Conservation Equations and Modeling of Chemical and Biochemical Processes

CONSERVATIONEQUATIONS ffND

MODELING OF CHEMICfiLfiND BIOCHEMICAL

PROCESSES

Said S. E. M. ElnashaieParag Garhyan

Auburn UniversityAuburn, Alabama, U.S.A.

M A R C E L

MARCEL DEKKER, INC. NEW YORK • BASEL

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Preface

We would like readers—instructors and students—to read this preface care-fully before using this book. This preface is classified into three parts:

1. Background and Basic Ideas explains the fundamentals of usinga system approach as a more advanced approach to teachingchemical engineering. It also discusses very briefly how thisapproach allows compacting the contents of many chemical engi-neering subjects and relates them with one another in a systema-tic and easy-to-learn manner. More details on this aspect of thebook are given in Chapter 1.

2. Review of Chapters and Appendices briefly describes the contentsof each chapter and the educational philosophy behind choosingthese materials.

3. Relation of the Book Contents to Existing Chemical Engineering

Courses shows how this book can be used to cover a number ofcourses in an integrated manner that unfortunately is missing inmany curricula today. The relation of the contents of the bookto existing courses is discussed. Although our frame of referenceis the curricula of the Chemical Engineering Department atAuburn University, the discussion can be applied to many curri-cula worldwide.

1. BACKGROUND AND BASIC IDEAS

We have adopted a novel approach in the preparation of this rather revolu-tionary undergraduate-level chemical engineering textbook. It is based onthe use of system theory in developing mathematical models (rigorous designequations) for different chemical and biochemical systems. After a briefintroduction to system theory and its applications, the book uses the gen-eralized modular conservation equations (material and energy balances) asthe starting point.

This book takes as its basis the vision of chemical engineering trans-formed, as expressed in the Amundson report of 1989, in which areas new tothe traditional subject matter of the discipline are explored. These new areasinclude biotechnology and biomedicine, electronic materials and polymers,the environment, and computer-aided process engineering, and encompasswhat has been labeled the BIN—Bio, Info, Nano—revolution. The bookaddresses these issues in a novel and imaginative way and at a level thatmakes it suitable for undergraduate courses in chemical engineering.

This book addresses one of the most important subjects in chemicalengineering—modeling and conservation equations. These constitute thebasis of any successful understanding, analysis, design, operation, and opti-mization of chemical and biochemical processes. The novel system approachused incorporates a unified and systematic way of addressing the subject,thus streamlining this difficult subject into easy-to-follow enjoyable reading.

By adopting a system approach, the book deals with a wide rangeof subjects normally covered in a number of separate courses—mass andenergy balances, transport phenomena, chemical reaction engineering,mathematical modeling, and process control. Students are thus enabled toaddress problems concerning physical systems, chemical reactors, and bio-chemical processes (in which microbial growth and enzymes play key roles).

We strongly believe that this volume strikes the right balance betweenfundamentals and applications and fills a gap in the literature in a uniqueway. It efficiently transmits the information to the reader in a systematic andcompact manner. The modular mass/energy balance equations are formu-lated, used, and then transformed into the design equations for a variety ofsystems in a simple and systematic manner.

In a readily understandable way, this book relates a wide spectrum ofsubjects starting with material and energy balances and ending with processdynamics and control, with all the stages between. The unique systemapproach shows that moving from generalized material and energy balanceequations to generalized design equations is quite simple for both lumpedand distributed systems. The same has been applied to homogeneous andheterogeneous systems and to reacting and nonreacting systems as well as to

steady- and unsteady-state systems. This leads the reader gracefully andwith great ease from lumped to distributed systems, from homogeneous toheterogeneous systems, from reacting to nonreacting systems, and fromsteady-state to unsteady-state systems.

Although steady-state systems are treated, we have provided enoughcoverage of transient phenomena and unsteady-state modeling for studentsto appreciate the importance of dynamic systems. While the early part of thebook is restricted to homogeneous systems, a later chapter introduces anovel systems approach and presents, in an easy-to-understand manner,the modeling of heterogeneous systems for both steady-state andunsteady-state conditions, together with a number of practical examples.

Chemical and biochemical units with multiple-input multiple-output(MIMO) and with multiple reactions (MRs) for all of the above-mentionedsystems are also covered. Nonreacting systems and single-input single-out-put (SISO) systems are treated as special cases of the more general MIMO,MR cases. The systems approach helps to establish a solid platform onwhich to formulate and use these generalized models and their special cases.

As the book covers both steady- and unsteady-state situations, itlogically includes a chapter on process dynamics and control that is anexcellent introduction to a more advanced treatment of this topic, withspecial emphasis on the industrially more relevant digital control systemsdesign.

Given that all chemical/biochemical engineering processes and systemsare highly nonlinear by nature, the book discusses this nonlinear behavior insome detail. All the necessary analytical and numerical tools required areincluded. Matrix techniques are also covered for large-dimensional systemsthat are common in chemical/biochemical engineering. The book alsocovers, in a manner that is clear and easy to understand for undergraduatechemical engineers, advanced topics such as multiplicity, bifurcation, andchaos to further broaden the student’s perspective. It is increasingly impor-tant for undergraduate students to think outside the conventional realm ofchemical engineering, and we have shown that these phenomena are relevantto many important chemical/biochemical industrial systems. It is also shownthat these phenomena cannot be neglected while designing these systems ortheir control loops. In the past these subjects—multiplicity, bifurcation, andchaos—have tended to be relegated to advanced research treatises. We treatthem here in a manner that undergraduate students can understand andappreciate.

In our fast-changing world the chemical/biochemical industry is alsorapidly changing. Today’s chemical/biochemical engineering graduatesshould be exposed to training in creativity as applied to these systems.Therefore a chapter on novel configurations and modes of operations for

two important processes is presented in the form of detailed exercises. Thisimportant chapter requires a special effort from the instructor to make it theexercise on creativity that it is meant to be.

