8 6.9 6.10

Flow of a Film on the Outside of a Circular Tube 160 Choice of Coordinate Systems for the Falling Film Problem 162 Annular Flow with Inner Cylinder Moving Axially 164 Flow Between Coaxial Cylinders and Concentric Spheres 166 Creeping Flow Between Two Concentric Spheres 169 Parallel-disc Viscometer 172 Momentum Fluxes for Creeping Flow into a Slot 174 Exercises 178

7. Models in Reaction Engineering 7.1 7.2 7.3 7.4 7.5

181

Chemical Reaction with Diffusion in a Tubular Reactor 181 Chemical Reaction with Heat Transfer in a Packed Bed Reactor 185 Gas Absorption Accompanied by Chemical Reaction 191 Reactors in Series-I 192 Reactors in Series-I1 196 Exercises 199

PART111 TREATMENT OF EXPERIMENTAL RESULTS 8. Error Propagation and Data Regression 8.1 Propagation of Errors 203 8.1.1 Propagation Through Addition 203 8.1.2 Propagation Through Subtraction 204 8.1.3 Propagation Through Multiplication and Division

203

204

8.2 Sources of Errors 204 8.2.1 8.2.2 8.2.3 8.2.4

Errors of Measurement 205 Precision Errors 205 Errors of Method 205 Significant Figures 205

8.3 Data Regression (Curve Fitting) 205 8.3.1 Theoretical Properties 205 8.3.2 Data-regression Methods 206 8.3.3 Problems on Data Regression 21 1

Exercises 216

PARTIV OPTIMIZATION 9. TraditionaI 0pt imizat ion Techniques 9.1 Limitations of Optimization 221 9.2 Applications of Optimization 222 9.3 Types of Optimization 224

221

9.3.1 Static Optimization 224 9.3.2 Dynamic Optimization 224

9.4 Methods of Optimization 225 9.4.1 Analytical Methods of Optimization 226 9.4.2 Optimization with Constraints (Lagrangian Multipliers)

227

Contents 9.4.3 Gradient Methods of Optimization 9.4.4 Rosenbrock Method 252 9.4.5 Other Methods 252

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231

Exercises 255

10. Non-traditional Optimization Techniques 10.1 Simulated Annealing 258

257

10.1.1 Introduction 258 10.1.2 Procedure 258 10.1.3 Algorithm 259

10.2 Genetic Algorithms 260 10.2.1 10.2.2 10.2.3 10.2.4 10.2.5 10.2.6 10.2.7 10.2.8

Introduction 260 Definition 260 The Problem 261 Techniques used in GA 261 Operators in GA 262 Comparison of GA with Traditional Optimization Techniques Applications of GA 265 Stepwise Procedure for GA Implementation 267

265

10.3 Differential Evolution 272 10.3.1 10.3.2 10.3.3 10.3.4 10.3.5 10.3.6 10.3.7 10.3.8 10.3.9

Introduction 272 XOR Versus Add 273 DE at a Glance 273 Steps Performed in DE 273 Choice of DE Key Parameters (NP, F,and CR) 275 Strategies in DE 275 Innovations on DE 276 Applications of DE 276 Stepwise Procedure for DE Implementation 283

10.4 Other Evolutionary Computation Techniques 286 10.4.1 10.4.2 10.4.3 10.4.4

Evolution Strategies 286 Evolutionary Programming 287 Genetic Programming 287 Other Population-based Search Algorithms

288

Exercises 290

PARTV SIMULATION 11. Modular Approaches and Equation-solving Approach 11.1 Modular Approaches to Process Simulation 295 1 1.1.1 Analysis Versus Design Mode

295

11.2 The Equation-solving Approach 300 11.2.1 11.2.2 11.2.3 11.2.4 11.2.5

Precedence-Ordering of Equation Sets Disjointing 305 Tearing a System of Equations 305 The SWS Algorithm 305 Maintaining Sparsity 307

Exercises 3 11

301

295

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Contents

12. Decomposition of Networks

31 3

12.1 Tearing Algorithms 3 13 12.2 Algorithms Based on the Signal Flow Graph 314 12.2.1 The Barkley and Motard Algorithm 12.2.2 The Basic Tearing Algorithm 319

315

12.3 Algorithms Based on Reduced Digraph (List Processing Algorithms) 321 12.3.1 Kehat and Shacham Algorithm 321 12.3.2 M&H Algorithms 323 12.3.3 Comparison of Various Tearing Algorithms

331

Exercises 332

13. Convergence Promotion and Physical and

337

Thermodynamic Properties 13.1 Convergence Promotion 337 13.1.1 13.1.2 13.1.3 13.1.4 13.1.5 13.1.6 13.1.7

Newton’s Method 338 Direct Substitution 339 Wegstein’s Method 341 Dominant Eigenvalue Method 346 General Dominant Eigenvalue Method Quasi-Newton Methods 347 Criterion for Acceleration 348

346

13.2 Physical and Thermodynamic Properties 349 13.2.1 Sources 349 13.2.2 Databanks 350 13.2.3 Modularity and Routing

352

Exercises 353

14. Specific-purpose Simulation and Dynamic Simulation 14.1 Auto-thermal Ammonia Synthesis Reactor 355 14.1.1 14.1.2 14.1.3 14.1.4 14.1.5

Ammonia Synthesis Reactor 355 Problem Formulation 357 Simulated Results and Discussion 361 Optimization 365 Conclusions 370

14.2 Thermal Cracking Operation 371 14.2.1 14.2.2 14.2.3 14.2.4 14.2.5

Thermal Cracking 371 Problem Description 372 Problem Reformulation 374 Simulated Results and Discussion Conclusions 376

375

14.3 Design of Shell-and-Tube Heat Exchanger 377 14.3.1 14.3.2 14.3.3 14.3.4

The Optimal HED Problem 377 Problem Formulation 378 Results and Discussion 380 Conclusions 395

355

Contents 14.4 Pyrolysis of Biomass 396 14.4.1 Stepwise Procedure 396 14.4.2 Problem Formulation 413 14.4.3 Method of Solution and Simulation 14.4.4 Results and Discussion 416 14.4.5 Conclusions 429 Exercises 430

415

432

15. Professional Simulation Packages 15.1 HYSIS-A 1 5 , l .1 15.1.2 15.1.3 15.1.4 15.1.5

xiii

Professional Software Package 432 Integrated Simulation Environment 433 HYSIS Products 433 Intuitive and Interactive Process Modelling 434 Open and Extensible HYSYS Architecture 435 Stepwise Methodology of HYSYS Usage for Problems

435

15.2 FLUENT-A Software Package for Computational Fluid Dynamics 479 15.2.1 Structure of the Program 480 15.2.2 Capabilities of FLUENT 481 15.2.3 Using FLUENT-An Overview 482 15.2.4 Physical Models in FLUENT 483 15.2.5 Stepwise Methodology of FLUENT Usage for a Problem Exercises 5 13

484

References

520

Index

532

CHAPTER 1

INTRODUCTION

The transformation of raw materials into desired products usually cannot be achieved in a single step in any chemical process. The overall transformation is broken down into a number of steps that provide intermediate transformations. These are carried out through reaction, separation, mixing, heating, cooling, pressure change, particle size reduction and enlargement, etc. So, the synthesis of a chemical process involves two broad activities. The first is the selection of individual transformation steps, and the second is finding the interconnections between these individual transformations to form a complete structure that achieves the required overall transformation, leading to a flowsheet, which is a diagrammatic representation of the process steps with their interconnections. The simulation of the process can be carried out after the flowsheet structure is defined. Simulation is a mathematical model of a process, which attempts to predict how the process would behave if it was constructed. After creating a model of the process, the flow rates, compositions, temperatures, and pressures of the feeds are assumed. The simulation model can then be used to predict the flow rates, compositions, temperatures, and pressures of the products. Simulation also allows the sizing of the individual items of equipment in the process and prediction of the amount of raw material, energy, etc. required. The performance of the design can then be evaluated. Once the basic performance of the design has been evaluated, changes can be made to improve the performance, i.e., optimization. These changes might involve the synthesis of alternate structures, i.e., structural optimization (Smith 2000). As energy consumption expands and resources of material and energy become more expensive and less certain, substantial changes will be required in many of

2

Process Plant Simulation

the conventional chemical processing plants (Husain 1986). These have to incorporate a high degree of energy integration and achieve greater efficiencies through process modifications, thus conserving material and energy resources. At the same time, safety standards are tightening, therefore lower temperatures and pressures are desired. It is obvious that higher reaction rates, fewer stages, and smaller plants will reduce the burden of the material in a process, leading to safer plants. In order to meet these challenges, greater effort on the part of the process engineers towards evaluation of the process flowsheets is necessary. Today, the emphasis is on tackling a process design problem from a broader perspective. This is facilitated by the use of computers and computer-aided design (CAD) software. The computer-aided design and simulation of a variety of systems is catching up very fast and being applied extensively in all fields of science, arts, engineering, and medicine. In addition, chemical process engineers need to carry out process plant simulation for visualizing plant processes, performing heat and material balances of process flowsheets, designing new plants, suggesting modifications and expansions of existing plants (retrofitting), helping engineers to develop a better understanding of how their plants really operate, reducing costs and increasing profits, predicting operating efficiencies and anticipating problems, troubleshooting process and control problems, assisting operators in planning for production changes and disruptions, training operators, etc. The design, development, and control of chemical plants and equipment are an essential part of chemical engineering. From an industrial standpoint, the economic feasibility and profitability of any process should be favourable for its implementation. It is therefore important for a chemical engineer to combine plant design and economics while formulating a prospective design plan. Process plant simulation constitutes the process analysis and optimization aspects of process design. Chemical engineering discipline has been evolving continuously over the years without any limits, both in depth as well as in breadth. The beauty of this discipline lies in its ability to be versatile and its adaptability to various interdisciplinary areas (environmental engineering, biochemical engineering, energy engineering, optimization, to name a few). In order to appreciate the aspects involved in process plant simulation and to understand the importance of this subject, we have to go back to the history of evolution of chemical engineering as a discipline. Until Drake discovered crude oil in 1859, the chemists and mechanical engineers used to do the work of chemical engineers. It was the discovery of crude oil which led to the feeling among the scientists and academicians that there was a need to have a separate discipline in order to cater to the needs of processing and purifying various useful products from crude oil. The sequence of events that led to the birth

Introduction

3

of chemical engineering as a discipline is as follows: The first course in chemical engineering was offered by Prof. Lewis M. Norton in 1888 at MIT, USA for chemical major students. A full-fledged course was offered for the first time by the University of Pennsylvania in 1908; but a methodology on which chemical engineering could develop as a distinct discipline was still lacking. Two other developments, which had a bearing on the growth of chemical engineering as a discipline, took place in the late nineteenth century and the early twentieth century. In 1915, Arthur D. Little introduced the unit operations (mechanical operations, fluid-flow operations, heat-transfer operations, mass-transfer operations, etc.) concept. Subsequently, in 1922, the unit processes (oxidation, hydrogenation, nitration, sulphonation, etc.) concept was introduced by Groggins. Therefore, chemical engineering is defined as a combination of unit operations and unit processes. The roots of chemical engineering are physics, chemistry, and mathematics. A broad-based definition for chemical engineering given by AIChE is ‘Application of the principles of physical sciences, together with the principles of economics and human relations, to fields that pertain directly to processes and process equipment in which matter is treated to effect a change in state, energy, content, or composition’. A working definition of a chemical engineer could be ‘one who can develop, design, construct, control, and manage any process involving physical and/or chemical changes (chemical changes include biochemical changes)’. Keeping the history and versatility of chemical engineering in the background, considering the AIChE and working definitions of improved and matured chemical engineering, and bearing the challenges ahead of a chemical engineer in mind, the complete process design can be viewed as being carried out in the following three stages: 1. Process synthesis 2. Process analysis 3. Optimization

1.1 Process Synthesis Process synthesis is the first stage of the process design activity, in which a flowsheet of the process is constructed and this includes all equipment and their interconnections. For this task, information from several sources has to be gathered. One of the most commonly used sources is experience gathered over the years in solving both simple and complex design problems in a particular area of a process industry. In other words, process synthesis involves the use of design heuristics. But this may not provide the best efficiency for a given system. There may be other flowsheets that give higher levels of efficiency. Therefore, in recent times, the

4

Process Plant Simulation

approach has been to develop a generalized and systematic approach to process synthesis. The major feature that distinguishes design problems from other types of engineering problems is that they are under-defined (i.e., only a very small fraction of the information needed to define a design problem is available from the problem statement) and open-ended (to supply the missing information, assumptions are to be made and there are numerous ways of achieving the same goal). Process synthesis deals with the conceptual design of chemical processes, following a systematic procedure (Douglas 1988). The objectives of conceptual design are to find the best process flowsheet (Lee,to select process units and their interconnections), and optimum design conditions. This is very difficult, as experience shows that there are to the tune of 104-105 possible alternatives to be considered in process synthesis activity. In addition, the success rate of new ideas and subsequent designs ever becoming commercialized is very less (in fact it is less than 1%). That means there are many possibilities to consider, with a small chance of success. In many cases the processing costs associated with the various process alternatives differ by an order of magnitude or more, so that short-cut calculations have to be used to screen the alternatives. At the same time, one should be certain that the solution is in the neighbourhood of the optimum design conditions for each alternative, in order to prevent discarding an alternative because of poor choice of design variables. So, cost studies must be used as an initial screening to eliminate ideas for designs that are unprofitable. If a process appears to be profitable, other factors such as safety, environmental constraints, controllability, etc. should be considered. Because of the under-defined and open-ended nature of design problems, and because of the low success rate, it is useful to develop a strategy for solving design problems. This process synthesis activity has to be carried out by establishing a hierarchy of design decisions. A systematic procedure has been laid down for screening the alternatives by following five levels of design decision hierarchy. With this approach, a very large and complex problem can be decomposed into a number of smaller problems that are much simpler to handle. By focusing on the decisions that must be made at each level in the hierarchy, one can identify the existing technologies that could be used to solve the problem without precluding the possibility that some new technology might provide a better solution. Moreover, by listing the alternate solutions that are proposed for each decision, a list of process alternatives can be systematically generated. In some cases, it is possible to use design guidelines (rules of thumb or heuristics) to make decisions about the structure of the flowsheet and/or to set the values of some of the design variables. Order-of-magnitude arguments are used to derive many of these heuristics, and a simple analysis of this type is used to identify the limitations of the heuristics. In cases where no heuristics

Introduction

5

are available, short-cut design methods are used as a basis for making decisions. The five levels of decision hierarchy in conceptual design are 1. Batch versus continuous production 2. Input-output structure of the flowsheet 3. Recycle structure of the flowsheet 4. General structure of the separation system (a) Vapour recovery system (b) Liquid recovery system 5. Heat-exchanger network synthesis (energy integration) A beginner can substitute the evaluation of a number of extra calculations for experience during the development of a conceptual design by following this hierarchical decision procedure. However, the penalty paid in the form of time required to screen more alternatives is not very high, as short-cut calculations are used. As a designer gains experience, it is possible to recognize alternatives that should not be considered for a particular type of process, and thereby improve efficiency. This entire process synthesis activity is carried out prior to the process analysis stage to find the best process alternative by systematically screening millions of possible process alternatives following the above-mentioned hierarchy of design decisions. But this process synthesis is beyond the scope of the present course on process plant simulation. Process synthesis itself is vast and is offered as a separate compulsory course named Process Design Decisions.

1.2 Process Analysis Process analysis is the next stage in process design after process synthesis. Once the process flowsheet is synthesized, an analysis is required for the following purposes: rn Solving material and energy balances for a steady-state process rn Sizing and costing the equipment rn Evaluating the worth of the flowsheet Chemical process simulation, also known as ‘flowsheeting’, is represented by a mathematical model in order to obtain information about the response of a plant to various inputs. The salient features and requirements of general purpose simulation are (Husain 1986) rn Modular approaches (sequential and simultaneous) to process simulation rn Equation-solving approach (unconstrained and constrained material balances) rn Decomposition of networks (partitioning, tearing algorithms, etc.)

6

Process Plant Simulation w w w w

Convergence promotion Physical and thermodynamic properties Specific-purpose simulation and Dynamic simulation

Simulation can be carried out either in the design mode or in the analysis mode. In the design mode the outputs are specified and the corresponding inputs are simulated using the system model equations for various units. On the contrary, in the analysis or performance mode, the inputs are specified and the corresponding outputs are simulated. Ideally, with an aim of using process simulation as a design tool, the system inputs andor design parameters should be calculated from the specified outputs (i,e,, the design mode should be used). But such a simulation in a design mode is numerically less stable than that in the analysis or performance mode. The latter is characterized by the fact that all system inputs and design parameters for the units are specified and the outputs are calculated using the given information. Thus, the information flow in the analysis mode is in the same direction as the energy and material flow in a chemical plant. Being numerically more stable, the analysis mode is more often used to perform design calculations iteratively for making case studies. Hence, a majority of application packages are written in this mode of simulation. Calculating heat and mass balances is the most tedious and repetitive problem of process design. As mentioned in the previous section on process synthesis, during the initial stages of a flowsheet study, of a new process or an existing one, simple material balances alone may suffice. At this level, specifications may be set from the plant data if an existing plant is being considered, or from engineering experience or pilot plant data if a new process flowsheet is under development. In the later stages of design, however, heat and mass balances must be calculated along with the equilibrium and rate equations, P-V-T relationships, and relations governing counter-current operations. All these equations are generally strongly non-linear. Moreover, in the simplest case, if a process is of sequential configuration, it is easy to proceed from the feed streams until the products are obtained, calculating sequentially for one process unit after the other. Unfortunately, most chemical plants are of complex configurations involving recycling of streams, mass and/or energy; they represent interlinked networks of units. To make the calculation procedure sequential in such situations, it is essential to decompose the network. This, in turn, requires convergence promotion. The necessity of incorporating constraints or specifications other than the natural parameters while modelling a flowsheet has, of course, been recognized for quite some time in computer-aided process design. The importance of simulating in the design mode compared to the analysis mode has been outlined above. This differentiates rating calculations from design calculations. In the former case, each

Introduction

7

unit is specified and its performance is calculated. In the latter case, performance is specified and each unit is designed to give the required performance. In the sequential modular approach, constraints can be accommodated using additional iteration loops around the module, provided these constraints involve stream quantities associated with that module. However, if the constraints involve streams not incident to the underspecified module, then iterations in the outer loops involving the entire flowsheet are needed; this procedure is quite cumbersome and tedious. For this reason, the modular approach has not gained popularity for simulation in the design mode. In the equation-solving approach, the mathematical model of a steady-state process is organized and handled as one large global set of equations representing the entire process. This is in contrast to subsets of equations, called modules, used in the modular approach according to the process units that appear in a given process. Hence, in the global approach of equation-solving, any number of constraints can be added in the form of equations to the set defining the problem. Because the global approach analyses all the equations representing an entire chemical process, it takes full advantage of the specific features of these equations, which are ignored in the modular approach. It also generates a tailored computer program for each new problem, while the same modular-oriented simulator is used for all the problems. The tailored computer program executes faster and uses less memory compared to the modular-oriented simulator. However, the major disadvantage of the global approach is the necessity of manually preparing a new input description for all the equations of each new problem. This preparation can become tedious and highly error-prone, since chemical processes can be modelled using hundreds and thousands of non-linear equations. Several design organizations started building libraries of computer programs for various unit operations in the early 1960s. It then became evident that many of these could be put into a system enabling one to direct calculations for an entire flowsheet, thus saving considerable engineering time. This resulted in the development of specific-purpose programs to simulate a particular plant or part of a plant; these were naturally more detailed in nature but at the same time rigid in structure. Upon examining these programs, it became clear that a major portion could be common to all types of plants, i.e., calculations involved in various unit operations, methods used to compute different physical and thermodynamic properties, decomposition and convergence algorithms, cost information, as well as a library of numerical routines. Furthermore, a common approach to diagnostics might also be beneficial. Ultimately, this gave rise to several general-purpose simulation packages. Over the last few years, ‘steady-state simulation’ has become a significant input to process analysis. It represents the system under consideration by a mathematical model and obtains information about the system response by applying different sets of inputs to the model. The model is often applied in the form of a computer

8

Process Plant Simulation

program for a steady-state deterministic system simulation. Dynamic simulation (unsteady-state simulation) is necessary during the startup and shutdown of a process plant.

1.3 Optimization This is the final step in the development of a process flowsheet. Many constrained and unconstrained optimization techniques are employed for this purpose. Of late, nontraditional optimization techniques such as genetic algorithms and differential evolution are being used. All these interrelated activities should obviously lead to an optimal design and safe operation of a chemical plant. The chemical industry has undergone significant changes during the past 15 years due to the increased cost of energy and increasingly stringent environmental regulations. Modifications of both plant design procedures and plant operating conditions have been made in order to reduce costs and meet the constraints. Most industry observers believe that the emphasis in the near future will be on improving the efficiency and profitability of existing plants rather than on plant expansion (Edgar & Himmelblau 1989). One of the most important engineering tools that can be employed in such activities is optimization. As a result of computers having become more powerful, the size and complexity of problems, which can be simulated and solved by optimization techniques, have correspondingly expanded. The goal of optimization is to find the values of the variables in a process which yield the best value of the performance criterion. This usually involves a trade-off between capital and operating costs. Typical problems in chemical engineering design or plant operation have many, and possibly infinite number of, solutions. Optimization is concerned with selecting the best among the entire set by efficient quantitative methods. Unfortunately, no single method or algorithm of optimization can be applied efficiently to all problems. The method chosen for any particular case will depend primarily on (i) the character of the objective function and whether it is known explicitly, (ii) the nature of the constraints, and (iii) the number of independent and dependent variables. The general objective in optimization is to choose, keeping in view the various constraints, a set of values of the variables that will produce the desired optimum response for the chosen objective function. There are two distinct types of optimization algorithms in use today. First, there are algorithms that are deterministic, with specific rules for moving from one solution to the other. These algorithms (also known as traditional methods) have been successfully applied to some of the engineering design problems. Second, there are algorithms that are stochastic in nature, with probabilistic transition rules. These are comparatively new and are gaining popularity due to certain properties which deterministic algorithms do not have. As stated earlier, most of the traditional optimization algorithms based on gradient methods are susceptible to getting trapped

Introduction

9

at local optima depending upon the degree of non-linearity and the initial guess. Unfortunately, none of the traditional algorithms are guaranteed to find the global optimal solution, but genetic and simulated annealing algorithms are found to have a better global perspective than traditional methods (Deb 1996). Moreover, when an optimal design problem contains multiple global solutions, designers are interested in finding not just one global optimum solution, but as many as possible for various reasons. First, a design suitable in one situation may not be so in another situation. Second, designers may not be interested in finding the absolute global solution. They might instead be interested in a solution that corresponds to a marginally inferior objective function value but is more amenable to fabrication. Thus, it is always prudent to know about other equally good solutions for later use. However, if the traditional methods are used to find multiple optimal solutions, they need to be applied a number of times, each time starting from a different initial guess and hoping to achieve a different optimal solution. During the past two decades there has been a growing interest in algorithms, that are based on the principle of evolution (survival of the fittest). A common term, coined recently, refers to such algorithms as evolutionary algorithms (EAs) or evolutionary computation (EC) methods. The best-known algorithms in this class include genetic algorithms, evolutionary programming, evolution strategies, and genetic programming. There are many hybrid systems which incorporate various features of the above paradigms and consequently are hard to classify, and can be referred to as just EC methods (Dasgupta & Michalewicz 1997). Simulated annealing (SA) is a probabilistic nontraditional optimization technique, which mimics the cooling phenomenon of molten metals to constitute a search procedure. Rutenbar (1989) gave a detailed discussion of the working principle of SA and its applications. Since its introduction, SA has diffused widely into many diverse applications. Genetic algorithms (GAS)are computerized search and optimization algorithms based on the mechanics of natural genetics and natural selection. They mimic the ‘survival of the fittest’ principle of nature to make a search process. The key control parameters in GA are N , the population size; pc,the crossover probability; and pm, the mutation probability (Goldberg 1989). Price and Storn (1997) have given the working principle of differential evolution (DE), which is an improved version of GA, along with its application to polynomial fitting problems. They have also suggested some simple rules for choosing the key parameters such as the population size NP, crossover constant CR, and the weight F applied to the random differential (scaling factor) of DE for any given application. The three interrelated steps mentioned above (process synthesis, process analysis, and optimization) thus help to generate an optimal process plant design for any chemical process starting from the initial stage of conception. The flowsheet shown in Fig. 1.1 describes the strategy for process engineering.

10

P m e s s Plant Simulation Define process objective Collect and store information

-------

I_______

I

I

Create alternate processconcept

-------

I_______

Synthesize alternate I flowsheets L---,----------I

Control analysis

Equipment

I

I

L---,--,,,------

I

Total R&D effort

I

> I

operating costs Optimization

Is the project

Safety analysis

+ + I

Detailed plant design

Construction and installation

+ 1

startup

1

Operation

1

Product

Fig. 1.1 Strategy for process engineering

1.4 Process Plant Simulation It constitutes process analysis and optimization stages of process design. This book deals with the various aspects of process analysis and optimization, in other words, process plant simulation.

1.5 Organization of the Text As has been discussed in the process analysis activity, modelling plays an important role. Any process plant comprises a number of interconnected units. These units may be reactors, heat exchangers, pumps, compressors, distillation columns, absorbers, adsorbers, evaporators, extractors, leaching units, driers, mixers, etc. The output of one unit becomes the input of another unit. So, in order to carry out process plant simulation, the outputs of the various units are to be predicted from

Introduction

11

the known inputs or from simulated outputs of the previous unit. This can be done using models of the individual units. Similarly, optimization is also a very important aspect in process design. With reference to some of the other process analysis aspects-such as modular approaches to process simulation, the equation solving approach, decomposition of networks, convergencepromotion, physical and thermodynamic properties-specific-purpose simulationand dynamic simulation are discussed in brief in this text for completeness of the subject under consideration.There has been no consolidatedliterature available on modelling and optimization aspects which covers the entire gamut of these primarily important process design aspects. Hence, the present text focuses on these two aspects of process plant simulation, and also covers the associated aspects on treatment of experimental results. Keeping these objectives in mind, the text is organized as follows. Part I deals with various aspects of modelling: Chapter 1 gives a brief introduction of the subject. Chapter 2 gives a brief introduction to modelling, deterministic versus stochastic processes, physical and mathematical modelling, model formulation principles, cybernetics, controlled systems, and principles of similarity. Chapter 3 discusses the classification of mathematicalmodelling, the black box principle, artificial neural networks, dependent and independent variables, parameters, and boundary conditions. Part I1 deals with chemical systems modelling, focusing on different areas of chemical engineering. The required mathematical techniques for solutions are emphasized as and when required. Chapters 4,5 , and 6 focus on models in the areas of mass-transfer, heat-transfer, and fluid-flow (momentum-transfer) operations. Chapter 7 deals with models on reaction engineering. Part I11 deals with the treatment of experimental results. Chapter 8 includes various types of errors and their propagation, and data regression techniques such as the method of averages and the method of linear least squares. Part IV focuses on various optimization techniques. Chapter 9 covers the traditional optimization techniques including analytical methods, constrained optimization (Lagrangian multiplier), gradient methods such as the method of steepest descent (ascent)and the sequential simplex method, random search methods such as the box complex and Rosenbrock methods, etc. Chapter 10 discusses the nontraditional optimization techniques such as simulated annealing, genetic algorithms, differential evolution, evolutionary strategies, etc. PartV deals with the remaining aspects of process analysis. Chapter 11 focuses on modular approaches to process simulation and the equation solving approach. Chapter 12 deals with the decomposition of networks and the corresponding tearing algorithms such as the Barkley and Motard algorithm, basic tearing algorithm, and list processing algorithms such as the Kehat and Shacham algorithm and various forms of the Murthy and Husain algorithm. Chapter 13 discusses the convergence

12

Process Plant Simulation

promotion, and physical and thermodynamic properties. Chapter 14 covers the specific-purpose simulation and dynamic simulation, discussing industrial problems as case studies. Chapter 15 discusses two professional simulation software packages (HYSYS and FLUENT),and the stepwise methodology for using these packages is demonstrated by solving four engineering problems.

EXERCISES 1.1 Discuss various stages of the development of chemical engineering as a discipline. 1.2 What are the various unit operations and unit processes? What are the analogies among these? 1.3 Describe various levels of decision hierarchy in process synthesis? 1.4 What are the aspects covered in process analysis? 1.5 What is the role of optimization in process design? 1.6 Discuss the strategy of process engineering with the help of a neat flow chart. 1.7 What is process plant simulation?

PART I

MODELLING

IN

THE NEXT two chapters we will understand the importance of modelling in general and the model formulation principles for applying to chemical engineering systems in particular.We will see the types of modelling (physical and mathematical), understand the relationship among cybernetics, computers, and mathematical modelling, and classify various processes and mathematical modelling. Then we will discuss the black box principle and artificial neural networks. We will also focus on the related aspects of modelling such as the controlled system, principles of similarity, dependent and independent variables,. parameters, and boundary conditions.

CHAPTERS 2. Modelling Aspects 3. Classification of Mathematical Modelling

CHAPTER 2

MODELLING ASPECTS

Most of the engineeringprocesses that demand accurate product quality, and proceed at high rates, high temperatures, and high pressures, are distinct for their utmost complexity. A simple change in one of the variables may bring about complex and non-linear changes in other variables. The external potential of information about any engineering process is extremely high. This complex situation can be handled diligently with very narrow channels of perception by gaining an insight into a particular process using models. A model is a simplified representation of those aspects of an actual process that are being investigated (Kafarov & Kuznetsov 1976). The flow of information is broken down into two stages. In the first stage, the model is compared with the real process and considered adequate if the discrepancy is negligible. In the second stage, the expectations are compared with the indications of the model. This procedure is called modelling. Modelling is subdivided into two groups: rn Physical modelling Mathematical modelling We will also focus on specific applicationsof mathematical modelling in chemical engineering, which is generally referred to as chemical systems modelling. Before actually going into the details of and differences among physical, mathematical, and chemical systems modelling, we need to have a good grasp of different types of processes, such as deterministic and stochastic processes, and their differences.

2.1 Deterministic Versus Stochastic Processes 2.1.1 Deterministic Process In this process the observables take on a continuous set of values in a well-defined (or definable) manner, while the output variable most representative of the process

16

Process Plant Simulation

is uniquely determined by the input variable. These processes can be adequately described by classical analysis and numerical methods. An example is the process that takes place in a simple continuous stirred tank reactor (CSTR).

2.1.2 Stochastic Process This is a process in which observables change in a random manner and often discontinuou.sly. The output variable is not directly related to the input variable. These processes are described in terms of statistics and probabilistic theory. Examples are the contact-catalytic process (packed beds) in which the yield of the product diminishes with decrease in the activity in the catalyst as it ages with time and the pulse properties (such as pulse frequency, pulse velocity, pulse height, base hold-up, and pulse hold-up) in trickle bed reactors (Babu 1993). A deterministic process is one whose behaviour (with respect to time, since it is of interest) can be predicted exactly. However, for stochastic processes, we can predict its response only approximately (generally invoking probabilistic notions). Simple CSTR and hydrocracking in trickle bed reactors are quoted as examples for deterministic and stochastic processes, respectively. However, it may be noted that any process, whether it is a process in CSTR or hydrocracking or fluid catalytic cracking (FCC), can be modelled as a deterministic or stochastic process, depending on how much trust we repose in the model we have constructed and how well we can describe all the inputs that affect the process behaviour. While a CSTR is a simpler process to describe than an FCC, this does not in any way imply that we can model CSTR behaviour exactly or that we know all the inputs that affect the response of a CSTR. It is more a matter of our faith than of the process itself.

2.2 Physical Modelling In physical modelling, the experiment is carried out directly on the real process. The process of interest is reproduced on different scales, and the effect of physical features and linear dimensions is analysed. The experimental data are reduced to relationships involving dimensionless groups made up of various combinations of physical quantities and linear dimensions. The relationships determined with this dimensionless presentation can be generalized to classes of events having these dimensionless groups, or similarity criteria. The resulting models are also known as ‘empirical models’. Physical modelling consists in seeking the same or nearly the same similarity criteria for the model and the real process. The real process is modelled on a progressively increasing scale, with the principal linear dimensions scaled up in proportion (the similarity principle). Thus, a physical model is restrained directly within the system where the real process of interest takes place. This approach requires that the process be modelled up to the commercial scale, along with the

Modelling Aspects

17

complex systems that one has to deal with in chemical engineering. For relatively simple systems (such as single-phase fluid-flow or heat-transfer systems) the similarity principle and physical modelling are justified because the number of criteria involved is limited. However, with complex systems and processes described by a complex system of equations, one has to deal with a large set of similarity criteria that are not simultaneously compatible and, as a consequence, cannot be realized. Let us consider the example of designing an industrial heat exchanger. For computing the heat-transfer coefficients that are required for designing a heat exchanger, the empirical correlations (of the form Nu = c Re"Pu", where Nu is the Nusselt number, Re is the Reynolds number, and Pr is the Prandtl number; c, m, and n are constant and the exponents are determined experimentally) developed at laboratory scale could be scaled up to industrial scale using geometric and dynamic similarities. Section 2.7 contains a detailed discussion on similarity principles. The similarity principle has proved its worth in the analysis of deterministic processes that obey the laws of classical mechanics and proceed in bounded singlephase systems (within solid walls as a rule). It is, however, difficult to apply physi.cal similarity to an analysis of probabilistic processes involving multivalued stochastic relationships between the events, an analysis of two-phase unbounded systems, and to processes complicated by chemical reactions.

2.3 Mathematical Modelling A mathematical model of a real chemical process is a mathematical description which combines experimental facts and establishes relationships among the process variables. Mathematical modelling is an activity in which qualitative and quantitative representations or abstractions of the real process are carried out using mathematical symbols. In building a mathematical model, a real process is reduced to its bare essentials, and the resultant scheme is described by a mathematical formalism (formulation) selected according to the complexity of the process. The resulting models could be either analytical or numerical in nature depending upon the method used for obtaining the solution. The objective of a mathematical model is to predict the behaviour of a process and to work out ways to control its course. The choice of a model and whether or not it represents the typical features of the process in question may well decide the success or failure of an investigation. A good model should reflect the important factors affecting a process, but must not be crowded with minor, secondary factors that will complicate the mathematical analysis and might render the investigation difficult to evaluate. Depending on the process under investigation, a mathematical model may be a system of algebraic or differential equations or a mixture of both. It is important that the model should also represent with sufficient accuracy both quantitative and qualitative properties

18

Process Plant Simulation

of the prototype process and should adequately fit the real process. For a check on this requirement, the observations made on the process should be compared with the predictions derived from the model under identical conditions. Thus, a mathematical model of a real chemical process is a mathematical description combining experimental facts and establishing relationships between the process variables. For this purpose, it uses both theory and experimentation (Babu & Ramakrishna 2002a). When one is lacking information about a process or a system, one begins with the simplest model, taking care not to distort the basic (qualitative) aspects of the prototype process. Mathematical modelling involves three steps: w formalization-the mathematical description of the process under investigation (mathematicalformulation) w development of an algorithm for the process w testing of the model and the solution derived from it In this method of analysis, the model itself is restrained through simulation on a computer, rather than with the real process or plant, as is the case with physical modelling. For this purpose, the variables that affect the course of a process are changed from the computer control console to a predetermined program (algorithm) and the computer represents the resultant outputs. With a conservative capital outlay, mathematical modelling coupled with present-day computers makes it possible to investigate various plant configurations in order to trace process behaviour under different conditions, and to find ways and means for improvement. Furthermore, this approach always guarantees an optimum solution within the framework of the model being used. However, it should be stressed that mathematical modelling is in no way set to oppose physical modelling. Rather, the former supplements the latter with its wealth of mathematical formulation and numerical analysis. The importance of lateral mixing from experimental evidence in thermal resistance models (Babu 1993; Shah et al. 1995) is a very good example of this combination. Mathematical modelling involves the simulation of a process on a computer by changing in the interlinkedvariables. Using this technique, all promising alternatives can be simulated in order to arrive at an optimum model and, as a consequence, to optimize the process itself within a relatively short time. Mathematical modelling is economic and less time consuming than physical modelling. Mathematical modelling also uses the principles of analogies, or correspondence between different physical phenomena, dekribed by analogous mathematical equations.An example is the analogy among energy, heat, mass and electricity transport as is demonstrated below.

Modelling Aspects

19

Energy or momentum transport (force of friction) Newton’s law of viscosity

z=

-.(%)

can be rearranged to

where v p is the momentum per unit volume. In addition, in terms of driving force hp itbecomes g, D M 2fpv2 = -

L

(2.3)

Heat transport (heat flux) Fourier’s law of heat conduction

Q =- k ( z ) A can be rearranged to

(2.4)

where pCpT is the heat per unit volume. Similarly, heat flux QLA in terms of driving force AT is -Q= - h A T A Mass transport (mass flux) Fick’s first law of diffusion qm = J = - D ( z )

can be rearranged to

where c is the mass per unit volume. Similarly, mass flux NA in terms of driving force Ac is N , =-k,Ac

(2.9)

20

Process Plant Simulation

Electricity transport (current) Ohm’s law

(2.10)

2.4 Chemical Systems Modelling Performing experiments and interpreting the results is routine in all applied sciences research. This may be done quantitatively by taking accurate measurements of the system variables, which are subsequently analysed and correlated, or qualitatively by investigating the general behaviour of the system in terms of one variable influencing another. The first method is always desirable, and if a quantitative investigation is to be attempted, it is better to introduce the mathematical principles at the earliest possible stage, since they may influence the course of investigation. This is done by looking for an idealized mathematical model of the system. The second step is the collection of all relevant physical information in the form of conservation laws and rate equations. The conservation laws of chemical engineering are material balances, heat balances, and other energy balances; whilst the rate equations express the relationships between flow rate and driving force in the fields of fluid flow, heat transfer, and diffusion of matter. These are then applied to the model, and the result should be a mathematical equation which describes the system. The type of equation (algebraic, differential, finite difference, etc.) will depend upon both the system under investigation and its model. For a particular system, if the model is simple, the equation may be elementary; whereas if the model is more refined, the equation will be more complex. Appropriate mathematical techniques are then applied to this equation and a result is obtained. This mathematical result must then be interpreted using the original model in order to give it physical significance. Most chemical engineering processes that proceed at high rates, high temperatures, and high pressures in multiphase systems are distinct in their utmost complexity. A change in one system variable may bring about non-linear changes in other variables. This complexity becomes still more formidable in the case of multiple feedback loops. In addition, random disturbances are superimposed on the process. The external potential of information about chemical engineering process is very high. An analysis of the system and system variables may therefore be carried out with very narrow channels of perception by gaining insight into a particular process through modelling. The development of new processes in the chemical industry is becoming more complex and increasingly expensive. If the research and development of a process can be carried out with confidence, the ultimate design will be more exact and the plant will operate more economically.

Modelling Aspects

21

Mathematics, which is the language of quantitative analysis, plays a vital role in all facets of such a project. Therefore, training in mathematical methods is of utmost importance to chemical engineers. The most important result of developing a mathematical model of a chemical engineering system is the understanding that is gained of what really makes the process work. This insight enables one to strip away from the problem many extraneous confusing factors and get to the core of the system. It is basically trying to find cause-and-effect relationships between the variables. Mathematical models can be useful in all phases of chemical engineering, from research and development to plant operations, and even in business and economic studies (Luyben 1990). In research and development: determining chemical kinetic mechanisms and parameters from laboratory or pilot-plant reaction data; exploring the effects of different operating conditions for optimization and control studies; aiding in scale-up calculations. In design: exploring the sizing and arrangement of processing equipment for dynamic performance; studying the interactions of various parts of the process, particularly when material recycling or heat integration is used; evaluating alternative process and control structures and strategies; simulating startup, shutdown, and emergency situations and procedures. In plant operations: troubleshooting control and processing problems; aiding in startup and operator training; studying the effects of and requirements for expansion (bottleneck-removal) projects; optimizing plant operation. It is usually much cheaper, safer, and faster to conduct the kinds of studies listed above on mathematical model simulations than experimentally on an operating unit. This is not to say that plant tests are not needed. As we will discuss later, they are vital for confirming the validity of a model and for verifying important ideas and recommendations that evolve from model studies.

2.4.1 Model Formulation Principles Basis The bases for mathematical models are the fundamental physical and chemical laws, such as the laws of conservation of mass, energy, and momentum. To study dynamics we will use them in their general form with time derivatives included. Assumptions Probably the most vital role an engineer plays in modelling is in exercising his engineering judgement as to what assumptions can be validly made. Obviously an extremely rigorous model that includes every phenomenon down to microscopic detail would be so complex that it would take a long time to develop and might be impractical to solve, even on the latest supercomputers. An engineering compromise between a rigorous description and getting an answer that is good enough is always required. This has been called ‘optimum sloppiness’. It involves making as many simplifying assumptions as are reasonable. In practice, this

22

Process Plant Simulation

optimum usually corresponds to a model, which is as complex as the available computing facilities will permit. The development of a model that incorporates the basic phenomena occurring in a process requires a lot of skill, ingenuity, and practice. It is an area where the creativity and innovativeness of the engineer is a key element for the success of the process. The assumptions that are made should be carefully considered and listed. They impose limitations on the model that should always be kept in mind when evaluating its predicted results.

Mathematical consistency of model Once all the equations of the mathematical model have been written, it is usually a good idea, particularly with big, complex systems of equations, to make sure that the number of variables equals the number of equations: The so-called degrees of freedom of the system must be zero in order to obtain a solution. If this is not true, the system is underspecified or overspecified and something is wrong with the formulation of the problem. This kind of consistency check may seem trivial, but experience shows that it can save many hours of frustration, confusion, and wasted computer time. This is required for a simulation exercise. For an optimization exercise, there should be some degrees of freedom available for optimizing, that is, an optimization problem is an underspecified problem. Checking to see that the units of all terms in all equations are consistent is perhaps another trivial and obvious step, but one that is often forgotten. It is essential to be particularly careful of the time units of parameters in dynamic models. Any unit can be used (seconds, minutes, hours, etc.), but these cannot be mixed. Dynamic simulation results are frequently in error because the engineer has forgotten a factor of ‘60’ somewhere in the equations. Solution of the model equations The available solution techniques and tools must be kept in mind as a mathematical model is developed. An equation without any way to solve it is not worth much. Verification An important but often neglected part of developing a mathematical model is proving that the model describes the real-world situation. At the design stage, this sometimes cannot be done because the plant has not yet been built. However, even in this situation there are usually either similar existing plants or a pilot plant from which some experimental dynamic data can be obtained. The design of experiments to test the validity of a dynamic model can sometimes be a real challenge and should be carefully thought out. 2.4.2 Fundamental Laws used in Modelling Some fundamental laws of physics and chemistry are required for modelling a chemical engineering system. These laws are reviewed in their general timedependent form in this section.

Modelling Aspects

23

2.4.2.1 Continuity equations

Total continuity equation (mass balance) The principle of conservation of mass applied to a dynamic system is [Mass flow into system] - [mass flow out of system] = [time rate of change of mass inside system]

(2.1 1)

The units of Eq. (2.11) are mass per time. Only one total continuity equation can be written for one system. According to the normal steady-state design equation we use, we say that ‘what goes in, comes out’. The dynamic version of this says the same thing with the addition of the word ‘eventually’. For any property in a system, if we assume that the property does not vary with respect to spatial location, we obtain an ordinary differential equation (ODE). Otherwise we obtain a partial differential equation (PDE). If we assume that the property does not change with time (steady-state assumption), we get an algebraic equation. The right-hand side of the above equation will be either a partial derivative or an ordinary derivative of the mass inside the system with respect to the independent variable t.

Component continuity equations (component balances) Unlike mass, chemical components are not conserved. Again, to be precise, the total mass for a reacting/ non-reacting system is conserved. However, in general, for a reacting system, the total number of moles is not a conserved quantity. For a reacting system, while the masses of individual elements are conserved, masses/moles of individual components are not conserved due to molecular rearrangement during a reaction. If a reaction occurs inside a system, the number of moles of an individual component will increase if it is a product of the reaction or decrease if it is a reactant. Therefore the component continuity equation of the jth chemical species of the system is [Flow of moles of jth component into system] - [flow of moles of jth component out of system] + [rate of formation of moles ofjth component from chemical reactions] = [time rate of change of moles ofjth component inside system] (2.12) The units of Eq. (2.12) are moles of componentj per unit time. The flows in and out can be both convective (due to bulk flow) and molecular (due to diffusion). We can write one component continuity equation for each component in the system. If there are NC components in a system, there are NC component equations. However, the total mass balance and these NC component balances are not all independent, since the sum of all the moles times their respective molecular weights equals the total mass. Therefore a given system has only NC independent continuity equations. We usually use the total mass balance and NC - 1 component balances. For example, in a binary (two-component) system, there would be one total mass balance and one component balance.

24

Process Plant Simulation

There are some exceptions to the above rule. Consider a non-reacting system first. If we consider a flow balance involving S streams connected to a process unit (either in terms of moles or mass), NC component balances and S ‘normalization’ relations (one for each stream), which describe how the total mass of a stream is related to the composition of the stream, then together these constitute NC + S + 1 equations, out of which one equation is redundant. Generally, it is not possible to work with NC - 1 components (and implicitly use the normalization equation for computing it) because the specified compositions may be distributed arbitrarily. The same holds for a reacting system, but we should note that the equations involve r parameters corresponding to the extents of reaction (with known stoichiometry) assumed to occur in the reactor. A further complication is that element balances are not equivalent to component balances unless some conditions are satisfied (Reklaitis 1983; Felder & Rousseau 1999). 2.4.2.2 Energy equation The first law of thermodynamics puts forward the principle of conservation of energy. Written for a general open system (where flow of material in and out of the system can occur) it is

[Flow of internal, kinetic, and potential energy into system by convection or diffusion] - [flow of internal, kinetic, and potential energy out of system by convection or diffusion] + [heat added to system by conduction, radiation, and reaction] - [work done by system on surroundings (shaft work and PVwork)] = [time rate of change of internal, kinetic, and potential energy inside system] (2.13) 2.4.2.3 Equation of motion According to Newton’s second law of motion, force F is equal to mass M times acceleration a for a system with constant mass M , i.e.,

F = - Ma gc

(2.14)

where g, is the conversion constant needed when FPS or MKS units are used = 32.2 (lb,ft)/lb,s2 in FPS (English engineering) units = 9.81 kg,m/kg,s2 in MKS units

= 1 kg,m/N s2 in SI units = 1 g,cm/dyns2 in CGS units This is the basic relationship used in writing the equations of motion for a system. In a slightly more general form, where mass can vary with time, (2.15)

Modelling Aspects

25

where vi is the velocity in the i direction and Fiiis thejth force acting in the i direction. From Eq. (2.15), we can say that the time rate of change of momentum in the i direction (mass times velocity in the i direction) is equal to the net sum of the forces pushing in the i direction. It can be thought of as a dynamic force balance, or more eloquently it is called the conservation of momentum. In the real world there are three directions: x, y, z. Thus, three force balances can be written for any system. Therefore, each system has three equations of motion (plus one total mass balance, one energy equation, and NC - 1 component balances). Instead of writing three equations of motion, it is often more convenient (and always more elegant) to write the three equations as one vector equation. The field of fluid mechanics makes extensive use of the conservation of momentum. 2.4.2.4 Transport equations

The equations discussed so far are the laws governing the transfer of momentum, energy, and mass. These transport laws all have the form of a flux (rate of transfer per unit area) being proportional to a driving force (a gradient in velocity, temperature, or concentration). The proportionality constant is a physical property of the system (such as viscosity, thermal conductivity, or diffusivity). However, for transport on a molecular level, the laws bear the familiar names of Newton, Fourier, and Fick [Eqs (2.1), (2.4), and (2.7),respectively]. Transfer relationships of a more macroscopic overall form are also used; for example, film coefficients and overall coefficients in heat transfer. Here the difference in the bulk properties between two locations is the driving force [Eqs (2.3), (2.6), and (2.9),respectively]. The proportionality constant is an overall transfer coefficient. 2.4.2.5 Equations of state

In mathematical modelling we need equations for physical properties, primarily density and enthalpy, as a function of temperature, pressure, and composition. Liquid density, (2.16) (2.17) (2.18) (2.19)

Occasionally, these relationships have to be fairly complex to describe the system accurately. But in many cases simplification can be made without sacrificing much

26

Process Plant Simulation

overall accuracy. Some of the simple enthalpy equations that can be used in energy balances are h = C,T

H = C,T

(2.20)

+ 2,

(2.21) The next level of complexity would be to make the Cp’sfunctions of temperature: (2.22) A polynomial in T is often used for Cp: C , ( T ) = A1 + 4 T

(2.23)

Then Eq. (2.22) becomes = A,(T - To)+ ! L ( T ~ - T~ 0 )

2

L

(2.24)

Of course, with mixtures of components, total enthalpy is needed. If heat-of-mixing effects are negligible, the pure-component enthalpies can be averaged:

c NC

xjhjMj

(2.25)

where xj is the mole fraction of the jth component, Mj is the molecular weight of the jth component, and hj is the pure-component enthalpy of the jth component (energy per unit mass). The denominator of Eq. (2.25) is the average, molecular weight of the mixture. Liquid densities can be assumed constant in many systems unless large changes in composition and temperature occur. Vapour densities usually cannot be considered invariant and some sort of PVTrelationship is almost always required. The simplest and most often used is the perfect-gas law: PV = nRT (2.26) where P is the absolute pressure (Wa), V is the volume (m3), n is the number of moles (kmol), R is a constant = 8.314 kPam3/kmolK, and T is the absolute temperature (K). Rearranging to get an equation for the density p, (kg/m3) of a perfect gas with molecular weight M, we get nM M P pv=y=i@-

(2.27)

Modelling Aspects

27

2.4.2.6 Equilibrium relations

The basis for the equations that give the conditions of a system when equilibrium conditions prevail is the second law of thermodynamics.

Chemical equilibrium Equilibrium occurs in a reacting system when NC

c v j p j=0

(2.28)

j=1

where vj is the stoichiometric coefficient of thejth component with reactants having a negative sign and products a positive sign, and p is the chemical potential of the jth component. The usual way to work with this equation is in terms of the equilibrium constant for a reaction. For example, consider a reversible gas-phase reaction of A to form B at a specific rate k , and B reaching back to A at a specific reaction rate k2. The stoichiometry of the reaction is such that va moles ofA react to form vb moles of B. (rate constant for reverse reaction is k2) From Eq. (2.28), equilibrium will occur when vbpb - vapa = The chemical potential for a perfect-gas mixture can be written as N& -.A,V

p j = py

V,B

+ RT lnPj

(2.29)

(2.30) (2.31)

where is the standard chemical potential (or Gibbs free energy per mole) of the jth component, which is a function of temperature only, Pj is the partial pressure of the jth component, R is the perfect-gas law constant, and T is the absolute temperature. Substituting into Eq. (2.30) and simplifying, we get in(

=

0

0

- 'bpB

RT

(2.32)

The right-hand side of this equation is a function of temperature only. The term in parentheses on the left-hand side is defined as the equilibrium constant Kp, which gives us the equilibrium ratios of products and reactants. (2.33)

Phase equilibrium Equilibrium between two phases occurs when the chemical potential of each component is the same in the two phases: (2.34)

28

Process Plant Simulation

where pj is the chemical potential of thejth component in phase I and py is the chemical potential of the jth component in phase 11. Since a vast majority of chemical engineering systems involve liquid and vapour phases, many vapour-liquid equilibrium relationships are used. They range from very simple to the very complex. Some of the most commonly used relationships are listed below. More detailed treatments are presented in many thermodynamics texts. Basically, we need a relationship that permits us to calculate the vapor composition if we know the liquid composition, or vice versa. The most common problem is a bubble point calculation: calculate the temperature T and vapour composition yp given the pressure P and the liquid composition xi. This usually involves a trial-and-error, iterative solution because the equations can be solved explicitly only in the simplest cases. Sometimes we have bubble point calculations that start from known values of xj and T and we want to find P and yi. This is frequently easier than when pressure is known, because the bubble point calculation is usually non-iterative.Dew point calculations must be made when we know the composition of the vapour y j and P (or 7J and want to find the liquid composition xj and T (or P). Flush calculations must be made when we know neither xj nor y j and must combine phase equilibrium relationships, component balance equations, and an energy balance to solve for all the unknowns. Let us assume ideal vapour-phase behaviour in our examples, i.e., the partial pressure of the jth component in the vapour is equal to the total pressure P times the mole fraction of thejth component in the vapour y j (Dalton’s law):

q=Pyj

(2.35)

Corrections may be required at high pressures. As far as the liquid phase is concerned, several approaches are being used, which are shown below.

Ruoult’s law Mixtures that obey Raoult’s law are called ideal mixtures. NC

P = Cx,P;

(2.36)

j= 1

xjp; y . =(2.37) ’ P where P,” is the vapour pressure of pure component j . Vapour pressures are functions of temperature only. This dependence is often described by lnPjS = >A+. Bj T

(2.38)

Modelling Aspects

29

Relative vohtiZity The relative volatility av of component i to component j is defined as (2.39)

Relative volatilities are fairly constant in a number of systems. They are convenient and hence are frequently used. In a binary system the relative volatility a of the more volatile component compared with the less volatile component is (2.40)

Rearranging,

m = 1+ (a- 1)x

(2.41)

K vahes Equilibrium vapourization ratios of K values are widely used, particularly in the petroleum industry. (2.42)

The K’s are functions of temperature and composition and, to a lesser extent, pressure.

Activity coencients For non-ideal liquids, Raoult’s law must be modified to account for the non-ideality in the liquid phase. The fudge factors used are called activity coeficients. NC

P =cxjp;r,

(2.43)

j=l

where r j is the activity coefficient for the jth component. The activity coefficient is equal to 1 if the component is ideal. The y’s are functions of composition and temperature. 2.4.2.7 Chemical kinetics

In modelling of chemical reactors, we must be familiar with the basic relationships and terminology used in describing the kinetics (rate of reaction) of chemical reactions.

Arrhenius temperature dependence The effect of temperature on the specific reaction rate k is usually found to be exponential:

k = Cre-E/RT

(2.44)

30

Process Plant Simulation

where k is the specific reaction rate, a is the preexponential factor, E is the activation energy [shows the temperature dependence of the reaction rate, i.e., the bigger E is, the faster the increase in k with increasing temperature (caVgmol) is], T is the absolute temperature, and R is the perfect-gas constant = 1.99 caVgmol K. This experimental temperature dependence represents one of the most severe non-linearities in a chemical engineering system. It may be noted that the apparent temperature dependence of a reaction may not be exponential if the reaction is mass-transfer limited and not chemical-rate limited. If both zones are encountered in the operation of the reactor, the mathematical model must obviously include both reaction-rate and mass-transfer effects.

Law of mass action Let us define an overall reaction rate 93,with the conventional notation, as the rate of change of moles of any component per volume due to the chemical reaction divided by that component's stoichiometric coefficient: (2.45)

The stoichiometric coefficients vj are positive for products of the reaction and negative for reactants. It may be noted that CH is an intensive property and can be applied to systems of any size. For example, assume we are dealing with an irreversible reaction in which components A and B react to form components C and D . (2.46)

(2.47)

From the law of mass action, the overall reaction rate CH will vary with temperature (since k is temperature-dependent) and with the concentrations of reactants raised to some powers: CH = k ( T ) ( c A > a

(CB>b

(2.48)

where C, is the concentration of component A and cB is the concentration of component B. The constants a and b are not, in general, equal to the stoichiometric coefficients v, and vba The reaction is said to be first-order in A if a = 1. It is second-order in A if a = 2. The constants a and b can be fractional numbers.

Modelling Aspects

31

As indicated earlier, the units of the specific reaction rate k depend on the order of the reaction. This is because the overall reaction rate 3 always has the same units (moles per unit time per unit volume). For a first-order reaction of A reacting to form B, the overall reaction rate 3,written for component A , would have units of moles of Nmin ft3. % = kCA (2.49) If cAhas units of moles of A/ft3, k must have units of min-'. If the overall reaction rate for the above system is second-order in A ,

3 = kci

(2.50)

3 still has units of molA/min ft3 (molA denotes moles of A). Therefore k must have units of ft3/minmolA. Considering the reaction A + B + C, if the overall reaction rate is first-order in both A and B , 3 = kCAC, (2.51) % still has units of molA/min ft3. Therefore k must have units of ft3/min moll?.

2.5 Cybernetics Cybernetics is a science dealing with any system capable of perceiving, storing, and processing information for the purposes of optimum control. Thus, cybernetics encompasses the concepts of a system, information, data storage and processing, system control, and system optimization. Mathematical modelling, computers, and cybernetics are interrelated. Computers are tools of cybernetics and mathematical modelling is the basic method of cybernetics.

2.6 Controlled System After having discussed the details of modelling, let us see how a simple controlled chemical system is represented schematically (Fig. 2.1). In the absence of disturbances in the system, the process runs smoothly, giving the desired designed outputs for the given inputs. But if there are any disturbances in the system, the outputs will change and the desired designed outputs cannot be obtained for the given inputs. Then, by means of manipulated variables, the effect of these disturbances can be nullified and the desired outputs can be obtained again, under controlled conditions.

32

Process Plant Simulation Disturbances

Inputs

Outputs

Manipulated variables

Fig. 2.1 A simple controlled chemical system

2.7 Principles of Similarity These are useful in scaling up operations in physical modelling. Similarity is used when models are either true or distorted ones. True models reproduce features of the prototype but at a scale-that is they are similar as per either any or all of the three similarity criteria, namely, geometric similarity, kinematic similarity, and dynamic similarity.

2.7.1 Geometric Similarity The solid boundaries of any flow system may be adequately described by a number of length dimensions such as L,, L2,L,, ..., L,. If these lengths are divided by L,, the system may be defined by 1, A,, A,,..., A,,,where A, = A , I L,, A3 = L, I L,, ..., and A , = L, / L, , Geometric similarity exists between a model and a prototype if the ratios of all corresponding dimensions in the model and the prototype are equal. (2.52)

where

A, is the scale factor for length. For area, (2.53)

All corresponding angles are the same.

2.7.2 Kinematic Similarity Kinematic similarity refers to the motion occumng in the system and considers the existing components of velocities. Kinematic similarity is the similarity of time as well as geometry. For kinematic similarity to exist in two geometrically similar systems, the velocities at the same relative point in each system must be related. It exists between the model and the prototype if (i) the paths of moving particles are geometrically similar and (ii) the ratios of the velocities of particles are similar. Some useful ratios are the following.

Modelling Aspects

33

For velocity, (2.54)

For acceleration, (2.55)

and for discharge,

(2.56) Similarly, the velocity gradients in each system will bear a similar relationship to each other. As a consequence, streamline patterns are the same.

2.7.3 Dynamic Similarity Dynamic similarity exists between geometrically and kinematically similar systems if the ratios of all the forces in the model and the prototype are the same. The force ratio is

(2.57) This occurs when the controlling dimensionless group on the right-hand side of the defining equation is the same for the model and the prototype.

Illustration 2.1 When a hydraulic structure is built it undergoes some analysis in the design stage. Often the structures are too complex for simple mathematical analysis and a hydraulic model is built. Usually the model is smaller than the full size, but it can be larger. The real structure is known as the prototype. The model is usually built to an exact geometric scale of the prototype, but in some cases-notably the river model-this is not possible. Measurements can be taken from the model and a suitable scaling law applied to predict the values in the prototype. To illustrate how these scaling laws can be obtained, we will use the relationship for the resistance of a body moving through a fluid. The resistance R depends on the following physical properties: r (ML-3), u (LT'), I (L), m (ML-'T')

34

Process Plant Simulation

So the defining equation isf(R, 5 - 3 = 2 p groups:

I-, u,

1, m ) = 0. Thus, m = 5 , n = 3, so there are

n, = pa1UbflC1R

n2 = p%b21c2p

(2.58)

For the p 1 group, = (ML-~ (LT-' lb1 ( L ) ~MLT-~ I

MOLOTO

(2.59)

leading to p 1 as R

n1 = pu212

(2.60)

(2.61)

(2.62) Notice how Up2is the Reynolds number. We can call this pas. So the defining equation for resistance to motion is f(Plt P2a) = 0 (2.63) We can write R

(2.64) This equation applies irrespective of the size of the body, i.e., it is applicable to the prototype and the geometrically similar model. Thus, for the model,

(2.65) and for the prototype,

(2.66)

Modelling Aspects

35

Dividing these two equations gives

(2.67)

At this point we can go no further unless we make some assumptions. One common assumption is to assume that the Reynolds number is the same for both the model and the prototype, i.e., Pmumlrn - PPuPl, pup Pm This assumption then allows the following equation to be written:

(2.68)

(2.69) which gives this scaling law for resistance force,

AR = ;&A

(2.70)

That the Reynolds numbers were the same was an essential assumption for this analysis. The consequence of this should be explained. Rem = Re,

(2.71)

(2.72) Substituting this into the scaling law for resistance gives (2.73)

36

Process Plant Simulation

So the force on the prototype can be predicted from the measurement of the force on the model, but only if the fluid in the model is moving with the same Reynolds number as it would in the prototype. That is to say, Rp can be predicted by (2.74) provided that PmPplm

up = um

P p~

(2.75)

m1p

In this case the model and prototype are dynamically similar. Formally, this occurs when the controlling dimensionless group on the right-hand side of the defining equation is the same for the model and the prototype. In this case, the controlling dimensionless group is the Reynolds number.

Example 2.1 An underwater missile, diameter 2 m and length 10 m is tested in a water tunnel to determine the forces acting on the real prototype. A 1120th scale model is to be used. If the maximum allowable speed of the prototype missile is 10 d s , what should be the speed of the water in the tunnel to achieve dynamic similarity? Solution For dynamic similarity, the Reynolds number of the model and the prototype must be equal: Rem = Re,

(F)m =(y) P

So the model velocity should be um = u p

Ppdp Pm P m dm Pp

As both the model and the prototype are in water, m, = m, and rm = rp, so

Note that this is a very high velocity. This is one reason why model tests are not always done at exactly equal Reynolds numbers. Some relaxation of the equivalence requirement is often acceptable when the Reynolds number is high. Using a wind tunnel could have been possible in this example. If this were the case, then the appropriate values of the r and m ratios need to be used in the above equation.

Modelling Aspects

37

Example 2.2 A model aeroplane is built at Moth scale and is to be tested in a wind tunnel operating at a pressure of 20 times atmospheric pressure. The aeroplane will fly at 500 km/h. At what speed should the wind tunnel operate to give dynamic similarity between the model and the prototype? If the drag measure on the model is 337.5 N, what will be the drag on the plane? Solution Earlier, we derived the equation for resistance on a body moving through air:

For dynamic similarity Re, = Rep, so Ppdp P

urn =

m

dm ~p The value of m does not change much with pressure, so m, = mp.The equation of state for an ideal gas is p = rRT. As the temperature is same, the density of the air in the model can be obtained from as follows: up p p

~ m PmRT - P, PpRT

Pp

P p

2OPP - p, -PP

P P

Pm =2O~p So the model velocity is found to be

1 1 urn = u --= 20 v10

0.5~~

urn = 250km/h The ratio of forces is found from

Rm - 20 (0.5)2 (0.1l2 _ = 0.05 R

p

l

l

1

So the drag force on the prototype will be 1 R, = -Rm 0.05

= 20 X 337.5 = 6750N

38

Pmcess Plant Simulation

EXERCISES 2.1 Classify modelling emphasizing on details of mathematical modelling. 2.2 Discuss the principles of similarity. 2.3 Describe deterministic and stochastic processes. 2.4 Discuss the various aspects involved in physical modelling. 2.5 What are the various model formulation principles? 2.6 Give a detailed account of the fundamental laws used in modelling. 2.7 Define cybernetics and discuss the controlled system by means of a neat schematic diagram. 2.8 What are the various similarity criteria? 2.9 Discuss physical modelling versus mathematical modelling.

CHAPTER 3 CLASSIFICATION OF MATHEMATICAL MODELLING

The choice of a particular type of mathematical model is determined by the specific conditions under which a given process can proceed in the processing unit of interest. Mathematical modelling is classified under three different bases: 1. Variation of various independent variables 2. State of the process 3. Type of the process A schematic diagram representing the classificationof mathematical modelling under different bases as discussed in detail below is shown in Fig. 3.1. It may be noted that a given model could be a possible combination of any of the three types of classification, namely, a static distributed parameter analytical rigid model or a dynamic lumped parameter numerical stochastic model or any other combination. Before going into the details of the classification of mathematical modelling, we need to understand the difference among independent variables, dependent variables, and parameters.

3.1 Independent and Dependent Variables, and Parameters Any solution of a differential equation is an algebraic equation involving symbols. These symbols fall into three classes.

Independent variables These are quantities describing a system that can be varied by choice during a particular experiment, independent of one another. Examples include time and coordinate variables.

40

Process Plant Simulation Mathematical modelling

Classification based on variation of independent variables

Classification based on the state of the process

(steady-state)

Analytical

Classification based on the type of the process

(unsteady-state)

I I Numerical I

Analytical

Numerical

Fig. 3.1 Classification of mathematical modelling

Dependent variables These are the properties of a system which change when the independent variables are altered in value. There is no direct control over a dependent variable during an experiment. The relationship between independent and dependent variables is one of cause and efect; independentvariables measure the cause and dependent variables measure the eflect. Examples for dependent variables include temperature, concentration and efficiency. Parameters This is by far the largest group, consisting mainly of the characteristic properties of the apparatus and physical properties of the materials. The term parameters is used because the properties remain constant during an individual experiment, but take different constant values during different experiments. Examples include overall dimensions of the apparatus, flow rates, heattransfer coefficients, thermal conductivity,heat capacity, density, initial or boundary values of dependent variables. Illustration 3.1 Unsteady-state single-stage extraction To demonstrate the differences among independent variables, dependent variables and parameters, the following example of unsteady-state single-stage solvent

Classifcation of Mathematical Modelling

41

extraction (Fig. 3.2) is considered (this model will be discussed in detail in Chapter 4).

Fig. 3.2 Unsteady-state single-stage extraction

The equilibrium relationship is y=mX

The mole fraction of the solute in the raffinate is x=

A[ 1 - exp( - R + m S e)] R+mS

V,

+ mV2

(3.2)

Eliminating x from Eq. (3.1) using Eq. (3.2), we get y=

*[

R + mS

1 - exp( - R + mS e)] V, +mV2

(3.3)

The symbols used in the above example can be classified as follows: 1. Independent variable: 8 2. Dependent variables: x, y 3. Parameters: m , R, S, c, V , , V,

Parametersare all fixed in value during the experiment,but can be varied between experiments for comparison purposes. 0 is the only variable whose value can be altered during an experiment. x and y vary during an experiment and between experiments;their values depend upon the choice of values for both the parameters and the independent variable. This classificationis mainly used in the interpretationof differentiationprocesses. We usuaily differentiate a dependent variable with respect to an independent variable and occasionally with respect to a parameter. By differentiating either Eq. (3.2) or Eq. (3.3)with respect to 8 , an expression giving the rate of change of concentration with time can be obtained. But differentiatingEq. (3.1) with respect to the dependent variable x gives dy/dx = m, an almost useless piece of information which does not throw any light on the behaviour of the apparatus.

42

Process Plant Simulation

3.2 Classification based on Variation of Independent Variables This classification of mathematical modelling (Fig. 3.1) is done based on whether the process variables vary with time as an independent variable or with both time and space coordinates as independent variables.

3.2.1 Distributed Parameter Models If the basic process variables vary with both time and space, or if these changes occur only with space of dimension exceeding unity (i.e., more than one space coordinate), such processes are represented by distributed parameter (DP) models, which are formulated as partial differential equations.

3.2.2 Lumped Parameter Models Processes in which the basic process variables vary only with time are represented by lumped parameter (LP) models, which are formulated as ordinary differential equations.

3.3 Classification Based on the State of the Process The primary objective of cybernetics is to control a given system or process. As a consequence, a complete mathematical model is expected to describe relationships between the basic process variables under steady-state conditions (a static model) and transient conditions (a dynamic model).

3.3.1 Static Model A static or steady-state model ignores the changes in process variables with time.

The construction of a static model involves the following steps: Step I Analysis of the process to establish its physical and chemical nature, its objective, the governing equations describing a given class of processes and also its specific features as a unit process. Step 2 Identification and ascertaining of the input and output variables of the process. The variables include (see Fig. 3.3) the following: 1. Control variables: Variables that change with the progress of the process CYj).

2. Manipulated variables: Variables that directly affect the course of the process and that can be measured and changed purposefully (xi). 3. Disturbances: Variables that directly affect the course of the process but that cannot be changed purposefully (q).

Classification of Mathematical Modelling

43

4. Intermediate variables: Variables related to the course of the process only indirectly (xni).

XI

9

x2

9

x7

9

.

.............,,.,9

.

*

X"

b

--

PROCESS

Y?

-.

Y3

:

t

Yl

r

!

.... ..

... ...

i

r

-

,,,.............w

Y"

Step 3 Identification and establishing of the relationships, constraints, and boundary conditions of the process. The static model of a unit process should allow for all possible modes of operation of the typical reactor.

3.3.2 Dynamic Model The construction of a dynamic or unsteady-state model reduces to obtaining the dynamic characteristics of the process, i.e., establishing relationships between its main variables as they change with time. Dynamic characteristics can be obtained by theory, experiment, or both. In obtaining dynamic characteristics experimentally, disturbances are applied to the input of a system and the time response to the disturbance is noted. Such experiments are based on the laws of signal flow dealt with in information arid control theory. The dynamic model of a process may take the form of any of the following: 1. A set of transfer functions relating the selected dependent variable to one or several independent variables. 2. Ordinary or partial differential equations derived theoretically and containing all the necessary dependent and independent variables. 3. Equations derived for the various elements of the unit process that may be analysed independently of one another.

44

Process Plant Simulation

3.3.3 The Complete Mathematical Model A complete mathematical model of a process combines the following: 1. The basic process variables 2. Relationships between the basic variables under static or steady-state conditions 3. Constraints 4. Optimality criteria 5. Objective functions 6. Relationships between the basic variables under dynamic or unsteadystate conditions

The schematic diagram representing various stages in the development of a complete mathematical model for a process is shown in Fig. 3.4.

I Unit process I I Process describing equations I

i;

~

Boundary conditions

I I

L--------------

Variable relationships

-: i

,

I

I

, , - - - - - - - - - - - - - - - ! AS , I

I

I I 1

I 6

I

Static optimization

!

I

I

transfer functions

Basic variable relationships

t As complete differential equations

As equations for process elements

I

Fig. 3.4 Stages in the development of a complete mathematical model for a process

Classwcation of Mathematical Modelling

45

3.4 Classification Based on the Type of the Process Depending on whether a given process is deterministic or stochastic, it may be represented by any one of the following mathematical models (refer to Fig. 3.1): 1. An analytical rigid model 2. A numerical rigid model 3. An analytical probabilistic model 4. A numerical probabilistic model It may be noted that in some practical cases a model cannot be classified under any of the above headings. The choice of a particular type of model is to a great extent determined by the simplicity of the solution it offers.

3.4.1 Rigid or Deterministic Models These models usually describe deterministic processes without the use of probability distributions. As mentioned above, the subclassification of rigid (or deterministic) models is analytical rigid models and numerical rigid models, depending on whether the solution is obtained analytically or numerically. This is not to say that the underlying phenomena, especially when a numerical rigid model is used, are not statistical in character. This implies that only the averages and not the whole distributions are dealt with here.

3.4.2 Stochastic or Probabilistic Models These models usually represent stochastic (random) processes. Here again, the subclassification of probabilistic models is analytical probabilistic (or stochastic) and numerical probabilistic (or stochastic) models, depending on whether the solution is obtained analytically or numerically. It may be noted that simulation corresponding to a numerical stochastic model is known as Monte Car10 simulation.

3.4.3 Comparison Between Rigid and Stochastic Models In constructing rigid models, we use different classical techniques of mathematics, namely, differential equations, integral equations, linear difference equations, and operators for conversion to algebraic models and so on. Probabilistic models represent the distribution of discrete and continuous variables and also sample distributions. A major advantage of statistical models (probabilistic) over analytical (rigid) ones is that these do not call for rough approximations and allow for a greater number of factors. But these models are difficult to analyse and grasp. From this point of view, it may be preferable to use less precise analytical models. The best results are obtained when both analytical and statistical models are used jointly in conjunction with each other. A simple analytical model will give an insight into the

46

Process Plant Simulation

basic behaviour of the prototype system and suggest further approaches in which statistical models of any complexity can be used. Residence time distribution (RTD) models are good examples of this approach.

3.5 Boundary Conditions An ordinary differential equation usually arises in any problem that involves a single independent variable. The general solution of this differential equation will contain arbitrary constants of integration, the number of constants being equal to the order of the differential equation. To complete the solution to a particular problem, these constants have to be evaluated. Special conditions are placed on the dependent variable at the end points of the range of independent variables. These are naturally called boundary conditions and are used to evaluate the arbitrary constants in the solution of differential equations. When time is the independent variable, the boundary condition is frequently called the initial condition since it specifies the starting state of the apparatus. The three most frequently used boundary conditions in heat transfer are 1. Boundary at a fixed temperature T = To 2. (a) Constant heat-flow rate through the boundary (isothermal condition): dt dx (b) Boundary thermally insulated (adiabatic condition):

- = A (constant)

dt

-=

O

dx 3. Boundary cools to the surroundings through a film resistance described by a heat-transfer coefficient: dr k - = h(T - T o ) dx Here, k is the thermal conductivity, h is the heat-transfer coefficient, and To is the temperature of the surroundings.

3.6 The Black Box Principle In cybernetics, the black box principle is employed when, in an analysis of complex processes, one cannot establish internal relationships within the system. As a result, the object under study appears to be enclosed in a container, the so-called black box, to which we, as observers, have incomplete access. By this principle, the mathematical description is concerned only with the dependence of outputs on the inputs and ignores the interior structure of the process.

Classification of Mathematical Modelling

47

The concept of the black box is a pivotal one in cybernetics, as it provides a useful mathematical approach to the behaviour of complex systems, i.e., to their responses to various external stimuli, without making any attempt to understand their internal structure. Many systems, especially large ones, are so complex that even with perfect information about the state of their components, it cannot practically be related to the behaviour of the system as a whole. In such cases, the matter can be greatly simplified by representing a complex system as a black box model which functions in the same manner as the system. Analysing the behaviour of the model and comparing it with that of the system, one can derive a number of conclusions as regards the properties of the system, and if the model adequately fits the prototype system, one can select a working hypothesis about the internal structure of the system. The black box principle proves extremely useful when one system has to be replaced with another functioning in a similar way. For example, in the automatic control of hazardous chemical manufacture the human operator needs to be replaced with an automatic device capable of performing the same functions. In recent times, artificial neural networks (ANNs), which also come under the category of models using the black box principle in general and the grey box principle in particular, emerged victorious with a wide range of applications in all fields of research (be it engineering, science, arts, or commerce). The following section deals with ANNs.

3.7 Artificial Neural Networks Neural networks have been used as an alternative to the traditional mathematical models to simulate complex patterns. The simplicity of their implementation makes them appropriate for modelling various complicated processes. Neural computing is largely motivated by the possibility of creating an artificial computing network similar to the brain and nerve cells in our body. Recent advances in neuroscience and in computers have sparked a renewed interest in neural network models for problem-solving. These networks are computing systems composed of a number of highly interconnected layers of simple neuron-like processing elements. The entire network collectively performs computations, with the knowledge represented as distributed patterns of activity all over processing elements. The collective activities result in a high degree of parallelism, which enables the network to solve complex problems. The distributed representation leads to greater fault tolerance and to graceful degradation when problems beyond the range of the network are encountered. In addition, these networks can adjust dynamically to environmental changes, infer general rules from specific examples, and recognize invariance from complex high-dimensional data (Lippmann 1987). Artificial neural networks (ANN), also called parallel distributed processing systems (PDPs) and connectionist systems, are intended for modelling the

40

Process Plant Simulation

organizational principles of the central nervous system. Its ability to learn, which makes it rich in knowledge, is the most important feature of an artificial network. Neural networks have the capability of simulating non-linear patterns. Their advantage relies on the fact that they demand less time for development than traditional mathematical models. Neural computation is an interdisciplinary field and has a significant role to play in many emerging areas of scientific researchpattern recognition, image processing, modelling, controlling, speech recognition and production, to name a few. Figure 3.5 depicts a simple structure of feed-forward network architecture. The circles represent neurons arranged in three layers: input, hidden, and output. Each hidden input is connected to each input and output unit. Each hidden and output unit is also connected to a bias. The bias provides a threshold for the activation of the neuron and is essential in order to classify network input patterns into various subspaces. Each connection has a weight associated with it. For the input layer, the input value is forwarded straight to the next hidden layer. The hidden and output layers carry out two calculations: first a weighted sum of the inputs is calculated and then the output is calculated using a nondecreasing and differential transfer function. Usually a sigmoidal function is used.

n

W

LHiddenlayerl Fig. 3.5 Structure of a simple artificial neural network

3.7.1 Network Training There are three ways in which the ANN learning takes place:

Supervised learning An input stimulus is applied to the network, which results in an output response. This is compared with the desired target response and an error signal is generated. The learning in back-propagation networks is supervised.

Classification of Mathematical Modelling

49

Unsupervised learning During training, the network receives different input excitations and arbitrarily organizes the patterns into categories. When a stimulus is later applied, the network indicates the class to which it belongs and an entirely new class of stimuli is generated. The learning in radial basis function networks is unsupervised. Reinforced learning In this case, the network indicates whether the output is matching with the target or not-a pass or fail indication. In other words, the generated signal is binary. This kind of learning is used in applications such as fault diagnosis.

3.7.2 Modes of Training For a given training set, the training may proceed in two ways:

Pattern mode Weight adjustment is made after the presentation of each training example. Consider a training set having Npatterns. The first pattern is presented to the network, and the whole sequence of forward and backward computations is performed, resulting in weight adjustment. Then the second pattern is presented and weights updated and so on until the Nth pattern. Batch mode Here, weight updating is done after the presentation of one full epoch. One complete presentation of the entire training set is called an epoch. The pattern mode of training is on-line, requires less storage, and is less likely to be trapped in a local minimum. The batch mode, on the other hand, is more accurate. The relative effectiveness, however, is finally problem-dependent. 3.7.3 Network Architecture Multilayered feed-forward networks typically consist of an input layer, one or more hidden layers, and an output layer. The input signal propagates through the network layer by layer. The principal unit of a neural network is the neuron. Consider a two-layered network. The output, hidden, and input layers are, respectively, denoted by i, j , and k. A neuron performs two functions: one is computing the weighed sum of the variables that feed it. This sum is called the activity value A . Depending on the activity value, the neuron gives a signal to the neurons of the following layer (in the simplest case, feed-forward). This signal depends on the transfer function. The transformation of the activity value through the transfer function is the second function performed by the neuron. Transfer functions have to meet the condition of being differentiable functions. Tan-sigmoid, log-sigmoid, and linear are the most common transfer functions. The number of neurons in the inner layers, called the hidden layers, and also the number of hidden layers are fixed by the designer of the network.

50

Process Plant Simulation

ANNs come in a variety of architectures, the most popular being the feedforward network trained by back-propagation (Rumelhart & McClelland 1986). Back-propagation is an example of a mapping network that learns an approximation to a function, y equal tof(x), from sample x , y pairs. The fact that the function to be learned is non-linear presents no problem to a back-propagation network (BPN). Representative applications of back-propagation include speech synthesis and recognition, visual pattern recognition, analysis of sonar signals, defence applications, medical diagnosis, and learning in control systems.

3.7.4 Back-propagation Algorithm Training is supervised and is based on the error in the back-propagation algorithm. It is a process in which the internal parameters to the network, the weighing factors W, and bias B, are adjusted. The bias is an adjusting parameter, which reduces the error in the system. Values of these parameters are calculated using multiple-variable optimization algorithms. The bias values are generally assigned based on experience. In the simple back-propagation algorithm, the change that has to be made to the weighing factors and bias is calculated using the derivative vector D and the input data to that layer according to the following rule: Wnew= Wold + l,DvT

pew = Bold+ 1,D where I, is the learning rate. This parameter acts as a gain in the change of the weighing factors and bias. If its value is large, then the learning is fast; but if it is too large, the learning can become unstable and the error can even increase. Thus, the back-propagation algorithm can be very slow because of the small learning rates that have to be used for stable convergence. Once the training is completed, it acts in a feed-forward manner only. Any network learns by making changes in the weights of the connections in accordance with the learning rule. There are a number of algorithms available for the above purpose, but the most widely used is the back-propagation algorithm (Masters 1990), the pseudocode of which is listed below: w Initialize the weights and offsets. Set all of them to low random values. w Present inputs and desired outputs. This is done by presenting a continuousvalued input vector and specifying the desired outputs. If the network is used as a classifier, all desired outputs are set to 1. The input could be new on each turn or one could use a cyclic pattern to train. w Calculate the actual outputs using the sigmoidal non-linearity. w Adapt weights using a recursive algorithm starting at the output nodes and working back. Adjust the weights using the formula WJ(t

+ 1) = W j ( t )+ @,x;

(3.4)

Classification of Mathematical Modelling

51

where Wiiis the weight from node i to nodej at time t, 77 is the gain term, and 6 is the error term for node j. If node j is an output node, then

6j = Yj(1- yj>(dj - Y j )

(3.5) where dj is the desired output of nodej and yj is the actual output. If node j is an internal hidden node, then

6 j = Xj(l - Xj)Z(6pVjk) (3.6) where k is the number of overall nodes in the layers above nodej. If a momentum term a is added, the network sometimes becomes faster and the weight changes are smoothed by

yj(t+ 1) = yk(t)+ q 6 j ~+; a[w,(r) - wV(t- l)]

(3.7)

where 0 c q c 1. rn Repeat step 2. 8

Stop.

3.7.5 ANN Applications Artificial neural networks based on BPNs have been applied to pattern classification problems in a number of fields, which include classification of sonar targets (Gorman & Sejnowski 1988), speech recognition (Lippmann 1989), and sensor interpretation (McAvoy et al. 1989). Application of BPNs to failure state recognition in chemical plants has been studied by Hoskins and Himmelblau (1988), Ferrada et al. (1989), Hoskins et al. (1988), Ungar et al. (1990), Venkatasubramanian & Chan (1989), Venkatasubramanian et al. (1990), and Watanabe et al. (1989; 1994). These studies have shown that BPN classifiers have the ability to learn the classifications of a set of training examples, and can often successfully generalize this knowledge to classify new cases of known failure types. Other applications of ANNs include dynamic modelling and control of chemical process systems (Bhat & McAvoy 1989), diagnosis using back-propagation neural networks (Kramer & Leonard 1990), equilibria prediction in ternary mixture extraction (Babu & Karthik 1997), estimation of heat-transfer parameters in trickle bed reactors using radial basis functional networks (Babu & Sangeetha 1998), modelling of adsorption isotherm constants (Babu & Ramakrishna 2003).

3.7.6 Example Problems Four chemical engineering problems have been considered as examples to illustrate the use of neural networks in process modelling and fault diagnosis. The first example (Illustration 3.2) deals with the work reported by Babu & Shailesh (2000) on fault diagnosis for unsteady-state processes in which faults occur.

52

Pmcess Plant Simulation

The second example (Illustration 3.3) discusses the modelling of pollution removal efficiency using an ANN (Babu & Ramakrishna 2001; Babu et al. 2002). The mathematical description of micro-organism cultures is a complex task that involves huge costs in development and computing. In most cases, the mathematical expressions describing these processes, as functions of certain reduced variables, are not known. Even if these variables are available, these often involve a great number of parameters that have to be fitted using non-linear optimization techniques. The use of neural networks for simulation of one such process has been dealt with in the third example (Illustration 3.4). Aqueous two-phase systems (ATPSs) are generally applied in separation and purification of macromolecules, particles, and biological materials, which are fragile and prone to get denatured. A neural network has been created to simulate this liquid-liquid equilibrium process and give the raffinate-extract concentrations for different feed values as its final output in the fourth example (Illustration 3.5).

Illustration 3.2 Fault diagnosis Babu and Shailesh (2000) used neural networks to do incipient fault diagnosis for unsteady-state processes in which faults occur (Fig. 3.6).

t

CSTR:A

Tho, 0°C

-

F,",, TW,Icwr

B (Exothermic reaction)

Fig. 3.6 A heat exchanger-CSTR

system

ClassiJcation of Mathematical Modelling

53

Illustration 3.3 Pollutant removal efficiency The pollutant removal efficiency (PRE) is the most significant output of adsorption studies with respect to waste water treatment. The variation of PRE depends on several factors such as adsorbent characteristics, contact time, adsorbate concentration, etc. Babu and Ramakrishna (2001) attempted to study some of the above factors v i s - h i s the PRE and found that such attempts were leading to contradictory conclusions. In addition, the results obtained by them are discouraging for the empirical models developed for predicting the PRE in adsorption studies. Then, taking the success of ANNs into account, Babu et al. (2002) applied these for the prediction of PRE using a large amount of data consisting as many as eight different pollutants and 25 different adsorbents. The database thus comprised heterogeneous data, which were systematically trained and tested using a BPN of an ANN, and succeeded with very encouraging results.

Illustration 3.4 Zymomonas mobilis CP4 batch fermentations Zymomonas mobilis is a gram-negative bacterium used for the fermentation of lignocellulosic hydrolyzates. These include substrates such as glucose, sucrose, and fructose. Fermentation of glucose-fructose mixtures leads to the formation of ethanol and biomasses such as sorbitol. The operational conditions include aeration, feed rate, and substrate conditions among others. These need to be monitored closely in order to enhance the system productivity. Using Zymomonas mobilis CP4, continuous fermentation of sucrose leads to the formation of biomass and ethanol when the culture reaches steady state. In the fermentation problem considered, high productivity of ethanol is preferred and Zymomonas mobilis is used for the same. Zymomonas mobilis has several attributes that make it advantageous for fermentation of lignocellulosic hydrolyzates: 1. High glucose uptake 2. High productivity 3. Low biomass yield 4. High product (ethanol) yield 5. High ethanol tolerance 6 . No addition of 0, required 7. Amenable to genetic manipulations The major disadvantage of Zymomonas mobilis is that it can work with a very limited substrate range (only sucrose, glucose, and fructose). The fermentation problem involves a high degree of complexity. Instead of trying to understand the physics of the system in order to predict the concentrations at the next stage of fermentation, we use ANNs to simulate the main state variables in fermentation processes (biomass, substrate, and product concentrations) as functions of the progress of the process.

Process Plant Simulation

54

Several input variables are considered. These are biomass concentration (xi), substrate concentration (Si), ethanol concentration (Ei) (these are given at the present time), the present time (ti), initial substrate concentrations (So), temperature (T), and the time interval (At = ti+l- ti) at the end of which the state of the system is desired to be predicted. The output variables (next stage) are: biomass concentration (xi 1), substrate concentration (Si 1), and ethanol concentration (El 1) at the end of the given time interval. +

+

+

p Y l Y //

x,: initial substrate

I

y

(Variable size: Ts denotes tan-sigmoid transfer function)

(Seven variables) I

I

Outer layer (Three variables; Ls denotes log-sigmoid transfer function)

I

Fig. 3.7 Schematic diagram of the black box neural network used to predict the next state of Zymomonas mobilis CP4 batch fermentations, given the present state and the operational conditions of the process

The schematic diagram of the ANN used is shown in Fig. 3.7. The configuration of the neural network considered has one hidden layer with neurons having the tan-sigmoid transfer function and an outer layer of three neurons having the logsigmoid transfer function. A neuron with a tan-sigmoid transfer function outputs values in the interval (-1, l), as its input goes from -a to + a .In our case, the values of the net outputs have to be positive, as these are concentrations, and this is why neurons with log-sigmoid transfer functions are used in the outer layer.

Classification of Mathematical Modelling

55

Illustration 3.5 Modelling of phase equilibria Liquid extraction, sometimes called solvent extraction, is the separation of the constituents of a liquid solution by contact with another insoluble liquid. If the substances constituting the original solution distribute themselves differently between two liquid phases, a certain degree of separation will result and this can be enhanced by the use of multiple contacts. The solution that is to be extracted is called thefeed and the liquid with which the feed is contacted is called the solvent. The component to be extracted is called the solute and the other component in the feed is known as the diluent. The solventrich product of the operation is called the extract and the residual liquid from which the solute has been removed is the raftinate. The present case involves the extraction of polyethylene glycol (PEG) from a liquid solution of PEG and potassium phosphate using water as a solvent, PEG is the solute while potassium phosphate is the diluent. 3.7.6.1 Thermodynamic models Because of the inherent complexity and cross interaction features of ATPSs, it is difficult to derive an equilibrium function just involving simple independent variables with precise definition. Many models have been developed to describe the mechanism of phase separation, but that based on statistical thermodynamics is the most effective mathematical model. However, the thermodynamic model has certain disadvantages as well:

A set of parameters measured from a system is valid exclusively for it. The adapted parameters measured from other systems lead to a poor agreement with experimental data. There must be a huge database for numerous systems of interest. The parameters need to be measured using special equipment, which is not always available. Keeping in mind the complexity of this system, a neural network was thus chosen to establish a design-oriented model suitable for a variety of applications for arbitrary systems with minimum number of input parameters. 3.7.6.2 Neural network model

A number of basic structures were first considered. These are Four outputs all in one network-all class one network (ACON) Four outputs divided into two subnetworks-partial ACON Four outputs divided into four subnetworks-one class one network (OCON) Previously carried out experiments indicate that through specifically divided subnetworks, the system errors encountered in the learning process diminished.

