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

(6.1