2. REVIEW OF CHAPTERS AND APPENDICES

This book presents a unified approach to the analysis of a wide range ofchemical and biochemical systems. It begins with a summary of the funda-mental principles governing thermodynamics and material and energy bal-ances and proceeds to consider the mathematical modeling of a range ofsystems from homogeneous steady state to heterogeneous unsteady state. Anovel feature is the inclusion of the concepts surrounding chaotic systems atundergraduate level—an area of growing importance but one sadlyneglected in most texts of this kind. The last chapter deals with two indus-trial processes—reforming and fermentation—in which the foregoing prin-ciples are applied and illustrated for novel configurations and modes ofoperation. The useful appendices deal with many of the mathematical tech-niques such as matrix algebra, numerical methods, and the Laplace trans-form that are utilized in the book.

Chapter 1: System Theory and Chemical/BiochemicalEngineering Systems

This chapter, one of the most important, introduces the main components ofthe philosophy governing the entire book. It covers in a simple manner themain ideas regarding system theory and its application to chemical andbiochemical systems. These systems are classified according to the principlesof system theory, and this more novel classification is related to the moreclassical classifications. This chapter also covers the main differencesbetween material and energy balances (inventory) and design equations,the concepts of rate processes together with their relation to state variables,and the general modeling of processes. The thermodynamic limitation ofrate processes in relation to modeling and simulation is examined. A briefdiscussion of the new approach adopted in this book in connection withrecent advances in the profession based on the Amundson report is alsopresented.

Chapter 2: Material and Energy Balances

This chapter addresses materials and energy balances for reacting (single aswell as multiple reactions) and nonreacting systems in a compact way. Italso covers SISO as well as MIMO systems. A generalized material andenergy balance equation for a MIMO system with MRs is rigorously devel-

oped. All other cases can be easily considered as special cases of this generalcase. A large number of solved illustrative examples are provided, andunsolved problems are given as exercises at the end of the chapter.Chapter 2 is sufficient for a solid course on material and energy balances.The modular system approach used in this chapter ultimately requires thereader to know only two generalized equations (material and energy bal-ances), with all other cases being special cases. This approach makes thesubject easy to comprehend and utilize in a short time, and will also proveextremely useful in preparing the reader to modify these equations intodesign equations (mathematical models).

Chapter 3: Mathematical Modeling (I): HomogeneousLumped Systems

This chapter covers in an easy and straightforward manner the transforma-tion of the material and energy balance equations to design equations(mathematical models). It explores closed, isolated, and open lumped homo-geneous systems. Steady-state as well as unsteady-state models are devel-oped and solved for both isothermal and nonisothermal systems. Bothchemical and biochemical systems are addressed. Again, generalized designequations are developed with all other cases treated as special cases of thegeneral one. This approach helps to achieve a high degree of efficiencyregarding rational transformation of knowledge in a concise and clear man-ner. We concentrate our efforts on reacting systems for two reasons: the firstis that for homogeneous systems the nonreacting systems are rather trivial,and the other is that the nonreacting system can be considered a special caseof reacting systems when the rates of reactions are set equal to zero. A goodnumber of solved and unsolved problems are given in this chapter.

Chapter 4: Mathematical Modeling (II): HomogeneousDistributed Systems and Unsteady-StateBehavior

This chapter covers the transformation of the material and energy balanceequations to design equations (mathematical models) for distributed sys-tems. Steady-state as well an unsteady-state models are developed andsolved for both isothermal and nonisothermal systems. Again, generalizeddesign equations are developed with all other cases treated as special cases ofthe general one, and this approach facilitates efficient transformation ofknowledge. We concentrate on reacting systems for the same reasons pre-viously discussed. Chapter 4 gives detailed coverage of the mathematicalmodeling and analytical as well as numerical solution of the axial dispersion

model for tubular reactors as an illustrative example for diffusion/reactionhomogeneous systems. The same example is extended to provide thesolution of the two-point boundary value differential equations and itsassociated numerical instability problems for nonlinear systems. Severalunsolved problems are provided at the end of this chapter.

Together, Chapters 3 and 4 provide systematic, easy-to-understandcoverage of all types of homogeneous models, both lumped/distributedand isothermal/nonisothermal systems. Both chapters can also be used asthe necessary materials for a thorough course on chemical reaction engineer-ing based on a well-organized approach utilizing system theory.

Chapter 5: Process Dynamics and Control

In the last 20 years, digital control has completely replaced analog control inindustry and even in experimental setups. It is our strong belief that theclassic complete course on analog control is no longer necessary. Controlcourses should be directed mainly toward digital control systems, which arebeyond the scope of this book. It is useful, however, for readers to have abasic background in analog control (only one well-chosen chapter, notnecessarily an entire course) to prepare them for a next course on digitalcontrol. Chapter 5 aims to do this by introducing the basic principles ofprocess dynamics and classical control, including the various forms of pro-cess dynamic models formulation, basic process control concepts, the use ofLaplace transformation and its utilization, the transfer function concepts,ideal forcing functions, block diagram algebra, components of the controlloop, and a limited number of simple techniques for choosing the controlconstants for PID controllers. All these important concepts are supplemen-ted with useful solved examples and unsolved problems.

Chapter 6: Heterogeneous Systems

Most chemical and biochemical systems are heterogeneous (formed ofmore than one phase). The modular system approach we adopt in thisbook makes the development of material and energy balances, as well asdesign equations for heterogeneous systems quite straightforward.Heterogeneous systems are treated as just a number of homogeneous sys-tems (each representing one phase), and these systems are connected toeach other through material and energy exchange. This approach provesto be not only rigorous and general but also easy to comprehend andapply to any heterogeneous system utilizing all the knowledge and experi-ence gained by the reader through the previous chapters on homogeneoussystems. Chapter 6 introduces these concepts and develops generalizedmaterial and energy balance equations as well as design equations for all

types of systems—isothermal/nonisothermal, lumped/distributed, andsteady-/unsteady-state. A number of chemical and biochemical examplesof varying degrees of complexity and unsolved problems are presented forbetter understanding of the concepts.

Chapter 7: Practical Relevance of Bifurcation, Instability,and Chaos in Chemical and BiochemicalSystems

This chapter covers the basic principles of multiplicity, bifurcation, andchaotic behavior. The industrial and practical relevance of thesephenomena is also explained, with reference to a number of importantindustrial processes. Chapter 7 covers the main sources of these phenom-ena for both isothermal and nonisothermal systems in a rather pragmaticmanner and with a minimum of mathematics. One of the authors haspublished a more detailed book on the subject (S. S. E. H. Elnashaieand S. S. Elshishini, Dynamic Modelling, Bifurcation and ChaoticBehavior of Gas-Solid Catalytic Reactors, Gordon & Breach, London,1996); interested readers should consult this reference and the other refer-ences given at the end of Chapter 7 to further broaden their understandingof these phenomena.