56

Process Plant Simulation

The fewer outputs the network had, the smaller the total deviation the subnetworks created. The major disadvantage of OCON is that the network took a longer time in the learning process than with ACON. However, since there appeared to be no significant difference between the outputs, and because of reduced error, OCON is selected as the fundamental structure.

3.7.6.3 Input and output parameters Three input parameters were chosen: rn Total weight proportion of PEG (polymer) in the phase system rn Total weight proportion of potassium phosphate (salt) in the phase system rn Normalized PEG molecular weight Because the molecular weight (MW) of PEG commonly applied for an ATPS is below 50,000, the normalized parameter is expressed by the following equation: PMW

= 10g(MW/lOO)/l0g(500) (MW<50,000)

Alternatively, if the highest (Id)and lowest ( L )molecular weights of PEG applicable for an ATPS are known, the normalized molecular weight parameter corresponding to the desired molecular weight (W) can be calculated as PMW = (W - L)/(H - L ) Four output parameters which completely predict the equilibrium phase diagram for the PEG/potassium phosphate/water system were selected: rn Mass fraction of PEG in the top phase rn Mass fraction of potassium phosphate in the top phase rn Mass fraction of PEG in the bottom phase rn Mass fraction of potassium phosphate in the bottom phase 3.7.6.4 Design and training of the ANN

A feed-forward neural network based on the back-propagation algorithm was designed. Consulting earlier work (Lee et al. 1966), it was observed that when the OCON model with two hidden layers was employed, the network converged very rapidly and the system error dipped lower than that in the model with one hidden layer. When the smallest error was reached during the learning process, the number of neurons in the hidden layers were referred to as the most suitable structure. The required data points for training and testing the network were obtained from experimental results (Lei et al. 1990). Phase diagrams at 4 "C and pH 7 for the PEG/potassium phosphate/water system, using PEG of molecular weight 400, 600, 1000, 1500, and 20,000 were obtained. All the graphs were enlarged so that the data points could be read clearly. Tie lines were drawn for each graph at different feed concentrations, and the points corresponding to the ends of the tie line were read as the raffinate, extract salt, and polymer concentrations, respectively.

Classvcation of Mathematical Modelling

57

A total of 60 data points were used for training the network and 12 were used for testing. The error percentage values indicate that the maximum error (44.13%) occurs while predicting the polymer concentration values in the bottom phase (raffinate). The PEG concentration in the bottom phase revealing a greater deviation from simulated equilibrium data might be attributed to the intrinsic error of the original experimental data. Since the PEG concentrations in the bottom phase were very low, down to about 3%-0.5% w/w, a slight error occurring in the weighing procedure, while conducting the experiment, could have a significant effect on the measured values.

Conclusion The network designed acts as a statistical and design-oriented model for an ATPS, giving promising predictability with minimum number of measured parameters. Training the network further with more data points can make the framework of the network an excellent design-oriented model, capable of predicting phase diagrams for a vast range of polymer molecular weights.

EXERCISES 3.1 Classify mathematical modelling in the form of a tree structure, and elaborate the various terms involved. 3.2 Describe various stages in the development of a complete mathematical model for a process using a neat flow diagram. 3.3 What is the black box principle? 3.4 Develop a computer code in C or C++ for artificial neural networks using the back-propagation algorithm. 3.5 Schematically represent the structure of a simple artificial neural network and explain the terms involved. 3.6 Explain how an ANN can be applied for process fault diagnosis. 3.7 What are the various promising areas of application of ANNs, especially in chemical engineering? 3.8 Distinguish among independent variables, dependent variables, and parameters with suitable examples. 3.9 Distinguish between boundary conditions and initial conditions with suitable examples. 3.10 Discuss the back-propagation network (BPN) versus radial basis function network (RBFN)in ANNs.

PART II

CHEMICAL SYSTEM MODELLING

IN

THIS PART we

will apply the modelling principles and the generalized approach to model various chemical engineering systems. The models for chemical engineering systems are divided into four groups: models in mass-transfer operations, models in heat-transfer operations, models in fluid-flow operations, and models in reaction engineering, This part also focuses on the mathematical techniques that will be used for solving the model equations developed (such as algebraic equations, ordinary differential equations, partial differential equations, finite difference equations, Laplace transformations, solution by series or method of Frobenius, Bessel equations, etc.).

CHAPTERS 4. Models in Mass-transfer Operations 5 . Models in Heat-transfer Operations 6. Models in Fluid-flow Operations 7 . Models in Reaction Engineering

CHAPTER 4 MODELS IN MASS-TRANSFER OPERATIONS

In this chapter, various models for systems in mass-transfer operations (steadystate single-stage solvent extraction, steady-state two-stage counter-current solvent extraction, steady-state two-stage cross-current solvent extraction, unsteady-state single-stage solvent extraction, unsteady-state mass transfer in a stirred tank, unsteady-state mass transfer in a mixing tank, unsteady-state mass transfer, steadystate multistage counter-current solvent extraction, multistage gas absorption, multistage distillation) are discussed. A generalized approach is developed for formulating the models. The models are chosen in such a way that various mathematical techniques for solutions of the proposed models are covered. Some models are simple algebraic equations, a few of them are ordinary differential equations, some are finite difference equations, and in some models Laplace transformations are applied for solution.

4.1 Steady-state Single-stage Solvent Extraction Illustration 4.1 Consider that benzoic acid (BA) is continuously extracted from toluene using water as the solvent. The two streams are fed into tank A (mixer) where they are stirred vigorously and the mixture is then pumped into tank B (settler) where it is allowed to settle into two layers (extract and raffinate). The upper toluene layer (raffinate) and the lower water layer (extract) are removed separately. Model the system to find out what fraction of BA has passed into the solvent phase. Solution Step 1 The first step in modelling of any system is to schematically represent the given system by understanding the actual process occumng in the system (Fig. 4.1).

62

Process Plant Simulation

Toluene

Water Mixer

Water + BA (aqueous layer)

€aToluene + BA (organic layer)

Settler (B)

Fig. 4.1 Single-stage mixer settler

Step 2 The second step is to see if there is a possibility of simplifying the schematic representation of the system under consideration. The above system can be idealized and simplified as follows by combining two stages into a single stage (Fig. 4.2), as we know that Input - output + generation - depletion = accumulation

c kg/m’ BA

s m’/s water

Idealized single-stage solvent extraction system

y kg/m3 BA

(Raffhate) R m3/s of toluene x kg/m’ of BA

s m3/swater 0 kg/m3of BA

(extract) Fig. 4.2 idealized system

The various streams have been labelled and two material balances have already been used: (a) Conservation of toluene and (b) conservation of water. These flow rates have been expressed on a solute-free basis to simplify the analysis. The concentration of BA in each stream has also been stated and this completes the mathematical model. Step 3 The third step is to make reasonable assumptions, and also justifying the assumptions made with reference to retaining accuracy. So far, it has been assumed that (1) all flow rates are steady and (2) toluene and water are immiscible (basic criterion for choosing a solvent). Further, it is assumed that (3) the feed concentration c remains constant. Now, have a look at the number of unknowns and the number of equations we can formulate with the known information so far. In fact, we have two unknowns

Models in Mass-transfer Operations

63

( x and y ) and one equation to be formulated based on the mass balance on benzoic acid [Eq. (4.2)]. That means we still need another equation to get the solution. Obviously, in the present problem, we are looking for another equation relating the two unknowns x and y . This being a model based on mass transfer, an equilibrium relation may do the job for us.

Step 4 Look for additional assumptions to be made in order to balance the number of unknowns and the number of equations required. For the present problem, we need to make another assumption that (4) the mixer is so efficient that the two streams leaving a stage are always in equilibrium with one another. This last fact can be expressed mathematically by the following expression: y=mX (4.1) where, m is distribution coefficient. The model equation is now derived by writing a mass balance for BA (component balance):

Input of BA = Rc (kg/s) Output of BA = Rx + Sy (kg/s) As the system is in steady state, i.e., BA must leave at the same rate as it enters, Rc=Rx+Sy (4.2) Equations (4.1) and (4.2) contain four known quantities (R, S, c, m ) and two unknown quantities (x, y ) and can be solved for the unknowns as follows: Rc = Rx + mSx (4.3)

+ Rc = ( R + mS)x

:. x = R +Rcm S

(4.4)

y = - mRc R + mS

(4.5)

~

Step 5 Always express all the quantities (both known and unknown, or a combination of both) in terms of dimensionless groups which explain the physics of the system. Therefore, the fraction of BA extracted is

SY

j-=

1

($)+1

(4.7)

64

Plvcess Plant Simulation

At this stage it can be seen that even in this simple problem, two dimensionless groups which are characteristic of the problem have arisen quite naturally as a result of the investigation: the ratio of the feed to the solvent flow rate at equilibrium,

and the fraction of BA extracted,

Substituting these two dimensionless groups in Eq. (4.7), we get

E=- 1 a+l and multiplying both numerator and denominator by (a- l), we get E=- a - 1

a2-1 Step 6 The final result should be expressed in a generalized form as far as possible, so that the method of induction principles can be applied for a cascade of stages. This means the fraction extracted is governed solely by the value of the dimensionless group a (the ratio of rates of feed to the solvent at equilibrium).

Example 4.1 Compute the fraction of solute that could be extracted in a singlestage solvent extraction using the numerical values of S = 12R, m = 118, and c = 0.1 kg/m3. Solution IfS =12R, m = 1/8, and c = 0.1, x=-

+

.*.

-

Rc

0.1R R( 1 +

mS

= 0.04

f)

1 8

y = m~ = - x 0.04 = 0.005

and

E = - Sy = Rc or

12R x 0.005 = 0.6 = 60% 0.1 x R

Models in Mass-transfer Operations

65

1 12 .I. E = -- - - - 0.6 = 60% 20 -8+ 1 12

4.2 Steady-state Two-stage Solvent Extraction Illustration 4.2 Develop a model and obtain an expression for the extraction of BA in a steadystate operation with two stages. The input data are same as in the previous problem. Each stage still consists of two tanks, a mixer and a settler, with counter-current flow through the stages. The idealized flow system is shown in Fig. 4.3, where the symbols have the same meaning as in the previous example, and the different concentrations in a particular phase are distinguished by suffixes.

Stage

Stage

t-

Fig. 4.3 Two-stage settler mixer

Solution In accordance with chemical engineering practices, the suffix denotes the stage from which the stream is leaving. The following assumptions are made again: (i) The feed concentration c remains constant. (ii) The mixer is so efficient that the two streams leaving a stage are always in equilibrium with each other. The equation y = w: is still valid for each stage separately, giving Y 1 = w:,

(4.8)

Y2 = m2

(4.9)

where m is the distribution coefficient. It is constant for the given system. A BA mass balance is now taken for each stage:

66

Pmcess Plant Simulation

Stage 1

Stage 2

Input of BA (kgh):

Re + Sy,

RXl

Output of BA (kgh):

RXl + SY,

k,+ SY,

Since the operation is carried out in steady state, the BA must enter and leave a stage at the same rate, i.e., Stage 1: Re + Sy, = Rx,

+ Sy,

(4.10)

(4.11) Stage 2: Rx, = Rx, + Sy, To get an expression for y , in terms of known quantities, substitute x1 and y , from Eqs (4.8) and (4.9) in Eqs (4.10) and (4.11). Equation (4.10) becomes

RYl + Sy, Re + mSx, = m

(4.12)

- ( R + mS) Y1

(4.13)

Ryl - - Rx2 + mSx,

(4.14)

m and Eq. (4.11) becomes m

m + y1 = -(Rx2 + mSx,)

(4.15) R Since x2 is unknown, it is to be expressed in terms of known quantities. Substituting y1 from Eq. (4.15) in Eq. (4.13), we get

Rc + mSx, =

( R + mS) m -(Rx, + mSx, ) m R

* R2e + mRSx, = R2x, + mRSx, + mRSx, + m2S2x, R2e = x,(R2

+ mRS + m 2 S 2 )

R2c , R +mRS+m2S2

*x2=

(4.16)

Substituting x2 from Eq. (4.16) in Eq. (4.15), we get mRc(R + mS) R +mRS+m2S2 Again, the proportion of BA extracted is

’1 -

SYl --

Re

2

(4.17)

mS(R + mS) R2 + mRS + m2S2

(4.18)

Models in Mass-transfer Operations

67

Multiplying and dividing the right-hand side (rhs) of Eq. (4.18) by m2S2, we get

-SY-l Rc

(WmS) + 1 (R2/m2S2)+ (WmS) + 1

(4.19)

Now introducing the dimensionless groups E = Sy,/Rc and a = WmS , we get a + l a2+a+1

E=

(4.20)

Multiplying and dividing the rhs of the above equation by (a- 1) , we get

E=-

a2- 1 a3- 1

(4.2 1)

By the method of induction, for N stages,

aN-1 = aN+l - 1

Example 4.2 Compute the fraction of solute that can be extracted in a two-stage counter-current solvent extraction using the same numerical values of S = 12R, m = 1/8, and c = 0.1 kg/m3, which were used for single-stage extraction, and compare the results. Solution x2=

R2c 2 R +mRS+m2S2 R2 (0.1)

R2 + 'R(12R) 8

R

'(

+ '(144R2) 64

12 144 1+-+8 64)

= 0.02105

and mRc(R + mS) y1 = R2 + mRS + m2S2

- vs(R)(O.l)[R + v8(12R)] R2 (1 + 1U8 + 144/64) -

68

Process Plant Simulation

-- R2(0.V8)(1+ 12/8)

R2(1+ 12/8 + 144/64)

= 0.00658 Therefore,

R a=--

0.66667

--=

(i)12R

- 1 - (0,66667)' - 1 -- -0.55555 a3- 1 (0.66667)3- 1 -0.703699

' E = -a

Therefore, E = fraction extracted = 0.789= 78.9%. A greater degree of extraction has thus been obtained with two stages (78.9%) than with one stage (60%), everything else being the same.

Example 4.3 160 cm3/s of a solvent S is used to treat 400 cm3/s of a 10% by weight solution ofA in B, where A is being extracted from B in a two-stage countercurrent liquid-liquid extractioncolumn. What is the compositionof the final &mate and the fraction extracted if m (distribution coefficient) = 3 and the densities of A , B, and S are 1200,1000,and 800 kg/m3, respectively? Solution S = 160 cm3/s = 1.6 x lo4 m3/s Mass flow of S = 1.6 x 10" x 800 = 0.128kg/s Considering the solution, 400 cm3/s = 4 x 104 m 3/s. Considering a m3/s of A and (4x lo4 - U) m3/s of B, Mass flow of A = 1200a kg/s Mass flow of B = (4x 10" - a ) x 1000 kg/s = (0.4- 1000a)kg/s Total mass flow = (0.4 + 200a) kg/s Concentration of the solution = 0.1 =

1200a 0.4 + 200a

.: .:

a = 3.39 x 10-~ m3/s Mass flow of A = 0.041 kg/s Mass flow of B = 0.366 kg/s

Now, R (volumetric flow rate of B) = 4 x 10" - 3.39 x 10-~ = 3.661 x 104 m3/s c (kg of A/m3 of B ) = (kg/s of A)/(m3/sof B )

c = 0.04V3.661x

lo4

= 112.02kg/m3

Models in Mass-transfer Operations

The composition of final raffinate ( x 2 ) = R 2 d ( R 2+ mRS

:.

69

+ m2S2)

x2 = 27.795 kg/m3

and

where a = WmS = 0.762

.*.E

= 0.7538

Therefore the conversion is 75.38%.

4.3 Steady-state Two-stage Cross-current Solvent Extractlon Illustration 4.3 Consider the idealized flow system shown in Fig. 4.4where the symbols have the same meaning as in the previous model.

Stage 1

s, 0

Stage 2

s, 0

Fig. 4.4 Idealized flow system

Solution The following assumptions are made: 1. Toluene and water are immiscible. 2. We are considering a pure solvent. 3. The mixer is so efficient that the leaving streams are always in equilibrium. 4. The concentration of BA in the feed is constant. 5. The solvent flow rate is the same in both stages. 6. All the quantities are expressed on a solute-free basis. 7. The composition of a stream leaving a stage is the same as that in the stage.

70

Process Plant Simulation

BA balance (mass component balance for each stage): Stage 1 Stage 2 Rc + 0 SY, + RX,

Input of BA (kgh): Output of BA (kgh): In steady state: Stage 1: Rc = Sy, Stage 2: Rx, = Sy,

+ Rx, + Rx,

RX,+O SY, + 4

(4.22) (4.23)

Since there are four unknowns ( x l ,yl, x2, y2) and only two equations, therefore two more equations are required and these are the equilibrium relations given by Y 1 = m, (4.24) (4.25) The objective in this problem is to find the total fraction of BA that can be extracted by the cross-current mode of extraction operation under steady-state conditions. The fraction of BA extracted, Y2

=m

2

Hence, we have to obtain the expressions for y1and y, in terms of known quantities by eliminating x1 and x2. So, from Eq. (4.24), we have

Y1 x1 = m From Eqs (4.22) and (4.26),

(4.26)

m mRc -

Y

l

=

G

(4.27)

Therefore, Re R+mS

x1 = -

(4.28)

From Eq. (4.25), we have Y2 x, = m

(4.29)

Models in Mass-transfer Operations

71

Substituting x1 and x2 from Eqs (4.26) and (4.29) in Eq. (4.23), we get

R--- Rc - Ry2 R+mS m

*

+ s y , = [--)Y2R + m S

mR2c y2 = ( R + mSI2

[’

*E=

mRc

mR2c + ( R + m+!7)2] Re

mRc

R

’( =)(’

+

=)

Rc

On multiplying and dividing both the terms in the parentheses by mS, we get

R

1

We know that

R mS

-=a

3“

*E =

-[

*E=

+1 (a+ I)*

1 1+ a+l a+l 2a

Example 4.4 Compute the fraction of solute that can be extracted in a two-stage cross-current solvent extraction using the same numerical values of S = 6R, m = 1/8, and c = 0.1 kg/m3, which were used for single-stage extraction, and compare the results. Solution If S = 6R (since 2s here = S in counter-current), m = 1/8, c = 0.1 kg/m3, R

R

8 6

= - = 1.3333

72

Pmcess Plant Simulation

*E=

2(1.3333)+ 1 = 0.6735 = 67.35% (1.3333 + 1)2

Comparison For counter-current steady-state two-stage extraction, a = 0.66667 (since S = 12R)

[: I

E = - = 0.789 = 78.9% Note As expected and as can be seen from the above comparison, with the same conditions the fraction of solute extracted in the counter-current operation is higher than that in the cross-current operation. This is attributed to the fact that the driving force available at any stage is higher in the case of the counter-current mode of operation, as no extract is taken out from any intermediate stage. However, in the cross-current operation, the extract is taken out from each of the stages and so the driving force is comparatively less in any given stage.

Exercise 4.1 How do you compare the co-current operation with counter-current and cross-current operations? Example 4.5 160 cm3/s of a solvent S is used to treat 400 cm3/s of a 10%by weight solution of A in B, where A is being extracted from B in a two-stage countercurrent liquid-liquid extraction column. The distribution coefficient m = 3 and the densities ofA, B, and S are 1200,1000,and 800 kg/m3,respectively.All the quantities are expressed on a solute-free basis. (a) What is the composition of the raffinate leaving from stage 1 and stage 2? (b) What is the final fraction extracted? Solution S = 160 cm3/s = 1.6 x 104 m3/s Mass flow of S = 1.6 x 1W x 800 = 0.128 kg/s Considering the solution, 400 cm3/s= 4 x lo4 m3/s.Considering a m3/s of A and (4 x lo4 - a) m3/sof B, Mass flow of A = 1200a kg/s Mass flow of B = (4 x lo4 - a ) x lo00 kg/s = (0.4 - 1OOOa) kg/s Total mass flow = (0.4 + 200a) kg/s Concentration of the solution = 0.1 =

:. :.

a = 3.39 x 10” m3/s Mass flow of A = 0.041 kg/s

1200a (0.4 + 200a)

Models in Mass-transfer Operations

73

Mass flow of B = 0.366 kg/s Now,

R (volumetric flow rate of B ) = 4 x 10-4 - 3.39 x 10” = 3.661 x lo4 m3/s c (kg of A/m3of B ) = (kg/s of A)/(m3/sof B) c = 0.04U3.661 x lo4 = 112.02 kg/m3 The composition of the extract from stage 1 XI

=

RC

(R+ mS)

= 48.456 kg/m3 The composition of the extract from stage 2 x2

=

R2C

( R + mSl2

= 20.9,692 kg/m3 and

E=- 2 a + l

(a+ 1)2

where a = WmS = 0.762 .*.E = 0.806 Therefore the fractional recovery is 80.6%.

Note Compare Examples 4.4 and 4.5 on two-stage steady-state counter-current and cross-current extraction. For the same numerical data, the fractional recovery obtained in counter-current and cross-current operations is 78.9% and 80.6% respectively. We expect the fractional recovery to be more in the counter-current operation than in the cross-current operation. Can you find out what is the reason of this contradictory result? The clue is, have a close look at the data given for S.

4.4 Unsteady-state Single-stage Solvent Extraction In the models considered so far, the system has been in a steady state, allowing the material entering the system to be equated to the material leaving the system, which has always given an algebraic equation. In unsteady-state problems, however, time enters as a variable and some properties of the system become functions of time. In the application of conservation laws, it is no longer true that the rate of input will equal the rate of output, since an allowance must be made for the material accumulating within the system. The general conservation law now becomes Input - output = accumulation

74

Pmcess Plant Simulation

Illustration 4.4 The single-stage steady-state example is reconsidered as follows: 1. Assume that the single-stage contains Vl m3 of toluene, V, m3 of water, and no benzoic acid (solute-free basis). 2. Assume, as before, that the mixer is so efficient that the compositions of the two liquid streams are in equilibrium at all times. 3. Assume that a stream leaving the stage is of the same composition at that phase in the stage. 4. c is constant in feed. This is illustrated in Fig. 4.5, which shows the state of the system at a general time 8 , where x and y are now functions of time. R m3/s c kg/m’

Sm3/s Y kg/m3

I-

w

v,,x vz,x

1-

-

R m3/s xkg/m3

s m3/s 0 kg/m3

Fig. 4.5 Time-dependent single-stage extraction (unsteady state)

Since the flow rates of water and toluene are constant (solute-free basis), V , and V , will remain constant (solute-free basis). The system being in unsteady state, the conditions are always changing, and so the material balance must be applied during a small time increment 68. Using Taylor’s theorem, a function of 8 can be expanded, and this allows the state of the system at a time 0 + 68 to be expressed in terms of its state at time 0 . In this case it is helpful to draw up a table (see Table 4.1). Mass balance for BA during a time interval 68: Input of BA = Rc&? kg, output of BA = average of Rr + Sy at 8 and 6 + 68. (The difference is not taken because the amount is varying with time. So the average has to be taken.) Therefore,

and

dx Accumulation = V, -88 dB

dY + V, -68

Since, Input - output = accumulation

d6

Models in Mass-transfer Operations

=

(y- 2+

v d20 9 0

[ :d“e ] - S[

Taking the limit as 60 + 0 , - Rx

:

y+--60

+Rc-R~+--60

Model: Rc

75

]

dx dY =V,-+V2d0 d0

dx + V2dY - Sy = V, d0

(4.30)

d0

Table 4.1 The system condition before and after a time increment

e

Property of the system

5.

Flow rate of toluene phase Flow rate of water phase Volume of toluene phase in stage Volume of water phase in stage Concentration of BA in entrance toluene

6.

Concentration of BA in exit toluene

1. 2. 3.

4.

X

e+&

x + -dx- e de

7.

Concentration of BA in water (extract)

8.

Amount of BA in the toluene layer

Amount of BA in the water layer

9.

V,X

v*y

v,x

dx + v, 6e de

V2Y

+ v2

dY 66 de

In stating the output, the arithmetic mean of the concentrations was taken, but the later calculations show that this is not necessary; the same Eq. (4.30) would have been obtained by using the concentrations at time 0 instead. Only the first two terms of Taylor’s series were taken for x and y , and these are all that are necessary since further terms which contain a factor (60)*will disappear when the limit 68 + 0 is taken, i.e., x

dx 1 d2x +60 + --(60)2

+

*-.

d0 2 do2 Equation (4.30)must now be solved in conjunction with the equilibrium relationship y=mX (4.31)

76

Process Plant Simulation

Eliminating y from Eq. (4.30) using Eq. (4.31), we get dx dx Rc - R x - S~.X= VI- + m v 2 d8 d8 Rearranging,

dx Rc - (R + mS)x On integrating,

do

V,

(4.33)

+ mV2

-ln[Rc - (R + mS)x] R+mS V,

(4.32)

+ mV,

+A

(4.34)

The constant of integrationA can be evaluated by using the given initial state of the system that the stage contains no BA initially, i.e., (4.35) Initial condition (IC): when 8 = 0, x = 0 . From Eqs (4.34) and (4.33, -ln(Rc) -=A R + mS

(4.36)

Therefore, ln(Rc) - ln[Rc - (R + mS)x] =

(R + mS)8 (V, + my,)

or in exponential form,

r

*1--

1

R

+ mS x = exp[RC

( R + mS)8 V, mV2

+

]

(4.37)

Models in Mass-transfer Operations

R R

+ mS

+ mS

77

(4.38)

1. This equation satisfies the boundary condition (BC), i.e., x = 0 at 8 = 0 (since e-" = l/eo = 1). 2. It satisfies the differential equation [model Eq. (4.30)] for any positive value of 8 , (find &/do ). 3. And when 8 j -,

Rc (4.39) R+mS When 8 + 00, the system reaches steady state, and hence this should be converted back to the steady-state model equation, which has already been obtained, as shown by Eq. (4.39). Therefore, for steady state, *X=-

x=-

Rc R +mS

For the unsteady state,

R

+ mS

R + mS V, +mV,

(4.40)

Therefore, the fraction of BA extracted = Sy/Rc

*E=-

a+ll Here, 8 is the independent variable, x and y are the dependent variables (static variables), and m, R , S, c, V,, and V, are parameters.

Example 4.6 160 cm3/s of a solvent S is used to treat 400 cm3/s of a 10% by weight solution of A in B, where A is being extracted from B under unsteady-state conditions in a single stage using the counter-current mode of operation in a liquidliquid extraction column. The volume of B(V,) is 100 dm3 and the volume of solvent (V,) is 75 dm3. The distribution coefficient m = 3 and the densities of A, B, and S are 1200, 1000, and 800 kg/m3, respectively. All the quantities are expressed on a

78

Process Plant Simulation

solute-free basis. (a) What is the composition of the final raffinate and fraction extracted after 1 min? (b) If 50% of A is to be extracted then what is the time required if V , and V, are same. Solution S = 160 cm3/s = 1.6 x 10- 4m3 /s Mass flow of S = 1.6 x lo4 x 800 = 0.128 kg/s Considering the solution, 400 cm3/s= 4 x lo4 m3/s. Considering a m3/s of A and (4 x lo4 - a) m3/sof B, Mass flow of A = 1200a kg/s Mass flow of B = (4 x lo4 - a) x 1000 kg/s = (0.4 - 1000a) kg/s Total mass flow = (0.4 + 200a) kg/s Concentration of the solution is = 0.1 =

1200a (0.4 + 200a)

.: :.

a = 3.39 x 10” m3/s Mass flow of A = 0.041 kg/s Mass flow of B = 0.366 kg/s Now, R (volumetric flow rate of B ) = 4 x lo4 - 3.39 x lo-’ = 3.661 x 1W m3/s c (kg of A/m3 of B ) = (kg/s of A)/(m3/sof B ) c = 0.041/3.661 x lo4 = 112.02 kg/m3

(a) The composition of the final raffinate

:.

x = 40.904 kg/m3

and

)j

E = J-[ 1 - exp( -(a + l)8 a+l (v, + rnV2)/mS where R mS :. E = 0.082 Therefore the fractional conversion is 8.2%.

a = -= 0.76271

(b) If 50% of A is to be extracted, E = 0.5. Substitutingthe values of E, a , V,, V,, m, and S in Eq. (l),we get 8 = time required for 50% extraction = 819 s = 13.65 min

Models in Mass-transfer Operations

79

4.5 Unsteady-state Mass Balance in a Stirred Tank Example 4.7 A tank contains 20 m3 of water. A stream of brine containing 2 kg/m3 of salt is fed into the tank at a rate of 8.33 x lo4 m3/s. Liquid flows from the tank at a rate of 5.56 x lo4 m3/s. If the tank is well agitated, what is the concentration of salt in the tank when the tank contains 30 m3 of brine? Solution The system is schematically represented as shown in Fig. 4.6. 8.88 x lo4 m3/s 2 kg/m3 kg/m3 2

I

I1

I

II 5.55 x lo4 m3/s

Fig. 4.6 Stirred tank

The system properties at time 8 and at incremental time B + 88 are tabulated and shown in Table 4.2. Table 4.2 The system condition before and after a time increment Property 1. 2.

At

e

At

8.88 x lo4 5.55 x 10-4 2

e + se

8.88 x 10-4 5.55 x 1oJ 2

3.

Rate of brine input (m3/s) Rate of brine output (m3/s) Input salt concentration (kg/m3)

4.

Volume of brine in tank (m3)

V

v + -dV m de

5.

Salt concentration in output stream (kg/m3)

X

x

6.

The amount of salt in tank (kg)

vx

dx + --6e de

(v +$ss)(

In a small interval 68, accumulation of brine in the tank = dV/d8 88 Volume balance: Input - output = accumulation

x

+ $Je)

80

Process Plant Simulation

dV (8.33 x m4)68- (5.55 x io4)6e = -68 de

(4.41)

dV = (8.33 - 5.55) x lo4

= 2.78 X lo4 (4.42) de On integrating, we get V = 2.78 x lo4 + C,, where C, is a constant. By applying the boundary condition that at 8 = 0, V = 20 m3,we get C, = 20. Therefore, (4.43) v = 2.78 x lo4 e + 20 Mass balance for salt in time interval of 68 : Input - output = accumulation (i.e., the difference of amount at 8 + 68 and 0 ) 3

(8.33 x 1 0 3 2 6 8 - (5.55 x lo4).&

= ( v + %6e)( x

+ $m) - v x

dx dV dV dx = vx + v-6e + --6e + ---(se)* - vx de de de de Simplifying and taking the limit as 68 + 0 , we get 16.66 x lo4

- 5.55 x 104x

dx

dV + -X de de

= V-

(4.44)

Substituting V and dV/de from Eqs (4.42) and (4.43), respectively, in Eq. (4.44), we get 16.66 x lo4 - 5.55 x 104x

= (2.78 x lo4

dx e + 20) + 2.78 x lo4 x

d0 The variables are separable, hence dx

16.66 x lo4

-

dx

8.33 x lo4

(4.45)

2.78 x

io4e + 20

(4.46)

(4.47) Boundary condition 1: At 8 = 0, x = 0 (pure water) We have to find out the value of x when the volume of brine in the tank = 30 m3, i.e., 8 = ? when V = 30 m3. From Eq. (4.43), v = 2.78 x 104 e + 20 j

30 = 2.78 x 10-~e + 20

+e=

10 = 35,971.22 = 35,972s (2.78 x lo4)

Models in Mass-transfer Operations

81

Boundary condition 2: At 8 = 35,972s, x = x kg/m3 (4.48) Integrating Eq. (4.46) using the suitable limits from Eqs (4.47) and (4.48), we get

[-

ln(16.66 x lo4

8.33 x lo4

[

-

x)

ln(2.78 x 104 e + 20)

= 2.78 x lo4

* -8.33 x1 lo4

- 8.33 x lo4

16.66 x lo4 - 833 x lo4 x 16.66 x lo4 In(

2.78 x lo4

In(;)

* - -31 ln(1 - 0 . 5 ~=) In (1 - O . ~ X ) - ” ~= 1.5

* 1 - 0 . 5 ~= 3x

(A) 3

= 0.2963

= 1.4074 kg/m3

Therefore the concentration of salt when the tank contains 30 m3 of water (after 35,972 s) = 1.4074 kg/m3.For 40 m3 of brine,

e=

20

2.78 x lo4

= 7 1,944s

40 * (1 - O . ~ X ) - ” ~=2 20

* X =

1 - 0.125 = 1.75 0.5

Therefore, x = 1.75 kg/m3

82

Process Plant Simulation

4.6 Unsteady-state Mass Balance in a Mixing Tank Illustration 4.5 A stream of solution containing dissolved salt flows at a constant volumetric flow rate Q m3/s into a tank of constant holdup volume V. The concentration of the salt in the entering stream, x kg/m3, varies with time. Develop a model for the outlet concentration, y kg/m3, when there is a sinusoidal input change in the inlet concentration. Assume the density of the solution to be constant. Solution The above system is schematically represented in Fig. 4.7. Q m’/s x ( t ) kg/m’

-@$-

Q m’/s Y (t) k d d

Fig. 4.7 Mixing tank

Since the holdup volume is fixed, and the solution density is constant, the inflow rate must be equal to the outflow rate (i.e., Qin= Q,,,). Mass Balance: Inflow rate of salt - outflow rate of salt = rate of accumulation of salt in tank

(4.49) Qx, - QY, = 0 Subtracting Eq.(4.50)from Eq. (4.49),we get

(4.50)

(4.51) Introducing the deviation variables X = x - x s , Y = y - y s , and dyldt = dY/dt, we have dY QX - QY = V dt

(4.52)

X-Y=(E)$

(4.53)

Models in Mass-transfer Operations

Putting

z = V/Q

83

= time constant, and taking the Laplace transform, we get

(4.54) It is clear from the above transfer function that the mixing process is also a firstorder process. For sinusoidal input change [i.e., X ( r ) = x - x, = Asinwt], A@ X(s)= s2

+ w2

(4.55)

whereA is the amplitude of variation, w is the radian frequency in rads. Eliminating X ( s ) from Eq. (4.54) using Eq. (4.55), we get

Y (s) =

vz (s2 + w 2 )(s + Vz) A@

(4.56)

This equation can be solved for Y(t)by means of partial fraction expansion, which gives A Awz e - t / r sinwt (4.57) coswt + (z2w2+ 1) (z2w2+ 1) (z202+ 1) Equation (4.57) can be written in another form by using the trigonometric identity, (4.58) pcosA + qsinA = rsin(A + 6)

Y (r) =

where r = d p 2 + q2 and tan8 = p/q. Applying the identity to Eq. (4.57), we get (4.59) As t + 00, the first term on the rhs of Eq. (4.59) vanishes where @ = tan-' (-ox). and leaves only the ultimate periodic solution, which is sometimes called the steadystate solution. (4.60) We know that the input X ( t ) = A sin wr (4.61) Comparing Eq. (4.61) for the input forcing function with Eq. (4.60) for the ultimate periodic response, we see that 1. The output is a sine wave with frequency 0 equal to that of the input

signal.

84

Process Plant Simulation

2. The ratio of the output amplitude to the input amplitude is VJz2w2+ 1

.

This is always smaller than 1. We often state this by saying that the signal is attenuated. 3. The output lags behind the input by an angle @ . It is clear that lag occurs, for the sign of 4 , the phase angle, is always negative.

I I

If the input peak occurs before the output peak, then @ < 0, i.e., phase lag. If the output peak occurs before the input peak, then @ > 0, i.e., phase lead.

Example 4.8 Consider an unsteady-state mass transfer in a mixing tank. The inlet stream is subjected to a sinusoidal input with an amplitude of 0.015 m3/min about a mean flow rate of 0.3 m3/min. The time constant for the flow process is taken to be 0.2 min. The frequency of the input sine wave is 4 cycles/min. Describe the output response, giving its frequency, amplitude, and by how many minutes it lags or leads the input response? Solution Input:f = cycle frequency of the input = 4 cycles/min, w = radian frequency = 2nf = 25.1 radmin, and x( t ) = A sin(wt) = 0.015 sin(25.lt). Here w = 25.1 radmin, z = 0.2 min, and A = 0.015. output: Y ( t )=

A [sin(wt + cp)] (1 + w2z2)0,5

@ = tan-' (-wz)= tan-' (-5.02) = -78.7'

1. The frequency of the output signal is same as that of the input signal. 2. Output amplitude = A/(1 + w222)0.5 = 0.003 and input amplitude = 0.015.

Therefore the ratio of the output amplitude to the input amplitude is 0.2. This ratio is less than 1. Hence the signal is attenuated. 3. The output lags the input by a phase of -78.7'. Input frequency = 4 cycle/ min. Therefore the period per cycle = period of 360' = 1/4 min. Hence time for -78.7' = 0.055 min. Hence the output has an amplitude of 0.003, a frequency of 25.1 radmin, and lags behind the input by 0.055 min.

Exercise 4.2 Consider an unsteady-state mass transfer in a mixing tank. The inlet stream is subjected to a sinusoidal input with an amplitude of 0.015 m3/min about a mean flow rate of 0.3 m3/min. The time constant for the flow process is taken to be 0.5 min. Find the frequency of the input sine in cycles/min. The output lags behind the input by 80'. Determine the number of minutes by which it lags or leads the input response.

Models in Mass-transfer Operations

85

4.7 Unsteady-state Mass Transfer (Fick's Second Law of Diffusion) Consider two-component diffusion: A w B The rates will be most conveniently described in terms of a molar flux or moll (area)(time). The area is measured in the direction normal to the diffusion. In a nonuniform solution containing only two constituents, however, both constituents must diffuse if uniformity is the ultimate result. This leads to the use of two fluxes to describe the motion of one constituent: N is the flux relative to a fixed location in space. It is important in the application to design the equipment. For example, a fisherman is most interested in the rate at which a fish swims upstream against the flowing current to reach his baited hook (analogous to N).J i s the flux of a constituent relative to the average mass velocity of all the constituents. It is characteristic of the nature of the constituent. The diffusivity, or diffusion coefficient, DABof a constituent A in a solution in B, which is a measure of its diffusive mobility, is defined as the ratio of its flux JA to its concentration gradient: (4.62) which is Fick's first law written for the z-direction. The negative sign emphasizes that diffusion occurs in the direction of the drop in concentration. The flux NAof A relative to a fixed position P is positive and the NB of B is negative. For a condition of steady state, the net flux is

(4.63) NA + N, = N The movement of A is made up of two parts, namely, (1) that resulting from the bulk motion N and the fraction xAof N, which is A, and (2) that resulting from diffusion JA: (4.64) NA = NxA + J A since, by definition,

i.e., the flux of A relative to the average molar velocity of all constituents. Here, W, = moles of A , W, + WE= total number of moles of A and B. Consider the volume element of a fluid shown in Fig. 4.8,where a fluid is flowing through the element. We shall need a material balance for a component of the fluid applicable to a differential fluid volume of this type.

86

Pmcess Plant Simulation Z

Ax

Y Fig. 4.8 Fluid volume element

The mass-flow rate of componentA into the three faces with a common comer at E is (4.65) MA ( ( N A , x )x A Y h + ( N A , y ) y Ax& + ( N A , z )z hxAY) where N,, signifies x-directed flux and (N,, ,>, signifies its value at location x. Similarly, the mass-flow rate out of the three faces with a common comer at G is M A ( ( N ~ , x ) x +AYAz b + ( N A , ~ ) ~AX& + A+ ~ ( N A , ~ )Ax&') ~ + ~ (4.66) The total component A in the element = (Ax Ay Az) p A and its rate of accumulation = (kAY &) ,+,/a. If, in addition, A is generated by a chemical reaction at the rate RA mol/(volume)(time), its production rate is MARAAx Ay Az massltime. Since, in general, Rate in - rate out = rate of accumulation - rate of generation 3 Rate out - rate in + rate of accumulation = rate of generation Then, M A { [ ( N A , ~ ) -x( N +A &, ~ ) ~ I A Y A Z [ ( N A , ~ ) ~ - +( NAA ~, y ) y l b k

dP*

+ [ ( N A , z ) z + & - (NA,z)zlhxAy~+ hxAyAz- at = MARAAxAyAz

(4.67)

Dividing by Ax Ay Az , and taking the limit as the three distances become zero gives

Models in Mass-transfer Operations

87

From Eq. (4.64), the counterpart of Eq. (4.64) in terms of masses and in the x-direction is MANA,x = M A N X A+ M A J A , ~

* M A N A ,=~ Yx P A + M A

J

A

~

(4.69)

~

where ux is the mass average velocity in d s . PUX

= uA,x P A + U B , PB ~ = MANA,+ ~ M BN B , ~

* ux

= uA,x PA +

(4.70)

PB

From Eq. (4.69)

(4.71) Equation (4.68) becomes

d2CA

dX2

+-+-I+d2cA

dy2

d2CA

dz2

'PA =MARA dt

(4.72)

which is the equation of continuity for substance A. For a solution of constant density, -d u+x Y d+ uZ = dOu ax ay dz

(4.73)

Dividing Eq. (4.72) by MA and substituting p A / M A = cA, we get

(4.74) In the special case where the velocity equals zero and there is no chemical reaction, Eq. (4.74) reduces to Fick's second law of difision: i.e., u x = uY = u Z = O and R,=O

00

Process Plant Simulation

Fick’s second law, therefore, is (4.75)

Illustration 4.6 The diffusion equation derived in the previous section can be extended for cylindrical coordinates. The diffusion equation for cylindrical coordinates can be obtained by doing an energy balance over a differential element, a procedure similar to that derived in the previous section. x=rcos$ y = r sin$

z=z $ = tan-’(y/x)

Using the chain rule we can carry out the partial differentiation of c with respect to rand $ as

or d c = ~ cos2+% cos $ dr dX

From Eqs (1)-(4) we get

Similarly,

+ sin4 cos4-d C A dY

(3)

Models in Mass-transfer Operations

89

If we multiply Eq. ( 5 ) with d c A / d x ,

Substituting the value of ( d c , / d x ) in Eq. (7)

d 2C A -cos$--

cos$-

dr

dX

sin $ d cA

dcA

sin$dcA - --

dcA dr

sin$dcA

or

dr

dX2

sin2 0 d c A +--+-r

dr

sin2 $ d 2 cA r2

+[

C O S sin$ ~

r2

-1

dcA d$

(9)

Similarly,

COS’ Q d 2 C A +--COS* $ dcA +---(

r dr Now, adding Eqs (9) and (10)

r2

C O S ~sin4

r2

dcA d$

or

We know that for Cartesian coordinates, Fick’s second law of diffusion is given by

If there is no chemical reaction and diffusivity is constant, then by substituting Eq. (12) in Eq. (13)

90

Process Plant Simulation

(14)

This is the final expression for Fick's second law of diffusion in cy coordinates.

rical

4.8 Steady-state N-stage Counter-current Solvent Extraction Illustration 4.7 Benzoic acid is continuously extracted from toluene using water as the solvent in a counter-current mode using N stages under steady-state conditions. Each stage consists of (1) a mixer where vigorous stirring of the contents takes place and (2) a settler where the mixture pumped from the mixer is allowed to settle into two layers. The upper toluene layer from the second stage and the lower water layer from the first stage are removed separately. Develop a mathematical model to find the final fraction of benzoic acid extracted by clearly mentioning the assumptions involved in the analysis. Solution The improved extraction leads to the consideration of more than two stages in the extraction system. In the single stage, the algebraic treatment was quite simple. There were two equations in two unknowns for finding the solution. In two stages, there were four equations in four unknowns for finding the solution. Following the same procedure for N stages, it would be necessary to solve 2N equations in 2N unknowns. This is too laborious and would require an individual solution for every integer value of N . Therefore, more advanced mathematical techniques are obviously needed to reduce the work. Among these; one method uses matrix algebra and another method uses finite differences. The general arrangement is as shown in Fig. 4.9. The flow rates of the two streams (feed and solvent) are R and S (on a solutefree basis). The benzoic acid concentrations in the extract and raffinate are y and x , respectively. The suffix notation is again used to distinguish between the different stages of each stream, the suffix denoting the stage which the stream has just left. The assumptions remain the same as those made in the single- and two-stage extraction problems. Now, a benzoic acid material balance is applied to the general stagep of the system: Input of acid (kg/s) = f i p - l+ SY,+~ Output of acid (kg/s) = Rx,

+ Syp

Models in Mass-transfer Operations S

91

R

I

Yp+I

A A" YN- 1

I

I

N-1

S

R

Fig. 4.9 Idealized N-stage extraction

At steady state, entry rate = exit rate hp-l

+ SYP+l = Rx, + SYP

(4.76)

The equilibrium equation 0,= mx) is still valid for any value of p . *Yp

(4.77)

=mxp

Then, Eq. (4.76) becomes [from Eq. (4.77)]

Rx,-,+ mSx,,,

=

( R + mS)x,

(4.78)

92

Process Plant Simulation

Dividing by mS on both sides, we get R

mS xp-1

+ Xp+l

(2

=

+ l)x,

(4.79)

We know that RImS = a,which is a dimensionless group.

* ax,-, +

XP+,

+

= (a 1)x,

(4.80)

Writing Eq. (4.80) in standard form,

- (a+ l ) x ,

+ ax,,,

=0

(4.81) This is a second-order linear finite difference equation. Putting Eq. (4.81) in the operational form and factorizing gives XP+,

[ E 2 - (1 + a ) E + a ] x , = ( E + 1)(E - a ) x , = 0

which has the solution xp = K,

+K2aP

(4.82)

where K , and K , are arbitrary constants. Now, xo = K ,

+ K 2 a 0 = K , + K2

where xo is the concentration of BA at the zeroth stage, which is the concentration of BA before entering the first stage = c +c=K,+K, + K ~ = c - K ~ Similarly,

+ K2aN xN+1= K1 + K2aN+l

(4.83)

xN = K ,

(4.84)

Here, xN is the concentration of the solute (BA) in the raffinate at the ( N + 1)th stage = 0 (because N + 1 is very large).

+ 0 = K , + K2 aN+l Substituting K , from Eq. (4.83) into Eq. (4.85),we get

K1

=

CaN+l aN+l

-1

(4.85)

Models in Mass-transfer Operations

93

Substituting K,and K2 in Eq. (4.82), the solution is

xp =

c(aN+l- aP) (aN+l - 1)

rnc(aN+'- aP) (aN+l - 1) Therefore, the proportion of BA extracted is (put p = 1 in the above equation) Y p

=

-a) E = - SYl - - - s rnc(aN+' Rc Rc (aN+l - 1)

For N = 2,

E = -a ' - 1 a3 - 1 which was derived earlier.

Example 4.9 Compute the fraction of solute that can be extracted in countercurrent solvent extraction with N stages (for values of N from 1 to 30) using the same numerical values of S = 12R, rn = 1/8, and c = 0.1 kg/m3, which were used for single-stage extraction, and compare the results. Solution If S = 12R and m = 1/8,

By putting various values of N in the above equation, the proportion extracted in N stages is given in Table 4.3. This table shows that more than 10 stages are wasteful, whereas one stage gives a poor degree of extraction. The most economical number of stages can be determined by optimization (trade-off between operating and capital costs).

94

Process Plant Simulation

Table 4.3 Percentage extraction as a function of number of stages

1 2 3 5 10 20 30

60 78.9 87.7 95.2 99.4 99.98 99.99

4.9 Multistage Gas Absorption (Kremser-Brown Equation) Chemical processes involving separation, purification, and chemical reactions are frequently carried out in a series of stages, and these processes are often analysed by stage-to-stage calculations or in simple examples by graphical methods. Thus distillation, gas absorption, and the extraction operation performed in plate columns involve stepwise changes in the concentration of each phase as it passes from plate to plate. In particular, in a distillation column, the vapours leaving any plate differ in composition from the vapours entering the same plate by a finite amount; thus an abrupt change in the vapour composition occurs on the plate. Therefore, in contrast with a packed distillation column where the vapour composition changes continuously throughout the packed height and where the height of the packing is estimated from the solution of a differential equation, a plate column should be analysed in such a way that these abrupt finite changes enter into the analysis. This is possible with the solution of finite difference equations.

Illustration 4.8

,

G N + kg moVs of a wet gas containing YN+ kg mol of solutekg mol of wet gas is fed into the base of a plate absorption column where the solute is to be stripped from the gas by absorption in Lokg moVs of lean oil, which is fed at the top of the column. The solute in the entering oil is Xo kg mol/kg mol of lean oil and the solute in the exit gas is Y , kg mol/kg mol of wet gas. The equilibrium constant K, is given as K, = y,/x,, where y , and x, are the mole fractions of the solute in the gas and liquid phases, respectively. Show that the performance of the absorber can be expressed in terms of the absorption factorA = (L,/KGN+I ) and the number of ideal stages (N) by the Kremser-Brown equation (which relates the fractional solute recovery to the absorption factor):

Models in Mass-transfer Operations

95

1

m-1

m

m+l

N

Fig. 4.10 Multistage absorption

y=-

Y

*Y=-

Y

(1 + y > (1 - Y> Considering Fig. 4.10 and taking material balance over plate m gives (4.86) Lo ( X m - X m - 1 ) = GN+1( Y m t l - Y m ) (The gas and liquid ratios on a solute-free basis are same on every plate, though the actual L and G may be different on each plate.) We have to eliminate X,,, and X,- from Eq. (4.86). But

Y m G m = YmGNt, xmLm= XmL,

(4.87)

96

Process Plant Simulation

We know that the equilibrium relation is Ym

= Kmxm

(4.88) Similarly, (4.89) Substituting the values of Xm andXm-,and from Eqs (4.88) and (4.89),respectively, in Eq. (4.86) gives

(4.90)

--Lm

- Am (absorption factor for plate rn)

KmGm Lm-1

-

- Am-l (absorption factor for plate m - 1)

K m - 1Gm-1

(4.91) (4.92) (4.93) Taking the mean value of the absorption factor between the top and the bottom of the column as A, and rearranging Eq. (4.93) in the form of a finite difference equation, Ym+l- (1 + A)Ym + AYm-, = 0

(4.94)

Models in Mass-transfer Operations

97

Putting Eq. (4.94) into the operational form and factorizing,

[ E 2 - E ( 1 + A ) + A]Ym = [ ( E- 1)(E - A)]Y, = 0 which has the solution

(4.95)

Y,,, = C, + C2Am-'

(4.96)

m = 0, Yo = C,+ C,

(4.97)

Put and

m = 1, Y, = C,+ C,A Similarly,

m =N

+ 1, YN+l= C, + C2AN+'

:. Y,+'

- Yo = C2 (AN+' - 1)

(4.98) (4.99) (4.100)

and YN+' - Yl = C2 (AN+' - A )

(4.101)

Dividing Eq. (4.101) by Eq. (4.100), we get

(4.102) Equation (4.102) is the well-known Kremser-Brown equation. The physical significanceof this equation is: The normalized fractional recovery of the solute in a multicomponentabsorber operated in a counter-currentmode of operation depends on the absorption factor.

Example 4.10 Some experiments are done on the absorption of air mixture in 2.5N caustic soda. In one experiment at atmospheric pressure, the results obtained are as follows (Fig. 4.1 1): Gas rate at the bottom of the tower = 0.34 kg/m2s Liquid rate at the top of the tower = 3.94 kg/m2s The CO, in the inlet gas is 315 ppm (parts per million) and in the exit gas it is 31 ppm. Find the mole fraction of CO, absorbed by NaOH. Solution At the bottom of the tower: y N + = 315 x lo4

G,,

= 0.34 kg/m2s

G N + = 0.34/29 = 0.0117 kmol/m2s

98

Process Plant Simulation y , = 31 x 10"

xo= 0

Lo = 3.94 kg/m2 s

y,,, =315 x 10"

G,

+

x,= ?

, = 0.34 kg/m- s

Fig. 4.1 1 Absorption of CO, from air using NaOH solution

At the top of the tower: y, = 31 x

xo = 0 Lo = 3.94 kg/m2s

Now, 2.5N NaOH contains 2.5 x 40 ( g L ) = 100 kg/m3 NaOH Mean molecular weight of the liquid 40) + (900 x 18) 1000 = 20.2 kgkmol

-

(100

x

3.94 20.2 By overall material balance Lo

= - =0.195

kmol/m's

-Y,)=Lo(xN-xo) 0 . 0 1 1 7 ( 3 1 5 ~1 0 - 6 - 3 5 ~10-6)=0.195(xN-O) %+I

bN+I

xN= 0.00001704 = 17.04 x loe6 ppm

4.1 0 Multistage Distillation Illustration 4.9 A continuous flow distribution column is fed with a binary mixture of A and B. The relative volatility a of the mixture is constant. By considering the plate above the feed plate, derive the difference equation that relates the liquid composition to the plate number n if the overall plate efficiency is 100%.Reduce the equation to a linear one with constant coefficients. Show that the axis translation required involves the points where the equilibrium and operating lines intersect.

Models in Mass-transfer Operations

99

Solution Let x, be the mole fraction of the more volatile component in liquid leaving plate n, xD be the mole fraction of the more volatile component in product D withdrawn from the condenser, y,* be the mole fraction of the more volatile component in the vapour in equilibrium with the liquid of composition xn, y, be the mole fraction of the more volatile component in the vapour leaving plate n (y, = y,* if the plate efficiency is loo%), 0 be the constant molar flow rate of liquid down the column, V be the (constant molar flow rate of vapour up the column, and D be the molar flow rate of the product from the condenser. ...........................................

I I I

I I

n+l

j

C (condenser) R (reboiler)

F, xF

4 3

2

-IFig. 4.12 Multistage distillation

Considering Fig. 4.12 and writing the material balance for the more volatile component around the section which includes the nth plate and the condenser, VY,-1 - OX, - Dx, = 0

(4.103)

100

Process Plant Simulation

This is known as the operating line for the enriching section. The definition of relative volatility is (4.104) This is the equation of the equilibrium line. Combining Eqs (4.103) and (4.104), we get x,x,-l

X +2 + a-1

Dx, (a- 1) - aV O ( a - 1)

Let a=-

1

Xn-1 -

DxD = 0 O(a - 1)

(4.105)

Dx, (a- 1) - aV 1 , b= , c = DXD a-1 O(a - 1) O ( a - 1)

Equation (4.105) becomes x , x , - ~+ ax, + bx,-,

+ c =0

This is known as Riccati equation. Put x , = x , (x,

(4.106)

+ 6:

+ 6)(x,-, + 6 )+ a(x, + 6 ) + b(x,-, + 6) + c = 0

jX,X,-~

(4.107)

+ x,(a + 6 ) + x , - ~( b + 6 )+ 6’ + (a + b)6 + c = 0

If the value of 6 is such that

6’

+ (a + b)6 +c = 0

(4.108)

we have X,X,-~

+ x,(a + 6 ) + xn-l(b+ 6 ) = 0

a+6 *1+-+Xfl-1

Put

b+6

=O

x,

v, = VX,, :. V,(b+ 6 ) + V,-l(a + 6 ) + 1 = 0 V,(b + 6 ) + V,-~(U+ 6) = -1

(4.109) (4.110)

(4.111) (4.1 12)

Putting this in the operational form, [E(b+ 6 ) + (a + S)]V,-I = - 1

(4.113) (4.114)

Models in Mass-transfer Operations

101

The complementary solution is (4.115) The particular solution is V,-I

=

-1 (b + 6)

+ (a + 6)

(4.116)

Hence the complete solution is (4.117)

(4.118) Since V, = V(x, - l), K = constant, (4.119)

Let the intersection of the equilibrium and operating lines occur at the point ( x i ,yJ: Vyi - Oxi - Dx, = 0

(4.120) (4.121)

On eliminating yi, we have from the above two equations, x;

+ (a + b)x; + c = 0

(4.122)

With the usual notation for a, b, and c,

6 = xi (4.123) Therefore, the translation of axis required to linearize the Riccati equation corresponds to the new origin of coordinates at the intersection of the equilibrium and operating lines. Example 4.11 A benzene-toluene feed containing 60 mol% of benzene is fed continuously to a distillation column. If there are nine plates between the reboiler and the feed plate, and the top product contains 98 mol% benzene whilst the liquid leaving the base of the column contains 2 mol % benzene, estimate the overall plate

102

Process Plant Simulation

efficiency of the column. The feed enters the column at its boiling point and the relative volatility of benzene to toluene can be considered to be constant at 2.3. The reflux ratio is 3. Solution Basis Consider 100 lbmol feed (F),D moles of the distillate, and W moles of the residue. Therefore the material balances are

(1)

1 0 0 = D +W and

60 = 0.98 D + 0.02 (100 - D)

(2) Solving Eqs (1) and (2), D = 60.4 lbmol and W = 39.6 lbmol. A mass balance between the bottom of the column and some plate n in the stripping section is

Lx,+ 1 = GY, + wxnI

(3) where L is the molar reflux down the column and G is the molar vapour rate up the column. The equation for the relative volatility is Yn

ax, 1 + (a - 1)x,

=

(4)

Substituting Eq. (4) in Eq. (3) gives k + l

=

aGx, 1 + (a - l)x,

+ wx,

or

This is the Riccati equation of the form

x, Now L = F

+

x,

+ ax,

+

bx,

-c

=0

+ RD = (100 + 3 x 60.4) = 281.21 lb molh

a=--- 1

(a - 1)

[

= aG

-

1 = 0.769 1.3

+ (a - l)Wxw L ( a - 1)

1

- (2.3 x 4 x 60.4) + (1.3 x 39.6 x 0.02) = 1.523 281.2 x 1.3

(7)

Models in Mass-transfer Operations

103

The complete solution of Eq. (7) is

where 6 is obtained from the equation

a2 - (a - b)6 - c = 0 62 - 0.5746 = 0.0022 The roots of this equation are 6 = 0.757 or -0.003 Selecting the root 6 = - 0.003 and inserting the boundary condition that when n = 0, x, = 0.02 gives K = 42.2. The value of n corresponding to x, = 0.6 is given by 1.66 = 42.2

[y:i;i]n+ -

1.32

As n = 7.0 ideal stages, subtracting the reboiler from this number gives the overall plate efficiency as 6.0/9 I 67%.

EXERCISES 4.1 A solute is continuously extracted from a diluent using a two-stage co-current mode of operation using a solvent. Each stage consists of (1) a mixer where vigorous stirring of the contents takes place and (2) a settler where the mixture pumped from the mixer is allowed to settle into two layers. The upper raffinate layer and the lower extract layer are removed separately. (a) Develop a mathematical model to find the total fraction of solute extracted under unsteady-state conditions by clearly mentioning the assumptions involved in the analysis. (b) Identify the dimensionless groups in the final expression and explain their physical significance. (c) Compare the performance of the co-current and counter-current modes of operation if m = 1/8, c = 0.1 kg/m3, S = 12 R, and 8 = 1 h. 4.2 Acetic acid is continuously extracted from benzene using a multistage cocurrent mode of operation using water as a solvent. Each stage consists of (1) a mixer where vigorous stirring of the contents takes place and (2) a settler where the mixture pumped from the mixer is allowed to settle into two layers. The upper raffinate layer and the lower extract layer are removed separately and sent to the subsequent stage.

104

Process Plant Simulation

(a) Develop a mathematical model to find the total fraction of solute extracted under unsteady-state conditions by clearly mentioning the assumptions involved in the analysis. (b) Identify the dimensionless groups in the final expression and explain their physical significance. (c) Compare the performance of the co-current and counter-current modes of operation for ten stages if m = 1/8, c = 0.1 kg/m3, S = 12R, and 8 = 1 h. 4.3 A falling film on a wetted wall column is brought into contact with a gas A for absorption of solute in the gas. The gas is consumed in the liquid phase by a firstorder reaction in terms of the gas concentration c. Develop a mathematical model for the concentration of A in the liquid phase in the form of a differential equation. 4.4 Benzoic acid is continuously extracted from toluene using water as the solvent in a counter-current mode using two stages. Each stage consists of (1) a mixer where vigorous stirring of the contents takes place and (2) a settler where the mixture pumped from the mixer is allowed to settle into two layers. The upper toluene layer from the second stage and the lower water layer from the first stage, which vary with time, are removed separately. Develop a mathematical model to find the final fraction of benzoic acid extracted by clearly mentioning the assumptions involved in the analysis. 4.5 A tank contains 20 m3 of water. A stream of brine containing 2 kg/m3 of salt is fed into the tank at a rate of 8.33 x 10- 4m3 /s. Liquid flows from the tank at a rate of 5.55 x 104 m3/s. If the tank is well agitated, what is the concentration of salt in the tank when the tank contains 30 m3 of brine? 4.6 A continuous extraction of benzoic acid from toluene is carried out in a twostage cross-current mode of operation under steady-state conditions using water as the solvent. Each stage consists of (1) a mixer where vigorous stirring of the contents takes place and (2) a settler where the mixture pumped from the mixer is allowed to settle into two layers. The upper toluene layer and the lower water layer are removed separately. (a) Develop a mathematical model to find the total fraction of benzoic acid extracted by clearly mentioning the assumptions involved in the analysis. (b) Identify the dimensionless groups in the final expression and explain their physical significance. (c) Compare the performance of the cross-current and counter-current modes of operation if m = 1/8 and c = 0.1 kg/m3, and S = 12R for counter-current operation. 4.7 Acetic acid is continuously extracted from benzene using water as the solvent in a counter-current mode using four stages. Each stage consists of (1) a mixer where vigorous stirring of the contents takes place and (2) a settler where the mixture pumped from the mixer is allowed to settle into two layers. The upper Benzene layer and the lower water layer from each stage are removed separately

Models in Mass-transfer Operations

105

under steady-stateconditions.Develop a mathematical model to find the final fraction of acetic acid extracted by clearly mentioning the assumptions made in the analysis. 4.8 A wetted wall column is to be used for the absorption of a gas A. The gas is consumed in the liquid phase by a first-order reaction in terms of the gas concentration c. Develop a mathematical model for the concentration of A in the liquid phase in the form of a differential equation. 4.9 Benzoic acid is continuously extracted from toluene using a two-stage cocurrent mode of operation under steady-state conditions using water as the solvent. Each stage consists of (1) a mixer where vigorous stirring of the contents takes place and (2) a settler where the mixture pumped from the mixer is allowed to settle into two layers. The upper toluene layer and the lower water layer are removed separately. (a) Develop a mathematicalmodel to find the total fraction of benzoic acid extracted by clearly mentioning the assumptions involved in the analysis. (b) Identify the dimensionless groups in the final expression and explain their physical significance. (c) Compare the performance of the cross-current and counter-current modes of operation, if m = 1/8, and c = 0.1 kg/m3, and for counter-current operation S = 12R. 4.10 GN+ kmol/s of a wet gas containing Y N + kmol of s o l u t e h o l of wet gas are fed into the base of a plate absorption column where the solute is to be stripped from the gas by absorption in Lo kmoVs of lean oil which is fed at the top of the column. The solute in the entering oil is X, kmoVkmol of lean oil and the solute in the exit gas is Yl kmoVkm01 of wet gas. The equilibrium constant K,,, is given as equal to y mh,,,,where ym and x,,, are the mole fractions of the solute in the gas and liquid phases, respectively. Show that the performance of the absorber can be expressed in terms of the absorption factor A = Lo /KG,+ and the number of ideal stages N by the Kremser-Brown equation as follows: 'N+1

-I'

-

AN+^ - A AN+^ - 1

YN+1 - yo 4.11 A well-agitated tank contains 10 m3 of salt solution of concentration 5 kg/m3. Fresh water enters the tank at the rate of 2 m3/h and the salt solution is withdrawn at the rate of 3 m3/h. What is the concentration of the salt in the tank when it contains 2 m3 of the solution? 4.12 Benzoic acid is continuously extracted from toluene using water as the solvent in a cascade of stages in series. Each stage consists of (1) a mixer where vigorous stirring of the contents takes place and (2) a settler where the mixture pumped from the mixer is allowed to settle into two layers. The upper toluene layer from the second stage and the lower water layer from the first stage are removed separately.

106

Process Plant Simulation

Develop a mathematical model to find the final fraction of benzoic acid extracted by clearly mentioning the assumptions involved in the analysis for the following cases: (a) Steady-state single-stage counter-current operation (b) Steady-state single-stage co-current operation (c) Steady-state single-stage cross-current operation (d) Unsteady-state single-stage counter-current operation (e) Unsteady-state single-stage co-current operation (f) Unsteady-state single-stage cross-current operation 4.13 Solve Problem 4.12 for the following cases: (a) Steady-state two-stage counter-current operation (b) Steady-state two-stage co-current operation (c) Steady-state two-stage cross-current operation (d) Unsteady-state two-stage counter-current operation (e) Unsteady-state two-stage co-current operation (f) Unsteady-state two-stage cross-current operation 4.14 Solve Problem 4.12 for the following cases (i) by the method of induction and (ii) by the finite difference equation approach and compare the results: (a) Steady-state N-stage counter-current operation (b) Steady-state N-stage co-current operation (c) Steady-state N-stage cross-current operation (d) Unsteady-state N-stage counter-current operation (e) Unsteady-state N-stage co-current operation (f) Unsteady-state N-stage cross-current operation 4.15 A benzene-toluene feed containing 60 mol% of benzene is fed continuously to a distillation column. If there are nine plates between the reboiler and the feed plate, and the top product contains 98 mol % benzene whilst the liquid leaving the base of the column contains 2 mol% benzene, estimate the overall plate efficiency of the column. The feed enters the column at its boiling point and the relative volatility of benzene to toluene can be considered to be constant at 2.3. The reflux ratio is 3.0. 4.16 A system consists of N-stirred tanks each of volume v m3 arranged in a cascade. If each tank initially contains pure water, and a salt stream of concentration xo kg/m3is fed to the first tank at a rate R m3/s,calculate what the output concentration from the last tank should be as a function of time if the stirring is 100%efficient. Use the result to compare the transient behaviour of all systems of total volume Nv = V = constant.

CHAPTER 5 MODELS IN HEAT-TRANSFER OPERATIONS

In the previous chapter, models in mass-transfer operations were discussed. This chapter deals with models in heat-transfer operations. Experience has shown that a mental visualization of the physical meaning of analytical expressions and statements (ie., equations) on the part of the analyst and programmer is absolutely invaluable (Franks 1972). We have seen this aspect in the models discussed so far. Most physical situations, when described as a set of equations, are non-linear and simultaneous; that is, the variables are defined implicitly, and the only practical method of obtaining a solution is using numerical methods that, although unsophisticated, are quite successful. This chapter covers these aspects. Let us recapitulate some important aspects of modelling. Based on the methodology we followed in the models discussed in Chapter 4,the generalized approach for mathematical modelling can be summarized as follows: 1. Recall the basic physical laws that govern the situation, and decide which quantities must be regarded as unknown variables. If every quantity can be expressed in terms of a single independent variable, an ordinary differential equation may be expected. 2. Formulate a differential equation according to the controlling physical laws. 3. Eliminate all but two variables from the equation formulated in step 2. A thorough understanding of the physical situation should ensure no difficulty in the first three steps. 4. Obtain the general solution by suitable integration of the differential equation in two variables resulting from step 3. This is primarily a mathematical problem.

108

Process Plant Simulation

5. Substitute the boundary conditions in the general solution, and determine the values of the constants for a particular solution applicable to the case at hand.

In this chapter, the following models are discussed: steady-state heat conduction through a hollow cylindrical pipe, unsteady-state steam heating of a liquid, unsteadystate heat loss through a maturing tank, heat transfer through extended surfaces (spine fin), counter-current heat transfer in two tanks in series, temperature distribution in a fin of triangular cross section, unsteady-state heat transfer in a tubular gas preheater, heat loss through pipe flanges, heat transfer in a thermometer system, and unsteady-state heat transfer by conduction. Apart from having a thorough understanding of the physical phenomena occurring in the system, there are many other important aspects to be taken into account in modelling: mental visualization, schematic representation, making suitable assumptions and justifying them, using appropriate terminology,identifying the right mathematical technique for solving the formulated model, thorough knowledge of numerical methods, ability to generalize the model for a wide range of applications, following the method of induction principles, etc. In addition, if we reach a situation where proceeding further becomes difficult after starting with some assumptions, we should know how to come out of the problem by making a suitable simplified assumption. All these important aspects of modelling are well demonstrated in the models discussed in this chapter.

5.1 Steady-state Heat Conduction Through a Hollow Cylindrical Pipe (Static Distributed Parameter Rigid Analytical Model) There are many problems in which the properties of the system are functions of position instead of time. The following example illustrates the application of the method to this type of problem.

Illustration 5.1 Two concentric cylindrical metallic shells are separated by a solid material. If the two metal surfaces are maintained at different temperatures, what is the steadystate temperature distribution within the separating material? Solution This is a steady-state problem, but some of the system properties depend upon the position at which they are measured. In this case, the temperature T and the heatflow rate per unit area Q are both functions of the radius r. The heat balance must be related to a space interval between Y and r + 6r as shown in Fig. 5.1. Following the same procedure as in the case of unsteady-state problems of Chapter 4,but considering variations of Y instead of 8 ,Table 5.1 can be compiled.

Models in Heat-transfer Operations

Fig. 5.1 Radial heat flow through a cylindrical conductor

Table 5.1 The condition of the system on either side of a space increment Sr Property of the system 1.

Temperature

2.

Heat-transfer aredunit length

3.

Radial heat flux density (W/m2)

4.

Total radial heat flow per unit length

At r

T 2nr

Q 2mQ

Considering the element of thickness Sr, the heat balance is Heat input to inner surface = 27rrQ Heat output from outer surface = 27r (r + Sr) Accumulation of heat = 0 (due to the steady state)

At r

+ Sr

dT Ti-& a?

2n(r + Sr) Q

dQ + -Sr dr

109

110

Process Plant Simulation

Since input - output = accumulation,

(::)+ -6r

2xrQ - 2 x ( r + 6r) Q

=0

Dividing by 27r throughout, we get rQ - rQ - r-6r dQ dr

- Q6r -

a(&)’ =0 dr

Cancelling out r e , dividing by &, and taking the limit 6r + 0, we get r -dQ +Q=O

dr

(5.3)

But Q (fludarea) is related to T by Fourier’s law of heat conduction dT (5.4) dr Since this is negative, T decreases with r. Differentiating Eq. (5.4),we get Q =-k-

(5.5) where k is the thermal conductivity. Substituting Eqs (5.4)and (5.5) in Eq. (5.3), we get

or r -d2T - - = OdT (5.7) dr2 dr Referring to Table 5.1, the change in the heat-flow rate across the element gives rise to two terms in this example: (1) one due to a change in the temperature gradient and ( 2 ) the second due to a change in the area of conduction. These are reflected in Eq. (5.7) as the second and first terms, respectively. The overall heat transfer by conduction in this problem is the combined effect of change in temperature and change in area. This problem can be solved in two ways: (1) by treating Eq. (5.7) as a homogeneous second-order differential equation and ( 2 )by solving Eqs (5.3) and (5.5) in succession. Following the second method, from Eq. (5.3), the variables are separable,

Models in Heat-transfer Operations

111

Integrating, we get 1nQ = -1nr + c (5.9) Here, c is the integrating constant.In order to make an antilogarithmic transformation in the above equation, c could be taken as lnA, where A is a constant (i.e., we can find A in such a way that 1nA gives the required value of c). Putting c = lnA, Eq. (5.9) becomes (5.10) 1nQ = -1nr + 1nA A r Putting this value of Q in Eq. (5.4), we get Q=-

(5.11)

(5.12) On rearranging this term and integrating, we get (5.13) lnr = - ( k T / A ) + B Equation (5.12) involves two constants of integration, A and B , and these can be evaluated from the fixed boundary temperatures. Boundary condition: r = a, At T = To At r = b, T = TI From Eqs (5.13)-(5.15), we get lna = - (kTo/A) + B lnb = -(kT,/A) + B Equations (5.16) and (5.17) give

(5.14) (5.15) (5.16) (5.17)

lna - lnb = k(T, - To)/A

(5.18)

A = k(T, - To) (lna - lnb)

(5.19)

Substituting A from Eq. (5.19) in Eq. (5.16), we get lna =

-kTo (lna - lnb) +B k(T, - To)

TI Inn - To h a B=

+ To h a - Tolnb = B(T, - To)

TI lna - To lnb

TI -To

(5.20) (5.21) (5.22)

112

Process Plant Simulation

Putting the values ofA and B from Eqs (5.19) and (5.22) in Eq. (5.13), we get lnr =

-kT(lna-lnb) k(T1 - To)

-4-

T,lna-Tolnb Tl - To

- To)lnr = (lnb - 1na)T + T, lna - To lnb + To lna - To lna (T, - To)lnr = (T - To) (lnb - lna) + (T, - To)lna (TI - To)(lnr - lna) = (T - To)(lnb - lna) T - To - lnr - lna -Tl - To lnb - lna (T,

(5.23) (5.24) (5.25) (5.26) (5.27)

Equation (5.27) is in dimensionless form and shows the variation of T with r.

Example 5.1 A hollow cylinder with an outer diameter of 10 cm and an inner diameter of 5 cm has an inner surface temperature of 200 "C and an outer surface temperature of 100°C. Determine the temperature of the point halfway between the inner and outer surfaces. If the thermal conductivity of the cylinder material is 70 W/mK, determine the heat flow through the cylinder per linear metre. Solution

Q=

2kLn(Ti - To) - (6.28 x 70 x 1)(200 - 100) 5 b In In a 2.5

= 63,434.3 W/m Halfway between b and a, the radius

r=

(5 + 2.5) ~

= 3.75 cm

L

Since T - To - lnr - lna -Ti -To l n b - lna

... T - To = =

lnr - lna lnb - lna

[3(loo)] 3.75

= 57.72"C

T = 57.72 + 100 = 142.28OC

Models in Heat-transfer Operations

113

Example 5.2 l b o concentric cylindrical metallic shells are separated by a solid material. A constant rate of heat q (in W/m3) is generated within the cylinder, while the boundary surfaces at r = ri and r = ro are maintained at different temperatures. Determine the steady-state temperature distribution and the radial heat-flow rate within the separating material? Solution The energy balance on a cylindrical shell of thickness dr is Qr

+ Qg = e r + d r

where Qq = q(2m)drL

dT dr

Q, = - k(2zr)L -

Substituting Eqs (2)-(4) in Eq. (l), we get d dT qr = - k- kdr[ dr] or

The mathematical formulation of this problem is -

:r[

Z] 7

k-

+--0,

q5r5ro

Boundary conditions: T = Ti at r = ri; T = To at r = ro. Integrating Eq. (5), we get

4r2 + C,lnr + C, T = -~ 4k Applying boundary conditions to Eq. (8), qr! T. -A = C, lnr, ' 4k

+ C2

114

Process Plant Simulation

Subtracting Eq. (9) from Eq. (lo), we get

:. c, =

(To- T i )

+ X(r: - q2) 4k In (ro/q)

c2= + 2 5 2 4k

(To - q ) + &(r:

- q2)

In (ro/q)

In q

Then the temperature distribution in the cylinder, T=

-& + [(To - T ) + x(r: - q2)]_L”’ + q, 4k 4k In (ro/q)

or ln(r/q) T - q =L(q2 -r0)+(To - Ti) + - ( ( T o 4k 4k In (ro/q )

]

-q )

(14)

or

Therefore, it is clear that if there is no heat generation, Eq. (15) gets reduced to T - q - ln(r/q) To - q ln(ro/q) which is the same as Eq. (5.27) corresponding to the temperature profile without

heat generation.

Models in Heat-transfer Operations

115

5.2 Unsteady-state Steam Heating of a Liquid (Dynamic Lumped Parameter Rigid Analytical Model) The illustrations considered so far have only used material or heat balances. In this illustration, the equation is derived from a rate equation.

Illustration 5.2 A closed kettle Fig. 5.2 of total surface area A m2 is heated through this surface by condensing steam at temperature T, K. The kettle is charged with M kg of liquid of heat capacity C, J k g at a temperature of To K. If the process is controlled by a heat-transfer coefficient h W/m2K, how does the temperature of the liquid vary with time?

h

u

Fig. 5.2 Steam heating of a liquid in a kettle

Solution In general C, in J k g is referred to as specific heat. In fact, the specific heat of a fluid (liquid or gas) is defined as the ratio of the heat capacity of that fluid (C ) to Pf the heat capacity of the reference fluid (C,,, i.e., water for liquids and air for gases). Hence, specific heat is a dimensionless quantity, and heat capacity (C,) has the units of Jkg. We follow this convention throughout this book. Consider a time interval 68: Heat input (J) = hA(T, - T)6B Heat output (J) = 0

dT

Heat accumulation (J) = MC -68 de Input - output = accumulation (5.28)

116

Process Plant Simulation

On separating the variables, we get

--dT --hAd8

(5.29)

T, - T

MC, On integrating, we get -ln(T,

- T) = (h4/MCP)8+ B

(5.30)

B can be evaluated by using the boundary conditions, i.e., when 8 = 0, T = To. Therefore, (5.31) -In(T, - To) = B Substituting B from Eq. (5.31) into Eq. (5.30), we get -ln(T,

- T) = (hA/MC,)8 - ln(T,

- To)

(5.32)

(5.33)

or (5.34)

This is the expression for variation of T with 8 (time).

Example 5.3 A closed vessel of total surface area 40 m2 is heated through this surface by condensing steam at a temperature of 100°C. The vessel is charged with 600 kg of liquid having a heat capacity of 25 12 J k g K at a temperature of 25°C. If the process is controlled by a heat-transfer coefficient of 142 W/m2K, model the system in dimensionless form and find out the temperature of the liquid after 1 h. Solution This is the case of unsteady-state steam heating of a liquid. Hence the modelling of the system is just the same as discussed above. It is given that Ts = 1OO"C, To = 25"C, Cp = 2512 J/kgK= 2512 Jkg"C,A =40 m2,M = 600 kg, 8 = 1 h = 3600 s, h = 142 W/m2 K = 142 W/m2"C 100 - T 100 - 25

40x142 ~ 3 6 0 0 ) 600 x 2512

100 - T = exp (-13.566878) 100 - 25

(5.35) (5.36)

Models in Heat-transfer Operations

117

loo - = 1.28226 x 100 - 25

(5.37)

100 - T = (100 - 25) x 1.28226 x

(5.38)

T = 100 - (9.5 x

(5.39)

T = 100°C

(5.40)

5.3 Unsteady-state Heat Loss Through a Maturing Tank (Dynamic Lumped Parameter Rigid Analytical Model) Illustration 5.3 An elevated horizontal cylindrical tank of 1.524 m diameter and 3.66 m length (Fig. 5.3) is insulated with asbestos lagging of thickness 1= 0.05 1 m and is employed as a maturing vessel for a batch chemical process. Liquid at 93.3 "C (366.3 K) is charged into the tank and allowed to mature over 5 days (120 h = 432,000 s). If the data given below applies, calculate the final temperature of the liquid in the tank and give a plot of the liquid temperature as a function of time. Solution Liquid film coefficient of heat transfer (hi) = 25.0 Btu/h ft2"F = 141.98 W/m2K Thermal conductivity of asbestos ( k ) = 0.1 Btu/hft"F = 0.1731 W/mK Surface coeficient of heat transfer by convection and radiation (h2) = 1.8 Btu/hft"F = 10.222 W/m2K Density of liquid ( p ) = 62.4 lb/ft3 = 1000 kg/m3 Heat capacity of the liquid ( Cp) = 0.6 Btu/lb"F = 2512 JkgK Atmospheric temperature t can be assumed to vary according to the relation t = 55 + 1 5 c o s ( n e m ) = 12.78

+ 8.33cos(3OOn8)

where 8 is in seconds. Atmospheric temperature at the time of charging = 70 "F = 294.1 K. Heat loss through supports is negligible. T is the bulk liquid temperature, T,,, is the inside wall temperature of the tank, and T, is the outside temperature of the lagging. Area of the tank ( A ) = nDL

2nD2 +4

118

Process Plant Simulation

= (n x 1.524 X 3.66)

+ 2 x [0.25 x n x (1.524)’]

= 21.171m’

Note that the thickness of lagging is not considered in the above calculation.

nD’1 n x 1.524’ x 3.66 = 6,6,6 Volume of the tank ( V ) = -4

m3

4

Rate of heat loss by the liquid = h,A(T - T,) Rate of heat loss through the lagging = k (A/Z)(T, - T,) Rate of heat loss from the exposed surface of the lagging = h,A(T, Considering the thermal equilibrium of the liquid:

- t)

Input rate of heat = 0 (since it is a batch process) Output rate of heat = h,A(T, - t ) dT dT Accumulation rate of heat = mCp-= V p C , de de Input - output = accumulation dT - h A ( T , - t ) = VpC de Substituting the values of 4, A, V , p , and C, in Eq. (5.41), we get dT

-hA

-dT- -

-10.22 X 21.171 (T, - t ) 6.676 x 1000 x 2512

- = -(T, d e Vpc,

de

-t)

(5.41)

(5.42)

(5.43)

dT = -1.29 x 10-~(T, - t ) (5.44) de The above equation consists of T, and t, which are to be expressed in terms of a known quantity, i.e., 8. We already know the relation of t as a function of 8, but now we have to express T, as a function of 8, which can be done as follows. At any given time 8, -

Rate of heat rate of heat transfer rate of heat loss (loss from liquid) = through lagging = (from surface

(

)

(5.45) (5.46)

Models in Heat-transfer Operations

119

On rearranging the first two terms of Eq. (5.46) and simplifying, we get (5.47)

1.524m(ID) T(@ T = 93.3"C

T 3.66 m

1

Fig. 5.3 Maturing tank (IDdenotes inner diameter)

Taking-the first and third terms of Eq. (5.46), h,(T - T,) = h,(T, - t ) h T, - t = " ( T - T , ) h2 Substituting T, from Eq. (5.47),

(5.48)

(5.49)

(5.50)

On simplification, we get (5.51)

(5.52)

(5.53)

120

Process Plant Simulation

T, = t +

hl&l

hlk

+ h,k + h,k

](T - t)

(5.54)

Substituting the values of h , , h,, 1, k in Eq. (5.54), we get T, = t +

141.98 x 0.1731 ](T - t) (141.98 x 10.222 x 0.051) + (141.98 x 0.1731) + (10.22 x 0.1731)

(5.55)

T, = 0.245T + 0.75% Substituting T, from Eq. (5.56) and simplifying, we get dT - + (3.161 x 10-6)T = (3.161 x 10-6)t do Substituting t = 12.78 + 8.33 cos (3001te), dT do

-

+ (3.161 x ~ o - ~ =) T(4.04 x

(5.56)

(5.57)

+ (2.633 x ~ O - ~ ) C O S ( ~ O O Z ~ )

(5.58) The above equation is a first-order linear differential equation, which can be solved 3.161x lo-% . The solution is by means of the integrating factor (IF) e T e 3 .i6 i ~i o4e

- 4.04x

+ 2.633 x

~e3.161x10ded~ je3.161x10-6e cos(300~o)de+ C, (5.59)

The first term on the right-hand side (rhs) is

(5.60) The second term on the rhs is

I

2.633 x 10-5 e3.161xl0-~0 cos (3oone)de = E

(5.61)

This can be integrated by parts. The definition of eie is cose + isin8, that is, cose is the real part of eie. Similarly, E can be written as E = 2,633 10-5

= Re[

I

e3.161X10"6e~e(ei942.58de)

2,633 10-5 (e3.i6i~io+e+i942.5e 3.161 x 10" + i942.5

(5.62)

(5.63)

Models in Heat-transfer Operations

121

On simplification, we get E = 2.96 x 10-11e3.161x10de r3.m x 10" cos(942.5e)

+ 942.5sin(942.58)]

(5.64)

Substituting this in Eq. (5.59), we get ~ ~ 3 . 1 6 1 X l O-~ 12.78e3.161X10de e 2-96 10-11e3.161X10-68 +

x [3.161 xlOd cos(942.58) + 942.5sin(942.58)] + C, (5.65)

T =12.78 + 9.357 x

cos(942.58)

sin(942.58) + Cle-3.161x10"e

+ 2.79 x

(5.66)

At 8 = 0, T = 93.3"C; hence, 93.3 = 12.78 + 9.357 x C,= 93.3 - 12.78 = 80.52 T = 12.78 + 9.357 x

+ 2.79 x

+ C, cos(942.58)

sin(942.58) + 80.52e-3.'61x'0de

(5.67)

The contribution of the second and third terms may be neglected, and we get T = 12.78 + 80.52e-3.'61x'0de

(5.68)

When 8 = 5 days = 120 h = 432,000 s T = 12.78

+

80.52e-3.161X10-6X432000

(5.69)

= 33.34OC (5.70) Using Eq. (5.68), the temperature profile is generated as a function of time and plotted as shown in Fig. 5.4 below. Temperature profile

0

1

2 Time (s)

3

4

Fig. 5.4 Variation of temperature with time in the maturing tank

122

Process Plant Simulation

Exercise 5.1 Refer to Eq. (5.45).Explain how this relation is valid for this model of the unsteady-state process. Example 5.4 In Illustration 5.3,the thickness of the lagging was not taken into account. Evaluate the error involved in the area calculation in neglecting the lagging using the data given in the illustration and compare the results. Solution

2nD2 4

Area of the tank (A)= nDL + = (nx 1.626 x 3.66)+ 2 x [0.25 x IC x (1.626)2] = 22.85m2

zD21- n x 1.6262 x 3.66 = 7.6m3 4 4 Substituting the values of h,, h,, I , k in Eq. (5.50),we get Volume of the tank (V) = --

T, = t +

[

141.98 x 0.1731 ](T - t ) (141.98 x 10.222 x 0.051) + (141.98 x 0.1731) + (10.22 x 0.1731)

= 0.245T + 0.755t Considering the thermal equilibrium of the liquid:

Input rate of heat = 0 (since it is a batch process) Substituting the values of h2, A, V, p, and C, in Eq. (5.54),we get

d T- - -10.22 x 22.85 (T, - t> d o 7.6 x 1000 x 2512 = -1.2232 x (T, - t) Substituting T, from Eq. (5.52)in the above equation and simplifying, we get

dT - + (2.996 x 10")T = (2.996 x 10")t do Substituting t = 12.78 + 8.33 cos(3OOn8) in the above equation, we get

dT - + (2.996 x 1O")T de

Models in Heat-transfer Operations

123

This equation is a first-order linear differential equation, which can be solved by means of the IF e3.161x10de.The solution is Te2.9%x10-'6 - 3.7828 x 10-5

I

+ 2.4956 x

e2.9%Xio-6@

d8

le2.996x10de cos ( 3 o o ~ e ) d e+ c,

The first term on the rhs is 3.7828 x

Ie

2.996xiode de

= 3.7828 x 2.996 x

e 2.996x1o4e

The second term on the rhs is 2.4956 x

Ie2.996x104e cos(300~8)d8= E

This can be integrated by parts. The definition of eie = cos8 + isin8, that is, cos8 is the real part of eie. Similarly, E can be written as

I

E = 2.4956 x = Re[

e2.996x10"e Re (ei942.5e d8)

2.4956 10-5 (e2.996x10-bB+i942.5f3 2.996 x 10" + i942.5

x [2.996 x 10" cos(942.58) + 942.5sin(942.58)] Substituting this in Eq. (5.59), we get Te2.996~ io"e - 12,626e2.996~ 1o"e +2.809 10-11e2.996~io"@ cos (942.58) + 942.5 sin(942.58)I + C,

x [2.996 x

T = 12626 + 8.4157 x

+ 2.6 x lo-'

cos(942.58)

sin (942.58) + Cle-2,996x lo"'

At 8 = 0, T = 93.3"C; hence, 93.3 = 12.626 + 8.4157 x lo-'' C, = 93.3 - 12.626 = 80.674

+ C,

T = 12.626 + 8.4157 x lo-'' ~ 0 ~ ( 9 4 2 . 5 8 )

+ 2.6 x lo-'

sin(942.58) + 80.674e-2.996x lo"'

124

Process Plant Simulation

The contribution of the second and third terms may be neglected, and we get T = 12.626 + 80.674e-2.996x10"B

When 8 = 5 days = 120 h = 432,000 s,

T = 12.626 + 80.674e-2.996~10" ~432000 = 34.738"C

Result As can be seen from this result, the temperature is 33.34 "C by neglecting the thickness of the lagging in the heat-transfer area calculation, as against 34.738 "C by considering the lagging thickness. This means a difference of 1.398 "C,which corresponds to 4.02% error C(34.738 - 33.34) x 100/34.738 = 4-02], which is insignificant and negligible for all practical purposes, as 5-10% error is allowed in all professional design calculations. Hence, the assumption that was made earlier (that the thickness of the lagging is negligible) is justified.

5.4 Counter-current Cooling of Tanks Illustration 5.4 1.26 kg/s of sulphuric acid of heat capacity 1508 J k g K is to be cooled in a twostage counter-current cooler of the following type. Hot acid at 174°C is fed to a tank where it is stirred well in contact with cooling coils. The continuous discharge from this tank at 88 "C flows to a second stirred tank and leaves at 45 "C. Cooling water at 20 "C flows into the coil of the second tank and then to the coil of the first tank. The water is at 80°C as it leaves the coil of the hot acid tank. To what temperature would the contents of each tank rise if, due to trouble in the supply, the cooling water is suddenly stopped for 1 h? Calculate also the intermediate water temperature before failure. The capacity of each tank is 4536 kg of acid and the flow rate of water is 0.975 kgh. Solution The steady-state conditions of the cooling system are shown in Fig. 5.5.

10,000 lb/h

*

174"

88"

45 O

Models in Heat-transfer Operations

125

Steady state: Considering a heat balance around tank 2, Rate of heat input by cooling water = rate of heat output by sulphuric acid Heat capacity of water = 1 cal/gm"C = 4187 J k g K 3

0.975 x 4187 x (T, - 20) = 1.26 x 1508 x (88 - 45)

T, =

1.26 x 1508 x (88 - 45) 0.975 x 4 187

+

2o

(5.71) (5.72)

T, = 40°C (5.73) When the cooling water supply is stopped: The rate of cooling water flow = 0. Thereafter, unsteady-state conditions prevail, as the acid is still flowing at the same rate as before. For the purpose of analysis, let the flow rate of acid be M = 1.26 kgh, heat capacity of acid be CpA= 1508 J k g K, feed acid temperature be To = 174 "C, capacity of each tank be W = 4536 kg, outlet acid temperature for tank 1 be T , "C, outlet acid temperature for tank 2 be T2"C, and time be 8 s. The heat balances are as follows for both tanks:

Input rate = output rate (in steady state) (5.74) dT2 Tank 2: MCpAT - MCpAT2= WcpAd8 From Eq. (5.74)

M(T0 - T I )= W -dT1 de

(5.75)

(5.76) (5.77)

-ln(T,

-

M T,) = -8

W

+ constant

When 8 = 0, T, = 88°C. :. Constant = -ln(T, -88) Substituting Eq. (5.79) in Eq. (5.78) and simplifying, we get

(3

(5.78)

(5.79)

T, = To - (To - 88) exp --8

(5.80)

= 174 - 86 exp(-2.78 x lo4@

(5.81)

126

Process Plant Simulation

When 8 = 3600 s (1 h), T , = 142.4"C. From Eq. (5.75)

M ( T , - T 2 ) = W - dT2 d8

(5.82)

dT, M M -+ -T2 = -[174 - 86exp(-2.78 x lo4@] (5.83) d0 W W This is a first-order linear differential equation that can be solved using an IF e2.78~io-4e.

M 174e-2.78~1~-4e M e-2.nxio"e T2 = - -(868) W (2.78 x lo4) W

+ constant

(5.84)

Boundary conditions: When 8 = 0, T2= 45 "C and MIW = 2.78 x lo4. Therefore, (5.85) 45 = 174 + constant (5.86) Constant = -129 From Eqs (5.84) and (5.86), we get T2 - 174 - e - 2 . 7 8 ~ 1 ~ - 4[;868 e

+ 1291

(5.87)

When 8 = 3600 s, MIW = 2.78 x lo4, So, T2 = 174 - 79.05 = 94.95 "C

(5.88)

Example 5.5 In Illustration 5.4, if the water supply is restored at a flow rate of 1.26kgls, calculate the acid discharge temperature T2after 1 h. The overall coefficient of heat transfer in the hot tank (tank 1) is 1135 Wlm2K and in the colder tank is 738 Wlm2K. The heat-transfer areas of the coils in tank 1 and tank 2 are 6.44 m2 (Al) and 8.9 m2 (A2), respectively. Solution Water flow rate m = 1.26 kgls. Let water supply temperature be to"C, outlet water temperature of tank 1 be t, "C, and outlet water temperature of tank 2 be t2 "C. The heat balances are as follows: Input output Accumulation Tank 1: (mCpwt2+ mC,To) - (mCq,t,

+ mCpAT,)= WC,

*Tl

(5.89) Tank 2: (mCpwto + mCpAT, ) - (mC, t2 + mCpAT2)= WC, dT2 (5.90)

Models in Heat-transfer Operations

127

The heat-transfer rate equations for the two tanks are (5.91)

(5.92) These equations simplify to (5.93)

(5.94) Equations (5.93) and (5.94) can be rearranged to simplify to (5.95)

T(1- a)= t, - at, and

(5.96) T2 (1 - P> = t , - Pto Equations (5.89), (5.90), (5.93, and (5.96) are the four equations which have to be solved simultaneously for T I ,T2,tl, and t2 This can be done by the ‘systematic elimination method’. This method is proposed for the elimination process that reduces the algebra involved to a minimum. Essentially, the method consists of setting up a table indicating the number of times and the form in which each variable appears in the simultaneous equations. Then the variable that appears in the simplest manner is first eliminated by making appropriate substitution. Following this, the second least frequent variable is removed from the remaining equations by appropriate substitution and this process is continued till only one equation remains. Generally, terms which appear as derivatives are left to the final steps of the substitution process. For the present problem, Table 5.2 gives the frequency of the appearance of the four variables. Table 5.2 Frequency of appearance of variables in the model equations TI 1 4-

T2

ti

1

1 1

+

1 1

1

3 plus +

2 plus 4-

t2

-

1 1 1 1 -

2

4

128

Process Plant Simulation

In Table 5.2, ' 1' indicates the appearance of a variable and '+' indicates its derivative. It is clear that tl should be eliminated first and then t2,because both T, and T2also appear in the differential form. After eliminating t, and t2, T,is to be eliminated (not T2, though its appearance is less than TI) because the relation required is T2 as a function of 8. Finally, a second-order differential equation is obtained for T2, which has the solution (5.97) T2 = Ae-C4e+ Be-Cse+ c6 For evaluating A and B, the final temperatures from the previous operations are to be taken as the boundary conditions. Then, at 8 = 1 h (3600 s) T2 = 48.9OC (5.98)

Example 5.6 10,000 k g h of acid having a specific heat of 0.35 enters a wellagitated tank containing 10,000 kg of acid and leaves at the same rate in a steadystate process. The inlet and outlet temperatures of the acid are 443 K and 363 K, respectively. This is maintained by cooling water, flowing at 8000 k g h having a heat capacity of 4187 J k g K. The inlet temperature of water is 298 K. (a) Calculate the outlet temperature of water. (b) If cooling water suddenly fails, what is the temperature of the acid in the tank after 30 min? Solution (a) Heat balance: For the steady state, the equation is Input = output .*. 10,000 x (0.35 x 4187) x (443 - 363) = 8000 x 4187 x (T - 298)

... T - 298 = 35 = 333 K

(b) For the unsteady state, the equation is Input - output = accumulation

:. (10,000 x 0.35 x 4187 x 443) - (10,000 x 0.35 x 4187T) 10,000 x 0.35 x 4187-

dT 443 - T = d6' Integrating this equation .*.

Models in Heat-transfer Operations

129

When 8 = 0 (that is, just before water fails), the system was at steady state and the temperature in the tank is the same as the temperature of the leaving stream = 363 K. Therefore, at 8 = 0 , T = 363 K. Solving Eq. (l), we get

(y )

= e-0

443 - T = 80e-@ T = 443 - 80e-0 Therefore, at 8 = 30 min = 0.5 h,

T = 443 -80e4'.'

:.

T = 394.48 K

5.5 Heat Transfer Through Extendeb Surfaces (Spllie Fin) Illustration 5.5 A graphite electrode 0.1524 m in diameter passes through a furnace wall into a water cooler (which takes the form of a water sleeve). The length of the electrode between the outside of the furnace wall and its entry into the cooling jacket is 0.3048 m; and as a safety precaution the electrode is insulated thermally and electrically in this section, so that the outside surface temperature of the insulation does not exceed 48.89"C. If the lagging is of uniform thickness and the mean overall coefficient of heat transfer from the electrode to the surrounding atmosphere is taken to be 1.701W/m2K of the surface of the electrode and the temperature of the electrode just outside the furnace is 1482.2 "C, estimate the duty of the water cooler if the temperature of the electrode at the entrance to the cooler is to be 148.89 "C. The following additional information is given: The temperature of the electrode may be assumed uniform at any cross section, the surrounding temperature is 21.1 "C, thermal conductivity of graphite = kT = k, - aT, where k, = 152.86 W/ mK and a = 0.561. Solution Referring to Fig. 5.6 the heat balance for a small element of length & at a distance x from the furnace is as follows. The cross-sectional area of the electrode 7r

7r

A = - d 2 = -(0.1524)2 4 4

130

Process Plant Simulation = 0.0182m2

Cold

Fig. 5.6 Heat transfer through an electrode

(3 (2)[: 3

Rate of heat input = - k,A -

(5.99)

Rate of heat output

= -kTA -

+ - -k,A-

SX + U ( d S x ) ( T - To) (5.100)

where U is the overall heat-transfer coefficient from the electrode to the surroundings, d is the electrode diameter, and To is the surrounding temperature. Rate of accumulation of heat = 0 (steady-state)

"[

= A(T dU

-k,A%]&

- To)&

dx

(5.101)

Substituting k , and simplifying by dividing throughout by Sx and letting d U / A = p, Eq. (5.101) becomes

:

(ko - aT)-

-

a(%)2 - P(T

-

To) = 0

(5.102)

Equation (5.102) is a second-order non-linear differential equation, which does not contain the independent variable explicitly. Thus, let p = -dT dx

(5.103)

dp dx - d2T 1 dp - -_ dT

dxdT

dx2 p

(5.104)

Models in Heat-transfer Operations

131

Therefore,

dp d 2 T p-=dT dr2 Equation (5.102) becomes

(5.105)

(ko - aT)p-dP - c ~2 p- P(T - To) = 0 (5.106) dT A careful observation of Eq. (5.106) shows that the variables p and T cannot be separated, and one might be tempted to introduce simplifying assumptions at this stage, or attempt to solve Eq. (5.102) by some other method. However, the way in whichp occurs in Eq. (5.106)suggests the substitution p2 = z. In addition, putting y = (T - To) gives dz [k, - aTo - ayl- - 2az - 2pY = 0 (5.107) dY Equation (5.107) is a first-order differential equation that can be solved by the IF method. The integrating factor is

(5.108) = (ko - aT0 -

CZY)~

The solution is

z ( k o - aTo - ay)* = 1 2 b ( k 0 - aTo - ay)2dy+ C

(5.109)

Converting back to the original variables, (ko

-

&)dT

2 C + P ( k o - aT)(T - To)2- -aP(T 3

1

(5.1 10)

Equation (5.110)cannot be integrated analytically because of the cubic polynomial under the square root sign, or graphically because the arbitrary constant C is completely unknown. Furthermore, there is no way of evaluating C other than perhaps by trial and error, which would not be very accurate. Hence, in order to obtain an estimate of the heat load on the cooler, it is assumed that the heat lost to the atmosphere (surroundings) from 1 m2 surface area of the electrode = Q = constant. Note This is an important aspect in modelling. As and when one gets stuck without getting any result in the middle of an analysis in spite of getting started well, one needs to go back to the original step and see if there is a possibility of making an

132

Process Plant Simulation

assumption that enables the solution. In the present case, let us assume that the variable T can be replaced by an average value, only for the purpose of calculating Q, in Eq. (5.101). Q = average value

= U(T - TO)

(5.111)

An average value of the temperature at the hot and cold ends of the extended surface would enable the heat load value to become a constant. In fact, it is this functional dependency of temperature on distance which resulted in the complex equation (5.110). Though the temperature is actually varying with distance, with reference to the heat-transfer rate value, taking an average temperature would not affect the accuracy of the result, and at the same time the analysis will become quite simple. Q = 1.701

1482.2 + 148.89 -

(5.112) (5.113)

= 1351.52 W/m2

Equation (5.101) can be written as d

dx

z]

[

-

(k, - aT)-

or

6c =

s& A

(k, - ~ Tdx) q - a ( s ) ~ = B

(5.114)

(5.115)

where, B = IrdQ/A. Put dT dp dp dx P=-=--=-dx’ dT dxdT

d2T 1 dx2 p

Therefore, dp d2T p- =dT dx2 Thus, Eq.(5.115) becomes (k,, - a T ) p -dP - w2= B dT

(5.1 16)

(5.1 17)

Separating the variables and rearranging, PdP

-

p+w2

dT k 0 - U

(5.118)

Models in Heat-transfer Operations

Put

133

P+w2= q , 2 a p d p = d q -dP =

dT

1 -1nq 2a

= --ln(k,

(5.119)

2~ k, -aT Integrating, we get 1

a

- aT) + 1nD

ln(P + ap2)v2= ln(k, On simplifying, we get

- aT)-’+ lnDa

P + op’ = E/(k, -

(5.120) (5.121) (5.122)

where E = D2a. Therefore,

Substituting p = dT/dx and taking only the negative root, since the temperature falls as distance x increases,

dr dx = -\/a(k,

E - -P - aT)’ a

(5.123)

Put (k, - aT)’ = r. Therefore, dr - = 2(k0 - aT)(-a)= - 2aJ; dT

(5.124)

dT dr dT - -dx dr dx

(5.125)

dT dr - 1 -dx d x - 2 a J Equation (5.123) becomes

(5.126)

(5.127) (5.128)

134

Process Plant Simulation

Integrating and simplifying, we get

(5.129) where B = &Q/A and E = DZa = constant. The boundary conditions (BC) at the ends of the electrode (furnace end at x = 0 and cooler at x = 0.3048 m) are

421.89 =b-

a

a B

- a B ( x + c2)2 at x = 0

(5.130)

- M ( x + cZ)'

(5.131)

at x = 0.3048

It is more convenient to determine how E and c2 occur in the required solution, rather than solve these equations directly. The duty of the water cooler = the rate at which heat arrives at x = 0.3048 m

=H Therefore, dT

H = -(ko - @ ) A

dxx=0.3048

(5.132)

Differentiating Eq. (5.129) for dT/a!x, dT = --1 -

- ~ M ( xc 2+)

(5.133) 2

dx dT

-

-

-aB(x +c2y

B(0.3048 + c 2 ) (5.134)

From Eqs (5.133) and (5.134),

H = AB(0.3048 + c 2 )

(5.135)

Substituting the value of AB in Eq. (5.135),

H

= - 647.08(0.3048

+ c2)

(5.136)

As can be seen, H depends on only one of the constants c2, and hence there is no need to solve for the other constant E. It is enough to know the value c2. From Eq. (5.130), I5

[ko - a(1755.2)]2 = - - &c: B

(5.137)

Models in Heat-transfer Operations

135

From Eq. (5,131), E (5.138) ~ ) ~ B Subtracting Eq. (5.138) from Eq. (5.137) and putting the values of k,, a,and B in the obtained equation and simplifying, we get c2 = -13.53 (5.139)

[k, = a (421.89)12 = - - ~ 2 (0.3048 3 +c

Substituting the value of c2 in Eq. (5.132), we get

H = 8557.76 W

5.6 Temperature Distribution in a Transverse Cooling Fin of Triangular Cross Section Illustration 5.6 The shape of the cooling fin is shown in Fig. 5.7. The radius of the pipe ( a ) is 20 mm, the radius of the rim of the fin (b) is 150 mm, and the coordinate x is measured inwards from the rim of the fin. (There are two natural origins for the coordinate x, but since the temperature distribution in the vicinity of the pipe axis is of no interest, the origin is taken on the rim instead.) The temperature variations normal to the central plane of the fin can be neglected, assuming that the fin is thin. The thermal conductivity of the fin ( k ) is 382 W/mK, and the surface heat-transfer coefficient is 11.36 W/mK. Denoting temperature by T"C (temperature decreases from the axis of the pipe towards the rim of the fin, i.e., transverse cooling) with TA"C representing the air temperature, develop a mathematical model for transverse cooling of the fin from the pipe axis in the form of a differential equation.

--

Fig. 5.7 Transverse cooling fin

Solution A = area of the cylinder with a radius of ( b - x) and a length of L = 21 = 2xtana

136

Process Plant Simulation

= 2nRL = 2n(b - x)(2x tana)

(5.140)

dT Heat input = - kA -

(5.141)

dx

dT dx

= - k4nx ( b - x) tana -

(5.142)

dT Heat output = - k4nx(b - x)tanadx dx

-k4nx(b

dT1 +

- x)tana-

dx

&

M,(T

- TA)

(5.143)

where A , = area of element 6c exposed from x = x to k = x

+&

= 2n(b - x)2p = 2n(b - ~ ) ( 2 & s e c a )

Accumulation = 0 (steady state) Input - output = accumulation dx

-k4nx(b

& - 4nh(b - x)(&seca)(T - TA) = 0

-

(5.144) On simplifying Eq. (5.144), we get d2T dT + -(b - 2X) - C,(b - X)(T - TA) = 0 dx2 dx Putting T - TA= y and simplifying, we get x ( b - X)-

(5.145)

x(b - x)-d2Y + -(b dY - 2x) - c,(b - x)y = 0 (5.146) dx2 dx The above equation is a second-order differential equation, which can be solved using a solution by series (method of Frobenius), described below. Method of Frobenius: If a general second-order differential equation can be expressed in the form d2Y + xF(x)-dy x2 dx2 dx where

+ G(x)Y= 0

137

Models in Heat-transfer Operations

and G(x) = Go + Glx

+ G2x2+

(3)

Its solution can be written in terms of an infinite series as follows: Do

y = &Zflxn+c

(a, # 0)

0 Do

dY = &(n

* *.

+ C)Xn+c-l

d x o

.. 2 = &(n + c)(n + c - l)Xn+c-* dx

0

Substituting Eqs (4)-(6) into Eq. (1) gives Do

+ c)(n+ c - l)Xfl+"

&(n 0

eo

+ (Go + G,x + G2x2+ * * * ) C a n x n=+ 0'

(7)

0

where the extra powers of x in Eq. (1) have been absorbed in the series for y and its derivatives. The lowest power of x which appears in Eq. ( 7 ) is x' (when n = 0), and coefficients of x' can therefore be equated as follows:

- 1) + F,%c

U~C(C

+ Go%

=0

:. c2 + (F, - 1)c + Go = 0

(8) This quadratic equation for the index of the first term c is called the indicial equation. We haven + 1 coefficients (i.e., ao,a l , a2, ...)uJ7 and one constant (i.e., c) totalling to n + 2 unknowns, which means we need to have n + 2 equations to solve for these n + 2 unknowns. Following the same procedure to obtain the indicial equation [Eq. (S)], we can x' 2, x" 3, ..., get n + 1 indicial equations by equating the coefficients of x" x"". Including Eq. (7),we have all the required n + 2 equations which are to be simultaneously solved for the above-listed n + 2 unknowns. By substituting these n + 2 unknowns in Eq. (4),we get the required solution. This is called solution by series or the method of Frobenius. +

+

+

138

Process Plant Simulation

5.7 Unsteady-state Heat Transfer in a Tubular Gas Preheater Illustration 5.7 A supply of hot air is to be obtained by drawing cool air through a heated cylindrical pipe. The pipe is 0.1 m in diameter and 1.22 m long, and is maintained at a temperature of T, = 3 16 "C throughout its length. The average values of the properties of air are as follows: heat capacity (C,)= 1005 J/kg"C, thermal conductivity ( k ) = 0.035 W/m"C, density ( p ) = 0.801 kg/m3, flow rate (u)= 7.87 x m3/s, inlet temperature = 21 "C, and overall heat-transfer coefficient ( h ) = 52.3x-'" W/m2"C, where x is the distance measured in metres from the pipe inlet. Assuming that heat transfer takes place by conduction within the gas in the axial direction, by mass flow of the gas in the axial direction, and by the abovementioned variable heat-transfer coefficient from the walls of the tube, develop a mathematical model for the temperature distribution in the axial direction in the form of a differential equation for the steady-state condition. Solution The system is shown in Fig. 5.8.

4 in. diameter

7ooFp-+~

Fig. 5.8 Gas preheater

The cross-sectional area A = z D 2 / 4 = 0.00785m'. The heat balance around the incremental length 6 x is as follows:

Heat input 1.

By conduction

2.

By mass flow

3.

Wall heat transfer

-kA-

dT dx

h(zD6x)(Tw- T )

Heat output

dx

dx

-

Models in Heat-transfer Operations

139

Input - output = 0 (steady state)

+ - kA-dT + upC,T + h(zDSx)(T, - T ) dx

+z&) dT (5.147)

= - k A -dT +d[-kA~]& dT+ u p C , ( T dx dx On simplifying, we get

(5.148) Putting T , - T = t and simplifying, we get

-d2t dx2

[- - - (-i;) UPC,) dt k A d x

t=O

(5.149)

Substituting the numerical values and h = 52.3x-'", we'get d2t - (2.31 x 104 )-dt -(5.98 x lO4)x-''t dx2 dx Putting x = z2 and simplifying, we get

(5.150)

=0

1 dt d2t - 1 d2t ___----dx2 4z2 dz2 4z3 dz Now Eq. (5.150) becomes

------( 1 d2t 4z2 dz2

1 dt 4z3 dz

(5.15 1 )

2.31 x lo4 dt 22

5.98 x l o 4

)z-[

z

)

t=o

(5.152)

Equation (5.152) can be simplified to give

where

z

= & and t = (T, - T ) .

5.8 Heat LossThrough Pipe Flanges Illustration 5.8 Two thin-walled metal pipes of 25 mm (a m) external diameter are joined by two flanges 12.5 mm thick (w m) and 100 mm in diameter (b m). The pipe is carrying steam at 120°C (T,"C). The conductivity of the flange metal is k = 380 W/mK, and the exposed surfaces of the flanges lose heat to the surroundings at TI = 16 "C

140

Process Plant Simulation

according to a heat-transfer coefficient h = 11.4 W/m2K. Develop a mathematical model for temperature in the differential form under steady-state conditions. Solution

Fig. 5.9 Pipe flange

The schematic diagram for the system is shown in Fig. 5.9.

dT Heat input = - kA dx

(5.154)

Heat output = (5.155)

Heat accumulation = 0 (at steady state) Equating Eqs (5.154) and (5.155) and simplifying, we get

d2T rdr2

dT +- cr(T - T I )= 0

dr

(5.156)

where c = h/wk. Putting r& = x and simplifying, we get (5.157)

Models in Heat-transfer Operations

141

Putting Eq. (5.157) in Eq. (5.156) and simplifying, we get

d2Y dY x-+x--x dx2 dx

2

y=o

(5.158)

Equation (5.158) is a modified Bessel equation of zeroth order.

5.9 Heat Transfer in a Thermometer System Illustration 5.9 A mercury-in-glass thermometer having a time constant of 0.05 min is at a steadystate temperature of 25 "C. At time t = 0, the thermometer is placed in a temperature bath maintained at 100°C. Develop a mathematical model for the response of the thermometer with respect to time and determine the time needed for the thermometer to reach 85 "C. (The time constant given in this problem applies to the thermometer when it is located in the temperature bath. The time constant for the thermometer in air will be considerably different from that given because of the lower heattransfer coefficient of air.) Solution Unsteady-state behaviour of mercury in a glass thermometer (development of the transfer function for a first-order system): Let x and y represent the surrounding temperature and the thermometer reading, respectively. The objective is to calculate the 'response' or the time variation of the thermometer reading y for a particular (step) change in x. We make the following assumptions: 1. All the resistance to heat transfer resides in the film surrounding the bulb (i.e., all the resistance offered by the glass and mercury is neglected). 2. All the thermal capacity is in the mercury. Furthermore, at any instant mercury assumes a uniform temperature throughout. 3. The glass wall containing mercury does not expand or contract during the transient response. (In an actual thermometer, the expansion of the wall has an additional effect on the response of the thermometer reading). 4. Initially, the thermometer is at steady state. This means that, before time zero (t c 0), there is no change in temperature with time. At time zero, the thermometer is subjected to some change in the surrounding temperature, x(t>. Applying the unsteady-state energy balance, Input rate - output rate = rate of accumulation

hA(x-y)-O=mC

dY dt

-

(5.159)

142

Process Plant Simulation

where A is the surface area of the bulb for heat transfer in m2, C, is the heat capacity of mercury in Jkg K, m is the mass of mercury in the bulb in m, t is the time in s, and h is the film coefficient of heat transfer. Equation (5.159) states that the rate of flow of heat through the film resistance surrounding the bulb causes the internal energy of mercury to increase at the same rate. The increase in internal energy is manifested through a change in temperature and a corresponding expansion of mercury, which causes the mercury column, or the reading of the thermometer, to rise. The coefficient h will depend on the flow rate and properties of the surrounding fluid and the dimensions of the bulb. In the present problem h is constant. Equation (5.159) is a first-order differential equation. This equation can be solved by Laplace transformation. ‘Deviationvariables’ are to be introduced before the transformation. (The characteristic of a Laplace transform is that the BC are introduced before the solution.) Prior to the change in x, the thermometer is at steady state and the derivative dy/ dt is zero. For the steady-state condition, Eq. (5.159) may be written as (5.160) hA(x, - y , ) = 0, t c 0 The subscripts is used to indicate that the variable is the steady-state value. Equation (5.160) simply states that y, = x,, or the thermometer reads the true bath temperature. Subtracting Eq. (5.160) from Eq. (5.159) gives (5.161) Notice that d (y - y,)/dt = dy/dt because ys is constant. If the deviation variables are defined to be the differences between the variables and their steady-state values, (5.162) x’ or X = x - x, y’ or Y = y - y , Equation (5.161) becomes

(5.163) dY dt

hA(X - Y ) = mC, -

(5.164)

Let mC,/hA = z, then Eq. (5.164) becomes dY dt Taking the Laplace transform of Eq. (5.165) gives X ( s ) - Y ( s ) = z[sY(s)- 01 x-Y=z-

(5.165) (5.166)

Models in Heat-transfer Operations

143

Rearranging Eq. (5.166) gives (5.167) The parameter z is called the 'time constant' of the system and has the units of time. The expression on right-hand side of Eq. (5.167) is called the transfer function of the system. Any physical system for which the relation between Laplace transforms of input and output deviation variables is of the form given by Eq. (5.167) is called a first-order system. Synonyms for first-order system are first-order lag and singleexponential stage. The naming of all these terms is motivated by the fact that Eq. (5.167) results from a first-order linear differential equation, Eq. (5.165).

Example 5.7 A thermometer, initially at 30 "C, is moved at time t = 0 into an air stream at 50°C. If the thermometer reads 4254°C at the end of 10 s, what is its time constant? When will the thermometer reading be 45 "C? Assume the steadystate gain to be unity. Solution Here, x,, = y,, = 30 "C, magnitude M = 50 - 30 = 20 "C. Since a thermometer is a first-order system and is subjected to step input, therefore the equation is

Here, Y(t) = y - y,, = 42.64 - 30 = 12.64. Therefore, Eq. (1) becomes 12.64 = 20(1- e-lW' and z (time constant) = 10 s. When the thermometer reads 45 "C, then Y(t)= 45 - 30 = 15"C and the time required to read 45 "C is 15 = 20( 1 - e-'"'), therefore t = 13.86 s. Note The input is subjected to a step change of magnitude M , i.e., X ( t ) = M and X ( s ) = M/s. The transfer function is

yo =-

X(s) Therefore

1

(1 + 7s)

1 M Y ( s )= -(1+ sz) s Separating by partial functions and solving, we get Y(t)IM = 1 - e-"'.

Exercise 5.2 A thermometer subjected to impulse change is initially at 30°C and is moved at time t = 0 into an air stream at 50°C. At what time with the thermometer read 42.64 "C, if the time constant is 10 s?

144

Process Plant Simulation

Hint Try this problem with the following output response function: Y(t)/M = Vze-"'.

Exercise 5.3 A thermometer subjected to ramp change is initially at 30 "C and is moved at time t = 0 into an air stream at 50 "C. If the thermometer reads 42.64 "C at the end of 10 s, what is its time constant? When will the thermometer reading be 45 "C? Assume the steady-state gain to be unity. Hint In this problem, the response function is Y(t)/M = (t - 7). Example 5.8 A thermometer having a time constant of 0.2 min is at a steadystate temperature of 0 "C.At time t = 0, the thermometer is placed in a temperature bath maintained at 100"C. Determine the time needed for the thermometer to read 40 "C. Solution

z = 0.2 min 100 X(s) = S

100 ... Y ( s )= = s

~

1 - 100 1 1 0.2s + 1 0.2 s s + 5

A(s + 5 ) + Bs = 1 Putting s = 0, A = 0.2; putting s = 0, B = -0.2.

(: $4

Y ( s ) = 100

:. ~ ( t=)(1 - e - 5 f ) We know that Y(t) = 40 - 0 = 40 "C :. 1- e-5t = 0.4 e-5r = 0.6 = ln(0.6)

:.

t = 0.102 min

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145

5.10 Unsteady-state Heat Transfer by Conduction Illustration 5.10 Consider a section of flat wall of thickness L m, whose height and length are both large compared to L. If the temperature distribution is uniform throughout the wall at t = 0 and heat is supplied at a fixed rate per unit area to one surface, develop a mathematical model for temperature distribution as a function of position and time in the differential form. Solution The system is schematically represented in Fig. 5.10.

I

pLbJ Fig. 5.10 One-dimensional heat transfer

Since the original temperature is uniform, and every part of each wall surface is subjected to the same conditions, no heat will travel parallel to the surface and the temperature will be constant in any plane parallel to the surface (Lea,heat is conducted in one dimension, only in the x-direction). Thus the temperature distribution can be specified in terms of a single coordinate, denoted by x in Fig. 5.10 and the discussion can be restricted to a section of unit area through the wall. Considering thermal equilibrium of a slice of the wall between a plane at a distance x from the heated surface and a parallel plane at x + tir from the surface gives the following balance (where T is the temperature in K and k is the thermal conductivity): Rate of heat input at a distance x and time t per unit area (5.168)

146

Process Plant Simulation

Rate of heat input at a distance x and time t

+ 6r

per unit area

(5.169) This second term is obtained considering that x is a constant (applying Taylor’s theorem). Similarly, two output rates can be determined from the corresponding input rates by applying Taylor’s theorem at constant t and variable x . Rate of heat output at a distance x + 6x and time t per unit area

(5.170) Rate of heat output at a distance x

+ 6x

and time t

+ 6t

per unit area

Heat content of an element at time t per unit area = (pGx)C,T

(5.172)

Heat content of an element at time r + 6t per unit area = pC,(T

+g6t)a.x

(5.173)

Accumulation of heat in time 6t

dT dt

= pc, -6t6x

(5.174)

Taking the average input and output rates during the time interval 6t gives Heat input = - k -

dT dx

1d dT + --k-6t 2&( dx

)

6t

(5.175)

Heat output =

(5.176)

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147

Using the general conservation law and simplifying, we get d2T

1 d3T + -k-St

kg2

dx2dt

dT pepat

(5.177)

Taking the limit of Eq. (5.176) as 6t + 0, the central term vanishes; introducing the thermal diffusivity, Eq. (5.177) becomes (5.178) which is the required model equation. The following remarks are in order: 1. In the above derivation, it will be seen that the same Eq. (5.178) results if the input and output rates at time t are used without the added complication of using average rates. 2. This simplification is consistent with the findings in the single-stage unsteady-state extraction model. But until we are confident about which simplifications are justified and which are not, it is advisable to allow for all possible correction terms as above. 3. Care must be taken in each term to write down the correct number of infinitesimals (SX, St) since these govern the terms that will be rejected in the step from Eq. (5.177) to Eq. (5.178).

Example 5.9 The temperature distribution across a large concrete 50-cm-thick slab heated from one side, as measured by thermocouples, approximates to the following relation: T = 60 - 50x + 12x2+ 20x3- 15x4,where'T is in degree Celcius and x is in metres. Considering an area of 5 m2, compute the following: (a) the heat entering and leaving the slab in unit time, (b) the heat energy stored in unit time, (c) the rate of temperature change at both sides of the slab, and (d) the point at which the rate of heating or cooling is maximum. Take the following data for concrete: thermal conductivity k = 1.2 W/m"C and thermal diffusivity a = 1.77 x m2/s. Solution Here, T = 60 - 50x + 12x2+ 20x3 - 1 5 ~ ~ dT .. = -50 + 24x + 60x2 - 60x3 dX *

d2T -= 24 + 1 2 0 ~1 8 0 ~ ~ dX2 (a) Q = -kA-

dT dX

= -kA(-50

+ 24x + 60x2 - 60x3)

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Process Plant Simulation

Heat entering the slab, Qo =

-u[g] X=O

= - (1.2 x 5)(-50) = 300W

Heat leaving the slab,

Q, =-a[$]x=OS = -(1.2 x 5)(-50

+ 12 + 15 - 7.5) = 183W

(b) Rate of heat storage = Q, - Q, = 300 - 183 = 117 W

(c) Now

= a(24 + 120x - 180x2)

at

[g]

= 1.77 x 10-3(24) = 4 2 . 4 8 ~10-30C/h

x=o

= 1.77 x

+ 60 - 45) = 69.03 x

"(31

(d) For the rate of heating or cooling to be maximum

120 - 3 6 0 = ~0

.:

x = 0.33m

EXERCISES 5.1 Summarize the steps involved in mathematical modelling. 5.2 A closed vessel of total surface area 50 m2 is heated through this surface by condensing steam at a temperature of 100°C. The vessel is charged with 1000 kg of liquid having a heat capacity of 25 12 J k g K at a temperature of 25 "C. If the

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149

process is controlled by a heat-transfer coefficient 142 W/m2K, model the system in dimensionless form and find the temperature of the liquid after 2 h. 5.3 20,000 kg/h of acid having specific heat of 0.5 enters a well-agitated tank containing 20,000 kg of acid and leaves at the same rate in a steady-state process. The inlet and outlet temperatures of the acid are 443 K and 363 K, respectively. This is maintained by cooling water, flowing at 16,000 k g h having a heat capacity of 4187 J k g K. The inlet temperature of water is 298 K. (a) Calculate the outlet temperature of water. (b) If cooling water suddenly fails, what is the temperature of the acid in the tank after 60 min? 5.4 Two concentric cylindrical metallic shells having radii u and b are separated by a solid material. If the two metal surfaces (inner and outer) are maintained at different constant temperatures To and T,, respectively, what is the steady-state temperature distribution within the separating material? Both the temperature and the heat-flow rate per unit area are functions of the radius. (a) If the temperature in the solid material is T at any radial position r, arrange the model equation in the following form: T - T o - ln(r/u) T, - To ln(b/a)

(b) Mention the significance of the dimensionless groups in the above model. 5.5 In a steady-state operation, 5000 k g h of sulphuric acid of specific heat 0.35 enters a well-agitated tank containing 5000 kg of H2S0, and leaves at the same rate. The inlet and outlet temperatures of acid are 170°C and 90"C, respectively. This is maintained by circulating cooling water at a flow rate of 4000 k g h through an immersed cooling coil. The inlet temperature of water is 25 "C. (a) Calculate the outlet temperature of water. (b) If cooling water suddenly fails, what is the temperature of the acid in the tank after 30 min? 5.6 A transverse fin of triangular cross section is used for enhancing cooling of a fluid flowing through a pipe. The radius of the pipe is 50 mm, the radius of the rim of the fin is 150 mm, and the coordinate x is measured inwards from the rim of the fin. The temperature variations normal to the central plane of the fin can be neglected, assuming that the fin is thin. The thermal conductivity of the fin is 382 W/mK, and the surface heat-transfer coefficient is 11.36 W/m2 K. Denoting temperature by T"C, with TA"C representing the air temperature, develop a mathematical model for transverse cooling of the fin from the pipe axis in the form of a differential equation. 5.7 A graphite electrode 0.2538 m in diameter passes through a furnace wall into a water cooler (which takes the form of a water sleeve). The length of the electrode between the outside of the furnace wall and its entry into the cooling jacket is 0.4048 m; and as a safety precaution the electrode is insulated thermally and electrically in this section, so that the outside surface temperature of the insulation

150

Process Plant Simulation

does not exceed 58.89"C. If the lagging is of uniform thickness and the mean overall coefficient of heat transfer from the electrode to the surrounding atmosphere is taken to be 1.701W/m2K of the surface of the electrode, and the temperature of the electrode just outside the furnace is 1582.2"C, estimate the duty of the water cooler if the temperature of the electrode at the entrance to the cooler is to be 158.89"C. The temperature of the electrode may be assumed uniform at any cross section. The surrounding temperature is 25.1 "C. The thermal conductivity of graphite can be estimated using the following relationship: k , = k, - a T where k, = 178.46 W/mK and a = 0.0628; T is in K. 5.8 Two thin-walled metal pipes of 25 mm external diameter are joined by two flanges of 12.5 mm thick and 100 mm in diameter. The pipe is carrying steam at 120°C. The conductivity of the flange metal is k = 380 W/mK and the exposed surfaces of the flanges lose heat to the surroundings at T , = 16"C according to a heat-transfer coefficient, h = 11.4 W/m2K. Develop a mathematical model for temperature in the standard form of a differential equation under steady-state conditions. 5.9 Consider a section of a flat wall of thickness L m, whose height and length are both large compared to L. If the temperature distribution is uniform throughout the wall at t = 0 and heat is supplied at a fixed rate per unit area to one surface, develop a mathematical model for temperature distribution as a function of position and time in the differential form. 5.10 Obtain a preliminary estimate of the diameter of the tubes to be installed in a fixed-bed catalytic reactor that is to be used for the synthesis of vinyl chloride from acetylene and hydrogen chloride. The tubes are to contain mercuric chloride catalyst deposited on 2.54-mm particles of carbon and the heat of the reaction is to be employed to generate steam at 121 "C for the remainder of the process. To do this, the temperature of the inside surface of the tubes should be constant at 149"C. The effective thermal conductivity of the bed is k, = 6.92 W/mK, the heat of reaction at bed temperature is AHR = 1.075 x lo8 J h o l , and the bulk density of the bed is P = 288 kg/m3. The rate of reaction is a function of temperature, concentration, and the various adsorption coefficients, but for the preliminary estimate, assume that the rate of reaction can be expressed as r = 5.89 x lo-' (1 + 0.02448T) kmol/(kg-catalyst s), where Tis the temperature in "C above a datum of 93.3 "C. The maximum allowable catalyst temperature to ensure a satisfactory life is 252°C. 5.11 Flanges 12.5 mm thick and 200 mm in diameter are used to join two thinwalled metal pipes of 50 mm external diameter. The pipe is carrying steam at (kgcatalyst s) 120°C. The conductivity of the flange metal is k = 380 W/mK and the exposed surfaces of the flanges lose heat to the surroundings at T I = 26 "C according

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151

to a heat-transfer coefficient, h = 11.4 W/m2K. Develop a mathematical model for temperature in the standard form of a differential equation under steady-state conditions. 5.12 The total surface area of A m2 of a closed kettle is heated through this surface by condensing steam at a temperature of T, K. The kettle is charged with M kg of liquid having a heat capacity of Cp J k g K at a temperature of To K. If the process is controlled by a heat-transfer coefficient h W/m2K, how does the temperature of the liquid vary with time? 5.13 In the unsteady-state heat loss through a maturing tank model (Sec. 5.3), the lagging area is neglected. If the lagging area is also included, how do the model equation and the solution change, and why? 5.14 A thermometer having a time constant of 0.05 min is at a steady-state temperature of 20°C. At time t = 0, the thermometer is placed in a temperature bath maintained at 90 "C. Determine the time needed for the thermometer to read 50 "C. 5.15 Heat is supplied at a fixed rate of Q W/m2 to one face of a large rectangular slab of density p kg/m3, specific heat Cp J/kg K, and conductivity k W/m K. Find the variation of surface temperature with time during the early stages of exposure. 5.16 Two concentric spherical metallic shells of radii a and b (where b > a) are separated by a solid thermal insulator. The inner shell is kept at a constant uniform temperature T , by a steady-state distributed source of heat. The temperature of the outer shell is found experimentally to vary with angular distribution from a fixed point on the surface. The distribution can be specified as T = T2 + T3 cosf3 + T4 cos28. How should T,, T2,T3,and T4 be plotted against heat input so that the thermal conductivity of the insulator can be found from a straight-line plot of results for different rates of heat input? 5.17 A solid rectangular slab at a uniform temperature To has its four edges thermally insulated. The temperature of one exposed face is raised to and maintained at T,, whilst the temperature of the other exposed face is held constant at To. Determine how the temperature varies with time.

CHAPTER 6 MODELS IN FLUID-FLOW 0 PERATIONS

In this chapter, systems of fluid-flow operations are modelled (starting from the simple continuity equation to the most complicated flow through a packed bed column). The focus is on model formulation for these systems. In addition, various models on momentum transfer such as laminar flow in a narrow slit, flow of a film on the outside of a circular tube, different coordinate systems for the classical falling film problem, annular flow with the inner cylinder moving axially, flow between concentric cylinders and concentric spheres, creeping flow between two concentric spheres, the working model equation for a parallel-disc viscometer, and momentum flux for creeping flow into a slot are discussed. The shell balance approach is used for formulating these models.

6.1 The Continuity Equation The continuity equation models the conservation of mass of a flowing fluid in the form of a differential equation. This important yet simple equation highlights the fact that in a flowing fluid the mass of the fluid is conserved. Consider a rectangular element of space the lengths of whose edges parallel to the three Cartesian axes are &, &, and & as shown in Fig. 6.1 below. The components of the fluid velocity are denoted by u , v, and w as shown, and the density of the fluid by p . The positive directions of u, v, and w must be the same as the positive directions of the axes so that fluid is considered to enter the element through faces ABCD, ABFE, and ADHE, and leave through the other three faces. The rate at which fluid enters through ABCD is proportional to the area of the face &&, the fluid density p , and the fluid velocity component u perpendicular to the face (i.e., m = pus).

Models in Fluid-flow Operations

153

Strictly, the velocity component has only the value u at point A , and the velocity through the face will vary slightly from u. This is a small correction, which will disappear at the end of the derivation, and will be ignored. C

G

I

-x

I

V’

W

Fig. 6.1 Volume element in Cartesian coordinates

If one wants to check the above statement, all velocity components can be considered as values averaged over the relevant face, and when the limit is taken as the size of the element shrinks to zero, each average velocity tends to its point value at A . Thus, Input rate through ABCD = pu @ 2% (6.1) Input rate through ADHE = pv & & (6.2) Input rate through AEFB = pw 6x @ (6.3) Once again, the output rates are obtained from the corresponding input rates by applying Taylor’s theorem:

d

Output rate through EFGH = Pu@& + Z ( P ~ @ & ) ~ X

(6.4)

d +h (PW & @>2%

(6.6)

Output rate through CDHG = PW & @

The mass of fluid contained in the element at time t = p 6 ~ @ &

d

The rate of increase of mass contained in the element = - ( p & dt

(6.7)

@&I (6.8)

Input - output = accumulation

154

Process Plant Simulation

(6.9) Dividing by the volume of the element (&6y&) gives

d (pu) + d (pv) + -(PI d +dP = 0 (6.10) dX dY dZ dt Taking the limit of Eq. (6.10), as the size of the element tends to zero, leaves it unaltered, because all surplus terms have been ignored at an earlier stage in anticipation of this result. Each term of Eq. (6.10) can be differentiated as a product and rearranged to give p-

au + p-av + p -dw +

dx

dy

dz

ap + v-ap + w-+ ap ap -= 0

u-

ax

dy

dz

dt

(6.1 1)

Since p is a function of x , y, I, and t, its substantive derivative can be taken as DP = u - dP + v- dP + w-dP + dP -

dx dy dz dt Combining Eqs (6.11) and (6.12) gives Dt

(6.12)

-lDP+dU+dv; dw-0 (6.13) p Dt d x d y d z which is the required continuity equation for a compressible fluid. For an incompressible (constant-density solution) fluid,

-DP =o Dt

.. -du + - + -dv = o dw *

(6.14) d y dz The following methods are used for solving partial differential equations (Jenson & Jeffreys 1963; Mickley et al. 1979; Antia 1991): 1. Method of images (error functions) 2. Method of separation of variables (orthogonal functions usage) 3. Laplace transformations 4. Finite Fourier transforms 5. Fourier transforms (discrete) 6. Hankel and Mellin transforms 7. Numerical methods (i) Step-by-step method for initial value problems (ii) Iteration (organized method of trial and error) ax

Models in Fluid-flow Operations

155

(iii) Liebmann’s method or method of squares for boundary value problems (iv) Relaxation methods for boundary value problems (v) Orthogonal collocation (method of weighted residuals) Select a suitable method from the above and solve the model equation (6.14).

6.2 Flow Through a Packed Bed Column Illustration 6.1 Liquid is distributed within a central core of radius a with velocity U, and flows through a packed column of radius R. At the position (r, z), the horizontal and vertical components of velocity are Vand U.It is also given that V = - D(bU/&), where D is a constant characteristic of the liquid and the packing material. Obtain a mathematical model in the differential form for U at steady state. Solution The schematic diagram of the system under consideration is shown in Fig. 6.2. a

-+ i, r

+ Sr

zr V

U

Fig. 6.2 Volume element in a packed bed column

Volumetric rate of input = 2nrSrU

+ 2nrS.V

a

= 2mSrU - D-2nrSz

&

156

Process Plant Simulation

Volumetric rate of output = 2mU6r

- 0 -dU 2m62

a

+ -((2mUSr)Gz a2

-

dr

(6.15)

At steady state, rate of input = rate of output

d * -(2mU6r)6z dZ

=

(6.16)

(6.17)

(6.18)

I,;

. dU d2U .. 3.7.= D[ 2

IdU

+

(6.19)

Find the solution for the above model equation (6.19).

6.3 Laminar Flow in a Narrow Slit Illustration 6.2 A Newtonian fluid is in laminar flow in a narrow slit formed by two parallel walls a distance 2B apart. It is understood that B<<W, so that ‘edge effects’ are unimportant. Make a differentialmomentum balance and obtain the following model equations: (a) Momentum flux and velocity distributions:

In these expressions P = p + pgh = p - pgz. (b) Ratio of the average velocity to the maximum velocity for the flow. (c) The slit analogue of the Hagen-Poiseuille equation. Solution The schematic diagram is shown in Fig. 6.3.

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157

I

4xz x

4UlZd

Fig. 6.3 Laminar flow in a narrow slit

(a) Momentum flux and velocity distributions: We postulate that v, = v,(x), v, = v y = 0, p = p(z). The rate of z-momentum in of t-momentum out across the surface across the surface at z = 0 is (WAX)@, a t z = ~ i (s w ~ ) @ , I , = ~oft-momentuminacrossthe , surfaceatxis (wL)cD,I,, of z-momentum in across the surface at x + Ax is (WL) @,, , h, and the gravity force acting on the fluid in the z-direction is ( W b L)pg. For steady flow, the momentum balance over a thin shell of fluid will be

lz=o,

I

Rate of momentum in by convective transport

+

rate of momentum out by convective transport

rate of momentum in by molecular transport

I

rate of momentum out by molecular transport

+i

force of gravity }=O acting on system

= (wL)(@xzl,

-@,,l,+AA

+ (WAxL)pg = 0

+ ( W ~ ) ( @ , l z = -@&J 0 (6.20)

158

Pmcess Plant Simulation

When this equation is divided by (W Ax L ) , and the limit taken as Ax approaches zero, we get

(6.22)

But,

ax,= ZXZ+ p v ,

v, = -p- dVZ + PVXVZ

(6.23)

dX

=p

+ 2 , + pv,v,

=p

- 2p- h

z

dz

+ pv,v,

So, from the postulates and Eqs (6.23) and (6.24),

(6.24)

m.(6.22) simplifies to (6.25)

-drxz - IP(0) - P d O ) } - I P W ) - P d L ) } L

dx

But, P = p

+ pgh = p - pgz

(6.26)

= modified pressure. (6.27)

(6.28)

Boundary condition: at x = 0, Z, = 0

:. c, = 0 Hence, (6.29)

By Newton's law of viscosity, 2, = - p - h z dX

(6.30)

Models in Fluid-flow Operations

159

From Eqs (6.29) and (6.30)

.=-(

)

P(0) - P(L) x 2 Lp 2 + c2

(6.31)

Boundary condition: at x = a,vz = 0 (6.32)

(6.33)

(b) Average velocity to maximum velocity ratio: Average velocity, =

(6.34)

(6.35)

(

= p(0:ipP(L))B2

(6.36)

Maximum velocity, Vz,max

-- vz x=o ~

(

= P(0:EIp(L))B2

(6.37)

From Eqs (6.36) and (6.37), we get - -2 3

(6.38)

'z,max

(c) The slit analogue of the Hagen-Poiseuille equation:

II W

Mass-flow rate ( w ) = = pW(2BW)

0

B

-B

pv,dxdy

(6.39) (6.40)

160

Process Plant Simulation

(6.41)

6.4 Flow of a Film on the Outside of a Circular Tube Illustration 6.3 In a gas absorption experiment a viscous fluid flows upward through a small circular tube and then downward in a laminar flow on the outside. Set up a momentum balance over a shell of thickness Ar in the film, as shown in Fig. 6.4. Note that the 'momentum in' and 'momentum out' arrows are always taken in the positive coordinate direction,even though in this problem the momentum is flowing through the cylindrical surfaces in the negative r direction. (a) Show that the velocity distribution in the falling film (neglecting end effects) is

(b) Obtain an expression for the mass rate of flow in the film. Solution r

r Ar 4

z=L Fig. 6.4 Film flow over a circular tube

(a) Velocity distribution in the falling film: We postulate that v, = vZ(r),v, = vg = 0, and p = p(r). The rate of z-momentum in across the surface at z = 0 is ( 2 m A r ) 0 z l z = 0 ,of z-momentum out across the surface at z = L is prnbr)azz = L , of z-momentumin across the cylindrical surface at r is ( 2 mL ) cDn ,, of z-momentum in across the cylindrical surface at r + Ar is (2mL)cDnIA,, and the gravity force acting on the fluid in the z-direction is (2nr Ar L)pg.

I

I

Models in Fluid-flow Operations

161

For steady flow, the momentum balance over a thin shell of fluid will be Rate of momentum in by convective transport

I-{

rate of momentum out by convective transport

rate of momentum in by molecular transport

1

rate of momentum out by molecular transport

force of gravity } = O acting on system

(~Z~LI(Q~I~

- Q n l r + A r ) + ( ~ ~ A ~ I ( Q ~ -~~ Iz zZ) z== O L)

+ (2mAr L)pg = 0

(6.42) When this equation is divided by 2~ Ar L, and the limit taken as Ar approaches zero, we get

(6.44)

But, cp, = z,

0, = p

+ pvrvz= -

d V

p L

dr

+ pvrvz

+ z, + pvzvz = p - 2 pd V3 + pvzvz

(6.45) (6.46)

d2

So, from the postulates and Eqs (6.45) and (6.46), Eq. (6.44) simplifies to (6.47)

:.

7,

Pgr =2

+

c, r

(6.48)

Boundary condition: at r = aR, z, = 0 (6.49)

162

Process Plant Simulation

Hence, = -pgr -

pg(aRI2 2r By Newton’s law of viscosity, rz

7, =

(6.50)

2

-p- avz

(6.51)

ar

From Eqs (6.50) and (6.5 l), we get

v, = -+ 4P Boundary condition: at r = R, v, = 0

[

c2= pgR2 - a2 (In R ) ] 4P

(6.52)

(6.53)

(6.54) (b) Mass-flow rate in the film:

(6.55)

Mass-flow rate (w)=

(6.56)

(6.57)

6.5 Choice of Coordinate Systems for the Falling Film Problem Illustration 6.4 Derive the velocity profile and the average velocity for Illustration 6.3 by replacing x by a coordinate X measured away from the wall, that is, X = 0 is the wall surface, and X = 6 is the liquid-gas interface. (a) Show that the velocity distribution is then given by

[ (^,I]

pg6* x - -1 v, = /.l 6 2 6

cosp

Models in Fluid-jlow Operations

163

(b) Use this distribution to get average velocity. Solution The schematic diagram of the system is shown in Fig. 6.5.

AI '

y = 0..,.." ,,,,.....' ,.... ....." %.'

z=o

Fig. 6.5 Coordinate systems for falling film problem

(a) Velocity distribution: It is given that

zF, = (pgc0sp)X

+ c,

(6.58)

Here, F = 0 at the wall surface and X = 6 at the liquid-gas interface. At X = 0,

zEz # 0. But, at X = 6, ,z,

= 0.

:. c, = - (pg cosp>s

(6.59)

:. ziz

(6.60)

(pgcosp)(X - 6) By Newton's law of viscosity, t-

xz

=

a,

= - p L

(6.61)

a

From Eqs (6.60) and (6.61), we get pgcosp

vz =

P

Boundary condition:

x2 - 6 i ] -t- c2

(6.62)

at X = 0, v, = 0

:. c2= 0

(6.63)

:. v, = pg62 [i - - -1 ( i)2] p

6

2 6

cosp

(6.64)

164

Process Plant Simulation

(b) Average velocity: rW r 8

Jl " rJv- v, , ddXdy Xdy Average velocity < v, > =

0

0

(6.65)

lw 0 l ' 0d T d y

(6.66)

.. = p g S 2 cosp

(6.67)

3P

6.6 Annular Flow with Inner Cylinder Moving Axially Illustration 6.5 A cylindrical rod of diameter kR moves axially with velocity vo along the axis of a cylindrical cavity of radius R as seen in Fig. 6.6. The pressure at both ends of the cavity is the same, so that the fluid moves through the annular region solely because of the rod motion. (a) Find the velocity distribution in the narrow annular region. (b) Find the mass rate of flow through the annular region. (c) Obtain the viscous force acting on the rod over the length L. Solution

z=o

L

z=L

Fig. 6.6 Annular flow with inner cylinder moving axially

(a) Velocity distribution in the narrow annular region: We postulate that v, = v,(z), vr = vo = 0, and P(0) = constant (the modified pressure is constant from z = 0 to z = L). The rate of z-momentum in across the surface at z = 0 is (2727 At-)0 ,=o, of z-momentum out across the surface at z = L is ( 2 m A r ) Q ) z I Z = Lof , z-momentum in across the cylindrical surface at r is (2727 L )@n r , and of z-momentum in across the cylindrical surface at r + Ar is

I

(2mL)

I

1

r=Ar*

165

Models in Fluid-flow Operations

For steady flow, the momentum balance over a thin shell of fluid will be Rate of momentum in by convective transport

rate of momentum out by convective transport

rate of momentum in by molecular transport

I-{

rate of momentum out by molecular transport

1

+-I

force of gravity }=O acting on system

- ' , I r + A r ) + (2nrAr)(o,/z=o-oulz=r) =

(2nrL)(o,)r

(6.68)

When this equation is divided by 2 z A r L and the limit taken as Ar approaches zero, we get (6.69)

I'

L

(6.70)

But, Qrz = 2,

dV + pvrvz= - p 2 + pv,vz dr

(6.71)

+ 2, + pv,v, = p - 2p-

vz + pvzvz (6.72) dZ So, from the postulates and Eqs (6.71) and (6.72), Eq. (6.70) simplifies to 0, = p

d(r7j-z) = O

dx

(6.73) (6.74)

By Newton's law of viscosity, (6.75) From Eqs (6.74) and (6.75), we get (6.76)

166

Process Plant Simulation

Boundary conditions: (a) at r = kR, v, = vo; (b) at r = R, v, = 0.

.:

P O c, = -

In ( Ilk)

(6.77)

(6.78)

.:

v, = vo In (r/R) - vo In (Wr) In (Ilk) In (k)

(6.79)

(b) Mass-flow rate through annular region:

(6.80) (6.81) - 2nPvo -ln(Ilk)

[rln(R) - rln(r)]dr kR

(6.82)

(6.83) (c) Viscous force acting on the rod: The viscous force acting on the rod over the length L is

(6.84) =2nRL

.*.

F=

[1;;)) A] ~-

2nLp vo ~

In( Ilk)

(6.86)

6.7 Flow Between Coaxial Cylinders and Concentric Spheres Illustration 6.6 The space between two coaxial cylinders is filled with an incompressible fluid at constant temperature. The radii of the inner and outer wetted surfaces are kR and

Models in Fluid-flow Operations

167

R, respectively. The angular velocities of rotation of the inner and outer cylinders are R, and R, , (a) Determine the velocity distribution in the fluid and torques on the two cylinders needed to maintain the motion. (b) Repeat part (a) for two concentric spheres. Solution (a) Velocity distribution in coaxial cylinders: We postulate that ve = ve ( r ) , V , = V , = 0, and p = p ( r , z ) With these postulates, the equation of continuity is identically satisfied and the components of the equation of motion become r-component :

ve2

dP

-7= -dr

Ocomponent: 0 = p - --(me) [ir(::r

)]

dP + pg z-component: 0 = -dZ For this problem we need only the B-component of the equation of motion, (6.87) By solving differential equation (6.87), we get C,r C2 ve=-+(6.88) 2 r Boundary conditions: (a) at r = kR, ve = R, (kR) ; (b) at r = R, vo = R, ( R ) . Applying these boundary conditions and solving for C , and C,, we get

c, =

2(R, - R , k 2 ) (1 - k 2 )

(6.89)

and

c, = R2(Ri - R , ) k 2

(6.90)

(1 - k 2 )

Therefore, from Eqs (6.88)-(6.90), we readily obtain 2

(Q, - Rik )-

r

kR

+ (R; - R,)

(6.91)

168

Process Plant Simulation

(b) Velocity in concentric spheres: We postulate that v B = v B ( r , 81, v, = V e = 0, and p = p ( r , 8) .With these postulates, equation of continuity is identically satisfied and the components of the equation of motion become r-component: 0 = --J P

(6.92)

JP 8 -component: 0 = --1 r%

(6.93)

dr

$-component: 0 = For this problem we need only the @component of the equation of motion. Suppose

v B = f(r)sin8 Then Eq. (6.94) becomes

(6.95)

(6.96) From the differential equation (6.96), we get (6.97) Equation (6.97) is an equidimensional equation. Assume a trial solution

f ( r )= rn

(6.98)

Therefore Eq. (6.97) becomes (6.99) It gives n = 1 or n = -2.

:. f ( r ) = C,r + c2 r

(6.100) (6.101)

Boundary conditions: (a) at r = kR, v9 = Q i ( k R ) ; (b) at r = R , v B = Q , ( R ) . Applying these boundary conditions and solving Eq. (6.101) for C, and C,, we get

c, = (Q,

- Qik3)

(1 - k 3 )

(6.102)

Models in Fluid-flowOperations

169

and c 2

R 3 ( R i- R , ) k 3 (1 - k 3 )

=

(6.103)

Therefore, from Eqs (6.101)-(6.103), we obtain Ve

:

r

= -(R, - R i k 3 ) (1 - k 3 ) kR

kR

+ (Ri - R,)

(6.104)

6.8 Creeping Flow Between Two Concentric Spheres Illustration 6.7 A very viscous Newtonian fluid flows in the space between two concentric pheres, as shown in Fig. 6.7. It is desired to find the rate of flow in the system as a function of the imposed pressure difference. Neglect end effects and postulate that V e depends only on rand 8, with the other velocity components being zero. (a) Using the equation of continuity, show that ve sin8 = u(r), where u(r) is a function of r to be determined. (b) Write the &component of the equation of motion for this system, assuming the flow to be slow enough that the (v.Vv) term is negligible. Show that this gives

(c) Separate this into two equations

dP

sine-

de

= B, B = ,U

where B is the separation constant, and solve the two equations to get

B=

4-4 21n( cot

);

, u(r) =

~

'

-p2

[(l-i)+(l-!)k]

(4p)ln[cot ( ~ / 2 ) ]

where P , and P, are the values of the modified pressure at 8 = E and 8 = n - E , respectively. (d) Use these results to get the mass-flow rate

w=- npR3 (P2 - ')(1- k 3 ) ( 1 2 , ~ ) In [cot ( ~ / 2 ) ]

170

Process Plant Simulation

Solution

2E

Fig. 6.7 Creeping flow between two concentric spheres. (Inner sphere radius = kR, outer sphere radius = R. Both spheres are stationary.)

We postulate that vg = vg (r, O), v, = v4 = 0, and P = P(@. (a) Velocity distribution: Let

v g sine = u(r) The equation of continuity is

* dt

+ -1- (2p r 2 dr

2

(6.105)

1 d 1 d Vr) + -- (pv, sin 6 ) + --(pv& = 0 r sin 6 dB r s i n 8 6’4 (6.106)

Using (6.107) and Eq. (6.106), we find that v, sin 8 = u(r) satisfies the continuity equation, where u(r) is a function of r to be determined. (b) Equation of motion: The @component of the equation of motion becomes

As v, = u(r)/sinO, Eq. (6.108) becomes 1 dP r de

(6.109)

Models in Fluid-jlow Operations

171

(c) Separation into equations: Equation (6.109) can be written as (6.110)

ae

= B (separation constant)

dP (i) B = sine-

(6.111) (6.112)

ae

(6.113)

(6.114)

(6.1 15)

.*.B = -p2 - 4

(6.1 16)

21n( cot:)

(ii) B = p

-r [ : ]f r)[: z

(6.1 17)

Solving the differential equation (6.1 17), we get (6.118) Boundary conditions: (a) at r = kR, u ( r ) = 0; (b) at r = R, u(r) = 0 (since both spheres are stationary). Applying these boundary conditions and solving Eq. (6.1 18) for C , and C,, we get

c, = -(BR2k)

(6.1 19)

c, = - BR( 1 + k )

(6.120)

(2p)

and (2p)

172

Process Plant Simulation

Therefore, From Eqs (6.1 18)-(6. l20), we obtain (6.121) But, from Eq. (6.1 16), p2-

B=

4

2 In( cot

): (6.122)

(d) Mass-flow rate: Mass-flow rate (w)= ~ 0 2 f f ~ krsinBdrdi$ ~pv, = 21cp

jkR4 r ) r sin 6 dr

(6.123) (6.124)

From Eq. (6.122), we get (6.125) 1cpR3 (PI - P2)(1- k)3 :. w = ( 1 2 ~ ) In [cot ( ~ / 2 ) ]

(6.126)

6.9 Parallel-disc Viscometer Illustration 6.8 A fluid, whose viscosity is to be measured, is placed in the gap of thickness B between two discs of radius R . One measures the torque T, required to turn the upper disc at an angular velocity R. Develop the formula for deducing the viscosity from these measurements. Assume creeping flow. (a) Postulate that for small values of R the velocity profiles have the form vr = 0, vz = 0, and V, = r f ( z ) : why does this form for the tangential velocity seem reasonable? Postulate further that P = P(r, z). Write down the resulting simplified equations of continuity and motion. (b) From the 6 -component of the equation of motion, obtain a differential equation forflz). Solve the equation forflz) and evaluate the constants of integration. This leads ultimately to the result v, = Qr(r/B).

Models in Fluid-flow Operations

173

Solution Disc at z = B rotates with angular velocity R

Fluid with viscosity p and density p is held in place by surface tension

Disc at

z

= 0 is fixed

Fig. 6.8 Parallel-disc viscometer. (Both discs have radius R; R >> B.)

We postulate that v, = @(z), vr = v z = 0, and P = P(r, z). (a) Simplified equations of motion: With these postulates, equation of continuity is identically satisfied and the components of the equation of motion become r-component:

dP

v;

-7 = -dr

@component: 0 = p z-component: 0 = --dP

dZ

(b) @component of equation of motion: The @component of the equation of motion is (6.127) Putting v, = rf(z) in Eq. (6.127), we get (6.128) On simplification, we finally get (6.129)

:. f(z) = c,z + c2

(6.130)

Hence, (6.131) r(C,z + C2) Boundary conditions: (a) at z = 0, vo = 0 (b) at z = B , r = R, v8 = RR. From the BC and Eq. (6.131), we obtain vg = @(z) =

c2 = 0

174

Process Plant Simulation

R c, = -

(6.132)

B

(6.133)

6.10 Momentum Fluxes for Creeping Flow into a Slot An incompressible Newtonian liquid is flowing very slowly into a thin slot of thickness 2B (in the y-direction) and width W (in the z-direction) Fig. 6.9. The mass rate of flow in the slot is w. It can be shown that the velocity distribution within the slot is

q(;I],

v x = 4BWp

v y = vz = 0

at locations not too near the inlet. In the region outside the slot, the components of the velocity for creeping flow are 2w

r

x2y

1

These equations are only approximate in the region near the slot entry for both x 2 0 and x I 0. (a) Find the components of the convective momentum flux inside and outside the slot. (b) Evaluate xx-component of p v v at x =-a,y = 0. (c) Evaluate xy-component of p v v at x = -a, y = +a. (d) Verify that the velocity distributions given in above equations satisfy the relation V . V= 0. (e) Find normal stress r X x at place y = 0 and on the solid surface x = 0. (0 Find shear stress T y x at x = 0. Solution

X

Fig. 6.9 Flow of liquid into a slot from a semi-infinite region x e 0

Models in Fluid-flow Operations

Inside the slot,

v, = W 4B wp [ l

-

($1)

v y = v, = 0

Outside the slot, x3

VX

=-"[ zwp ( x

+ y2)2

1,

(6.134)

1)

2

=-&[ nwp ( x 2 x+ y2)2

Vy

175

=o

V,

(6.135)

(a) Components of the convective momentum flux: (i) Inside the slot,

P X V X

=P[

&[

1-

pv,vy = p,v, = 0

(*:

($11

2

[;

= 1

3w

v y = v, = 0)

1-

I}$(

2

136) (6.137)

p y v , = p y v y = p y v , = 0 (.: vy = v, = 0)

(6.138)

pvzvx = p,v, = pv,v, = 0 (*: vy = vz = 0)

(6.139)

(ii) Outside the slot,

I]

pvxv, = l[i.[.x3 p

7cw

(x

2

(6.140)

+ y2)2

1

pv,vy = p v Y v X = - x5y p

11'

-

[ 2nw w ( (x2 + y2)2

]]

(6.141)

2

pvyvy = 1 [%( 2x2y p ZW (x + y2)2

(6.142)

p , v z = 0 (.: v, = 0)

(6.143)

pvyv, = 0 (.: v, = 0)

(6.144)

PVZV',

= P , V y = P , V Z = 0 (.: v, = 0)

(b) xx-component of p v v at x = -a, y = 0:

(6.145)

176

Pmcess Plant Simulation

(i) Inside the slot,

(6.146)

-

9w2 16W2B2p

(6.147)

(ii) Outside the slot, (6.148)

-

4w2 7 r W2 a2 p

(6.149)

2

(c) xy-component of pvv at x = -a, y = +a: (i) Inside the slot, ~ x v ~ I x = - ,y=+a ,

=O

(6.150)

(ii) Outside the slot, 1

(6.151)

p v x v y lx = -a,y= + a x=-a,y=+a

2 2 - -w a -p7r2W2

(6.152)

(d) Verification of the relation v . V = 0.

a ax

v . v = -(v,)

a aY

+ -(VJ

a

+ ---(vz)

(6.153)

az

(i) Inside the slot,

ax

a + -(O) + -(O) ay

4BWp

= 0 [*.' V , = f(y) and vY =

V,

= 01

az

(6.154)

(6.155)

Models in Fluid-flow Operations

(ii) Outside the slot,

V..=”-?L[

x3

177

1)

nwp (x2 + y 2 ) 2

ax

+ 2 [ - z [x 2 y

]]+-(O) d nwp (x2 + y 2 ) 2 dz

ay

(6.156)

1 1 (e) Normal stress 7, = 4

7 ,

+o=o

(6.157)

at plane y = 0 and on the solid surface x = 0:

2 2 1

(6.158)

(i) Inside the slot,

z,

[

= - p 2-:[4;p[l-(;,l]}]=o 7,~y=o = 7~,=,0

=0

L.’v,=f(Y>l

(6.159) (6.160)

(ii) Outside the slot,

(6.161) (6.162) (6.163)

178

Process Plant Simulation

( f ) Shear stress 2 , at x = 0:

(6.164)

(i) Inside the slot, (6.165)

(6.166) (ii) Outside the slot, 2,

1) $[-$[

= - P [ $ [ - ” [ nwp (x 2 x+ 2 yy 2 ) 2

+

(x2 :;)2])} (6.167) (6.168)

EXERCISES 6.1 Develop a model in the form of a differential equation for the conservation of mass of a flowing fluid considering the flow in three dimensions, and arrange it in the following form if the velocities in the x-, y-, and z-direction are u, v, and w, respectively:

-l D p + -du + _ +dv - = od w p Dt a x d y dz 6.2 A gas containing entrained mist of non-volatile tar is located inside the cylinder of a reciprocating compressor. It is desired to determine the work required to compress the gas adiabatically and reversibly from its present pressure of 0.33 atm to a pressure of 1.0 atm. The following information is available: The gas: (a) Molecular weight: 24 k g h o l (b) Heat capacity at constant volume (constant): 25,960 J k o l K

Models in Fluid-jlow Operations

179

(c) Initial temperature: 76.7 "C (d) Initial pressure: 3.34 x lo4 N/m2 (e) Final pressure: 1.01325 x lo5 N/m2 (f) Obeys the perfect-gas equation of state. The tar: (a) The tar is always present as mist in the ratio 0.2 tar per lb of tar-free gas. (b) The volume of the tar may be neglected in comparison with the volume of the associated tar-free gas. (c) The temperature of tar is always the same as the temperature of associated tarfree gas. (d) Heat capacity of tar: 2093.5 J k g K The compression cylinder: (a) Initial cylinder volume: 0.01 13 m3 (b) Intial cylinder pressure: 3.34 x lo4 N/m2 (c) Final cylinder pressure: 1.01325 x lo5 N/m2 (d) The compression process is reversible and adiabatic in the sense that friction and heat transfer to or from the cylinder, piston, and associated machinery may be neglected. 6.3 A solid sphere of radius R is placed in an incompressible, inviscid fluid of infinite extent. The flow was initially uniform and parallel to the z-axis, flowing with a speed V, in the negative z-direction. Determine the velocity of the fluid after the sphere is placed in the flow and steady state has been achieved. 6.4 A fluid is flowing through a constant-diameter cylindrical pipe in turbulent regime. The density and velocity change in the axial direction of the pipe. Deduce the final working equation for this system by making suitable assumptions, starting from the total continuity equation. 6.5 An incompressible liquid is pumped into a gravity-flow tank at a variable volumetric flow rate (F, m3/s). The height of the liquid in the vertical cylindrical tank is h m and the flow rate out of the tank is F m3/s. Develop a mathematical model for the rate of change of velocity with time. 6.6 Develop a mathematical model for the laminar flow of an incompressible, Newtonian liquid in a cylindrical pipe starting from the force balance on a microscopic system. 6.7 A fluid of constant density p kg/m3 is pumped into a cone-shaped tank of total volume HnR2/3. The flow out of the bottom of the tank is proportional to the square root of the height h of the liquid in the tank. Develop the model equation describing the system. 6.8 A perfect gas with molecular weight M flows at a mass-flow rate W , into a cylinder through a restriction. The flow rate is proportional to the square root of the pressure drop over the restriction:

180

Process Plant Simulation

Wo = KO JP, - P where P is the pressure in the cylinder and Po is the constant upstream pressure. The system is isothermal. Inside the cylinder, a piston is forced to the right as the pressure P builds up. A spring resists the movement of the piston with a force that is proportional to the axial displacement x of the piston (F, = Kp).The piston is initially at x = 0 when the pressure in the cylinder is zero. The cross-sectional area of the cylinder is A. Assume the piston has negligible mass and friction. (a) Develop the model equation describing the system. (b) What will the steady-state piston displacement be?

CHAPTER 7 MODELS IN REACTION ENGINEERING

So far, we have discussed the models corresponding to transport phenomena (mass, momentum, and heat transfer). Though it is important to have a physical understanding of these three modes of transfer, ultimately it is the reaction that is the most important aspect in chemical engineering processes. In the earlier chapters on modelling, the reaction term is taken to be zero, focusing only on laws of conservation of mass, momentum, and heat. Depending upon the reaction kinetics (order of reaction and reaction-rate constant) and the type of reactor, the reaction term in the conservation laws contributes significantly. Henceforth, various possible mechanisms and ways by which the reaction term contributes to the overall understanding of the physics of the system need to be fucused upon. Keeping these objectives in mind, this chapter focuses on modelling of reaction engineering systems, where along with the reaction, mass, heat, or momentum transfer also takes place simultaneously. The models discussed are the chemical reaction with diffusion in a tubular reactor, chemical reaction with heat transfer in a packed bed reactor, gas absorption accompanied by chemical reaction and two different problems on reactors in series.

7.1 Chemical Reaction with Diffusion in a Tubular Reactor Illustration 7.1 A tubular chemical reactor of length L and cross section 1 m2 is employed to carry out a first-order chemical reaction in which a material A is converted to a product B. The chemical reaction can be represented as A-+B and the specific reaction-rate constant is k s-'. The feed rate is u m3/s, the feed concentration of A is c,,, and the diffusivity of A is assumed to be constant at

182

Process Plant Simulation

.

D m2/s. Assume that there is no volume change during the reaction and steadystate conditions are established. Develop a mathematical model in the form of a differential equation for the concentration of A as a function of length along the reactor, and discuss the boundary conditions. Solution Take a coordinate x to specify the distance of any point from the inlet of the reactor section.

I

A1-

, , t

Reaction section

Entry section

sx

Fig. 7.1 Tubular reactor analysis

Let c denote the variable concentration of A in the entry section (x c 0) and y denote the concentration of A in the reactor section (x > 0), as shown in Fig. 7.1. The concentration of A varies in the inlet section due to diffusion, in the reactor section due to diffusion and chemical reaction, but does not vary in the section following the reactor. A material balance can be taken over an element of length Sx at a distance x from the inlet in the reactor section At x At x + S x Bulk flow of A Diffusion of A

-D-dY dx

dx

dx

As steady-state conditions are established, the accumulation in this case is zero. But the input must exceed the output to supply the reaction taking place within the element. Rate of removal of A by reaction = rate of depletion of A = kysx

(since the reactor is of unit cross-sectional area). Alternatively, this depletion term may be considered as either an output or an accumulation, and the general conservation equation applied directly:

Models in Reaction Engineering

183

Rate of input - rate of output = rate of depletion (7.1) Dividing by Sx and simplifying, and rearranging, gives

Similarly, since there is no chemical reaction, the material balance in the entry section leads to D -d2c - U-dc = 0

(7.3) dx2 dx Equation (7.3)can also be obtained from Eq. (7.2)by removing the reaction term and putting y = c. Equations (7.2)and (7.3)are both second-order linear differential equations with right-hand sides being zero, thus the complementary function is the complete solution in both cases. The auxiliary equation of Eq. (7.2)is Dm 2 - u m - k = O

(7.4)

f ,/u2 + 4kD = ~ (fla)/2D 20 Similarly, for Eq. (7.3),

:.

m =

u

(7.5)

Dn2 - un = 0

a n(Dn-

U)

(7.6)

=0

Since n # 0, (Dn - u ) = 0 n=u/D Therefore the solution of Eqs (7.2)and (7.3) is

;[

y = Aexp -(l

1

+a)

;[

+ Bexp

-(1

-a)

1

(7.9)

where a=

d

1 + -4;:

(7.10)

and c=a

+ Pexp(ux/D)

(7.11) which contains four arbitrary constants: A, B, a, P. The four boundary conditions are as follows:

184

Process Plant Simulation

At x = - m O 3 C,

(7.12)

= C ~

Atx=O, c = y

(7.13)

At x = 0, d d d x = dy/dx

(7.14)

At x = L, dy/dx = 0

(7.15)

1. The first condition specifies the state of the feed stream. 2. The second ensures continuity of composition 3. The third, together with the second, is necessary to conserve the material at the boundary assuming that the diffusivities are equal in both sections. 4. The fourth forbids diffusion out of the reactor and is necessary as a conservation law for the section following reactor.

Example 7.1 A tubular chemical reactor of length 2 m and cross section 1 m2 is employed to cany out a first-order chemical reaction in which a material A is converted to a product B. The chemical reaction can be represented as A + B. The m3/s. s-l and the feed rate is 1 x specific reaction-rate constant is 6.33 x The feed concentration of A (c,) is 0.1 kg/m3 and the diffusivity of A is assumed to be constant at 0.26 x lo4 m2/s. Assume that there is no volume change during the reaction and the steady-state conditions are established. Find the concentration of A at 1.4 m from the entrance with respect to the inlet concentration? Solution We know that the concentration along the length of the reactor is governed by the following equations:

[z 1+ [z 3

y = Aexp - ( l +

a)

Bexp -(1

-a)

and c=a

; ;1 J

+ Pexp

-

where a = d l + 4kD/u2 = 81.143. These equations contain four arbitrary constants: A, B, a, p. The four boundary conditions are as follows: At x = - w ,

C = C ~

At x = O , c = y dc dy At x = O , - = dx dx

(3) (4)

Models in Reaction Engineering

185

Equations (1)-(4) give, respectively,

a = co a+p=A+B

2p = A(l + a ) + B(l - a) (10) Eliminating a and

p

from Eqs (7)-(9) gives

2 ~ 0= A(l - a) + B(l + U ) Solving Eqs (10) and (1 1) for A and B gives

A=

2c0(a- 1)

K

[ g]

exp -

B = 2c0(a - 1) exp[%]

K where

(14) Putting these values of A and B into Eq. (1) gives the final result as

(15) Therefore the concentration of A at a distance of x = 1.4m from the entrance of the reactor is Y = 1.037 x lo-" CO

Therefore, substituting c,, = 0.1 kg/m3, we get y = 1.037 x

kg/m3

7.2 Chemical Reaction with Heat Transfer in a Packed Bed Reactor Illustration 7.2 Using the data given below, obtain a preliminary estimate of the diameter of the tubes to be installed in a fixed bed catalytic reactor which is to be used for the

186

Process Plant Simulation

synthesis of vinyl chloride from acetylene and hydrogen chloride. The tubes are to contain mercuric chloride catalyst deposited on 2.54-mm particles of carbon and the heat of the reaction is to be employed to generate steam at 121 "C for the remainder of the process. To do this, the temperature of the inside surface of the tubes should be constant at 149 "C. The effective thermal conductivity of the bed, k, = 6.92 Wlm"C, the heat of reaction at bed temperature, AHR = 1.075 x lo8 J h o l , and the bulk density of the bed, P = 288 kglm3. The rate of reaction is a function of temperature, concentration, and the various adsorption coefficients, but for the preliminary estimate assume that the rate of reaction can be expressed as r = r, (1 + A T ) where ro = 0.12, A = 0.024, and T is the temperature in degrees Fahrenheit above a datum of 200°F (93.3"C). The maximum allowable catalyst temperature to ensure a satisfying life is 252 "C (Lea, T = 252 - 93.3 = 158.7 "C above the datum temperature). Solution The reaction rate based on unit mass of solid in the fluid-solid system is Y = 0.12 (1 + 0.0247'). Table 7.1 shows the data generation. Table 7.1 T ("C)

0

100

121

141

149

252

93.3

r x 10" (kmoVs kg-catalyst)

0.589

2.031

2.33

2.614

2.73

4.214

1.933

r=

c,(1 + C J )

(i) At T = 0 "C, r = 5.89 x

kmoUskg-catalyst

C, = 5.89

10-5 (ii) At T = 100 "C, Y = 2.031 x lo4 kmoUs kg-catalyst

a C, = 0.02448 Therefore the present relation between r and T is ( 1 + 0.024487') kmoUskg-catalyst r = 5.89 x where A = 0.02448, r0 = 5.89 x and T is the temperature in "C above the datum of 93.3 "C. Consider Fig. 7.2 which illustrates the following additional symbols describing the process: G is the mass-flow rate of fluid in the reactor section in kgls m2, R is the tube radius in m,x is the radial coordinate, z is the axial coordinate from inlet, and Cp is the heat capacity of the gas in J k g "C.

Models in Reaction Engineering

187

Fig. 7.2 Fixed bed reactor

Taking a heat balance for an element of thickness Sx at a radial distance of x from the tube axis, and a distance z from the entrance of the tube, we get Input = - k e ( 2 n x & ) -

dT

dx

+ G(2nx6x)CPT

Rate of reaction heat = (2nx6xSz)p.4HRr The unit of each term is J/s = W. The output is given by

(7.16) (7.17)

(7.18) Accumulation = 0 (because the process is at steady state) Input - output = accumulation - heat of reaction Dividing throughout by 2 r k c x S z 6 x gives (7.19) Equation (7.19) is a well-known partial differential equation expressing the relationship between the temperature and the dimensions of the reactor (twodimensional pseudo-homogeneous model for packed beds). If, for a preliminary estimate, it is assumed that the temperature of the bed reaches a maximum at all radii, at the same distance z from the reactor entrance

188

Process Plant Simulation

(i.e,, with axial dispersion neglected), then at this radial section of the bed, d T / d z will be zero even though the actual temperatures will vary with x. The temperature will be maximum at the tube axis and this must not exceed 252°C (to ensure a satisfactory life of the catalyst). Hence at the tube axis, d T / d z will also be zero (symmetry condition at x = 0), but will have a finite value at all radii greater than zero. Hence the partial differential equation (7.19) becomes an ordinary differential equation at this particular value of z. That is, (7.20) Equation (7.20) can be easily converted to a Bessel equation of zeroth order, x 2 d2Y + x- dY - ( x 2 + k 2 ) y = 0 d x h

1

and solved by the method of Frobenius (solution by series using indicial equations). But in the present problem, it will be solved using Laplace transformation. Equation (7.20) can be rewritten as (multiply with x ) (7.21) Taking Laplace transforms, we get (7.22)

I:[ L

- = sT(s) - T ( 0 )

(7.23)

(7.24)

(7.25)

(7.26) T'(S)[s'

+

'TroA]

+sT(s) = PMR% kes2

(7.27)

Models in Reaction Engineering

189

Let P = P m R r , / k , and Q = PmRr,A/k,

T’(s)[s’ + Q ] + sT(s) = P / s 2 Dividing by (s2 + Q), we get .’(s)

S +T ( s )=

s2+Q

(7.28)

P (s2

(7.29)

+ Q)s2

d * -“T(s)] + -T ( s ) = P ds s2 + Q (s2 + Q)s2 S

(7.30)

Equation (7.31) is of the form dyldx + My = N , which can be solved by the integrating factor (IF) method.

IF = exp(1Md.x) = exp(,t.) s

+Q

Hence the solution of Eq. (7.30) is s2(s2

+ Q)

J G d s

+ C,

(7.3 1)

(7.32) Let P

I =JS2JfidS Put s = &tanB=

ds = &sec2Bd6

-P 1 ... I = --

=-

Q sine

-PJs2+e Q s

Now, Eq. (7.32) becomes -P d s 2 J G T ( S ) =-

Q

+Q s

+

c3

(7.33)

(7.34)

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Process Plant Simulation

Since P/Q = UA, (7.35)

Inverting the Laplace transforms,

which is a Bessel function of zeroth order and first kind. Equation (7.35) becomes 1 T = C3J0(x&) - A where C3 is an arbitrary constant, which can be evaluated as follows: BC: at x = 0, T = 158.7 "C j158.7 =

(7.36)

1 A

C,Jo(O)- -

J,(O) = 1 j158.7 =

1 C3 - A

But A = 0.00248

a 158.7 = C, -

1 0.002448

s C3 = 158.7 + 40.85 = 199.55 Now, pAH,roA - 288 x (1.075 x 10') x (5.89 x 6.92 kt? = 6450.92

Q=

>

x

0.02448

Jis = 80.32

Now Eq. (7.36) becomes T = 199.5550(80.32~)- 40.85 At the boundary, x = R, T = 149 - 93.3 = 55.7 "C

55.7 = 199.5550 (80.3213) - 40.85

5, (80.32R) = (55.7 + 40.85y199.55 = 0.4835

(7.37) (7.38)

Models in Reaction Engineering

191

Using the table of Bessel functions of zeroth order, we get R = 1.W80.32 = 0.01929 m = 19.29 mm. For safety, the diameter of the tubes that would be installed in the reactor should be less than 38.58 mm, as for this diameter the cooling rate would be just sufficient to prevent the temperature of the catalyst at the centre of the tube from exceeding 252 "C.Hence tubes of 36.7 mm diameter are recommended (5% safety).

Exercise 7.1 Solve Eq. (7.20) of the above model using the method of Frobenius and compare the end result.

7.3 Gas Absorption Accompanied by Chemical Reaction Illustration 7.3 A wetted wall column is to be used for the absorption of a gas A. The gas is consumed in the liquid phase by a first-order reaction in terms of the gas concentration of A in the liquid phase in the form of a differential equation. Solution The schematic diagram of the system is shown in Fig. 7.3, which shows a section through the liquid film, where z is the distance from the liquid inlet and x is the coordinate measured inward from the surface of the film. X-

L

L

V

Gas phase

Fig. 7.3 Wetted wall column

We make the following assumptions: 1. The film is assumed to be of uniform thickness L. 2. Although it is a poor assumption which depends upon inlet conditions (end effects), the parabolic velocity distribution v(x) will be assumed to exist right at the inlet and remain unchanged throughout the length of the column (z) (fully developed flow).

192

Process Plant Simulation

3. The gas is assumed to be pure so that there will be no gas phase resistance and the surface concentration of A in the liquid will be constant at c., 4. Variations of v and c in the y-direction are neglected. For any element of the liquid (SxSz), as shown above, to be in equilibrium, the amount of A carried in the z-direction by the bulk flow plus the amount difising in the x-direction must be sufficient to supply the chemical reaction proceeding within the element. Expressing this fact symbolically (mass balance for unity), considering a differential region Sx S. in the liquid phase, Input rate of A = Output rate of A

Consumption rate of A (depletion) by a first-order reaction in differential volume is given by kc8xS.z mol/s, where k is the first-order reaction-rate constant. At steady state, Input rate of A = output rate of A - consumption rate of A Substituting,

v is independent of z (assumption) and D (diffusivity) is constant. Hence,

(7.39) Substituting v = V[1 - ( x / L > ’ ] , Vis the maximum velocity (at x = 0), (7.40)

7.4 Reactors in Series-I Illustration 7.4 In a proposed chemical process, 1.26 kg/s of a pure liquid A is to be fed continuously to the first of a battery of two equal-sized stirred tank reactors (CSTR) operating in series. If both vessels are to be maintained at the same constant temperature so that the chemical reaction A k ’ B k 2 C takes place, estimate the size of the

Models in Reaction Engineering

193

vessels that will give the maximum yield of the product B. The specific reactions - l and k, = 8.33 x lo4 s-l at the temperature of rate constants are k, = 1.67 x the reactors, and the fluid density is constant at 961 kg/m3. Solution The schematic diagram of the system of reactors in series is shown in Fig. 7.4.

Q= 1.311x

m3/s

Fig. 7.4 Reactors in series

Volumetric flow rate of fluid, Q = 1.26/961 = 1.31113 x m3/s. Let us assume that, in general, the concentration of A leaving the vessel n is cA,,. The material balance over stage n for component A: QC.4,n-l

- Q c A ,=~ VklCA,n

(7.41)

where V is the volume of each tank.

(7.42) - CA,n = klCA,ne where 8 = V/Q is the mean residence time or holding time in each reactor. Similarly, for component B, the material balance is CA,n-l

(7.43) - CB.n = k2CB,n8 -klCA.ne The solution of Eq. (7.42) is obtained by the finite difference method. Thus, let a = k18 and p = k,B. Then Eq. (7.42) becomes ‘B,n-l

(l + a)CA,n - CA,n-l = Putting Eq. (7.44) into operational form, [E(1 + a ) - l]CA,n-l = O

(7.44)

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Process Plant Simulation

Let p1 = V(1 + a). Therefore, the solution is (7.45)

= Ml/$-’

CA,n-l

where M , is constant. Similarly, ‘A,n

(7.46)

= Mlpy

Substituting cA, from Eq. (7.46) in Eq. (7.43) gives

+ P> - CB,n-l = &1p;

cB,n(l

(7.48)

Putting it into operational form, we get

[JW+ PI - l l c B , n - l

= ffM1p;

(7.49)

Therefore, the complementary solution of Eq. (7.49) is (7.50)

= M2p’;-’

cB,n-l

where p2 = V(1 + P) and M , is constant. The particular solution of Eq. (7.49) is given by

(7.51) The complete solution is a sum of the complementary solution and the particular solution. n-1

CB,n-l

= M2p2

+

“M,p;-’

P-ff

(7.52)

(7.53)

Since the initial feed to the reactor is pure A at n = 0, cA= cA,o, and cB = 0 (for constant M , and M,).From Eq. (7.46), ‘A.0

= Mlp! ‘A,O

=

From Eq. (7.53),

(7.54)

Models in Reaction Engineering

195

(7.55) (7.56) =- acA,O

P-a

[P; - p;1

(7.57)

We know that a = k16 and p = k26. Therefore, p1 = V(1+ k,B) and p 2 = V(l + k26). (7.58)

In the present problem, n = 2 (two reactors in series) and k , ~ , , ~ / ( k-, kl ) f 0, and hence (7.60)

4i2 3 2 - 1

ae= k,

K

- k2 3

k , = 1.67 x e = 421.208 s

2

s-l

and k, = 8.33 x lo4

s-l I

= 0.5523m3 V = Q6 = 421.208 x 1.31113 x Therefore the volume of each vessel = 0.5523 m3 = 552.3 L.