Chapter 8: Novel Designs for Industrial Chemical/Biochemical Systems

As discussed in the introduction of this preface, it is now important todevelop creative talents in chemical engineers. Chapter 8 aims to do thisby offering two examples of novel processes: one for the efficient productionof the ultraclean fuel hydrogen and the other for the production of the cleanfuel ethanol through the biochemical path of utilizing lingo-cellulosicwastes. Readers can expect to use the tools provided earlier in this bookin order to develop these novel processes and modes of operation withoutthe need of the expensive pilot plant stage.

Appendices

Although it is difficult to make a book completely comprehensive, we triedto make this one as self-contained as possible. The six appendices cover anumber of the critical mathematical tools used in the book. Also included isa short survey of essential available software packages and programmingenvironments. These appendices include analytical as well as numerical toolsfor the handling and solution of the different types of design equations,

including linear and nonlinear algebraic and ordinary differential and partialdifferential equations.

3. RELATION OF THE BOOK CONTENTS TOEXISTING CHEMICAL ENGINEERINGCOURSES

Chapters 1 and 7 should always be included in any usage of this book.Chapter 2 can be used for a course on material and energy balance

(CHEN 2100, Principles of Chemical Engineering, which covers the appli-cation of multicomponent material and energy balances to chemical pro-cesses involving phase changes and chemical reactions).

Chapter 3 can be used as the basis for CHEN 3650, ChemicalEngineering Analysis (which covers mathematical modeling and analytical,numerical, and statistical analysis of chemical processes). Statistical processcontrol (SPC) is not, of course, covered in this book and the course readingshould be supplemented by another book on SPC (e.g., Amitava Mitra,Fundamentals of Quality Control and Improvement, Prentice Hall, NewYork, 1998).

Chapters 4, 5, 6, and 8 are suitable for a senior class on modeling ofdistributed systems and process dynamics and control (CHEN 4160, ProcessDynamics and Control, which covers steady-state and dynamic modeling ofhomogeneous and heterogeneous distributed chemical processes, feedbacksystems, and analog controller tuning and design) prior to the course ondigital control (CHEN 6170, Digital Process Control).

Chapters 3 and 4 and the first part of Chapter 8 can be used for anundergraduate course on chemical reaction engineering (CHEN 3700,Chemical Reaction Engineering, which covers design of chemical reactorsfor isothermal and nonisothermal homogeneous reaction systems).

Acknowledgments

I would like to express my appreciation and thanks to many colleagues andfriends who contributed directly and indirectly to the successful completionof this book, namely: Professor Robert Chambers, the head of the ChemicalEngineering Department at Auburn University, and Professors MahmoudEl-Halwagi and Chris Roberts of the same department. I also appreciate thesupport I received from Professor Nabil Esmail (Dean of Engineering,Concordia University, Montreal, Canada), Professor John Yates(Chairman of the Chemical and Biochemical Engineering Department,University College, London), Professor John Grace (University of BritishColumbia, Canada), and Professor Gilbert Froment (Texas A & M

University). I also thank Professor A. A. Adesina (University of New SouthWales, Australia) and Professor N. Elkadah (University of Alabama,Tuscaloosa).

Last but not least, I express my love and appreciation for the extensivesupport and love I receive from my wife, Professor Shadia Elshishini (CairoUniversity, Egypt), my daughter Gihan, and my son Hisham.

Said Elnashaie

I would like to express my sincere thanks to Dr. Said Elnashaie for givingme the opportunity to work with him as his graduate student and lateroffering me the chance to be the coauthor of this book. I express my gra-titude to my grandfather, Shri H. P. Gaddhyan, an entrepreneur fromChirkunda (a small township in India) for always being an inspiration tome. Without the motivation, encouragement, and support of my parentsSmt. Savita and Shri Om Prakash Gaddhyan, I would have not been ableto complete this book. A special note of thanks goes to my brother Anuragand his wife Jaishree. Finally, I express my love and thanks to my wifeSangeeta for her delicious food, endurance, and help; her smile alwayscheered me up and provided the impetus to continue when I was busyworking on the manuscript.

Parag Garhyan

Administrator

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Contents

Preface

1 System Theory and Chemical/Biochemical Engineering Systems

1.1 System Theory1.1.1 What Is a System?1.1.2 Boundaries of System

1.2 Steady State, Unsteady State, and ThermodynamicEquilibrium1.2.1 The State of the System1.2.2 Input Variables1.2.3 Initial Conditions

1.3 Modeling of Systems1.3.1 Elementary Procedure for Model Building1.3.2 Solution of the Model Equations1.3.3 Model Verification

1.4 Fundamental Laws Governing the Processes in Termsof the State Variables1.4.1 Continuity Equations for Open Systems1.4.2 Diffusion of Mass (Transport Law)

1.4.3 Energy Equation (Conservation of Energy, FirstLaw of Thermodynamics for an Open System)

1.4.4 Equations of Motion1.4.5 Equations of State1.4.6 Rate of Reaction1.4.7 Thermodynamic Equilibrium

1.5 Different Classifications of Physical Models1.6 The Story of Chemical Engineering in Relation to

System Theory and Mathematical Modeling1.7 The Present Status of Chemical Industry and

Undergraduate Chemical Engineering Education1.8 System Theory and the Mathematical Modeling

Approach Used in This Book1.8.1 Systems and Mathematical Models1.8.2 Mathematical Model Building: General Concepts1.8.3 Outline of the Procedure for Model Building

1.9 Modeling and Simulation in Chemical Engineering1.10 Amundson Report and the Need for Modern Chemical

Engineering Education1.11 System Theory and Mathematical Modeling as Tools for

More Efficient Undergraduate Chemical EngineeringEducation

1.12 Summary of the Main Topics in this Chapter1.12.1 Different Types of Systems and Their Main

Characteristics1.12.2 What Are Models and What Is the Difference

Between Models and Design Equations?1.12.3 Summary of Numerical and Analytical Solution

Techniques for Different Types of ModelReferencesProblem

2 Material and Energy Balances

2.1 Material and Energy Balances2.1.1 A Simple, Systematic, and Generalized Approach2.1.2 Development of Material Balance Relations

2.2 Single and Multiple Reactions: Conversion, Yield, andSelectivity2.2.1 Single Reactions2.2.2 Degrees-of-Freedom Analysis

2.2.3 Relations Among Rate of Reaction, Conversion,and Yield

2.3 Generalized Material Balance2.3.1 Sign Convention for the Stoichiometric Numbers2.3.2 The Limiting Component2.3.3 Reactions with Different Stoichiometric Numbers

for the Reactants2.3.4 Multiple Reactions and the Special Case of

Single Reaction2.3.5 The Algebra of Multiple Reactions (Linear

Dependence and Linear Independence of MultipleReactions)

2.3.6 The Most General Mass Balance Equation(Multiple-Input, Multiple-Output, and MultipleReactions)

2.4 Solved Problems for Mass Balance2.5 Heat Effects

2.5.1 Heats of Reactions2.5.2 Effects of Temperature, Pressure, and Phases on

Heat of Reaction2.5.3 Heats of Formation and Heats of Reaction2.5.4 Heats of Combustion and Heats of Reaction

2.6 Overall Heat Balance with Single and Multiple ChemicalReactions2.6.1 Heat Balance for Multiple Reactions and the

Special Case of a Single Reaction2.6.2 The Most General Heat Balance Equation

(for Multiple Reactions and Multiple-Input andMultiple-Output System Reactor with MultipleReactions)

2.7 Solved Problems for Energy BalanceReferenceProblems

3 Mathematical Modeling (I): Homogeneous Lumped Systems

3.1 Mathematical Modeling of Homogeneous LumpedProcesses3.1.1 Basic Concepts for the Mathematical Modeling

of Processes3.1.2 Systems and Mathematical Models

3.1.3 What Are Mathematical Models and Why DoWe Need Them?

3.1.4 Empirical (Black Box) and Physical(Mathematical) Models

3.2 Mathematical Model Building: General Concepts3.2.1 Classification of Models3.2.2 Difference Between Modeling and Simulation3.2.3 Design Equations and Mathematical Models3.2.4 Simplified Pseudohomogeneous Models Versus

Rigorous Heterogeneous Models3.2.5 Steady-State Models Versus Dynamic Models3.2.6 A Simple Feedback Control Example

3.3 Generic and Customized Models3.3.1 Practical Uses of Different Types of Models3.3.2 Steady-State Models3.3.3 Dynamic Models3.3.4 Measures for the Reliability of Models and

Model Verification3.4 Economic Benefits of Using High-Fidelity Customized

Models3.4.1 Design and Operation3.4.2 Control

3.5 Incorporation of Rigorous Models into FlowsheetSimulators and Putting Mathematical Models intoUser-Friendly Software Packages

3.6 From Material and Energy Balances to Steady-StateDesign Equations (Steady-State Mathematical Models)3.6.1 Generalized Mass Balance Equation3.6.2 Isothermal Reactors (Temperature Is Constant)3.6.3 Nonisothermal Reactors

3.7 Simple Examples for the General Equations3.8 Modeling of Biochemical Systems

3.8.1 Modeling of Enzyme Systems3.8.2 Modeling of Microbial Systems

ReferencesProblems

4 Mathematical Modeling (II): Homogeneous Distributed

Systems and Unsteady-State Behavior

4.1 Modeling of Distributed Systems4.1.1 Isothermal Distributed Systems

4.1.2 Nonisothermal Distributed Systems4.1.3 Batch Systems (Distributed in Time)

4.2 The Unsteady-State Terms in Homogeneous andHeterogeneous Systems4.2.1 Lumped Systems4.2.2 Distributed Systems4.2.3 Nonisothermal Systems

4.3 The Axial Dispersion Model4.3.1 Formulation and Solution Strategy for the

Axial Dispersion Model4.3.2 Solution of the Two-Point Boundary-Value

Differential Equations and NumericalInstability Problems

Problems

5 Process Dynamics and Control

5.1 Various Forms of Process Dynamic Models5.2 Formulation of Process Dynamic Models

5.2.1 The General Conservation Principles5.2.2 Conservation of Mass, Momentum, and Energy5.2.3 Constitutive Equations5.2.4 The Laplace Transform Domain Models5.2.5 The Frequency-Response Models5.2.6 Discrete Time Models5.2.7 SISO and MIMO State-Space Models5.2.8 SISO and MIMO Transform Domain Models5.2.9 SISO and MIMO Frequency-Response Models

5.2.10 SISO and MIMO Discrete Time Models5.3 State-Space and Transfer Domain Models5.4 Introductory Process Control Concepts

5.4.1 Definitions5.4.2 Introductory Concepts of Process Control5.4.3 Variables of a Process5.4.4 Control Systems and Their Possible

Configurations5.4.5 Overview of Control Systems Design

5.5 Process Dynamics and Mathematical Tools5.5.1 Tools of Dynamic Models

5.6 The Laplace Transformation5.6.1 Some Typical Laplace Transforms5.6.2 The Inverse Laplace Transform

5.6.3 The Transform of Derivatives5.6.4 Shift Properties of the Laplace Transform5.6.5 The Initial- and Final-Value Theorems5.6.6 Use of Laplace Transformation for the Solution

of Differential Equations5.6.7 Main Process Control Applications of Laplace

and Inverse Transformations5.7 Characteristics of Ideal Forcing Functions5.8 Basic Principles of Block Diagrams, Control Loops, and

Types of Classical Control5.9 Linearization

5.10 Second-Order Systems5.10.1 Overdamped, Critically Damped, and

Underdamped Responses5.10.2 Some Details Regarding the Underdamped

Response5.11 Components of Feedback Control Loops5.12 Block Diagram Algebra

5.12.1 Typical Feedback Control Loop and theTransfer Functions

5.12.2 Algebraic Manipulation of the Loop TransferFunctions

5.12.3 Block Diagram and Transfer Functions5.13 Some Techniques for Choosing the Controller Settings

5.13.1 Choosing the Controller Settings5.13.2 Criteria for Choosing the Controller Settings

from the Time Response5.13.3 Cohen and Coon Process Reaction Curve

MethodSolved ExamplesProblems

6 Heterogeneous Systems

6.1 Material Balance for Heterogeneous Systems6.1.1 Generalized Mass Balance Equations6.1.2 Two-Phase Systems6.1.3 The Equilibrium Case6.1.4 Stage Efficiency6.1.5 Generalized Mass Balance for Two-Phase