Example 7.2 In a proposed chemical process, 1.26 kg/s of a pure liquid A is to be fed continuously to the first of the battery of n equal-sized stirred tank reactors (CSTR) operating in series. If all the tanks are to be maintained at the same constant temperature so that the chemical reaction A - b B k Z C

196

Process Plant Simulation

will take place, estimate the minimum number of tanks required for maximizing the yield of the product B. The volume of each tank is 750 L and the specific -4 -1 s-l and k2 = 8.33 x 10 s at the reaction-rate constants are k, = 1.67 x temperature of the reactor, and the fluid density is assumed to be constant at 96 1 kg/m3, Solution V=QB

0.6 = 1.31113 x 9 = 457.62s

- k19)-" - (1 - k29)-"1 k2 - kl Differentiating Eq. (1) with respect to n (to determine the minimum number of stages required), we get CB,n

= -[(1 kl

-- klcAo [-(l -dn k2 - kl dCB,"

- kle)-"log(1-

+ (1 - k2@"

k19)

log(1 + k29) = 01

(2)

to get the minimum number of reactors required, dc,,,/dn = 0. Therefore, [-(1 - k19)-"log(1 - k19)+ (1 - k29)-" log(1 + k2B)] = 0

(3)

(1 - k18)-"log(1 - k19)= (1 - k,@" log(1 + k2.9)

(4)

Putting the value of k , , k2, and 9, we get the minimum number of reactors required, n = 1.87 e 2 .

7.5 Reactors in Serks-II Illustration 7.5 A battery of N stirred tank reactors is arranged to operate isothermally in series. Each reactor has a volume of V m3 and is equipped with a perfect agitator so that the composition of the reactor effluent is the same as the tank contents. If initially the tanks contain pure solvent only, and at time t = 0, q m3/s of a reactant A of concentration co kmol/m3 is fed to the first tank, estimate the time required for the concentration of A leaving the Nth tank to be cNkmol/m3.The stoichiometry of the reaction is represented as

Models in Reaction Engineering kl

k3

k2

k4

197

A o B o C All the reactions are of first-order and the feed to the reactor does not contain any product B or C, but is assumed to contain a catalyst that initiates the reaction as soon as the feed enters the first reactor. Solution The mass balance for A around tank n

For B

For C

Dividing Eqs (7.61)-(7.63) by V and referring to V/q =

e as the holding time, (7.64)

dt

CB,n

e

+ klCA,n + k4CC,n

(7.65)

(7.66) Writing Eqs (7.64)-(7.66) in matrix form, we get

(i+k1) -kl

0

4

(i+

2

k3 + k 2 ) 4

3

(7.67)

198

Process Plant Simulation

or more concisely, d? - -1[ t f l - l ] - G?,, nA

(7.68)

dt e where c?~ and ? f l - , are column vectors and constant elements. For the first reactor,

is a third-order square matrix with

d?l - ?o - - - G?, dt B

.*,

(7.69)

eG1?l= j $ - e G t d t + constant =

At t = 0, t1= 0. Hence, constant = Hence,

E

eGl

B G

+ constant

-(?o/e)6-1. (7.70)

where

i is a unit matrix. For the second reactor, (7.71)

Substituting the values of cI1 and solving with the initial condition that at t = 0, E, = 0, 1 -

t, = ?e[ Z - e

-GI

-6

+

,-dr

]&2

to

(7.72)

For the third reactor, (7.73)

For the Nth reactor,

(7.74)

Equation (7.74) can be rearranged as

(i = 1)

Models in Reaction Engineering

199

Equation (7.75) is the solution for the system considered, which gives the variation of concentration as a function of time.

EXERCISES 7.1 A tubular chemical reactor of length L and cross section 1 m2 is employed to carry out a first-order chemical reaction in which a material A is converted to a product B according to chemical reaction A + B and the specific reaction-rate constant is k s-'. The feed rate is u m3/s, the feed concentration of A is c,,, and the diffusivity of A is assumed to be constant at D m2/s. If there is a volume change during the reaction, develop a mathematical model in the form of a differential equation for the concentration of A as a function of length along the reactor. 7.2 In a continuous stirred tank reactor, a pure liquid A is fed continuously at 2 kg/ s. If the vessel is maintained at constant temperature so that the chemical reaction

A~'B-&\c will take place, model and estimate the size of the vessel that will give the maximum yield of the product B. The specific reaction-rate constants k , and k2 are equal to 0.2 s-l and 0.09 s-l, respectively, and the fluid density is constant at 1000 kg/m3. 7.3 Liquid benzene is to be chlorinated batchwise by sparging chlorine gas into a reaction kettle containing benzene. If the reactor contains such an efficient agitator that all the chlorine that enters the reactor undergoes chemical reaction, and only the hydrogen chloride gas liberated escapes from the vessel, estimate how much chlorine must be added to give the maximum yield of monochlorobenzene. The reaction can be assumed to take place isothermally at 55 "C when the ratios of the specific reaction-rate constants are k,/k2 = 8.0 and k2/k3= 30.0, where k , , k,, and & refer to reactions (l),(2), and (3), respectively, given below:

+ C&@ + HC1

(1)

Cl2 + C6H4cl2+ HC1

(2)

C,jH,j + c12 C,&Cl+

C6H4C12+ Cl2 + C6H3C13+ HC1 (3) 7.4 1.0168 kg/s of tallow mixed with 0.286 kg/s of high-pressure hot water is fed into the base of a spray column operated at a temperature of 232.2 "C and a pressure of 41.37 x lo5 N/m2. 0.5091 kg/s of water at the same temperature and pressure is sprayed into the top of the column and descends in the form of droplets through the rising fat phase. Glycerin is generated in the fat phase by hydrolysis and is extracted by the descending water so that 0.7 kg/s of final extract containing 12.16%

200

Process Plant Simulation

glycerin is withdrawn continuously from the column base. Simultaneously, 1.1214 kg/s of fatty acid raffinate containing 0.24%glycerin leaves the top of the column. If the effective height of the column is 21.94 m and the diameter 0.6604 m, the glycerin equivalent tallow 8.53%and the distributibn ratio of glycerin between the water and the fat phase at the column temperature and pressure is 10.32, estimate the concentration of glycerin in each phase as a function of column height. Also find the fraction of the tower height required principally for the chemical reaction. The hydrolysis reaction is pseudo-first-order, and the specific reaction-rateconstant is 0.17 s-'. 7.5 0.126 kg/s of ethyl alcohol is to be esterified by reacting it with 0,1071 kg/s of acetic acid in a battery of continuous stirred tank reactors, each of 0.85 m3capacity and maintained at 100°C. If the equilibrium relation is such that 75.2%of the acid will be esterified, estimate the number of reactors required for 60% conversion. At lOO"C, the specific reaction rate for esterification is 7.93 x 10- 6m3 kmols, and that for the hydrolysis of the ester is 2.72 x 10- 6m3 kmols. The density of the reaction mixture can be assumed constant at 865 kg/m3. 7.6 Acetic anhydride is to be hydrolysed in a laboratory stirred tank reactor battery m3 of anhydride solution consisting of three vessels of equal size. 1.8 x of concentration 0.21 kmol/m3 is charged into each vessel at 40 "C, and 1 x 10-5 m3/s of a solution containing 0.137 km0Ym3 is continuously fed to the first reactor. Estimate the time required for the reactor system to settle down to steady-state operation. The specific reaction-rate constant for the hydrolysis at 40°C is 6.33 x 10-3s-'.

PART 111

TREATMENT OF EXPERIMENTAL RESULTS

IN

PROCESS PLANT simulation, a combination of analytical and empirical approaches

is followed in modelling, for the prediction of good results matching the actual plant data. As mentioned in Part I, the analytical models can be fine-tuned by incorporating empirical correlations based on experimentally obtained plant data into the models. There is a need to treat the experimental results, as there is a possibility of error propagation through various mathematical operations such as addition, subtraction, multiplication, and division. Similarly, one has to be conversant with the data-regression techniques used for proposing the empirical correlations based on the experimental data. This part deals with these two aspects, error propagation and data regression, in detail. Two of the most widely used dataregression techniques (method of averages and method of linear least squares) are described and explained using examples.

CHAPTER 8. Error Propagation and Data Regression

CHAPTER 8 ERROR PROPAGATION AND DATA REGRESSION

Once the data are generated from the model equation using simulation, the developed model is to be validated with experimentally generated data. Before comparison, the experimentally generated data, which are subject to error, have to be treated to avoid the propagation of these errors.

8.1 Propagation of Errors Any experimentally determined quantity is subject to error, and hence any calculated result, which is based on experimental evidence, is limited in accuracy. The determination of the derived error from the observed error is a calculation of some importance since it helps identify where the experimental technique should be improved upon. The way in which errors accumulate is governed by different rules depending upon the type of calculation performed (i.e., addition, subtraction, multiplication, etc.).

8.1.1 Propagation Through Addition If a quantity z is calculated from z=x+y (8.1) where x and y are two measured variables, then any error in determining x or y will result in an error in z. If x is the experimental reading for the variable x, then the error in the determination of x is given by I X ’ - X I < sx (8.2) where ax is the absolute error in x. Sx is usually assigned by the observer, with due regard to the manner in which the measurement was made. A similar argument leads to the definition of 6 y , the absolute error in y.

204

Process Plant Simulation

sz = sx + s y

(8.3)

8.1.2 Propagation Through Subtraction If a quantity z is calculated from z=x-y (8.4) then sz = s x + s y (8.5) Thus, the absolute error in the result of any calculation involving only addition and subtraction is the sum of the absolute error in the constituent parts. absolute error ( 6 x ) result ( x ’ ) The relative error is a dimensionless value and is expressed as a percentage. Relative error =

8.1.3 Propagation Through Multiplication and Division For multiplication or division, if a quantity z is calculated from z = x y or

z = -X

Y the relative errors are defined as

SX and yr = SY xr = X’ Y’

(8.7)

If SZ

z,

=-

z’

= x’y’

Z’

where

(8.9)

then (8.10) Yr Thus the relative error in multiplication and division is the sum of the relative errors in the constituent parts. Zr = X r +

8.2 Sources of Errors There are many sources of errors, which should be assessed before quoting the accuracy of any determination. These can be classified as 1. Errors of measurement

Error Propagation and Data Regression

205

2. Precision errors 3, Errors of method

All these errors can only be estimated, and rarely measured; consequently an error is usually given upto only one or two signijicantfigures. A brief account of these sources of errors is given below.

8.2.1 Errors of Measurement Errors of measurement are due to physical limitations of reading a scale. Without a Vernier attachment, a length of 2 cm could easily be in error by 0.5 mm, which is 2.5%.

8.2.2 Precision Errors Precision errors are the built-in errors of the apparatus. For example, the scales on mercury in a glass thermometer assume that the bore is uniform, so that a reading of 25 OC could easily be in error by 1 "C (i.e., 4%) due to an uncalibrated scale.

8.2.3 Errors of Method Errors of method include faults such as neglecting heat losses, assuming constant molal overflow, or neglecting back-mixing in a tubular reactor.

8.2.4 Significant Figures A significant figure is any digit, except zero when it is positioned after the decimal point. It is usual to imply the accuracy of a reading by the number of significant figures quoted. That is, if a reading is known to be in error by about 2%, three significant figures are barely justified; whilst an error of 10% justifies only two significant figures. To quote too many significant figures is misleading since it implies a false sense of accuracy; whilst quoting too few squanders the accuracy which has been dearly bought. One extra significant figure should always be carried through a calculation to smooth the calculation error due to rounding off.

8.3 Data Regression (Curve Fitting) First, we discuss various theoretical properties, which are very important in dataregression techniques.

8.3.1 Theoretical Properties If a set of experimental data is to be represented by an empirical equation, the equation must possess the correct theoretical properties, such as 1. The origin

206

Process Plant Simulation

2. Starting gradient 3. Efficiencies 8.3.1.1 The origin

1. The pressure drop Ap across a length of pipe varies with the volumetric flow rate Q.But it is certain that AP = 0 at Q = 0. Hence, the origin is more important than any single experimental point. 2. Similarly, when time is the independent variable, the intercept at zero time is a boundary condition (i.e., initial condition). 3. In the same way, if a chemical reaction is being studied by plotting the concentration of one of the products as a function of time (c versus t), the origin must be a point on the curve, that is, c = c, at t = 0. 8.3.1.2 Starting gradient In most problems involving axial symmetry, the gradient of the curve tends to zero: dv - dT - -dc = O a t r = O d r dr dr

Examples of this situation are 1. Velocity profiles in cylindrical pipe flow. 2. Temperature profiles within a cylindrical conductor. 3. Variations of a local mass-transfer coefficient near the front pole of a dissolving sphere. 8.3.1.3 Efficiencies

A lot depends upon the context, but most of the curves (when plotting efficiency as a function of any dependent variable such as volumetric flow rate, etc.) either approach a limit asymptotically (increase up to some extent and become constant afterwards) or pass through a maximum value before decaying to zero (increase initially and decrease subsequently to zero). It is certain, however, that an efficiency curve cannot be a straight line over an appreciable range.

8.3.2 Data-regression Methods There are two most widely used methods on data regression: 1. Graphical methods 2. Analytical methods (i) Method of averages (ii) Method of least squares

Error Propagation and Data Regression

207

8.3.2.1 Graphical methods

Assume that an equation of the form y = a + bx" is known to fit a set of results for x and y ; the values of a, b, and n are to be'estimated. The best way of finding these values graphically is to assume a value of a and plot y - a against x on a log-log graph paper.

y

- a = bx"

ln(y - a ) = lnb + n lnx

1n(Y - a )

1: a too small 2: a correct 3: a too large

In x Fig. 8.1 Log-log plot Of y = a + bfl

If the value of a assumed is correct, the result will be a straight line of slope n and intercept In b, but this is unlikely. Too large a value for a will give a curve which is convex upwards, and too small a value for a will give a curve which is concave upwards as shown in Fig. 8.1. One should be careful and should take a wide range of data for fitting an equation. Note Whether the curve is fitted by the graphical method or the analytical method

(method of averages or least squares), it is necessary to choose the appropriate type of curve by physical modelling (dimensional analysis).

Example 8.1

2) P

Nu = c(R e)m(P r )n( = c1(Re)" (Pr)"

(wall effect: p = 0.14)

(if wall effects are neglected)

= In (Nu) = In c1 + rn(Re) + n(Pr)

200

Process Plant Simulation

Example 8.2

8.3.2.2 Analytical methods The following are the two most widely used analytical methods for data regression: 1. Method of averages 2. Method of linear least squares Method of averages Once the type of curve desired to fit to the experimental data is chosen, the problem reduces to determining the values of the constants (certain parameters) in the equation so that the bestfit is obtained. In graphical methods, the best fit is a matter of opinion, but in the present method a set of average points is defined and the best-fit curve is one that passes through the average points. Suppose that nine experimental values of a variable Z are available at nine different known values of x , and the best-fit curve of the type Z = A + Bx + Ce" (8.11) is to be chosen to represent them. As the values of Z are experimental, it is unlikely that any set of values (xn,2,) will satisfy the best-fit curve exactly [Eq. (8.1l)]. Therefore, the experimentalresults are substituted into Eq. (8.1 l),and an extra term is included for the unknown error. Thus Z , - A - Bx, - Ce"' = R, Z 2 - A - B x 2 - CeX2= R 2 Z , - A - B x , - Ce"' = R , Z , - A - B x , - CeX4= R , (8.12) Z 5 - A - B x , - Ce"' = R , Z 6 - A - B x 6 - Cex6= R 6 Z , - A - B x , - Ce"' = R , Z , - A - B x , - CeX8= R , Z , - A - B x , - Cex9= R , where R, are the error terms. In order to determineA , B, and C anz..,jtically, three equations are required, and these can be obtained from the set of nine equations [Eq. (8.12)] by assuming relationships between the error terms. Assuming that R, + R, + R, = 0

and summing the first three equations of Eq. (8.12) gives

(8.13)

Error Propagation and Data Regression 3

3

3

c Z n - 3A - B C x , - C x e X n= 0 fl=l

n=l

209

(8.14)

n=l

Dividing by the number of points used ( 3 ) gives

(

z, + z, + z, )-A-B(x1':+x3)-C[

ex' + e x 2 + ex' 3

(8.15) Equation (8.15) can be obtained directly from Eq. (8.11) by using an average point to represent the three actual points. It should be noted that the average point is determined in a special manner, which depends on the functions arising in the type of Eq. (8.1 1) chosen. Two further equations, similar to Eq. (8.15), can be obtained by arranging the remaining points in two groups. Thus,

(8.16)

(

z, + z, + z,

ex7

)-A-13(x7+;+xq)-C[

+ ex* + ex9 3

(8.17) Since every symbol with a suffix is a known experimental result, only A , B, and C are unknown in the three equations [Eqs (8.15)-(8.17)], and hence these can be determined (there are three unknowns and three equations). In applications of the method of averages, the following points must be observed: 1. Points must be arranged in ascending values of independent variable x. 2. The number of groups must equal the number of unknown parameters. 3. Groups should contain approximately equal number of points: 9 points + 3 groups x 3 points 8 points + 3 points + 2 points + 3 points 10 points -$ 3 points + 4 points + 3 points 4. Each experimental point should be used only once. 5. The appropriate average must be taken.

Method of least squares Usually, this method is used to fit a set of data to linear or non-linear equations depending on the data. Hence, the formulae will be derived by determining the values of m and c in case of a linear relationship (or by determining the values of the constant and the exponents in case of a non-linear relationship), which give the best least-mean-squares fit of the equation.

210

Process Plant Simulation

Let us consider a linear relationship between the dependent and independent variables (i.e., y and x ) of N pairs of data represented by the following equation: y = m + c (8.18) Similar to the method of averages, each pair of results is substituted into Eq. (8.18) and an extra term is introduced to allow for the unknown error: y n - mn- c = R,

(8.19)

Rnmay be either positive or negative. Therefore, to obtain a positive representation for the errors, Eq. (8.19) is squared to give y,' + m2x; + c2 - 2 m n y n - 2cyn + 2 c m n = R,2 (8.20) Equations (8.20) for the individual experimental points are then summed to give

n=l

n=l

n=l

n=l

(8.21) n= 1

n=l

Equation (8.21) gives an expression for the sum of the squares of the error terms. The method ofleast squares defines the best straight line as the one for which the sum of the squares of the error terms is a minimum. The left-hand side of Eq. (8.21) will have a turning value (0) when its partial derivatives with respect to both m and c are zero. Thus,

i.e., N

N

N

n=l

n=l

n=l

2 m C x , ' - 2 C xnyn + 2 c C x n = o

(8.22)

and N

2Nc

-2

N

C y n+2mCxn=0 n=l

(8.23)

n=l

Solving Eqs (8.22) and (8.23) [i.e., multiplying Eq. (8.22) by Nand Eq. (8.23) by Ex,, and subtracting to eliminate c], we get (8.24)

Error Propagation and Data Regression

211

Substituting this expression for m in Eq. (8.23) and solving for c, we get (8.25) where all sums must be taken over all experimental points. The best way to use formula (8.25) is to construct a table with four columns headedx,, y,, x,y, and complete it for the experimental results. The sums of the columns will then give all the values required by Eqs (8.24) and (8.25).

Comparison of analytical methods for curve fitting The methods of averages and least squares are based on different definitions of the best-fit curve, which fits the data. If a computing machine is available, the method of least squares should always be used, since it is more soundly based. Without the aid of a machine, however, the calculation is rather tedious and liable to arithmetical error; and the method of averages, which is shorter, should be used to give an adequate result.

8.3.3 Problems on Data Regression Example 8.3 Using method of least squares, fit the data given in Table 8.1 to the y = m +c type of equation, and compare the calculated values with the actual values of y . Table 8.1 Data on x and y X

1.1

2.0

3.2

4.0

6.1

7.0

Y

10

13.2

16

20

25.2

27.9

Solution The data in Table 8.1 were generated with some deviations from y = 3x Table 8.2).

+ 7 (see

Table 8.2 Source data for Table 8.1 X

1

2

3

4

5

6

7

Y

10

13

16

19

22

25

28

We may use the method of least squares:

212

Process Plant Simulation

The terms required for computing m and c using the above equations, such as c x,, c y n , x: , and c x,y, , are determined by tabulating the variables as shown in Table 8.3.

c

Table 8.3 Computing the required summation terms

1.1 2.0 3.2 4.0 6.1 7.0

CX,

.:

1.21 4.0 10.24 16.0 37.21 49.0

11.0 26.4 51.2 80.0 153.72 195.30

CX,' = 117.66

C X , ~=, 517.62

10 13.2 16 20 25.2 27.9

= 23.4

m=

Cy,

= 112.3

(6 x 517.62) - (23.4 x 112.3) (6 x 117.66) - (23.4)'

- 3 105.72 - 2627.82 705.96 - 547.56 = 3.017 and c=

( 1 17.66)( 1 12.3) -

(5 17.62)(23.4) 158.4

= 6.95 y = 3.017~+ 6.95 Table 8.4 shows a comparison of yactand ycal. Table 8.4 Comparison of ye,, and ycal X Yac1

Ycal

1.1 10.0 10.26

2.0 13.2 12.98

3.2 16.0 16.6

4.0 20.0 19.02

6.1 25.2 25.3

7.0 27.9 28.06

Example 8.4 The reaction-rate constant for the decomposition of a substituted dibasic acid has been determined at various temperatures as given in Table 8.5. Use the method of least squares to determine the activation energy E in the equation. k = Ae-E'RT,where T is measured in degrees Kelvin.

Error Propagation and Data Regression

213

Table 8.5 Data on temperature and rate constant

T (“C)

50

70.1

89.4

101

k x lo4 (h-’)

1.08

7.34

45.4

138

y=c--mT

Taking the data given in the problem and making the mathematical transformations shown above, the data on the transformed dependent and independent variables 01 and x) are shown in Table 8.6. Table 8.6 Linearly transformed data on dependent and independent variables

X

T 6) 1/T

In k

Y

323 3.096 x -9.1333

343.1 2.915 x lo-’ -7.217

362.4 2.759 x -5.3 948

374 2.674 x lo-’ 4.2831

The terms required to estimate the slope (m)and intercept (c)of the linear equation, such as the summations of x,y , x2,xy, are determined by tabulating these variables along with their summations, as shown in Table 8.7. Table 8.7 Computation of required summation terms Y

X2

XY

-9.1333 -7.217 -5.3948 4 . 2 8 31

9.585 x lo-’ 8.497 x 7.612 x lo-’ 7.150 x lo-’

-0.02 83 -0.0210 -0.0149 -0.01 15

Zyy-26,0282

xx2=32.844 x lo-’

C~y=-0.0757

X

3.096 x 2.915 x 2.759 x 2.674 x

lo-’ lo-’ lo-’ lo-’

C X =11.444 x 10”

4(-0.0757) - (11.444 x 10-3)(-26.0282) m = N C x , y , - C X ,Cy, N C X~ (Cx,Y 4(32.844 x - (11.444 x

-

-0.3028 + 0.2979 1.31376 x - 1.30965 x

= - 11,926.087

214

Process Plant Simulation

-

(32.844 x 10-6)(-26.0282) - (-0.0757)(11.444 x 4.10864 x lo-'

= 27.8457 = 1nA A = 1.23946 x 10l2

=j

:.

E = R x 11,962.087 R = gas constant = 8314 NmkmolK = 1.987 caVgmolK = 82.06 atm cm3/gmolK :. E = 8314 x 11,926.087 = 99.153 x lo6 Nmkmol = 1.987 x 11,926.087 = 23.697 x lo3 caVgmol = 82.06 x 11,926.087 = 97.865 x lo4 atmcm3/gmol

Example 8.5 The thermal conductivity (W/mK) of methyl chloride (gas) at 1O"C, 37.8 "C, 65.6 OC, 93.3 "C, 121.1 "C, 148.9 "C, 176.7 "C, and 204.4 "C is 0.00986, 0.0119, 0.014, 0.0159, 0.018, 0.023, 0.022, and 0.0241, respectively Estimate the thermal conductivity of methyl chloride at 463 K and 353 K. Solution Let us solve this example using both the methods of data regression. Method of averages k=A+BT k -A - BT = 0 (A and B are two unknowns) k,

+ k, + k3 + k, 4

- A - B (TI + T2 + T3 + T4) = O

4

(8.26) (8.27)

Solving Eqs (8.26) and (8.27) 0.0081 - 111.1B = 0 3

B = 0.0081/111.1 = 7.291 x

Substituting B in Eq. (8.26), we get 0.0129 - A - (51.7 x 7.291 x lod5)= 0 3

A = 0.0129 - 3.76944 x

= 0.009131

215

Error Propagation and Data Regression

Therefore, the relation between k and T is (Tin "C,k in W/mK) k = 0.009131 + (7.291 x 10-5)T

(8.28)

The value of k at 463 K (190 "C) is 0.02298 W/m K and at 353 K (80 "C) is 0.01496 W/m K Method of least squares (8.29)

(8.30)

Table 8.8 shows the values of T,,, k,, T:, and T,,k, and their summations. Table 8.8 Computation of required summation terms

10 37.8 65.6 93.3 121.1 148.9 176.7 204.4

0.00986 0.01 19 0.014 0.0159 0.018 0.02 0.022 0.024 1

100 1,428.84 4,303.36 8,704.89 14,665.21 22,171.21 3 1,222.89 41,779.36

0.0986 0.44982 0.9184 1.48347 2.1798 2.978 3.8874 4.92604

Total = 8N CT,,= 857.8 Ck,=0.13576 CT:= 124,375.76 CT,,k,,= 16.92153

.:

B=

(8 x 16.92153) - (857.8 x 0.13576) (8 x 124,375.76) - (857.8)'

- 135.37224 - 116.45492 995,006.08 - 735,820.84 - 18.917312 259,185.24 -

= 7.29876 x

216

Process Plant Simulation

A=

(124,375.76 x 0.13576) - (16.92153 x 857.8) 259,185.24 16,885.253- 14,515.288 259,185.24

The relation between k and T is (Tin "C, k in W/mK)

k = 0.0091439 + (7.299 x IO-')T

(8.3 1)

Therefore, the value of k at 463 K (190 "C) is 0.023 W/m K and at 353 K (80 "C) is 0.01498 W/m K.

EXERCISES 8.1 Discuss the ways in which the propagation of errors occurs. 8.2 What are the various sources of errors? 8.3 Describe various data-regression techniques briefly. 8.4 Compare the graphical and analytical methods of data-regression techniques. 8.5 Compare the method of averages and the method of linear least squares. 8.6 Regress the data given in Table 8.9 to the form y = a( 1 + x * ) ~using the method of averages and the method of linear least squares. Table 8.9 Data on x and y X

0.0

y

1.000

0.5

1.0

0.894 0.707

1.5

-2.0

0.555 0.447

2.5

3.0

3.5

4.0

0.371

0.316

Oi274

0.242

8.7 Use the method of least squares and the method of averages to fit the best equation of the type Nu = aPrn to the data given in Table 8.10. Table 8.10 Experimental data on Pr and Nu to be regressed Pr

Nu

Pr

Nu

Pr

Nu

0.46 0.53 0.63 0.74 4.2 5.6 3 .O 5.0

24.8 26.5 28.5 30.0 60.3 69.0 58.4 70.7

10.0 17.7 18.6 25.3 41.0 37.0 58.5 95.0

84.5 115 115 150 170 165 193 245

32 31.6 70.3 93 185 340 590 55

140 127 189 245 315 380 480 195

Error Propagation and Data Regression

217

8.8 The top product leaving a batch distillation column is being continuously reprocessed so that it is difficult to measure directly the amount produced. However, the flow rate can be continuously measured and samples taken at hourly intervals giving the results shown in Table 8.11. (a) What is the total amount produced during a period of 8 h? (b) What proportion of this amount is the more volatile component? Table 8.1 1 Data on samples taken in hourly intervals Time (h) 0 90 More volatile component (%) Flow rate (lblmin) 60

1 92

2 91

3 91

62

63

62

4 89

5 87

6 84

7 80

8 75

62

60

59

57

57

8.9 The experimental data obtained on power-law fluids for shear stress as a function of rate of shear (velocity gradient) are given in Table 8.12. (a) Regress the data into the power-law form correlation using the method of least squares. (b) Also propose the correlation using the method of averages. Table 8.12 Data on x and y to be regressed ~~

~

X

0.2

0.5

1 .o

2.0

3 .O

4.0

Y

3.2

3.7

4.1

8.1

13.7

22.6

8.10 From the data given in Table 8.13 calculate the value of the constants a and b in the equation y 2 + a2x2- 2abx - sby - 2u.q + b2 = 0. Table 8.13 Data on x and y to be regressed x

2.24

4

23

56.1

90

144

210

272

324

y

376

360

256

182

132

81

42.2

25

9

8.11 The data shown in Table 8.14 were reported for the vapour-liquid equilibrium for binary mixtures of benzene and toluene, where x and y represent the mole fraction of benzene in the liquid and vapour, respectively. Indicate how these data should be plotted to give a straight line if the relative volatility is constant. Fit the best straight line to the points on such a graph by (a) visual inspection of the best straight line through the points and (b) the method of averages. Determine the average relative volatility of the mixture using the straight lines obtained.

218

Process Plant Simulation

Table 8.14 Data on x and y to be regressed X

0.167

0.333

0.500

0.667

0.833

Y

0.320

0.550

0.710

0.830

0.930

8.12 Apply the data-regression technique to the data given in Table 8.15 (48data points) on adsorption studies in waste water treatment by Babu and Ramakrishna (2002b). What are your observations in terms of the limitations of the technique for such problems? Table 8.15 Experimental data on adsorption studies (Babu & Ramakrishna 2002b) 1

2

3

4

5

6

7 ~

~

8 ~~

9

10

11

12

0.4 6.5 4 120

0.2 6.5 2 60

0.85 10 32 300

0.85 10 16 960

21

22

23

24

0.845 10 5 5

0.9 10 5 15

0.936 10 5 30

34

35

36

~

0.5 0.7692 0.7692 0.6 6.5 6.5 3 6.5 8 16 2 32 120 120 240 30

5

0.5 3 16 7.5

0.5 3 8 22.5

0.5 3 4 25

13

14

15

16

17

18

19

PRE 0.5 ci 10 cd 32 t, 5

0.5 10 16 30

0.5 10 8 90

0.5 20 32 15

0.5 20 16 30

0.5 20 8 80

0.5 0.5 0.5 30 30 30 3 2 1 6 8 30 60 270

25

26

27

28

29

30

31

PRE c, cd f,

0.5 3 32

PRE 0.944 0.944 0.951 0.951 0.951 0.951 c, 10 10 10 10 10 10 cd 5 5 5 5 5 5 f, 45 60 100 120 150 180

37

PRE 0.68 ci

cd t,

10 2.5 120

38

39

0.68 0.68 10 10 2.5 2.5 150 180

40

41

42

0.423 0.453 0.506 10 10 10 1 1 1 5 15 30

20

32

33

0.544 0.597 0.634 0.634 0.649 10 10 10 10 10 2.5 2.5 2.5 2.5 2.5 5 15 30 45 60

43

44

45

46

47

0.68 10 2.5 100

48

0.513 0.529 0.544 0.544 0.544 0.544 10 10 10 10 10 10 1 1 1 1 1 1 120 150 180 100 45 60

The required correlation should be of the form PRE = Kc,mc:t:, where PRE is the pollutant removal efficiency, ciis the initial concentration of the pollutant in mgL, cd is the adsorbent (carbon) dosage in g/L, and t, is the contact time in min.

PART IV

0 PTIMlZATl0N

THIS

PART DEALS with various traditional and non-traditional optimization techniques. Chapter 9 covers the traditional optimization techniques such as analytical methods, the constrained optimization (Lagrangian multiplier) method, gradient methods of optimization (steepest descent), the sequential simplex method, Rosenbrock method, and other methods (golden section search, Brent’s method, quasi-Newton methods, direction set methods, linear programming, etc.). A comparison of the various methods is made and the advantages and limitations of each of these methods are described with suitable examples. The limitations of traditional optimization techniques are exposed; especially, the problem of getting trapped into local optima (incapability of ensuring global optimal solutions) is demonstrated in Chapter 10, and novel (non-traditional) optimization techniques that ensure optimalhear-optimal solutions are discussed. Chapter 10includes various non-traditional optimization techniques such as simulated annealing, genetic algorithms, differential evolution, and other evolutionary computation techniques (evolution strategies, evolutionary programming, genetic programming, and other population-based search algorithms). The stepwise procedures of these evolutionary computation techniques are demonstrated by solving numerical examples.

CHAPTERS 9. Traditional Optimization Techniques 10. Non-traditional Optimization Techniques

CHAPTER 9

TRADITIONAL OPT1MlZATlON TECHNIQUES

Optimization is defined as the method of minimizing or maximizing a function of several variables, i.e., finding those values of the coordinates for which the function takes on the minimum or maximum value. The maximum of a functionflx) is the negative of the minimum of -Ax). Symbolically, it can be represented as max { f ( x ) }= - min { - f ( x ) }

(9.1)

9.1 Limitations of Optimization The point at which the function takes its minimum value is ,called the minimizer. Computer users often misuse minimization methods, because a variety of problems can be formulated in terms of optimization, even though that is rarely the best method of attacking these problems (Antia 1991). For example, we can take a set of non-linear equations and produce a positive function by taking the sum of the squares of the left-hand sides (the right-hand sides being zero). The solution of this set of equations will coincide with the minimizer of this function. However, this technique should be used only as a last resort, when all other methods for solving the system of non-linear equations have failed. Unfortunately, it is very often the first thing that users try. The problem here is that the amount of computations required using this approach could be an order of magnitude larger than what would be required by a method for solution of a system of non-linear equations. Further, the accuracy of the computed minimizer is, in general, much lower than that of the calculated roots. The reliability of optimization methods is also rather low. In fact, till very recently there was no practical method for finding the global minimizer of a function of several variables. Here, by the global minimizer we mean the point where the function assumes its lowest value out of the entire range of coordinate values allowed.

222

Process Plant Simulation

9.2 Applications of Optimization Optimization problems arise in almost all fields where numerical information is processed (science, engineering, mathematics, economics, commerce, etc.). In science, optimization problems arise in data fitting, variational principles, solution of differential and integral equations by expansion methods, etc. Engineering applications are in design problems, which usually have constraints in the sense that variables cannot take arbitrary values. For example, while designing a bridge, an engineer will be interested in minimizing the cost, while maintaining a certain minimum strength for the structure. Optimizing the surface area for a given volume of a reactor is another example of constrained optimization. While most formulations of optimization problems require the global minimum to be found, mosi of the methods are only able to find a local minimum. A function has a local minimum, at a point where it assumes the lowest value in a small neighbourhood of the point, which is not at the boundary of that neighbourhood. To find a global minimum we normally try a heuristic approach where several local minima are found by repeated trials with different starting values or by using different techniques. The different starting values may be obtained by perturbing the local minimizers by appropriate amounts. The smallest of all known local minima is then assumed to be the global minimum. This procedure is obviously unreliable, since it is impossible to ensure that all local minima have been found. There is always the possibility that at some unknown local minimum, the function assumes an even smaller value. Further, there is no way of verifying that the point so obtained is indeed a global minimum, unless the value of the function at the global minimum is known independently. On the other hand, if a point i s claimed to be the solution of a system of non-linear equations, then it can, in principle, be verified by substituting in equations to check whether all the equations are satisfied or not. Of course, in practice, the round-off error introduces some uncertainty, but that can be overcome. Owing to these reasons, minimization techniques are inherently unreliable and should be avoided if the problem can be reformulated to avoid optimization. However, there are problems for which no alternative solution method is known and we have to use these techniques. The following are some examples. 1. Not much can be said about the existence and uniqueness of either the global or the local minimum of a function of several variables. 2. It is possible that no minimum of either type exists, when the function is not bounded from below [e.g.,Ax) = XI. 3. Even if the function is bounded from below, the minimum may not exist [e.g.,Ax) = e"]. 4. Even if a minimum exists, it may not be unique; for exarnple,Xx) = sin x has an infinite number of both local and global minima.

Traditional Optimization Techniques

223

5. Further, infinite number of local minimum may exist, even when there is no global minimum [e.g.,Ax) = x + 2 sin x ] . 6. If the function or its derivative is not continuous, then the situation could be even more complicated. For example,Ax) = & has a global minimum at x = 0, which is not a local minimum [i.e.,Ax) = 01.

Optimization in chemical process industries infers the selection of equipment and operating conditions for the production of a given material so that the profit will be maximum. This could be interpreted as meaning the maximum output of a particular substance for a given capital outlay, or the minimum investment for a specified production rate. The former is a mathematical problem of evaluating the appropriate values of a set of variables to maximize a dependent variable, whereas the latter may be considered to be one of locating a minimum value. However, in terms of profit, both types of problems are maximization problems, and the solution of both is generally accomplished by means of an economic balance (trade-@ between the capital and operating costs. Such a balance can be represented as shown in Fig. 9.1, in which the capital, the operating cost, and the total cost are plotted againstf, which is some function of the size of the equipment. It could be the actual size of the equipment; the number of pieces of equipment, such as the number of stirred tank reactors in a reactor battery; the frames in a filter press; some parameter related to the size of the equipment, such as the reflux ratio in a distillation unit; or the solvent-to-feed ratio in a solvent extraction process. Husain and Gangiah (1976) reported some of the optimization techniques that are used for chemical engineering applications.

f (size of equipment) Fig. 9.1 Capital-operating cost trade-off

224

Process Plant Simulation

9.3 Types of Optimization Optimization in the chemical field can be divided into two classes: 1. Static optimization 2. Dynamic optimization

9.3.1 Static Optimization Static optimization is the establishment of the most suitable steady-state operation conditions of a process. These include the optimum size of equipment and production levels, in addition to temperatures, pressures, and flow rates. These can be established by setting up the best possible mathematical model of the process, which is maximized by some suitable technique to give the most favourable operating conditions. These conditions would be nominal conditions and would not take into account the fluctuations in the process about these nominal conditions. With steady-state optimization (static Optimization), as its name implies, the process is assumed to be under steady-state conditions, and may instantaneously be moved to a new steady state, if changes in load conditions demand so, with the aid of a conventional or an optimization computer. Steady-state optimization is applicable to continuous processes, which attain a new steady state after a change in manipulated inputs within an acceptable time interval. The goal of static optimization is to develop and realize an optimum model for the process in question.

9.3.2 Dynamic Optimization Dynamic optimization the establishment of the best procedure for correcting the fluctuations in a process used in the static optimization analysis. It requires knowledge of the dynamic characteristics of the equipment and also necessitates predicting the best way in which a change in the process conditions can be corrected. In reality, it is an extension of the automatic control analysis of a process. As mentioned earlier, static optimization is applicable to continuous processes which attain a new steady state after a change in manipulated inputs within an acceptable time interval. With unsteady-state (dynamic)optimization, the objective is not only to maintain a process at an optimum level under steady-state conditions, but also to seek the best path for its transition from one steady state to another. The optimality function then becomes a time function, and the objective is to maximize or minimize the time-averaged performance criterion. Although similar to steadystate optimization in some respects, dynamic optimization is more elaborate because it involves a time-averaged function rather than individual quantities. The goal of control in this case is to select at any instant of time a set of manipulated variables that will cause the controlled system to behave in an optimum manner in the face of any set of disturbances.

Traditional Optimization Techniques

225

Optimum behaviour is defined as a set of output variables ensuring the maximization of a certain objective or return function, or a change in the output variables over a definite time interval such that a predetermined functional value of these output variables is maximized or minimized. As mentioned earlier, the goal of static optimization is to develop and realize an optimum model for the process in question, whereas dynamic optimization seeks to develop and realize an optimum control system for the process. In other words, static optimization is an optimum model for the process, whereas dynamic optimization is the optimum control system for the process.

9.4 Methods of Optimization Optimization is categorized into the following five groups: 1. Analytical methods: (a) Direct search (without constraints) (b) Lagrangian multipliers (with constraints) (c) Calculus of variations (examples include the solution of the Euler equation, optimum temperatureconditions for reversible exothermic reactions in plug-flow beds, optimum temperature conditions for chemical reactors in the case of constraints on temperature range, Multilayer adiabatic reactors, etc.) (d) Pontryagin’s maximum principle (automatic control) 2. Mathematical programming: (a) Geometric programming (algebraic functions) (b) Linear programming (applications include the manufacture of products for maximum return from different raw materials, optimum utilization of equipment, transportation problems, etc.) (c) Dynamic programming (multistage processes such as distillation, extraction, absorption, cascade reactors, multistage adiabatic beds, interacting chain of reactors, etc.; Markov processes, etc.) 3. Gradient methods: (a) Method of steepest descent (ascent) (b) Sequential simplex method (applications include all forms of problems such as optimization of linear and non-linear functions with and without linear and non-linear constraints, complex chemical engineering processes, single and cascaded interacting reactors) 4. Computer control and model adaptation: All forms (complex chemical engineering systems) 5. Statistical optimization: (a) Regression analysis (non-deterministic systems)

226

Process Plant Simulation

(b) Correlation analysis (experimental optimization and designs: Brandon method, Box-Wilson method, etc.) Though, for completeness, all the methods of optimization are listed above, let us restrict our discussion to some of the most important and widely used methods of optimization.

9.4.1 Analytical Methods of Optimization There are classical methods for maximizing or minimizing a function, used when the function to be optimized is specified analytically and the number of independent variables is limited. When a process can be described by a small number of variables, it is possible to find the maxima or minima by differentiating, and equating the resulting expressions to zero. These are then solved in order to find the optimum conditions. However, as the number of variables increases, the solution of the resulting equations becomes prohibitive. With a large number of variables, one runs into the curse of dimensionality, and analytical methods become difficult to use. Constraints, too, may curb the usability of analytical methods. Because of this, analytical methods find limited use in practice. Then one has to go for gradient methods (e.g., the steepest descent method). 9.4.1.1 Problems involving two variables (one independent variable) It is possible to establish the optimum conditions for a process or operation by determining the maxima of an equation containing one independent and one dependent variable, though it is rare. For example, the optimum number of stirred tanks for a given time can be evaluated for the ‘reactors in series’ model. The technique used is differentiating and equating to zero to find the maxima or minima (e.g., dc$dn = 0). However, it must be emphasized that such an optimum will, in all probability, be a curtailed optimum, and should be treated cautiously, because one or two parameters have been specified arbitrarily. 9.4.1.2 Problems involving three variables (two independent variables) Many simple engineering operations can be solutions of an equation containing two independent variables. Let an operation be represented by the equation z =Ax, y ) (9.2) where x and y are two independent variables. Equation (9.2) can be expressed in three dimensions by a surface in which the maximum of z will be represented by the peak of a mountain. Similarly, the minimum will be the lowest point of the bowl in the surface. Mathematically, this means that for z to be maximum at a point (x, y), it is obvious that the functionfix, y) must be a maximum for displacements away from the point in any direction. That is, the conditions

Traditional Optimization Techniques

227

must be satisfied simultaneously, which means that the tangent plane to the surface expressed by Eq. (9.2) is horizontal. 9.4.1.3 Problems involving four variables (three independent variables) The maximum of minimum values of w = f ( x , Yl 2) can be located by differentiating Eq. (9.4) and equating it to zero. Thus

(9.4)

when the conditions

are satisfied simultaneously. 9.4.1.4 Problems involving n independent variables

The above analytical procedure can be extended to ( n + 1) variables (i.e., n independent variables) as follows. In an analytical search for a maximum or minimum value of a continuous objective function, w = f(x,,

(9.7) where xl, x2, ...,x, are n independent variables. Its partial derivatives are set equal to zero: dw =

x 2 1 * * * ,

C-dr, df dXi ?I

x,)

=0

i=l

Since the differentials dri may be chosen independently, (9.9) Note In problems containing four or more variables (i.e., three or more independent variables), the analytical solution becomes increasingly complex and impracticable. Therefore, other types of optimization techniques must be applied.

9.4.2 Optimization with Constraints (Lagrangian Multipliers) This category of problems is also referred to as optimization problems with a restrictive condition.

228

Process Plant Simulation

In some optimization problems, a relationship must be satisfied by the independent variables. For example, in vapour-liquid equilibria, the temperature, pressure, and compositions of the liquid and vapour are related by the boilingpoint condition. One variable, say, temperature, can always be eliminated analytically, but this may not be algebraically convenient, and an alternate method is available. Thus, consider (9.10) w = f( x , y , 2 ) with the restrictive condition (9.1 1) g(x, y , z ) = 0 Differentiating Eqs (9.10) and (9.1 1) gives, respectively, g d x dx

+ -8.f dy+ dy

-dz df dz

=0

(9.12)

and (9.13) At the optimum point, Eqs (9.12) and (9.13) must be satisfied for displacements in any direction, and in particular for dx = 0. Hence Eqs (9.14) and (9.15) must be consistent: g d y dy

+ -dz df

=0

(9.14)

%dy

+ -dz 8s

=0

(9.15)

ay

dz

az

The consistency of Eqs (9.14) and (9.15) is possible only if (9.16) and (9.17) By relaxing the condition dx = 0 and imposing the condition d y = 0, it follows that

af -+a-=o ax

ax

Equations (9.16)-(9.18) can be obtained by optimizing the function P = f( x , y , z ) + y, z )

(9.18)

(9.19)

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229

when the optimum point is given by the solution of Eqs (9.11) and (9.16)-(9.18), which contain the four unknowns (x, y , z, A),where A is the Lagrungiun multiplier. Note The above problem can be extended to two or more restrictive conditions by introducing two or more arbitrary multipliers such as A,,A,, A,, ..., A,, etc. For example, if we have a two-constraint problem as follows:

(9.20) P = f ( x , y, Z) + Aig(x, y , Z) + A,h(x, y , Z) Then the optimum point is given by the solution of the following equations:

af + A,-88 dX

dX

dh + A, =0 dX

(9.21)

df

88

dh + A, =0

(9.22)

dh + A,-88 + A, =0

(9.23)

- + A,8Y dY

df

-

dZ

dZ

8Y

dZ

(9.24) (9.25) We have five unknowns (x, y , z, A l , and

A,) and five equations.

Illustration 9.1 Lagrangian multiplier Determine the dimensions of an open rectangular tank with a capacity of 1000 m3, which will minimize the surface area of the bottom and walls. The tank volume is V = Zbh (constraint or restrictive condition), where 1 is the length, b is the width, and h is the depth. Solution We seek to minimize the surface area, i.e, the function is (9.26) f (1, b, h ) = 2bh + Zb + 2hZ Constraint: (9.27) g ( l , b, h) = 0 i.e., g(Z, b, h ) = lbh - V (9.28) The auxiliary function has the form P(Z, b, h, A ) = f (I, b, h ) + Ag(1, b, h ) = 2bh + lb + 2hl + A(lbh - V ) (9.29)

230

Process Plant Simulation

Taking partial derivatives with respect to 1, b, h, and 1,and setting them equal to zero, we get

_ " - b-+ 2 h + & h = O dl

(9.30)

dP _ - 2h + 1 + Alh = 0

(9.3 1 )

_ " - 2b+21+Ub=O

(9.32)

db

dh dP _ lbh - V = 0

dA

(9.33)

Solving these four equations simultaneously, from Eq. (9.30),we get

+ 2h) bh Substituting A from Eq. (9.34) into Eq. (9.31) and solving it for 6 , A=

-(b

2h + 1 + l h (-b T ) - 2h = 0 2bh + lb - lb - 21h = 0

(9.35) (9.36) (9.37)

b=l From Eq. (9.32), -2(b + 1) - -2(b + b ) = --4 lb b2 b From Eqs (9.34) and (9.38),

A=

-4 _ --- b - 2 h b b = 2h

(9.34)

b2

(9.38)

(9.39) (9.40)

Thus V = lbh

(9.41) Eliminating 1 and h from these equations and substituting b from Eqs (9.37) and (9.401,

(3 :

V = (b)(b) - = - = 1000m3

(9.42)

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231

b = 1/2000 = 12.6m

(9.43)

1 = b = 12.6m

(9.44)

h = -b 2

(9.45)

and

12.6

- 6.3m

Therefore, 1 = 12.6 m, b = 12.6 m, and h = 6.3 m.

9.4.3 Gradient Methods of Optimization These are numerical methods of the search type.They are universal and well adapted to digital computers, and prove most efficient in most cases where the search is for external values of constrained and unconstrained non-linear functions, and also when a function is not known analytically altogether. This is why gradient search methods are widely used in practical applications. In essence, these methods reduce to finding the values of independent variables generating a maximum change in the objective function. As a rule, this is achieved by moving along a gradient, which is a vector orthogonal to the performance contour surface at a given point. The various search methods mainly differ in the manner of the following aspects: 1. Ascertaining the direction of movement 2. The length or size of the step 3. Time required for search along the selected direction 4. Criteria for search termination 5. Algorithms 6 . Adaptability to different digital computers The technique of search for external (maximum or minimum) values is based on computations, which readily ascertain the direction that ensures the fastest approach to the optimum. If the criterion is defined by F = F ( x , , x 2 , ..., x , ) then its gradient at a point ( x x2, ..., x,) is defined by a vector

(9.46)

(9.47)

232

Process Plant Simulation

The partial derivative dF/dxi is proportional to the cosine of the angle between the gradient vector and the i-axis of coordinates. Then (9.48)

In addition to the direction of the gradient vector, a very important problem in gradient methods is to decide on the step length in the direction of the gradient. The step size along gradF is to a considerable degree dependent on the nature of the performance surface. If the step is too short, lengthy calculations will result. On the other hand, if the step is too long, the optimum may well be missed. The step size Axi should satisfy the condition that all steps from the base point must be in the direction of the gradient at the base point. For each variable xi, the step size is found from the values of the partial derivatives at the base (starting) point:

dF

K-

(9.49) = i=l

where K is a constant defining the step size and is the same in each of the i coordinate directions. It is only at the base point that the gradient is precisely orthogonal to the surface. If the step is too long in each of the i coordinate directions, the vector from the base point will not be orthogonal to the surface at the new point. With a properly selected step size, the derivative at the next point will be substantially close to that at the base point. For linear functions, the direction of the gradient is independent of the position of the point on the surface for which it is calculated. If the surface has the form (9.50) F = clxl + c2x2+ c3x3 then

dF

dF

- = c1, -=

c2,

dF

- = cj

(9.51)

8x1 8x2 8x3 and the gradient component in the ith coordinate direction will be C:

Jc;

+ c; + c;

(9.52)

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233

For non-linear functions,the direction of the gradient vector depends on the position of the point on the surface for which it is calculated. 9.4.3.1 Methodical procedurefor optimization Although there is a good deal of difference among the various gradient methods, the procedure for seeking the optimum in most cases is about the same, and is as follows: 1. A base point is selected. 2. The direction of descent (or ascent) from the base point is ascertained. 3. The step size is found. 4. The next point of search is determined. 5. The value of the objective function at a given point is compared with that at the previous point. 6 . The direction of movement is ascertained. 7. The procedure is repeated until the optimum is reached. 9.4.3.2 The method of steepest descent (ascent) The method of steepest descendascent is one of the best known methods of optimization when the number of variables involved is large. The procedure was developed by Box et al. (1969). Essentially, the method consists in defining the criteria to be optimized as the response of the process. This may be the profit, yield, or the output from a process and will depend on a number of factors such as temperature, concentration, pressure, etc. The relationship between the response and these factors can be written in the form (9.53) = f(x,, x21 x,) which is called the response function. The response function can be plotted in terms of two of the factors to give what is known as the response suflace. Such a surface will be undulating, and reaching the optimum will correspond to climbing the highest hill on the surface. Generally speaking, the greatest gradient will be in the direction of the maximum (minimum), and therefore if the steepest ascent (descent) can be located on the response surface, it will lead in the direction of the optimum response. However, it is very difficult to represent a surface on paper; consequently it is advantageous to plot the surface in the form of lines of equal response on a graph whose coordinates are two of the variables. These lines are called response contours and are shown in Fig. 9.2, which is called a response contour map. The line of steepest descent (ascent) starting from any point is obtained by trial, by drawing lines perpendicular to the contour. These lines will show the extent to which different factors must be varied to give a maximum (minimum) increase in the response. This procedure will only direct the way to the optimum conditions,

w

* * a ,

234

Process Plant Simulation

since at the maximum (minimum) the gradient is zero, and as the optimum is approached, the line of steepest ascent (descent) becomes more difficult to predict. Hence this method alone will not locate the optimum, but will rapidly show the engineer the region of optimum conditions.

z 60

Fig. 9.2 Response contour map

Illustration 9.2 As an example of a gradient method, we will describe a procedure for minimizing functions of two variables, F = F(x,, x2), subject to constraints H,( x , , x2) I 0 and H,( X I , x2) 0. Solution This principle can be generalized to any number of variables and any number of constraints. Consider the (x,,x2) plane (Fig. 9.3). According to the equation F = F(x,, x2), each point on this plane corresponds to a particular value of F. The closed curves on this plane represent contours ofconstunt F around the point M* at which F has a maximum (minimum) value. The method of steepest descent (ascent) operates as follows. To begin with, initial values (guesses) are provided for x1 and x2 such that they are at a starting (base) point M,, as well as initial incremental change for these parameters. The search process is started with a series of explanatory steps or moves. > 0, while x2 is held constant At first, x1 is increased by a small amount and the change SF, in F , resulting from the manipulation of xl, is measured. This change may be assumed to be proportional to the product derivative: dF =zl= dF constant = SF, dx,

ax,

(9.54)

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235

(provided Sx, is the same always). Then x2 is increased by 6x2 > 0 while x1 is held constant. The resultant change SF, in F is a measure of the partial derivative:

dF = dx, dx,

=

constant = SF,

(9.55)

I

gradF

Response surface

Fig. 9.3 Response contour map (for Illustration 9.2)

The partial derivatives given by Eqs (9.54) and (9.55) define a vector with coordinates dF/dxl and dF/dx,, which is called the gradient of F symbolized as

{.: }:

gradF = -,

-

(9.56)

The vector extends in the direction of the maximum increase of F at a given point (see Fig. 9.3). The negative of the gradient is in the direction of steepest descent in F a t that point. After the gradient component defining a successful direction of search has been ascertained, exploratory steps or moves are discontinued and pattern moves or steps are made in the negative direction of the gradient, the step size being in proportion to the magnitude of gradF. This condition is satisfied if the pattern step sizes Ax, and Ax, are proportional to the partial derivatives obtained before:

dF

Axl = a - ,

Ax,

dF

=a-

h l

where a is a positive constant.

8x2

(9.57)

236

Process Plant Simulation

After each pattern move, the change A F in F is estimated. If the previous change has caused a decrease in F, the next move will be taken in the same direction, MOM,. If the change at M , has caused an increase in F(i.e., A F > 0), no pattern move is taken further, and exploratory moves are resumed. Then again gradF is estimated at the new point M , , and pattern moves are resumed at the newly established direction of steepest descent, i.e., along the line M1M2,etc. This is why the technique is called method of steepest descent (in case of minimization) and method of steepest ascent (in case of maximization). When the quantity E = IdF/dx,I + I dF/dx, I drops to a small value, which is an indication that the system is in the vicinity of the minimum, a more conservative search technique is applied, called the gradient method. It differs from the method of steepest descent, in that after gradF is determined, only one pattern move at a time is made, and exploratory moves are resumed at the new point. With this search technique, the minimum is located more accurately than with the method of steepest descent, although the latter secures the fastest approach to the minimum. If, during the search, the point M reaches the boundary of the feasible area and at least one of the quantities M , or M2 changes sign, the search procedure is modified so that the point Mmoves along the boundary. The efficiency of the method of steepest descent depends on the scale of the variables used and the nature of the response surface. The fastest approach to the optimum is attained in the case of a surface with spherical contours. 9.4.3.3 Demerits of the method of steepest descent

The following are some of the demerits of the method of steepest descent: Extrapolation is limited. In moving along the gradient, we are guided by the extrapolation of the derivatives of the objective function with respect to appropriate variables. However, the shape of the response surface may change, thereby necessitating a change in the direction of search. In other words, we cannot move on the surface for any considerable length of time (see the examples discussed in Sec. 9.4.3.5 below).

A global optimum is dificult toJind. The method is applicable only to find local optima. 9.4.3.4 Method of steepest descent (ascent) with constraints

When some constraints are imposed on changes in the parameters of the objective function, the base point is chosen such that it will lie within the limits set by the constraints, and the search is conducted by the method of steepest descent (ascent). After the next point is calculated, a test is run to see if the constraints have been violated. If there is no violation, the search is resumed. If any constraint has been violated, the gradient is calculated so as to allow for the constraint.

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237

As an alternative, a method may be used in which the violation of a constraint causes the point to be returned to the constraint line. If more than one constraint has been imposed and a new constraint is violated at each new instant of time, the method requires that the points be moved to the new constraint. In this method, it is assumed that the optimum line lies on the constraint. 9.4.3.5 Problems on method of steepest descent Example 9.1 Find the optimum by the method of steepest descent (ascent) for an objective function of the form y = x: + 2x22 + x1x2.The starting point of the search has the coordinates xl0= 2 and x20= 2. Solution The direction of the gradient is ascertained by finding the partial derivative along the respective coordinate axes.

y = x12 +2x,2

+ X1X*

3= 2x, + x2

(9.58)

dy = 4x2 + x1 and 3x2

3x1

(9.59)

At starting point M (2,2), the partial derivatives are

3 = 2(2) + 2 = 6 2x1

dy = 4(2) + 2 = 10 and dX2

Both partial derivatives are positive and so the objective function changes in the same direction as x1and x2 (as y is also positive). If the function is to be minimized, x1and x2should be reduced until a maximum value is reached for y in the proportion

2. 3= 6:lO = 0.6 = Sx,: 6x2 dx, h2 *

Let the first move cause x2to change by 0.5 (step change). Accordingly, x1 will change by 0.6 x 0.5 = 0.3. Then we may write

xl0= 2, x~~= 2, yo = 16 ~ 1 =1 ~ 1 -0

~ 2= 1 ~ 2 -0

Hence,

SX,= 2 - 0.3 = 1.7 6x2 = 2 - 0.5 = 1.5

y1 = (1.7)2 + 2( 1.5)2 + (1.7)( 1.5) = 9.94 On moving along the gradient, the objective function will decrease progressively and we may write (ax, = 0.3, SX, = 0.5) xI2= 1.4, x*2 = 1.0, y2 = 5.36 ~ 1 =3 1.1, x23= 0.5, y3 = 2.26 ~ 2= 4 0.0, y4 = 0.64 xi4 = 0.8,

238

Process Plant Simulation ~ 2 = 5 - 0.5,

~ 1 =5 0.5,

ys = 0.5 y6 = 1a84

x26 = -1 .o,

x16 = 0.27

The line MN is the path of the line of steepest descent from the starting point M . The point N represents x16 and Y6. Since we have passed the minimum of the function at ys, we return to the point P for x1 = 0.5 and x2 = - 0.5 [i.e., P (0.5, - 0.5)]:

ay = 2x1 + ~2

= 1.0 - 0.5= 0.5,

3x1

dy = 4x2+ x1 = - 2.0 + 0.5= - 1.5

~

3x2

From the above equalities, it follows that Gy/Sx, is positive and Sy/Sx, is negative and so the change in y is opposite to that in x2. 1. In other words, in order to minimize the objective function, x2 should be increased. 2. Since we are now in the vicinity of the optimum, the step size AX, should be reduced to 0.3. S X , / ~ =X (~d y / d ~ 1 ) / ( d y / d ~ ,=) 0.5/- 1.5 = - 1.3 Since the ratio of 6 x l to Sx, is -1/3, the corresponding change in xl will be 0.1. Thus the following new values will be obtained for the sixth step ( 6 x l = 0.1, 6x2 = -0.3): x16 = x15 - 0.1 = 0.5 - 0.1 = 0.4

+ 0.3 = - 0.5 + 0.3 = - 0.2

~ 2= 6~ 2 5

Hence

y6 = (0.4)2 + 2(-2)2

+ (0.4)(0.2) = 0.16

On moving along the new gradient path, we find x17 = 0.3, = 0.2,

XI8

~ 2 = 7 0.1, x28 = 0.4,

y7 = 0.14 (min) y8 = 0.44 (increases)

The last point shows that we have passed the minimum again because y8 > y7. Now we reverse the direction again and resume the calculations until we reach the optimum, which in our case is the minimum value of y. Using point 7 ( x l = 0.3, x2 = 0,i)

dy = 4x2 + X I = 0.4 + 0.3= 0.7 -dy- - 2x1 + ~2 = 0.6 - O.1= 0.7, 3x1

h 2

6 ~ 1 / 6 ~=2( d y / d ~ ~ ) / ( d y / d x ,=) 0.7/0.7 = 1.0 6 x 2 is reduced, as we are nearing the optimum, to 0.05. Hence, ~ 1 =8 xi7 X2g

- 6 x 1 = 0.3 - 0.05 = 0.25

= X27 - Sx, = 0.1 - 0.05 = 0.05

Traditional Optimization Techniques

ys = x18 2 + 2x;,

+x

~ = (0.25)’ ~ x

239

+ 2(0.05)’ ~ ~ + (0.25)(0.05) = 0.08

This procedure has to be continued till y = 0.004.

Example 9.2 Compute the optimum to an accuracy of 0.01 by the method of steepest descent for an objective function of the form Y = (x: + x2 - 11)’ + (xl + x i - 7)’. The starting point for the search is (1,l). Solution The direction of the gradient is ascertained by finding the partial derivatives along the respective coordinate axes: y = (xl2 = Xf

+ x2 - 11)* + (xl + x; - 7 y + 2x,2x2 - 22x2 - 2 2 4 + x; + 121+ x12 + x24

+ 49 + 2XlX; - 4x22 - 14X1 dy/dxl = 4x: + 4XlX2 - M x , + 2x1 + 2x22 - 14 = 4x: + 4x,x2 - 42x1 + 2x22 - 14 dy/dX, = 2X: + 22 + 2x2 + 4x23 + 4X1X2 + 28x2 = 4x23 + 2X: - 22 4- 4X1X2 - 26x2 At the starting point (1, l), the partial derivatives are (dy/dx,)l,, = 4

+ 2 + 4 - 42 - 14 = -46

+ 2 + 4 - 26 - 22 = - 38 (Y)l,l = 1 + 2 - 22 - 22 + 1 + 121 + 1 + 1 + 49 + 2 - 14 - 14 = 106

(dy/dX;!)1,1 = 4

Both the partial derivatives are negative and so the objective function changes in the same direction as x1 and x2. The function is to be minimized, so x1 and x2 should be increased until a minimum value is reached for y in the proportion (dy/dx,):(dy/dx2) = (-46):(-38) = 1.211 = 6 ~ 1 : 6 ~ 2

t.e., ax, = 1.21(6x2)

Let the first move cause x2 to change (increase) by 0.5 (step change). Accordingly, x1 will change by 1.211 x 0.5 = 0.6055. Then we may write ax, = 0.6055, ax2 = 0.5

xl0 = 1, x~~ = 1, yo = 106 xll = xl0

+ axl = 1 + 0.6055 = 1.6055

240

Process Plant Simulation = x,,,

+ Sx,

=1

+ 0.5 = 1.5

Then Yl = ( x : ~+ x , ~- 11)2 + ( x l l + x i l - 7)* = 57.808 On moving along the gradient, the objective function will decrease progressively and we write X12 = 2.21 1, x22= 2.2, y2 = 17.526 x13= 2.8165, ~ 2 = 3 2.5, y3 = 4.5923 y4 = 43.1621 ~ 1 =4 3.422, ~ 2= 4 3.0, Since we have passed the minimum of the function at y,, we return to the point (~13,~ 2 3 = ) (2.8165, 2.5) and find the gradients again at the point to verify the direction and step change: dy/dx, = 2(2.8165)3 + 2(2.5)2 + 4(2.8165)(2.5) - 42(2.8165) -14 = -2.2585 dy/dx, = 2(2.8165)2 + 4(2.5)3 + 4(2.8165)(2.5) -26(2.5) -22 = 19.5303 From the above equations, it follows that Sy/Sxl is still negative but Sy/Sx, is positive and so the change in y is opposite to that in x2 (i.e., positive). In other words, in order to minimize the objective function, x1 has to be still increased but x2 has to be decreased. SY/SX, = ( d y / d ~ , ) / ( d y / d ~ 2=)-2.2585119.5303 = - 0.1156

Hence 8x1 = (-0.1 1 5 6 ) s ~ ~

We have reached the vicinity of the minimum, so we reduce the step change. Let Sx, = -0.3, i.e., Sx, = ( 4 . 1 156)(-0.3) = 0.03468 + 0.03468 = 2.85118, ~ 2 =4 2.5 - 0.3 = 2.2, y4 = 0.927666 x15= 2.88586, x , ~= 1.9, y5 = 0.849851 x16= 2.92054, x26= 1.7, y6 = 2.008402

~ 1 =4 2.8165

So we again return to ( ~ 1 5~, 2 5 and ) find dy/dx, and dy/dx2 at that point: dy/dx, = - 9.9 176, dy/dx, = - 5.375 1 Sx,/Sx, = 1.845 3

SX, = 1.845(6x2)

Now both dy/dxl and dy/dx, are negative, hence the step change has to be added. Again we are in the vicinity of the minimum, so we reduce the step: Let SX, = 0.1

Traditional Optimization Techniques

241

a Sx, = 1.845(0.1) = 0.1845 xi6 = 3.0706, x26 = 2.0, y6 = 0.187374 which is in the vicinity of the minimum, so we reduce the step again. Let ax, =0.05

a Sx, = 1.845(0.05)= 0.09225 xi7 = 3.0706 + 0.09225 = 3.16261, ~ 2 =7 2.05, y7 = 1.240224 The y-value increases again, so we return to ( X l 6 , x26): d y / d x , = 5.3863, d y / d x , =1.4171 Both these are positive, so both step changes are to be subtracted: Sx,/Sx, = 5.836311.4171 = 3.801 Let Sx, = 0.02

a Sx,= 3.801(0.02) = 0.07602

so ~ 1 = 7 3.07036 - 0.07602 = 2.99434, ~27'

x18 = 2.99434-0.07602=2.91832,

2 - 0.02 = 1.98, Y = 0.01077

1.98-0.02 7 ~ = 1.96, ~=0.331595 We return to ( ~ 1 7~, 2 7 and ) find d y / d x , and d y / d x , at that point: ~

2

d y / d x 1 = - 0.8164, d y / d x , = - 0.7831 Both gradients are negative, so the step changes are to be added: SX,/SX, = - 0.8164/- 0.7831 = 1.0425

S X , = 1.0425 (6x2)

Let Sx, = 0.01 Sx, = 1.0425(0.01)=0.010425 Then x18 = 2.99434 + 0.010425 = 3.004765, x28 = 1.98+0.01 = 1.99 y = 0.001581 Now we can stop as the required accuracy 0.001 is attained in the objective function. So the optimum values of x 1 and x2 are x1 = 3.004, x2 = 1.99, y = 0.0015 9.4.3.6 The sequential simplex method

A number of studies proposing the use of regular simplex designs in the descent (ascent) phase have appeared. A regular simplex is defined as a set of k + 1 equidistant points in a k-dimensional space. A simplex is a geometric figure. In one-dimensional space, i.e., for one factor (one independent variable), a regular simplex is a straight line (two points). For two factors (two independent variables),

242

Process Plant Simulation

a regular simplex is an equilateral triangle (three points). Similarly, for three factors (independent variables), a regular simplex is a regular tetrahedron (four points). In the search for the optimum, trial values of the experimental factors can be selected at points in multidimensionalfactor space located at the vertices of a regular simplex. After the first cycle of observations, the vertex associated with the conditions yielding the poorest response is identified. This point is now deleted and a new point, which is a mirror reflection from the oppositefuce of the simplex, is added (afuce is a set of k points of a k-dimensional simplex). This new point and the remaining old points form a new regular simplex, termed reflection, in which the centroid is displaced in the direction of the poorest response point towards the centroid of the remaining points. In the general case, this direction is not the path of the steepest descent (ascent); yet it points towards better performance of the process. Continuation of this procedure, always deleting the point that yields the poorest response and replacing it with a mirror image, leads to a nearly stationary region. The number of moves needed to reach the optimum region may be small when compared to the steepest descent method. This is because k + 1 observations have to be made only at the beginning; after that every move involves one additional observation under conditions selected from an analysis of the previous observations. This method is an extension of the steepest descent (ascent) method and was proposed by Himsworth (1962). It differs from the steepest descent method, in that no attempt is made to find the lines of steepest descent, but rather a rapid determination is made of a direction which is steep, although not necessarily the steepest, so that moves are made frequently in a favorable direction. For a process involving two independent variables the procedure is as follows. In Fig. 9.4, three points are selected fairly close together in such a manner that they form the vertices of an equilateral triangle. Let these points be 1, 2, and 3. Following this, three further points are selected such that each forms an equilateral triangle with any two of the first three points. If the response surface is locally a plane or nearly so, one of the possible new points will give a higher result than the other two; furthermore, point 4 will be a mirror image of the lowest point in the first triangle. Hence, after the first three trials, the point giving the lowest result is rejected and replaced by the mirror image point forming the second triangle. This construction process is repeated until the summit is reached. Thus, as long as the surface is sloping and reasonably plane over the area of a particular triangle, each step of the construction leads to a more favourableregion. The path taken (trajectory) from the starting point will oscillate about the line of steepest descent. For a system containing more than two variables, the procedure will be similar, except that (k + 1) trial points will be made, where k is the number of variables. The most favourable of these points is retained to replace the appropriate member of the original set, thus completing a move towards the optimum. The (k + 1)

Traditional Optimization Techniques 243 points correspond to the vertices of a regular simplex in k dimensions. Hence this is called the sequential simplex method.

- x

Fig. 9.4 The sequential simplex method. The dashed line denotes the descent method: the triangles denote the sequential simplex method.

Advantages of the sequential simplex method 1. The calculations are elementary, requiring no mathematical or statistical knowledge. 2. Each move is determined by the previous result and so human judgement is eliminated. 3. It is possible to add a variable at any time. 4. A numerical measure of the response is not needed. It is only necessary to rank the results and discard the lowest. Reflection, contraction, and expansion In the simplex method, at any stage we have a simplex with n + 1 vertices where the function value is evaluated. The basic iteration consists of changing the simplex to reduce the function values. Several transformations are possible. The simplest is the one where the vertex at which the function value is the largest is reflected in the hyperplane formed by the other n vertices. The function value at the new vertex is then evaluated and the process can be repeated. However, it may turn out that the function value is maximum at the new vertex, in which case, if it is reflected, we will get back to the old simplex and no progress can be made as shown in Fig. 9.5. It is also possible that the reflection process leads to a cyclic process (i.e., after a few iterations, we will get back to the original simplex) as shown in Fig. 9.6. In such cases, either another point is selected for reflection or the simplex is extended or contracted in some

244

Process Plant Simulation

direction. The iteration proceeds in such a manner that the volume of the simplex keeps on decreasing once appropriate conditions are satisfied, and ultimately the volume may shrink to the required accuracy and the process may be terminated. At this stage, the vertex with the lowest value of the function will hopefully give a good approximation to the minimizer.

XI

Fig. 9.5 Reflection not leading to a new simplex

x2

1

XI

Fig. 9.6 Reflection leading to a cyclic process

Traditional Optimization Techniques

245

Let xh be the vertex corresponding to the highest value of the objective function (i.e., the poorest), X, be the reflection point, and X, be the centroid of all points xi except i = h.

x,

=

-cxi 1 n+l n i=l

(9.60)

i+ h

Mathematically, the reflected point is given by

x, = (1 + a)X,

- ax,

(1) a = reflection coefficient =

(9.61) distance between X, and X, distance between xh and X,

(2) p = contraction coefficient =

distance between X, and X, distance between X, and X,

(9.62) (9.63)

ifflX,)>flXJ for all i except i = h andflx,)
(3) y = expansion coefficient =

distance between Xe and X, distance between X, and X,

(9.64)

if AX,) l

(9.65)

[ AXl) is the vertex corresponding to the minimum function value]. (This means we are moving in the right direction in search of the optimum value, and hence it is better to increase the step size, which is possible by expanding it. Hence it is better to go for expansion.)

x, = (1+ a)xo - ax, x, = PX, + (1+ PIX,

(9.66)

Xe = YX, + (1 + Y)X, Test for convergence:

(9.68)

(9.67)

(9.69) Figure 9.7shows the flow diagram for the sequential simplex method (Rao 1991).

246

Process Plant Simulation Start with any set of (n + 1) points xI, x,, ...,x,, , defining the current simplex. Specify the values for a,p, y, and E.

4-

Find xh and X,such thatfix,,) = max HX,)] and AX,) = min [AX,)] 1 n+'

Find centroid X, using X, = -

n

I

i=1 i#h

xi

I

1 [

I Find X, by reflection

-.

No

No Yes

I

I

Find X, by expansion

I

I No

4

Replace X, !xi,l

yes x2, ...,x,

Yes Replace X, in xI, x,,

...,x,, by X,

1

No

Is the minimum reached (Q < E)?

I Yes

Set Xi, = X,

r. Replace X, in x,, x2, ..., x,

+

, by

t Isf(x,) >AX,)?

Yes

Fig. 9.7 Flow diagram for sequential simplex method

I

by X,

I

Traditional Optimization Techniques

247

Example 9.3 Minimize Ax,, x,) = x1- x2 + 2 ~ :+ 2 xlx2+ xi, taking the points defining the initial simplex as

x,=

{

4-0}, 4.0

x, =

{

4 5*0}, .0

x, =

{

4*0} 5.0

a = 1.0, p = 0.5, and y = 2.0. For convergence, take

E

= 0.2.

Solution

Iteration 1: 1. The value of the function at each of the vertices of the current simplex is given by fi =AXl) = 4 - 4 + 2(16) + 2(16) + 16 = 80 (lowest) f 2 =AX2)= 5 - 4 + 2(16) + 2 (20) + 16 = 107 (highest) f3 =AX,)= 4 - 5 + 2(16) + 2(20) + 25 = 96

2. The centroid X, is obtained by

x,

= W(X,

+ X,)= v 2

{

4.0 + 4.0 4.0 + 5,0} =

{:::}

and f(X,) = 87.75

3. The reflection point is found as X, = (1 + a)X, - ax, X r = 2 XO -x -

{i:} {:::} {:::} -

=

and f(X,) = 3 - 5 + 2(9) + 2(15) + 25 = 71.0

4. ASAX,)
x, = 2x, = x,

{ }:I;{-

= 10.0 6*o}

=

{;::}

and f(X,) = 2 - 5.5 + 2(4) + 2(11) + 30.25 = 56.75

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Process Plant Simulation

5. SinceAX,)
6 . To test for convergence, we compute Q:

(80 - 87.75)2 + (56.75 - 87.75)2 + (96.0 - 87.75)2 3 =

19.06

>&

Hence we do the next iteration. Iteration 2 : 1.

and

x 1 -- x 2

-

{;:;}

2. The centroid is

x, = V2(X1 + X2) = v 2

4.0 + 2.0 4.0 + 5.5}

{

=

{

:;5}

and f(X,) = 67.31

3. Reflection point X ,

= 2X0 -

X, =

{:::}{:::}{:::} -

and f ( X , ) = 2.0 - 4.5+ 2 ( 4 ) + 2(9) + 20.25 = 43.75 4. A s AX,) < AX/) we find X,:

=

Traditional Optimization Techniques

x, = 2xr - x,

4.0

= {9.0} -

I

1i 1

(4.75 3.0 = 4.25

and

f(X,)= 1.0 - 4.25 + 2(1.0) + 2(4.25) + 18.0625 = 25.3125 5. As AX,)
6 . For convergence, we compute Q:

n = j(80.0

+ (56.75 - 67.31)2 + (25.3125 - 67.31)2\"

- 67.31)*

" I

J

3

= 26.1 > &

Hence we do the next iteration. Iteration 3: 1. Here

and

f(X,) = 80.0, f(X,) = 56.75, f(X3) = 25.3125 Thus

x, = x, =

{I::}, x,

= x3 = { 5;}

2. The centroid is X, = (l/2)(X2 + X3) = (U2)

and

f(X,) = 39.50

3.

x, = 2x0 - Xh = and f(X,) = 16.8125

{

93705)

-

{:::}{;.4:} =

5.5 + 4.25

249

250

Process Plant Simulation

4. AsfiXr)
and f(Xe) = 11.89 5. Sincefix,)
6.675

5.50

4.25

6. Convergence:

Q= =

(11.89 - 39.5)2 + (56.75 - 39.5)2 + (25.3125 - 39.5)2 [ 3

)”

20.5

> &

Hence we do the next iteration. We have to continue this procedure till the convergence criterion is satisfied.

Example 9.4 method:

Minimize the function given below using the sequential simplex

Y = (x; + x2 - 11)2+ (XI + x i - 7 ) 2 Take the points defining the initial simplex as (3, 3), (4, 3), and (3, 4). For convergence, take E = 0.2 and the coefficients of reflection, contraction, and expansion as 1.0, 0.5, and 2.0, respectively. Perform at least two iterations. Solution

Iteration 1: 1. The value of the function at each of the vertices of the current simplex is given by fl =AXl) = (9 + 3 - 11)2+ (3 + 9 - 7)2 = 26 (lowest) f2 =fix2) = (16 + 3 - 11)2+( 4 + 9 - 7 ) 2 = 100 f 3 =AX3) = (9 + 4 - 11)*+ (3 + 16 - 7)’ = 148 (highest) X, = X3 = (3,4), AX,) = 148 X, = XI = (3, 3), AX,) = 26 2. The centroid X o is obtained as xo= 1/2 (XI+ X 2 ) = (1/2) (3 + 4, 3 + 3) = (3.5, 3) AXo) = 48.31

Traditional Optimization Techniques

251

3. The reflection point is x, = (1 + a)X, - ax, X, = 2x0 - X, [(7, 6) - (3,4)] = (4, 2) and AX,) = 50 4. As AX,) >AXl), we check ifAX,) >AXi)for all i f h, which is not the case, i.e.,JgX,) >AXj) for all i # h [asAX, >f(x,) butAX,) < f ( X 2 ) ] .So we replace X,, in (X,, X,, X3)by X, , 5. The vertices of the new simplex after replacing X, by X, are x, = (3, 3), x, = (4, 3), x3= x, = (4,2) AX,) = 26, AX,) = 100, AX3>= 50 6 . Test for convergence: compute Q, (26 - 48.3)'

+ (100 - 48.31), + (50 - 48.3)'

Q={

3

= 32.52 >&

Hence perform next iteration. Iteration 2: 1. ASAX,) = 26,AX2)= 100,AX3) = 50

x, = (4, 3) and Xl = x, = (3, 3), x3= (4, 2) 2. The centroid is X, = (1/2) (X, + X 3 ) = (1/2) (7, 5) = (3.5, 2.5) and AX,) = (3.52+ 2.5 - 11), + (3.5 + 2.5, - 7)2 = 14.06 + 7.56 = 21.62 3. x, = 2x0-x, = ~ 7 ~-5(4,311 )

= (32)

and AX,) = (9 + 2 - 11)2+ (3 + 4-7),

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Process Plant Simulation

=O


Therefore, the optimum point is (3, 2). 3 XI = 3, x2 = 2, A x ) = 0

9.4.4 Rosenbrock Method In this method (Rosenbrock 1960), the objective function is modified by the introduction of multipliers. Each time a variable violates a constraint, the multiplier becomes zero. That is, the objective function is multiplied by zero and is therefore equal to zero. If the variable remains within the feasible area, the multiplier is unity and the objective function takes on its full value. However, when the variable drops to the limits prescribed by the boundary region, the multiplier changes parabolically from 0 to 1 within the boundary region and the objective function varies from 0 to its full value.

9.4.5 Other Methods Some of the other methods (Antia 1991) include the Golden section search, Brent’s method, quasi-Newton methods, direction set methods, linear programming, etc. A brief description of these methods is given below: 9.4.5.1 Golden section search

A zero of a non-linear function can be bracketed between two values, wnere the function value has opposite signs. To bracket a minimum we need more information. For example, if the first derivative of the function has opposite signs at the two end points, the function will have a stationary point inside the interval, provided it is continuous over the interval. This point could be either a minimizer or a maximizer. It may be noted that the first derivative does not change sign at a point of inflection. If this point is known to be a minimizer, we can apply the simple method of bisection to determine the minimizer more accurately. Of course, here we have assumed that the first derivative of the function can be calculated at every point. If it is not possible to calculate the derivative easily, then we need a third point inside the interval to determine if the function has a minimum or not. The ratio of the two segments is called the golden ratio (also occurs in Fibonacci numbers), which gives the optimal method for dividing the interval, and the corresponding algorithm is called the golden section search. In this algorithm, at each stage the new point is selected to be a fraction T as measured from the central point and into the larger of the two subintervals. Even if

Traditional Optimization Techniques

253

we start out with a triplet whose segments are not in the golden ratio, this procedure will ultimately lead to intervals with the golden ratio. Once the bracketing triplet has achieved the golden ratio, at each step the bracketing interval reduces by a factor of 1 - r 0.618034. Hence, this method converges linearly to the minimizer. Further, this ratio is somewhat larger than the factor of 1/2 in the method of bisection. Hence, the convergence of golden section search will be slower than that of bisection. Once we are close to the minimum, we can switch to some iterative method, which converges superlinearly. 9.4.5.2 Brent’s method

Brent (1973) has given an algorithm for finding the minimum of a function of one variable. This method combines the golden section search with parabolic interpolation in such a way that the iteration is guaranteed to converge to the specified accuracy in a finite number of iterations, provided, of course, that the round-off error is neglected. In the presence of round-off error, it may be meaningless to continue the iteration beyond a certain stage, which unfortunately cannot be detected by this method. 9.4.5.3 Quasi-Newton methods

The simplest method based on a quadratic model is Newton’s method (Antia 1991) for determining the zeros of a gradient vector. If we can calculate the second derivatives of a function, then we can approximate the function using the first three terms of the Taylor series. The main drawback of Newton’s method is that far from a minimum the matrix need not be positive definite and the approximating quadratic may not have a minimum. In fact, the value of the function may not even decrease. This is eliminated in Newton’s method with line search, where the Newton correction is only used to generate a direction of the line search. As in the case of the solution of a system of non-linear equations, the main disadvantage of Newton’s method, even when modified to ensure global convergence, is the need to calculate all the n(n + 1)/2 second derivatives. Hence, it may be better to approximate the Hessian matrix using the value of the function and the gradient vector. The simplest technique is to use a finite difference approximation, which may not yield a symmetric matrix. Further, evaluating the finite difference approximation requires n + 1 evaluations of the gradient vector, which could be very expensive. These disadvantages can be avoided if some updating procedure similar to that of Broyden’s method (Antia 1991) can be given for the Hessian matrix or its inverse, such that the matrix is assured to be positive definite. Such methods incorporating the above-mentioned updating procedure are referred to as quasi-Newton methods and these enjoy the following advantages over Newton’s method: 1. Only first derivatives are required.

254

Process Plant Simulation

2. The Hessian matrix is assured to be positive definite, which ensures that the search is in the descent direction. 3. Each iteration requires only O(n2)floating-point operations, as opposed to 0(n3)for the Newton’s method. 9.4.5.4 Direction set methods Quasi-Newton methods require the calculation of derivatives, which may not be possible if the function is rather complicated. Direction set methods do not require the calculation of derivatives. These methods are not as efficient as the methods described earlier in terms of the number of function evaluations required, and unless the derivative calculation is extremely difficult, the methods discussed earlier are more efficient. If the derivative can be evaluated, then it is also possible to use the conjugate gradient methods (Antia 1991), but these require much less memory and may be considered if the number of variables is very large. In the absence of even the first derivatives, it is difficult to approximate the quadratic terms, unless a search is made along n independent directions. The basic idea of all direction set methods is to start with an arbitrary set of orthogonal vectors and update this set using information about the function value and attempt to come up with a set which has non-interfering directions such that a minimization along one direction is not spoiled by subsequent minimization along another. If such a set can be found, then after one cycle the process will converge to the minimizer. Such a set of directions may not even exist for arbitrary functions. However, it can be shown that for a quadratic function, such a set of directions exists. Hence, such methods should terminate after a predetermined finite number of iterations for a quadratic function. This is a useful property of any optimization method and is referred to as quadratic termination. Since a general function can be approximated reasonably well by a quadratic function in the neighbourhood of a minimum, we can expect such methods to converge very fast, once we are close to a minimum. 9.4.5.5 Linear programming In many practical optimization problems, there are constraints on the independent variables. However, finding a minimum of a non-linear function of several variables in the presence of constraints is a.dificult problem. Hence, we restrict ourselves to only linear functions with linear constraints. Such problems are usually referred to as linear programming (LP) problems. Even if our problem is non-linear, it may be approximated by a linear model to make it more tractable. The principal application of LP has been in industries, where it is used to determine the optimum allocation of resources subject to various constraints imposed by the prevailing situation. Because of the economic importance of LP, a considerable effort has been put in to develop sophisticated algorithms for solving large LP problems on computers. Problems involving tens of thousands of variables and thousands of

Traditional Optimization Techniques

255

constraints have been successfully solved. There is a vast amount of literature available on this issue. Because of its importance, an elaborate quasi-economic terminology has been developed, which, to some extent, obscures the basic numerical process from the non-experts. In numerical methods, the principal problems of functional approximation can be formulated as LP problems. Rabinowitz (1968) has discussed various applications of LP to numerical analysis.

EXERCISES 9.1 Compute the optimum to an accuracy of 0.001 using the method of steepest descent for an objective function of the form Y = (x: + x2 - 11)’ + (xl + x i - 7)’. The starting point of the search has the coordinates (1, 1). 9.2 Compute the optimum to an accuracy of 0.001 using the method of steepest descent for an objective function of the form f ( x ) = lOO(x, - x:)’ + (1 - x ~ ) ~ . The starting point of the search has the coordinates (3, 2). 9.3 Using the method of sequential simplex, minimize the function y = (x: + x2 - 11)2 + (xl + x i - 7)2. Choose the initial simplex of three points as (O,O), (2,0), and (1, 1). Set a = 1.0, p = 0.5, y = 1.5, and E = Perform at least two iterations. 9.4 Using the sequential simplex method, minimize the Himmelblau function f ( x l , xz) = (x: + x2 - 11)2 + (xl + x i - 7)2. Choose the initial simplex of three points as (3, 3), (4, 3), and (3, 4). Set a = 1.0, ,8 = 0.5, y = 0.2, and E = 0.1. Perform at least two iterations. 9.5 Minimize the function Y = (x: + x2 - 11)’ + (x, + x i - 7)’ using the sequential simplex method. Take the points defining the initial simplex as (3, 3), (4,3), and (3,4). For convergence, take E = 0.2, and the coefficients of reflection, expansion, and contraction are a = 1.0, p = 0.5, and y = 0.2, respectively. Perform at least two iterations. 9.6 Compute the optimum to an accuracy of 0.001 using the method of steepest descent for an objective function of the form Y = (x: + x2 - 11)’ + (xl + x i - 7)2. The starting point of the search has the coordinates (1, 1). Take the initial step change in x 2 = 6 x 2 = 0.5. 9.7 The functional relationship between the dependent and independent variables of a chemical process is expressed as z = x2 + xy + y2. Using the optimization routine of the method of steepest descent, find the optimum value of the dependent variable, starting with an initial guess of (2, 2) to an accuracy of 0.0045. 9.8 Using the Lagrangian multiplier optimization technique, find out the dimensions and the optimum surface area of a closed cylindrical tank with a capacity of 500 m3. 9.9 Find out the dimensions and the optimum surface area of a closed cylindrical tank with a capacity of 500 m3 using the method of Lagrangian multipliers.