Systems

6.2 Design Equations (Steady-State Models) for Isothermal,Heterogeneous Lumped Systems

6.3 Design Equations (Steady-State Models) for Isothermal,Distributed Heterogeneous Systems

6.4 Nonisothermal Heterogeneous Systems6.4.1 Lumped Heterogeneous Systems6.4.2 Distributed Systems6.4.3 Dynamic Terms for Heterogeneous Systems

6.5 Examples of Heterogeneous Systems6.5.1 Absorption Column (High-Dimensional Lumped,

Steady-State, and Equilibrium Stages System)6.5.2 Packed-Bed Absorption Tower6.5.3 Diffusion and Reaction in a Porous Structure

(Porous Catalyst Pellet)6.6 Dynamic Cases

6.6.1 The Multitray Absorption Tower6.6.2 Dynamic Model for the Catalyst Pellet

6.7 Mathematical Modeling and Simulation ofFluidized-Bed Reactors6.7.1 Advantages of Freely Bubbling Fluidized Beds6.7.2 Disadvantages of Fluidized Beds6.7.3 Mathematical Formulation (Steady State)

6.8 Unsteady-State Behavior of Heterogeneous Systems:Application to Fluidized-Bed Catalytic Reactors

6.9 Example: Simulation of a Bubbling Fluidized-BedCatalytic Reactor

6.10 A Distributed Parameter Diffusion-Reaction Modelfor the Alcoholic Fermentation Process6.10.1 Background on the Problems Associated with

the Heterogeneous Modeling of AlcoholicFermentation Processes

6.10.2 Development of the Model6.10.3 Solution Algorithm6.10.4 Comparison Between the Model and

Experimental/Industrial DataReferencesProblems

7 Practical Relevance of Bifurcation, Instability, and Chaos in

Chemical and Biochemical Systems

7.1 Sources of Multiplicity

7.1.1 Isothermal Multiplicity (or ConcentrationMultiplicity)

7.1.2 Thermal Multiplicity7.1.3 Multiplicity Due to the Reactor Configuration

7.2 Simple Quantitative Discussion of the MultiplicityPhenomenon

7.3 Bifurcation and Stability7.3.1 Steady-State Analysis7.3.2 Dynamic Analysis7.3.3 Chaotic Behavior

References

8 Novel Designs for Industrial Chemical/Biochemical Systems

8.1 Novel Reforming Process for the Efficient Productionof the Ultraclean Fuel Hydrogen from Hydrocarbonsand Waste Materials8.1.1 Introduction8.1.2 Literature Review8.1.3 Limitations of Current Reforming Technologies8.1.4 Main Characteristics of the Suggested Novel

Ultraclean/Efficient Reforming ProcessConfiguration

8.1.5 Components of the Suggested Novel UltracleanProcess for the Production of the UltracleanFuel Hydrogen

8.1.6 Main Tasks for the Exercise8.2 A Novel Fermentor

8.2.1 Introduction8.2.2 Basic Research Description8.2.3 Tasks for the Exercise

References

Appendix A Matrices and Matrix Algebra

Appendix B Numerical Methods

Appendix C Analytical Solution of Differential Equations

Appendix D Table of Laplace Transform of Some Common

Functions

Appendix E Orthogonal Collocation Technique

Appendix F Some Software and Programming Environments

1

System Theory and Chemical/Biochemical Engineering Systems

1.1 SYSTEM THEORY

1.1.1 What Is a System?

The word system derives from the Greek word ‘‘systema’’ and means anassemblage of objects united by some form of regular interaction or inter-dependence. A simpler, more pragmatic description regarding systemsincludes the following:

. The system is a whole composed of parts (elements).

. The concept of a system, subsystem, and element is relative anddepends on the degree of analysis; for example, we can take theentire human body as a system, and the heart, the arms, the liver,and so forth as the elements. Alternatively, we can consider theseelements as subsystems and analyze them with respect to smallerelements (or subsystems) and so on.

. The parts of the system can be parts in the physical sense of theword or they can be processes. In the physical sense, the parts ofthe body or of a chair form a system. On the other hand, forchemical equipment performing a certain function, we considerthe various processes taking place inside the system as the elementswhich are almost always interacting with each other to give thefunction of the system. A simple chemical engineering example is a

chemical reactor in which processes like mixing, chemical reaction,heat evolution, heat transfer, and so forth take place to give thefunction of this reactor, which is the changing of some reactants tosome products.

. The properties of the system are not the sum of the properties ofits components (elements), although it is, of course, affected bythe properties of its components. The properties of the systemare the result of the nonlinear interaction among its components(elements). For example, humans have consciousness which is nota property of any of its components (elements) alone. Also, masstransfer with chemical reaction has certain properties which arenot properties of the chemical reaction or the mass transferalone (e.g., multiplicity of steady states, as will be shown later inthis book).

This is a very elementary presentation of system theory. We will revisit thesubject in more detail later.

1.1.2 Boundaries of a System

The system has boundaries distinguishing it from the surrounding environ-ment. Here, we will develop the concept of environment. The relationbetween the system and its environment gives one of the most importantclassifications of a system:

1. An Isolated System does not exchange matter or energy with thesurroundings. Thermodynamically it tends to the state of thermo-dynamic equilibrium (maximum entropy). An example is a batchadiabatic reactor.

2. A Closed System does not exchange matter with the surroundingsbut exchanges energy. Thermodynamically it tends to the state ofthermodynamic equilibrium (maximum entropy). An example isa batch nonadiabatic reactor.

3. An Open System does exchange matter and energy with the sur-roundings. Thermodynamically, it does not tend to the thermo-dynamic equilibrium, but to the steady state or what should becalled the ‘‘stationary non equilibrium state,’’ characterized byminimum entropy generation. An example is a continuous stirredtank reactor.

This clearly shows that the phrase we commonly use in chemical engineer-ing, ‘‘steady state,’’ is not really very accurate, or at least it is not distinctiveenough. A better and more accurate phrase should be ‘‘stationary non-equilibrium state,’’ which is a characteristic of open systems and distin-

guishes it from the ‘‘stationary equilibrium state,’’ associated with anisolated and closed systems.