256

Process Plant Simulation

9.10 What are the limitations of optimization as a technique? 9.11 Give a brief account of the range of applications of optimization. 9.12 Discuss the types of optimization. 9.13 Categorize and describe the methods of optimization. 9.14 Give short notes on analytical methods of optimization. 9.15 Give short notes on gradient methods of optimization. 9.16 Compare the method of steepest descent and the method of sequential simplex. 9.17 Draw a neat sketch of the flow diagram for the sequential simplex method. 9.18 Discuss the following optimization techniques: (a) Rosenbrock method (b) Golden section search (c) Brent’s method (d) Quasi-Newton methods (e) Direction set methods (0 Linear programming 9.19 Solve the following equations using the steepest descent method by taking the starting point as (0, 0, 0):

+ x2 = 4 + 2x2+ x3 = 8 x2 + 3x3 = 11 2x,

XI

9.20 Minimize f = 4x: + 3x22 - 5x1x2- 8x1 starting from the point (0,O) using the method of steepest descent and the sequential simplex method. 9.21 Minimize f(x,, x2) = xf - 2x:x2 + x: + x i - 2x, + 1. 9.22 Minimize f = x: + 3x22 + 6x: by taking the starting point as (2, -1, 1). 9.23 Minimize f = 2x: + x i by using the steepest descent method with the starting point (1, 2). 9.24 An electric power of 5 x lo4 kW generated at a hydroelectric plant is to be transmitted 400 km to a step-down transformer station for distribution at 11 kV. The power dissipated due to the resistance of conductors is Z2C-l,where Z is the line current in amperes and C is the conductance in mhos. Based on the cost of the delivered power, the resistance loss can be expressed as Rs 0.263(Z2C1).The power transmitted ( K ) is related to the transmission line voltage at the power plant ( E ) by the relation K = h E Z , where E is in kV. The cost of conductors is Rs 3.9 x 106C and the investment in equipment needed to accommodate the voltage E is Rs 103E. Find the values of E and C to minimize the cost of the transmission system.

CHAPTER 10 NON-TRADITIONAL OPTIMIZATION TECHNIQUES

Most of the traditional optimization techniques based on gradient methods have the possibility of getting trapped at local optima, depending upon the degree of non-linearity and the initial guess. Hence, these traditional optimization techniques do not ensure the global optimum and also have limited applications. In the recent past, non-traditional search and optimization methods based on natural phenomena (evolutionary computation) such as simulated annealing (SA) (Kirkpatrick et al. 1983), genetic algorithms (GA) (Goldberg 1989), and differential evolution (DE) (Price & Storn 1997), to name a few, have been developed to overcome this problem. These algorithms are stochastic in nature, with probabilistic transition rules. These are comparatively new and are gaining popularity due to certain properties which the deterministic algorithms do not have. These are found to have a better global perspective than the traditional methods (Deb 1996). Floudas et al. (1999) listed many test problems on local and global optimization in their handbook. Moreover, when an optimal design problem contains multiple global solutions, designers are not interested in finding just one global optimum solution, but as many as possible for various reasons. First, a design suitable in one situation may not be suitable in another situation. Second, designers may not be interested in finding the absolute global solution. Instead they may be interested in a solution which corresponds to a marginally inferior objective function value but is more amenable to fabrication. Thus, it is always prudent to know about other equally good solutions for later use. However, if the traditional methods are used to find multiple optimal solutions, they need to be applied a number of times, each time starting from a different initial guess and hoping to achieve a different optimal solution. During the last two decades there has been a growing interest in algorithms based on the principle of evolution (survival of the fittest). A common term, coined recently, refers to such techniques as evolutionary algorithms (EA) or evolutionary

258

Process Plant Simulation

computation (EC) methods. The best-known algorithms in this class include genetic algorithms, evolutionary programming, evolution strategies and genetic programming. There are many hybrid systems which incorporate various features of the above paradigms and consequently are hard to classify, which can be referred to just as EC methods (Dasgupta & Michalewicz 1997). A brief review of the evolutionary computation techniques was also presented by Babu (2001). Some of these most widely used non-traditional optimization techniques (simulated annealing, genetic algorithms, differential evolution, and other evolutionary computation techniques) are discussed in this chapter.

10.1 Simulated Annealing 10.1.1 Introduction This optimization technique resembles the cooling process of molten metals through annealing. The atoms in molten metal can move freely with respect to each other at high temperatures. But the movement of the atoms gets restricted as the temperature is reduced. The atoms start to get ordered and finally form crystals having the minimum possible energy. However, the formation of crystals mostly depends on the cooling rate. If the temperature is reduced at a very fast rate, the crystalline state may not be achieved at all, instead, the system may end up in a polycrystalline state, which may have a higher energy than the crystalline state. Therefore, in order to achieve the absolute minimum energy state, the temperature needs to be reduced at a slow rate. The process of slow cooling is known as annealing in metallurgical parlance.

10.1.2 Procedure The simulated annealing procedure simulates this process of slow cooling of molten metal to achieve the minimum value of a function in a minimization problem. The cooling phenomenon is simulated by controlling a temperature-like parameter introduced with the concept of the Boltzmann probability distribution. According to the Boltzmann probability distribution, a system in thermal equilibrium at a temperature T has its energy distributed probabilistically according to P(E) = exp (-EAT), where k is the Boltzmann constant. This expression suggests that a system at high temperature has almost uniform probability of being in any energy state, but at low temperature it has a small probability of being in a high energy state. Therefore, by controlling the temperature T and assuming that the search process follows the Boltzmann probability distribution, the convergence of an algorithm can be controlled. Metropolis et al. (1953) suggested a way to implement the Boltzmann probability distribution in simulated thermodynamic systems. This can also be used in the function minimization context. Let us say, at any instant the

259

Non-traditional Optimization Techniques

current point is x'') and the function value at that point is E(t) =f(x(')). Using the Metropolis algorithm, we can say that the probability of the next point being at x" ') depends on the difference in the function values at these two points or on AE = E ( f+ 1) - E ( f ) and is calculated using the Boltzmann probability distribution: P(E(t + 1)) = min[l, exp(-hE/kT)] (10.1) If c\E I0, this probability is 1 and the point x(' ') is always accepted. In the function minimization context, this makes sense because if the function value at x" ') is better than that at x('), the point x(' ') must be accepted. An interesting sjtuation results when AE > 0, which implies that the function value at x(' ') is worse than that at x('). According to many traditional algorithms, the point x'" ') must not be chosen in this situation. But according to the Metropolis algorithm, there is some finite probability of selecting the point x('+ even though it is worse than the point x('). However, this probability is not the same in all situations. This probability depends on the relative magnitude of AE and T values. If the parameter T is large, this probability is more or less high for points with largely disparate function values. Thus, any point is almost acceptable for a large value of T. On the other hand, if the parameter T is small, the probability of accepting an arbitrary point is small. Thus, for small values of T, the points with only a small deviation in the function value are accepted. Simulated annealing (Laarhoven & Aarts 1987) is a point-by-point method. The algorithm begins with an initial point and a high temperature T. A second point is created at random in the vicinity of the initial point and the difference in the function values (AE)at these two points is calculated. If the second point has a smaller function value, the point is accepted; otherwise the point is accepted with probability exp (-AE/T). This completes one iteration of the simulated annealing procedure. In the next generation, another point is created at random in the neighbourhood of the current point and the Metropolis algorithm is used to accept or reject the point. In order to simulate the thermal equilibrium at every temperature, a number of points (n)are usually tested at a particular temperature before reducing the temperature. The algorithm is terminated when a sufficiently small temperature is obtained or a small enough change in function values is found. +

+

+

+

+

10.1.3 Algorithm Step 1 Choose an initial point X('), a termination criterion E. Set T sufficiently high, number of iterations to be performed at a particular temperature n, and set t = 0. Step 2

Calculate a neighbouring point x"' the neighbourhood is created.

')

= N(x(')).Usually, a random point in

Step 3 If AE = E(x(' ") - E(x('))c 0, set t = t + 1; else create a random number r in the range (0,l). If r 5 exp (-AE/T), set t = t + 1; else go to step 2. +

260

Process Plant Simulation

Step 4 If Ix('+ - x(')l c E and T is small, terminate; else if ( t mod n) = 0, then lower T according to a cooling schedule. Go to step 2; else go to step 2.

The initial temperature T and the number of iterations ( n )performed at a particular temperature are two important parameters which govern the successful working of the simulated annealing procedure. If a large initial T is chosen, it takes a number of iterations for convergence. On the other hand, if a small initial T is chosen, the search is not adequate to thoroughly investigate the search space before converging to the true optimum. A large value of n is recommended in order to achieve a quasi-equilibrium state at each temperature, but the computation time is more. Unfortunately, there are no unique values of the initial temperature and n that work for every problem. However, an estimate of the initial temperature can be obtained by calculating the average of the function values at a number of random points in the search space. A suitable value of n can be chosen (usually between 20 and 100) depending on the available computing resources and the solution time. Nevertheless, the choice of the initial temperature and the subsequent cooling schedule still remain an art and usually require some trial-and-error efforts.

10.2 Genetic Algorithms 10.2.1 Introduction Genetic algorithms (GAS) are used primarily for optimization purposes. They belong to the g o u p of optimization methods known as non-traditional optimization methods. I Non-traditional optimization methods also include optimization techniques such as neural networks (simulated annealing, in particular). Simulated annealing is an optimization method that mimics the cooling of hot metals, which is a natural phenomenon. Through this mode of simulation, this method is utilized in optimization problems. w Just like simulated annealing, GAStry to imitate natural genetics and natural selection, which are natural phenomena. The main philosophy behind GA is 'survival of the fittest'. As a result of this, GAS are primarily used for maximization problems in optimization. w GAS do not suffer from the basic setbacks of traditional optimization methods, such as getting stuck in local minima. This is because GAS work on the principle of natural genetics, which incorporates large amounts of randomness and does not allow stagnation. GAS were first introduced by Holand (1975).

10.2.2 Definition Genetic algorithms are computerized search and optimization algorithms based on the mechanics of natural genetics and natural selection.

261

Non-traditional Optimization Techniques

10.2.3 The Problem Consider a maximization problem Maximize f ( x ) x;l)

(10.2)

5 xi 5 xi(u)

i = 1, 2, ..., N This is an unconstrained optimization problem. Our aim is to find the maximum of this function by using GAS.

10.2.4 Techniques used in GA For the implementation of GAS, there are certain well-defined steps. 10.2.4.1 Coding Coding is the method by which the variables xi are coded into string structures. This is necessary because we need to translate the range of the function in a way that is understood by a computer. Therefore, binary coding is used. This essentially means that a certain number of initial guesses are made within the range of the function and these are transformed into a binary format. It is evident that for each guess there might be some required accuracy. The length of the binary coding is generally chosen with respect to the required accuracy. For example, if we use an eight-digit binary code. (0000 0000) would represent the minimum value possible and (1 111 1111) would represent the maximum value possible. A linear mapping rule is used for the purpose of coding: (/)

xi = xi

+

x y - x!/)

2Il - 1

(decoded value)

(10.3)

The decoded value (of si,the binary digit in the coding) is given by the rule /.I - I

~

Decoded value = C 2 ' s i i=O

where s; E (0, 1) For example, for a binary code (0111), the decoded value will be

(10.4)

(0111) = (112~+ (112~+ (1p2 + (012~= 7 It is also evident that for a code of length 1, there are 2' combinations or codes possible. As already mentioned, the length of the code depends upon the required accuracy for that variable. So, it is imperative that there be a method to calculate the required string length for the given problem. So, in general, Eq. (10.5) is used for this purpose:

262

Process Plant Simulation

(10.5)

So, adhering to the above-mentioned rules, it is possible to generate a umber of guesses or, in other words, an initial population of coded points that lie in the given range of the function. Then, we move on to the next step, that is, calculation of the fitness. 10.2.4.2 Fitness

As has already been mentioned, GAS work on the principle of ‘survival of the fittest’. This in effect means that the ‘good points’ or the points that yield maximum values for the function are allowed to continue in the next generation, while the less profitable points are discarded from our calculations. GAS maximize a given function, so it is necessary to transform a minimization problem to a maximization problem before we can proceed with our computations. Depending upon whether the initial objective function needs to be maximized or minimized, the fitness function is hence defined in the following ways: F(x) = A x ) for a maximization problem

-

1 1+ f(x>

for a minimization problem

(10.6) (10.7)

It should be noted that this transformation does not alter the location of the minimum value. The fitness function value for a particular coded string is known as the string’s fitness. This fitness value is used to decide whether a particular string carries on to the next generation or not. GA operation begins with a population of random strings. These strings are selected from a given range of the function and represent the design or decision variables. To implement our optimization routine, three operations are carried out: 1. Reproduction 2. Crossover 3. Mutation

10.2.5 Operators in GA 10.2.5.1 Reproduction

The reproduction operator is also called the selection operator. This is because it is this operator that decides the strings to be selected for the next generation. This is the first operator to be applied in genetic algorithms. The end result of this operation is the formation of a ‘mating pool’, where the above average strings are copied in a probabilistic manner. The rule can be represented as

263

Non-traditional Optimization Techniques

Probability of selection a (fitness of string) into mating pool

(10.8)

The probability of selection of the ith string into the mating pool is given by (10.9)

where Fi is the fitness of the ith string. Fj is the fitness of thejth string, and n is the population size. The average fitness of all the strings is calculated by summing the fitness of individual strings and dividing by the population size, and is represented by the symbol F:

-

(10.10) F = -i = l n It is obvious that the string with the maximum fitness will have the most number of copies in the mating pool. This is implemented using ‘roulette wheel selection’. The algorithm of this procedure is as follows. Roulette wheel selection Step I Using Fi calculate pi. Step 2

Calculate the cumulative probability Pi.

Step 3 Generate n random numbers (between 0 and 1). Step 4 Copy the string that represents the chosen random number in the cumulative probability range into the mating pool. A string with higher fitness will have a larger range in the cumulative probability and so has more probability of getting copied into the mating pool. At the end of this implementation, all the strings that are fit enough would have been copied into the mating pool and this marks the end of the reproduction operation. 10.2.5.2 Crossover

After the selection operator has been implemented, there is a need to introduce some amount of randomness into the population in order to avoid getting trapped in local searches. To achieve this, we perform the crossover operation. In the crossover operation, new strings are formed by exchange of information among strings of the mating pool. For example,

264

Process Plant Simulation

oo/ooo

-

00~111

lljlll

Crossover point Parents

Children

Strings are chosen at random and a random crossover point is decided, and crossover is performed in the method shown above. It is evident that, using this method, better strings or worse strings may be formed. If worse strings are formed, then they will not survive for long, since reproduction will eliminate them. But what if majority of the new strings formed are worse? This undermines the purpose of reproduction. To avoid this situation, we do not select all the strings in a population for crossover. We introduce a crossover probability (p,). Therefore, (loop,)% of the strings are used in crossover. ( 1 - p,)% of the strings is not used in crossover. Through this we have ensured that some of the good strings from the mating pool remain unchanged. The procedure can be summarized as follows: Step I

Select (loop,)% of the strings out of the mating pool at random.

Step 2 Select pairs of strings at random (generate random numbers that map the string numbers and select accordingly). Step 3 Decide a crossover point in each pair of strings (again, this is done by a random number generation over the length of the string and the appropriate position is decided according to the value of the random number). Step 4 bits.

Perform the crossover on the pairs of strings by exchanging the appropriate

10.2.5.3 Mutation Mutation involves making changes in the population members directly, that is, by flipping randomly selected bits in certain strings. The aim of mutation is to change the population members by a small amount to promote local searches when the optimum is nearby. Mutation is performed by deciding a mutation probability p m and selecting strings, on which mutation is to be performed, at random. The procedure can be summarized as follows: Step 1 Calculate the approximate number of mutations to be performed by

Approx no. of mutations = nlp,

(10.11)

Step 2 Generate random numbers to decide whether mutation is to be performed on a particular population member or not. This is decided by a ‘coin toss’. That is, select a random number range to represent true and one to represent false. If the outcome is true, perform mutation, if false, do not.

Non-traditional Optimization Techniques

265

Step 3 If the outcome found in step 2 is true for a particular population member, then generate another random number to decide the mutation point over the length of the string. Once decided, flip the bit corresponding to the mutation point. With the end of mutation, the strings obtained represent the next generation. The same operations are carried out on this generation until the optimum value is encountered.

10.2.6 Comparison of GA with Traditional OptimizationTechniques 1. In GA, the coding discretizes the search space even if the function is continuous. This means that even discontinuous functions can be optimized using GA. But, in most traditional methods, differentiation is to be carried out and this is not possible for discontinuous functions. 2. GAS work with a population of points instead of a single point, as in most traditional methods. 3. Using GAS, multiple optimal solutions can be reached; whereas, using traditional methods, only a single optimum can be obtained. 4. For implementing GAS, no auxiliary information about the function is required. 5. GAS are probabilistic, while most traditional methods are deterministic.

10.2.7 Applications of GA The various applications of GAS include the following: Process design and optimization (Androulakis & Venkatasubramanian 199l), computer-aided molecular design (Venkatasubramanian et al. 1994), optimal design of the ammonia synthesis reactor (Upreti & Deb 1997), estimation of heat-transfer parameters in tdckle bed reactors (Babu & Vivek 1999), automated design of heat exchangers using artificial intelligence based optimization (Babu & Mohiddin 1999), optimal design of heat exchangers (Manish et al. 1999), etc. Apart from these, there are many applications in different disciplines (Goldberg 1989; Deb 2000). Some of the above-mentioned applications are discussed here in brief. 10.2.7.1 Process design and optimization (Androulakis & Venkatasubramanian 1991)

A general optimization framework for discrete and continuous problems based on genetic algorithmic techniques is presented. The proposed framework exhibits a structured information exchange leading to an efficient search procedure. The utility of the method is demonstrated by applying it to the design of an unsplit heat exchanger network. The basic idea of genetic algorithms to develop a new method, called the extended genetic search (EGS), is extended for optimization in continuous

266

Process Plant Simulation

spaces. The performance of the EGS for several test cases of non-linear unconstrained and constrained optimization problems is also presented. The performance of this technique with others, including simulated annealing, is compared. The proposed framework turns out to be quite robust and is able to locate global optimum solutions for problems where gradient-based algorithms fail. The results obtained were comparable to those obtained with simulated annealing. The implementational simplicity of the EGS framework makes it particularly useful as a tool for problems where approximate solutions are needed quickly. 10.2.7.2 Optimal design of ammonia synthesis reactor (Upreti & Deb 1996) This paper presents an optimal design procedure of an ammonia synthesis reactor using GAS-search and optimization techniques based on principles of natural genetics. GASare chosen as an optimization tool simply because of their successful application to many engineering optimization problems. The optimal problem requires maximization of an objective function subject to a number of equality constraints involving the solution of coupled differential equations. Although there exist at least a couple of other studies on the optimal design of the ammonia synthesis reactor, they have ignored some terms in the formulation of the objective function, for which the reported optimal solution does not match the solution obtained using an enumerative search technique. This paper includes those terms, and applies GAS to find the optimal reactor length. Simulation results with GAS find optimal reactor lengths at various feed-gas temperatures at the top of the reactor. The results of this paper are also found to agree with the reactors used in industries. The successful application of GAS in the ammonia reactor design suggests GAS’ immediate application to other reactor design or modelling. 10.2.7.3 Heat-transfer parameters in trickle bed reactors (Babu & Vivek 1999) The estimation of heat-transfer parameters in a trickle bed reactor (TBR), namely, k,, and h,, using the two-dimensional two-parameter pseudo-homogeneous model is attempted using the non-traditional optimization algorithm-genetic algorithm. The initial population of points is subjected to operators’ reproduction, crossover, and mutation to get the desired optimum. The iterative calculations are carried out as long as the sum of the squares of the errors is greater than 0.03, using the experimentally measured temperature profile at three different radial positions in a trickle bed reactor. There is a good agreement between the values of k,, and h, estimated using genetic algorithms and the well-proven radial temperature profile (RTP) method.

Non-traditional Optimization Techniques

267

10.2.7.4 Automated design of heat exchangers (Babu & Mohiddin 1999) Heat exchangers are very important in any process industry, and their optimal design is of crucial importance in terms of performance and economy. Due to several design variables and different alternatives for each variable, the design of a shell-and-tube heat exchanger can be posed as a large-scale discrete optimization problem. A genetic algorithm is a non-conventional optimization technique which scores over conventional techniques. In this study, genetic algorithm is applied to the optimal design of the shell-and-tube heat exchanger, and found to converge in very few (10) generations considering six design variables with a total of 4608 configurations.

10.2.8 Stepwlse Procedure for GA Implementation The following are the steps involved in solving an optimization problem using GA. Consider an objective function to be minimized in the interval 0 I xl, x2 I 6 as given below. The true solution for this problem is (0, 0.5). f ( x , , x2) = x1 - x2

+ 2x,2 + 2x1x2 + x22

Step 1 As a first step we use binary coding to represent the variables x1 and x2. In the calculations shown, we have used 10 bits for each variable (see columns 2 and 3 of Table lO.l), thereby making the total string length 20 for both the variables together. With 10 bits, we can get an accuracy of 1/21° of the search space, i.e., a solution accuracy of (6 - 0)/(21° - 1) or 0.006 in the search space of (0,6). Here, we have chosen roulette wheel selection, single-point crossover, and a bitwise mutation operator. The crossover and mutation probabilities @, and p,) are taken to be 0.8 and 0.05, respectively. We have used 20 points in the population. (The thumb rule or heuristic available for the population size Np is 5-10 times of the dimension of the problem. In the present case, we have two variables, x1 and x2, which means we are dealing with a two-dimensional problem. It implies we may choose anywhere between 10 and 20 points as the population size. We have chosen arbitrarily 20 as the population size. The random population is generated by a uniform random number generator with a random seed equal to 5 and shown as in columns 2 and 3 in Table 10.1. Finally, we have set the termination condition t = 30 and initialized the generation counter as t = 0.

The next step is to evaluate each string in the population. We calculate the fitness of the first string. The first substring (1 110111011) decodes to 29 + 28 + 27 + + 24 + 23 + 2l + 2' or 955. Thus, the corresponding parameter value is equal to [0 + (6 - O)] x 955/1023 or 5.601. The second substring (1001010000) decodes to 29 + 26 + 24 or 592. Thus the corresponding parameter value is equal to [0 + (6 - O ) ] x 592/1023 or 3.472. Thus the first string corresponds to the point Step 2

Table 10.1 Evaluation and reproduction x, (binary)

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

1110111011 1101001101 1100001111 0110110100 1000111111 1011010011 1111101110 1010100100 101OOOOOO1 0000OOO0OI 1011101100 111oo00110 0010101001 1001001011 0001111110 1010100000 0000101010 1111001000 0010101010 111010OOOO

x2 (binary)

10010 1OOOO I11 1100011 010101001 1 1010100111 1111010101 1111101011 1101001110 1011011101 I100101111 1 lo001 1000 0001110110 0010101010 0000001010 1111100011 1101110111 0011100001 0100011100 1001011000 0010101011 10100 10100

x,

5.601 4.956 4.592 2.557 3.372 4.240 5.900 3.965 3.760 0.006 4.387 5.290 0.991 3.443 0.739 3.941 0.246 5.677 0.997 5.443

x2

3.472 5.836 1.988 3.982 5.754 5.883 4.962 4.299 4.780 4.645 0.692 0.997 0.059 5.836 5.202 1.320 1.666 3.519 1.003 3.87 1

flx)

115.827 140.145 66.999 47.880 92.278 118.818 153.739 83.678 86.038 16.993 48.740 71.812 3.017 95.552 31.382 45.834 2.297 118.967 4.988 1 17.943

F(x)

0.009 0.007 0.015 0.020 0.011 0.008 0.006 0.012 0.01 1 0.056 0.020 0.014 0.249 0.010 0.031 0.021 0.303 0.008 0.167 0.008

Cum prob.

Rand no.

N,

pi

0.009 0.007 0.015 0.021 0.011 0.008 0.007 0.012 0.012 0.056 0.020 0.014 0.252 0.010 0.03 1 0.022 0.307 0.008 0.169 0.009

0.009 0.0 16 0.031 0.051 0.062 0.07 1 0.077 0.089 0.101 0.157 0.178 0.191 0.443 0.454 0.485 0.507 0.814 0.822 0.991 1.ooo

0.612 0.441 0.713 0.923 0.245 0.651 0.885 0.775 0.348 0.812 0.745 0.029 0.164 0.786 0.887 0.326 0.734 0.639 0.998 0.008

16 12 16

Prob.

S

Mating pool x2 (binary)

x, (binary)

18

12 16 18 16 12 16 16 2 10 16 18 12 16 16 19 0

1 0 1 0 0 0 0 0 0 0 1 0 4 0 0 0 9 0 3 1

0000101010 0010101001 0000101010 0010101010 0010101001 0000101010 0010101010 0000101010 0010101001 0000101010 0000101010 1100001111 1011101100 0000101010 0010101010 0010101001 0000101010 0000101010 1110100000 1110111011

0100011100 0000001010 0100011100 0010101011 000000 1010 0100011100 0010101011 0100011100 0000001010 0100011100 0100011100 0101010011 0001110110 0100011100 0010101011 0000001010 0100011100 0100011100 1010010100 1001010000

Non-traditional Optimization Techniques

269

(5.601, 3.472). These values are shown for all the population points as x1 and x2in columns 4 and 5 , respectively, of Table 10.1. These values can now be substituted in the objective function expression to obtain the function valueflx). The function value at this point was found to be 115.827. These values are shown in column 6 of Table 10.1. Since this is a minimization function, the fitness function value is given by F(x) = 1/[1 +Ax)]. Therefore, the fitness function value is 1/(1 + 115.827) = 0.00855966514, which is shown as 0.009 in the table. Similarly, other strings in the population are evaluated and their fitness values are calculated as shown in column 6 of Table 10.1. These values are used in the reproduction operation. Step 3 Here we check for termination. Since this is the first iteration t = 0, we proceed to step 4. We continue like this till t = 30 (i.e., the maximum number of generations which is pre-specified. Step 4 At this step, we select the good strings that must form the mating pool. In order to use the roulette wheel selection procedure, we first calculate the probability of selection of the given population point. The probability of selection of the ith string into the mating pool is given by Eq. (10.9) as shown below:

n

X F j = sum of the fitness values of all the strings j=l

= 0.009 + 0.007 + 0.015 + 0.020 + 0.011 + 0.008 + 0.006 + 0.012 + 0.011 + 0.056 + 0.020 + 0.014 + 0.249 + 0.010 + 0.031 + 0.021 + 0.303 + 0.008 + 0.167 + 0.008 = 0.986

Then, for the first string, p 1 = 0.009/0.986 = 0.009127 This procedure is followed for all the 20 population strings and shown as pi (probability of each string being copied in the mating pool) in column 8 of Table 10.1. Next the cumulative probabilities are calculated and represented in column 9 of Table 10.1.In order to form the mating pool, we create random numbers between 0 and 1 (column 10 of Table 10.1) and identify the particular string which is specified by each of these random numbers. For example, if the random number generated is 0.612, the 16th string gets a copy in the mating pool because that string occupies the interval (0.507, 0.814). Column 11 shows the selected string. Similarly, other strings are selected according to random numbers shown in column 10.After this selection procedure is repeated n times, the number of selected copies

270

Process Plant Simulation

for each string is counted. This is shown in column 12 of Table 10.1. Columns 8 and 12 reveal that the theoretical expected count and the true count of each string more or less agree with each other. Thus we can see that the reproduction operator probabilistically eliminates inferior points from further consideration. As can be seen, the 16th string is selected nine times, while the 12th and 18th strings are selected four and three times, respectively. The Oth, 2nd, loth, and 19th strings are selected only once, and all the remaining strings are not selected at all. Notice that not all selected points are better than all rejected points. Though roulette wheel selection is easier, it is noisy, so we can also use other reproduction techniques such as ‘tournament selection’.All the selected population points are shown as the mating pool in columns 13 and 14 of Table 10.1. Step 5 In this step we perform crossover. In single-point crossover, two strings are selected at random and crossed at a random site. First, we have to choose a site for crossover. For this, a random number between 1 and 20 is generated and the respective position is taken as crossover site for each string. Here we perform a crossover between adjacent strings, and since we are performing single-point crossover, the crossover site is taken to be same for adjacent strings. Not all strings take part in the crossover. It again depends on the crossover probability we have chosen, i.e., if it is 0.8 then 80% of the strings take part in the crossover. Now, which pairs of strings take part in the crossover is again decided by generating a random number between 0 and 1; and if it is less than 0.8 (crossover probability) we go for the crossover, otherwise we do not. Let us take the first two strings in the mating pool: 00001010100100011100and 0010101001000OOO1010. The crossover site was randomly generated as 11 (as shown in column 4 of Table 10.2). So we swap all the bits to the right of the 1lth position (i.e., 12th bit onwards) in the first two strings to get the children strings, i.e.,

Parent strings: 00001010100100011100 00101010010000001010 Children strings: 00001010100000001010 00101010010100011100 Like this, we generate two new strings for every pair of substrings in the mating pool (the crossover site for strings 0 and 1 is 11, for strings 2 and 3 is 0, for strings 4 and 5 is 9, for strings 6 and 7 is 1, for strings 8 and 9 is 0, for strings 12 and 13 is 12, for strings 14 and 15 is 10, for strings 16 and 17 is 17, and for strings 18 and 19 it is 1) and form the intermediate population, which is shown in columns 5 and 6 of Table 10.2.

Table 10.2 Crossover and mutation. H is the crossover site.

Mating pool XI

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

(binary)

0000101010 001010 1001 0000101010 0010101010 0010101001 0000101010 0010101010 0000101010 0010101001 0000101010 0000101010 1100001111 1011101100 000010 1010 0010101010 0010 101001 0000 1010 10 0000101010 11 10100000 1110111011

After crossover

(binary) 010001 1100 00000010 10 010001 1100 0010101011 000000 10 10 010001 1100 0010101011 010001 1100 000000 1010 010001 1100 0100011100 0101010011 0001 110110 0100011100 001010101 1 0000001010 010001 1100 010001 1100 1010010100 10010 10000

After mutation

H

x1 (binary)

x2 (binary)

x , (binary)

x2 (binary)

11 11 0 0 9 9 1 1 0 0

oo0010 1010 0010101001 0010101010 o000101010 001010I 000 oo00101011 0000 101010 0010101010 0000101010 00 10101001 0000101010 1100001111 1011101100 000010 1010 0010101010 001010 1001 m101010 0000101010 1110111011 1110100000

00m1010 Olo0011100 0010101011 Olo0011100 0100011100 0000001010 01o0011100 0010101011 Olo0011100 00oO001010 Olo0011100 0101010011 0000011100 0101110110 0000001010 0010101011 0100011100 0100011100 1001010000 1010010100

0000101010 00lO!Ol001 0010101010 0oO0101010 0010101oO0 0000101011 0000101010 001o001010 oooo101010 0010101011 1o00101010 110oO01110 1011101100 0011101010 0010101010 0010101001 0oO0101011 oo00101010 1110111011 1110100000

OOOOOOlOlO 01oO011 1 0 0 lOlOlOI00l 010101I100 01oO011100 OOOOOO1010 Olo0011100 001010101 1 Olo0011100

N N 12 12 10 10 17 17 1 1

OOOOOOIOOO 01o0010100 1101010011 01oO011100 0101010110 m001000 001o001011

ooooo11 100 010001 lo00 100101m 1010010100

X,

Xz

0.246 0.991 0.997 0.246 0.985 0.252 0.246 0.809 0.246 1.003 3.249 4.587 4.387 1.372 0.997 0.991 0.252 0.246 5.601 5.443

0.059 1.666 3.994 2.041 1.666 0.059 1.666 1.003 1.666 0.047 1.619 4.991 1.666 2.006 0.047 0.815 0.164 1.642 3.472 3.871

Ax)

0.341 7.367 22.909 3.498 7.318 0.354 2.297 3.746 2.297 3.064 35.886 112.364 58.604 12.663 3.034 4.422 0.325 2.231 115.827 117.943

F(x)

0.746 0.120 0.042 0.222 0.120 0.739 0.303 0.21 1 0.303 0.246 0.027 0.0°9 0.017 0.073 0.248 0.184 0.755 0.309 0.009 0.008

3

z? g. 3

0" 3'

2 0 J L

272

Process Plant Simulation

Step 6 By performing a crossover alone we cannot guarantee better child strings. Sometimes a situation may arise when a neighbouring string is better. For example, if we want a child string with 1 in the second position, performing crossover on the parent strings (00001010100000001010, 0010101001010001 1100) will not help unless the bit in the second position of either string is 1. This can be done by mutation, which is nothing but flipping of the bits (0 becomes 1, and 1 becomes 0). So we again randomly find out, i.e., generate a random number between 0 and 1 for each and every bit. If the value of the random number is less than 0.05 (i.e., pm, the mutation probability), we flip that bit, otherwise we do not. Like this, we repeat the process for each and every string. Like the crossover operator, mutation has created some points better and some points worse than the original points. The new points after mutation are shown in columns 7 and 8 of Table 10.2. The cross over sites are shown in bold, i.e., the sites where the bits are shown in bold in columns 5 and 6 are flipped and columns 7 and 8 are generated. And this actually enables GA to explore the search space. Note that the mutation is applied at very few places. In fact, we want it to happen in this limited way and that is the reason why the mutation probability is always kept to a very low value (0.05 in this case).

Step 7 The resulting population after crossover and mutation becomes the new population for the next generation (columns 7 and 8 of Table 10.2). We now evaluate each string as before, by first identifying the substrings for each variable and mapping the decoded values of the substrings in the chosen intervals. These are shown in columns 9 and 10 of Table 10.2. This completes one iteration (or one generation) of the genetic algorithm. We increment the generation counter to t = 1 and proceed to step 3 for next iteration (generation). Like this, we proceed till r becomes 30.

10.3 Differential Evolution 10.3.1 Introduction Differential evolution (DE) is a generic name for a group of algorithms that are based on the principles of GA but have some inherent advantages over GA. DE algorithms are very robust and efficient in that they are able to find the global optimum of a function with ease and accuracy. DE algorithms are faster than GAS. GAS evaluate the fitness of a point to search for the optimum. In other words, GAS evaluate a vector’s suitability. In DE, this vector’s suitability is called its cost or profit depending on whether the problem is a minimization or a maximization problem.

Non-traditional Optimization Techniques H

H

273

In GAS only integers are used and the coding is done in binary format; in DE, no coding is involved and floating-point numbers are directly used. If GASare to be made to handle negative integers, then the coding becomes complicated since a separate signed bit has to be used; in DE, this problem does not arise at all.

10.3.2 XOR Versus Add In GASwhen mutation is performed, bits are flipped at random with some mutation probability. This is essentially an XOR operation. In DE, direct addition is used. For example, In GA, 16 = 10,000 and 15 = 01111 (after mutation). In DE, 15 = 16 + (-1). As is evident, the computation time for flipping four bits and encoding and decoding the numbers in GAS is far more than the simple addition operation performed in the case of DE. The question that remains is how to decide how much to add? This is the main philosophy behind DE.

10.3.3 DE at a Glance As already stated, DE in principle is similar to GA. So, as in GA, we use a population of points in our search for the optimum. The population size is denoted by NP. The dimension of each vector is denoted by D. The main operation is the NP number of competitions that are to be carried out to decide the next generation. To start with, we have a population of NP vectors within the range of the objective function. We select one of these NP vectors as our target vector. We then randomly select two vectors from the population and find the difference between them (vector subtraction). This difference is multiplied by a factor F (specified at the start) and added to a third randomly selected vector. The result is called the noisy random vector. Subsequently, crossover is performed between the target vector and the noisy random vector to produce the trial vector. Then, a competition between the trial vector and the target vector is performed and the winner is replaced into the population. The same procedure is carried out NP times to decide the next generation of vectors. This sequence is continued till some convergence criterion is met. This summarizes the basic procedure carried out in differential evolution. The details of this procedure are described below.

10.3.4 Steps Performed in DE Assume that the objective function is of D dimensions and that it has to be minimized. The weighting constant F and the crossover constant CR are specified. Refer to Fig. 10.1 for the schematic diagram of differential evolution.

274

Process Plant Simulation Population size = NP

Target vector

Xi

-7I-y vector

Crossover

Trial vector

.'-> cost of

x,'x,

I

////

Fig. 10.1 Schematic diagram of differential evolution

Step 1 Generate NP random vectors as the initial population: Generate (NP x D) random numbers and linearize the range between 0 and 1 to cover the entire range of the function. From these (NP x 0)random numbers, generate NP random vectors, each of dimension D, by mapping the random numbers over the range of the function. Step 2 Choose a target vector from the population of size NP: First generate a random number between 0 and 1. From the value of the random number decide which population member is to be selected as the target vector ( X i )(a linear mapping rule can be used). Step 3 Choose two vectors at random from the population and find the weighted difference: Generate two random numbers. Decide which two population members are to be selected (X,,Xb). Find the vector difference between the two vectors (X,- Xb). Multiply this difference by F to obtain the weighted difference.

Weighted difference = F(X, - Xb)

(10.12)

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Step 4 Find the noisy random vector: Generate a random number. Choose a third Add this vector to the weighted difference random vector from the population (X,). . to obtain the noisy random vector (Xf) (10.13) Noisy random vector (Xf) = X , + F(X, - Xb) Step 5 Perform crossover between Xiand Xi to find X,,the trial vector: Generate D random numbers. For each of the D dimensions, if the random number is greater than CR, copy the value from Xiinto the trial vector; if the random number is less than CR, copy the value from Xf into the trial vector. Step 6 Calculate the cost of the trial vector and the target vector: For a minimization problem, calculate the function value directly and this is the cost. For a maximization problem, transform the objective function Ax) using the rule, F(x) = 1/[1 +Ax)] and calculate the value of the cost. Alternatively, directly calculate the value ofAx) and this yields the profit. In case cost is calculated, the vector that yields the lesser cost replaces the population member in the initial population. In case profit is calculated, the vector with the greater profit replaces the population member in the initial population. Steps 1-6 are continued till some stopping criterion is met. This may be of two kinds. One may be some convergence criterion that states that the error in the minimum or maximum between two previous generations should be less than some specified value. The other may be an upper bound on the number of generations. The stopping criterion may be a combination of the two. Either way, once the stopping criterion is met, the computations are terminated.

10.3.5 Choice of DE Key Parameters (NP, F, and CR) NP should be 5-10 times the value of D, that is, the dimension of the problem. Choose F = 0.5 initially. If this leads to premature convergence, then increase F. The range of values of F is 0 c F c 1.2, but the optimal range is 0.4 c F c 1.0. Values of F c 0.4 and F > 1.0 are seldom effective. CR = 0.9 is a good first guess. Try CR = 0.9 first and then try CR = 0.1. Judging by the speed, choose a value of CR between 0 and 1.

10.3.6 Strategies in DE Several strategies have been proposed in DE. A list of them is as follows: (a) DE/ best/l/exp, (b) DE/best/2/exp, (c) DE/best/l/bin, (d) DE/best/2/bin, (e) DE/randto-best/l/exp, (0 DE/rand-to-best/l/bin, (g) DE/rand/l/bin, (h) DE/rand/2/bin, (i) DE/rand/l/exp, and Q) DE/rand/2/exp. The notations that have been used in the above strategies have distinct meanings. The notation after the first '/', i.e., best, rand-to-best, and rand, denotes the choice

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of the vector X,. ‘Best’means that X , corresponds to the vector in the population that has the minimum cost or maximum profit. ‘Rand-to-best’ means that the weighted difference between the best vector and another randomly chosen vector is taken as X,. ‘Rand’ means X , corresponds to a randomly chosen vector from the population. The notation after the second ‘/’, i.e., 1 or 2, denotes the number of sets of random vectors that are chosen from the population for the computation of the weighted difference. Thus ‘2’ means the value of X ; is computed using the following expression: (10.14) x: = xc + F ( X a l - x b l ) + F ( X a 2 - xb2,) where Xal, x b l , Xa2, and x b 2 are random vectors. The notation after the third ‘/’, i.e., exp or bin, denotes the methodology of crossover that has been used. ‘Exp’ denotes that the selection technique used is exponential and ‘bin’ denotes that the method used is binomial. What has been described in the preceding discussion is the binomial technique where the random number for each dimension is compared with CR to decide from where the value should be copied into the trial vector. But, in the exponential method, the instant a random number becomes greater than CR, the rest of the values are copied from the target vector. Apart from the above-mentioned strategies, there are some more innovative strategies that are being worked upon by the author’s group (discussed in Sec. 10.3.7) and implemented on some application (Sec. 10.3.8).

10.3.7 Innovations on DE The following additional strategies are proposed by the author and his associates, in addition to the ten strategies listed before: (k) DE/best/3/exp, (1) DE/best/3/bin, (m) DE/rand/3/exp, and (n) DE/rand/3/bin. Very recently, a new concept of ‘nested DE’ has been successfully implemented for the optimal design of an auto-thermal ammonia synthesis reactor (Babu et al. 2002). This concept uses DE within DE wherein an outer loop takes care of optimization of key parameters (NP is the population size, CR is the crossover constant, and F is the scaling factor) with the objective of minimizing the number of generations required, while an inner loop takes care of optimizing the problem variables. Yet, the complex objective is the one that takes care of minimizing the number of generations/function evaluations and the standard deviation in the set of solutions at the last generatiodfunction evaluation, and trying to maximize the robustness of the algorithm.

10.3.8 Applications of DE DE is catching up fast and is being applied to a wide range of complex problems. Some of the applications of DE include the following: digital filter design (Storn

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1995), fuzzy-decision-making problems of fuel ethanol production (Wang et al. 1998), design of a fuzzy logic controller (Sastry et al. 1998), batch fermentation process (Chiou & Wang 1999; Wang & Cheng 1999), dynamic optimization of a continuous polymer reactor (Lee et al. 1999), estimation of heat-transfer Parameters in a trickle bed reactor (Babu & Sastry 1999), optimal design of heat exchangers (Babu & Munawar 2000; 2001), Optimization and synthesis of heat integrated distillation systems (Babu & Singh 2000), optimization of an alkylation reaction (Babu & Chaturvedi 2000), optimization of non-linear functions (Babu & Angira 2001a), optimization of thermal cracker operation (Babu & Angira 2001b), scenariointegrated optimization of dynamic systems (Babu & Gautam 2001), optimal design of an auto-thermal ammonia synthesis reactor (Babu et al. 2002), global optimization of mixed integer non-linear programming (MINLP) problems (Babu & Angira 2002a), optimization of non-linear chemical engineering processes (Babu & Angira 2002b; Angira & Babu 2003), etc. A detailed bibliography on DE covering the applications on various engineering and other disciplines is available in literature (Lampinen 2003). Let us discuss some of them in brief. 10.3.8.1 Digital filter design (Storn 1995) The task of designing an 18-parameter infinite impulse response (IIR) filter which has to meet stringent specifications for both magnitude response and group delay is discussed. This problem must normally be tackled by specialized design methods and requires expertise in digital signal processing for its solution. The use of the general-purpose minimization method DE, however, allows one to perform the filter design with a minimum knowledge about digital filters. 10.3.8.2 Fuzzy-decision-makingproblems of fuel ethanol production (Wang et al. 1998) A fuzzy-decision-making procedure is applied to find the optimal feed policy of a fed-batch fermentation process for fuel ethanol production using a genetically engineered Succhuromyces yeast 1400 (pLNH33). The policy consisted of feed flow rate, feed concentration, and fermentation time. The recombinant yeast 1400 (pLNH33) can utilize glucose and xylose simultaneously to produce ethanol. However, the parent yeast utilizes glucose only. A partially selective model is used to describe the kinetic behaviour of the process. In this study, this partially selective fermentation process is formulated as a general multiple-objective optimal control problem. By using an assigned membership function for each of the objectives, the general multiple-objective optimization problem can be converted into a maximizing decision problem. In order to obtain a global solution, a hybrid method of differential evolution is introduced to solve the maximizing decision problem. A simple guideline is introduced in the interactive programming procedures to find a satisfactory solution to the general multiple-objective optimization problem.

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10.3.8.3 Batch fermentation process (Chiou & Wang 1999) A hybrid algorithm of evolutionary optimization, called hybrid differential evolution (HDE), is developed in this study. The acceleration phase and migration phase are embedded into the original algorithm of DE. These two phases are used to improve the convergence speed without decreasing the diversity among individuals. With some assumptions, this hybrid method is shown as a method using N p parallel processors of the two-member evolution strategy, where Np is the number of individuals in the solution space. The multiplier updating method is introduced in the proposed method to solve the constrained optimization problems. The topology of the augmented Lagrange function and the necessary conditions for the approach are also investigated. The method is then extended to solve the optimal control and optimal parameter selection problems. A fed-batch fermentation example is used to investigate the effectiveness of the proposed method. For comparison, several alternate methods are also employed to solve this process. 10.3.8.4 Estimation of heat-transfer parameters in a trickle bed reactor (Babu & Sastry 1999) A new non-sequential technique is proposed for the estimation of effective heattransfer parameters using radial temperature profile measurements in a gas-liquid co-current down flow through packed bed reactors (often referred to as trickle bed reactors). The orthogonal collocation method combined with a new optimization technique (DE) is employed for estimation. DE is an exceptionally simple, fast, and robust, population-based search algorithm which is able to locate near-optimal solutions to difficult problems. The results obtained from this new technique are compared with those of the RTP method. Results indicate that orthogonal collocation augmented with DE offers a powerful alternative to other methods reported in the literature. The proposed technique takes less computational time to Gonverge when compared to the existing techniques, without compromising on the accuracy of the parameter estimates. This technique takes an average 10 s on a 90-MHz Pentium processor as compared to 30 s taken by the RTP method. This new technique also assures convergence from any starting point and requires a less number of function evaluations. 10.3.8.5 Optimal design of heat exchangers (Babu & Munawar 2000) We all must accept that ‘nature knows the best!’ and we try to mimic nature and its natural phenomena in our efforts for the development of technology and resources. Few of the outcomes of such efforts are the evolution of non-traditional optimization techniques such as SA, GA, and DE. In the recent past, these algorithms have been successfully applied to solve complex engineering optimization problems and are widely known to diffuse very close to the global optimum solution. The simple

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279

GA was later modified to enable it to work with real variables, followed by the evolution of strategies for faster convergence. Differential evolution DE, an improved version of GA, is an exceptionally simple evolution strategy that is significantly faster and robust at numerical optimization and is more likely to find a function’s true global optimum. This paper presents the application of differential evolution for the optimal design of shell-and-tube heat exchangers. A primary objective in the heat-exchanger (HE) design is the estimation of the minimum heat-transfer area as it governs the overall cost of the HE. However, many discrete combinations of design variables are possible. Hence a design engineer needs an efficient strategy to search for the global minimum HE cost. In the present study, for the first time, differential evolution has been successfully applied with 161,280 design configurations by varying the design parameters such as the tube’s outer diameter, tube pitch, tube length, number of tube passes, baffle spacing, and baffle cut. For comparison, SA and GA are also applied to the above problem. Genetic algorithms consume less computational time relative to SA and have the advantage over other methods in obtaining multiple solutions of the same quality, thus providing more flexibility to a designer. Among DEs advantages are its simple structure, ease of use, speed, and robustness. Significant savings in the HE cost and computational time have been achieved using DE, proving it to be a potential tool for accurate and faster optimization. 10.3.8.6 Optimization and synthesis of heat integrated distillation systems (Babu & Singh 2000)

A new evolutionary technique for optimization and synthesis of heat integrated distillation systems is proposed. This evolutionary technique, differential evolution, inherits basic features of a genetic algorithm. Synthesis and optimization of heat integrated distillation systems is complex due to the huge search space in which global optimum may be located. Traditional techniques do not locate the global optimum when heat integration is also taken into consideration. Differential evolution is a population-based search algorithm that is able to locate near optimum solution to difficult problems of multidimensional search space. Differential evolution requires a less number of function evaluations and assures convergence from any starting point. 10.3.8.7 Optimization of an alkylation reaction (Babu & Chaturvedi 2001)

The alkylation process is an important reaction used in refineries to upgrade light olefins and isobutene into a much more highly valued gasoline component. The primary alkylation reaction involves the reaction of isobutene with a light olefin, such as butylene, in the presence of a strong acid catalyst to form the high octane. A long-standing problem is to determine the optimal operating conditions for the

280

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alkylation process. In order to achieve this, differential evolution has been employed. An improved scheme of genetic algorithms, differential evolution is exceptionally simple, fast, and robust. It is primarily a search algorithm that is able to locate near-optimal solutions to difficult problems. An analysis of the results obtained on varying the parameters of the problem is also presented. 10.3.8.8 Optimization of non-linear functions (Babu & Angita 2001)

Global optimization is important if the function has a number of optima, of which some are local and some global. Since many real-world problems contain multiple optima, traditional methods may not be adequate to solve such problems due to the possibility of getting trapped at local optimum. Hence they do not ensure global optima. In this paper, a non-linear function (Himmelblau function) with three local optima and one global optimum has been solved by using DE. The results obtained from DE are compared with that of GA. The Himmelblau function is given by

f ( x ) = ( x ; + x2 - 11)2 + (XI + x; - 7)2 Figure 10.2 shows the F A , DEIFA, GA versus NINP for GA and DE [strategy 2 (DE2) and strategy 7 (DE7)], where F A , DE is the average number of function evaluation using DE, FA,GA is the average number of function evaluation using GA, N is the population size in the case of GA, and NP is the population size in the case of DE. 700

+GA

*DE7 0

20

30

40

Population size ( N I N P )

Fig. 10.2 Comparison of GA and DE

From Fig. 10.2, it is clear that F A , DE c F A , GA. Strategy 7 (that takes minimum number of function evaluations) is found to be the best for the present problem. Hence it can be concluded that the performance of DE is better than that of GA. This successful application of DE over GA for the Himmelblau function indicates that DE has great potential and can be applied to advantage in all the highly nonlinear and complex engineering problems.

Non-traditional Optimization Techniques

281

10.3.8.9 Optimal design of shell-and-tube heat exchangers (Babu & Munawar 2001)

This paper presents the application of DE for the optimal design of shell-and-tube heat exchangers. The main objective in any heat-exchanger design is the estimation of the minimum heat-transfer area required for a given heat duty, as it governs the overall cost of the heat exchanger. Millions of configurations are possible with various design variables such as the outer diameter, pitch, and length of the tubes; tube passes; baffle spacing; baffle cut, etc. Hence a design engineer needs an efficient strategy in searching for the global minimum. In the present study, for the first time DE, an improved version of GAS, has been successfully applied with different strategies for 161,280 design configurations using Bell’s method to find the heattransfer area. In the application of DE, 9680 combinations of the key parameters are considered. For comparison, GAS are also applied to the same case study with 1080 combinations of its parameters. For this optimal design problem, it is found that DE, an exceptionally simple evolution strategy, is significantly faster compared to GA and yields the global optimum for a wide range of the key parameters. 10.3.8.1 0 Optimization of thermal cracker operation (Babu & Angira 2001)

This paper presents the application of DE, an evolutionary computation method, for the optimization of thermal cracking operation. The objective in this problem is the estimation of optimal flow rates of different feeds to the cracking furnace under the restriction on ethylene and propylene production. Thousands of combinations of feeds are possible. Hence an efficient optimization strategy is essential for searching the global optimum. In the present study, the LP simplex method and DE have been successfully applied with different strategies to find the optimum flow rates of different feeds. In the application of DE, various combinations of the key parameters are considered. It is found that DE is significantly faster and yields the global optimum for a wide range of the key parameters. The results obtained from DE are compared with that of LP simplex method. 10.3.8.11 Scenario-integratedoptimization of dynamic systems (Babu & Gautam 2001)

In any chemical process industry, it is necessary to assess the consequences of unexpected events such as equipment failures, change of raw material prices. and product demand. In this work, these unexpected events are called ‘scenarios’, which allow the determination of operational strategies that are economically optimal and keep the plant within a chosen regime in the state space. This is achieved by the consideration of several scenarios. The formulations in this paper are based on a unifying model framework containingboth discrete and continuous dynamics. Taking the case of a single scenario, differential evolution is used for the scenario-integrated

282

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dynamic optimization of the formulated problem. The applicability of this technique is demonstrated with an example. 10.3.8.12 Optimal design of an auto-thermal ammonia synthesis reactor (Babu et al. 2002) In the present study, DE is applied to the optimal design of an auto-thermal ammonia synthesis reactor. This paper also presents the new concept of ‘nested’ DE (DE is also used to find the best combination of key parameters of DE itself). The main objective in the optimal design of an auto-thermal ammonia synthesis reactor is the estimation of the optimal length of a reactor for different top temperatures with the constraints of energy and mass balance of reaction and feed-gas temperature and mass-flow rate of nitrogen for ammonia production. Thousands of combinations of feed-gas temperature, nitrogen mass-flow rate, reacting gas temperature, and reactor length are possible. This paper presents the application of four methods, namely, Euler’s method, Runge-Kutta method (both variable and constant step size), and Gear’s method in combination with DE, and verifies the contradictory results reported using GA in earlier literature. A software package POLYMATH is also used to solve the three equality constraints, i.e., three coupled differential equations. Apart from determining the optimal reactor length, the comparison of results obtained from different methods is presented. DE is found to be a robust, fast, and simple evolutionary computation technique for optimization problems. 10.3.8.1 3 Global Optimization of MINLP problems (Babu & Angira 2002a) The global optimization of MINLP problems is an active research area in many engineering fields. In this work, DE is used for the optimization of non-convex MINLP problems and a comparison is made among the algorithms based on hybrid of simplex and simulated annealing (M-SIMPSA), GA, and DE. It is found that DE is significantly faster and yields the global optimum for a wide range of the key parameters. Results indicate that DE is more reliable, efficient, and hence a better approach to the optimization of non-convex non-linear problems. DE is found to be the best evolutionary computation method in all the problems studied. 10.3.8.14 Global optimization of non-linear chemical engineering processes (Babu & Angira 2002b; Angira & Babu 2003) The first paper (Babu & Angira 2002b) presents the application of DE for the optimization of non-linear chemical processes. Two test problems optimum fuel allocation in power plants and optimization of the drying process for a throughcirculation dryer, taken as case studies, are solved using DE. Comparison is made with a traditional algorithm based on direct search and systematic search region reduction. It is found that DE is significantly faster and yields the global optimum for a wide range of the key parameters.

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The second paper (Angira & Babu 2003) deals with the DE application to two highly non-linear chemical engineering processes, namely, the heat-exchanger network problem and reactor network problem. DE is found to be exceptionally simple, significantly faster, and robust at numerical optimization and it is more likely to find a function’s true global optimum. Comparison is made with the aBB algorithm (which is based on a branch-and-bound approach). The results indicate that performance of DE is better than the aBB algorithm.

10.3.9 Stepwise Procedure for DE Implementation The following are the steps involved in solving an optimization problem using DE. Let us consider the same objective function that is used for explaining the GA methodology in Sec. 10.2.7. The function to be minimized is given below in the interval 0 I x l , x2 I 6. The true solution for this problem is (0,0.5).

f ( x , , x 2 ) = x1 - x2

+ 2x12 + 2x,x2 + x22

(10.15)

The DE key parameters are chosen as NP = 20 (10 times the dimension of the problem, i.e., 10 x 2 = 20), CR = 0.5, F = 0.8. Step 1 Population initialization Initialize population (20 points) randomly between the given upper and lower bounds for all the parameters (second and third columns of Table 10.3). Table 10.3 Initial population. x, and x, are variables and cost is the objective function x1

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

3.677337 2.338032 0.528440 1.187137 3.133475 3.216689 0.447552 4.642 198 2.412028 5.517039 4.42203 1 3.36038 1 4.689647 4.020225 4.645876 4.023861 5.000684 3.093430 4.498183 5.460903

x7

0.222648 3.289922 5.870270 4.523740 4.677932 2.511420 1.098901 1.944511 0.662900 5.075050 4.354949 3.322859 2.206954 3.727019 2.039149 3.51 1431 5.525994 4.681739 2.8 10695 3.884746

cost

32.187374 36.188379 35.8809 12 30.686806 69.292287 43.863578 1.940472 67.632426 17.022193 143.072049 96.656809 55.995388 72.038578 76.475217 68.88044 1 73.484507 135.292475 68.434256 75.340828 118.738774

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Step 2 Cost evaluation Calculate the objective function value [using Eq. (10.15)] for initial population (fourth column in Table 10.3). Step 3 Mutation and crossover In Table 10.4, i is population counter i = (0, 1, 2, ..., 19) (first column). 1. Randomly choose three population points a , b, and c such that i f a f b C (see the second, third, and fourth columns in Table 10.4). 2. Select randomly a parameterj for mutation (fifth column in Table 10.4), 0’ = 0, 1). 3. Generate a random number [0,1] (sixth and seventh columns in Table 10.4).

*

If this is less than CR (0.5), i.e., the crossover constant mutates the parameter as shown in the equation below: trial [ j ] = x l [ c ] [ j ]

+ F ( x l [a] [ j ] - x l

[b] [ j ] ) ,

where F = 0.8 for this problem. If this is not less than CR, take the parameter as shown in the equation below: trial [ j ] = xl [i] [ j ]

(eighth and ninth columns in Table 10.4). Check for bounds; if bounds are violated, then randomly generate the parameter as shown below: trial [ j ] = lower limit

+ rand.no. [0, 13 (upper limit - lower limit);

(Columns 10 and 11 in Table 10.4) (lower limit = 0.0 and upper limit = 6.0). Repeat 3 til.1 all parameters are mutated. Here we have two parameters. Step 4 Evaluation Calculate the objective function value for the vector obtained after mutation and crossover (see column 12 in Table 10.4). Step 5 Selection Select the least cost vector for next generation, if the problem is of minimization as is the case with the present problem (see columns 2-4 in Table 10.5). Step 6 Repeat Repeat steps 3 to 5 for a specified number of generations, or till some termination criterion is met. The optima is at x1 = 0.0 and x2 = 0.5 withflx) = - 0.25.

Table 10.4 Mutation and crossover i

a

b

c

0 1 2 3 4

8 14 4 2 15 19 15 1 3 19 16 15 2 9 8 17 11 8 9 0

14 15 6 7 11 15 10 19 18 16 11 2 6 19 3 11 14 14 7 9

7 3 16 6 13 0 14 3 13 17 7 9 10 17 5 18 4 7 13 6

5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

j

1 1 0 1 1 1 1

0 1 0 0 1 0 1 1 1 1 1 1 1

rand-no.[0, 11

0.4602 1 0.894391 0.31743 0.381619 0.058299 0.659134 0.1 15894 0.086053 0.095357 0.006763 0.214108 0.845929 0.781496 0.0628 14 0.564136 0.24459 0.005858 0.604323 0.889923 0.127835

rand-no.[0, 11 0.699338 0.5914 0.260761 0.315992 0.882461 0.978825 0.613161 0.824881 0.457 I95 0.796946 0.073992 0.354696 0.255899 0.146946 0.54419 0.71658 1 0.551914 0.201341 0.197326 0.064069

x,

x2

2.855120 1.684749 7.149422 -2.84345 4.551009 4.826971 4.327340 -1.31 1I6 1.371389 3.461605 5.954440 8.3 13375 4.689647 3.138339 4.196601 4.284621 2.105079 2.855120 4.720098 -1.0242 1

0.843512 3.289922 8.389219 4.23951 3.877877 2.511420 1.364335 4.04788 1 5.097455 3.36874 1 3.707019 3.322859 8.172045 5.633982 2.039149 3.897799 5.704900 4.681739 2.810695 -2.78302

XI

1.935486 4.197434

2.563926

3.239810

3.081357

x2

cost

23.843 199 25.980564 1.109460 13.843822 88.758375 92.430927 79.467 166 54.084030 48.805758 40.000583 58.729204 131&I6566 53.481945 2.577234 76.912755 84.307141 58.653499 85.6968 13 61327322 63.129346 80.901567 3.496450 52.347228

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Table 10.5 Next generation population XI

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

2.855 120 1.684749 1.935486 1.187137 3.133475 3.216689 0.447552 2.563926 2.412028 3.46 1605 4.42203 1 3.239810 4.689647 4.020225 4.196601 4.023861 2.105079 2.855120 4.498 183 3.08 1357

x2

0.843512 3.289922 1.109460 4.523740 4.677932 2.51 1420 1.098901 4.047881 0.662900 3.368741 4.354949 3.322859 2.206954 3.7270 19 2.039149 3.51 1431 5.704900 4.681739 2.8 10695 3.496450

cost

23.843 199 25.980564 13.843822 30.686806 69.292287 43.863578 1.940472 48.805758 17.022193 58.729204 96.656809 53.48 1945 72.038578 76.475217 58.653499 73.484507 61.827322 63.129346 75.340828 52.347228

10.4 Other Evolutionary Computation Techniques 10.4.1 Evolution Strategies Evolution strategies (ES) (Schwefel 1985) are also search procedures that mimic the natural evolution of the species in natural systems. These work directly with the real representation of the parameter set, searching from an initial population (a set of points). Like GAS,these only require data based on the objective function and constraints, and not derivatives or other auxiliary knowledge. However, the transition rules are deterministic and the constraints are handled normally using an elimination mechanism (the non-feasible points are eliminated). The search for new points is based on one operator, namely, the mutation operator. We will consider two distinct types of evolution strategies differing basically in the selection procedure: (a) the ( p + A) evolution strategy and (b) the (p, A) evolution strategy, where p represents the number of parents of a generation and il is the number of offspring of a generation. In the ( p , A) evolution strategy, at a given generation, there are p parents, and A offspring are generated by mutation. Then, the ( p + A) members are sorted according to their objective function values. Finally, the best p of all the ( p + A) members become the parents of the next generation (i.e., the selection takes place between the ( p + A) members).

Non-traditional Optimization Techniques

287

In the ( p , A) evolution strategy, at a given generation, there are p parents, and A offspring are generated by mutation (assuming that A > p ) . Next, the A members are sorted according to their objective function values. Then, the p best of A members generated become the parents of the next generation (i.e., the selection takes place between the A members). For detailed algorithms, refer to the work by Costa and Oliveira (2001).

10.4.2 Evolutionary Programming Evolutionary programming (EP) has recently been applied to real-valued continuous optimization problems and is virtually equivalent in many cases to the procedures used in evolution strategies. However, the theoretical and empirical comparison between these mechanisms is an open area for research. As implemented currently, there are two essential differences between evolution strategies and evolutionary programming. 1. ESs rely on strict deterministic selection. EP typically emphasizes the probabilistic nature of selection by conducting a stochastic tournament for survival at each generation.The probability that a particular trial solution will be maintained is made a function of its rank in the population. 2. ESs typically abstracts coding structures as analogues of individuals. EP typically abstracts coding structures as analogues of distinct species (reproductive populations). Therefore, evolution strategies may use recombination operations to generate new trials, but EP does not, as there is no sexual communication between species.

10.4.3 Genetic Programming Genetic programming (GP) is a branch of genetic algorithms. The main difference between genetic programming and genetic algorithms is the representation of the solution. Genetic programming creates computer programs in the lisp or scheme computer languages as the solution. Genetic algorithms create a string of numbers that represent the solution. Genetic programming uses four steps to solve problems. 1. Generate an initial population of random compositions of the functions and terminals of the problem (computer programs). 2. Execute each program in the population and assign it a fitness value according to how well it solves the problem. 3. Create a new population of computer programs: (a) Copy the best existing programs (b) Create new computer programs by mutation (c) Create new computer programs by crossover 4. The best computer program that appears in any generation, the best-so-far solution, is designated as the result of genetic programming (Koza 1992).

288

Process Plant Simulation

10.4.3.1 Fitness function The most difficult and most important concept of genetic programming is the fitness function. The fitness function determines how well a program is able to solve the problem. It varies greatly from one type of program to the next. For example, if one were to create a genetic program to set the time of a clock, the fitness function would simply be the amount of time that the clock is wrong. Unfortunately, few problems have such an easy fitness function;most cases require a slight modification of the problem in order to find the fitness. Genetic algorithms do not appear to be appropriate for symbolic regression problems, where the structure and parameters of a model are to be determined simultaneously. This is because GAS generally use fixed length binary strings to code potential solutions to a problem. Clearly this is unsuitable for symbolic regression, where the model structure is allowed to vary during evolution. However, GP is a closely related approach that does lend itself to the implementation of symbolic regression. Genetic programming differs from GAS in the following points: 1. Tree structure variable length chromosomes (rather than fixed length and structure). 2. Chromosomes coded in a problem-specific fashion (that can usually be executed in their current form) rather than binary strings. 3. Genetic operators that preserve the syntax of the tree structured chromosomes during ‘reproduction’.

The steps involved in GP are shown as a flow chart in Fig. 10.3.

10.4.4 Other Population-based Search Algorithms Continued investigation into local search and population-based optimization techniques, and a consequent growing regard for their ability to address realistic problems well, has inspired much investigation of new and novel optimization methods. Onwubolu & Babu (2004) gave a detailed account of some of these population-based search algorithms along with engineering applications. Memetic algorithms were invented by Moscato (1999). These represent one of

the more successful emerging ideas in the ongoing research effort to understand population-based and local search algorithms. The idea itself is simply to hybridize the two types of method. Memetic algorithmsrepresent a particular way of achieving the hybridization, one which has chalked up a considerable number of successes in recent years. Scatter search was invented by Glover (1995, 1999). Scatter search focuses the use of previously generated candidate solutions in the generation of new ones. Similar to differential evolution, scatter search emphasizes the production of new

289

Non-traditional Optimization Techniques

solutions by a linear combination of the ‘reference’ solutions, although guided instead by combining intensification and diversification themes, and introducing strategies for rounding integer variables.

Create initial random population

-

i. Termination criterion satisfied?

yes

~

Designate result

1

No Select genetic operation probabilistically

Reproduction

Mutation

1 Select one individual based on fitness

A Perform reproduction

population

based on fitness

I

Performcrossover

I

offspring into new population

i. I Individuals = individuals + 1I

1

I

Select one individual based on fitness

I

Performytation

I

1 J

Insert mutant into new population I

I Individuals = individuals + 11

t

Individuals = individuals + 2 V

4

v

Fig. 10.3 Flow chart for genetic programming

Ant colony optimization was developed by Dorigo (1992). It has drawn inspiration from the workings of natural ant colonies to derive an optimization method based on ant colony behaviour. This has lately been shown to be remarkably successful on some problems, and there is now a thriving international research community

290

Process Plant Simulation

worldwide investigating further the technique. Essentially, the method works by equating the notion of a candidate solution with the route taken by an ant between two (possibly the same) places. Ants leave a trail as they travel, and routes that correspond to good solutions will leave a stronger trail than routes that lead to poor solutions. The trail strength affects the routes taken by future ants. Essentially, previously generated solutions (routes taken by past ants) affect (via trail strengths) the solutions generated by future ants. However, it makes more sense to view it for what it is a new and novel idea for optimization inspired by a process which occurs in nature.

Self-organizing migrating algorithms (Zelinka & Lampinen 2000; Zelinka 2001) originally proposed by Zelinka have drawn inspiration from existing algorithms such as evolutionary algorithms and genetic algorithms. However, in general, the design of a self-organizing migrating algorithm is unique and original. Its principle is both interesting and novel. It offers a novel approach to optimization and combines global search (the attraction of individuals towards the fitter ones) and local search (the multiple jumps of each individual during one migration loop), hence it covers both exploration and exploitation. The temporary replacement of the cost function with a virtual cost function seems to be a unique approach to overcome the difficulties when local optima are populating search space. The idea of the search algorithm is certainly a contribution containing new scientific knowledge.

EXERCISES 10.1 In a four-variable problem of optimization with genetic algorithms, the following variable bounds are specified: 0.1 I w 20.9 -7 I x I 12 0.0009 I y I 0.007

lo2 I I lo5 Compute the minimum string length of any point (w,x , y , z ) coded ~ in a binary string to achieve three-significant-digits accuracy in the solution. 10.2 We would like to use genetic algorithms to solve the following NLP problem: Minimize (xl - 1.5)2 + ( x 2 - 4)2, subject to 4 . 5 ~+ ~xf - 18 S O , 2x1 - x2 - 1 2 0, and 0 Ixl, x 2 I4. We decide to have three and two decimal places accuracy for variables x1 and x2 respectively. (a) How many bits are required for coding the variables. (b) Write down the fitness function which you would be using in reproduction.

Non-traditional Optimization Techniques

291

10.3 List the steps involved in a DE algorithm with the help of a neat schematic diagram. (a) How do you choose the values of NP, F, and CR in DE? (b) List and briefly explain the 10 DE strategies proposed by Price & Storn. 10.4 Discuss the steps involved in GA with the help of a neat flow chart. (a) Explain the choice of the values of the parameters N , p i , and p c in GA. (b) Compare GA with traditional optimization techniques. 10.5 Discuss the following: (a) simulated annealing, (b) genetic algorithms, and (c) differential evolution, and (d) also compare these three non-traditional optimization techniques. 10.6 What is the purpose and significanceof various operators such as reproduction, crossover, and mutation, employed in GA and DE. What are the guidelines for choosing various key parameters of GA and DE? 10.7 Compare GA with traditional optimization techniques. 10.8 Explain the working principle of DE by using a neat flow diagram. 10.9 What are the steps carried out in DE? Demonstrate with suitable examples. 10.10 Discuss the DE strategies and elaborate the nomenclature used. 10.11 Use GA for solving the following NLP problem: Minimize (xl - 1.5)2 + (x2 - 4)2 subject to 4 . 5 ~ + x~ i - 18 I 0, 2x, - x2 - 1 2 0, and 0 I xl, x 2. <- 4. Choose three and two decimal places accuracy for variables x1 and x2, respectively. (a) How many bits are required for coding the variables? (b) Write down the fitness function that you would be using in reproduction. (c) Which schemas represent the following regions according to your coding: (i) 0 I x1 5 2, (ii) 1 I x1 I 2, and x2 >2. 10.12 Consider the following population of binary strings for a maximization problem: Strings

01101

11000

10110

00111

10101

00010

Fitness

5

2

1

10

3

100

Find the expected number of copies of the best string in the above population in the mating pool under (a) roulette wheel selection, and (b) binary tournament selection without replacement. If only the reproduction operator is used, how many generations are required before the best individual occupies the complete population under each selection operator? 10.13 In a simulated annealing algorithm, a neighbouring point is created with the following probability distribution: p(x, 0)=

a cr2

+ x2

What must be the value of a in order to have a valid probability distribution? The current point is x = 10.0 and a variation 0 = 2.0 is used. If a random number 0.723 is chosen to create a point, what is the new point?

292

Process Plant Simulation

10.14 In a GA simulation, the following population is observed at generation 4: Strings

0101101

1001001

1010110

0111110

Fitness

10

8

14

20

1110111 0111100 2

12

If roulette wheel selection,single-point crossover with probability 0.9, and a bitwise mutation with a probability 0.01 are used, how many copies of the schema (* 1* 11**) are expected in generation 5? 10.15 In order to solve the maximization problem

c2x

if 1.5 - 3x if f(4= 5 x - 2.5 if 12.4 - 2x if

x 50.3

1

0.3 < x < 0.5 0.5 < x < 0.8

t

x > 0.8

J

using simulated annealing, we begin with an initial point do)=0.2. A standard normal probability distribution is used to create a neighbouring point. (a) Calculate the probability of creating a point in the global basin. (b) If a temperature T = 10 is used, what is the overall probability of accepting a point in the global basin? 10.16 Minimize the function given in Problem 10.11 using DE and compare the results obtained with GA. 10.17 Minimize the function given in Problem 10.13 using DE and compare the results obtained with SA. 10.18 Minimize the function given in Problem 10.15 using DE and compare the results obtained with SA. 10.19 Using the DE code, simulate the results for a ammonia synthesis reactor (see Sec. 10.3.8) and compare the results obtained with those reported by Babu et al. (2002). 10.20 Using the DE code, simulate the results for a alkylation problem (discussed in Sec. 10.3.8), and compare the results obtained with those reported by Babu and Chaturvedi (2000).

PART V

SIMULATlON

THIS

PART COMPRISES five chapters. Chapter 11 covers the sequential modular, simultaneous modular, and equation-solving approaches, which are essential for process plant simulation. The structural analysis with substitution algorithm is demonstrated using an example. Chapter 12 deals with decomposition of networks, which is needed to find out the cut-set (the streams to be teared to have a sequential calculations in simulation of process plants). Various proven algorithms are presented in this chapter in the form of flow charts, so that the reader can make computer codes using these flow charts. The working of these algorithms is demonstrated by applying them to problems from process industry. In Chapter 13, various convergence promotion techniques, the sources of physical and thermodynamic properties are discussed. The modularity and routing concept is demonstrated using a flow diagram. Finally, Chapter 14 takes up the simulations (specific-purpose and dynamic) of actual industrial process plant problems starting from model formulation, application of numerical methods for solutions, and incorporating the optimization techniques for optimal design purposes. Case studies on an autothermal ammonia synthesis reactor, thermal cracking operation, pyrolysis of biomass, and the design of shell-and-tube heat exchanger are covered in this chapter. In Chapter 15, the capabilities and use of two most widely used professional software packages, namely, HYSYS and FLUENT are discussed. The problems are solved using these software packages.

CHAPTERS 11. Modular Approaches and Equation-solving Approach 12. Decomposition of Networks 13. Convergence Promotion and Physical and Thermodynamic Properties 14. Specific-purpose Simulation and Dynamic Simulation 15. Professional Simulation Packages

CHAPTER 11 MODULAR APPROACHES AND EQUATlON=SOLVlNGAPPROACH

In this chapter, sequential and simultaneous modular approaches are described, which are essential for process plant simulation. Analysis and design modes are distinguished.Another commonly used simulationapproach,called equation-solving approach, is also discussed. Precedence ordering of equation sets is demonstrated using an example. The concepts of disjointing and tearing a system of equations are discussed. The structural analysis with substitution (SWS) algorithm is described. Finally, maintaining sparsity, block triangular form (BTF), bordered block triangular form (BBTF), and block triangular splitting (BTS) concepts are explained.

11.I Modular Approaches to Process Simulation 11.1.1 Analysis Versus Design Mode Ideally, with the aim of process simulation to be a design tool, the system inputs and/or design parameters should be calculated from the specified outputs. But such a simulation in the design mode, in general, is numerically less stable than that in the analysis or performance mode. However, it may be noted that the stability is related to the solution technique, in particular, and not the purpose or mode. The ‘design mode is numerically unstable’ if a code written for a performance mode is used iteratively to solve the design problem (as is the case with sequential modular flowsheet approaches). However, in the equation-orientedapproach, since the two modes are indistinguishable,the stability of the numerical method is the only issue. The correct point of view (in either simulation strategies) is whether the numerical procedure is stable or not.

296

Process Plant Simulation

The two contrasting approaches are described in Fig. 11.1: SYSTEM

w

~

Inputs

Inputs (specified)

outputs

?

1

?

I

Outputs ?

(b) Fig. 11.1 (a) Design mode. (b) Analysis mode.

Calculating the material and energy balances is the most tedious and repetitive problem of process design. Carrying out simple material balances is enough in the initial stages of the flowsheet study. However, in the later stages of the design study, heat and mass balances must be calculated along with the equilibrium and rate equations, P-V-T relations, and those governing counter-current operations. Generally, all these equations are strongly non-linear. Moreover, in the simplest of the cases, if a process is of a sequential configuration, as shown below [Fig. 11,2(a)], it is easy to proceed from the feed streams until the products are obtained calculating sequentially for one process unit after the other. However, majority of the chemical plants are of complex configurationconsisting of recycle of streams such as mass [Fig. 11.2(b)] or energy [Fig. 11.2(c)] or both. They represent the interlinked networks of units. In such cases, it is essential to decompose the network in order to make the calculation procedure sequential, which in turn requires convergence promotion. Taking the above complexities in simulation into account, during the 1960s several design organizations started building libraries of computer programs for various unit operations. It then became evident that many of these could be put into a system enabling one to direct calculations for an entire flowsheet, thus saving considerableengineering time. This resulted in the development of specific-purpose prognms to simulate a particular plant or part of a plant. These were more detailed in nature and rigid in structure. On close examination of these programs, it became clear that a major portion of the simulations could be common to all types of plants (i.e., calculations involved in various unit operations, methods to compute different physical and thermodynamic properties, decompositionand convergencepromotion algorithms, cost information, as well as a library of numerical subroutines. Ultimately, it gave rise to several general-purpose simulation packages.

Modular Approaches and Equation-solving Approach

Heater

-

Reactor

297

Separator

(a) Sequential configuration (simple) Unreacted feed

t

Make-up feed Reactor

Separator

+
(b) Recycle of mass (complex)

Reactor

Feed

Product

(c) Recycle of energy (complex) Fig. 11.2 Possible configurations of chemical plants

There are two methods of dealing with the modular approach to simulation. These are (1) the sequential modular approach and (2) the simultaneous modular approach. 11.1.1.1 Sequential modular approach

This approach involves collecting equations and constraints for each process unit into a separate computational subroutine or module. This concept is known as ‘unit computation’. Thus, each module calculates values pertaining to the output streams for the given input conditions and parameters for that process unit or equipment, irrespective of the source of input information or the sink of output information.A diagrammatic representation of the sequential modular approach is given in Fig. 11.3. Using this concept, it is possible to create a library of modules and use them for a variety of flowsheets. The sequential part of the approach involves carrying out calculations from module to module, starting with the feed streams until products are obtained. In order to make it totally sequential, it is then necessary to identify recycle loops of units in a given flowsheet and to ‘tear’ certain streams. Tearing of streams implies assumption of some value for that stream. For example, in Fig. 11.4, tearing streams 2 and 7, it is possible to make the calculations sequential in order of the modules (l), (4), (3), (3,and (2).

290

Process Plant Simulation Physical properties

Streams in

~

9

Unit

- K -

7 Streams out ~

Fig. 11.3 Sequential modular approach

m

3

Team streams: (2), (7) Fig. 11.4 Tearing of streams

Therefore, a general-purpose sequential modular program has the following structural components: 1. A store of physical and thermodynamic properties for pure chemical species and their mixtures. 2. A system-working database derived from the store as well as the problem input. 3. Individual module database derived from the working database. 4. Modules or subroutines for individual process units, computational procedures including input/output for each module, and internally iterated variables, if any. 5. Flowsheet topology to be used by the system executive to order computational sequence. This is usually defined by a ‘process matrix’ which specifies standard unit modules for use in simulation of a particular process along with identification numbers, positive for input streams and negative for the output streams. 6. Routines for the recycle calculations and convergence analysis. 7. Other mathematical service routines.

Modular Approaches and Equation-solving Approach Nodes rn, (rn + 11, ..., (rn t 1)

Feed

299

Nodes ’

*-

+

l) O u t p u t

+

Recycle loop

Feed stream values, equipment parameters

Guess all flukfor nodes rn, (rn + l), (rn + 2), ..., (m+ 1)

I b

Solve Eqs (11.2) and (1 1.3) for yij for nodes m,(m + 1), ..., (m+ 1)

for yij for nodes m,(m+ l), ..., (rn + 1)

I

1

t Recalculate

pijkfor each node

Yes

Recalculate for node (m + 1 + 1)

1 Exit Fig. 11.5 Simultaneous modular approach

11.1.1.2 Simultaneous modular approach In this type of approach, the unit modules are similar to those in the sequential modular approach. However, the main difference is that for each unit, an additional module is written, which approximately relates each output value by a linear

300

Process Plant Simulation

combination of all input values. Thus for the given input x i l ,xi2, ..., xin for module i and equipment parameters ui; the exact model for module i could be written as Yij = J j ( x i p

4 2 , xi3,

a * * ,

(11.1)

xin, Ui)

and the linear models for module i could be represented by the following two connecting equations: n

(11.2) (11.3)

Xsr = Y p q

In the above equations, y denotes output values and P coefficients of linear equations. Let us assume that for a given system of nodes, the feed stream values are specified. This means we represent the actual process by two models: (a) the exact model and (b) the linear model. The unknowns of the linear model ( P i j k are determined by the simultaneous modular approach described below. The flow diagram for the simultaneous modular approach is given above (Fig. 11S ) . The simultaneouscalculations are carried out using the method described above. On comparing these two approaches,the sequentialmodular approach seems easier to write; but in the simultaneous modular approach, there is no decomposition required. Also, the convenience of the two approaches depends upon the type of the process and the nature of recycle streams. This comparison is described in Table 11.1. Table 11.1 Comparison of sequential and simultaneous modular approaches 5 p e of process

Approach recommended

Without recycles

With recycles

Sequential Linear Medium non-linear Non-linear in heat balance Strongly non-linear

Simultaneous Simultaneous Simultaneous Sequential

11.2 The Equation-solving Approach An unconstrained material balance problem is one in which all the parameters are assigned known values. These parameters constitute the set of natural simulation specifications,namely, the species, or stream split fractions, stoichiometry,and key reactant conversion for all the reactors and the species flow rates in all the external process streams.

Modular Approaches and Equation-solving Approach

11.2.1 Precedence-ordering

301

of Equation Sets

In general, the steady-state simulation of a process requires solving a set of nonlinear simultaneous equations expressed in the form

J(x,u)=O,i € I , (11.4) in which x is a vector of n unknown state variables and u a vector of r known decision variables. In an iterative procedure to solve Eq. (1 1.4), one has to start with an initial vector xo and generate successive approximation to it xl, x2, ..., so that the solution is obtained within a prescribed accuracy. For this purpose, in applying Newton’s method excessive computational efforts are needed in order to evaluate an nth order Jacobian matrixf, and to find its inverse fx-’; this has necessitated to develop more efficient algorithms. The major drawback in most iterative techniques is that the whole set of n unknown variables are to be iterated upon, which requires an nth order convergence matrix. When n becomes large with the growing complexity of the process, simulation needs larger storage space and excessive computer time. It is, therefore, desirable to break up a large system of equations into smaller subsets, which only need to be solved simultaneously. A systematic strategy to identify such subsets and to arrange them in a proper sequence is known as precedence ordering.The equations are arranged in a certain order by first arranging them into diagrams. The first of them is the Boolean ‘incidence’ (or ‘occurrence’) matrix, in which the rows correspond to system equations and the columns correspond to a system variable. Then the system is represented by a ‘digraph’ or directed graph or its associated ‘Boolean adjacency matrix’. The output set assignment is not unique. Nevertheless, same results for irreducible sets and precedence ordering are obtained. In this adjacency matrix, the rows correspond to the vertices from which the flow originates and the columns correspond to the vertices to which the flow is directed. This method of dealing with a system of equations is known as path tracing method (PTM). Consider the following example.

Example 11.1 Find (a) the incidence matrix, (b) the digraph, and (c) the associated adjacency matrix for a process represented by a set of the following equations in functional form:fl (xl,x2)= 0,f2(x4) = 0,f3 (x3,xg) = 0,f4 (x4,x 5 ) = 0,f5 (xl,xg) = 0, and& (x2, x3, x5) = 0. Solution (a) Incidence matrix In the incidence matrix, the process functions and their relationship with the process variables are represented as rows and columns of the matrix, respectively, using the binary coding (Lea,1’s and 0’s). For example,f, is a function of x1 and x2 only and so, for rowfl against the columns x1 and x2, ones (1’s) are put while against the remaining columns of the variables x3,x4, x5, and xg,

302

Process Plant Simulation

zeros (0’s) are put. Similarly, all the other functions and their relationship with the variables are represented with 1’s and 0’s. The corresponding incidence matrix for the above system is as follows:

A

0 f3 0 f4 0 fs 1 f 6 .0 f2

-

1 1 0 0 0 0

0 0 0 0 1

0 1 0 0 1

1 0 1 0 0

0 0 1 0 1

0 1 0 1 0-

(b) Digraph In a digraph, the flow of information (i.e., the sequence of calculations) is given. From the equation f2(x4) = 0, the value of x4 can be calculated. Using this x4 and the equationf4 (x4, x5) = 0, the value of xs can be obtained. Using this x5, and with an assumed value of x3, the equation f6 (x2, x4, x5) = 0 gives the value of x2. Using this value of x2 and the equation fl (xl, x2) = 0, the value of x1 could be computed. Subsequently, using this value of x1 in combination with the equation f5 (xl, x6) = 0, the value of x6 could be obtained. Finally, using this x6 and the equationf3(x3, x6) = 0, the value of xgis recalculated. This value should be equal to the earlier assumed value of x3. If not, the iterations are continued till both the values are same. This entire flow of information is represented in the form of a digraph. The digraph for the system is shown in Fig. 11.6. As can be seen from the figure and the earlier description, the values of x4 and x5 can be obtained independently, while the values of the remaining variables xl, x2, x3, and x6 are calculated iteratively. This is evident from Fig. 11.6, as these variables are in a loop.

Fig. 11.6 Digraph for the system

Modular Approaches and Equation-solving Approach

303

(c) Associated adjacency matrix In the above digraph, the equations are numbered as the nodes. In the adjacency matrix, rows correspond to the vertices from which the flow originates and columns correspond to the vertices to which the flow is directed; it can be obtained directly from the incidence matrix as well. Rules of Boolean multiplication and union apply to these matrices. The corresponding associated adjacency matrix for the system can be written using the digraph as 1 2 ‘ 0 0 0 0 0 0 0 0 0 0 ( 1 0

3 0 0 0 0 1 0

4 0 1 0 0 0 0

5 1 0 0 0 0 0

6 0 0 1 1 0 0

The rows in the above matrix represent the vertices from which the flow is directed, and the columns of the above matrix represent the vertices to which the flow is directed. When each equation feeds information to the next equation in sequence and the last one feeds to the first, a loop of information flow exists, requiring the whole system to be solved simultaneously. If a set of equations is part of a larger set of equations which itself forms a larger or maximal loop, the smaller subset must be solved together with the bigger subset. The problem of partitioning is thus reduced to finding all the maximal loops through an adjacency matrix and ordering them into a proper sequence. A sound decomposition strategy should be applicable to any system of equations whether linear or non-linear; algebraic, differential, difference, and integral equations; continuous or discrete variables. Hence, several methods have been proposed for partitioning. These generally fall into two broad categories, namely, path tracing methods (PTM) and those based on powers of adjacency matrix (PAM). 11.2.1.1 Path tracing method

The methods proposed by Sargent and Westerberg (1964), Steward (1962; 1965), Lee et al. (1966), Billingsley (1967), and Christensen and Rudd (1969) come under the category of PTM. According to this method (PTM), the path is traced in two phases as described below.

First phase In the first phase, all columns that contain only zero elements and their corresponding rows are removed from the associated adjacency matrix; these

304

Process Plant Simulation

columns belong to equations that are not fed from any of the other equations (see column 2 in the adjacency matrix in the above example), consequently they can be solved independently. The equationf, (x4) = 0 corresponds to this situation of the first phase. Next, there may be equations that contain, besides their own output variables, only the output variables of the equations that are already precedence ordered; the appropriate rows and columns pertaining to such equations are deleted to obtain a further reduced matrix. The equationf, (x4, x s ) = 0 corresponds to this situation of the first phase. So, in Example 11.1, first remove column 2 (all zeros) and corresponding row 2 from the associated adjacency matrix as x4 is independent. As can be seen, after doing so, in the remaining 5 x 5 matrix, now column 4 contains all zeros. Hence, next remove column 4 and to. 4 as x5 now could be computed using the known x4. This further reduces the associated adjacency matrix to a 4 x 4 matrix as shown below.

3

0

0

0 1

5

0

1

0 0

Second phase A path of information is traced in the final reduced matrix starting with any equation and going to the others that feed it. This is continued until an equation is encountered repeatedly, which means a path has been traced around a loop of information flow. In the present problem, one of the possible sequences is as follows: f3 (x39 x6)

=0

f 6 (x29 x 3 ,

xs> =

fi ( X I 9 X J

=0

fj

f3

XI^) = 0 (x39 x6>

=

The above sequence of equations forms a loop. Hence, the path of information is traced.

Modular Approaches and Equation-solving Approach

305

11.2.1.2 Powers of adjacency matrix

This is simple and easy to program. The inherent shortcoming of the PTM is the difficulty of programming,which is overcome by the simplicity of the matrix notation in the powers of adjacency matrix (PAM). However, a major drawback of the PAM is the requirement of considerable storage space (memory requirement).The methods proposed by Norman (1965), Ledet and Himmelblau (1970), and Himmelblau (1973) come under this category of path tracing methods. A method that is the combination of both PTM and PAM has been suggested by Jain and Eakman (1975).

11.2.2 Disjointing In addition to partitioning, the analysis should identify possible subsets of equations among which no intersection exists. In formulating a model of a large process, for instance, a whole chemical plant, there is a possibility that the subset of the system of equations does not contain any variables in common with the rest of the equations in the system. Such a system of equations, called a disjoint, can be solved independently, thus reducing the dimensionality of the whole system to that of the largest disjoint subset.

11.2.3 Tearing a System of Equations Once a system of equations is partitioned into irreducible subsets of simultaneous equations, it is desirable to decompose them further in order to consume less computation time in solving them compared to solving the entire block of equations simultaneously. It is achieved by tearing certain variables and applying an iterative process in which convergence is obtained on the values of torn variables. One such algorithm for this purpose is the structural analysis with substitution (SWS) algorithm.

11.2.4 The SWS Algorithm This algorithm is described using the following illustration.

Illustration 11.1 Consider a set of equations to be solved: (a> x1x4 + x,2/x4 - 4 = o (b) X2X5 - 3x6 = 0 (c) x , / x 2 + ln(x3/x4)- 2 = 0 3 2 (d) x3 + 2x3 - 2x3 = 0 (e) x2 + x4 - 3 = 0 (f) x3(x3 f x6) - 7 = 0

306

Process Plant Simulation

Solution There are two algorithms that are put to use while solving the problem: 1. Rudd’s algorithm: It locates those variables that occur in the fewest equations. 2. Steward’s algorithm: It isolates equations with the.fewest variables to be solved independently. After the application of these two algorithms, the rest of the equations are solved using the SWS algorithm. At first, it is seen that equation (d) can be solved directly for x3. With the value of x3 obtained from equation (d), equation (f)can be solved for x6. By using Rudd’s algorithm, it is seen that x5 is the variable that occurs in the fewest number of equations. Therefore, the set of equations can be precedence ordered such that equation (b) is solved last. The remaining three equations are then solved using the SWS algorithm. The maximal set of equations are arranged one by one to express one of its variables, the output variable, in terms of the others. For example, equation (e) is well suited because of its linear form, thus giving the output variable as x2: X2=3-X4 (11.5) All occurrences of x2 in the remaining equations (a) and (c) are then substituted by the above expression (1 1.5). In this manner, one reduces the number of unknown variables in the maximal set; the remaining equations are as follows: x1xq

+ x6/x4 2

-4=0

(11.6)

1‘ + 1n(x3/x4)- 2 = o (11.7) 3 - x4 Now the substituted variable x2 will be evaluated after the remaining unknowns x1 and x4 in the maximal set are determined. Equation (11.6) can be rearranged to isolate xl,

=

4 - x;/x4

(11.8) x4 where x1 is its output variable. The occurrence of x1 in the remaining Eq. (1 1.7) is substituted by Eq. (1 1A), thus giving XI

4 - x;/x4

+ ln(x,/x,) - 2 = 0 (11.9) x4 (3 - x4) Although one may be able to solve a system of non-linear equations analytically as illustrated in this illustration, for computer implementation, the equations have to be solved iteratively by linearizing the equations (a variant of the NewtonRaphson solution technique).

Modular Approaches and Equation-solving Approach

307

Table 11.2 describes the order of the solution of the set of equations (a)-(0. Table 11.2 Order of the solution for the given set of equations (a)-(f)

Order of solution

Equation

Final form

output variable

4 - x,2/x,

3

x4

4

(4)

X1

x4 (3 - x4 1

+ ln(x,/x4)

x, =

-2=0

4 - x;/x4 x4

For preparing a sequential modular steady-state flowsheet, we need (a) a partitioning algorithm and (b) a tearing algorithm and a numerical scheme for solving linear equations (which can be used iteratively to solve non-linear algebraic equations). For preparing an equation-oriented flowsheet we need (a) an output set assignment followed by partitionjnglprecedenceordering and a numerical method to solve algebraic equations.

11.2.5 Maintaining Sparsity There are a few problems in terms of difficulties involved in solving large sets of non-linear equations simultaneously in the above-discussed method of equationsolving approach. Hence, the acceptance of this global approach of equation-solving in the design mode simulation is rather slow than anticipated. For solving such a system of equations, Newton-Raphson or related techniques are very effective. These techniques, in turn,require a periodic solution of a set of linear equations written in the form Ax=b

( 11* 10)

where A is an n x n matrix, x E Rnis an unknown vector, and b E R"is a vector of constants. The set of equations [Eq. (11.lo)] in practical problems is characterized by a large value of n (several hundreds), and interestingly at the same time it is sparse (i.e., the number of non-zero elements A is very small, usually less than 5%). That is, more than 95% of the elements of the sparse matrix are zeros.

308

Process Plant Simulation

11.2.5.1 Sparse systems A sparse system of equations can be solved in three stages: (1) partitioning, (2) tearing, and (3) numerical solution. The first two stages require only logical operations with least round-off error propagation, but consume more time. In fact, the time required for the first two stages is twice the time required for the third stage. The third stage of solving the sparse systems is carried out using (a) direct methods (variants of Gauss or Gauss-Jordan elimination) and (b) indirect methods (iterative methods). The disadvantages of direct methods are twofold:

1. During the elimination process (i.e., during algebraic matrix operations), the structure of matrix A is changed since new non-zero elements are generated Cfill-ins),which results in loss of sparsity. This, in turn, increases the storage space and computational time. 2. Both row-wise and columnwise scanning of the matrix is to be carried out, which needs a larger and more complicated storage scheme. Fill-ins can be minimized by tearing the equations (Lin & Mah 1977). The tearing is risky, however, since it may prevent pivoting, which is necessary for reducing round-off error propagation. The indirect methods (iterative methods), such as the Gauss-Seidel method, successive over relaxation method, conjugate gradient method, etc., have often been preferred for solving sparse set of equations. The advantages of these methods are twofold: 1. Matrix A remains unchanged during the computation. 2. Only row-wise scanning of matrix is required in these methods. These advantages lead to a simpler computer code and reduction in the overhead time necessary for index manipulation. However, the limitation of these indirect methods is that these cannot be used for all types of problems, since unless matrix A has certain special properties, the convergence may be slow or unattainable. So, in spite of having disadvantages, the direct methods are preferred over the indirect methods because of the above-mentioned limitation of iterative methods. But, the focus is on finding a way in which the fill-ins could be reduced. In recent years, attention has been diverted towards reducing the fill-ins associated with the direct solution methods (Tewarson 1973; Duff 1977). These techniques can be divided into two categories: (1) those that involve pivoting (reducing round-off error propagation) and (2) those that reorder matrix A prior to elimination, so that the fill-ins can be limited to certain areas of the matrix. The pivoting techniques can be subdivided into two local strategies: 1. Strategies in which both columns and rows are reordered during the process of elimination.

Modular Approaches and Equation-solving Approach

309

2. A priori strategies in which columns are reordered before elimination and rows during elimination (or) vice versa.

Some of the most widely used pivoting techniques are described below. 11-2.5.2 BTF, BBTF, and BTS The techniques that reorder matrix A prior to elimination involve permuting A to a block triangular form (BTF) or bordered block triangular form (BBTF). These forms, in turn, can be so factored that fill-in is limited to the diagonal blocks of a BTF, and to the diagonal blocks and one border of a BBTF. These forms are explained below.

Block triangular form A matrix A can be in BTF as follows:

All A22

*

*

*

AIN

*

*

A2N

. . . . .

A=

-

ANN

where A, is a square matrix of order mi. The matrix which can be permuted to BTF is reducible and that which cannot be is irreducible. For a reducible matrix, BTF is unique in the sense that N and the particular rows and columns represented in each block are fixed. Among the several techniques suggested for permuting to BTF, the algorithm given by Tarjan (1972) seems to be the most effective. A system in BTF can be readily solved by block back substitution. Since only diagonal blocks of A need to be inverted, fill-in is limited to those areas only. Algorithm of block back substitution Step 1 Read n, mi, i = 1,2, ..., It, and b, where n is the order of the matrix A, mi are orders of diagonal matrices of A, and b is the vector of constants. Step 2

Read I, A (where I is the identity matrix and A is the BTF of A).

Step 3 Set the counter k = 1. Step 4

Compute Ek and Fk as follows: Ek is a n x mk matrix identical to the

[l + 1;:;mi]th to

[Xi"=, mi]th columns of the unit matrix I. Fk is a n x mk matrix mi]th to [xi"=, mi]th columns of the matrix A. identical to the [ 1 + Xi"=;' Step 5

Compute

Kk

as follows:

Ak = I, + (

E -~ F ~ ) A E:. ;~

310

Process Plant Simulation

Step 6

k = k + 1. If k I n, go to step 4.

- -

Srep 7 Compute A-' as follows: A-' = A,, A,,

&, ..., &.

Step 8 Compute X = A-' eb.

Bordered block triangular form If A is irreducible, it can be permuted to BBTF. For this purpose, let us denote various parts of the matrix A as follows:

A=[$] where M is the m x m BTF, R , S are non-zero borders, and T = t x r matrix which may have either zero or non-zero elements. A system in BBTF may be solved by block elimination, by factoring A as follows:

A=[=][ SII

ZIM-'R SIT - SM-'R

]

where Z is the identity matrix. Since both the factors in the above expression are BTF, the system is readily solved by block substitution. In this process, fill-in occurs in the diagonal block of M as it is inverted, in R and T as they are replaced by M ' R and T - S M ' R , respectively, and finally in the filled-in T as T - SM-'R is solved during block substitution.

Discussion There is no clear-cut advantage in either of the two general approaches, as outlined above (BTF and BBTF) in maintaining sparsity. The pivoting techniques seem to be better suited for general-purpose user, since it is not always feasible to permute a matrix to BTF or BBTF with small borders. On the other hand, when permutations are possible (i.e., reducible to BTF or BBTF forms), which is often the case in chemical engineering design problems, savings can be appreciable.

Block triangular splitting A new sparsity preservation technique, called block triangular splitting (BTS), has been suggested by Stadtherr (1979), which is outlined as follows. If A is BBTF in the following system:

[

A = =]x MIR

=

[i]

=b

where Cis an m x 1 and D is a t x 1 matrix. Then the goal is to reduce A to BTF so that it can be solved by block-back substitution. This can be done by subtracting the column vector [S/SM-'R]x from both sides of the above equation. Further, from the above equation it can be noted that

[MIRIx = [q Multiplying both sides by SM-', it follows that

[SM-'MISM-'R]x = [SM-'c]

Modular Approaches and Equation-solving Approach 3

311

[SISM-'R]x = [SM-'c]

In this way, the original system is reduced to the block triangular system

A* ex = b* where

A*=[

MIR OIT - S W ' R

b*=[

]

]

C

D - SM-'C This is equivalent to splitting A into the sum of a BTF A* and a matrix of low rank, i.e.,

[

A=A*-+ Si'R] Thus, in performing BTS, fill-in occurs in the diagonal blocks of M as it is inverted, and in T as it is replaced by T - SM-'R. The solution is readily obtained by computing b* and inverting A* as follows: x = [A*]-' b*. Since M has already been inverted, only T - SM-'R has to be inverted to invert A*.

EXERCISES 11.1 A chemical process is represented by the following set of equations: f1

(x39 x4) = 0, f 2 (x59 x2) = 0, f 3 (x6) = 0,

x2) = 0, f 6 (xq, x59 X I ) = 0 (x6, = 0, f5 Determine the following: (a) Associated incidence matrix with output-set encircled. (b) Digraph of the process. (c) Associated adjacency matrix. 11.2 Compae the sequential and simultaneous modular approaches used for process plant simulation. 11.3 Discuss the following: (a) Analysis versus design mode. (b) Process matrix. (c) Types of processes against the modular approach recommended (sequential versus simultaneous). 11.4 A chemical process is represented by the following set of equations: f4

(x33

f' ( X I , x4) = 0, f2

(x2, x3)

= 0, f 3

(x5)

= 0,

x2) = 0, fs (x3, x4, x6) = (x59 x6) = 0, f 5 Determine the following: (a) Associated incidence matrix with output-set encircled. (b) Digraph of the process. (c) Associated adjacency matrix. f4

312

Process Plant Simulation

11.5 Below given figure represents a ‘mixer’ (Mix) followed by a ‘splitter’ (Split) returning part of the output to the mixer. In the splitter the material entering the unit is divided according to the fraction a: A=@

B

C D

(1 - a)B

Compare the sequential and simultaneous modular approaches of simulation calculations of a recycle set for the above process, by taking a as 0.33 and 0.90. Comment on results. 11.6 A chemical process is represented by the following set of equations: 2 ~ 1 x 4+ ( ~ 6 1 x 4 ) 4 =0 x2x5 + 3x6 = 0 x1/x2+ In (x3/x4)- 2 = 0 x; + (2X,2/X4) - 2x, = 0 x2 + x4 - 3 = 0 xg (x3 + x6) - 7 = 0 Precedence order these equations using the SWS (structural analysis with substitution) algorithm along with Steward and Rudd algorithms. 11.7 Suggest an algorithm in the form of a flow chart for the simultaneous modular approach in process plant simulation. 11.8 Using a block back substitution algorithm, solve the following set of equations: 1 2 3 4/3

x

7/3

[:: :][:I=[5;j 11.9 Write down the pseudocode of the algorithm and the steps involved in solving a matrix of block triangular form (BTF) using the block back substitution method. 11.10 Discuss the three stages of operations involved in solving a sparse system of equations. 11.11 Discuss BTF (block triangular form) and BBTF (bordered block triangular form).

CHAPTER 12 DECOMPOSITION OF NETWORKS

This chapter deals with various algorithms for decomposition of networks (determining the streams to be teared) in the form of flow charts. These algorithms are classified into two groups: (1) Algorithms based on signal flow graph such as the B&M algorithm (Barkley & Motard 1972), the basic tearing algorithm (Pho & Lapidus 1973a, 1973b), etc., and (2) algorithms based on reduced digraph (list processing algorithms) such as the K&S algorithm (Kehat & Shacham 1973), M&H-1 algorithm (Murthy & Husain 1981), M&H-2 algorithms (Schemes I and 11) (Murthy & Husain 1983), etc. Each of these algorithms is explained with examples.

12.1 Tearing Algorithms In order to be able to make calculations sequential in a process plant with recycle and bypass streams, it is necessary to tear those streams or, in other words, assume certain values for streams such that the calculations become sequential. To optimize this process of assuming values for streams, it is imperative to use tearing algorithms. This warrants the introduction of certain terms. The criteria of tearing procedures (Husain 1986) are either a minimum number of variables associated with the tear streams or a minimum number of tear streams. The former criterion is tantamount to choosing stream weights equal to the number of stream parameters and the latter to choosing all stream weights equal to unity. An alternative is to weigh each stream with a number reflecting the probable degree of difficulty which can be encountered in an iterative procedure to converge, if that particular stream were torn. Then the best cut-set is one in which the sum of weights assigned to the tear streams is minimum. However, this criterion suffers from the disadvantage that proper weighting factors are not readily known for all types of networks to be decomposed.

314

Process Plant Simulation

The earlier algorithms proposed for tearing networks used either minimum sum of variables or minimum sum of weights associated with tear streams as criteria for efficiency.

12.2 Algorithms Based on the Signal Flow Graph A signal flow graph (SFG) is an alternate representation of a digraph. It was first proposed by Barkley and Motard (1972) by interchanging the nodes and edges of a digraph. Consider the following example. The digraph of a process is as follows (Fig. 12.1). The numbers within brackets represent the process units [(l), (2), (3), (4),( 5 ) ] called nodes, and the numbers without brackets (1,2, 3,4,5,6 , 7 , 8) are the process streams interconnecting the units called edges: 5

1

3 11 2

41

I

-

7

, 7,

,, I

/ I

a Tear streams: (2), (7) Fig. 12.1 Digraph of a process flowsheet

The corresponding SFG for the above digraph is shown in Fig. 12.2 (i.e., interchanging the nodes and edges):

Fig. 12.2 Signal flow graph of a process flowsheet

Decomposition of Networks

315

12.2.1 The Barkley and Motard Algorithm Barkley and Motard (B&M) (1972) suggest an alternate representation of a digraph by interchanging nodes and edges known as a ‘signal flow graph (SFG)’. It contains as nodes what are the edges in a digraph and vice versa. The decomposition procedure is then applied on the SFG by tearing or cutting nodes in place of edges on a digraph. The following steps are involved in decomposition of a network using the B&M algorithm: Step 1 Convert the digraph into an equivalent SFG. Step 2 Eliminate any node with a single precursor, since it belongs to that precursor. Replace such a node whenever it appears as a precursor by its header node. Step 3 If any self-loops appear, cut those nodes and assign them to the cut-set. Eliminate such nodes, which may render the graph reducible again, i.e., nodes may now appear with a single precursor. Step 4 Two-way edges can also prevent complete reduction of the SFG. A twoway edge is shown in Figs 12.3 and 12.4. In the figure, the situation presented is for thejth node. Before the SFG is totally reduced, two-way edges as shown below may appear; in such a situation, assign either the ith or thejth node to the cut-set.

Fig. 12.3 Simple two-way edge

I

I

Fig. 12.4 Compound two-way edges

Step 5 If all other conditions fail to appear, select the node with the maximum number of output edges for cutting. Step 6 When SFG is totally reduced, the cut-set is found.

Illustration 12.1 Using the Barkley and Motard (1972) method (B&M algorithm) for the decomposition of a maximal cyclic net (MCN), find out the streams that are to be

316

Process Plant Simulation

teared (i.e., cut-set) for the digraph (Fig. 12.1) of a process: (decomposition is by a minimum number of tear streams, all of them assigned unit weight). Solution First reduction Eliminate any node with a single precursor, since it belongs to that precursor. Then, replace such a node whenever it appears as a precursor by its header node. Nodes

Precursors

2

%, 4

3

2

4

'AX2

a

6 7

4

2&

' A6

v

Note Replace the nodes which are eliminated with their precursors. In the above table, (1) eliminate each node with only one precursor (i.e., remove nodes 1, 3, 5 , 8 since these belong to that precursor) and (2) replace such nodes in all their occurrences in the list of precursors by their precursors, i.e., we need to replace node 3 with its precursor 2. Similarly, we replace node 5 with its precursor 7 , node 1 with its precursor 2, and node 8 with its precursor 7. The remaining nodes are 2,4, 6, and 7. (If nodes 1,3, 5 , 8 are removed.) The remaining nodes are shown above in which two 'selfloops' are formed at nodes 2 and 7 . Now cut these nodes, remove them from the above table and repeat the same procedure. Second reduction Nodes

-

Precursors . -7

4

xz

6

A 4

q

C I T

I

I ,

u

Third Reduction Nodes 4 6

Precursors -

6

As a result, the SFG is totally reduced and the cut-set found consists of nodes 2 and 7 (where self-loops are formed) or equivalent streams 2 and 7 in the digraph of Fig. 12.1.

Decomposition of Networks

317

Illustration 12.2 Using the Barkley and Motard (1972) method (B&M algorithm) for the decomposition of MCN find out the streams that are to be teared (i.e., cut-set) for the digraph of the process shown in Fig. 12.10. Solution First reduction Nodes

Precursors

-

Precursors after reduction (*indicates eliminated)

* * k

* *

6 7 8

7. 18

7, 18 8, 19 6, 10

10 11

8, 26 l9 8, 26 l9

8, 19 8, 19

13

8, 20 l9

8, 19

17 18 19

21

21

8, 20 l9

23 24 25 26

21

28

30 31

22,25

?&25 24, 28 24, 21 28 21

n

* *

* * *

21, 25 24 6, 17 21, 25 24

*

8, 19

* *

21, 25 24 24, 28 * 24, 28 *

IV IV V

*

Zf, 13, 2.3 21, 26, 31

11, 13, 21, 26 24, 2f

30

Id,13, d 21, 26, 31

11, 13, 21, 26 28,30 *

30

** @, 30

Pairs

24,

2f

v, VI VII VI, VII

After the first reduction, seven pairs were found as shown above. For node 6 the precursors are 7, 18. We check whether for any of the nodes 7 and 18 the precursor is 6 to know whether it forms a pair. For node 7 the precursors are 8, 19 (not 6). So node 7 does not pair with 6. But for node 18 the precursors are 6, 17 (6 is there), which means it forms a pair with 6. Nodes 6 and 18 form the first pair.

318

Process Plant Simulation

Similarly, nodes 8 and 10 form the second pair, nodes 19 and 21 form the third pair, nodes 24 and 25 form the fourth pair, nodes 26 and 28 form the fifth pair, nodes 28 and 31 form the sixth pair, and nodes 30 and 31 form the seventh pair. Nodes 28 and 31 have two common pairs (two-way edges of type 2). But out of these two nodes, node 28 is cut and eliminated from the list of precursors, since it has the largest number of output edges. Therefore, node 28 is the precursor for nodes 25,26, and 31. Node 31 is the precursor for nodes 28 and 30. Node 28 has the largest number of output edges. The graph is further reduced, as nodes become single precursor nodes after elimination of node 28. The nodes 25, 26, and 31 become single precursor nodes and as such can be eliminated. This is the advantage of putting the node having maximum output edges into the cut-set. But cutting such nodes of SFG,more number of outputs can be estimated, i.e., the reduction is more effective (i.e., the digraph can be reduced to the smallest size possible). Second reduction Nodes 6 7 8 10 11 13 17 18 19 21 24 28 30

Precursors 7, 18 8, HZ1 6, 10 8, M z l 8, M z l 8, 19 21 21,x 6,M 21 21,% 8, J-9 21 21, 24

C 11, 13, 21, 24, 30

Precursors after reduction 7, 18 8, 21 6, 10 8, 21 8, 21 8, 21 * (eliminate as 24 is in cut-set) 6, 21 * (eliminate as 24 is in cut-set) 8, 21 C (self-loop) C (self-loop) C (self-loop)

Third reduction [C = 24, 28, 301 Nodes 6 7 8 10 11 13 18 21

Precursors X s ,18 8, 6,Ms 8, 8,H 8,Zf 6, X 8, 21

Precursors after reduction 8, 6

*

6, 8 * (21 is in the cut-set)

* * *

21 (self-loop)

Now the cut-set includes the nodes 24, 28, 30, 21.

Decomposition of Networks

319

Fourth reduction [C = 24, 28, 30, 211 Nodes 6 8

Precursors 8, 6 6, 8

Precursors after reduction C (self-loop) C (self-loop)

Finally, the cut-set obtained is [C = 24, 28, 30, 21, 6, 81 or [C = 6, 8, 21, 24, 28, 301 Conclusion By tearing the streams (cutting the edges) 6, 8, 21, 24, 28, and 30 the original digraph computations can be carried out sequentially. It took four reduction steps for this illustration.

12.2.2 The Basic Tearing Algorithm This is also based on the concept of SFG. Pho and Lapidus (1973a, 1973b) presented an alternative algorithm which offers the advantage of assigning any arbitrary weights (weigh each stream with a number reflecting the probable degree of difficulty that can be encountered in an iterative procedure to converge) to the edges or streams of a digraph. It locates the cut-set of tear streams by minimizing the sum of their weights. This is equivalent to finding minimum weights of tear streams when all of them are assigned unit weights. The features of the basic tearing algorithm (BTA) are as given below: 1. The reduced digraph (one that excludes feed and product streams) is first converted into an equivalent SFG. 2. Information on the SFG is provided to the algorithm in the form of an adjacent matrix. 3. Ineligible nodes are found according to the following criterion: If the weight of node i is greater than or equal to the sum of weights of non-zero elements in the first rows of the matrix, then that node is ineligible. 4. The reduction of ineligible node I is performed as follows: Shift the elements of column I to those columns in row I containing non-zero elements forming a Boolean sum. Make row I and column I null. 5. Primarily the cut node is found by locating a self-loop. 6 . If no ineligible nodes are found, then two-way edges are searched. 7. When all attempts fail, the branch and bound method is ultimately applied to totally reduce the SFG. The algorithm is shown in Fig. 12.5.

320

Process Plant Simulation Signal flow

If any row iis null, make column I

Form an adjacency matrix

i=i+ 1

Scan the matrix row-wise starting from node (i), i = 1

Apply ineligible node reduction. Make the row and column of node (i) null. i=i+l No

I'

1 Yes

Yes

t

Search the matrix for two-way edges.

Apply the self-loop node to the cut-set and make the corresponding row and column null.

two-way edges?

two-way edges? Yes

two-way edge reduction by cutting one or more nodes and make the row and column of the cut node null.

Fig. 12.5 Basic tearing algorithm

with lower weight as cut node and make the row and column of the cut node null.

-

Decomposition of Networks

321

12.3 Algorithms Based on Reduced Digraph (List Processing Algorithms) The two algorithms described above (that is, B&M and BTA) use a cumbersome approach of converting a digraph into a SFG and later forming an adjacency matrix. This section will describe three more algorithms, which are based on a single approach of starting directly with a reduced digraph, listing all the nodes (i.e., process modules) and their input and output streams, and processing these lists. These algorithms are simple enough to be applied by hand calculation for sizeable flow sheets.

12.3.1 Kehat and Shacham Algorithm The K&S algorithm was reported by Kehat and Shacham (1973) (Fig. 12.6). It searches a minimum number of tear streams, all of unit weight, such that the cutset found affects the maximum number of output streams from those nodes to which the tear streams are the inputs. This strategy is not sound enough since more computer time is consumed in simulating a flowsheet with the cut-set found.

Example 12.1 Using K&S algorithm for the decomposition of MCN find out the streams that are to be teared (i.e., cut-set) for the digraph of a process shown in Fig. 12.1. Solution Step 1 List the input and output streams of each node. Nodes

Input 2 6, 8 1, 5 7 4, 3

(1)

(2) (3)

(4) (5)

output 1, 3 7

4 5, 8 2, 6

Step 2 Search the list for nodes with minimum input and maximum output streams. As seen, there are two nodes which have one input and two outputs (nodes 1 and 4). Now considering first node 1, make a trail cutting by eliminating the input streams of this node in both the lists. Nodes (1)

(2) (3) (4) (5)

4

Input

output

. I /

1

L

1, J

6, 8 5 7

7 4

4 6

Z 6

-

5, 8

So it is seen that three nodes are affected. Now, we consider node 4.

322

Process Plant Simulation

Reduced digraph

i. List input and output streams of each node.

i Search the list for nodes with minimum input and maximum output streams.

1 this node to the cut-set and reduce the list after making these streams void in both the lists.

Take each node one by one separately from these nodes.

eliminating the input streams of this node in both the lists.

Is this list completely reduced?

1

, Rationalize the list

When inputloutput is zero, take off that node as well as its outputlinput streams appearing in the inputloutput list.

1

List the number of nodes affected as a result of trial cutting this node.

I Restore the original list

1 maximum number of other nodes in the list of its input streams were tom.

Fig. 12.6 The K&S algorithm Input 2

c o v, u

1 3

4, 3

output 1, 3

.,

Y

4

1

Decomposition of Networks

323

As seen for node 4,four nodes are affected. Assign the input stream of this node to the cut-set (as this node affects the maximum nodes) and reduce the list. Nodes (1)

(3) (5)

Input 2 1 4, 3

output 1, 3

4 2

Now node 1 has minimum input and maximum output; trail cutting is done by eliminating the input streams of this node in both the lists. Nodes (1) (3) (5)

Input

+ x A

output 1

9

& 1

9 I

It is seen that this trial cutting reduces the digraph completely and the input stream of node 1 (i.e., 2) goes to the cut-set. Final cut-set = [2, 71.

12.3.2 M&H Algorithms These algorithms were developed by Murthy and Husain (1981; 1983). The first of these two algorithms, M&H-1, uses a criterion of finding a minimum number tear streams, all of unit weight. The second algorithm, M&H-2, finds the cut-set by minimizing sum of weights and offers a facility of assigning weights to various streams according to a definite stream and not arbitrarily as assigned in the BTA. In contrast with K&S both M&H-1 and M&H-2 locate cut-sets, which affects a minimum number of output streams from those nodes to which the tear streams are the inputs (and hence the computational time required is less).

1. From elimination point of view (and from reducing the size of SFG point of view), tearing those streams which affect a maximum number of output streams is advantageous (more computational time). 2. But as proved by Murthy and Husain later, from the computational time of view, tearing those streams which affect a minimum number of output stream is advantageous. 12.3.2.1 The M&H-l algorithm

The following steps are involved in the decomposition of a network using the M&H- 1 algorithm:

1. The algorithm starts with a reduced diagraph by preparing a list of nodes and their input and output streams. 2. Any node with single input-output streams is removed. 3. The single output stream in all its occurrences in the input list is replaced by its precursor.

324

-

Process Plant Simulation

Reduced digraph

List nodes, input, and output streams.

I

t

Replace the single output in the input list by precursor.

Assign the successor digit (SD) to nodes with single output.

t-

Assign output of nodes with SD2 2 successively to the cut-set, if the same is single input to

Fig. 12.7 Flow chart of the M&H-l algorithm

4. Any self-loops formed are assigned to the cut-set and the nodes are deleted from the list. 5. Further processing is done by identifying nodes with a single output stream and assigning them ‘successor digit (SD)’, which is equal to the number of its successor output streams.

Decomposition of Networks

325

6. Then the nodes are eliminated, one by one starting with SD = 1 and SD 2 2. 7. Every time self-loops are assigned to the cut-set. 8. The procedure is continued till the total list is reduced. The overall procedure is shown as a schematic flow chart in Fig. 12.7.

Example 12.2 Using the M&H-1 algorithm for the decomposition of MCN find out the streams that are to be teared (i.e., cut-set) for the digraph of a process shown in Fig. 12.10. Solution First reduction of the digraph shown in Fig. 12.10 ( M 6 H - 1 ) Nodes (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (1 1) (12) (13) (14) (15) (16) (17) (18) (19)

Input

output

1 I

?I

Input after reduction

*

L

*

,

n 1 1

I1

K6

%-3,X1O 8,28 l1

H, 13, 23, 26, 31

,?I

IL.

*

1, 3,4 8 7, 10, 11, 13, 21 28, 30

6 1, 3, 10 8, 19 11, 13, 23, 26, 31

I+

Lt

*

I

U

J

7, 18

6 28 22, 23 31

in I,

21 29, 30

* *

7, 18

*

21 29, 30

*

1%

-&A,

1J

r lu-

*

4dKM, 17 22, 25 24, 27 28

18 17, 19, 24 25, 26 27, 29

4, 17 22, 25 24, 27 28

e

Second reduction of the digraph Nodes Input

output

(4)

1, 3, 4 8 7, 10, 11, 13, 21 28, [301 6 22, 23 31 18 17, 19, 24 25, 26 27, 29

(5)

(6) (7) (10) (12) (13) (16) (17) (18) (19)

6 1, 3, 10 8, 20 11, 13, 23, 2 6 / n 7 , a 4,17 21 29, 30 4, 17 22, 25 24, 27 28

29,30

Successor digit (SD)

Input after reduction

5

11, 13, 23, 26, 29, [31] 3 2 1

* *

326

Process Plant Simulation

SD (successor digits) are assigned to single output streams 8, 6, 31, and 18: SD of 18 [of node (16)] = 1 [as for input 18, there is only one output 6, i.e., node (lo)] Similarly, SD of 31 [of node (13)] = 2 [as for input 18, there is only one output 6, i.e., node (1011 SD of 6 [of node (lo)] = 3 [as for input 18, there is only one output 6, i.e., node (1011 SD of 8 [of node ( 5 ) ] = 5 [as for input 18, there is only one output 6, i.e., node (1011 Now, eliminating node (16) of SD =1, replace node (10) with 4, 17. Now, no single input-output stream and no self-loops are present. Now for SD = 2 [i.e., for node (13)] output is 3 1, which is not a single input to any node. So, 3 1 cannot be put in the cut-set. So, we now eliminate node (13) of SD = 2. The inputs for 3 1 are 29 and 3 1. So, replace input 3 1 [see node (13)] of node (7) with 29 and 30. The corresponding output of (7) is 28 and 30, self-loop is formed in 30. Numbers within square brackets represent self-loops. So, cut stream = { 30j. Now, input of single output 31 = 29 and 30 = 29 (as 30 is put in the cut-set) Input after reduction 6 1, 3, 10 8, 19 (7) (10) (12) (17) (18) (19)

11, 13, 23, 26,29 7,4,17 21 21

z, ~ 4 1

24, .HZ8 28

28 6 22,z 17, 19, [24] 25,X 27,W

2 3

2

Eliminate both nodes

4,7, 17

*

21

* *

Eliminate node (7) of SD = 2. Input of 28 is 11, 13, 23, 26, and 29. Assign outputs of nodes with SD 2 2 successively to the cut-set, if the same is single input to any other node. For SD = 2 [node (7)], output is 28, which is the same as the single input to node (19). Hence, put 28 into the cut-set, and eliminate node (7). The inputs of node (7) are 11, 13, 23, 26, 29. These inputs appear as outputs in nodes (6), (12), (18), and (19). So, cut them. Then nodes (12) and (18) become single input-output nodes, so eliminate them. The precursor of 25 is 24, therefore in node (17), there is a self-loop of 24. Therefore, put 24 into the cut-set. Hence cut-set streams = 28, 24.

Decomposition of Networks

327

Fourth reduction of the digraph Nodes

Input

(4)

6

(5) (6)

8, 19

(10)

4,7, 18

(17)

21

10

output

Successor digit (SD)

1, 3, 4 8 7 , M , 21 6 N,19

?I

Eliminate bothnodes

3

Input after reduction

*

6, 10 8, 19

*

21

Now the single output stream nodes are ( 5 ) and (6).

SD of ( 5 ) = 3 (as for input 8, there are three outputs, i.e. 7, 10, 21) SD of (10) = 3 (as for input 6, there are three outputs, i.e., 1, 3, 4)

Eliminate node (lo), having SD = 3 (as 8 is not a single input to any node, whereas 6 is a single input to node 4).Then eliminate nodes (4) and (7). Input of node 10 = 4,7, 17 which is eliminated so cut 7 and 17 from nodes (6) and (17), respectively. Cut stream = 6 . Fijlh reduction of the digraph Nodes

Input

(5)

10

(6) (17)

‘O/8;-€9

21

output

21

8 [lo, 211 19

Successor digit (SD) 3

Input after reduction

* [lo, 211

*

Eliminate nodes ( 5 ) and (17), since these are single input-output streams. Replace 8 and 19 of node ( 6 ) with precursors 10 and 21, respectively. Then 10, 21 form self-loops. Put them in cut streams. Therefore cut-set = 6, 10, 21, 24, 28, 30. The number of streams affected = 9 (i.e., 1 + 1 + 2 + 2 + 2 + 1). 12.3.2.2 The M&H-2 algorithm

The flow chart of the M&H-2 algorithm is shown in Fig. 12.8. This algorithm is superior to the M&H-1 algorithm. There are two schemes that are employed to assign weights to the streams: Scheme I: This scheme allots a unit weight or any arbitrary weight to each stream. Scheme 11: The scheme allots to each input stream of a node a weight equal to the number of output streams of that node. The node denominations (ND) are to be calculated in this case: ND = (sum of weights of output streams)

- (sum of weights of input streams)

328

Process Plant Simulation Reduced digraph I

I

List nodes, input, and output streams.

I

Assign weights to streams according to scheme I or 11.

1 Find node denominations (ND) and list.

innovations (1)-(4). Rationalize list.

Replace the single output by the precursor.

No

No

Identify self-loops, assign streams to cut-set, and I

Yes I

Fig. 12.8 Flow chart of M&H-2 algorithm

The nodes with single input-output and ND = 0 are first deleted by replacing the single input-output stream in all its occurrences in the input list by its precursor. Any self-loops formed are assigned to the cut-set and the nodes are deleted. Further processing is done for nodes with negative NDs. Every time the list is rationalized and NDs are readjusted. When magnitudes of NDs are not substantially different from each other, certain innovations are necessary. These innovations are as follows: 1. Replace zero denomination of a node by a positive one when the total number of input streams to that node is greater than that of the outputs.

Decomposition of Networks

329

2. During reduction, retain the node for further consideration having a tear

stream (already identified) as one of its inputs, in case its denomination is zero or negative. 3. In the process of reduction for ND I 0, do not replace an input stream by its precursor in case the same stream is an input to another node with a positive denomination. 4. In the process of reduction for ND I0, do not replace an input stream by its precursor in case the latter is a single input to a forward node with zero denomination.

Example 12.3 Using the M&H-2 (scheme 1) algorithm for the decomposition of MCN, find out the streams that are to be teared (i.e., cut-set) for the digraph of the process shown in Fig. 12.10. Solution The nodes which are denoted by A to S in Fig. 12.10 have been changed to (1)(19), respectively, for solving this problem. First reduction of the digraph of Fig. 12.10 (M&H-2) Nodes

(1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14) (15) (16) (17) (18) (19)

Input

output

1 1 1, 3, 4 %3f110 8 8, XI9 7, 10, 11, 13, 21 "E,13, 23, 26, 31 28, 30

K6

7, 18

6

21 29, 30

22, 23 31

4m16, 22, 25

17

2592224, 27 28

18 17, 19, 24 25, 26 27, 29

ND

Input after reduction (* Eliminate)

0 0 0 2 -2 3 -3 0 0 -1 0 1 -1 0 0 -1 1 0 1

* * *

6 1, 3, 10 8, 19 11, 13, 23, 26, 31

*

* 7, 18

*

21 29, 30

* *

4, 17 22, 25 24, 27 28

For node 18, though ND = 0, it is not eliminated. The reason being innovation no. 2, which says that during reduction retain the node for further consideration having a tear stream (already identified) as one of its inputs, in case its denomination is zero or negative. Stream 25 is identified to be a tear stream, as it forms a self-loop. If node 18 is eliminated, we might not have been able to put 25 in the cut stream, which forms a self-loop.

330

Process Plant Simulation

Second reduction of the digraph Nodes

Input

6 8, 1 9 , 6 4 ~ ~ 198 ,

, ,

%: 19 11, 13, 23, 2 6 , X 29330 4

1

9

q 4.17 ~

21 29, 30 4, 17 22, M 22,25

s, ~

2

8

28

output

ND

X K 4 [81 7, M, -M, +3-,21 28, [301 6 22,a 31 18 17, 1 9 , H [251, 26 27,29

2 -2 3 -3 -1 1 -1 -1 1 0 1

Input after reduction 6 6, [81, 19 19 19, 21, 24,2',3$, [30] 4,17, 19 21

* *

22 22, [25], 28 28

Numbers enclosed within square brackets denote self-loops. Cut streams: 8, 25, and 30. Third reduction of the digraph Nodes

Input

output

ND

Input after reduction

* *

0 0 -2 -2 0 +1 0

19, [28] * 19, [6], 17 *

*

[191*

*

Cut streams = 6, 19, 28. Cut-set = {6, 8, 19, 25, 28, 30}.

Example 12.4

Using the M&H-2 (scheme 11) algorithm for the decomposition of MCN, find out the streams that are to be teared (i.e., cut-set) for the digraph of the process shown in Fig. 12.1, Solution Step 1 List the input and output streams of each node with their corresponding weights. Calculate the corresponding ND. Numbers enclosed within square brackets denote self-loops. Nodes

Input

(1) (2) (3) (4)

2 6,%"1 1, 5 7 4, 3

(5)

Weight of input 2 1, 1 1, 1 2 2, 2

output

1, 3 [71 4 5, 8 2, 6

Weight of output 3 2 2 1, 1 2, 1

ND

1 0 0 0 -1

Decomposition of Networks

331

Step 2 For all nodes with zero or negative ND, replace input streamsby precursors. If all the nodes with ND = 0 are eliminated from the above table [i.e., nodes (2), (3), and (4)], then there will be a problem in node (5) while replacing the eliminated outputs. There is a continuous loop of ‘7’to be replaced with ‘8’, and ‘8’ in turn to be replaced with ‘7’, which is never ending. In order to avoid this problem, node (2) should not be eliminated (also as per innovation no. 2). There is a self-loop in [7] from node (2), as shown in step 1. Hence, the list in step 1 reduces to Nodes

Input

(1) (2)

2 [*]/r:5, 3

Weight of input 2 1, 1, 1

output

Weight of output 2 2

1, 3 [21

ND 0 -1

Step 3 Node (1) is eliminated, as the ND of this node is zero. Now, replace the input stream ‘1’ of node (2) by its precursor 2 as shown above. Then again there is a self-loop in [2]. Therefore the final cut-set is { 2, 7}. Table 12.1 Comparison of results for the digraph shown in Fig. 12.10 Number of output streams affected

Computational time

6, 8, 21, 24, 28, 30

13

Unit Unit Unit Unit

8, 18, 22, 25, 27, 31 5 , 8, 20, 25, 28, 31 6, 10, 21, 24, 28, 30 6, 8, 19, 25, 28, 30

16 20 9 13

Less but more cumbersome More More Less Less

Scheme Weight

6, 10, 21, 24, 28, 30

9

Less

Algorithm

Weights

B&M

Unit

BTA K&S M&H- 1 M&H-2 Scheme-I M&H-2 Scheme-I1

Cut-set

The performance of the various algorithms described so far is compared in Table 12.1 for the digraph of Fig. 12.10. Basis of comparison: The number of output streams affected from those nodes to which the tear streams are the inputs. This criterion is equivalent to the sum of weight according to Scheme I1 of M&H-2 algorithms. From Table 12.2, it is clear that both algorithms M&H-1 and M&H-2, the latter using weights according to scheme 11, produce the best cut-set affecting the least number of output streams.

12.3.3 Comparison of Various Tearing Algorithms Table 12.2 shows a comparison of various tearing algorithms,

332

Process Plant Simulation

Table 12.2 Comparison of various tearing algorithms. SD stands for successor digit; ND for node denomination. Algorithm

Diagram

Weight

1

B&M

SFG

Unit

2

BTA

SFG

Unitfarbitrary

3

K&S

Reduced digraph

Unit

4

M&H-l

Reduced digraph

Unit

5

M&H-2 scheme I M&H-2 scheme I1

Reduced digraph Reduced digraph

Unitfarbitrary

6

Definite scheme weight

Reduction criterion Self-loop two-way edges Weights assigned reflecting the probable degree of difficulty in an iterative procedure to converge Cut-set affects maximum number of output streams Minimum number of tear streams Minimizing sum of weights Cut-set affects minimum number of output streams

Weights assigned

Arbitrary

SD = 1, SD 2 1 eliminate ND ND

EXERCISES 12.1 For the reduced digraph given below (Fig. 12.9), find the following: (a) Signal flow graph. (b) Cut-set using the M&H-l algorithm.

Fig. 12.9 Digraph of a process

12.2 Using the B&M algorithm for the decomposition of MCN, find the cut-set for the digraph of a process plant (Fig. 12.10).

Decomposition of Networks A

I

I

C

-

A' 5

H

B

333

6

1

14 1

N

15

0

16

Flg. 12.10 Digraph (Sargent 84 Westergerg 1964)

12.3 The reduced digraph of a chemical process plant is shown in Fig. 12.9. Using the algorithm for decomposition of networks proposed by Murthy and Husain-2 (M&H-2), find the following: (a) Cut-set containing the tear streams. (b) Total number of output streams affected with the cut-set. 12.4 Using the B&M algorithm for the decomposition of MCN, find the cut-set for the digraph of a process shown in Fig. 12.9. 12.5 The reduced digraph of a chemical process plant is shown in Fig. 12.10. Using the algorithms for decomposition of networks proposed by Murthy and Husain (M&H-1 and 2 with both the schemes), find the following: (a) Cut-set containing the tear streams. (b) Total number of output streams affected by the cutset. 12.6 The reduced digraph of a chemical process plant is shown in Fig. 12.11. Using the algorithm for decomposition of networks proposed by Murthy and Husain (M&H-1 and 2 with both the schemes) and the B&M algorithm, find the following: (a) Cut-set containing the tear streams. (b) Total number of output streams affected by the cut-set.

334

Process Plant Simulation

Fig. 12.1 1 Digraph (Christensen & Rudd 1969)

12.7 The reduced digraph of a chemical process plant is shown in Fig. 12.12. Using the algorithms for decomposition of networks proposed by Murthy and Husain (M&H-1 and 2 with both the schemes) and the B&M algorithm, find the following: (a) Cut-set containing the tear streams. (b) Total number of output streams affected by the cut-set. 8

~~

~

33 Fig. 12.12 Digraph (Christensen & Rudd 1969)

Decomposition of Networks

335

39 60

I

39

I

42

61

65

10

6

q+$ 29

/

40) \

42

40

38

43

47

43

29

64

I

1

69

64

18

21 51

30

I Fig. 12.13 Digraph: Sulphuric acid plant

53

336

Process Plant Simulation

12.8 The reduced digraph of a sulphuric acid plant is shown in Fig. 12.13. Using the algorithms for decomposition of networks proposed by Murthy and Husain (M&H-1 and 2 with both the schemes) the B&M algorithm,basic tearing algorithm, K&S algorithm, find the following: (a) Cut-set containing the tear streams. (b) Total number of output streams affected by the cut-set. (c) Compare the results of various algorithms.

CHAPTER 13

CONVERGENCE PROMOTION AND PHYSICAL AND THERMODYNAMIC PROPERTIES

In this chapter, first we will discuss the importance and methods that can be used for convergence promotion. Then, we focus on the retrieving methodologies for physical and thermodynamic properties that are required for any process plant simulation. The sequential modular approach discussed in Chapter 11 is generally executed by decomposing the networks (tearing certain streams, as discussed in Chapter 12) and estimating the values of the teared streams. A number of competing methods are available in literature for convergence promotion of the material and energy balances.

13.1 Convergence Promotion In convergence promotion or acceleration, the objective is to find a vector such that starting the calculations with it leads to exactly the same vector as the result. Such a problem can be regarded formally as of finding the solution of x = g(x), where x represents the vector of guessed variables and g is the computational algorithm which when applied to x produces a vector generally different from x, unless the solution has been obtained. Hence, the problem is expressed in the following form: (13.1) fix> = x - g(x> wheref, g, and x are all n x 1 vectors. The following methods are reported to promote convergence of Eq. (13.1) tofix) = 0.

338

Process Plant Simulation

13.1.1 Newton’s Method The iterative formula for this method is as follows: (13.2)

where i represents the cycle of calculation or the iteration number. The method is characterized by fast convergence when the initial guess xo is close to the solution. However, it requires evaluation of an n x n matrix of the function derivatives (the Jacobian) as well as solving n simultaneous linear equations. The other methods suggested can be thought of as particular ways of approximating the Jacobian of the non-linear functionsAx).

Example 13.1 Solve the equation x2 - 2 = 0 with an initial guess of x = 2. Solution The function is x2 - 2 = 0. The starting point for Newton’s method is 2.0. The derivative of the function is 2x. Newton’s method for this function is as follows: x(i + 1) = x(i) -

x2(i>- 2 2x(i)

= 2.0 = 1.5 = 1.41666 = 1.41422 = 1.41421 = 1.41421356237309504880168962 = 1.41421356237309504880168872420969807856967187537723 Root = 1.41421356237309504880168872420969807856967187537694 8073 17668.. .

x[O] x[l] x[2] x[3] x[4] x[5] x[6]

The estimated solution is 1.41421356237309504880168872420969807856967187537723

There may be more than one solution.

Example 13.2 Solve the equation x3 - 4x2 + 6x - 5 = 0 with an initial guess of x=4.

Solution The values shown in Table 13.1 are obtained in various iterations which finally converge to x = 2.35320996419932.

Convergence Promotion and Physical and Thermodynamic Properties

339

Table 13.1 Iteration number

x(i) x[l] x[2] x[3] x[4] x[5] x[6] x[7] x[8] x[9]

= 3.13667761101986

= 2.62594320362587 = = = = = = =

2.4005817996535 2.35497912232491 2.35321391356784 2.35320996740109 2.35320996420191 2.35320996419933 2.35320996419932

13.1.2 Direct Substitution The Jacobian is approximated by the identity matrix in the simplest of these methods, leading to xi + 1 = x i -f i (13.3) It is the most widely used method, because it is simple and requires little storage space. However, its convergence can be very slow, unless the Jacobian is indeed close to the identity matrix.

Illustration 13.1 Minimizef=flX) subject to gj(X) = 0, j = 1, 2, ..., m where

and m < n to make the problem well defined. Solution We can solve this problem by direct substitution; in other words, we solve simultaneously the m equality constraints and express any set of m variables in terms of the remaining n - m variables. The new objective function is not subjected to any constraint; hence the optimum can be found by using any unconstrained optimization techniques. But, in practice, it is not often possible to express m variables in terms of remaining n - m variables in the case of non-linear constraints. Thus a more general method is needed.

340

Process Plant Simulation

Solution by the method of Lugrange multipliers In order to avoid the deficiencies encountered by the direct substitution method, we use the Lagrange multipliers method, which has been widely used in areas such as dynamics and controls. The above-mentioned optimal problem with equality constraints is equivalent to the problem formulated in the form of the Lagrange multiplier: L ( X , A) = f ( X > + Ag(W The necessary conditions for its extremum are given by dL -(X, dX

dL A ) = 0, - ( X ,

dA

A) = 0

In the case of multiple constraints: equality, inequality, side, etc., we can always convert inequality constraints into equality ones by adding slack variables, e.g., 2 g,(X) I0 , j = 1 , 2 , ..., m after adding slack variables y; turns out to be gj(X) + Y j = O , j = 1, 2, .,,, m, where the values of the slack variables are yet unknown. Applying the Lagrange method to the new function L ( X , Y . A ) = f ( X ) + Cy==l A j G j ( X ,Y ) , we get

dL -(x, dXj

Y , A) = G ~ ( xY, ) + gj(x)+ y: = 0, j

dL -(X,

Y , A ) = 2Ajyj = 0, j = 1, 2, ..., m

=

1,2, ... m )

*j

Fig. 13.1 Simplification of active and inactive constraints

Convergence Promotion and Physical and Thermodynamic Properties

341

The above third equation implies that either A = 0 or yj = 0. The former case means that thejth constraint is inactive and hence can be ignored. Otherwise means that the constraint is active at the optimum point. We can further simplify the Lagranges according to active and inactive constraints. Take a two-constraint problem for instance; the above equations can be represented as (Fig. 13.1):

-Vf = A,Vg, + A,Vg, Let S be a feasible direction at the optimum point. Premultiplying both sides of this equation by ,'S -STVf = A,STVg, + A2STVg, as S is a feasible direction, it satisfies the relations STVg, < 0 and STVg2 < 0. If A > 0, then STVf > 0. In other words, we are not able to find any direction in the feasible domain along which the function value can be decreased further.

Kuhn-Tucker conditions As shown above, the conditions to be satisfied at the constrained minimum point X* are -a +f+ j , , , A j - = 0%8, i = 1 , 2

axi

,..., n

Ai

A > 0, j

E J,,where J, is

the active constraint.

These are called the Kuhn-Tucker conditions.

13.1.3 Wegstein's Method In this method (Wegstein 1958), the Jacobian is approximated by a diagonal matrix whose elements are i

i-1

xj - xj

, j = l , 2,..., n

The iteration formula thus becomes (13.4)

In other words, a secant step is taken for each variable separately. In this way, the method ignores the interaction among variables; changes made in each component of x do not take into account those made in the other components. This may result sometimes into oscillations in the iterative procedure or cause divergence instead of convergence.

342

Process Plant Simulation

Example 13.3 Write a pseudocode for convergence promotion using Wegstein’s method. Solution function x=vweg2(vl ,g,tmin,tmax,nmax,erc) % vweg2: vector solution of n equations in n unknowns using Wegstein’s method % function x=vweg2(vl ,g,tmin,tmax,nmax,erc) % Vector solution of n equations in n unknowns using the Wegstein method % of finding roots % Use VWEGl to do each Wegstein step % Argument Gives a character vector naming a function that tells how to % g find a better value for the unknown % vector x given an approximate set: v l . % The vector must be called x in g. % % vl the initial guess for x. % tmin min allowed change in a step. % tmax max allowed change in a step. % nmax max iterations allowed. % erc desired error. % Typical (default) values are: nmax=lO, erc=.001, % tmax=5 and tmin=.lO % You may give the function 2, 4 or 6 arguments. % If you give it only two, the default values are used % for the rest. % If you give it 4, the default values for nmax and erc % will be used. % Example: >> vweg2(zeros(1,7),’g312(x)’,-10,5,10,0.001) % revised help by TYLC if narginc=4 nmax= 10; erc=.001; if narginc=2 tmi n=- 10; tmax=5; end end x=vl; v2=eval(g);

Convergence Promotion and Physical and Thermodynamic Properties

343

for k=l:nmax x=vwegl(vl ,v2,g,tmin,tmax); if max(abs(x-v2))<erc return end vl=v2; v2=x; end disp(‘vweg2 did not converge’) prntmx(‘v1 and v2’,[vl;v2],10,4)

Example 13.4 Make a module that simulates a two-phase equilibrium process. The diagram for the two-phase equilibrium process is shown in Fig. 13.2.

Phase 2

u

NC components

Phase 1

Fig. 13.2 Two-phase equilibrium process

Solution There is one feed stream that is split into two phases at a constant temperature T and pressure P . The feed stream is defined by its flow rate F,, composition (component mole fractions, zi), P , and TF. It is necessary to calculate the exit stream flow rates and compositions for specified T and P . Note that in an equilibrium process the exit streams are at equilibrium with each other. That is, at the specified T and P , the two exit streams are at equilibrium. The model to be used is based only on mass balance. The component and total mass balance equations are (NC stands for the number of components in a multicomponent mixture).

344

Process Plant Simulation

yiF2 + xiF3 = ziFl, i = 1, 2, ..., NC

F2 + F3 = F , The equilibrium relation is given by, yi = Ki xi, i = 1, 2, ..., NC

(3) The expressions for Ki have been given in the description of the problem for a vapour-liquid equilibrium system. The definition of mole fractions requires that (4) EXi - cyi = 0

czj

=1 (5) We have 3NC + 3 equations with 4NC + 5 variables (x, y , z , K, T, P, F,, F2, F3). This means that we need to specify NC + 2 variables. The variables to specify are zi, i = 1, 2, ...,NC - 1, F,, T, and P. Note that there are NC equations for K .

Solution procedure By combining Eqs (1) and (3), it is possible to eliminate yi, and we get (Ki F2 + F3) xi = ziF1,i = 1, 2, ..., NC

Based on Eq. (6), the following solution procedure is proposed. 1. Given values of zi, i = 1, 2, ..., NC - 1, F,, T and P. 2. Using Eq. ( 5 ) , calculate zNC = 1 - (zl + z2 + z3 + + zNC3. Assuming ideal gas law, calculate K i = e S / P 4. Guess F2/F1;calculate F2 from (F2/F1)F,. 5. Calculate F3 from Eq. (2). 6. Calculate xifrom Eq. (6). 7. Using the calculated values of x i and Ki, calculate yi from Eq. (3). 8. Check whether Eq. ( 5 ) is satisfied. If yes, stop. Otherwise, assume a new value for F2/F1and repeat from step (4). Note that the above procedure is valid for a composition-independent expression for Ki. If the composition dependence should be considered, the above procedure can be employed to generate an initial estimate for xi and yi. Then with these values, the following solution procedure is proposed. Given values of zi, i = 1, 2, ..., NC - 1, F,, T and P ; and initial estimates for x and y. Using Eq. ( 5 ) , calculate zNC= 1 - (z, + z2 + z3 + . * * + z N C - ,). Guess F2/F1;calculate F2 from (F2/F1)F,. Calculate Ki using an appropriate expression. Calculate F3 from Eq. (2).

Convergence Promotion and Physical and Thermodynamic Properties

345

Calculate xi from Eq. (6). Using the calculated values of x i and Ki, calculate yi from Eq. (3). Check whether Eq. (5) is satisfied. If yes, stop. Otherwise, assume a new value for F2/F1and repeat from step (4)(also normalize values of x and y). Alternative methods can also be found in Separation Processes by King (1980). Convergence criteria The residual function for Eq. (5) may be written as R = C X -~C y i (7) A plot of R versus F2/F1 (for the ideal system) will show that R will have zero value between F2/F1= 0 and F2/F, = 1. At F2/F1= 0, there is only one phase (F3), for F2/F, = 1, there is also only one phase (F2).This means that at values of F2/F1 very close to unity, R is positive, while at F,IFl very close to zero, R is negative. The solution for F2/F1therefore must lie between 0 and 1. If for both values of F2/F1,R is either positive or negative, it indicates the feed mixture at the specified T and P is either vapour or liquid, respectively. Convergence is achieved when R is less than abs (lo4)or lower. To update the value of F2/F1,Wegstein's method can be employed once two estimates on either sides of the solution have been found. Alternatively, the Newton method can be employed. Let us consider p = F2/F1.Then new updates with the Newton method are obtained from

p=p-WJ (8) where J = dW& Note that for the non-ideal case, for convergence, the changes in x and y between iterations should also be monitored. Liquid-liquid systems The above procedure can also be used for a liquid-liquid system by simply changing the expression for Ki. For a liquid-liquid equilibrium system, the condition of equilibrium is given by,

Example 13.5 Consider a mixture of n-pentane, n-hexane, and n-heptane. The values for the specified variables are z1 = 0.65, z2 = 0.20, F , = 100, T = 330 K, and P = 1 atm. The converged F2/F1value is approximately 0.7645 with the ideal gas law for a vapour-liquid system in equilibrium. Use the vapour-pressure data and other necessary data from the supplied database.

Process Plant Simulation

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Solution If the SRK model (Soave modification of the Redlich-Kwong equation of state) is used for the above the system, the following results are obtained: PT-flash: Temperature (K) = 330.00000 Pressure (atm) = 1.00000 Number of phases = 2 The mole fractions in feed and each phase are as follows: Phase type 1 2 3

Feed 0.65000 0.20000 0.15000

Gas 0.72857 0.18 100 0.904299e - 01

Liquid 0.38802 0.26334 0.34864

Fraction of feed: 0.76929, 0.2307 1

13.1.4 Dominant Eigenvalue Method If the Taylor series approximation is used to first-order terms about an arbitrary point x' in the neighbourhood of xi, then xi+l= Ax' + b

(13.5)

where A=(%)

(13.6) x=x,

b =Ax') -Ax'

(13.7)

Now labelling eigenvalues A of A in descending order of the absolute magnitude, it can be shown that the necessary and sufficient condition for Eq.(13.5) to converge is IA, I < 1.0 (13.8) when A, is the dominant eigenvalue. This is the basis of the dominant eigenvalue method (DEM) proposed by Orbach and Crowe (1971), which claims major advantage over Newton's method that neither evaluation nor inversion of a large matrix is necessary.

13.1.5 General Dominant Eigenvalue Method As an extension of the DEM to two or more dominant eigenvalues, Crowe and Nishio (1975) presented the general dominant eigenvalue method (GDEM). In this method, the interaction of the variables is taken into account very nicely by the following iteration formula:

Convergence Promotion and Physical and Thermodynamic Properties

xi+l

= g'

+

i

347

(13.9)

where coefficients p are determined by the least square approximation i

(13.10) subject to (13.11)

Po = 1

or (13.12) j=O

The use of Eq. (13.12) has been suggested by Kaniel and Stein (1974). Crowe (1978) showed that this leads to more efficient convergence than the use of Eq. (13.11).

13.1.6 Quasi-Newton Methods The iteration formula in this case is xi+l

= x i - p. f. '

(13.13)

where Hi is a matrix approximating the inverse of the Jacobian and is updated in every iteration as follows: (13.14) where qi = x i + l - x i yi = f i + 1 - f i

'

(13.15) (13.16)

The vectorp' is arbitrarily chosen; therefore, the convergence rate depends strongly on its choice as well as on the initial guess 9. Broyden (1965) made the following choice for p i , with 9 = I: pi

= H'Tqi

(13.17)

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Process Plant Simulation

In the modification suggested by Metcalfe and Perkins (1978), the matrix is updated in each iteration unless the newly estimated value is an order of magnitude worse in the sense of a sum of squares of residuals than the current point. Soliman (1978) chose pi as follows:

Pi -- g'-lqi

=fi

(13.18)

with

@=I

(13.19)

or

Ho

=(%)

-1

(13.20) x=xo

Four more algorithms were presented by Soliman (1979), which incorporate the identity matrix approximation for the Jacobian.

13.1.7 Criterion for Acceleration While using both sequential and simultaneous modular approaches, Kluzik (1979) observed that for easy problems (small A1), Wegstein's method works reliably and rapidly. But for hard problems ( A1 near l), Wegstein's method faces difficulty in convergence. He further noted that quasi-Newton methods converge rapidly near the solution when in the linear region, and minimum loop tearing does not help convergence when full matrix methods are used. Since no single method seems to work universally well over a broad range of problems, combinations of methods have been tried. These include direct substitution between acceleration attempts. Assuming that a positive A, case is generally encountered in the process simulation, Rosen (1979) presented' the following criterion for accelerating convergence between the bounded Wegstein method (Kleisch 1967) and the direct substitution method: (13.21) In Eq. (13.21), the acceleration criterion 7, is a function of Ai alone. If 7, < (1 - A,)/(l + A,),acceleration takes place; otherwise direct substitution must be used. The net effect is that with low A,,the easy problems will always accelerate, but the hard problems ( A, near 1) will accelerate only 7, is small. Crowe (1984) modified the DEM such that an acceleration step is taken at every iteration. A relationship is derived between the class of rank-1 quasi-Newton methods and the modified DEM.

Convergence Promotion and Physical and Thermodynamic Properties

349

13.2 Physical and Thermodynamic Properties The facility to retrieve physical and thermodynamic properties of the components and their mixtures involved is essential and the basic requirement of any process plant simulation. The commonly needed properties are density, compressibility, heat capacity, enthalpy, entropy, and K values. The less common are the thermal expansion coefficient, heat of mixing, Gibbs free energy, heats of formation and reaction, free energy of reaction, and surface tension. A second category of properties needed is the transport properties such as viscosity, thermal conductivity, and diffusion coefficients.

13.2.1 Sources The two sources of data on properties are those experimentally measured and those already reported in the published literature. Bulks of the data are available for pure components and for mixtures of usually not more than two or three components over limited ranges of temperature and pressure. The commonly measured properties are vapour pressures; temperature and enthalpy changes for phases at equilibrium; density, volume, heat capacity, and enthalpy of a single phase in equilibrium with a second phase; critical constants; enthalpy changes for chemical reactions; equilibrium constants; electrical cell potentials; etc. Additional properties measured for the mixtures are volume and enthalpy changes on mixing and composition of phases at equilibrium. Some other properties which are less often measured are volume, heat capacity, and enthalpy of a single phase as a function of temperature, pressure, and composition; velocity of sound; JouleThompson coefficient; and adiabatic compressibility. The following properties often tabulated are usually derived from the experimental data: enthalpy, volume, entropy, fugacity, Gibbs energy, and Helmholtz energy of a single phase as functions of temperature and pressure; thermodynamic properties of ideal gases; enthalpy and Gibbs energy of formation of compounds and components of mixtures in the standard state; excess thermodynamic functions, activity coefficients, and osmotic coefficients of mixtures. The demand for thermodynamic properties will probably always exceed than what can be obtained from experimental measurements. This can be met only by calculating or estimating properties. Such estimation methods range from those based on rigorous statistical mechanics to those based entirely on empirical approaches. The most successful use of statistical mechanics is the prediction of ideal gas properties from molecular properties. Reid and Sherwood (1958; 1966) presented a critical review of procedures for estimating the properties of gases and liquids, which has been broadened and expanded by Reid et al. (1977). Reonick (1981) showed how the various physical and thermodynamic properties of a single

350

Process Plant Simulation

compound can be estimated based on the information on its structural formula and its normal boiling point. The other useful sources for data are ACS (1961), Canjar and Manning (1967), Starling (1973) and Yaws (1977). Simple and efficient means for computer estimation of thermodynamic properties of real gases have been proposed by Llinas and Freze (1976). Several kinds of properties can be successfully estimated as a sum of bond or group contributions [Somayajulu and Zwolinski (1972, 1974, 1976), Benson (1976), Fredenslund et al. (1977), Yoneda (1975)l. The sources for vapour-liquid equilibrium data are Air Liquide (1976), Chu et al. (1950; 1956), Gmehling and Onken (1977), Ohe (1976), and Wichterle et al. (1973; 1976).

13.2.2 Databanks A databank is an electronic analog of a file cabinet in which data are stored for direct retrieval.The design of a thermodynamicproperty databank requires decisions to be made on (i) the kind and number of substances to be stored and their properties, (ii) types of representation for the data, and (iii) organization of the information in total. These questions are closely related in the choice of algorithms for searching the data files, interpreting and processing the data located and procedures for quality control and for file expansion and revision. The data on properties to be retrieved can be classified as follows (Wilhoit 1980): 1. Raw data: results of direct experimental measurements. 2. Normalized data: raw data converted to uniform set of units and reference states. 3. Selected data: single best value for each property of a substance. 4. Smoothed data: selected data represented as smooth functions of independent variables, usually requiring interpolation or extrapolation. 5. Correlated data: experimental data represented by parameters characteristic of molecular structure and dynamics. It is difficult to organize the different types of data into a single file. They have to be necessarily stored in separate files. For most applications, selected and smoothed data are preferable. In some cases, the estimated value of a property might be more accurate than the measured one. Hence, the use of molecular parameters is a potential means of generating physical property data. The estimation procedures can be better incorporated into a retrieval software and invoked whenever a needed property is missing from the data file. For instance, by the use of group contributions,properties of a large number of pure species and their mixtures can be calculated from a small set of group values. Thus, such a procedure will be more efficient to estimate properties as needed rather than storing them in a file. The group contribution values can, therefore, be stored in an auxiliary

Convergence Promotion and Physical and Themtodynamic Properties

351

file. Empirical methods are restricted to certain classes of compounds or limited ranges of independent variables; hence, they should be used with caution when embedded in a retrieval software. The following kinds of specifications define the range and scope of the information to be stored and retrieved: 1. Substances (pure species or mixtures or both) 2. Classes of compounds (organic, inorganic, metals, polymers, electrolytes, etc.) 3. Phases (solid, liquid, gas, plasma) 4. Kinds of properties and their units 5. Kinds of independent variables and their ranges 6. Measures of accuracy Adequate documentation of the data in the file is essential. The immediate source of data should also be recorded and retrieved. If a property value is estimated, it is helpful to identify estimating method. If it is a measured value, reference to the original source should be available. The choice of properties to be stored and retrieved is the principal factor determining the utility of the databank. From the process simulation point of view, these will probably include the following. 1. Phase equilibrium data: for pure compounds such as vapour pressures, melting, and boiling points, enthalpies of phase transition. 2. Properties of single phase: density and volume, enthalpy and entropy, heat capacity. 3. Thermochemical properties: enthalpy and Gibbs energy of formation for various phases, heat of combustion (many of these properties could be given for mixtures). 4. Additional mixture properties: solubility and composition of phases at equilibrium, enthalpy and volume changes on mixing, Henry’s law constants, activity coefficients. 5. Transport properties: viscosity, thermal conductivity, diffusion coefficients. About 100 pure compound properties significant in practical applications can be listed; mixtures of non-electrolytes will add to the list by 30-40 and of electrolytes additional 10-20. A critical review of available property systems both stand alone and those embedded in simulation programs is given by Reid and Evans (1970). An exhaustive list has been compiled by Hughson and Steymann (1971; 1973). However, a complete survey of existing databanks was made by Rose (1979) and Edmonds (1983). According to Rose (1979), there are three categories of databanks available, which are as follows.

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Process Plant Simulation

Category I In this category are the databanks developed by industrial organizations for their own use. These organizations recover the development cost by selling these databanks. They generally have most of the compounds with an adequate range of properties and an accuracy guarantee. Important banks are those developed by ICI, Monsanto, Solvay, DSM, and Uhde. Category 2 In this category are those banks that are developed as a service for marketing. These can be divided into two subgroups. Banks in the first subgroup are operated as a service specifically for chemical engineers, such as Institution of Chemical Engineers (UK), DECHEMA, CHEMSHARE, Liege University, RRL Hyderabad, etc. The bank’s objective being to make well-known data conveniently available, they contain the range of properties usually required by design engineers, particularly VLE data. In the second subgroup are services not limited to chemical engineering. These have data in different available ranges; some of which may be useful to chemical engineers. They provide predominantly data from sources that otherwise would be difficult to locate and cover thousands of components for the limited set of properties. Some of the organizations and the databanks developed by them are NPL (VP & VLE), University of Sussex (Computer Analyzed Thermochemical Data), NPL/Association Thermodata (MTDATA-GEMT), NPL (ALLOYDATA), Dortmund University (DDB), TRC/API (TRC Databank), NEL (APPES), Tokyo University (EROTICA), Purdue University (CINDAS), etc. Category 3 In this category are small banks, mostly developed for teaching purpose with limited number of components. These include Edinburgh University (VLE), Manchester University (Physical Property Data System), Milan Polytechnic (PHYSCO), EURECHA-European Universities (CHEMCO), University of Houston (CHESS), CADC (CONCEPT), SSI (SSI/lOO), RRL Hyderabad (TPACK), etc.

13.2.3 Modularity and Routing As envisaged by Seider et al. (1979), modularity and routing are highly important and useful features in a thermodynamic properties package. These are concerned with the choice of an appropriate method among several alternatives available to calculate a property. A typical case is the prediction of vapour-liquid equilibrium (VLE) data so often required in the design of multicomponent distillation columns. Here, a variety of models is available for calculating fugacity and activity coefficients and other related parameters for both the phases. For the large number of mixtures handled in the industry under a wide range of conditions, a comprehensive VLE package is needed. Husain (1986) discussed such a package developed by him and his associates, which is totally modular in structure and offers the user to specify his own route of calculation. This option in a VLE package is particularly desired due to several reasons, namely, accuracy, range of applicability, availability of any

Convergence Promotion and Physical and Thermodynamic Properties

353

experimental data for the system under study, computational effort required, etc., for providing different modules to perform a single task. As shown by the tree structure in Fig. 13.3, the VLE package has two types of modules-task and method modules represented by circles and rectangles, respectively. Each task module has a choice among various method modules provided to accomplish it. A task module may also invoke another task module. The nomenclature used in Fig. 13.3. are as follows: CP is the fugacity, Y is the activity coefficient,p is the vapour pressure, v is the molar volume,fis the number of moles, I is the integral v/RT dp, and K is the equilibrium constant. Circles denote task modules and rectangles denote method modules.

I,

32values

YenWood

RedlichKwong

I

V

Curve fit

Laar

UNIFAC

NRTL

RedlichKwong

Curve fit

Seader

Antoine

Fig. 13.3 Tree structure of the VLE package

EXERCISES 13.1 Discuss modularity and routing for the prediction of vapour-liquid equilibrium data. 13.2 Discuss the following: (a) Newton’s method of convergence promotion. (b) Classification of data on properties to be retrieved. 13.3 Discuss the sources and databanks for physical and thermodynamicproperties.

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Process Plant Simulation

13.4 Discuss various convergence promotion methods. 13.5 Compare dominant eigenvalue method and general dominant eigenvalue method. 13.6 Discuss the quasi-Newton methods of convergence promotion. 13.7 Discuss the criterion for acceleration. 13.8 Discuss and compare the following convergence promotion methods: (a) Newton’s method, (b) direct substitution, (c) Wegstein’s method.

CHAPTER 14

SPECIFIC-PURPOSE SIMULATION AND DYNAMIC SIMULATION So far we discussed various aspects of simulation in different chapters. In this chapter, all the simulation aspects, starting from model formulation, optimization, and application of numerical methods for solution, are applied to some real case studies. Simulation is categorized into two groups: specific-purpose simulation and general-purpose simulation. General-purpose simulation leads to professional packages. These packages can be applied to all kinds of problems, as they are much generalized and incorporate huge databanks. The examples include ASPEN Plus, HYSIS, ASCEND, PROSIM, CONCEPT, FLUENT, etc. On the other hand, specific-purpose simulation is used to the specific cases/problems in order to be able to incorporate all the complexities of a chosen case/problem (FLOWTRAN, OPERA, PACER 245, HX, etc). Dynamic simulation is used for unsteady state problems. In this chapter, four case studies reported by the author and his associates, covering specific-purpose simulation and dynamic simulation, are discussed in detail starting from the model formulation, solution techniques used, and the methodology adopted for obtaining the solution, and optimization routines. The case studies include (1) the auto-thermal ammonia synthesis reactor, (2) thermal cracking operation, (3) design of a shell-and-tubeheat exchanger, and (4) pyrolysis of biomass.

14.1 Auto-thermal Ammonia Synthesis Reactor 14.1.1 Ammonia Synthesis Reactor A typical ammonia production process consists of production of the synthesis gas from the petroleum feed stock, compression of the gas to the required pressure, and the synthesis loop in which the conversion to ammonia takes place. The ammonia

356

Process Plant Simulation

converter is part of the synthesis loop, and its operation is quite crucial in the overall control strategy of the plant. In the converter, the following catalytic reaction takes place at elevated temperatures and pressures releasing a large amount of heat:

N, + 3H, e 2NH,, AH = -22.0 kcal (14.1) This heat has to be removed to obtain a reasonable conversion as well as to protect the catalyst life. At the same time, the released heat energy is utilized to heat the incoming feed gas to proper reaction temperature. A typical ammonia synthesis reactor is shown in Fig. 14.1,

Product gas (NH,)

a

'

Heat exchanger Bottom Feed gas (200-350 M)

Fig. 14.1 Ammonia synthesis reactor The reaction zone (shaded) contains the catalyst. A number of cooling tubes are inserted vertically through the reaction zone. The feed gas comes in from the lower part of the reactor and flows up through the top of the reactor. Then, changing direction, it flows down through the reaction zone and heat exchanger to the outlet. As with all reversible exothermic reactions, the temperature at which the reaction rate is maximum decreases as the conversion increases. Even though one would like to maximize the reaction rate at each instant, it is impossible to obtain the ideal temperature profile by control of available design variables. The countercurrent flow does, however, cause the temperature to decrease in the bottom part of the reactor because of heat transfer between the reacting and feed gas. Babu et al. (2002) carried out detailed simulationsincorporatingthe optimization for the above problem.

Specific-purpose Simulation and Dynamic Simulation

357

14.1.2 Problem Formulation The problem formulation by Babu et al. (2002) is similar to that by Murase et al. (1970) including the modifications mentioned by Upreti and Deb (1997). Feed gas contains 21 -75 mol % nitrogen, 65.25 mol% hydrogen, 5 mol % ammonia, 4 mol % methane, and 4 mol % argon. In a typical ammonia reactor, feed gas enters the bottom of the reactor. The production of ammonia depends on the temperature of feed gas at the top of the reactor (henceforth called top temperature), the partial pressures of the reactants (nitrogen and hydrogen), and the reactor length. 14.1.2.1 Assumptions The following assumptions are used in modelling the ammonia synthesis reactor (Murase et al. 1970):

1. The rate expression is valid. 2. The model is one-dimensional, i.e., the temperature and concentration gradients in the radial direction are neglected. 3. Heat and mass diffusion in the longitudinal direction are negligible. 4. The temperature of the gas flowing through the catalyst zone is equal to that of the reacting gas and the catalyst particles. 5. The heat capacities of the reacting gas and the feed gas are constant. 6. The catalyst activity is uniform along the reactor and equal to unity. 7. Pressure drop across the reactor is negligible compared to the total pressure of the system. 14.1.2.2 Objective function

The objective function is the economic return based on the difference between the value of the product gas (heating value and the ammonia value) and the value of feed gas (as a source of heat only) less the amortization of reactor capital costs. Other operating costs are omitted. As shown by Upreti and Deb (1997), the final consolidation of the objective function terms is f ( x , N N 2Tf, , T') = 1.33563 x lo7 - 1.70843 x 104NN2

+ 704.O9(Tg - 7'')

-699.27(T' - To)

(14.2) - [3.45663 x lo7 + 1.98365 x 109 x] 112 It is clear from the above expression that the objective function depends on four variables: the reactor length x, proportion of nitrogen N,, the top temperature Tg, the feed gas temperature Tf.

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Process Plant Simulation

14.1.2.3 Constraints There are mistakes in modelling equations reported in Murase et al. (1970), which were corrected and some of them were reported by Upreti and Deb (1997). Hence, for the sake of clarity, all the corrected modelling equations are presented below.

Energy balances Feed gas Refemng to Fig. 14.2, an energy balance on the feed stream yields the following equation:

w

Tube

Fig. 14.2 Energy and material balance on control volume

(14.3)

where Uis the overall heat-transfer coefficient in kcaUm2hK,S,is the surface area of cooling tubes per unit length of reactor in m,T, is the temperature of reacting gas in K, W is the total mass-flow rate in kgh, CPfisthe specific heat of feed gas in kcal/kgK, and x is the distance along the axis in m.

Reacting gas Similarly, for the reacting gas, (14.4)

where AH is the heat of reaction in kcallkg mole of N,, S, is the cross-sectional area of the catalyst zone in m2, - d N N 2 / d xis the reaction rate in kg moles of N,ih m3, and Cpgis the heat capacity of reacting gas in kcaVkg K. On the rhs of Eq. (14.4), the first term represents the heat transfer from reacting gas to feed gas and the second term represents the heat released due to reaction.

Material balance Considering the incremental distance in the catalyst zone (Fig. 14.2) and performing a N, material balance yields (14.5)

Specijkpurpose Simulation and Dynamic Simulation

359

wherefis the catalyst activity; p ~, PH,, , P N Hare ~ the partial pressures of N,, H,, and NH,; and K, and K2 are rate constants.

I?:-[ I?:-[

K, = 1.78954 x lo4 exp -

(14.6)

K, = 2.5714 x 1016exp -

(14.7)

In order to maintain the energy and material balance of the reaction, the above three coupled differential equations [Eqs (14.3)-( 14.5)] must be satisfied. It turns out that three of the above variables (Tr Tg,NN,) can be eliminated by satisfying these energy and material balance equations. Thus, practically, there is only one design variable for the given top temperature. The partial pressures appearing in the differential equations are computed as follows. Feed gas contains 21.75 mol% nitrogen, 65.25 mol% hydrogen, 5 mol% ammonia, 4 mol % methane, and 4 mol % argon. Moles of N, in feed (Ni2) = 701.2 kmoVhm2 and total pressure (P) = 286 atm. The partial pressures appearing in the differential equations are computed as follows:

N2 + 3H2 e 2NH, From the stoichiometry of the above reaction, it is clear that mole fractions of nitrogen, hydrogen, and ammonia are given by

PN -moles of 2

N, total moles

P

-pH, - - 3p,,

P

PNH, -P

-

moles of NH, total moles

where Total moles = moles of N, + moles of H, + moles of NH, + moles of CH, + moles of Ar Moles of nitrogen = NN, (1) Moles of hydrogen = 3NN, (2) Moles of argon + moles of methane = 257.92 kmoVhm2 (3) Moles of ammonia in feed = 161.195 kmoVh m2

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Process Plant Simulation

Moles of ammonia formed = 2(Ni2 - NN2) Moles of ammonia = moles of ammonia in feed + moles of ammonia formed

(4)

Total moles = (1) + (2) + (3) + (4) = 2.598Ni2 2", Therefore, the partial pressures of nitrogen, hydrogen, and ammonia are given by the following equations:

pN2

286NN2 - 2.598Ni2 4- 2",

(14.8) (14.9)

PH, = ~ P N ,

286(2.23Ni2- 2", PNH3 =

)

2.598Ni2 4-2",

(14.10)

The boundary conditions are Tf=Toatx=O

(14.11)

Tg = Tfat x = 0

(14.12)

= 701.2 k m o n m 2 at x = o

(14.13)

The three inequality constraints that limit the values of three of the design variables are as given below:

0.0 I NN, I 3220

(14.14)

400 I Tf I 800

(14.15)

(14.16) 0.0 I x I 10.0 Since the reaction gas temperature Tgdepends on the nitrogen mass-flow rate NN,, feed-gas temperature Tp and reactor length x, explicit bounds on Tgare not necessary. From the system model, we have three differential equations and four variables, making the degrees of freedom equal to 1. We specify the length of the reactor, calculate the remaining variables using the system model, and then pass these variables to the optimization algorithm. The computation procedure for the optimization camed out is shown in Fig. 14.3.

Specgc-purpose Simulation and Dynamic Simulation

361

Reactor length

equation solver

Optimization routine Fig. 14.3 Computation procedure

14.1.3 Simulated Results and Discussion 14.1.3.1 Temperature and flow rate profiles First, the system equations (14.3)-( 14.5) were solved using Runge-Kutta fixed step size method (RKFS). Figure 14.4(a) shows the profiles obtained. It is clear from the graph that the profiles of NN, and Tfintersect at a reactor length of 4.933 m, and the profiles of NN, and Tgintersect at a reactor length of 8.915 m. It is evident from the graph that the profiles are very smooth and there are no such spikes as those reported by Upreti and Deb (1997). Therefore, in order to check if it is the limitation of RKFS because of which spikes are not obtained, the other numerical methods, namely, Euler’s method (EULER), Runge-Kutta variable step size method (RKVS) and Gear’s method (GEAR) are also used for simulating the results. Apart from the above-mentioned numerical methods, the POLYMATH (a software package) is also used to simuIate the results for profiles. Figures 14.4 (b)-(e) show the profiles obtained using the above four methods. The profiles obtained are quite smooth without any spikes and similar to that of RKFS. The profiles are same qualitatively, but they differ slightly quantitatively. To illustrate the exact difference quantitatively, the above observations and comparison of various numerical methods used for simulating the results are presented in Table 14.1 for the reactor lengths of 10 m. Tables 14.2 and 14.3 show the reactor length for which the variables NN, and Tgand NN, and Tfintersect, respectively.

362

Process Plant Simulation

800

800

700

700

as 600

z*

600

2

2

500

500

hS 400

Lq 400

300

300

200

I

0

. . . . . . . . . 2

200

4 6 8 1 Reactor length (m)

0

0

2

(a) RKFS

.......

800 700

500

k

&$oo

(b) EULER

-

NN,

400

h i

300

300 \

0

2

0

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

600

2

4 6 8 1 Reactor length (m)

4 6 8 1 Reactor length (m)

200{ 0

0

,

,

2

(c) POLYMATH

,

,

,

,

\; ,

,

4 6 8 1 Reactor length (m)

(d) RKVS

.......

....

700

P 2

600 500

k

3001,, 200

0

2

\ ,

,

,

,

,

\

4 6 8 1 Reactor length (m)

0

(e) GEAR

Fig. 14.4 The profiles obtained using different numerical methods

,

0

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14.1.3.2 Comparison of results

From Tables 14.1-14.3, it is evident that all the numerical methods (stated above) are equally good, barring a few of the following differences. There is good agreement between RKFS and POLYMATH, which can be explained with the fact that POLYMATH software is based on the RKFS algorithm. Similarly, GEAR and EULER are giving same results though slightly different from RKFS and POLYMATH. The difference in the prediction is less than 2.3% between GEAR/ EULER and RKFSPOLYMATH. Similarly, the difference in the prediction of intersections is less than 5.0% between RKVS and GEAREULER methods. RKVS is the latest and has the ability to adopt its step size and varies as per requirement at every move. Based on the above observations, it can be substantiated that all the five methods are equally good both qualitatively, giving the same results, and quantitatively with slight difference. Hence, any of the above numerical methods can be used for the solution of three coupled differential equations. Table 14.1 Comparison of different numerical methods at a reactor length of 10 m

10.00 490.55 422.00 183.23

X "2

TB Tf

EULER

POLYMATH

RKVS

GEAR

10.00 499.48 432.89 204.23

10.00 490.46 422.25 183.38

10.00 499.68 462.86 235.71

10.00 499.46 432.69 204.73

Table 14.2 Reactor length at which variables N N , and Tgintersect RKFS

X 9

Tg

8.915 490.48

EULER

8.915 499.56

POLYMATH

8.916 490.56

RKVS

GEAR

9.355 499.68

8.895 499.57

Table 14.3 Reactor length at which variables N N , and T, intersect RKFS

X

N N , , Tf

4.933 501.525

EULER

4.900 511.50

POLYMATH

4.935 501.64

RKVS

GEAR

5.156 511.75

4.900 511.55

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Process Plant Simulation

Upreti and Deb (1997) reported that the reverse reaction predominates the forward reaction at the top temperature of 664 K. In this study, RKVS, which is well accepted for many engineering problems, is used to generate temperature and flow rate profiles at that temperature. Surprisingly,there is no such trend observed in the profiles obtained, as can be seen in Fig. 14.5(a).To see the presence of any reverse reaction effect, the program is executed for temperatures even below 664 K with intervals of 10 K up to 600 K. But no such trend is observed. Qpical results obtained at top temperatures of 640 K and 600 K are shown in Figs 14.5(b) and 14.5(c), respectively. As is evident from the figures, there is no reverse reaction effect even at a top temperature as low as 600 K. Again, to check if it is the limitation of RKVS, the other methods are applied but none of them could show this effect. Also, Upreti and Deb (1997) reported that the three differential equations (14.3)(14.5) become unstable at the top temperature of 706 K. However, using the abovestated numerical methods, it is observed that the equations are not unstable even at a top temperature as high as 800 K. It may be noted that Upreti and Deb (1997) used NAG library's subroutine D02EBF [which is now replaced by D02EJF (NAG 2002)]. .....,.

'O0

.............,_

650 r" 600 550

?

h". 450 400 350 300

0

2

\

4 6 8 1 Reactor length (m)

450 500

1

\

I , , , , , , , , , ;

0

0

2

4 6 8 1 Reactor length (m)

(b) 640 K

(a) 664 K

Reactor length (m)

(c)600 K Fig. 14.5 The profiles obtained using RKVS for different top temperatures

0

Specijic-purpose Simulation and Dynamic Simulation

365

14.1.4 Optimization 14.1.4.1 'Nested' DE-a new concept The key parameters of differential evolution (DE), which was discussed in Sec. 10.3, are the population size (NP), the scaling factor (F),and the cross over constant (CR). Choosing the values for NP, F, and CR depends on the specific problem to which is DE is applied, and is often difficult. But some general guidelines are available. Normally, NP should be about 5 to 10 times the number of parameters in a vector. As for F , it lies in the range 0.4-1.0. Initially F = 0.5 can be tried, then F and/or NP is increased if the population converges prematurely. A good first choice for CR is 0.1, but in general CR should be as large as possible (Price & Storn 1997). The best combination of these key parameters of DE for each of the strategies mentioned earlier is again different. Price and Storn (2003) have mentioned some simple rules for choosing the best strategy as well as the ranges of corresponding key parameters. 14.1-4.2 Effect of DE key parameters The key parameters are generally found by trial-and-error method. In this method one parameter is kept constant while varying the others in steps. In this process, we may m i s s out the optimum combination that gives global optima, as it is very difficult to cover the entire range of all the key parameters, however small the incremental step may be. The comparison of results obtained using different combination of DE key parameters (CR and F) are shown in Figs 14.6(a) and 14.6(b). These graphs show us how difficult it is to find out which combination can be the most suitable one for our problem as there is no specific trend observed. 50 45 40 -

,,___ --,

- CR = 0.75

, ..\ , .,'

-F=0.8

,", ,

,----,,' 35: 30 ,,' 25 20 -' ,I'

15 10 5 '

0"

'

1

'

'

'

I

'

1

'

I

'

I

Fig. 14.6 The effect of Fand CR on the number of generations

I

366

Process Plant Simulation

By looking at the problems mentioned above and the results obtained in this study in choosing the right combination of the DE key parameters, it was felt that why not use DE itself for finding the optimum key parameters along with the optimum variables of the actual problem formulation [i.e., using DE within DE wherein the outer loop takes care of optimization of key parameters (NP, CR, F) with the objective of minimizing the number of generations required, while the inner loop takes care of optimizing the problem variables]. A yet complex objective can be one that takes care of minimizing the number of generationdfunction evaluations and the standard deviation (SD) in the set of solutions at the last generatiodfunction evaluation, and to try to maximize the robustness of the algorithm. Initially, the tuning parameters for outer DE (NP-OUTER, CR-OUTER, and F-OUTER) were chosen on the basis of general heuristics. The pseudocode for ‘nested DE’ used in the present study is given below: 1. Choose the DE strategy to be used for both the inner and the outer DE. 2. Initialize the variable D for both the inner [D-INNER = 1, the reactor length isx] and outer (D_OUTER=2, CR-INNER, and F-INNER for inner DE) DEs 3. Initialize the tuning parameters NP-OUTER, CR-OUTER, and F-OUTER for the outer DE. 4. Initialize the initial population of NP-OUTER (DE) points within the defined limits for the variables CR-INNER and F-INNER as follows: for(iouter = 1; iouter <= NP-OUTER; iouter++) {

innerCR[iouter] = CR-INNER-LOWER-LIMIT + Rando m-N u m ber * (C R-I N N ER-LOW ER-L I M IT C R-I N N ER-LOW ER-L I M IT); inner F[ i outer] = F-I N N ER-LO W ER-L I M IT + Random-Number * (F-INNER-LOWER-LIMIT F-I N N ER-LOWER-LI M IT);

1 5. Calculate the number of generations required for the inner DE to converge to the global optima using the recent population of innerCR[i] and innerF[i] tuning parameters for all NP-OUTER points of the recent generation. for(iouter = 1; iouter <= NP-OUTER; iouter++) {

Generations-Required[iouter] i n ner F[ i o ute r]);

1

= innerDE(innerCR[iouter],

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367

5a. Initialize the initial population for the inner DE: for(iinner = 1; iinner c= NP-INNER; iinner++) { ~ [ i i n n e r ]= X-LOWER-LIMIT + Random-N u m ber*(X-LOW ER-LI M IT - X-LOW ER-LI M IT);

5b. Calculate the value of the profit function for all x's by performing simulation using the chosen differential equation solver (e.g., RKFS, EULER, POLYMATH, RKVS, D02EJF, GEAR, etc.). for(iinner = 1; iinner c= NP-INNER; iinner++) {

[T,[iinner], T,[iinner], NN2[iinner]] = simulate-profiles(x[iinner]); Profit[ i inner] = prof i t-f unct ion(x[ i i nner], T,[ i inner], Tg[ i i nner], NN2[iinner]);

1 5c. Form the next generation for inner DE by performing mutation, recombination, and selection (specific to the chosen DE strategy) with the tuning parameters innerCR[iinner] and innerF[iinner]. For the present study, the seventh DE strategy (DE/rand/l/bin) was used. The steps involved are randomly choosing three vectors (a, b, c; distinct from the target vector[i] against which the trial vector will be checked) from the N P-I N NER vectors of the present generation and then continuing as follows (let i be the index of the target vector): x-trial = x[a] + innerF[iouter]*(x[b].x[c]); [T,-trial, T,trial, N,,_trial] = simulate-profiles(x-trial); , ria I,t ,T ria I, N,,-tria I); Prof it-t ria I = prof i t-f u nct ion(x-t rial,t T if( Profit[i] c Prof it-trial) pass the trial vector i n t i the next generation to replace the target vector else retain the target vector for the next generation

5d. Check if the standard deviation [using Eqs (14.17) and (14.18)] for the present generation has gone below the maximum allowable limit (SD). If this condition is met, go to the next step, else go to step 5b. 5e. Report the number of generations required for the inner DE to reach the SD to the outer DE. 6. Form the next generation for outer DE by performing mutation, recombination, and selection (specific to the chosen DE strategy) with the tuning parameters CR-OUTER and F-OUTER. As explained in step 5c,

360

Process Plant Simulation

this was done using the seventh DE strategy. Similar calculations will follow (let i be the index of the target vector). innerCR-trial = (Random-Number c CR-OUTER ? (i nnerCR[ a] + F-0 UTE R*( inne rCR[ b]-in nerC R[ c])): in ne rCR[ i]); innerF-trial = (Random-Number c CR-OUTER ? (i nner F[a] + F-OU TE R*( i nne r F[ b]-inne r F[ c])): inner F[ i]); Generat ions-Req u i red-t rial = innerDE( innerCR-t ria I, inner F-t ria I); if(Generations-Required[i] c Generations-Required-trial) pass the trial vector into the next generation to replace the target vector else retain the target vector for the next generation

7. Check if the termination criteria (either reaching the maximum number of generations allowed or reaching the standard deviation below SD) for the outer DE is satisfied. If yes, go to the next step, else go to step 5. 8. Print the results.