1.2 STEADY STATE, UNSTEADY STATE, ANDTHERMODYNAMIC EQUILIBRIUM

As briefly stated above, the steady state and unsteady state are conceptsrelated to open systems (almost all continuous chemical engineering pro-cesses are open systems). Steady state is when the state of the system doesnot change with time, but the system is not at thermodynamic equilibrium(i.e., the process inside the system did not stop and the stationary behaviorwith time is due to the balance between the input, output, and processestaking place in the system). The thermodynamic equilibrium is stationarywith time for isolated and closed systems because all processes have stopped,and the nonequilibrium system is changing with time but tending to thethermodynamic equilibrium state. We will come back to these conceptswith more details later in this book.

1.2.1 The State of the System

We have used the term ‘‘state of the system’’ many times; what is the state ofa system? The state of a system is rigorously defined through the statevariables of the system. The state variables of any system are chosen accord-ing to the nature of the system. The state of a boiler can be described bytemperature and pressure, a heat exchanger by temperature, a nonisother-mal reactor by the concentration of the different components and tempera-ture, an isothermal absorption tower by the concentration of differentcomponents on different plates, a human body by blood pressure and tem-perature, flow through a pipe by the velocity as a variable varying radiallyand axially, and so on.

Thus, state variables are variables that describe the state of the system,and the state of the system is described by the state variables.

1.2.2 Input Variables

Input variables are not state variables; they are external to the system, butthey affect the system or, in other words, ‘‘work on the system.’’ For exam-ple, the feed temperature and composition of the feed stream to a distillationtower or a chemical reactor or the feed temperature to a heat exchanger arethe input variables. They affect the state of the system, but are not affectedby the state of the system (except when there is a feedback control, and inthis case, we distinguish between control variables and disturbances or inputvariables).

Later, we will discuss the distinction between different types of vari-ables in more detail; for example, design and operating variables and thedifference between variables and parameters.

1.2.3 Initial Conditions

These are only associated with unsteady-state systems. An unsteady-statesystem is a system in which its state variables are changing with time.Unsteady-state open systems will change with time, tending toward the‘‘stationary nonequilibrium state’’, usually called the ‘‘steady state’’ in thechemical engineering literature. On the other hand, for closed and isolatedsystems, the unsteady-state behavior tends toward thermodynamic equili-brium.

For these unsteady-state systems, whether open or closed, the systembehavior cannot be defined without knowing the initial conditions, or thevalues of the state variables at the start (i.e., time¼ 0). When the initialconditions are defined, the behavior of the system is uniquely defined. Themultiplicity (nonunique) phenomena that appear in some chemical engineer-ing systems are related to the steady state of open systems and not to theunsteady-state behavior with known initial conditions. The trajectorydescribing the change of the state variables with time starting at a specificinitial condition is unique.

1.3 MODELING OF SYSTEMS

The simplest definition of modeling is the following: putting physical realityinto an acceptable mathematical form.

A model of a system is some form of mathematical representation thatgives, in the final analysis, a relation between the inputs and outputs of thesystem. The simplest and least reliable are empirical models, which are basedmore or less on the black-box concept. As shown in Figure 1.1, an empiricalmodel may have the following form:

O ¼ f Ið Þ

Figure 1.1 The black-box concept.

where O is the output and I is the input. On the other hand, a physicalmodelis a model based on the understanding of what is happening inside thesystem and the exchange between the system and the surrounding. Thereis no model which is completely empirical and there is no model which iscompletely physical, but we name the model according to its dominantfeature.

1.3.1 Elementary Procedure for Model Building

An elementary procedure for model building is as follows:

1. Define the boundaries of the system (Fig. 1.2).2. Define the type of system: open, closed, or isolated.3. Define the state variables.4. Define the input variables (sometimes called input parameters).5. Define the design variables (or parameters).

Figure 1.2 Boundaries of a system and subsystems.

6. Define the nature of the interaction between the system and thesurroundings.

7. Define the processes taking place within the boundaries of thesystem.

8. Define the rate of the different processes in terms of thestate variables and rate parameters, and introduce the neces-sary equations of state and equilibrium relations between thedifferent phases. Note: Usually, equilibrium relations betweencertain variables are used instead of rates as an approxima-tion when the rate is quite high and the process reachesequilibrium quickly.

9. Write mass, heat (energy), and momentum balance equations toobtain the necessary equations (or model equations) relating theinput and output through the state variables and parameters.These equations give the variation of state variables with timeand/or space.

1.3.2 Solution of the Model Equations

The model developed should be solved for certain inputs, design parameters,and physicochemical parameters in order to obtain the output and thevariation of state variables within the boundaries of the system. To solvethe model equations, we need two main items:

1. Determination of the model parameters (to be determined experi-mentally)

2. A solution algorithm, the complexity of it, and whether it isanalytical or numerical depends on the complexity of the system

Of course, we will discuss the solution of the different types of model in fulldetail later in this book.

1.3.3 Model Verification

To be able to use the developed model in design, control, andoptimization, it has to be verified against the real system. The behaviorof the system is compared with the behavior of a real system (labora-tory scale, pilot plant, or industrial units). This important aspect ofdeveloping model (design) equations of real value by verifying its repre-sentation of real units will be discussed and analyzed in full detaillater.

1.4 FUNDAMENTAL LAWS GOVERNING THEPROCESSES IN TERMS OF THE STATEVARIABLES

Here, we give a very simple presentation of the necessary components fordeveloping model (design) equations for chemical/biochemical processes.Full details and generalization will be given in Chapter 3.