The optimum reactor length obtained using GEAR with nested DE is 6.79 m (at top temperature of 694) with the objective function value of US $4,848,383.0 per year. Using RKFS with DE the values are 6.59 m (at top temperature of 694 K) and US $5,006,814.58 per year. To find the best combination of these parameters for the present problem, we vary CR and F, but keep NP to the higher side (NP = lOD, to be on the safer side). The algorithm followed for the nested DE operation in combination with GEAR is shown in Fig. 14.7. The optimum DE parameters using nested DE (with the strategy DE/rand/l/ bin) with the GEAR method for various standard deviations are shown in Table 14.4. For population of xi’s (i = 1, 2, 3, ..., NP) we define NP

(X - X i ) 2

j=l

(Np- 1)

Standard deviation (SD) =

(14.17)

where NP

( 14.18) i=l

From Table 14.4, it is clear that irrespective of the values of SD (1.0,0.5,0.25,0.1 0.01, and 0.01) and the corresponding CR and F values, we obtained almost same values of the objective function (US $4,848,383.0/year) and reactor length (6.79 m). It consolidates the robustness of the DE algorithm. The more wider the range of values of CR, F, and SD for which same values of the objective function and reactor length are obtained, the more robust is the algorithm.

Specific-purpose Simulation and Dynamic Simulation

369

-IOptimize reactor length

t Solve the system model equation using GEAR

1 Perform mutation, crossover, and recombination on the reactor length till convergence

Print the optimum values of CR, F, and reactor length

Fig. 14.7 The solution strategy for nested DE Table 14.4 Optimum DE parameters using nested DE with GEAR for various SD values Standard Reactor Objective function deviation length (m) (US $/year) 1.o 0.5 0.25 0.1 0.01 0.001

6.79 6.79 6.79 6.79 6.79 6.79

4,848,382.5 4,848,382.5 4,848,382.5 4,848,383.0 4,848,383.0 4,848,383.0

CR 0.745547 0.939156 0.995696 0.8686 0.902649 0.793 195

F 0.520328 0.485891 0.7012 0.553093 0.434454 0.43707

Gmin

1 2 4 6 10 18

Also, the code of nested DE was run with another strategy DE/rand/l/exp, and we found that the results are exactly the same. This also proves DE’s power and robustness. Now that we have already tried the effect of DE parameters and strategy on speed and accuracy and found that it has little or no effect on accuracy, we can say that the NP in a generation, if taken within the limits, would not have any effect on the result.

370

Process Plant Simulation

Table 14.5 shows the results obtained from different methods and their comparison with those obtained by Murase et al. (1970), Edgar and Himmelblau (1989), and Upreti and Deb (1997) using Pontryagin’s maximum principle (PMP), Lasdon’s generalized reduced-gradient method (LGRG), and genetic algorithm (GA), respectively. Table 14.5 Results of various numerical methods and their comparison Ph4P LGRG D02EBF EULER (Murase (Edgar& withGA withDE Himme(Upreti & et al. 1970) lblau 1989) Deb 1997) Optimum

5.18

2.58

5.33

6.8

RKFS withDE

GEAR withDE

RKVS withDE

6.59

6.79

7.16

x (m)

Not 1.29 x lo6 4.23x lo6 4.84x lo6 Objective function reported (US $/year)

5.00 x lo6 4.84x lo6 4.84x lo6

FromTable 14.5,we observe that an optimum reactor length of 2.58 m is reported by Edgar and Himmelblau (1989) and 5.18 m by Murase et al. (1970), both of which are wrong due to the errors in their problem formulations as pointed out by Upreti and Deb (1997).An optimum reactor length of 5.33 m and the corresponding objective function value is 4.23 x lo6US $/year, reported by Upreti and Deb (1997), are also not correct as found in this study due to a possible error in the software they used (D02EBF-NAG Library subroutine).This may also be wrong, because simple GA does not ensure a global optimum. Among other methods, GEAR, EULER and RKVS have the same objective function value though the optimum reactor length is slightly different in each case. RKVS is an improved method over the other methods and is the well-accepted one. Also, this study establishes the accuracy and robustness of DE. Hence, the correct optimum reactor length can be considered 7.16 m with an objective function value of 4.841 x lo6 US $/year.

14.1.5 Conclusions In this problem, DE is used for the optimal design of an auto-thermal ammonia synthesis reactor. The new concept of ‘nested’DE is introduced. Results indicate that the profiles of temperature and flow rate are smooth and there is no reverse reaction effect irrespective of the numerical method used for the solution of differentialequations.The optimum reactor length depends upon the top temperature. Also, the power and robustness of DE is brought out using the new concept of nested DE. This successfulapplication of DE for the optimal design of the ammonia

Specijk-purpose Simulation and Dynamic Simulation

371

synthesis reactor indicates that DE has great potential and can be applied to advantage in all the highly non-linear and complex engineering problems.

14.2 Thermal Cracking Operation This problem (Babu & Angira 2001) presents the application of DE for the optimization of thermal cracking operation. The objective in this problem is the estimation of optimal flow rates of different feeds to the cracking furnace under the restriction on ethylene and propylene production. Thousands of combinations of feeds are possible. Hence an efficient optimization strategy is essential in searching for the global optimum. In this problem LP simplex method and DE have been successfully applied with different strategies to find the optimum flow rates of different feeds. In the application of DE, various combinations of the key parameters are considered.

14.2.1 Thermal Cracking It is defined as the thermal decomposition, under pressure, of large hydrocarbon molecules to form smaller molecules. Lighter, more valuable hydrocarbons may thus be obtained from such relatively low-value stocks as heavy fueVgas oils (boiling up to 540 "C) and residues. This is conducted without any catalyst. Thermal cracking is normally carried out at temperatures varying from 450 "C to 750 "C and pressures ranging from atmospheric to 1000 psig (Hobson 1975). The important reactions occurring are the following 1. Decomposition and destructive condensation 2. Hydrogenation and dehydrogenation 3. Polymerization 4. Cyclization The first two reactions are endothermic, whereas polymerization is exothermic. Coke formation is an additional reaction, which plays an important role in thermal cracking. Although the mechanism by which coke is formed is not entirely understood. It is thought, however, that coke results from extensive degradation of relatively heavy molecules to form increasing quantities of light hydrocarbon gases (dry gas) and polycyclic compounds having low hydrogen-to-carbon ratios. The rate at which hydrocarbons crack is strongly dependent on temperature. Cracking reactions begin about 3 15-370 "C, depending on the type of material being cracked (Hobson 1975). Depending upon the pressure and temperature employed for the cracking and the characteristics of feed, there are various thermal cracking processes in which the product yields and characteristics are different (Gupta 1994). Mainly, there are

372

Process Plant Simulation

four commercial processes employed for thermal cracking in oil refineries. They are as follows: m Dubbs thermal cracking process m Pyrolysis Visbreaking Coking Pyrolysis or mild thermal cracking is done mainly for the production of lighter products, mainly unsaturated such as olefins (ethylene, propylene) and naphthene polymers, diolefins, benzene and toluene, etc. (Sourander et al. 1984). It is carried out at high temperature (650-700 "C) and low pressure.

14.2.2 Problem Description This problem (Edgar & Himmelblau 1989)deals with maximization of profit while operating within furnace and downstream process equipment constraints. Figure 14.8 lists various feeds and corresponding product distribution for a thermal cracker that produces olefins.

Fuel

,,

-

-Thermal cracker

(Ethane) -

(Propane) Methanel I

The capacity to run gas feeds through the cracker is 200,000 lbhtream h (total flow based on an average mixture). Ethane uses the equivalent of 1.1 lb of capacity per pound of ethane, propane uses 0.9 lb of capacity per pound of propane, gas oil uses 0.9 lb/lb, and DNG has a ratio of 1.0. Based on the plant data, eight products are produced in varying proportions according to the following matrix (Table 14.6).

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Table 14.6 Yield structure (weight fraction) Feed Product Methane Ethane Ethylene Propane Propylene Butadiene Gasoline Fuel oil

Ethane

Propane

Gas oil

DNG

0.07 0.40 0.50

-

-

0.10 0.04 0.20 0.01 0.15 0.04 0.25 0.21

0.15

0.01 0.01 0.01

0.25 0.06 0.35 0.10 0.15 0.02 0.07

-

0.05

0.25 0.01 0.18 0.05 0.30 0.01

Downstreamprocessing limits exist of 50,000 lb/streamh on ethylene and 20,000 lbhtream h on propylene. The fuel requirements to run the cracking system for each feed-stock type are as follows: Ethane Propane Gas oil DNG

8364 Btu/lb 5016 Btuflb 3900 Btu/lb 4553 Btuflb

Methane and fuel oil produced by the cracker are recycled as fuel. All the ethane and propane produced is recycled as feed. The heating values are as follows: Natural gas 21,520 Btuflb Methane 21,520 Btuflb Fuel oil 18,000 Btuflb Because of heat losses and the energy requirements for pyrolysis, a fixed fuel requirement of 20.0 x lo6 Btuhtreamh occurs. The price structure on the feeds and products and fuel costs is (all values are in cents per pound) as follows: Feeds

Ethane Propane Gas oil DNG Products Methane Ethylene Propylene Butadiene Gasoline Fuel-oil

6.55 9.73 12.50 10.14 5.38 (fu 1 ralue) 17.75 13.79 26.64 9.93 4.50 (fuel value)

374

Process Plant Simulation

Assume an energy (fuel) cost of US $2.5/106 Btu. The variables to be optimized are the amounts of the four feeds (namely, gas oil, propane, ethane, and DNG). This problem was solved using both the linear programming simplex method and DE. The assumption used in formulating the objective function and constraints are as follows: 1. 20.0 x lo6 Btuh fixed fuel requirement (methane) to compensate for the heat loss. 2. Propane and ethane are recycled with the feed, and methane and fuel oil is recycled as fuel.

14.2.2.1 Objective function The objective function for the profit is given by (14.19) f = 234x1 - 0.22~2- 3.33~3+ 1.09X4 + 939x5 + 9.51~6 wherefis the profit function (centsh), x1 is the fresh ethane feed (lbh), x2 is the fresh propane feed (lbh), xj is the gas-oil feed (lbh), x4 is the DNG feed (lbh), x5 is ethane recycle (lbh), x6 is propane recycle (lbh), and x7 is the fuel added (lbh).

14.2.2.2 Constraints

(a) Cracker capacity of 200,000 lbh: 1.1x1 + O.9X2 + O.9X3 + 1.0x4 + 1h5 + 0.9x6 (b) Ethylene processing limitation of 50,000 lbh:

< 200,000

(14.20)

OSX1 f O.35X2 + 0.2X3 + O.25X4 + O S X , + 0.35x6 5 50,000 (c) Propylene processing limitation of 20,000 l b h :

(14.21)

0.01x1 + 0.15~2+ O.I5X3 0 . 1 8 ~+~0.01x5 (d) Ethane recycle:

(14.22)

0.4X1 + 0.06x2 + 0.04x3 (e) Propane recycle:

+ 0.15x6 < 20,000

+ 0.05X4 - 0 . 6 ~+ ~0.06x6 = 0

0.1+ ~ 0~ . 0 1 +O.Olx, ~~ -0 . 9 ~ =~ 0 (f) Heat constraints:

(14.23) (14.24)

-6857.6~~ + 3 6 4 +~ 2032x3 ~ - 1145x4 - 6 8 5 7 . 6 ~ ~

+ 364x6 + 21,520~7= 20,000,000

(14.25)

14.2.3 Problem Reformulation When DE was applied to the above problem, it was found that the equality constraints were difficult to deal with. Hence, the problem was reformulated by eliminating the equality constraints and incorporating them in inequality constraints, thereby

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reducing the number of constraints and parameters. The reformulated problem is as follows. 14.2.3.1 Objective function

Maximizef= 9.1X,+ 1 . 8 8 ~ ~ 2 .-5 8 7 9 ~+~1 . 9 8 8 6 ~ ~

(14.26)

14.2.3.2 Constraints

(a) Cracker capacity of 200,000 lbh: 1 6 . 5 ~+~1 0 . 1 + ~ 8~ . 8 6 1 ~ + ~9 . 9 2 6 ~I~ 18,00,000

(14.27)

(b) Ethylene processing limitation of 50,000 l b h : 7.5X1 + 4.0X2 + 2 . 1 4 ~+~2 . 6 6 5 ~I~ 450,000

(14.28)

(c) Propylene processing limitation of 20,000 l b h : 0.15x1+ 1 . 5 1 +~ 1 ~ . 3 7 1 1+ ~ ~1 . 6 4 2 6 ~I~ 180,000

(14.29)

14.2.4 Simulated Results and Discussion The reformulated problem was solved using both the DE and the LP simplex method and the following results were obtained. Table 14.7 shows the results obtained by both DE and LP Simplex methods. It may be noted that the maximum possible amount of ethylene is produced. As the ethylene production constraint is relaxed, the objective function value increases. Once the constraint is raised above 90,909.0909 lbh, the objective function remains constant at 67,6018.1 875 centsh. The LP simplex solution was cross-checked using a software package named TORA (Taha 1997), and the same results were obtained as shown in Table 14.7. Table 14.8 presents the comparison in terms of the number of objective function evaluations, CPU time, and proportion of convergences to the optimum, between the different DE strategies. In this table, FA, NRC, and CPU time represent, respectively, the mean number of objective function evaluations over all the 10 executions, the percentage of convergences to the global optimum, and the average CPU time per execution (key parameters used are NP = 40, CR = 0.9, F = 0.6, accuracy = 0.0001%). From Table 14.8 it is evident that strategy 7 is the best strategy. It takes least average CPU time, maximum NRC, and minimum FA. However, the best key parameters for strategy 7 are NP = 40, CR = 0.8, F = 0.5, giving CPU time of 0.113 s, NRC = 95, and FA = 2656.

376

Process Plant Simulation

Table 14.7 Results of LP simplex and DE methods Stream Fresh ethane feed ( x , ) Fresh propane feed (x,) Gas-oil feed (x,) DNG feed (x4) Ethane recycle (x,) Propane recycle (x,) Fuel added (x,) Ethylene Propylene Butadiene Gasoline Methane Fuel oil Objective function (centsk)

Flow rate (lbih) using DE

Flow rate (lbih) using LP simplex

60,000 0 0 0 40,000 0 32,795.539 50,000 1,000 1,000 1,000 7,000 0 369,560.00

60,000 0 0 0 40,000

0 32795.539 50,000 1,000 1,000 1,000 7,000 0 369,560.00

Table 14.8 Results of DE with all ten strategies Strategy 1 2 3 4 5

6 7 8 9 10

DE/rand/ l/bin DE/best/l/bin DE/best/2/bin DE/rand/2/bin DE/rand-to-best/ l/bin DE/rand/l/exp DE/best/ l/exp DE/best/2/exp DE/rand/2/exp DE/rand-to-best/l/exp

FA

6,268 3,168 9,076 11,696 6,052 5,252 2,796 10,132 12,600 6,536

CPU time

NRC

0.28 0.145 0.418 0.539 0.28 0.22 0.126 0.44 0.55 0.275

100 100 100 100 100 100 100 100 100 100

14.2.5 Conclusions The best key parameters for the present problem are NP = 40, CR = 0.8, and F = 0.5. The strategy that took minimum CPU time with highest NRC is strategy 7. The results obtained by the two methods (namely, DE and LP simplex) are same and match with that reported in literature. Differential evolution exhibits difficulties in dealing with equality constraint problems; but, in general, they are the most efficient in terms of function evaluations.

Specijic-purpose Simulation and Dynamic Simulation

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14.3 Design of Shell-and-Tube Heat Exchanger This problem (Babu & Munawar 2001) presents the application of DE for the optimal design of shell-and-tubeheat exchangers. The main objective in any heatexchangerdesign (HED) is the estimation of the minimum heat-transferarea required for a given heat duty, as it governs the overall cost of the heat exchanger. Millions of configurations are possible with various design variables such as the outer diameter, pitch, length of the tubes, tube passes, baffle spacing, baffle cut, etc. Hence the design engineer needs an efficient strategy in searching for the global minimum. In this study, for the first time DE has been successfully applied with different strategies for 161,280 design configurations using Bell’s method to find the heat-transfer area. In the application of DE, 9680 combinations of the key parameters are considered. For comparison, GA is also applied to the same case study with 1080 combinations of its parameters. For this optimal design problem, it is found that DE is significantly faster compared to GA and yields the global optimum for a wide range of the key parameters.

14.3.1 The Optimal HED Problem The proper use of basic heat-transfer knowledge in the design of practical heattransfer equipment is an art.The designer must be constantly aware of the differences between the idealized conditions for which the basic knowledge was obtained versus the real conditions of the mechanical expression of his design and its environment. The result must satisfy process and operational requirements (such as availability, flexibility, and maintainability) and do so economically. Heat-exchanger design is not a highly accurate art under the best of conditions (Perry & Green 1993). 14.3.1.1 Generalized design procedure

The design of a process heat exchanger usually proceeds through the following steps (Perry & Green 1993). Process conditions (stream compositions, flow rates, temperatures, pressures) must be specified. Required physical properties over the temperature and pressure ranges of interest must be obtained. The type of heat exchanger to be employed is chosen. A preliminary estimate of the size of the exchanger is made, using a heattransfer coefficientappropriateto the fluids, the process, and the equipment. w A first design is chosen, complete in all details necessary to carry out the design calculations. The design chosen is now evaluated or rated as to its ability to meet the process specifications with respect to both heat duty and pressure drop.

378

Process Plant Simulation

Based on this result, a new configuration is chosen if necessary and the above step is repeated. If the first design was inadequate to meet the required heat load, it is usually necessary to increase the size of the exchanger, while still remaining within the specified or feasible limits of pressure drop, tube length, shell diameter, etc. This will sometimes mean going to multiple exchanger configurations. If the first design more than meets heat load requirements or does not use the entire allowable pressure drop, a less expensive exchanger can usually be designed to fulfil process requirements. rn The final design should meet process requirements (within the allowable error limits) at the lowest cost. The lowest cost should include operation and maintenance costs and credit for ability to meet long-term process changes as well as installed (capital) cost. Exchangers should not be selected entirely on a lowest first cost basis, which frequently results in future penalties. The flow chart given in Fig. 14.9 (Sinnott 1993) gives the sequence of steps and the loops involved in the optimal design of a shell-and-tube heat exchanger.

14.3.2 Problem Formulation The objective function and the optimal problem of shell-and-tube HED of this study are represented, similar to the problem formulation of Manish et al. (1999), as Minimize C(X) or A(X) E

x2, x3, x4,x5,x,5,x7}

where

( 1 , 2, ..., 12) x2 = 11,2} x3 = 1 1 , 2, 3941 x4 = { 1, 2, ...)5 } x5 = { 1 , 2, ..., 8 } x6 = { 1, 2, ..., 6 } x7 = { 1, 2, ..., 7 ) XI =

subject to feasibility constraints (pressure drop). The objective function can be minimization of HE cost C(X) or heat transfer area A(X) and X is a solution string representing a design configuration. The design variable x1 takes 12 values for the outer tube diameter in the range of 0.25”-2.5” (0.25”, 0.375“, OS”, 0.625”, 0.75”, 0.875”, LO”, 1.25”, 1.5”, 1.75”, 2”, 2.5”). x2 represents the tube pitch-either square or triangular-taking two values represented by 1 and 2. x3 takes the shell head types: floating head, fixed tube

Specipc-purpose Simulation and Dynamic Simulation Specification: Define duty, make energy balance if needed to calculate unspecified flow rates or temperature Collect physical properties:p, p , C,, k

c

Assume overall coefficient, U,,,,,,

tube passes. Calculate At,,, correction factor F,,and henceAT,

Calculate the overall heat-transfer

4

1

Determine heat-transfer area required: A, = Uo, Auu,)]

[a(

~6

-

I

1 Yes Estimate tube-side and

Decide type, tube size, material layout. Assign fluids to shell or tube side

1

Pressure drops within the

Calculate number of tubes

L Calculate the shell diameter

Estimate the tube-side heat-transfer coefficient

i Decide baffle spacing and estimate the shell-side heat-transfer coefficient

1

I Estimate the cost of the exchanger I

Yes I

to reduce cost?

Accept the design

Fig. 14.9 Flow chart for optimal design of shell-and-tube heat exchanger

379

380

Process Plant Simulation

sheet, U tube, and pull through floating head represented by the numbers 1, 2, 3, and 4, respectively. x4 takes the number of tube passes 1-1, 1-2, 1-4, 1-6, 1-8 represented by numbers from 1 to 5 . The variable xs takes eight values of the various tube lengths in the range 6’-24’ (6’, 8’, lo’, 12’, 16’, 20’, 22‘, 24’) represented by numbers 1 to 8. x6 takes six values for the variable baffle spacing, in the range 0.2-0.45 times the shell diameter (0.2, 0.25, 0.3, 0.35, 0.4, 0.45). x, takes seven values for the baffle cut in the range 15%-45% (0.15, 0.2,0.25, 0.3, 0.35, 0.4, 0.45). In the present problem, the pressure drop on the fluids exchanging heat is considered to be the feasibility constraint. Generally a pressure drop of more than 1 bar is not desirable for the flow of fluid through a HE. For a given design configuration,whenever the pressure drop exceeds the specified limit, a high value for the heat-transfer area is returned so that, as an infeasible configuration, it will be eliminated in the next iteration of the optimization routine. The total number of design combinations with these variables is 12 x 2 x 4 x 5 x 8 x 6 x 7 = 161,280. This means that if an exhaustive search is to be performed, it will take at the maximum 161,280 function evaluations before arriving at the global minimum heat-exchanger cost. So the strategy that takes few function evaluations is the best one. Considering minimization of heat-transfer area as the objective function, the differential evolution technique is applied to find the optimum design configuration with pressure drop as the constraint.For the case study considered, the performance of DE is compared with GA and the results are discussed in Sec. 14.3.3.

14.3.3 Results and Discussion The algorithm of differential evolution as given by Price and Storn (2003) is in general applicable to continuous function optimization. The upper and lower bounds of the design variables are initially specified. Then, after mutation, because of the addition of the weighted random differential, the parameter values may even go beyond the specifiedboundary limits. So, irrespective of the boundary limits initially specified, DE finds the global optimum by exploring beyond the limits. Hence, when applied to discrete function optimization, the parameter values have to be limited to the specified bounds. In the present problem, since each design variable has a different upper bound when represented by means of integers, the same DE code given by Price and Stom (2003) cannot be used. We have used the normalized values for all the design variables and randomly initialized all the design variables between 0 and 1. Whenever it is required to find the heat-transfer area using Bell’s method for a given design configuration, these normalized values are converted back to their corresponding boundary limits. The pseudocode of the DE algorithm used in this study for the optimal heat-exchanger design problem is given below. Choose a strategy and a seed for the random number generator.

Spec@-purpose Simulation and Dynamic Simulation

381

w Initialize the values of D, NP, CR, F, and MAXGEN. w Initialize all the vectors of the population randomly. Since the upper bounds

are all different for each variable in this problem, the variables are all normalized. Hence generate a random number between 0 and 1 for all the design variables for initialization. for i = 1 to N P { f o r j = 1 to D xi,j = random number } w Evaluate the cost of each vector. The cost here is the area of the shell-and-

tube heat exchanger for the given design configuration, calculated by a separate function cal-a rea() using Bell’s method. for i = 1 to N P Ci = cal-area() w Find the vector with the lowest cost, i.e., the best vector so far.

Cmin= C, and best = 1 for i = 2 to N P { if (Ci < Cmin) then Cmi, = C, and best = i } w Perform mutation, crossover, selection, and evaluation of the objective

function for a specified number of generations. While (gen < MAXGEN) { f o r i = 1 to N P w For each vector Xi (target vector), select three distinct vectors X, X, and X,

(select five, if two vector differences are to be used) randomly from the current population (primary array) other than the vector Xi Do { r, = random number * N P r2 = random number * N P r3 = random number * N P } while (r, = i) OR (r2 = i) OR (r3 = i) OR (r, = r2) OR (r2 = r,) OR 0-1 = r3) w Perform crossover for each target vector Xi with its noisy vector X,,, and create a trial vector, Xt, i. The noisy vector is created by performing mutation. If CR = C, inherit all the parameters from the target vector Xi, except one which should be from X, i. For exponential crossover,

382

Process Plant Simulation

{ p = random number * 1 r = random number * D n=O do { Xn, i = Xa, i + F (Xb, i - Xc, i) r=(r+l)%D increment r by 1 } while ((pcCR) and (rcD))

1

/* add two weighted vector differences for two vector perturbation. For best/random vector perturbation the weighted vector difference is added t o the best/random vector of the c u r rent popu I at io n . */

For binomial crossover,

*1 r = random number * D for n = 1 t o D { if ((pcCR) or (p = D-1)) xn, i = xa, i + (x b, i - c, i) r = (r + 1)%D} { p = random number

1 if (Xn, > 1) Xn, = 1 if (Xn, c 0) Xn, = 0

/* change at least one parameter if CR=O */

/* for discrete function optimization check the values t o restrict t o the limits */ /* 1 - normalized upper bound; 0 - normalized lower bound */

Perform selection for each target vector Xi by comparing its cost with that of the trial vector X, i; whichever has the lowest cost will survive for the next generation. Ct, = cal-area() if (Ct, c Ci) new Xi = Xt, new Xi = X, } else

/ * f o r i = l t o NP */

1 Print the results. The entire scheme of optimization of shell-and-tube heat-exchanger design is performed by the DE algorithm, while intermittently it is required to evaluate the heat-transferarea for a given design configuration.This task is accomplishedthrough the separate function cal-area(), which employs Bell’s method of heat-exchanger design. Bell’s method gives accurate estimates of the shell-side heat-transfer coefficient and pressure drop compared to Kern’s method, as it takes into account the factors for leakage, bypassing, flow in the window zone, etc. The various

Spec@-purpose Simulation and Dynamic Simulation

383

correction factors in Bell's method include the temperature correction factor, tubeside heat transfer and friction factor, shell-side heat transfer and friction factor, tube row correction factor, window correction factor for heat transfer and pressure drop, bypass correction factor for heat transfer and pressure drop, friction factor for cross-flow tube banks, baffle geometrical factors, etc. These correction factors are reported in the literature in the form of monographs (Sinnott 1993; Perry & Green 1993).In this study, the data on these correction factors from the monographs are fitted into polynomial equations and incorporated in the computer program. As a case study the following problem for the design of a shell-and-tube heat exchanger (Sinnott 1993) is considered. 20,000 kgih of kerosene leaves the base of a side-stripping column at 200°C and is to be cooled to 90°C with 70,000 kgih light crude oil coming from storage at 40°C. The kerosene enters the exchanger at a pressure of 5 bar and the crude oil at 6.5 bar. A pressure drop of 0.8 bar is permissible on both the streams. Allowance should be made for fouling by including fouling factor of 0.00035 (W/m2"C)-' on the crude stream and 0.0002 (W/m2"C)-' on the kerosene side. By performing the enthalpy balance, the heat duty for this case study is found to be 1509.4kW and the outlet temperature of crude oil is found to be 78.6"C. The crude is dirtier than the kerosene and so is assigned through the tube-side and kerosene to the shell-side. Using a proprietary program (HTFS, STEPS) the lowest cost design meeting the above specifications is reported to be a heat-transfer area of 55 m2 based on the outside diameter (Sinnott 1993). Considering the result of the above program as the base case design, in this study DE is applied to the same problem with all the ten different strategies. As a heuristic, the pressure drop in a HE normally should not exceed 1 bar. Hence, the DE strategies are applied to this case study separately with both 0.8 bar and 1 bar as the constraints. In both the cases, the same global minimum heatexchanger area of 34.44 m2 is obtained using DE, as against 55 m2 reported by Sinnott (1993). But, in the subsequent analysis, the results for 1 bar as the constraint are referred here in drawing the generalized conclusions. A seed value for the pseudo-random-number generator should be selected by trial-and-error.In principle, any positive integer can be taken. Integers 3,5,7, 10, 15, and 20 are tried with all the strategies for a NP value of 70 (10 times 0). The F values are varied from 0.1 to 1.1 in steps of 0.1 and CR values from 0 to 1 in steps of 0.1, leading to 121 combinations of F and CR for each seed. When the DE program is executed for all the above combinations, the global minimum HE area for the above heat duty is found to be 34.44 m2 as against 55 m2 for the base case design-indicating that DE is likely to converge to the true global optimum. For each seed, out of the 121 combinations of F and CR considered, the percentage of the combinations converging to this global minimum (CDE)in less than 30 generations is listed. The average C,, for each seed as well as for each strategy

384

Process Plant Simulation

can be consideredto be a measure of the likelihood of achievingthe global minimum. It was found that the individual values of C D E cover a wide range from 28.1 to 8 1.8-indicating that DE is more likely to find the global optimum. Combining this above observation with the earlier one, it can be concluded that DE is more likely to find a function’s true global optimum compared to the base case design. The average C D E for each seed ranges from 40.7 to 64.8. In this range, the average C D E for seeds 5,7, and 10 is above 54 and so these seeds are good relative to others. With a benchmark of 40 for the individual C D E values as well, seeds 5, 7, and 10 stand good from the rest. Thus with these seeds, there is more likelihood of achieving the global minimum. Whereas with seeds 3, 5, and 20 there are more of CDEvalues below 40 and hence not considered to be good from the likelihood point of view. The average C D E for different strategies varies from 44.2to 59.4 and hence all are good if an average CDEof 40 is considered to be the benchmark. With the same benchmark for C D E values also taken, then, excepting strategies 2, 7, and 8, all other strategies are good. Considering ‘speed’ as the other criteria to further consolidate the effect of strategies on each seed and vice versa, the best combinations of F and CR-taking the minimum number of generations to converge to the global minimum (G~,,)-are listed in Table 14.9. The criteria for choosing a good seed from the ‘speed’point view could be the following: (1) It should yield the global minimum in less number of generations and (2) it should yield the same over a wide range of F and CR for most of the strategies. Following the first criterion with a benchmark of-achieving the global minimum in not more than two generations-seed 3 is eliminated. Satisfying the second criterion also, only seeds 7 and 10 remain. But comparing the overall performance of seeds 7 and 10, seed 10 yields the global minimum in two generations for more number of combinations of F and CR (18 times) than for seed 7 (1 1 times). Also, when random numbers are generated between 1 and 10 using seed 10, it is observed that it generates more l’s, which is required in the design configuration leading to the global minimum. Though seed 5 was good from the ‘more likelihood’point of view, but with ‘speed’as the other criterion, it got eliminated. Similarly with seed 15, using DE/rand/2/exp, with F = 0.6 and CR = 0.9, though the global minimum is achieved in one generation itself, still it is not considered, as more number of generations are taken to converge with other strategies and other combinations of F and CR. From the speed point of view, it can be observed from Table 14.9 that strategies 2,5, 7, and 10 are good. Hence, from both more likelihood and speed points of view, for different seeds, strategies 1, 3,4, 6, and 9 are good. From the above results, it is also observed that, if for a given seed the random numbers generated are already good, then by using DE/best/l/ ... (strategies numbers 1 and 6), the global minimum is achieved in few generations itself. For example,

Table 14.9 Effect of seed on DE strategies with respect to

G,, NP = 70 and MAXGEN = 30

Strategy 1 DE/best/l/exp

2 DE/rand/l/exp

3 DE/rand-to-best/l/exp

4 DE/best/2/exp

Seed = 3 F CR Gmin

0.7 0.9 3

1.1 0.1 3

1.0 0.9 4

0.4 0.9 4

Seed = 5 F CR Gmin 0.9 1.0 3 1.0 1.0 3 1.1 1.0 3

1.0 0.6 6 1.1 0.6 6

1.0 0.2 5 1.1 0.2 5

1.1 1.0 2

Seed = 7 F CR Gmin

5 5 5 5 5 5 5 5 5 6 6

0.8 0.9 0.9 1.0 1.1

0.9 1.0 2 1.0 1.0 2 1.1 1.0 2

0.6 0.8 0.9 1.0 1.0

0.4 0.4 0.5 0.4 0.6 0.4 0.7 0.4 0.8 0.4 0.9 0.4

0.9 1.0 1.1 0.9 1.0

0.7 0.4 0.4 0.9 0.9

0.8 0.7 2

0.6 0.9 12 1.1 0.7 12

0.8 0.8 1.0 1.0 1.0

2 2 2 2 2

Seed = 15 F CR Gmin 0.6 0.3 4 0.7 0.3 4

1.0 0.5 6 1.1 0.5 6

0.8 0.9 2

Seed = 20 F CR Gmin 0.8 0.4 3 0.9 0.4 3 1.0 0.4 3

?

5 2a

s

0.9 0.9 7

0.8 0.9 1.0 0.7 0.6

4 4 4 4 4

0.9 0.4 7 1.0 0.4 7 1.1 0.4 7

1.0 0.5 6 1.1 0.5 6

0.7 0.9 3

1.0 0.7 4 1.0 0.9 4

1.0 1.0 3

$ E,

f? 6' 3

Ba b

Y

3

0.8 0.8 4

1.1 0.6 4

0.7 0.9 4 0.9 1.0 4 1.1 0.7 4

5 DE/rand/2/exp

Seed = 10 F CR Gmin

0.7 1.0 2 0.8 1.0 2

0.6 0.9 1 1.0 0.3 9

0.2 1.0 6

3ii'

p g 0 3

w (contd)

%

Table 14.9 (contd) NP = 70 and MAXGEN = 30

Strategy

6 DElbest/l/bin

Seed = 3 F CR G,, 0.4 0.6 7 1.0 0.9 7

Seed = 5 F CR Gmin 0.9 1.0 3 1.0 1.0 3 1.1 1.0 3

Seed = 7 F CR G,,

Seed = 10 F CR G,,

Seed= 15 F CR G,,

1.0 0.8 3

0.5 0.7 2 0.6 0.4 2 0.6 0.5 2

0.9 0.8 5

Seed = 20 F CR G,, 0.4 0.7

5

1.0 0.9 5

0.6 0.9 2 0.7 0.4 2 0.9 1.0 2 1.0 0.9 2 1.0 1.0 2 1.1 0.8 2 1.1 0.9 2 1.1 1.0 2 7 DE/rand/l/bin

8 DWrand-to-best/l/bin

1.0 0.5 9

0.7 0.1 6 1.1 0.9 6

9 DElbestlUbin

0.9 1.0 4

10 DE/rand/Ubin

0.6 0.8 6

0.1 0.7 0.8 0.7 0.8 1.1

0.3 7

0.6 0.7 1.1 0.4 1.1

0.8 0.8 1.0 0.8 0.5

0.5 7 0.8 7 0.4 5

0.5 5 0.4 5

2 2 2 5 5

0.9 0.9 5

0.4 0.5 7

0.7 0.9 3

0.9 0.6 0.9 0.7 0.9 0.8 0.9 1.0 1.0 1.0 1.1 1.0 1.1 0.8

0.9 0.7 3 0.9 0.8 3

0.7 0.8 5 0.7 1.0 5

1.0 0.5

3

0.5 0.5 2 0.6 0.5 2

0.7 0.4 5

0.7 1.0 2 0.8 1.0 2

0.9 1.0 6

1.0 1.0 3

1.1 0.4 9

0.2 0.9 4 0.8 0.9 4

2

2 2 2 2 2

0.6 0.7 10

3

Specific-purpose Simulation and Dynamic Simulation

387

with seeds 3 and 10 the global minimum is achieved in two generations itself with DE/best/l/ ... strategies. To summarize the effect of seed on each strategy, from Table 14.9, the best seeds and the number of combinations of F and CR in which the global minimum is achieved are found for each strategy along with Gfin. The seed value of 10 works well for four out of the ten strategies considered for a total of 19 combinations of F and CR. Seed 7 does well with three strategies, but for a total of 10 combinations of F and CR only. From this table also, it is clear that the seed value of 10 is the best of all the seeds considered. Once a good seed is chosen, the next step is to study the effect of the key parameters of DE to find the best combinations of NP, F, and CR for each strategy. For this, the DE algorithm for the optimal HED problem is executed for all the 10 strategies for various values of NP, F and CR with MAXGEN = 15. The population size NP is normally about five to ten times the number of parameters in a vector D (Price & Storn 1997). To further explore the effect of the key parameters in detail, the NP values are varied from 10 to 100 in steps of 10 and F , and CR values are varied again as before. The reason for taking MAXGEN = 15 can be explained as follows. As can be seen from Table 14.9, at the maximum 12 generations are taken to converge to the global minimum with NP = 70. Hence, to study the effect of NP, the MAXGEN is restricted to only 15 so that with other NP values if more number of generations are taken then NP = 70 itself can be claimed to be the best population size. With MAXGEN = 15, and for the selected seed value of 10, the percentage of the combinations converging to the global minimum CDE are found for each strategy for different NP values. Though the NP values are varied from 10 to 100, the results for NP = 10,20, and 30 are not considered in the subsequent analysis for the following reasons: (1) The global minimum is not achieved for any of the combinations of F and CR with some strategies for these NP values and (2) to have the same basis for comparison with GA, for which the N values are varied from 32 to 100 (the reasons for taking the above range of N for GA is explained later). From these results, it is seen that the individual CDE values cover a wide range from 0.8 to 66.1. The average CDEfor each NP as well as for each strategy is also computed. This average is again a measure of the likelihood of achieving the global minimum. The average CDE for different NP varies from 6.3 to 31.3. With a benchmark of 10 for this average CDE, it can be observed that, NP value of 70 and above stand good from the rest. From the more likelihood point of view, with a benchmark of 20 for CDE values, again NP values of 70 and above are good. But for NP values of 100 and above, it is observed that the likelihood decreases. The average CDE for different strategies varies from 6.0 to 32.1. With a benchmark of 15 for this average CDE, from the more likelihood point of view, strategies 1, 3, 4, 6, 8, and 9 stand good from the rest.

388

Process Plant Simulation

Considering speed again as the other criterion, the best combinations of F and CR-taking the least number of generations to converge to the global minimum G~n-are determined for various strategies with different NP values. It was found that for certain combinations of F and CR, DE is able to converge to the global minimum in a single generation also. From the speed point of view also, again it is evident that the NP values of 70 and above are good-indicating that at least a population size of 10 times D is essential to have more likelihood of achieving the global minimum. It is in agreement with the simple rules proposed by Price and Storn (1997). And strategies 1,4,6,and 9 are good. Hence, from the more likelihood as well as speed points of view, for different NP values, strategies 1,4,6, and 9 are good. Combining the results of variations in seed and NP, from the more likelihood as well as speed points of view, it can be concluded that DE/besV... (strategies 1,4, 6, and 9) are good. Hence, for this optimal HED problem, the best vector perturbations either with a single or two vector differences are the best with either exponential or binomial crossovers. The number of function evaluations (NFE) is related to the population size NP as NFE = N P ( G ~ ,+ 1) (+1 corresponds to the function evaluations of the initial population). It can be inferred from the simulated results (using NP and Gminvalues to compute NFE) that NFE varies from only 100 to 1300, out of the 161,280 possible function evaluations considered. Hence, the best combination corresponding to the least function evaluations is found to be for NP = 50 and DE/best/l/exp strategy, with 100 function evaluations as it converges in one generation itself. For this combination, DE took 0.1 of CPU time on a 266-MHz Pentium-I1 processor. Now, to study the effect of F and CR on various strategies, the DE algorithm is executed for the present optimal problem, for different values of F and CR. For the selected seed value of 10 and NP = 70, the F values are varied from 0.5 to 1.0 in steps of 0.1 and CR values from 0 to 1 in steps of 0.1. For the 10 different strategies considered in the above range of F and CR, 21 plots are made for the number of generations versus CR values with F as a parameter; 18 plots for the number of generations versus F values with CR as a parameter. Typical graphs are shown in Figs 14.10-14.13. The numbers given in the legend correspond to the strategy. Whenever the global minimum is obtained, it is observed that DE converges in less than 70 generations for strategies 1, 4, 6, and 9, and in less than 100 generations for the rest of the strategies. This implies that for any particular combination of the key parameters, if convergence is not achieved in less than 100 generations, then it is never achieved. So the points at 100 generations in the graphs correspond to the misconvergence and hence represent local minima (other than 34.44 m2 and obviously higher values of area).

SpeciJic-purpose Simulation and Dynamic Simulation

389

110 100 90 80

2 f

d

70

I

60 50 40

\:

I

30 20 10 0

0

0.3

0.5

I

0.7

\

! /

\

\

0.8

0.9

1.0

CR Fig. 14.10 Comparison of DE strategies for best vector perturbation with F a s a parameter ( F = 0.5)

8 '-

8

110 100 90 80

-

f,

- 4.DE/best/Yexp

6. DE/best/l/bin -.e9. DUbest/2/bin ...A-

70 60 50

40 30 20 10 0

0

0.3

0.5

0.7

0.8

0.9

1.0

CR

Fig. 14.11 Comparison of DE strategies for best vector perturbation with F as a parameter ( F = 0.9)

For studying the effect of F and CR on each strategy, the criterion considered is speed. So, it should yield the global minimum in less number of generations. From the speed point of view, it can be seen that with F as a parameter, in general (excluding the misconvergence points), DE/best/l/ ... (strategies 1 and 6) are better than DE/best/2/ ... (strategies 4 and 9) unlike from the more likelihood point of view, where DE/best/2/ ... is better than DE/best/l/ ... as the former takes less number of generations. Similar trends are observed for CR also as a parameter. This means that if the best vector perturbation is to be tried, thkn, from the speed point of view, it is worth trying it with single vector perturbation first to see quickly if there is any convergence. If misconvergence occurs, DE/best/2/ ... can be tried, as it has more likelihood of achieving the global minimum. With DE/best/l/ ... it is observed that

390

Process Plant Simulation

at low values of F binomial crossover is better than exponential crossover; as F values are increased there is no marked difference in the performance of exponential and binomial crossover. From the speed point of view this observation is again clearly evident from the results obtained. But for DE/best/2/ ... binomial crossover seems to be better than that of exponential, as it yields the global minimum in less number of generations at medium to higher values of F.Similar trends are observed with CR also as a parameter.

y1

8

'1

6

5

20

---

0

2.1

....&.. 7. ]

10

0.5

0.6

0.7

0.8

0.9

1.0

F Fig. 14.12 Comparison of DE strategies for random vector perturbation with CR as a parameter (CR = 0.5)

-

2

+- 2. DE/rand/l/exp - - 5. DE/rand/2/exp -A- 7. DE/rand/l/bin - e - 10. DE/rand/2/bin

80 70

10 0

\

0

0.5

0.6

0.7

0.8

0.9

1.0

F Fig. 14.13 Comparison of DE strategies for random vector perturbation with CR as a parameter (CR = 1.O)

If misconvergence occurs with DE/bes t/..., either DE/rand-to-best/l/ ... or DE/ rand/... seem to be the immediate good options. From the speed point of view, DE/ rand/2/ ... is seen to be better than DE/rand/l/ ... at high values of F and CR. This indicates that if random vector perturbation is to be used, then, from the speed point of view, it is better to use it with two vector differences. With DE/rand/l/ ...,

Spec@c-purpose Simulation and Dynamic Simulation

391

exponential crossover seems to be a better option. For the optimal HED problem considered, it is concluded from the preceding discussions that for variations in seed and NP, from the more likelihood as well as speed points of view, DE/bestl... strategies are better than DE/rand... . From these results it is also observed that the DE strategies are more sensitive to the values of CR than to F.Extreme values of CR are worth to be tried first. It was observed that there is no difference in the performance of exponential and binomial crossover at CR = 1.0, which is quite obvious from the nature of these operators. The priority order of the strategies to be employed for a given optimal problem may vary from what is observed above. Selection of a good seed is indeed the first hurdle before investigating the right combination of the key parameters and the choice of the strategy. Some of these observations made in the present study form the supportive evidence of the recommendations and suggestions made for other applications by Price and Storn (2003). For comparison, genetic algorithms with binary coding for the design variables are also applied to the same case study with roulette wheel selection, single-point crossover, and bit-wise mutation as the operators for creating the new population. The GA algorithm is executed for various values of the population size N , the crossover probability p c and the mutation probability pm.With a seed value of 10 for the pseudo-random-numbergenerator, N is varied from 32 to 100 in steps of 4, p c from 0.5 to 0.95 in steps of 0.05, and p m from 0.05 to 0.3 in steps of 0.05, leading to a total of 1080 combinations. The reasons for not considering the N values of 30 and below are same as those already explained for NP while discussing the results with DE. N/2 has to be an even number for single-point crossover and hence the starting value of 32 and the step size of 4 are taken. The step size is smaller for GA compared to DE, as can be seen later that GA has less likelihood, so more search space is required. For each population size, 60 combinations of p c and pm are possible in this range. For the case study considered, the same global minimum heat-transfer area is obtained (34.44 m2)by using GA also. The minimum ,, in number of generations required by GA to converge to the global minimum G the above range of the key parameters is listed in Table 14.10 along with the NFE. For each combination of N and p c listed in this table, GA is converging to the global minimum heat-transfer area of 34.44 m2 for all the six values of p m from 0.05 to 0.30 in steps of 0.05. While executing the GA program, it is observed that more number of generations is taken by GA to converge and, hence, the maximum number of generations (MAXGEN) is specifiedas 100.For a given N , the percentage of the combinations converging to the global minimum C, in less than 100 generations, out of the 60 possible combinations of p c and p m considered, is also , ranges from 10 to 50 only, as against the listed in Table 14.10.As can be seen, C individual CDE,which varies from 0.8 to 66.1. Comparing the individual C, range for one of the best strategy DE/best/2/bin, which varies 11.6 to 66.1 (except one

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Process Plant Simulation

that is 3.3), C G A is relatively less. The average C G A is calculated to be 20.9, whereas the average C , is above 22 for four out of the ten strategies (strategies 1,4,6, and 9). It is interesting to note that had the basis of MAXGEN been same (i.e., 100) for both GA and DE, then it is quite obvious that the C,, values would have been very high (may be close to 100)-indicating that DE has more likelihood of achieving the global optimum compared to GA-as it has a wide range of the individual C,, values. Also DE has more strategies to choose from, which is an advantage over GA. As a measure of likelihood, another criterion is identified and defined-the percentage of the key parameter combinations converging to the global minimum, In Table 14.10, out of out of the total number of combinations considered (C,,,). the 1080 combinations of the key parameters considered with GA, only 138 combinations (i.e., C,, = 12.8) converge to the global minimum in less than 100 generations. Whereas in DE, out of the total of 9680 possible combinations of key parameters considered, 1395 combinations (i.e., C,,,= 14.4) converge to a global minimum in less than 15 generations itself. Table 14.10 GA parameters converging to the global minimum (for pm= 0.05-0.30)

Seed = 10 and MAXGEN = 100

1

44

2 3

48 52

4

60

5

64

6

68

7 8 9

72 76 80

10

84

11

100

0.55 0.60 0.75 0.75 0.90 0.65 0.70 0.50 0.60 0.80 0.65 0.85 0.50 0.75 0.60 0.75 0.80 0.85 0.95 0.75 0.90 0.60 0.70

20 10 20 20 30

20 10 10 50

20 20

53 63 5 67 19 71 14 59 43 25 64 6 47 90 23 18 39 35 46 97 70 13 4

2376 2816 288 3536 1040 4320 900 3 840 2816 1664 4420 476 3456 6916 1920 1520 3200 2880 3760 8232 5964 1400

500

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393

The relation between NFE and the number of generations in GA also remains the same as in DE [NFE = N(G,, + l)]. Using GA, with a seed value of 10, NFE varies from 288 to 8148 as against a small range of 100 to 1300 only for DE, which is an indication of the tremendous speed of the DE algorithm. The above two observations clearly demonstrate that for the case study taken up, the DE algorithm is significantlyfaster, has more likelihood of achieving the global optimum and so is more efficient compared to GA. It is also evident from Table 14.10 that, the best combination corresponding to the least function evaluations is for N = 48, p c = 0.75, and p , 0.05 to 0.30 (entire range of p,), with 288 function evaluations as it converges in five generations itself. For this combination of parameters, GA took 0.46 s of CPU time (on a 266-MHz’Pentium-I1 processor) to obtain 34.44 m2 area. The summary of the results from the preceding discussion for the selected seed value of 10 is listed in Table 14.11. The performance of DE and GA can be compared from this table with respect to the various criteria identified and defined. Out of the entire range of key parameter combinations considered, the range of the individual key parameter values converging to the global minimum heat-exchanger area of 34.44 m2, along with the best strategies recommended, are listed for DE and GA. For comparison of DE and GA, the characteristic criteria identified are the ‘likelihood’ and the ‘speed’ in achieving the global minimum. As a measure of likelihood, the following parameters are defined: C D E and average CDE for DE; C,, and average C,, for GA; C,,,for both GA and DE. Similarly, as a measure of speed, the parameters defined are the following: G-, NFE, and CPU time. From the range of the values of these identified parameters for the above criteria listed in Table 14.11, it is evident that DE has shown remarkable performance within 15 generations itself from both likelihood and speed points of view, which GA could not show even in 100 generations. Hence, it can be concluded that DE is significantly faster at optimization and has more likelihood of achieving the global optimum. The results shown in this comprehensive table consolidate all the observations made and the conclusions drawn from the preceding discussion. So, the authors recommend the above range of the key parameter values (as listed in Table 14.11) for the optimal design of a shell-and-tube heat exchanger using differential evolution. The performance of DE and GA is compared for the present problem in Table 14.12, with respect to the ‘best’ parameters-parameter values converging to the global minimum out of the entire range considered. For NP = 50, with the DE/best/l/exp strategy, CR = 0.7, and F = 0.8 to 1.1 (any value in steps of O.l), as already mentioned, DE took one generation, 100 function evaluations, and 0.1 s of CPU time. But with GA, for N = 48, p c = 0.75, and p m = 0.05 to 0.30 (any value in steps of 0.05) it took five generations, 288 function evaluations, and 0.46 s of CPU time.

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Table 14.11 Comparison of DE and GA with respect to various criteria for the entire range of parameters Seed = 10 Criteria

DE

GA 100

1

MAXGEN

15

2

Global minimum heat-exchanger area (m’)

34.44

34.44

Parameter values converging to the global minimum

NP: 50-100 F 0.3-1.1 CR: 0.1-1.0

N: 44-100

3

(a) key parameters

(b) strategy

4

Measure of likelihood in achieving the global minimum

5

Measure of ‘speed’ in achieving the global minimum

(a) Gmin (b) NFE (c) CPU time (s)

P,: 0.05-0.30 P,: 0.50-0.95

DE/best/l/exp DE/best/ 1/bin DE/best/2/exp DE/best/2/bin CDE:0.6-66.1 Avg CDE:6.0-32.1 Clot:12.8

CGA: 10-50 Avg CGA:20.9 Ctol: 14.4

1-12 100-1 300 0.1-2.23

4-97 288-8148 0.46-15.29

Table 14.12 Comparison of DE and GA with respect to various criteria for the best parameters Seed = 10 Criteria

DE

GA

1

Global minimum heat-transfer area (m2)

34.44

34.44

2

Best parameter values (a) key parameters converging to the global minimum (b) strategy

NP: 50 F 0.8 - 1.1 CR: 0.7 DEhestll /exp

pm:0.05-0.30 pc: 0.75

N : 48

3

Gmin

1

5

4

NFE

100

288

5

CPU time (s)

0.1

0.46

From Table 14.12, it can be seen that DE is almost 4.6 times faster than GA. By using DE, there is 78.3% saving in the computational time compared to GA. Comparing the results of the proprietary program (HTFS, STEPS) with these algorithms (both DE and GA), there is 37.4% saving in the heat-transfer area for

Specijic-purpose Simulation and Dynamic Simulation

395

the case study considered. For the optimal shell-and-tube HED problem considered, the best population size using both DE and GA is seen to be around seven times the number of design variables with about 70%crossover probability. Hence, for the heat duty of the case study taken up, the best design configuration of the shelland-tube heat exchanger with respect to the design variables considered corresponds to the global minimum heat-exchanger area of 34.44 m2,The best design variables are listed in Table 14.13 along with the best key parameters of the DE algorithm used for this optimization. Table 14.13 Summary of the proposed final design for the case study Parameters

1

Heat duty (kW)

2

Best design variables

1509.4 (a) tube outer diameter (inch) (b) tube pitch (c) shell head type (d) tube passes (e) tube length (ft) ( f ) baffle spacing (g) baffles cut

1/2 5/8” triangular Fixed tube sheet Single 24 20% 15%

3

Heat-exchanger area (m2)

34.44

4

Pressure drop (bar)

(a) tube side (b) shell side

0.67 0.3 1

5

DE parameters

(a) strategy (b) seed (c) NP (d) F (el CR

DE/best/ 1/exp 10

50 0.8-1 * 1 0.7

14.3.4 Conclusions We have demonstrated the first successful application of differential evolution for the optimal design of shell-and-tubeheat exchangers.A generalized procedure has been developed to run the DE algorithm coupled with a function that uses Bell’s method of heat-exchangerdesign to find the global minimum heat-exchanger area. For the case study taken up, application of all the 10 different working strategies of DE is explored. The performance of DE and GA is compared. From this study we conclude the following. 1. The population-basedalgorithms such as GAS and DE provide significant improvement in the optimal designs, by achieving the global optimum, compared to the traditional designs.

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Process Plant Simulation

2. For the present optimal shell-and-tube HED problem: I The best population size, using both DE and GA, is about seven times the number of design variables. I From the more likelihood as well as speed points of view, DE/best/ ... strategies are better than DE/rand...for the selected seed value of 10. I DE/best/l/... strategy is found to be the best out of the presently available 10 strategies of DE. 3. Differential evolution, a simple evolution strategy, is significantly faster compared to GA. 4. DE achieves the global minimum over a wide range of its key parametersindicating the likelihood of achieving the true global optimum. 5. DE proves to be a potential source for accurate and faster optimization.

14.4 Pyrolysis of Biomass Pyrolysis of biomass is a promising route for the production of solid (charcoal), liquid (tar and other organics), and gaseous products (H2, CO,, CO). These products are of interest as they are possible alternate sources of energy. Pyrolysis is a process in which a biomass feedstock is thermally degraded in the absence of oxygenlair. The study of pyrolysis is gaining increasing importance, as it is not only an independent process, but also a first step in the gasification or combustion process. The basic phenomena that take place during pyrolysis are (1) heat transfer from a heat source leading to an increase in temperature inside the fuel, (2) initiation of pyrolysis reactions due to this increased temperature, leading to the release of volatiles and the formation of char, (3) outflow of volatiles resulting in heat transfer between hot volatiles and cooler unpyrolysed fuel, (4) condensation of some of the volatiles in the cooler parts of the fuel to produce tar, and ( 5 ) autocatalytic secondary pyrolysis reactions due to these interactions. Studies have been carried out on pyrolysis of biomass and other substances by several researchers (Srivastava & Jalan 1994; Srivastava et al. 1996; Koufopanos et al. 1989).The actual reaction scheme of pyrolysis of biomass is extremely complex because of formation of over hundred intermediate products. Pyrolysis of biomass is, therefore, generally modelled on the basis of apparent kinetics. Ideally, the chemical kinetics model should account for primary decomposition reactions as well as secondary reactions. Significant contributions have been made on kinetic modelling (Jalan & Srivastava 1999; Pyle & Zaror 1984).

14.4.1 Stepwise Procedure Before actually going into the simulated results in this area, let us understand the systematic method by which the finite difference method can be applied to solve the partial differential equations by considering the following illustration.

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397

Illustration 14.1 Pyrolysis reactions can be described by means of the following scheme as used by Babu and Chaurasia (2003b). Virgin biomass B(nl order decay)

I\ Reaction 2

Reaction 1

Reaction 3 (Volatile + gases), + (char)l (volatile + gases)2 + (char), (n2 order decay) (n, order decay) This kinetic scheme indicates that the biomass decomposes to volatiles, gases, and char. The volatiles and gases may further react with char to produce different types of volatiles, gases, and char, where the compositions are different. Therefore, the primary pyrolysis products participate in secondary interactions (reaction 3), resulting in modified final product distribution. The kinetic equations for the mechanism shown above along with heat transfer model equation and initial and boundary conditions are as follows:

(ii) (iii)

a r

at

dr2

Initial conditions at t = 0: CB

= CEO,

cG1

= cC1 =

cG2 = c C 2 = 0, T ( r >O ) = To

(vii)

Particle boundary conditions: At t > 0, Y = 0, (viii)

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Process Plant Simulation

At t > 0, r = R,

(5)

= h(Tf - T )

+ m(Tf4 - T 4 )

r=R

Step 1 Derivation of the model equation The partial differential equation given in the problem formulation is

k [ y $ + g ] + ( - M ) ( - $ ) = Eat( CPPT Starting from the basic conservation and rate laws, the above model equation can be obtained as follows (Holman 1997): Fourier's law of heat conduction is given by q = - M -d;T dX where q is the heat-transfer rate and d;T/dx is the temperature gradient in the direction of heat flow. The positive constant k is called the thermal conductivity of the material, and the minus sign is inserted so that the second principle of thermodynamics is satisfied, i.e., heat must flow downhill on the temperature scale. For the element of thickness ax, the following energy balance may be made (see Fig. 14.14 below representing the elemental volume for one-dimensional heat conduction analysis):

kx-id. tFig. 14.14 Elemental volume for one-dimensional heat conduction

SpecFc-purpose Simulation and Dynamic Simulation

399

Energy conducted in the left face + heat generated within the element = change in internal energy + energy conducted out of the right face (3) These energy quantities are given as follows: Energy in the left face qx =

-u-m dx

(4)

Energy generated within the element = qAdx

m

Change in internal energy = pC A-ak

dz

d

Energy out of the right face = qx+h= 4, + x(qx)h

(7)

After substituting the value of q, and rearranging, we get

where q is the energy generated per unit volume per unit time in W/m3, Cp is the specific heat in J/kg"C, and p is the density in kg/m3. Substituting the above expressions in Eq. (3), we get

Equation (9) can be written as

Equation (10) is for the unsteady state because temperature is changing with time, i.e., dT/dz. This is the one-dimensional heat conduction equation. To treat more than one-dimensional heat flow, we need to consider only the heat conducted in and out of a unit volume in all three coordinate directions. The general three-dimensional heat conduction equation is

For the constant thermal conductivity, we get 8 T + -8f T ax2 dy2

8 T f -q - 1 6 T &' k a dr

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Process Plant Simulation

where the quantity a = k/pCp is called the thermal diffusivity of the material. In this derivations, the expression for the derivative x + dr has been written in the form of a Taylor-series expansion with only the first two terms of the series employed for the development. Equation (12) may be transformed into cylindrical coordinates using the standard calculus techniques:

For a cylinder with length very large compared to the diameter, it may be assumed that the heat flows only in a radial direction, so that only the space coordinate needed to specify in the system is r. The axial, azimuth terms are neglected as the cylinder is sufficiently large. So Eq. (13) becomes 8 T 1 8 T + -4= -18T +-dr2 r d r k a d z

(14)

Equation (14) can be written as

8T

1di'

8T

(15)

p and Cp are considered functions of time. So Eq. (15) can be rewritten as

where q is the energy generated per unit volume per unit time in W/m3. It can be written as q = (-AH)-d(-p)

at

where AH is the heat of reaction in J k g , p is the density in kg/m3, and t is the time in s. After the substitution of units of all the terms in Eq. (16), the units of the lhs and rhs are found to be the same. Equation (16) can be rewritten as

Consider Cp to be independent of time; however, actually Cp is a function of temperature and temperature is a function of time. Therefore, indirectly Cp is a function of time. However, the above consideration that Cp is independent of time is considered for facilitating solution of the equation. Subsequently, Cp values are taken as a function of temperature.

Specific-purpose Simulation and Dynamic Simulation

401

So Eq. ( 1 8) becomes

Equation (21) can be modified for different geometries as

k[& L

dr

+

y$] +

[(-AH)

+ CpT]-3 d-ip')

di'

= pc dt

A

where b is the geometry factor (slab = 1, cylinder = 2, sphere = 3). Equation (22) is based on the following initial and boundary conditions: Initial condition at r = 0:

T(r, 0 ) = To Boundary conditi ons : Att>O,r=O,

(23)

At t > 0, r = R, = h(Tf - T )

k($)

+ m(Tf4 - T 4 )

r=R

In Eq. (25), external heat transfer is considered to occur by a combination of conductive and radiative mechanisms. The dimensionless form of Eq. (22) is given by

Step 2 Obtaining the dimensionlessform of the partial differential equation The initial and boundary conditions in dimensionless forms are as follows: Initial condition at r = 0

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Process Plant Simulation

Boundary conditions: At z > 0, x = 0,

de _ -0 dX

At z > 0, x = 1,

The following groups are introduced to obtain Eqs (26)-(29):

a=- k @P

x=-

r R

t

[h + .5a(T3+ T 2 T f + Tf’T + T;)]

Q=

(-AH)

(34)

+ C,T

PC,(TO - Tf 1

(35)

Note How to obtain the dimensionless forms of Eqs (26)-(29) using the dimensionless groups given by Eqs (30)-(35) is demonstrated below: We have

x = -r R dx 1 =-=d r R

~ d r = R d x

dr2 = R2dx2

j

Specijbpurpose Simulation and Dynamic Simulation

403

Now,

r = -at R2 * - =drdt

a

R2

= at = -R2 8d7 a Now,

=$

6T = -(Tf - T o ) d e

jaZT

= -(Tf - T0)aZe

After substituting dt, dr, dr2, 8T and 8 T in Eq. (22) and simplifying, we get Eq. (26). Similarly Eqs (27)-(29) can be obtained from Eqs (23)-(25). However, in order to solve Eq. (26), it is essential to obtain the value of (-dp/dt) from the chemical kinetics model (Babu & Chaurasia 2003b). From Eq. (32), we have

rR2

t =-

a

Hence, R2dr dt =-

a

(37)

We know from kinetic equations that (Babu & Chaurasia 2003b) (38) which is equal to dp/dt. Hence (39)

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Process Plant Simulation

From Eqs (37) and (39), we get k,R2c2 --dp - -

at

a

Introducing Eq. (40) in Eq. (26) gives dB = -b - 1 6 8 8’8 +dr x dx 8x2

+ QR2k,cz a

Treatment of condition at the centre, i.e., at x = 0: A close look at Eq. (41) reveals that at the centre, i.e., at x = 0, the first term on the right-hand side will become infinity, which is physically not possible and so not acceptable. However, this difficulty can be alleviated by making the use of l’H6pital’s rule. If and

then

Therefore, invoking l’H8pital’s rule for the centre condition, we obtain 1 % - a2e lirnX+O -ax

xdx

Hence, at x = 0, Eq. (41) becomes

-dB- - b 8T0+ QR2k,cz dr dx a Now, Eq. (41) can be solved easily by using the initial and boundary conditions given by Eqs (27)-(29). Step 3 Method of solution and writing the equation inPnite difference form Equation (41) along with initial and boundary conditions given by Eqs (27)-(29) is solved by the finite difference method using a pure implicit scheme. A pure implicit scheme is an unconditionally stable scheme, that is, there is no restriction on time step in sharp contrast with the Euler and Crank-Nicholson method as discussed by Ghoshdastidar (1998). The finite difference forms of Eqs (41), (44), and the equation obtained by using boundary condition, i.e., Eq. (29), are as follows:

Spec$c-purpose Simulation and Dynamic Simulation

- (b - i)e$+lH + QRzk,c; a

405

(47)

Rearranging Eqs (45)-(47) will give the following sets of equations, which would facilitate numerical solution:

Hence, we have a set of M linear simultaneous algebraic equations and M unknowns, which can be easily solved by standard numerical methods. The tri-diagonal matrix algorithm (TDMA), also known as the Thomas Algorithm, is used to solve the set of Eqs (48)-(50). Step 4 Numerical solution of the problem [applying the tri-diagonal matrix algorithm (Ghoshdastidar 1998) to Eqs (48)-(50)]. Zteration 1 Let us assume the following values (Tables 14.14 and 14.15):

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Process Plant Simulation

Table 14.14 Values of parameters used in the numerical solution of the model Property Convective heat-transfer coefficient Wood specific heat Char specific heat Wood thermal conductivity Char thermal conductivity Heat of reaction Initial density of wood Initial thermal diffusivity of wood

Value h = 8.4 W/mZK Cp = 1112.0 + 4.85 (T- 273) J k g K Cp = 1003.2 + 2.09 ( T - 273) J k g K k = 0.13 + 0.0003 (T- 273) W/mK k = 0.08 - 0.0001 (T - 273) W/mK = -255,000 J k g Po = 650 kg/m3 a = 1.79 x lo-' m2/s

Table 14.15 Nominal values of parameters employed in the present study Property Geometry factor Order of reaction 1 Order of reaction 2 Order of reaction 3 Particle radius range Initial temperature Final temperature range Emissivity coefficient Stefan-Boltzmann constant

Value

b = 2, dimensionless nl = 0, dimensionless n2 = 1.5, dimensionless n3 = 1.5, dimensionless R = 0.005 m To= 473 K T'= 873 K E = 0.95, dimensionless d = 5.67 x lo4 W/mZK4

k, = AlexP[(D,m + (L,f12>1

(51)

k, = 4 e x P w2m + (L,f12 11

k,

=

A,exp [(-E,/RT)]

and A, = 9.973 x 10-5s-', A2 = 1.068 x lO-,s-', A, = 5.7 x 105s-'

D, = 17,254.4K, D2 = 10,224.4 K L,

= - 9,061,227 K2, L2 = - 6,123,081 K2

E, = 81 kJ/mol Let t =1.4 s. So, from Eq. (32), we get z = 0.010024. Az will be same as z. So A T = 0.010024. Let us consider that there are five equations. So, Ax = 0.25, x1 = 0, x2 = 0.25, x3 = 0.5,x4 = 0.75, and x5 = 1 .O. Using the values given in Tables 14.14

Spec fit-purpose Simulation and Dynamic Simulation

407

and 14.15, we get k = 0.13 and Cp = 2082. From Eq. (34), H = 1.88945. From Eq. (33, Q = 2.2903 x From Eq. (51), we have k, = 0.17864 x From Eq. (48), we have 2bAz

From Eq. (49), we have

-). (-

Az(b- 1) - A T 2(WX,

-),

Ar(b - 1)

2A r

b2 =

b3 =

(-

2A z

+ 1)

(W2

(Ax)

-$$)I

+ 1)

(W2

(Ax)

Az(b- 1) - A T 2(h)x3

A z QR2klc;

2bA z

+

d4=84+[

From Eq. (50), we have

d 5 = 8 5 + [ At-QR2 a k,c$

1

AT QR2klc;J' a

j

After substituting the values, we get bl = 1.6416 = -0.6416 d1 = 1.000004 a2 = -0.0804 b2 = 1.3208 c2 = -0.2404 d2 = 1.000004 a3 = -0.1204 b, = 1.3208 c3 = -0.2004 d3 = 1.000004 ~4 = -0.1337 b4 = 1.3208 ~4 = -0.1871 d4 = 1.000004 US = -0.3208 b5 = 1.49126 d5 = 1.000004 8, = e2= e3 = 8, = 8, = 1.0 (boundary condition)

408

Process Plant Simulation

The above values can be arranged in matrix form as follows:

Now based on the above values let us calculate the other values:

P1= b, = 1.6416 P2 = b2 - -a2c1 - - 1.2894 P1 ac P3 = b3 - 2-2 = 1.2894 P2

ac P4= b4 - 3 = 1.3002 P3

ac Ps = b, - 2 = 1.4451 P4

dl

- 0.6092

=Fl-

Y1

Y2 =

d2 - % '"

= 0.81354

P2

d, - a y3=

3Y2 =

P3 d, - a

y4

=

4Y3

=0.8561

P4

d5 y5 =

0.84562

= 0.88204

P5

e, = Y, = 0.88204 C

8, = y 4 - Qe, P4

= 0.98303

Specific-purpose Simulation and Dynamic Simulation

409

P3

C

8, = y 1 - r

PI

e 2= 0.99984

Now,

i.e., T = 873 - 4008. So we will get T I = 473.064 T2 = 473.204 T3 = 474.064 T4= 479.788 T, = 520.184 The above are the temperatures at five different points of the cylinder along the radius from the centre to the surface after 1.4 s as we have taken the time steps of 1.4 s. Iteration 2 The temperatures calculated in iteration 1 at different points will be used to calculate the concentrations by using Runge-Kutta fourth-order method from Eqs (i)-(v). Let us assume that we obtain the results given in Table 14.16. Table 14.16

473.064 473.204 474.064 479.788 520.184

0.999993 0.999993 0.999992 0.999988 0.999822

0.000003 0.000003 0.000003 0.000005 0.000102

0.000005 0.000005 0.000005 0.000008 0.000076

0.000000 0.000000 0.000000 0.000000 0.000000

Now calculate ff,, for which thermal conductivity at the last point (i.e., point 5 ) (k,) is required, which has to be calculated at temperature T,. Substituting the values in the respective equation, we get k, = 0.2042 and H, = 1.90552. Now we have to calculate Q , which is changing at different points due to change in density p and specific heat Cp.At different temperatures, the different values of Cp are

410

Process Plant Simulation

Cpl = 2082.31 Cp2= 2082.99

Cp3= 2087.16 Cp4= 21 14.922 Cp5= 2310.842 Similarly, p = 650 (cB+ ccl + c,), where 650 is the initial value of density. So, the different values of densities are as follows: p1 = 649.9987

p2 =649.9987 p3 =649.9981 p4 =649.9974 ps =649.9337

The different values of Q are as follows: Q, = 2.2905 x Q2 = 2.2909 x

Q3= 2.2932 x Q4= 2.3091 x lou3 Q, = 2.4251 x

The values of rate constant k , at different points are as follows:

k , , = 0.17972 x k,, = 0.1821 x lod5 k , , = 0.1974 x k14

= 0.3327 x

k15 = 7.2639 x

lo-,

After substituting the values, we get U,

43 44 45

= -0.0804 = -0.1204 = -0.1337 = -0.3208

b1 = 1.6416 b2 = 1.3208 b3 = 1.3208 b4 = 1.3208 b5 = 1.49126

C, = -0.6416 C,

~3 ~4

d, = 0.99984 = -0.2404 d2 = 0.99949 = -0.2004 d3 = 0.99734 = -0.1871 d4 = 0.98303 d5 = 0.8820402

8, = 0.9984 8, = 0.99949 8, = 0.99734 0 4 = 0.98303 O5 = 0.88204

Specific-purpose Simulation and Dynamic Simulation

411

The above values can be arranged in matrix form as follows:

Note The values of a, b, and c are the same as that of first iteration and these will be the same for all the iterations. The value of d is changing because the value of Q, which is a function of density p and specific heat Cp, is changing at all the points. Also the value of the rate constant k, is changing at different points. Now based on the above values, let us calculate the other values:

PI = b,

P3

= 1.6416

= b3 -

--

- 1.2894

P2

P4 = b4 P5 =

--

-

1.3002

b5 - - -

-

1.44655

-

P3

P 4

dl

- 0.6091

yl=Fld2 - a2y1 = 0.81314 y2 =

P2

412

Process Plant Simulation

8, = y 5 = 79666

c4 8 , = y 4 - -8,

= 0.95744

P4

8,

=y3

C

-2

e 4 = 0.9913

P3 C

8, = y 2 - -9,= 0.99796 P2

Now,

i.e., T = 873 - 4008. So we will get T , = 473.344 T2 = 473.816 T3 = 476.48 T4 = 490.024 T, = 554.336

These are the temperatures at five different points of the cylinder along the radius from the centre to the surface after 2.4 s as we have used the values of iteration 1 (i.e., after 1.4 s) to calculate it. Similarly, we can use the values of iteration 2 (i.e., after 2.4 s) to calculate the values after 4.2 s. The values of iterations 1 and 2 are listed in Table 14.17. Table 14.17

T4

T5

Iteration 1

Iteration 2

473.064 473.204 474.064 479.788 520.184

473.344 473.816 476.48 490.024 554.336

Speci$c-purpose Simulation and Dynamic Simulation

413

14.4.2 Problem Formulation It is generally accepted that the most important parameters affecting the process of pyrolysis of biomass are temperature, concentration, time, and heating conditions. Babu and Chaurasia (2003a), in their initial study, found the optimum values of these parameters for both non-isothermal and isothermal heating conditions. And in their subsequent study (Babu & Chaurasia 2003b), they examined the combined effect of kinetics and heat transfer. The results of both the studies are presented below. Koufopanos et al. (1991) proposed the following two-step-mechanism scheme for describing the kinetics of the pyrolysis of biomass: Virgin biomass B(n, order decay)

I\ -

Reaction 1

Reaction 2

Reaction 3 (Volatile + gases), + (char), (volatile + gases), (n, order decay) (n, order decay)

+ (char),

This model indicates that the biomass decomposes to volatiles, gases, and char. The volatiles and gases may further react with char to produce different types of volatiles, gases, and char, whose compositions are different. Therefore, the primary pyrolysis products participate in secondary interactions (Reaction 3), resulting in modified final product distribution. 14.4.2.1 Chemical kinetics model

The kinetic equations for the mechanism shown above are as follows: (14.30) (14.3 1) (14.32) (14.33) (14.34)

414

Process Plant Simulation

where

k, = A,exp[(D,/T) + (L1/T2)]

+ (L2/T2)J

k,

= A,exp[(D,/T)

k,

= A,exp[(-E,/RT)]

and A, = 9.973 x

s-,, A, = 1.068 x lo-, s - ~ , A, = 5.7 x lo5 s-,

D, = 17,254.4 K, D, = 10,224.4 K

L,= -9,061,227 K2, L, = -6,123,081 K2 E3 = 81 kJ/mol Srivastava et al. (1996)assumed that in thermo-gravimetric analysis, temperature, and time have a linear relationship, and consequently it is appropriate to describe the above phenomenon as follows: T = (HR)t + To (14.35) where To is the initial temperature, HR is the heating rate in Ws, and t is the time in s. 14.4.2.2 Heat-transfer model

Consider a cylinder of radius R, length 1, and radial thickness dr. Assume the length of cylinder to be so large that heat transfer takes place in the radial direction only. It is assumed that heat flows in one dimension and that heat transfer inside the particle occurs by conduction only. Based on the above assumptions, the equation representing the energy balance is given by +[(-AH)+C,T]

dt

The initial and boundary conditions for Eq. (14.36) are as follows: Initial condition: At t = 0,

T(r, 0) = To

(14.37)

Boundary conditions: At t > 0, r = 0, (14.38)

At t > 0, r = R,

-k(

: 1

= h(Tf

=R

- T ) + ~ r ~ ( T-f 4T 4 )

(14.39)

Specijic-purpose Simulation and Dynamic Simulation

415

In Eq. (14.39), external heat transfer is considered to occur by a combination of conductive and radiative mechanisms.

14.4.3 Method of Solution and Simulation Equations (14.30)-( 14.35) are used for non-isothermal conditions with an initial temperature To = 773 K and with different heating rates of 25 Ws,40 Ws,50 Ws, 60 Ws, and 80 Ws. Also the heating rates beyond 80 Ws are used in simulation over a wide range between 125 Ws and 360 Ws with a heating rate interval of 15 Ws (step size) for finding out the optimum heating rate. In addition, the simulation is also carried out at the optimum heating rate corresponding to minimum pyrolysis time to find the corresponding optimum parameters. Equations (14.30)-( 14.34) are used for isothermal conditions ( T = To = constant) with different temperatures of 773 K, 873 K, 973 K, 1023 K, and 1073 K. Similarly, in this case also, simulation is carried out with a wide range of temperatures ranging from 773 K to 1773 K with a maximum step size of 100 K initially to a very small step size of 0.1 K to ensure not to miss out the optimum parametric values. In addition, the model equations are solved at the optimum temperature also to find the corresponding optimum parameters. For both the non-isothermal and isothermal cases, the corresponding equations are solved numerically using the fourth-order R u n g e Kutta method and the results are simulated. The pyrolysis process is very slow below a temperature of 773 K (Srivastava et al. 1996), hence the lower values of the initial temperature are not considered in the simulations of the present study. The following initial conditions are used for solving the coupled ordinary differential equations: At t = 0, cB = 1.0, cG, = cC, = cG2= cC, = 0 The final dimensionless concentration of biomass was assumed to be equal to 0.03, because beyond that concentration, pyrolysis is found to be very slow and of little practical importance. The iterations are continued till cB 2 0.03. The optimum values of the parameters are found using n2 = n3 = 1.5 with two values of nl (Le., for nl = 0 and nl = 1) for both the non-isothermal and isothermal conditions. In order to study the combined effect of heat transfer and chemical kinetics model, the following method of solution is adopted. Equation (14.36) along with the initial and boundary conditions given by Eqs (14.37)-( 14.39) is solved by the finite difference method using a pure implicit scheme. A pure implicit scheme is an unconditionally stable scheme, that is, there is no restriction on the time step in sharp contrast with the Euler and Crank-Nicholson method as discussed by Ghoshdastidar (1998). The tri-diagonal matrix algorithm (TDMA), also known as the Thomas algorithm, is used to solve the heat-transfer model equation. In order to fix the values of Ar and A x , the results are obtained for various values of M ranging

416

Process Plant Simulation

from 26 to 176. For AT = 0.01, and values of M = 151 and M = 176, the results are found to be same. So, in order to save the computer time, M = 151, i.e., ( A x = 1/M) is used in this study. The temperature profiles obtained from the heat-transfer model based on initial conditions serve as an input to the chemical kinetics model. Equations (14.30)(14.34) are solved by the Runge-Kutta fourth-order method and the Runge-Kutta variable step-size method. It is found that the Runge-Kutta variable step size is faster than the Runge-Kutta fourth-ordermethod as discussed by Babu et al. (2002). But the Runge-Kutta variable step-size method does not give the solution for a particular and fixed interval of time. Hence, to obtain the solution at a particular interval of time, which is required to compare results of the present study with those reported in literature, the Runge-Kutta fourth-order method with a fixed step size is used in the present study.

14.4.4 Results and Discussion 14.4.4.1 Non-isothermal conditions (Babu 81Chaurasia 2003a)

The concentration profiles for a particular type [wood sawdust (0.30-0.85 mm)] of biomass with nl = 1 and n2 = n3 = 1.5 for the heating rates of 25 Ws and 60 Ws are shown in Figs 14.15 and 14.16. Temperature (K)

773

823

873

923

973

1023

1(

1.2 Legend 1 - Initial biomass 2 -Volatile 1 3-Ch1 4 - Char 2 or Volatile 2 Present study - Srivastava et al. (1996)

1.o h

2. 0.8

-6

0.6

8

0.4

2

.3

H

-

0.2 0.0

4 0

4 2

6 Time (s)

4,

8

10

12

Fig. 14.15 Concentration vs timehemperahre for pyrolysis under non-isothermal condition (HR = 25 Ws, n, = 1, n, = n, = 1.5)

Specijic-purpose Simulation and Dynamic Simulation

417

Temperature (K) 773

893

1013

1133

1254

1373

Legend 1 -Initial biomass 2 -Volatile 1 3-Charl 4 -Char 2 or Volatile 2 Present study - Srivastavaet al. (1996)

-

'

0.0

A

0

4

2

4

I

1

3

6

8

10

Time (s)

Flg. 14.1 6 Concentration vs time/ternperaturefor pyrolysis under non-isothermalcondition (HR = 60 Ws, n, = 1, n, = n3 = 1.5)

By substituting units of various quantities in Eqs (14.30)-( 14.34), the unit of pyrolysis time is found to be in seconds, but, probably by mistake, it was reported to be in minutes (Srivastava et al. 1996). Accordingly, the unit of heating rate should be in Ws and not Wmin as reported (Srivastava et al. 1996). The trends observed in figures are qualitatively same as those reported (Srivastava et al. 1996). But, quantitatively there is slight difference in the final values. It may be due to the different methods adopted for the solution and the assumptions made. Srivastava et al. (1996) used the fourth-order Runge-Kutta predictor-corrector method to simulate the model and assumed the maximum time for simulation to be 20 s, while in this study the simulation is carried out not based on stopping criteria of time of 20 s but the final dimensionless concentration of the biomass of 0.03. The final pyrolysis time, final temperature and final concentrations for the above orders of reactions and for heating rates of 40, 50 and 80 Ws obtained in this study are compared with those reported (Srivastava et al. 1996) as given in Table 14.18. It is seen from Table 14.18 that, as the heating rate increases, the final pyrolysis temperature increases, and final concentrations of volatile 1 and char 1 decrease while the final concentrations of volatile 2 and char 2 increase. The trends obtained in the final temperature and final concentrations of biomass in this study are the same as those reported (Srivastava et al. 1996). But the trend in the final pyrolysis time is not matching. It is seen from Table 14.18 that for the heating rates of 40 Ws and 50 Us,the final pyrolysis time decreases. It is noted that for 80 Ws,the final pyrolysis time increases, but Srivastava et al. (1996) reported a decreasing value. To have a clear and better understanding of this contradiction, the model is simulated for different values of heating rates with smaller step sizes in the range 50-80 Ws.

418

Process Plant Simulation

Table 14.18 Comparison of the simulated results obtained (Babu & Chaurasia 2003a) for pyrolysis under non-isothermal conditions (4with those reported by Srivastava et al. (1996) (R)(n, = 1, n, = n, = 1.5)

Final values t

T CB

cc, cGl

CC* CG*

40 Ws

P

80 Ws

50 Ws

R

9.707 9.58 1161.28 1156.2 0 0.030010 0.0002 0.001 13 1 0.692126 0.671 0.133 0.138367 0.133 0.138367

P

R

9.532 9.28 1249.6 1237.0 0 0.030005 0 0.000731 0.666 0.686190 0.135 0.141538 0.135 0.141538

P

R

10.792 9.08 1636.36 1494.6 0.030001 0 0.000168 0 0.648 0.663411 0.143 0.153210 0.153210 0.143

The final pyrolysis time versus heating rates is plotted and shown in Fig. 14.17. It is found that the final pyrolysis time first decreases with increasing the heating rate, reaches the optimum value and then increases as the heating rate is further increased. For the above orders of reactions, the optimum values of the heating rate and the final pyrolysis time are found to be 5 1 Ws and 9.53 s, respectively. As Srivastava et al. (1996) did not carry out the simulation of the model in between the heating rates of 50 Ws and 80 Ws, they could not notice this decreasing and increasing trend. In order to explain this interesting decreasing and increasing trend in the final pyrolysis time with the heating rate or temperature, the simulated data for the optimum heating rate (51 Ws)and the heating rate of 80 Ws are used as shown in Tables 14.19 and 14.20, respectively.

2

g

t=

11.4 11.2 11.0 10.8 10.6 10.4 10.2 10.0 9.8 9.6 9.4 9.2 9.0

Legend

Present study - Srivastava

et al. (1996)

Fig. 14.17 Time vs heating rate for pyrolysis under non-isothermal condition (n, = 1, n, = n, = 1.5)

Specijk-purpose Simulation and Dynamic Simulation

419

Table 14.19 Simulated results (Babu & Chaurasia 2003a) with the optimum heating rate (51 Ws) for pyrolysis under non-isothermalcondition (n,= 1, n*= n,= 1.5) t

0.000000 0.100000 1.800000 1.900000 2.000000 2.200000 2.300000 2.400000 2.500000 3.100000 3.200000 3.300000 4.000000 5.000000 6.000000 7.000000 8.000000 9.000000 9.530000

T 773.000000 778.100000 864.800000 869.900000 875.000000 885.200000 890.300000 895.400000 900.500000 93 1.100000 936.200000 941.300000 977.000000 1028.00000 1079.OOOOO 1130.00000 1181.00000 123 2.00000 1259.03000

cGl

1.oooooo 0.984444 0.666499 0.646787 0.627204 0.588554 0.569549 0.550794 0.5323 16 0.428431 0.412433 0.396842 0.2995 18 0.195522 0.126164 0.08 1744 0.053790 0.036213 0.030004

0.000000 0.013370 0.283709 0.299580 0.3 15098 0.344865 0.359027 0.372664 0.385754 0.452937 0.462433 0.471527 0.526486 0.586404 0.628144 0.655061 0.671731 0.68 1929 0.685557

0.000000 0.002186 0.042532 0.044 153 0.045522 0.047397 0.047861 0.047994 0.047793 0.04056 1 0.038653 0.036651 0.022800 0.010409 0.005079 0.002697 0.001519 0.000895 0.000699

0.000000 0.000000 0.003630 0.004740 0.006088 0.009592 0.01 1781 0.014274 0.017068 0.039036 0.04324 1 0.047490 0.075598 0.103833 0.120306 0.130249 0.136480 0.140482 0.14 1870

0.000000 0.000000 0.003630 0.004740 0.006088 0.009592 0.011781 0.014274 0.017068 0.039036 0.043241 0.047490 0.075598 0.103833 0.120306 0.130249 0.136480 0.140482 0.141870

It is seen that as pyrolysis proceeds (i) temperature increases, (ii) the concentration of biomass decreases and reaches the final value, (iii) concentration of volatile 1 increases, (iv) concentration of char 1 increases up to a certain value of time and then decreases, and (v) concentrationsof volatile 2 and char 2 increase. It may be noted that the decrease in the concentration of biomass is more for 80 Ws up to a certain time (7 s). The maximum concentration of char 1 is obtained at 2.4 and 1.9 s for heating rates of 51 and 80 Ws, respectively. After 1.9 s, the concentration of char 1 starts decreasing for the heating rate of 80 Ws, so the rate of consumption of biomass also decreases. After 7 s, the decrease in the concentration of biomass is less for 80 Ws as compared to 51 Ws, because the char 1 left for 80 Ws is less as compared to 51 Ws. So the reaction between volatile 1 and char 1 is slow for 80 Ws as compared to 51 Ws time, and hence it takes more time to reach the desired final concentration of biomass. Hence the final pyrolysis time is more for 80 Ws as compared to 5 1 Ws.The optimum values of the parameters for the above orders of reactions are shown in Table 14.21. Similar trends as shown in Figs 14.15-14.17 can be obtained for orders of reactions of nl = 0 and n2 = n3 = 1.5 for different values of heating rates. The optimum values of the parameters for the above orders of reactions are also given in Table 14.21. From Table 14.21, it is observed that pyrolysis is much faster

420

Process Plant Simulation

(optimum pyrolysis time is 2.636 s) for the case of (n, = 0, n2 = n3 = 1.5), because in this case the model equations (14.30)-( 14.32) become independent of the initial biomass concentration. However, the optimum pyrolysis time is much more (9.53 s) for the case of (n, = 1; n2 = n3 = 1.5) than the earlier case, because for this case the model equations (14.30)-( 14.32) depend on the initial biomass concentration. The optimum temperature for both the orders of reactions are found to be the same, i.e., 1059.03 K, as it is independent of orders of reactions. Table 14.20 Simulated results (Babu & Chaurasia 2003a) with the heating rate of 80 Ws for pyrolysis under non-isothermal condition (n, = 1, n2 = n, = 1.5)

0.000000 773.000000 0.100000 781.000000 1.800000 917.000000 1.900000 925.000000 2.000000 933.000000 2.200000 949.000000 2.300000 957.000000 2.400000 965.000000 3.100000 1021.00000 3.200000 1029.OOOOO 3.300000 1037.0OOOO 4.000000 1093.00000 5.000000 1173.OOOOO 6.000000 1253.00000 7.000000 1333.00000 8.000000 1413.00000 9.000000 1493.00000 10.0000001573.00000 10.7920001636.36000

1 .000000 0.984002 0.619522 0.597060 0.574944 0.531948 0.511 153 0.490875 0.365117 0.349578 0.334646 0.246239 0.160908 0.108788 0.076824 0.056744 0.043710 0.034947 0.030001

0.000000 0.013754 0.318944 0.335530 0.351302 0.380276 0.393489 0.405905 0.476451 0.484943 0.493153 0.543283 0.593360 0.623367 0.640868 0.651157 0.657332 0.661127 0.663411

0.000000 0.002244 0.044262 0.044785 0.044778 0.043187 0.041675 0.039773 0.022603 0.020413 0.018399 0.008955 0.003752 0.001831 0.000985 0.000571 0.000352 0.000228 0.000168

0.000000 0.000000 0.000000 o.ooo0oo

0.008636 0.011313 0.014488 0.022295 0.026841 0.031724 0.067914 0.072532 0.076901 0.100761 0.120990 0.133007 0.140662 0.145764 0.149303 0.151849 0.153210

0.008636 0.011313 0.014488 0.022295 0.026841 0.031724 0.067914 0.072532 0.076901 0.100761 0.120990 0.133007 0.140662 0.145764 0.149303 0.151849 0.153210

Table 14.21 Optimum parametric values for different orders of reactions for pyrolysis under non-isothermal condition (Babu & Chaurasia 2003a) Optimum parameters Heating rate (Ws) Final pyrolysis time (s) Final pyrolysis temperature (K) Final concentration of initial biomass (-) Final concentration of char 1 (-) Final concentration of volatile 1 (-) Final concentration of char 2 (-) Final concentration of volatile 2 (-)

nl = 1, nz = n3 = 1.5 51

9.53 1259.03 0.030004 0.000699 0.685557 0.141870 0.141870

n, = 0, n2 = n3 = 1.5

184.38 2.636 1259.03 0.030045 0.007357 0.662166 0.150216 0.150216

Specijk-purpose Simulation and Dynamic Simulation

421

14.4.4.2 Isothermal conditions (Babu 81Chaurasia 2003a)

The concentration profiles for isothermal conditions with orders of reactions of n1 = 1 and n2 = n3 = 1.5 for the isothermal temperatures of 873 K and 1023 K are shown in Figs 14.18 and 14.19.