1.4.1 Continuity Equations for Open Systems

Total continuity equations (mass balances):

Mass flow into systemð Þ � Mass flow out of systemð Þ� �¼ Time rate of change of mass inside the system� �

Component continuity equations (component mass balances):

ðFlow of moles of jth component into the systemÞ� Flow of moles of jth component out the systemð Þþ ðRate of formation of moles of jth component by

chemical reactionsÞ¼ ðTime rate of change of moles of jth component inside

the systemÞ

1.4.2 Diffusion of Mass (Transport Law)

Fick’s law gives that the mass transfer (diffusion) is proportional to theconcentration gradient (Fig. 1.3):

NA ¼ �DA

dCA

dZ

where, NA is the molar flux in moles=ðunit areaÞðunit timeÞ, DA is thediffusion coefficient, dCA is the concentration driving force, and dZ is thedistance in the direction of diffusion. The same equation over a certainthickness (film, interface, etc.) can be approximated as

NA ¼ KL�CA

where KL is the overall mass transfer coefficient and NA is the flux.Hence, we can write

KL ¼ DA

�

1.4.3 Energy Equation (Conservation of Energy, First Law ofThermodynamics for an Open System)

Flow of internal,

kinetic and potential

energies into the

system by convection

or diffusion

0BBBB@

1CCCCA�

Flow of internal,

kinetic and potential

energies out of the

system by convection

or diffusion

0BBBB@

1CCCCA

þ

Heat added

to the system

by conduction,

radiation and

reaction

0BBBB@

1CCCCA�

Work done

by the system

on surroundings,

i.e. shaft work

and PV work

0BBBB@

1CCCCA

¼ fTime rate of change of internal, kinetic and potentialenergies inside the systemg

In most chemical engineering systems that we will study, the above generalform reduces to essentially an enthalpy balance, as will be shown later.

Heat Transfer

Fourier’s law describes the flow of heat in terms of temperature gradient asfollows:

q ¼ ��dT

dZ

Figure 1.3 Concentration gradient for diffusive mass transfer.

The above relation can be approximated in terms of the temperature differ-ence between two points as follows:

q ¼ h�T

where q is the heat flux in units of J/cm2s, � is the thermal conductivity, dTis the temperature driving force, dZ is the distance in the direction of heattransfer, and h ¼ �=� (analogous to the mass transfer case shown above andin Fig. 1.3).

1.4.4 Equations of Motion

The force is given by

F ¼ 1

gCðMaÞ

where F ¼ force (lbf ), gC is the conversion constant needed to keep unitsconsistent [32.2 (lbm ft/lbf s

2)], M is the mass (lbm), and a is the acceleration(ft/s2). Alternatively, we can write that

Force ¼ Mass � Acceleration

In this case, the mass is considered to be constant.When the mass varies with time, the equation will have the following

general form:

1

gC

d Mvið Þdt

¼XNj¼1

Fji

where vi is the velocity in the i direction (ft/s or m/s) and Fji is the jth forceacting in the i direction (lbf or N).

Momentum Transfer (Fig. 1.4)

�xy ¼ Momentum Flux ¼_PPy

Ax

where _PPy is the time rate of change of momentum caused by the force in they direction Fy (the momentum transfer is in the x direction) and Ax is thearea perpendicular to x, where x is the direction of momentum transfer.

Shear Stress

�xy ¼Fy

Ax

where Fy is the scalar y -component of the force vector and Ax is the areaperpendicular to the x axis. We can write

F ¼ k _PP

For one particular direction,

Fy ¼ k _PPy

where

k ¼ 1

gC

� �for the British system

¼ 1 for the SI system

For shear stress, we can write

�xy ¼k _PPy

Ax

¼ k�xy

and

�xy ¼1

k�xy

where �xy is the momentum flux.

Newton’s Law of Viscosity (or Momentum Transfer)

This is given by

�xy ¼1

k�xy ¼ ��

@vy@x

Figure 1.4 Momentum transfer.

1.4.5 Equations of State

To write mathematical models, in addition to material and energy balancesand rate of different processes taking place within the boundaries of thesystem, we need equations that tell us how the physical properties, primarilydensity and enthalpy, change with temperature.

Liquid density ¼ �L ¼ f1 P;T; xið ÞVapor density ¼ �V ¼ f2 P;T; yið ÞLiquid enthalpy ¼ h ¼ f3 P;T; xið ÞVapor enthalpy ¼ H ¼ f4 P;T; yið Þ

For simplicity, h is related to CpT and H is related to CpT þ �V .If Cp is taken as function of temperature and we consider that the

reference condition is To (at which h ¼ 0), then

h ¼ðTT0

Cp Tð Þ dT ¼ sensible enthalpy change

If

Cp Tð Þ ¼ A1 þ A2T

then

h ¼ A3 þ A4T þ A5T2

where A3, A4, and A5 are defined in terms of T0, A1 and A2.

For Mixture of Components (and Negligible Heat of Mixing)

The enthalpy of the liquid mixture can be expressed as

h ¼PJ

j¼1 xjhjMjPJj¼1 xjMj

where xj is the mole fraction of the jth component, Mj is the molecularweight of the jth component (g/gmol), and hj is the enthalpy of the purecomponent j (J/gmol). The denominator depicts clearly the average mole-cular weight of the mixture.

Density

Liquid densities are usually assumed constant (unless large changes incomposition and temperature occur).

Vapor densities can be obtained from the relation

PV ¼ nRT ð1:1Þ

which gives

� ¼ MP

RT

where P is the absolute pressure, V is the volume, n is the number of moles,R is the universal gas constant, T is the absolute temperature, and M is themolecular weight.

Notes

1. The reader must be cautious about the use of consistent units forR and other variables.

2. For a high-pressure and/or high-temperature system, the com-pressibility factor (z factor) should be introduced, which isobtained from the knowledge of the critical temperature andcritical pressure of the system (the reader is advised to refer toa thermodynamics book; for example, Ref. 1).

3. For an open (flow) system, Eq. (1.1) becomes

Pq ¼ �nnRT

where q is the volumetric flow rate (L/min) and �nn is the molar flowrate (mol/min).

1.4.6 Rate of Reaction

Homogeneous reaction rates are, in general, functions of concentrations,temperature, and pressure:

r ¼ f T;C;Pð Þ

where r is the rate of reaction (gmol/L s), T is the temperature, usually thedependence of the rate of reaction constant on the temperature has the formk ¼ k0e

ð�E=RT Þ, P is the pressure (usually used for gas-phase reactions), andC represents the concentrations.

In many cases, we can write the rate of reaction as the product of threefunctions, each a function in one of the variables (temperature, concentra-tions, or pressure.)

r ¼ f1 Tð Þ f2 Cð Þ f3 Pð Þ

Gas–solid catalytic reactions will have the same form, but will refer to theweight of catalyst rather than the volume:

r ¼ f T;C;Pð Þwhere r is the rate of reaction (gmol/g catalyst s). Note the difference in theunits of the rate of reaction for this gas–solid catalytic reaction comparedwith the homogeneous reaction.