Legend 1 -Initial biomass 2 -Volatile 1 3-Char1 4 -Char 2 or Volatile 2 Present study -Srivastava et al. (1996)

1.0 h

L 0.8 -

5

1

0.6

8

0.4

-

'a

0.2 0.0

-

I 12

1

2

0

4

6

a

10

Time (s)

Fig. 14.18 Concentration vs time for pyrolysis under isothermal condition ( T = 873 K, n, = 1, n2 = n3 =1.5)

The obtained trends are the same as reported (Srivastava et al. 1996).The results obtained for 773 K, 973 K, and 1073 K are compared with those reported (Srivastava et al. 1996) as shown in Table 14.22. 1.2 1 .o h

d

0.8

2

0.2

\ 3

0.0

0

3

2

4 Time (s)

6

Legend 1 - Initial biomass 2 -Volatile 1 3-Char1 4 - Char 2 or Volatile 2 Present study -Srivastava et al. (1996)

-

1

a

Fig. 14.19 Concentration vs time for pyrolysis under isothermal condition ( T = 1023 K, n, = 1, n 2 = n3= 1.5)

422

Process Plant Simulation

Table 14.22 Comparison of the simulated results obtained (P) (Babu 81Chaurasia 2003a) for pyrolysis under isothermal condition with those reported by Srivastava et al. (1996) (R) (n, = 1, n2 = n3 = 1.5) 773 K

Final values

R 23.54 0.030003 0.013130 0.709398 0.123735 0.123735

t CB CCI CGl

cc2 cG*

1073 K

973 K

20.22 0 0.012 0.690 0.122 0.122

P

R

P

R

8.545 0.030008 0.003234 0.686146 0.140306 0.140306

8.38 0 0.001 0.658 0.140 0.140

7.99 0.030006 0.001832 0.653198 0.157257 0.157257

7.88 0 0 0.625 0.158 0.158

The trend in the final pyrolysis time is the same as that reported (Srivastava et al. 1996). But when the model is simulated for a wide range of temperatures ranging from 773 K to 1773 K, it is found that the optimum value of the final pyrolysis time is 7.987 s at a temperature of 1066 K as shown in Fig. 14.20. This can be explained on similar lines as explained for the non-isothermal case. The optimum values for the above orders of reactions are given in Table 14.23. It may be noted that Srivastava et al. (1996) simulated results only in a small temperature range between 773 K to 1073 K with a step size of 100 K from 773 K to 973 K and subsequently a step size of 50 K from 973 K to 1073 K, because of which they missed out this decreasing and increasing trend.

26

-

24 22 h

v

8

F

Legend Present study Srivastava et al. (1996)

20 18 16 14 12 -

10 86 '

I

I

I

I

I

800

lo00

1200

1400

1600

1800

Heating rate (Ws)

Fig. 14.20 Time vs temperature for pyrolysis under isothermal condition (n, = 1, n2 = n3 = 1.5)

Specijic-purpose Simulation and Dynamic Simulation

423

Similarly,the concentration profiles can be obtained for the orders of reactions of nl = 0 and n2 = ng = 1.5. The optimum values of the parameters are also given in Table 14.23. The optimum temperature for both the orders of reactions is found to be the same, i.e., 1066 K as in the previous case, because it is independent of orders of reactions. It may be noted that, in this study, a wide range of temperatures is used for simulating the model equations ranging from 773 K to 1773 K with a maximum step size of 100 K initially to a very small step size of 0.1 K to ensure not to m i s s out the optimum parametric values. Table 14.23 Optimum parametric values for different orders of reactions for pyrolysis under isothermal condition (Babu & Chaurasia 2003a) Optimum parameters Final pyrolysis time (s) Final pyrolysis temperature (K) Final concentration of initial biomass (-) Final concentration of char 1 (-) Final concentration of volatile 1 (-) Final concentration of char 2 (-) Final concentration of volatile 2 (-)

n1 = 1, n2 = n3 = 1.5 7.987 1066 0.030008 0.001900 0.656206 0.155943 0.155943

n , = 0, n2 = n3 = 1.5 2.209 1066 0.030250 0.017863 0.672005 0.139941 0.139941

14.4.4.3 Experimental validation and comparison (Babu & Chaurasia 2003b) The simultaneous kinetics and heat transfer model developed was compared with the experimental data reported by Pyle and Zaror (1984), Barnford’s model used by Pyle and Zaror (1984), and the model developed by Jalan and Srivastava (1999). The model is in better agreement with the experimental data of Pyle and Zaror (1984). vpical temperature profiles are shown in Figs 14.21-14.23. Figure 14.21 shows the temperature profile as a function of time at the centre (i.e., x = 0) of the cylindrical pellet of radius 0.003 m. This was compared with profiles obtained by Jalan and Srivastava (1999) and the experimental data obtained by Pyle and Zaror (1984) at the centre of the cylindrical pellet. It was found that the model developed was in excellent agreement with the experimental data better than the agreement with the Jalan and Srivastava model (1999). Figures 14.22 and 14.23 show the temperature profiles as a function of the dimensional radial distance for the final temperatures of 643 K and 753 K, respectively. The temperature profiles obtained were in much better agreement with the experimentaldata of Pyle and Zaror (1984) when compared to the other two models.

424

Process Plant Simulation

600

1

/ -_--- -

I

1''

h

x_

B

BB

6-

500-

Legend Pyle and Zaror (experimental) Present model _--- Jalan and Srivastava model

0

400-

300-

0

50

100

150

200

250

Time (s)

Fig. 14.21 Temperature profile as a function of time at the centre of the cylindrical pellet ( R = 0.003 m, To= 303 K, T,= 643 K, x = 0) 600 0

I

Legend Pyle and Zaror (experimental) Present model ---- Jalan and Sriva$tava model -Bamford's model

300

'

I

0.0

0.2

0.4

0.6

0.8

1.o

Radial distance (-)

Fig. 14.22 Temperature profile as a function of radial distance (R=0.011 m, T0=303K, T,=643K, t=4min)

The values used in Figs 14.21-14.23 are given in Tables 14.24-14.26 also for quantitative comparison of the Babu and Chaurasia model (2003b) with the earlier models and the literature data. In all the cases, it was found that the average percentageerror and the standard deviation from experimental data were significantly less in the model developed by Babu and Chaurasia (2003b) as compared to the other models.

Specific-purpose Simulation and Dynamic Simulation

'--

300

425

I

1

---0

-

100

50

0

Pyle and Zaror (experimental) Present model Jalan and Srivastava model Bamford's model 150

200

250

Radial distance (-)

Fig. 14.23 Temperature profile as a function of radial distance ( R = 0.011 m, To = 303 K, T,= 753 K, t= 3 min) Table 14.24 Comparison of models for various stages of pyrolysis (time) at the centre of cylindrical pellet ( R = 0.003 m, To= 303 K, T, = 643 K) Time (s)

Temperature Pyle and Zaror (1984) (experimental)

0 20 40 60 80 100 150 200

303 397 493 54 1 581 609 641 648

Average percentage error Standard deviation

Babu and Chaurasia model (2003b)

Jalan and Srivastava model (1999)

303 493 552 588 610 634 640

303 387 478 533 574 602 630 639

0.75 0.0106

1.56 0.0189

400

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Process Plant Simulation

Table 14.25 Comparison of models for pyrolysis time = 4 min (R = 0.01 1 m, To = 303 K, Tf= 643 K) Temperature

X

0 0.3 0.5 0.7 0.9

Pyle and Zaror (1984) (experimental)

Babu and Chaurasia model (2003b)

Jalan and Srivastava model (1999)

Bamford model (1946)

483 493 510 525 580

482 493 5 12 537 568

469 480 499 524 556

473 485 495 520 550

0.94 0.0156

2.40 0.0305

2.55 0.0329

Average percentage error Standard deviation

Table 14.26 Comparison of models for pyrolysis time = 3 min (R = 0.01 1 m, To = 303 K, T, =753 K) Temperature

X

0 0.3 0.5 0.7 0.9

Pyle and Zaror (1984) (experimental)

Babu and Chaurasia model (2003b)

Jalan and Srivastava model (1999)

Bamford model (1946)

500 520 570 600 658

496 515 548 594 649

487 506 539 584 639

480 500 540 570 620

1.60 0.0220

3.26 0.0384

4.78 0.0540

Average percentage error Standard deviation

14.4.4.4 Convective heat-transfer coefficient (Babu 81Chaurasia 2003b)

Figure 14.24 shows the temperature profiles as a function of the radial distance at various times of progression of pyrolysis of 2 s, 10 s, 20 s, 30 s, and 40 s for the particle radius of 0.0075, considering both convection and radiation on the wall (surface of the particle). It is observed that as pyrolysis time increases, temperature increases at a given radial position. The rate of increase is high at the radial position close to the wall compared to that at the centre of the particle. In the initial stages of pyrolysis, the temperature profile is very steep near the wall (refer the temperature profile corresponding to 2 s in Fig. 14.24),and, as time progresses, the steepness

Specijk-purpose Simulation and Dynamic Simulation

427

in the temperature profile near the wall decreases. This can be explained by the fact that when the heat transfer takes place by both the mechanisms of convection and radiation from the wall surface, the resistance offered to heat transfer near the wall at initial stages of pyrolysis is very high. On the contrary, when heat transfer from the wall surface takes place only by convection and with no radiation, the resistance offered to heat transfer near the wall is not as high as in the above case, which is clearly demonstrated by plotting the simulation results as shown in Fig. 14.25.It is also observed that the increase in temperature at various radial positions at different times of progression of pyrolysis is not significant, which is obvious due to the reasons mentioned above. 800 -

( -

Legend

Convection and radiation on wall surface

2s

10s

'0°-

8

600-

e!

g

500400-

300 200 -

100-

,

I

I

428

Process Plant Simulation

to get the same extent of conversion of biomass and with lesser pyrolysis time under controlled conditions by increasing convective heat transfer at much lower operating temperatures which are much safer than at higher operating temperatures leading to combined convective and radiative heat-transfer mechanisms which are not safe. Studies have also been carried out by these authors to examine the effect of the heat-transfer coefficient and orders of reaction (Babu & Chaurasia 2002a), the effect of the particle size and temperature (Babu & Chaurasia 2002b), heat of reaction number (Babu & Chaurasia 2003c), thermal and thermodynamic properties (Babu & Chaurasia 2004a; Chaurasia & Babu 2003a; 2003b), optimization of pyrolysis of biomass (Babu & Chaurasia 2003d), dominant design variables in pyrolysis of biomass particles (Babu & Chaurasia 2004b), improvised models for pyrolysis of biomass (Babu & Chaurasia 2004c), shrinking of biomass particles (Babu & Chaurasia 2004d), etc.

-

IW

Legend

goo-

-

2s

,@J-

10s

-t 20s

3

I 3

cp

30s

600-

-40s

500-

400-

200 300

,

>""

0.2

0.0

0.4

0.6

0.8

1.0

Radial distance (-)

Fig. 14.25 Temperature profile as a function of radial distance considering convection only with no radiation on the wall ( R = 0.0075, To= 303 K, T,= 900)

,""

, 0.0

0.2

0.4

0.6

0.8

1.0

Radial distance (-1

Fig. 14.26 Temperature profile as a function of radial distance considering tenfold convection with no radiation on the wall ( R = 0.0075, To= 303 K, T,= 900)

Specific-purpose Simulation and Dynamic Simulation

429

14.4.5 Conclusions rn Based on the importance and promising nature of the pyrolysis of biomass

rn

rn

rn

rn rn

rn

rn

rn

process, the modelling and simulation has been carried out for finding out the optimum parameters of the pyrolysis process. The qualitative trends obtained for concentrations of initial biomass, volatile 1, char 1, and volatile 2 are found to be the same as those reported by earlier investigators. The small quantitative differences are attributed to the numerical methods used and the assumptions made. A wide range of the heating rate and temperature is used to simulate the model equations with small step sizes for both isothermal and nonisothermal pyrolysis process conditions with different orders of reactions. Some interesting trends have been obtained, especially with respect to the effect of the net heating rate and temperature on the final pyrolysis time. The reported literature results indicate a decrease in the final pyrolysis time as the net heating rate or temperature is increased. The range of operating conditions used for simulating the model equations is small in the case of earlier investigators’ work. But the results obtained using a wide range of operating conditions in this study show that the final pyrolysis time initially decreases and then increases as the net heating rate or temperature is increased, giving an optimum final pyrolysis time corresponding to the net heat rate or temperature. The above interesting trends are observed for both isothermal and nonisothermal conditions and also with different orders of reactions. These interesting trends that were not reported earlier could be well explained from the kinetics by systematically generating huge data upon simulating the model equations. For non-isothermal conditions with the orders of reactions of nl = 1, n2 = n3 = 1.5 and n , = 0, n2 = n3 = 1S , the optimum values of the final pyrolysis time are found to be 9.53 s and 2.636 s, respectively, while the optimum values of the heating rate are found to be 51 K/s and 184.38 Ws, respectively. The optimum values of temperature are same (1059.03 K) for both the orders of reactions. For isothermal conditions with the orders of reactions of n1 = 1, n2 = n3 = 1.5 and nl = 0, n2 = n3 = 1.5, the optimum values of the final pyrolysis time are found to be 7.987 s and 2.209 s, respectively. The optimum values of temperature are same (1066 K) for both orders of reactions. The simulated results obtained from the model developed in the present study are in excellent agreement with the experimental data of Pyle and Zaror (1984), in comparison with the mathematical model of Jalan and Srivastava (1999) and Bamford’s model (1946) used by Pyle and Zaror.

430

Process Plant Simulation

The main conclusion resulting from this study concerns the possibility of modelling pyrolysis of a single biomass particle by coupling the heattransfer equation with the pyrolysis chemical kinetics equation. The pyrolysis rate has been simulated by two parallel primary reactions and a third secondary reaction between the volatile and gaseous products and char. The secondary reactions are responsible for carbon enrichment of the final residual. The model developed in the present study is used for wide range of temperature from 303-1000 K and of the particle diameter from 0.00050.026 m. As the particle size increases, the time required for completion of pyrolysis at a certain pyrolysis temperature and the effect of secondary reactions increase. A simple model as developed in the present study with very few realistic and restrictive assumptions, combined with the thermal properties variation with temperature, can describe the overall progress of a set of processes of much complexity such as pyrolysis. The model developed can be utilized to predict the temperature and concentration profiles for different types of biomass for a wide range of particle dimensions and temperatures. The work carried out in this study is important and useful for optimal design of the biomass gasifiers, reactors, etc.

EXERCISES 14.1 Discuss the following: (a) Importance of specific-purpose simulation (SPS) (b) The reasons with justification for using SPS for ammonia converters (c) Auto-thermal process (d) Neat schematic of Casale converter (e) Description and operating procedure of Casale converter 14.2 Using the concepts of specific purpose simulation, formulate the model equations, mention the method of solution, and discuss the simulation results obtained by Yoon et al. (1978) for the coal gasification process. 14.3 A conventional distillation column contains 14 equilibrium stages, a partial condenser, and a reboiler. The feed and other operating conditions are as given. Feed conditions: The feed, which is a saturated liquid, enters the column at stage 8 with a flow rate of 100 kmolh at a temperature of 108 "C. The mole fractions of various fractions C,, C,, C,, C,, and C, are 0.03, 0.20, 0.37, 0.35, and 0.05, respectively. Operating conditions: The pressure is 17 atm and the reflux and distillaterates are 154.3kmoVs and 22.6kmolh, respectively. It is desired to calculate

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431

the temperature, vapour flow late, and composition profiles of all the components. Assume that the polynomial coefficients for vapour-liquid equilibrium constants and the enthalpies of the internal streams are available. Starting from the initially assumed temperature and vapour rate profiles, develop a MATRIX algorithm for finding the converged profiles and show the flow chart schematically. 14.4 Describe and discuss the ammonia synthesis loop in ammonia production. 14.5 Formulate the model equations, simulate the results, and discuss the methanol synthesis process. 14.6 Formulate the model equations, simulate the results, and discuss the Casale converter used in ammonia synthesis. 14.7 Discuss the coal gasification modelling and simulation. 14.8 Describe and discuss the dynamic distillation modelling and simulation. 14.9 Formulate the model equations, simulatethe results, and discuss the sulphuric acid process optimization. 14.10 Using the information given by Babu and Chaurasia (2003a), formulate a model and simulate the results for the estimation of optimum parameters in pyrolysis of biomass. 14.11 Taking the information given by Babu and Gautam (2002), formulate a model and simulate the results for scenario-integrated optimization of dynamic systems. 14.12 Using the information given by Babu andhgira (2001b), formulate a model and simulate the results for optimization of thermal cracker operation. 14.13 Taking the information given by Babu and Munawar (2001), formulate a model and simulatethe results for optimal design of shell-and-tubeheat exchangers using different strategies of differential evolution. 14.14 Using the information given by Babu and Sastry (1999), formulate a model and simulate the results for estimation of heat-transfer parameters in trickle-bed reactors. 14.15 Taking the information given by Babu and Shailesh (2000), formulate a model and simulate the results using adaptive networks (back-propagation networks of ANN) for fault diagnosis and process control.

CHAPTER 15 PROFESSIONAL SIMULATION PACKAGES

So far we have discussed the various aspects and general methodology of modelling and simulation in the earlier chapters. In this chapter, two of the most widely used professional software packages, namely, HYSYS and FLUENT, are discussed in detail. The stepwise procedures of simulations using these two packages (HYSYS and FLUENT) corresponding to four engineering problems (multicomponent absorption, multicomponent distillation, refrigeration system, and flow through a circular pipe) are demonstrated.

15.1 HYSIS-A

Professional Software Package

As mentioned earlier, there are a few general-purpose simulation software packages available for both commercial and educational purposes. Out of 'these, ASPEN Plus (of AspenTech) and HYSIS (of HyproTech) are widely used round the globe. Very recently (3 1 May 2002), HyproTech merged with AspenTech and is now part of a new engineering business unit within AspenTech that will focus on engineering and asset life-cycle solutions, including simulation, engineering, evaluation, and optimization technologies. HYSYS is a product that continually extends the bounds of process engineering software. With HYSYS, one can create rigorous steady-state and dynamic models for plant design and troubleshooting. Through the completely interactive HYSYS interface, it is easy to manipulate process variables and unit operation topology, as well as to fully customize a simulation using its OLE (object linking and embedding) extensibility capability. HYSYS has been developed with Hyprotech's overall vision of the ultimate process simulation solution in mind. The vision has led to create a product that is

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rn integrated rn intuitive and interactive rn open and extensible

15.1.1 Integrated Simulation Environment In a truly integrated simulation environment, all of the necessary applications work is performed within a common framework, eliminating the tedious trial-and-error process. Such a system has a number of advantages: rn Information is shared, rather than transferred, among applications. rn All applications use common thermodynamic models. rn All applications use common flowsheet topology. rn Users have to familiarize themselves with just one interface. rn It is possible to switch between modelling applications at any time, gaining the most complete understanding of the process. The plant life cycle might begin with building a conceptual model to determine the basic equipment requirements for a process. Based on the conceptual design, one could build a steady-state model and perform optimization to determine the most desirable operating conditions. Next, some sizing and costing calculations are done for the required equipment, and then some dynamic modelling to determine appropriate control strategies. Once the design has *becomea reality, one might perform some online modelling using actual plant data for ‘what-if’ studies, troubleshooting, or even online optimization. If a change at any stage in the design process affects the common data, the new information is available immediately to all the other applications-no manual data transfer is ever required. While this concept is easy to appreciate, delivering it in a useable manner is difficult. Developing this multi-application information-sharing software environment is realistically only possible using object-oriented design methodologies, implemented with an object-oriented programming language.

15.1.2 HYSIS Products 15.1.2.1 HYSYS.Process

Process design-HYSYS.Process provides the accuracy, speed, efficiency required for process design activities. The level of detail and the integrated utilities available in HYSYS.Process allows for skilful evaluation of design alternatives. 15.1.2.2 HYSYS.Plant

Plant design-HYSYS.Plant provides integrated steady-state and dynamic simulation capability, and offers rigorous and high-fidelity results with a very fine

434

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level of equipment geometry and performance detail. HYSYS.Plant+ provides additional detailed equipment configurations, such as actuator dynamics. 15.1.2.3 HYSYSRefinery

Refinery modelling-HYSYS.Refinery provides truly scalable refinery-wide modelling. Detailed models of the reaction processes can be combined with detailed representations of separation and heat integration systems. Each hydrocarbon stream is capable of predicting a full range of refinery properties based on a refinery assay matrix. 15.1.2.4 HYSYS.OTS

Operations training system-HYSYS.OTS provides real-time simulated training exercises that train operations personnel and help further develop their skills for performing critical process operations. Increased process understanding and procedural familiarity for operations personnel can lead to an increase in plant safety, and improvements in process performance. 15.1.2.5 HYSYSRTO

Real-time optimization-HYSYS.RT0 is a real-time optimization package that enables the optimization of plant efficiency and management of production rate changes and upsets in order to handle process constraints and maximize operating profits. 15.1.2.6 HYSYSConcept

Conceptual design application-HYSYS.Concept includes DISTIL, which integrates the distillation synthesis and residue curve map technology of Mayflower with data regression and thermodynamic database access. HYSYS.Concept also includes HX-Net, which provides the ability to use pinch technology in the design of heat-exchanger networks. Conceptual design helps enhance process understanding and can assist in the development of new and economical process schemes.

15.1.3 Intuitive and Interactive Process Modelling The role of process simulation is to improve the understanding of the process so that one can make the best process decisions. The HYSYS solution has been, and continues to be, interactive simulation. This solution has not only proven to make the most efficient use of simulation time, but also enabled complete understanding of the simulation by building the model interactively-with immediate access to results. HYSYS uses the power of object-oriented design, together with an eventdriven graphical environment, to deliver a completely interactive simulation environment where

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rn calculations begin automatically whenever new information is supplied,

and rn access to the information one needs is in no way restricted.

At any time, even as calculations are proceeding, one can access information from any location in HYSYS. As new information becomes available, each location is always instantly updated with the most current information, whether specified by the user or calculated by HYSYS.

15.1.4 Open and Extensible HYSYS Architecture The integrated simulationenvironmentand the fully object-oriented software design has paved the way for HYSYS to be fully OLE compliant, allowing for complete user customization. Through a completely transparent interface, the OLE extensibility lets the user: rn develop customized steady-state and dynamic unit operations rn specify proprietary reaction kinetic expressions rn create specialized property packages With seamless integration, new modules appear and perform like standard operations, reaction expressions or property packages within HYSYS. The automation features within HYSYS expose many of the internal objects to other OLE-compliant software such as Microsoft Excel, Microsoft Visual Basic, and Visio Corporation’sVisio. This functionalityenables one to use HYSYS applications as calculation engines for customized applications. By using industry standard OLE automation and extension, the custom simulation functionality is portable across Hyprotech software updates. The open architecture allows the user to extend his simulation functionality in response to his changing needs.

15.1.5 Stepwise Methodology of HYSYS Usage for Problems Illustration 15.1 Multicomponent absorption An absorber which contains 20 trays and operates at 60 psia charges a wet gas of composition given in Table 15.1 at 90 OF. The lean oil can be assumed to have the properties of normal octane, and at present has a maximum circulation rate of 0.905 times the wet-gas rate. An inexpensive bottleneck-removal project would increase the oil circulation rate to 1.104 times the wet-gas rate. At this expected rate, the lean oil would enter the absorber at about 90°F and contain n-C, and n-C, to the extent of about 2 and 5 mol %, respectively. Estimate the recovery of each of the gas components at the new oil rate, and calculate the corresponding product rates and compositions. Assume that the absorber stage efficiency is 20%for all components.

436

Process Plant Simulation

Table 15.1 Composition of wet gas Component

Mole fraction

c,

0.285 0.158 0.240 0.169 0.148 1.000

c2

c3

n-C, n-C5 Total

Fig. 15.1 Multicomponent absorption

Solution First let us solve this problem manually and then show how the HYSYS software package can be used to solve it. Basis: 100 mol of wet gas. Oil rate = 100 x 1.104 = 110.4 mol The amounts of each component entering the column and the initial assumptions for the various recoveries are given in Table 15.2.

Professional Simulation Packages

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Table 15.2 Amounts of each component entering the column with initial assumption for recovery Component

V,po

LN

+

xN

+

= Sx, Cl c2 c3

n-C, n-C, Oil (n-C,) Total

q, = SxjA

Total moles ( A)

Assumed

Llxl = f A

V, yN = (1 - f ) A

2.85 3.16 14.40 16.24 19.30 102.67

25.65 12.64 9.60 2.87 1.02 0

158.62

51.78

f

28.5 15.8 24.0 16.9 14.8 0

0 0 0 2.21 5.52 102.67

28.5 15.8 24.0 19.11 20.32 102.67

100.0

110.4

210.4

0 0 0 0.1156 0.2717 1.o

0.1 0.2 0.6 0.85 0.95 1.o

The residue will leave the tower at the temperature of the top stage. For the initial trial, the top stage temperature will be assumed to be 5 OF above that of the entering lean oil, or 95 OF in this example. The temperature of the exiting rich oil (also the temperature of stage 1) is calculated by the overall heat balance. Enthalpy data can be obtained from Figs B-4 and B-5 of Design of Equilibrium Stage Processes by Smith (1963). The enthalpy value of various components at 90 O F and 95 OF are given in Table 15.3. Stage efficiency = (theoreticalstages or equilibrium stages)/(actualstages) i.e., 0.2 = N120 i.e., N = 4 Table 15.3 Enthalpy of various components at 90 "F and 95 O Component

to = 90 O F

Hv 5,600 9,900 13,700 c3 n-C, 17,600 n-C, 2 1,800 Oil (n-C,) 32,400 Cl

c2

Total

(4.08 atm)

tN = 95 O F (4.08 atm)

HV(VOY0)

HV

159,600 156,000 328,100 298,400 323,000 0

5,600 9,900 13,700 17,780 22,000 32,700

1,265,500

HV ('N

F

t N + I = 90°F

YN)

144,500 125,600 132,100 5 1,000 22,500 0 475,700

4,300 6,600 7,200 8,900 10,500 15,100

0 0 0 19,600 58,100 1,547,700

1,624,700

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Process Plant Simulation

Substituting the values, we get

HL(L,x , ) = 2,414,500 i.e., the enthalpy of rich oil is 2,414,500Btd100 mol of wet gas. Let us assume t, = 135 O F , then the enthalpies of the various components are given in Table 15.4. Table 15.4 Enthalpy of various components at 135 O F

Component

HL

HL(L1 xi)

4,800 7,600 8,600 10,600 12,700 17,800

Cl c2 c3

n-C, n-C, Oil (n-C,)

13,680 24,O16 123,840 172,144 245,110 1,827,526

Total

2,406,316

Now, C H L ( L , x , ) = 2,406,316,which is nearly equal to 2,414,500.So t ,

= 135°F. The phase-rate and temperatureprofiles are estimated by the following equations:

(15.1) L,,,

v, + L, - v,

(15.2)

vo - vfl ( t l - t N + l ) t, - -

(15.3)

=

tn+, =

‘0

- vN

In Eqs (15.1)-(15.3), V , = 51.78, V,, = 100,L , =158.62, and t, = 135. The calculations are shown in Table 15.5. Table 15.5 Values of V, L, and t at various stages

Stage Wet gas (bottom) n=l n=2 n=3 n=4 Lean oil (top)

Vn 100 84.83 72.0 61.0 51.8

Ln 158.62 143.4 130.6 119.6 110.4

VJL, 0.535 0.502 0.467 0.433

v, - vn 15.2 28.0 39.0 48.2

t,

90 135 121 109 99 90

The data on VIL versus n, and t versus n from Table 15.5are plotted in Figs 15.2 and 15.3,respectively. To calculate the effective stripping factor S, first the value

Professional Simulation Packages

439

of the ratio ( N - n + 1)IN for all the components is assumed, as shown in the second column of Table 15.6. Then n is calculated for these components using the known N = 4 (third column in Table 15.6). Using n, the corresponding values of VIL and t are calculated from Figs 15.2 and 15.3, respectively (fourth and fifth columns in Table 15.6). Then K is obtained from the DePriester nomographs given in Figs B-3a and B-3b of Design of Equilibrium Stage Processes by Smith (1963), and subsequently S is calculated using K and VIL (sixth and seventh columns in Table 15.6). Table 15.6 Effective stripping factors for various components

Component

(N - n + 1)lN

n

VIL

1.o 0.9 0.8 0.7 0.6

1.o 1.4 1.8 2.2 2.6 4.0

0.535 0.522 0.508 0.495 0.482 0.433

Cl CZ c3

n-C, n-C, Oil (n-C,)

0.0

t

135 130 124 119 114 99

s = KVIL

K 43.0 10.0 3.6 1.16 0.37 0.0155

0.55 -

0.50

s

-

0.45 I

I

I

I

I

1

2

3

4

Number of stages (n)

Fig. 15.2 VIL as a function of the number of stages (n)

23.0 5.22 1.83 0.574 0.178 0.0067

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Process Plant Simulation

90

'

I

I

I

I

1

2

3

4

Number of stages ( n )

Fig. 15.3 Temperature as a function of the number of stages (n)

The equation off for this problem is given by

f=

(1 - S N ) + q s ( S N - S) (1 - S N + ' )

Substitution of the stripping factor in the above equation gives the component recoveries and product rates for trial 1 shown in Table 15.7. Table 15.7 Component recoveries and product rates for the first trial Component Cl c2 c3

n-C, n-C, Oil (n-C,) Total

f 0.0435 0.191 0.524 0.893 0.951 0.993

Ll x1

VNY N

1.24 3.02 12.6 17.1 19.3 101.9

27.3 12.8 11.4 2.0 1.o 0.7

155.2

55.2

The new overall material balances can be used to calculate a new rich oil temperature. The tN values calculated in the first trial should be used now for the temperature of the residue gas. The enthalpies of V, and LN+ are unchanged from trial 1. The enthalpy of the new V, at 99 O F (Table 15.5) is 501,300 Btu/lOOmol of wet gas as shown in Table 15.8. The corresponding temperature of rich oil is found to be 135OF, unchanged from trial 1. Using Eqs (15.1)-( 15.3) with the values V, = 55.2, V, = 100, L , = 155.2, t , = 135, the calculations are shown in Table 15.9.

,

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Table 15.8 Enthalpy of components at 99 O F

Cl c2

c3

n-C, n-C, Oil (n-C,)

-

Total

501,300

Table 15.9 Values of V, L, and t at various stages Stage

Vn

Wet gas (bottom) n=l n=2 n=3 n=4 Lean oil (top)

100 86.2 74.3 64.0

Ln 155.2 141.4 129.6 119.2 110.4

55.2

V,IL,

v, - v,

t,

13.8 25.7 35.9 44.8

90 135 121 109 99 90

0.555

0.526 0.495 0.463

Repeat the steps in trial 1 and calculate the stripping factor. The values of effective stripping factors when substituted in the desired equation off give the final component recoveries, product rates, and compositions shown in Table 15.10. For comparison purposes, the effective stripping factors obtained by the Edmister equations

+ 1) + 0.25 - 0.5 l/S = A = ,/A,(A, + 1) + 0.25 - 0.5

(15.4)

S = ,/S,(S,

(15.5)

are listed in Table 15.11 beside the Horton-Franklin values. Table 15.10 Component recoveries and product rates for the second trial

c, CZ

c3

n-C, n-C, Oil (n-C,) Total

23.9 5.43 1.91 0.601 0.188 0.00718

0.042 0.184 0.504 0.884 0.948 0.993

1.2 2.9 12.1 16.9 19.3 101.9

27.3 12.9 11.9 2.2 1.1 0.7

0.008 0.019 0.078 0.109 0.125 0.661

0.486 0.230 0.212 0.040 0.019 0.013

154.3

56.1

1.ooo

1.ooo

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Process Plant Simulation

Table 15.11 Effective stripping factors obtained by the Horton-Franklin and Edmister methods

Component

Horton-Franklin

Eq. (15.4)

Eq. (15.5)

20.5 4.56 1.60 0.482 0.15 0.007

23.8 5.4 1.9 0.595 0.20 0.0119

23.9 5.43 1.91 0.601 0.188 0.007 18

Cl c2

c3

n-C, n-C, Oil (n-C,) ~~

~~~

~~

The sample calculation for component C,by the Edmister method is done as follows:

For component C,, (Kl)iYp:i = 43 and for the first plate V , / L ,= 0.555 (from the second trial).

S, = 43 x 0.555 = 23.9 S, = S4 = K4-v4

L4

For component C,,(K4)z:i:a = 38.5 and for the fourth plate V4/L, = 0.463 (from the second trial). S4 = 38.5 x 0.463 = 17.83

S = ,/17.83(23.9+ 1) + 0.25 - 0.5

= 20.58 Now, A, = 1/S, = U23.9 = 0.04184 A, = A 4 = l/S4 = U17.83 = 0.05609

l/S = A = ,/0.04184(0.05609)+ 0.25 - 0.5

1/S = A = 0.042389

i.e., S = 23.59 Now, let us demonstrate how this problem can be solved using HYSYS in a stepwise manner.

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1. Click on the Start button, select the Programs menu, select the Hyprotech menu, and then click on the Hysys program icon (i.e., execute hysys.exe) 2. From the File menu, select New Case (or press the keys ALT+N). 3. The Simulation Basis Manager window shown in Exhibit 15.1 appears.

Exhibit 15.1

4. From the File pull-down menu, save as MulticomponentAbsorption. 5. Click on the Fluid Pkgs tab. Exhibit 15.2 appears.

Exhibit 15.2

6 . Click on the Add button to get Exhibit 15.3.

444 Process Plant Simulation

Exhiblt 15.3

7. Choose the required property package from the Property Package Selection menu. Let us choose Peng Robinson for this problem. Exhibit 15.4 appears.

Exhibit 15.4

8. Close this window; the again original Simulation Basis Manager window appears as shown in step 3 (Exhibit 15.1). In that window, click on the Components tab; Exhibit 15.5 appears.

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Exhibit 15.5

9. Highlight Component List-1 in Exhibit 15.5 and press the View button. Exhibit 15.6 appears.

Exhibit 15.6

10. Select the components one after another by clicking on the Add Pure button. Exhibit 15.7 appears for the present problem after selecting all the components.

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Exhibit 15.7

11. Close this window; the Simulation Basis Manager appears (Exhibit 15.8).

Exhibit 15.8

12. Click on the Enter Simulation Environment button. Exhibit 15.9 appears.

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Exhibit 15.9

13. Along with Exhibit 15.9, the object palette consisting of symbols of various equipment and other accessories will appear by the side of it. From this palette, drag the required unit on to Exhibit 15.9. For this problem, the absorber unit is dragged, resulting in Exhibit 15.10.

Exhibit 15.10

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14. Double-click on the absorber unit and enter the following data: Absorber Column name: Lean oil Top stage inlet: Wet gas Bottom stage inlet: Ovhd vapour outlet: Residue gas Bottom liquid outlet: Rich oil n (no. of stages): 4 Click on the Bottom Up radio button; click on the Next button and enter the following data: 60 psia (automatically converted Top stage pressure: into 413.7 kPa) 60 psia (automatically converted Bottom stage pressure: into 413.7 kPa)

Click on the Next button and then click on Done, as there is no optional data for this problem. Exhibit 15.11 appears.

Exhibit 15.11

15. Close this window; Exhibit 15.12 appears.

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Exhibit 15.12

16. Double-click on the Wet gas stream for entering the corresponding properties. Exhibit 15.13 appears.

Exhibit 15.13

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17. Enter the following data in Exhibit 15.13. Temperature: 90 O F Pressure: 60 psia Molar flow: 100 lb molh Exhibit 15.14 appears.

Exhibit 15.14

18. On the Worksheet menu, click on the Composition tab. Exhibit 15.15 appears. 19. Entering the data results in a window having options for entering the composition in mole fractions, mass fractions, liquid volume fractions, mole flows, mass flows, and liquid volume flows. In the present problem, the data are entered in mole fractions. After entering the data, press the Normalize button and click on the OK button, which gives Exhibit 15.16.

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Exhibit 15.15

Exhibit 15.16

20. Close Exhibit 15.16; this leads to Exhibit 15.17 [note that the arrow colour of the Wet gas stream has changed to dark blue (on your screen), indicating that data for the stream has already been entered].

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Exhibit 15.17

21. Double-click on the Lean oil stream to enter the corresponding data, which results in Exhibit 15.18.

Exhibit 15.1 8

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22. Enter the following data corresponding to the Lean oil stream in Exhibit 15.18. Temperature: 90 O F Pressure: 60 psia Molar flow: 110.4 lb moVh Exhibit 15.19 appears.

Exhibit 15.19

23. Move to Composition change the basis to Molar Flows, and enter the following data: 0 1bmoVh Methane: Ethane: 0 lb moVh Propane: 0 lb moVh n-butane: 2.21 1bmoVh n-pentane: 5.52 lb moVh n-octane: 102.67 1bmoVh All the above data in lb moVh are automatically converted into kg moVh. Now change the basis to Mole Fractions, Click on the Normalize button and then on OK.Exhibit 15.20 appears. 24. Close Exhibit 15.20. Exhibit 15.21 appears [note that the colour of the Lean oil stream has also changed to dark blue (on your screen), indicating that it is completely defined].

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Exhibit 15.20

Exhibit 15.21

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25. Double-click on the Absorber of Exhibit 15.21, which leads to the appearance of Exhibit 15.22 (note that the problem has not converged yet, i.e., it is unconverged).

Exhibit 15.22

26. Click on the Run button. Exhibit 15.23 appears.

Exhibit 15.23

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27. Highlight the Monitor button on the Design menu to see the iterations for convergence. In this problem, as the number of stages is less (n = 4), convergence is obtained in a few (five) iterations, as shown in Exhibit 15.24. This is the final result of this problem.

Exhibit 15.24

Illustration 15.2 Multicomponent distillation A continuous fractionating column having 17 stages with a total condenser and reboiler is to be designed to separate 27,522.2 kmoVh of a mixture of 28.19% methanol, 46.26%n-butane, and 25.55%i-butene on molar basis. A,reflux ratio of 10 mol to 1 mol of the product is to be used. The feed enters at bubble point at 1172 kPa and at the 10th stage. The overhead product rate coming out of the column is 2043.0 kmoVh. The pressure at the condenser and the reboiler is 11 15 kPa and 1216 kPa, respectively. Simulate the distillation column assuming no pressure drop across the reboiler and condenser to find the (1) conditions of the feed, distillate, and bottom streams; (2) properties of the feed, distillate, and bottom streams; (3) composition of the distillate and bottom products; (4)temperature, liquid rate, and vapour rate profile throughout the column; (5) composition profiles of all the components throughout the column; and (6) heat duties required for the condenser and reboiler, and the temperature of the condenser and reboiler. Column specifications Total number of trays: Feed tray location:

17 10th tray

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Feed flow rate: Distillate flow rate: Feed compositions: Methanol: n-butane: i-butene: Feed condition: Reflux ratio: Reboiler pressure: Condenser pressure: Pressure drop in condenser:

457

2752.2 kmollh 2043.0 kmovh 0.2819 (mole fraction) 0.4626 (mole fraction) 0.2555 (mole fraction) Saturated liquid at 1172 kPa 10 1216 kPa 1115 kPa 0 kPa

Solution 1. Open the file as a new case; Exhibit 15.25 named Simulation Basis Manager will be displayed.

Exhibit 15.25

2. Click on the Add button to select the components and fluid package to be used for the simulation of the column (Exhibit 15.26).

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Exhibit 15.26

3. Select the components and add them from the list by clicking on the Add Pure button. Here, the components selected are methanol, i-butene, and n-butane (Exhibit 15.27).

Exhibit 15.27

4. Close the window; the Simulation Basis Manager window appears. Click on Fluid Pkgs. Exhibit 15.28 is displayed.

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Exhibit 15.28

5. Click on the Add button and select the fluid package to be used for the simulation of the column (Exhibit 15.29).

Exhibit 15.29

6. Select the required fluid package from the list (here, we have selected the Peng Robinson fluid package as shown in Exhibit 15.30.

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Exhibit 15.30

7. Close the window and return to the Simulation Basis Manager (Exhibit 15.31).

Exhibit 15.31

8. Click the Enter Simulation Environment button. Exhibit 15.32 is displayed.

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Exhibit 15.32

9. Select the distillation column by double-clicking on it in the Case (main) window. Name the different streams feed, distillate, bottoms, Qcondenser,

ereboiler,

etc. Enter the value of number of trays, feed tray location, etc. and then click on the Next button. The window related to the pressure will be displayed. Enter the condenser pressure, reboiler pressure, and condenser pressure drop values and click on the Next button. The window related to the temperature at the condenser and reboiler will be displayed, which is optional. Again after clicking on the Next button, enter the value of the reflux ratio and distillate flow rate. Then click on the Done button. A window showing all the information ahout the column we have added will be displayed (Exhibit 15.33). 10. Close the window; the next window will display the distillation column with Feed, Distillate, and Bottom streams (Exhibit 15.34).

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Exhibit 15.33

Exhibit 15.34

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11. Close this Window and double-click on the Feed stream to specify the various conditions of the feed (Exhibit 15.35).

Exhibit 15.35

12. Select Composition and add the known compositions for various components in the feed. Here the molar composition of the feed is (Exhibit 15.36). Methanol: 0.28 19 n-butane: 0.4626 i-butene: 0.2555 13. Select Conditions and add the known properties of the feed. Here the feed properties are Feed flow rate: 2752.2 kmovh Vapour/phase fraction: 0.0 (as the feed is a saturated liquid) Pressure: 12 atm With this information, HYSYS will automatically simulate for the rest of the feed properties (Exhibit 15.37).

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Exhibit 15.36

Exhibit 15.37

Previous Page

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14. Close the feed window, double-click on the Distillation Column, all the specifications of the distillation column will be displayed (Exhibit 15.38). At this stage also we can change the specifications if needed. Now click on the Run button to simulate the distillation column.

Exhibit 15.38

Results

1. Conditions of the feed, distillates, and bottoms: Click on the Worksheet button, select Conditions. The next window will display the conditions of the feed, distillate, and bottom products (Exhibit 15.39). 2. Properties of the feed, distillate, and bottoms: Click on Properties, the next window will display the properties of the feed, distillate, and bottom products (Exhibit 15.40).

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Exhibit 15.39

Exhibit 15.40

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3. Composition of the distillate and bottom products: Click on Compositions, the compositions of the feed, distillate, as well as the bottom products are displayed (Exhibit 15.4 1).

Exhibit 15.41

4. Temperature, vapour, and liquid profiles throughout the column: Click on the Performance button, select Column Profiles,the window will display the temperature, vapour rate, and liquid rate in all trays inside the distillation column (Exhibit 15.42). 5. Composition profiles: Click on Plots, select Composition, and click on View Tables (or on View Graphs), the window will display the compositions of all components in each tray inside the column in the form of a table (or graph) (see Exhibits 15.43-15.45). Similarly we can view plots for liquid, vapour, and temperature profiles throughout the column after clicking on the View Graph button for the respective profiles. For example, Exhibit 15.46 shows the result of the temperature profile. 6. Boiler and condenser duties and temperatures: Click on Feeds/Products, the window will display all the loads and properties related to the condenser and reboiler (Exhibit 15.47).

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Exhibit 15.42

Exhibit 15.43

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Exhibit 15.44

Exhibit 15.45

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Exhlblt 15.46

Exhibit 15.47

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Illustration 15.3 Refrigeration system The refrigerator shown in Fig. 15.4 uses R-134a as the working fluid. The mass flow through each component is 0.1 kg/s and the power input to the compressor is 5 kW. The heat lost to the compressor is 0.21 kW. The following state data are known, using the state notation (denoted by encircled numbers) given in Fig. 15.4: P, = 100 kPa, TI = -2O"C, P2 = 800 kPa, x3 = 0.0 (vapour fraction), and T4 = -25 "C.

Expansion valve I

~

Evaporator

Fig. 15.4 Refrigeration system

State 1: Cold vapour State 2: Warm vapour State 3: Warm liquid State 4: Cold liquid + vapour Determine the (1) quality at the evaporator inlet, (2) rate of heat transfer to the evaporator, (3) mass-flow rate of cold water in the condenser and evaporator if temperature differences are from 10 to 15"C and 85 to 45 "C, and (4) pressure and temperature in the all four states. Solution 1. Click on the Hysys 3.1 icon. 2. Use ALT+F or open a New Case from the menu. The window shown in Exhibit 15.48 will open.

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Exhibit 15.48

3. Click on the Add button and select the component R-134a and water as shown in Exhibit 15.49.

Exhibit 15.49

4. Now close the above window. In the Simulation Basis Manager, press the Fluid Pkgs tab. In the Fluid Package window, select the Peng Robinson package as shown in Exhibit 15.50.

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Exhibit 15.50

5. Now close the Fluid Package: Basis1 window. 6. In the Simulation Basis Manager window, click on Enter Simulation Environment.A new window named PFD Case (Main) will open along with the Case (Main) window. 7. From the Case (Main) window, select the compressor by double-clicking on it. The window shown in Exhibit 15.51 appears.

-

Exhibit 15.51

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8. In the K-100 window, enter cold vapour, warm vapour, and work in the Inlet, Outlet, and Energy streams, respectively. 9. Click on the Worksheet tab and fill the data for cold vapour, warm vapour, and work as shown in Exhibit 15.52.

Exhibit 15.52

10. Click on Composition and fill the data shown in Exhibit 15.53.

Exhibit 15.53

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11. As can be seen, the compressor (K-100) is solved and the outlet warm vapour temperature is found to be 48.26"C. 12. Select the next component, heat exchanger, by double-clicking on it and fill the data shown in Exhibit 15.54.Here, vapours are getting condensed, and hence one has to choose the heat exchanger. If we select the cooler, then the solver will automatically assume that no phase change occurs during cooling, which is not the case in the present problem.

Exhibit 15.54

13. Click on Parameters and enter 0 Wa for AP. Click on the Worksheet tab and select Composition. Enter the fraction 1.00 for H20 and make CW, (the composition of water vapour in) and CW,,, (the composition of water vapour out) 100% H20. 14. Now click on Conditions and enter the vapour fraction 0 for warm liquid. Then enter the CW, and CW,, temperatures and its vapour fraction. The solver will calculate the mass-flow rate as shown in Exhibit 15.55. 15. The condenser has been solved and the PFD Case (Main) window will look as shown in Exhibit 15.56.

-

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Exhibit 15.55

Exhibit 15.56

16. Now select the expander by double-clicking on it and enter the stream names as shown in Exhibit 15.57.Actually, we require an expansion valve (Joule-Thompson) in this problem, but this feature is not included in the

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Case (Main) window of this software (HYSYS). However, it is possible to take care of such problems in HYSYS. Though all the components required for various problems are not directly available, with some approximations and assumptions, equivalent components can be found. In the present problem, an expander [which is available in the Case (Main) list] with the options of no work and no heat loss can be approximated by the Joule-Thompson expansion valve.

Exhibit 15.57

17. Click on the Worksheet tab and enter the data for cold mixture temperature and zero work. One can see that all the other quantities have been solved. However, there is still a warning for the liquid inlet in the expander, but we are using a Joule-Thompson valve with no work and no heat transfer instead of an expander. 18. Select the heat exchanger by double-clicking on it, which would work as an evaporator, and fill the stream names as shown in Exhibit 15.58. 19. Click on the Worksheet tab and enter the data for the cold water inlet and outlet temperatures and give its composition. The problem is solved and the results are displayed (Exhibit 15.59).

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Exhibit 15.58

Exhibit 15.59

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-

20. Finally, the PFD Case window will look as shown in Exhibit 15.60.

Exhibit 15.60

15.2 FLUENT-A Software Package for Computational Fluid Dynamics FLUENT is a state-of-the-art computer program for modelling fluid flow and heat transfer in complex geometries. It provides complete mesh flexibility, solving flow problems with unstructured meshes that can be generated about complex geometries with relative ease. Supported mesh types include two-dimensional (2D) triangular/ quadrilateral, 3D tetrahedral/hexahedral/pyramid/wedge,and mixed (hybrid) meshes. FLUENT also allows one to refine or coarsen the grid based on the flow solution. This solution-adaptive grid capability is particularly useful for accurately predicting flow fields in regions with large gradients, such as free shear layers and boundary layers. In comparison to solutions on structured or block-structured grids, this feature significantly reduces the time required to generate a ‘good’grid. Solutionadaptive refinement makes it easier to perform grid refinement studies, and reduces the computational effort required to achieve a desired level of accuracy, since mesh refinement is limited to regions where greater mesh resolution is needed.

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FLUENT is written in the C computer language and makes full use of the flexibility and power offered by the language. Consequently, true dynamic memory allocation, efficient data structures, and flexible solver control are all possible. In addition, FLUENT uses a clientherver architecture, which allows a simulation to run as separate simultaneous processes on client desktop workstations and powerful computer servers,for efficient execution,interactivecontrol, and complete flexibility of the machine or operating system. All functions required to compute a solution and display the results are accessible in FLUENT through an interactive, menu-driven interface. The user interface is written in a language called SCHEME, a dialect of LISP. An advanced user can customize and enhance the interface by writing menu macros and functions.

15.2.1 Structure of the Program The FLUENT package includes the following products: FLUENT, the solver. prePDF, the preprocessor for modelling non-premixed combustion in FLUENT. GAMBIT, the preprocessor for geometry modelling and mesh generation. TGrid, an additional preprocessor that can generate volume meshes from existing boundary meshes. filters (translators) for importing surface and volume meshes from CAD/ CAE packages such as ANSYS, I-DEAS, NASTRAN, PATRAN, and others. Figure 15.5 shows the organizational structure of these components. Note that a ‘grid’is the same as a ‘mesh’; the two words are used interchangeably here and throughout this chapter. One can create the required geometry and grid using GAMBIT. One can also use TGrid to generate a triangular, tetrahedral, or hybrid volume mesh from an existing boundary mesh (created by GAMBIT or a third-party CAD/CAE package). It is also possible to create grids for FLUENT using ANSYS (Swanson Analysis Systems, Inc.) or I-DEAS (SDRC); or MSC/ARIES, MSCPATRAN, or MSC/ NASTRAN (all from the MacNeal-Schwendler Corporation). Interfaces to other CAD/CAE packages may be made available in the future, based on customer requirements, but most CAD/CAE packages can export grids in one of the above formats. Once a grid has been read into FLUENT, all the remaining operations are performed within the solver. These include setting boundary conditions, defining fluid properties, executing the solution, refining the grid, and viewing and postprocessing the results.

Professional Simulation Packages

GAMBIT 0 Geometry set-up 0 2D/3D mesh generation

Geometry packages

\ 0

PDF files

Boundary

FLUENT 0 Mesh import and adaption 0 Physical models Boundary conditions 0 Material properties 0 Calculation 0 Postprocessing Mesh

I Boundary andor volume mesh

2D/3D mesh

Calculation of PDF look-up tables

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Mesh

TGnd 0 2D triangular mesh 0 3D tetrahedral mesh 0 2D or 3D hybrid mesh

I

Fig. 15.5 Basic program structure

Note that preBFC and GeoMesh are the names of FLUENT preprocessors that were used before the introduction of GAMBIT.

15.2.2 Capabilities of FLUENT The FLUENT solver has the following modelling capabilities: w flows in 2D or 3D geometries using unstructured solution-adaptive triangularhetrahedral, quadrilateralhexahedral, or mixed (hybrid) grids that include prisms (wedges) or pyramids. (Both conformal and hangingnode meshes are acceptable.) incompressible or compressible flows w steady-state or transient analysis w inviscid, laminar, and turbulent flows w Newtonian or nowNewtonian flow w convective heat transfer, including natural or forced convection coupled conductiodconvective heat transfer radiation heat transfer inertial (stationary) or non-inertial (rotating) reference-frame models w multiple moving reference frames, including sliding mesh interfaces and mixing planes for rotor/stator interaction modelling

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and surface deposition reaction models w arbitrary volumetric sources of heat, mass, momentum, turbulence, and

chemical species w Lagrangian trajectory calculations for a dispersed phase of particles/

w

w w w

dropletshbbles, including coupling with the continuous phase phase-change models flow through porous media lumped parameter models for fans, pumps, radiators, and heat exchangers multiphase flows, including cavitation free-surface flows with complex surface shapes

These capabilities allow FLUENT to be used for a wide variety of applications, including the following: w process and process equipment applications w power generation and oiVgas and environmental applications aerospace and turbomachinery applications w automobile applications w heat-exchanger applications w electronics/HVAC/appliances w material processing applications w architectural design and fire research Summarizing, FLUENT is ideally suited for incompressible and compressible fluid- flow simulations in complex geometries.Fluent, Inc. also offers other solvers that address different flow regimes and incorporate alternative physical models. Additional computational fluid dynamics (CFD) programs from Fluent, Inc. include Airpak, FIDAP, Icepak, MixSim, and POLYFLOW.

15.2.3 Using FLUENT-An

Overview

FLUENT uses unstructured meshes in order to reduce the amount of time required for generating meshes, simplify geometry modelling and the mesh generation process, model more complex geometries than one can handle with conventional, multiblock structured meshes, and lets the adaptation of the mesh resolve flowfield features. FLUENT can also use body-fitted, block-structured meshes (e.g., those used by FLUENT 4 and many other CFD solvers). It is capable of handling triangular and quadrilateral elements (or a combination of the two) in 2D, and tetrahedral, hexahedral, pyramid, and wedge elements (or a combination of these) in 3D. This flexibility allows one to pick mesh topologies that are best suited for a particular application.

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One can adapt all types of meshes in FLUENT in order to resolve large gradients in the flow field, but one must always generate the initial mesh (whatever the element types used) outside of the solver, using GAMBIT, TGrid, or one of the CAD systems, for which mesh import filters exist.

15.2.4 Physical Models in FLUENT FLUENT provides comprehensive modelling capabilities for a wide range of incompressible and compressible, laminar and turbulent fluid-flow problems. Steady-state or transient analyses can be performed. In FLUENT, a broad range of mathematical models for transport phenomena (such as heat transfer and chemical reactions) is combined with the ability to model complex geometries. Examples of FLUENT applications include laminar non-Newtonian flows in process equipment; conjugate heat transfer in turbomachinery and automotive engine components; pulverized coal combustion in utility boilers; external aerodynamics; flow through compressors, pumps, and fans; and multiphase flows in bubble columns and fluidized beds. To permit the modelling of fluid flow and related transport phenomena in industrial equipment and processes, FLUENT provides several useful features. These include porous media, lumped parameters (fan and heat exchanger), streamwise periodic flow and heat transfer, swirl, and moving-reference-frame models. The moving-reference-frame family of models includes the ability to model single and multiple reference frames. A time-accurate sliding mesh method, useful for modelling multiple stages in turbomachinery applications, for example, is also provided, along with the mixing plane model for computing time-averaged flow fields. Another very useful group of models in FLUENT is the set of free-surface and multiphase flow models. These can be used for analysis of gas-liquid, gas-solid, liquid-solid, and gas-liquid-solid flows. For these types of problems, FLUENT provides volume-of-fluid (VOF), mixture, and Eulerian models, as well as the discrete phase model (DPM). The DPM performs Lagrangian trajectory calculations for dispersed phases (particles, droplets, or bubbles), including coupling with the continuous phase. Examples of multiphase flows include channel flows, sprays, sedimentation, separation, and cavitation. Robust and accurate turbulence models are a vital component of the FLUENT suite of models. The turbulence models provided have a broad range of applicability, and include the effects of other physical phenomena, such as buoyancy and compressibility. Particular care has been devoted to addressing issues of near-wall accuracy via the use of extended wall functions and zonal models. Various modes of heat transfer can be modelled, including natural, forced, and mixed convection with or without conjugate heat transfer, porous media, etc. The set of radiation models and related submodels for modelling participating media is

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general and can take into account the complications of combustion. A particular strength of FLUENT is its ability to model combustion phenomena using a variety of models, including eddy dissipation and probability density function models. A host of other models that are very useful for reacting flow applications are also available, including coal and droplet combustion, surface reaction, and pollutant formation models.

15.2.5 Stepwise Methodology of FLUENT Usage for a Problem Illustration 15.4 Flow through a circular pipe Consider a fluid (air) flowing through a circular pipe at an inlet velocity of 1 m/s. The inlet temperature is 300 K and the pipe wall outlet temperature is 600 K. The length and diameter of the pipe are 50 mm and 1 mm, respectively (Fig. 15.6). Find the temperature, pressure, and velocity contours with the following assumptions: (1) the flow is steady and incompressible, and (2) the fluid viscosity and density are constant.

To= 300 K

t-

L=50mm

-i

Fig. 15.6 Flow through a circular pipe

Solution Step 1 Starting FLUENT on a Windows system Click on the Start button, select the Programs menu, select the Fluent.Inc menu, and then select the FLUENT 6 program. The opening screen should look something like that shown in Exhibit 15.61. Step 2 Choosing the solverfomulation FLUENT provides three different solver formulations: m Segregated m Coupled implicit m Coupled explicit

All three solver formulations provide accurate results for a broad range of flows, but in some cases one formulation may perform better than the others. The segregated and coupled approaches differ in the way that the continuity, momentum, and energy and species equations are solved: the segregated solver solves these equations sequentially, while the coupled solver solves them simultaneously.

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Exhibit 15.61

User inputs for solver selection:To choose one of the three solver formulations, you will use the Solver panel (Fig. 15.7). Define + Models + Solver (Exhibit 15.62). To use the segregated solver, retain the default selection of Segregated under Solver. To use the coupled implicit solver, select Coupled under Solver and Implicit (by default) under Formulation. To use the coupled explicit solver, select Coupled under Solver and Explicit under Formulation. For the present problem, the segregated 2D steady-state absolute model solver has been selected.

Exhibit 15.62

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Fig. 15.7 The solver panel

Step 3 Reading meshjles Mesh files, also known as grid files, are created with the GAMBIT grid generator. A grid file is-from FLUENT’S point of viewsimply a subset of a case file. The grid file contains the coordinates of all the nodes, and connectivity information that tells how the nodes are connected to one another forming faces and cells. GAMBIT grid files: GAMBIT is used to create 2D and 3D .structured/ unstructuredhybrid grids. It is a software package designed to help analysts and designers build and mesh models for CFD and other scientific applications. It receives user inputs by means of its graphical user interface (GU1):The GAMBIT GUI makes the basic steps of building, meshing, and assigning zone types to a model. To create any of these meshes for FLUENT, the following procedure is used: click on the exe file of GAMBIT. The opening screen should look like Exhibit 15.63. Creating the Geometry: When you click on the Geometry command button on the Operation toolpad, GAMBIT opens the Geometry subpad, which contains command buttons that allow one to create, move, copy, modify, summarize, and delete vertices, edges, faces, and volumes. It also contains a command button that allows one to perform operations involving groups of topological entities. The symbols associated with each of the Geometry subpad command sets are shown in Exhibit 15.64.

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Exhibit 15.63

Exhibit 15.64

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GAMBIT modelling forms require one to specify a location in space relative to a specified coordinate system. For example, the Create Real Vertex form shown in Fig. 15.8 requires one to specify the three coordinates describing the point at which the vertex is to be created.

Fig. 15.8 Creating a real vertex

After specifyingthe coordinatesin the Create Real Vertex form, the four vertices look as shown in Exhibit 15.65.

Exhibit 15.65

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Creating edges: Exhibit 15.66 shows the commands available in the Geometry/ Edge subpad.

Exhibit 15.66

To open the Create Straight Edge form (see Exhibit 15.67), click on Create Straight Edge to open the command button on the Geometrymdge subpad.

Exhibit 15.67

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To create an edge using the Straight Edge command, one must specify two vertices that comprise the end points of the edge. The edge starts from the first (start) vertex and ends at the second (end). For example, a straight-line edge is created between vertex 1 and vertex 4, as shown in Exhibit 15.68. All other edges are created in a similar way (Exhibit 15.69).

Exhibit 15.68

Creating a face: The Create Face From Wireframe command allows one to create a face from three or more existing edges. Exhibit 15.70 shows the commands available in the Geometrymace subpad.

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Exhibit 15.69

Exhibit 15.70

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To create a face using this command, one must specify the following parameters: edges that define the wireframe face type-real or virtual. The face is created as shown in Exhibit 15.71. Exhibit 15.72 shows the final face window.

Exhibit 15.71

Meshing the model: Upon clicking on the Mesh command button on the Operation toolpad, GAMBIT opens the Mesh subpad, which contains command buttons that allow one to perform mesh operations involving boundary layers, edges, faces, volumes, and groups. Exhibit 15.73 shows the symbols associated with each of the Mesh subpad command sets.

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Exhlblt 15.72

Exhibit 15.73

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Mesh edges: The Mesh Edges command allows one to grade or mesh any or all edges in the model. When we grade an edge, GAMBIT applies the mesh node spacing specifications but does not create mesh nodes on the edge. When we mesh an edge, GAMBIT creates mesh nodes according to the specifications. To perform a grading or meshing operation, one must specify the following parameters: edge@)to which the grading specifications apply the grading scheme mesh node spacing (number of intervals) edge meshing options. These operations are shown in Exhibit 15.74.

Exhibit 15.74

Mesh faces: The Mesh Faces command allows one to create meshes for one or more faces in the model. When we mesh a face, GAMBIT creates mesh nodes on the face according to the currently specified meshing parameters. To mesh a face, one must specify the following parameters: face(s) to be meshed meshing scheme

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530

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Srivastava,V.K. and Jalan, R.K. (1994).‘Predictionsof concentration in the pyrolysis of biomass materials-I,, Energy Conversion and Management, 35 ( 1 2), 103 1 . Srivastava, V.K., Sushil, and Jalan, R.K. (1996). ‘Predictions of concentration in the pyrolysis of biomass materials-11’ , Energy Conversion and Management, 37 (4), 473 * Stadtherr, M.A. (1979). ‘Maintainingsparsity in process design calculations’,American Institute of Chemical Engineers Journal, 25 (4), 609. Starling, K.E. (1973). Fluid Thermodynamic Properties for Light Petroleum Systems, Gulf Publishing Company, Houston, Texas. Steward, D.V. (1962). ‘On an approach to techniques for the analysis of the structure of large system of equations’, SIAM Review, 4 (4), 321. -(1965). ‘Partitioning and tearing systems of equations’, SIAM (Society for Industrial andApplied Mathematics) Journal of Numerical Analysis, B2,345. Stom, R. (1995). ‘Differential evolution design of an IIR-filter with requirements for magnitude and group delay’, International Computer Science Institute, Technical Report no. TR-95-018. Taha, H.A. (1997).Operations Research-An Introduction, Prentice-Hall of India, New Delhi. Tarjan, R. (1972). ‘Depth-first search and linear graph algorithms’, SIAM Journal of Computers, 1, 146. Tewarson, R.P. (1973).Sparse Matrices, Academic, New York. Ungar, L.H., Powell, B.A., and Kamens, S.N. (1990). ‘Adaptive networks for fault diagnosis and process control’, Computers and Chemical Engineering, 14 (4/5), 561-572. Upreti, S.R. and Deb, K. (1997). ‘Optimal design of an ammonia synthesis reactor using genetic algorithms’, Computers and Chemical Engineering, 21 (I), 87-92. Venkatasubramanian,V. and Chan, K. (1989).‘A neural network methodology for process fault diagnosis’, American Institute of Chemical Engineers Journal, 35 (12), 19932002. Venkatasubramanian, V., Chan, K., and Caruthers, J.M. (1994). ‘Computer-aided molecular design using genetic algorithms’, Computers and Chemical Engineering, 18 (9), 833-844. Venkatasubramanian, V., Vaidyanathan, R., and Yamamoto, Y. (1990). ‘Process fault detection and diagnosis using neural networks. 1 . Steady-stateprocesses’, Computers and Chemical Engineering, 14 (7), 699-712. Wang, F.S. and Cheng, W.M. (1999). ‘Simultaneous optimization of feeding rate and operation parameters for fed-batch fermentation processes’, Biotechnology Progress, 15 ( 5 ) , 949-952. Wang, F.S., Jing, C.H., andTsao, G.T. (1998).‘Fuzzy-decision-making problems of fuel ethanol production using genetically engineered yeast’, Industrial and Engineering Chemistry Research, 37 (8), 3434-3443.

References 531 Watanabe, K., Hirota, S . , Hou, L., and Himmelblau, D.M. (1994). ‘Diagnosis of multiple simultaneous fault via hierarchical artificial neural networks’,American Institute of Chemical Engineers Journal 40 (5), 893-848. Watanabe, K., Matsuura, I., Abe, M., Kubota, M., and Himmelblau, D.M. (1989). ‘Incipient fault diagnosis of chemical processes via artificial neural networks’, American Institute of Chemical Engineers Journal, 35 (1 l), 1803-1812. Wegstein, J.H. (1958). ‘Acceleratingconvergence of iterativeprocesses’, Communications of the Association for Computing Machinery, 1(6), 9-13. Wichterle, I., Linek, J., and Hala, E. (1973). Vapor-Liquid EquilibriumData Bibliography, Elsevier, Amsterdam. -(1976). Vapor-Liquid Equilibrium Data Bibliography-Supplement I, Elsevier, Amsterdam. Wilhoit, R.C. (1980). Journal of Chemical Information and Computer Science, 20,138. Yaws, C.L. (1977). Physical Properties, Chemical Engineering Publications, McGrawHill, New York. Yoneda,Y. (1975). ‘A proposal of an estimation and retrieval system EROICA for physical properties of organic compounds by chemco inputs’, in S . Fujiwaya and H.B. Mark (eds), Information Chemistry, University of Tokyo Press, Tokyo. Yoon, H., Wei, J., and Denn, M.M. (1978). ‘A model for moving-bed coal gasification reactors’, American Institute of Chemical Engineers Journal, 24,885-903. Zelinka, I. (2001). ‘Prediction and analysis of behavior of dynamical systems by means of artificial intelligence and synergetic’, PhD thesis, Department of Information Processing, Lappeenranta University of Technology, Finland. Zelinka, I. and Lampinen, J. (2000). ‘SOMA: Self-organizing migrating algorithm’, 3rd International Conference on Prediction and Nonlinear Dynamic. Nostradamus2000),Zlin, Czech Republic.

INDEX

Index Terms

Links

A acceleration criterion

348

Air Liquide

350

algorithms

8

231

257

263

272

278

288

295

301

305

313

360

363

368

380

387

394

ant colony

289

Barkley and Motard (B&M)

315

basic tearing (BTA)

319

K&S

313

321

M&H

313

323

memetic

288

scatter search

288

self-organizing migrating

290

signal flow graph (SFG) based

314

tearing

313

tri-diagonal matrix (TDMA)

405

alkylation reaction, optimization

279

ammonia

357

ammonia synthesis reactor

355

assumptions

357

constraints

358

359

370

This page has been reformatted by Knovel to provide easier navigation.

Index Terms

Links

ammonia synthesis reactor (cont.) numerical methods

363

objective functions

357

optimal design

265

problem formulation

357

Runge−Kutta fixed step size method (RKFS)

361

simulated results

361

temperature and flow rate profiles

361

amortization

367

370

357

annular flow with inner cylinder moving axially

164

ant colony optimization

289

argon

357

ASCEND

355

ASPEN Plus

355

auto-thermal ammonia synthesis reactor

355

Artificial Neural Networks (ANN)

360

432

47

aqueous two-phased systems (ATPSs)

52

back-propagation algorithm

50

back-propagation network (BPN)

50

design and training

56

input and output parameters

56

learning

48

modes of training

49

network architecture

49

network training

48

This page has been reformatted by Knovel to provide easier navigation.

Index Terms

Links

Artificial Neural Networks (ANN) (cont.) neural network model

55

thermodynamic models

55

260

B Barkley and Motard (B&M) algorithms

315

basic tearing algorithms (BTA)

319

Bessel equation of zeroth order

141

biomass

396

concentration profile

416

188

190

419

429

413

417

black box principles of mathematical modelling

46

boiling

371

Boltzmann probability distribution

258

boundary conditions of mathematical modeling

46

Box−Wilson method of optimization

226

Brandon method of optimization

226

Brent's method of optimization

253

Broyden method of optimization

253

bypassing

382

C cavitation

482

char

396

chemical reaction

181

with diffusion in a tubular reactor

181

gas absorption accompanied by

191

heat transfer in a packed bed Ractor

186

419

430

This page has been reformatted by Knovel to provide easier navigation.

Index Terms chemical system modelling

Links 20

cause-and-effect relationship of system variables

21

chemical equilibrium

27

chemical kinetic

29

component balances

23

conservation laws and rate equations

20

continuity equations

23

energy equation

24

equilibrium relations

27

fundamental laws used in modelling

22

law of mass action

30

model formulation principles

21

motion equation

24

optimum sloppiness

21

phase equilibrium

27

Raoult’s law

28

state equations

25

system variables

20

transport equations

25

chromosomes

288

coaxial cylinders and concentric spheres, flow between

167

coke formation

371

combustion

480

composition

351

compressibility

349

compression

355

CONCEPT

355

482

483

This page has been reformatted by Knovel to provide easier navigation.

Index Terms

Links

conservation

398

cooling computation techniques

1

256

260

273

286

337

169

179

evolutionary programming (EP)

287

evolutionary strategies

286

genetic programming

287

computer-aided molecular design

265

computer control and model adaptation

225

continuity equation

152

controlled systems of modelling

358

31

contraction

243

convergence

245

249

253

258

273

275

279

296

298

301

375

388

392

convergence promotion

337

acceleration criterion

348

criteria

345

direct substitution

339

dominant eigenvalue method (DEM)

346

348

general dominant eigenvalue method (GDEM)

346

Jacobian

338

liquid-liquid systems

345

Newton’s method

338

quasi-Newton methods

347

separation processes

345

Wegstein’s method

341

341

348

coordinate systems for falling film problem, choice

162

This page has been reformatted by Knovel to provide easier navigation.

Index Terms

Links

counter-current cooling of tanks

124

counter-current operations

296

cross-current solvent extraction, steadystate, two-stage model for masstransfer operations cut-set

69 315

cybernetics

31

system control

31

system optimization

31

cyclic process

318

244

254

205

434

D data regression analytical methods

208

graphical methods

207

methods

206

theoretical properties

205

deviation variables

142

decomposition of networks

296

313

371

Barkley and Motard (B&M) algorithm

315

basic tearing algorithm (BTA)

319

K&S algorithm

313

321

M&H algorithms

313

323

maximal cyclic net (MCN)

315

self-loops

315

319

324

328

signal flow graph (SFG)based algorithms

314

tearing algorithms

313

331

This page has been reformatted by Knovel to provide easier navigation.

Index Terms

Links

dehydrogenation

371

density

152

193

349

193

196

272

365

283

351

411

374

380

484 Design of Equilibrium Stage

437

deterministic versus stochastic processes

15

continuous stirred tank reactor (CSTR)

16

deterministic processes

15

differential evolution (DE)

257 389

alkylation reaction, optimization

279

applications

276

convergence

273

digital filter design

276

fermentation process

277

fuel ethanol production

277

heat exchangers

355

optimal design

278

281

Himmelblau function

280

370

implementation, stepwise procedure

283

infinite impulse response (IIR)

277

innovations

276

key parameters

275

365

376

380

391

387

MINPL problems, global optimization

283

mutation

367

nested

369

non-linear chemical engineering processes, global optimization

282

This page has been reformatted by Knovel to provide easier navigation.

Index Terms

Links

differential evolution (DE) (cont.) rand-to-best

275

recombination and selection

367

robustness

368

scaling factor (F)

365

scenario-integrated optimization of dynamic systems

281

shell-and-tube heat exchangers design

281

steps performed

273

strategies

275

377

synthesis of heat integrated distillation systems

279

thermal cracker operation

281

vector dimension

273

XOR versus add

273

differential equations

360

diffusion coefficients

351

direction set methods of optimization

254

distillation

225

multi-stage Riccati equation

381

363

98 100

divergence

341

DNG

373

dynamic lumped parameter rigid analytical model dynamic optimization

115 224

281

E Edmister equations

441

This page has been reformatted by Knovel to provide easier navigation.

Index Terms energy

Links 1

399

energy balance

296

358

359

enthalpy

349

351

437

ethane

372

ethylene

373

entropy

349

equality constraints

340

376

295

300

block triangular form (BTF)

295

309

block triangular splitting (BTS)

310

Boolean adjacency matrix

301

303

295

309

digraph

301

319

disjoining

305

Gauss-Jordan elimination

308

Guass−Seidel method

308

incidence matrix

301

maintaining sparsity

307

Newton’s method

301

numerical solutions

308

partitioning

307

path tracing method (PTM)

301

303

pivoting

308

310

powers of adjacency matrix (PAM)

303

305

precedence-ordering of equation sets

301

440

equation-solving approach to process simulation

bordered block triangular form (BBTF)

321

306

structural analysis with substitution (SWS) algorithm tearing

295

305

297

305

307

This page has been reformatted by Knovel to provide easier navigation.

Index Terms

Links

equilibrium process

343

error propagation

203

through addition

203

through subtraction

204

error, sources

204

measurement errors

204

precision errors

205

significant figure

205

Euler equation

225

361

363

404

415

483

exothermic reactions

356

expansion

243

367

370

6

296

298

432

479

extended genetic search (EGS) method of optimization

265

extraction

225

extrapolation

236

F feed gas

356

fermentation process

53

Fick’s second law of diffusion

85

flow charts flowsheet

FLOWTRAN

277

294 1

3

307

433

355

FLUENTa software package for computational fluid dynamics

294

Airpak

482

ANSYS

480

355

This page has been reformatted by Knovel to provide easier navigation.

Index Terms

Links

FLUENTa software package for (cont.) capabilities

481

CAD/CAE packages

480

Create Real Vertex

488

Create Straight Edge

489

FIDAP

482

flexibility

479

482

GAMBIT

480

486

492

494

497

GeoMesh

481

graphic user interface (GUI)

486

I−DEAS

480

Icepak

482

mesh

479

486

492

494

497

methodology for usage

484

MixSim

482

moving-reference-frame model

483

NETRAN

480

PATRAN

480

physical models

483

POLYFLOW

482

PreBFC

481

PrePDF

480

turbulence models

483

TGrid

480

using

482

169

179

fluid-flow operations, models continuity equation

152 152

creeping flow between two concentric spheres

169

This page has been reformatted by Knovel to provide easier navigation.

Index Terms

Links

fluid-flow operations, models (cont.) differential equation

167

171

flow of a film on outside of a circular tube

160

mass-flow rate

159

162

166

momentum flux

156

160

165

170

172

162

167

momentum flux for creeping flow into a slot

174

motion equation

169

172

Newtonian fluid

156

179

through a packed bed column

155

velocity component

152

155

172

174

fouling

383

free energy of reaction

349

Frobenius method

136

fuel ethanol production

277

fuel oil

373

fugacity

349

fundamental laws used in modeling

188

159

191

352

22

G gas absorption accompanied by chemical reaction

160 191

multi-stage (Kremser−Brown equation)

94

gas oil

373

gaseous products (H2, CO, CO)

396

430

GEAR

361

363

369

This page has been reformatted by Knovel to provide easier navigation.

Index Terms genetic algorithms (GAs)

Links 9

257

260

279

282

286

370

380

387

391

applications

265

coding

261

265

267

391

crossover

263

266

270

272

286

365

381

390

fitness

262

267

288

heat exchangers, automated design

267

implementation, stepwise procedure

267

269

272

mating pool

262

269

mutation

264

272

284

operations

262

randomness

263

267

272

reproduction

262

266

270

roulette-wheel selection

263

267

269

symbolic regression

288

392

395

Gibbs free energy

349

global minimum

387

284

286

288

golden section search method of optimization gradient methods of optimization

252 231

266

adaptability

231

derivative

232

vector orthogonal

231

graphite, thermal conductivity

129

gravity force

157

160

Hageni−Poiseuille equation

156

159

heat capacity

349

253

165

H

This page has been reformatted by Knovel to provide easier navigation.

Index Terms

Links

heatting conditions

413

415

isothermal

413

415

421

non-isothermal

413

415

429

278

283

434

413

heat diffusion

357

heat duty

377

heat exchangers

267

automated design

267

optimal design

278

281

shell-and-tube

281

377

heat of mixing

349

heat-transfer operations

107

281

396

479

481

483

380

383

394

area

429

423

by conduction through cylindrical pipe, steady-state

108

conductive and radiative mechanisms

401

by conduction, unsteady-state

144

convective coefficients

426

400

through extended surfaces (spine fin)

129

Fourier’s law of heat conduction

110

heat balance

109

124

138

296

heat flow

108

112

124

126

117

150

139

150

266

278

heat loss through a maturing tank unsteady-state heat loss through pipe flanges parameters in trickle bed reactors (TBR)

This page has been reformatted by Knovel to provide easier navigation.

Index Terms

Links

heat-transfer operations (cont.) systematic elimination method

127

temperature profile

114

117

138

140

thermal conductivity

110

thermal diffusivity

146

thermal equilibrium

119

in thermometer system

139

121

130

132

129

135

138

151

122

145

in tubular gas preheater, unsteadystate

108

heating

357

heats of formation

349

Helmholtz energy

349

Hessian matrix

253

138

Himmelblau function of non-linear optimization

280

HX

355

hybrid differential evolution (HDE)

278

hybridization

288

hydracarbons

371

hydrogen

357

359

HYSIS

294

355

concept

434

DISTIL

434

HyproTech

432

information access

433

integrated simulation environment

433

432

435

intuitive and interactive process modelling methodology of usage for problems

434 435

This page has been reformatted by Knovel to provide easier navigation.

Index Terms

Links

HYSIS (cont.) multicomponent absorption

435

object-oriented design

433

open and extensive architecture

435

operations training system (OTS)

434

plant

433

process

433

products

433

real-time optimization (RTO)

434

refinery modelling

434

I independent and dependent variables parameters

39

inequality constraints

340

360

iteration

155

249

347

411

1

3

10

227

340

interconnections

251

266

301

J Joule-Thompson coefficients

349

K K&S algorithm

321

Kremser-Brown equation, see gas absorption, multi-stage

L Lagrangian multipliers (with constraints)

225

This page has been reformatted by Knovel to provide easier navigation.

Index Terms

Links

Lagrangian trajectory calculations

482

laminar flow in a narrow silt

156

non-Newtonian

483

Laplace transformation

142

188

190

261

281

Lasdon’s generalized reduced-gradient method (LGRG) leakage

370 382

linear programming method of optimization

254

liquid (tar and other organics) production LP simplex

396 371

375

M M&H algorithms

323

Markov processes

225

mass balance

79

197

mass-transfer operations

61

counter-current flow

65

67

induction principles method

64

67

296

343

mixing tank, unsteady-state mass balance schematic representation

82 61

solvent extraction, steady-state, single-stage model for masstransfer operations

61

solvent extraction, steady-state, twostage model for mass-transfer operations

65

This page has been reformatted by Knovel to provide easier navigation.

Index Terms

Links

mass-transfer operations (cont.) solvent extraction, unsteady-state, single-stage model for masstransfer operations

73

steady-state N-stage counter-current solvent extraction

90

stirred tank, unsteady-state mass balance

79

unsteady-state

85

mathematical modelling classification

17

483

39

classification based on variation of independent variables

42

classification based on state of process

42

complete mathematical model

44

distributed parameters models

42

dynamic model

43

electricity transport (current)

20

energy or momentum transport (force of friction)

19

formalization

18

heat transport (heat flux)

19

lumped parameter models

42

mass transport (mass flux)

19

static model

42

theory and experimentation

18

mathematical programming dynamic

482

225 225

This page has been reformatted by Knovel to provide easier navigation.

Index Terms

Links

mathematical programming (cont.) geometric

225

linear

225

maximum number of generations (MAXGEN)

387

391

melting

351

memetic algorithms

288

methane

357

360

373

minimization

222

226

236

258

272

366

297

299

modelling fundamental laws used

254

256

10 22

modular approaches to process simulation

295

analysis versus design mode

295

exact model

300

linear model

300

numerical method

295

performance mode

295

process matrix

298

sequential modular approach

295

simultaneous modular approach

295

momentum flux

156

convective

174

for creeping flow into a slot

174

multicomponent absorption

435

N NAG library subordinate

364

370

This page has been reformatted by Knovel to provide easier navigation.

Index Terms

Links

Newton’s law of viscosity

162

Newton−Raphson solution technique

306

nitrogen

357

359

nodes

314

321

327

non-electrolytes

351

1

8

21

219

339

355

360

365

380

393

231

257

263

272

165

non-linear chemical engineering processes, global optimization

282

non-linear equations

306

O OLE (object linking embedding)

432

OPERA

355

optimization

432 algorithms

8

analytical methods

225

ant colony

289

applications

222

Brandon method

226

Brent’s method

253

Broyden method

253

computations

231

278

286

continuity equation

152

169

179

contraction

243

convergence

245

249

253

273

278

differential evolution (DE)

257

272

365

374

380

257

389 direction set methods

254

This page has been reformatted by Knovel to provide easier navigation.

Index Terms

Links

optimization (cont.) dynamic

22

281

methods

9

257

fluctuations

224

evolutionary computation (EC)

genetic algorithms (GAs)

9

257

260

279

282

286

370

380

387

391

global

280

375

387

golden section search method

252

gradient methods

231

257

266

hybrid differential evolution (HDE)

278

227

340

254

258

Lagrangian multipliers (with constraints)

225

limitations

221

linear mapping rule

261

methods

225

minimization

222

274

226

236

272 non-linear functions

280

non-traditional techniques

257

numerical

279

optimum control system

224

population-based search algorithms

288

point-by-point method

259

quasi-Newton methods

253

282

365

radial temperature profile (RTP) method

266

278

reflection process

243

251

response function

233

This page has been reformatted by Knovel to provide easier navigation.

Index Terms

Links

optimization (cont.) Rosenbrock method

252

sequential simplex method

241

simulated annealing (SA)

9

statistical

224

steepest descent method

226

257

282

233

synthesis of heat integrated distillation systems

279

thermal cracking operation

281

371

traditional techniques

221

265

optimum control system

224

osmotic coefficients

349

279

P PACER

245

parallel-disc viscometer

172

parallelism

47

perturbation

382

petroleum

355

physical modelling heat-transfer coefficients physical and thermodynamic properties

388

15 17 294

databanks

350

modularity and routing

352

sources

349

pollutant removal efficiency (PRE)

355

296

349

363

367

428

53

POLYMATH

361

Pontryagin’s maximum principle (PMP)

370

population-based search algorithms

288

This page has been reformatted by Knovel to provide easier navigation.

Index Terms

Links

precedence-ordering

301

pressure drop

377

380

383

probability

256

258

263

391

395

484

5

10

problem-solving

47

process analysis

2

convergence promotion

6

design mode

6

equilibrium and rate equations

6

inputs and outputs

6

modular-oriented simulator

7

P-V-T relationships

6

system model equations

6

process conditions

377

process constraints

434

process design and optimization steady-state simulation process modelling process synthesis

2

269

272

296

296

6

296

265 7 434 3

design decision, hierarchy

4

design heuristics

3

environmental constraints

4

processing costs

4

safety

4

process type, mathematical modelling classification based on

45

rigid or deterministic models

45

stochastic or probabilistic models

45

This page has been reformatted by Knovel to provide easier navigation.

Index Terms

Links

PROMIS

355

propane

373

pseudocode

342

381

pyrolysis of biomass

355

396

chemical kinetic model

413

convective heat-transfer coefficients

426

481

Crank-Nicholson method

404

415

experimentation validation and comparison

423

finite difference method

396

404

heat transfer

397

404

413

isothermal conditions

413

415

421

kinetic equations

397

413

l’Hôpital‘s rule

404

non-isothermal conditions

413

415

partial differential equations

396

402

procedure

396

412

thermogravimetric analysis

414

429

429

Q quasi-Newton methods of optimization

253

R radiation

427

radial temperature profile (RTP) method of optimization

266

reacting gas

358

reaction engineering

181

reactors in series-I

193

This page has been reformatted by Knovel to provide easier navigation.

Index Terms

Links

reaction engineering (cont.) reactors in series-II

197

see also chemical reaction reflection process

243

response function

233

Rosenbrock method of optimization

252

Rudd

251

306

Runge−Kutta fixed step size method (RKFS)

361

367

409

417

367

370

416

Runge−Kutta predictor−corrector method

417

Runge−Kutta variable step size method (RKVS)

363

Saccharomyces

277

scatter search algorithm

288

self-organizing migrating algorithms

290

S

sequential simplex method of optimization shell-and-tube heat exchangers, design

241 281

Bell’s method

381

Kern method

383

377

optimal heat-exchanger design (HED) problem

377

problem formulation

378

process conditions

377

387

395

signal flow graph (SFG) based algorithms

314

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Index Terms similarity

Links 32

dynamic

33

geometric

32

kinematic

32

principles of

32

simulated annealing (SA) simulation sizing

9

257

282

296

351

355

415

1

5

21

433

solid (charcoal) production

396

solubility

351

standard deviation (SD)

366

432

424

static distributed parameter rigid analytical model of heat transfer

108

statistical optimization

224

Stefan-Boltzmann constant

406

steepest descent method of optimization

226

with constraints

236

demerits

236

steward stochastic processes fluid catalytic cracking (FCC) hydroracking stoichiometry reactors

5

233

306 16 16 16 197

300

359

295

305

74

146

153

297

305

307

structural analysis with substitution (SWS) algorithm

T Taylor’s theorem tearing

313

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Index Terms

Links

tearing algorithms

313

temperature distribution in a transverse cooling fin temperature and flow rate profiles

108

135

361

370

377

356

409

413

438

440

145

258

482

486

thermal coefficients

349

thermal conductivity

351

398

thermal cracking operation

281

371

coking

372

condensation

371

constraints

374

cyclization

371

endothermic reaction

371

exothermic reaction

371

hydrogenation

371

polymerization

371

pyrolysis

372

simulated results and discussion

375

visbreaking

372

thermal diffusivity

400

thermal equilibrium

119

thermochemical properties

351

thermodynamic systems

256

258

thermodynamics

398

433

thermometer system

141

Thomas algorithm

405

122

400

see tri-diagonal matrix algorithms (TDMAs) topology

298

433

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Index Terms

Links

TORA

375

tri-diagonal matrix algorithms (TDMAs)

405

415

trickle bed reactors (TBR)

266

278

tuning parameters of differential evolution

366

V velocity

152

155

159

162

172

174

349

484

viscosity

162

165

351

484

VLE package

352

volatiles

396

413

417

419

volume

351

167

430

W wet gas

435

Z Zvmomonas mobilis

53

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