1.4.7 Thermodynamic Equilibrium

Chemical Equilibrium (for Reversible Reactions)

Chemical equilibrium occurs in a reacting system whenXJ

j¼1�j�j ¼ 0

where �j is the stoichiometric coefficient of jth component with the signconvention that reactants have negative sign and products have positivesigns (this will be discussed in full detail later) and �j is the chemical poten-tial of the jth component.

Also, we have the following important relation:

�j ¼ �0j þ RT lnPj

where �0j is the standard chemical potential (or Gibbs free energy per mole)

of the jth component, Pj is the partial pressure of the jth component, R isthe universal gas constant, and T is the absolute temperature.

For the reaction

�aA ,k1k2

�BB

with the forward rate of reaction constant being k1 and the backward beingk2. Then, at equilibrium,

�B�B � �A�A ¼ 0

which can be written as

�B �0B þ RT lnPB

� �� A �0A þ RT lnPA

� � ¼ 0

which gives

lnPB � lnPA ¼ �A�0A � �B�

0B

RT

which leads to

lnPB

PA

� �|fflffl{zfflffl}

Kp

¼ �A�0A � �B�

0B

RT|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl}Function of temperature only

where, Kp is the equilibrium constant; we can write

lnKp ¼ f Tð Þ:

Finally, we can write

Kp ¼ e f Tð Þ ¼ Kp0e� �H=RTð Þ

This equilibrium constant of a reversible reaction is very important in orderto determine the limit of conversion of reactants under any given design andoperating conditions.

Phase Equilibrium

Equilibrium between two phases occurs when the chemical potential of eachcomponent is the same in the two phases:

�Ij ¼ �II

j

where, �Ij is the chemical potential of the jth component in phase I and �II

J isthe chemical potential of the jth component in phase II.

For Vapor–Liquid Systems

This phase equilibrium leads to the satisfaction of our need for a relation-ship which permits us to calculate the vapor composition if we know theliquid composition or vice versa when the two phases are at equilibrium. Wewill give an example regarding the bubble-point calculation.

Bubble-Point Calculation (Fig 1.5)

Given the pressure P of the system and the liquid composition xj, we cal-culate the temperature of the system T and the vapor composition yj.This calculation usually involves trial and error (for some other cases, thesituation will be that we know xj and T and want to find P and yj, or weknow P and yj and want to find xj and T).

For Ideal Vapor-Phase Behavior (Needs Correction at HighPressures)

Dalton’s Law

Dalton’s law applies to the ideal vapor-phase behavior and states that thepartial pressure of component j in the vapor phase is the mole fraction of thecomponent multiplied by the total pressure:

pj ¼ Pyj

where pj is the partial pressure of component j in the vapor phase, P is thetotal pressure, and yj is the mole fraction of component j in the vapor phase.

Rault’s Law

Rault’s law states that the total pressure is the summation of the vaporpressure of each component ðp0j Þ multiplied by the mole fraction of com-ponent j in the liquid phase:

P ¼XJj¼1

xjp0j

We can write the partial pressure in the vapor phase of each component asrelated to the vapor pressure as follows:

pj ¼ xjp0j

Hence,

P ¼XJj¼1

pj

where P is the total pressure, xj is the mole fraction of component j in liquid,and p0j is the vapor pressure of the pure component j.

Figure 1.5 Calculation of bubble point.

Now, because the relations

pj ¼ xjp0j and pj ¼ Pyj

hold, we can equate the two relations to get

xjp0j ¼ Pyj

Therefore, we get

yj ¼p0j

Pxj

The vapor pressure p0j is a function of temperature only, as shown by therelation

ln p0j ¼Aj

Tþ Bj

Therefore, the vapor–liquid equilibrium computation can be performedaccording to the above relation.

The Utilization of Relative Volatility

The relative volatility of component i to component j is defined as

�ij ¼yi=xiyj=xj

¼ Volatility of i

Volatility of j

For binary systems, we have

� ¼ y=x

1� yð Þ= 1� xð Þwhich gives

y ¼ �x

1þ ð�� 1Þx

The K-Values

Equilibrium vaporization ratios or K values are widely used, particularly inthe petroleum industry:

Kj ¼yj

xj

Kj is a function of temperature and, to a lesser extent, of composition andpressure.

Activity Coefficient

For nonideal liquids, Rault’s law may be easily modified to the form:

pj ¼ xjp0j �j

Also,

P ¼XJj¼1

xjp0j �j

where �j is the activity coefficient (a function of composition, pressure, andtemperature).

The Z-Factor

For nonideal gases, the z factor (compressibility factor) is introduced. It is afunction of the critical temperature and pressure.

This very brief review of phase equilibrium gives the reader theprinciples for dealing with two-phase systems. More background onthe thermodynamic equilibrium of multiphase systems will sometimesbe needed; it is not covered in this book. The reader is advised toconsult a multiphase thermodynamics book for that purpose.

1.5 DIFFERENT CLASSIFICATIONS OF PHYSICALMODELS

Before we move to a more intellectual discussion of the history ofchemical engineering and the role and position of system theory andmathematical modeling, we present the different basis for classification ofmathematical models.

I. Classification according to variation or constancy of the state

variables with time

. Steady-state models: described by algebraic equations orODEs (ordinary differential equations) or PDEs (partialdifferential equations).

. Unsteady-state models: described by ODEs or PDEs.II. Classification according to the spatial variation of the state

variables

. Lumped models (usually called lumped parameter models,which is wrong terminology because it is the state variablesthat are lumped together not the parameters): described byalgebraic equations for the steady state and ODEs for theunsteady state.

. Distributed models (usually called distributed parametermodels, which is wrong terminology because it is the statevariables that are distributed, not the parameters):described by ODEs or PDEs for the steady state andPDEs for the unsteady state.

III. According to the functional dependence of the rate governing laws

on the state variables

. Linear models: described by linear equations (can be alge-braic equations, ODEs, or PDEs). These models can besolved analytically.

. Nonlinear models: described by nonlinear equations (can bealgebraic equations ODEs, or PDEs). These models need tobe solved numerically.

IV. According to the type of processes taking place within the bound-

aries of the system

. Mass transfer (example: isothermal absorption).

. Heat transfer (example: heat exchangers).

. Momentum transfer (example: pumps or compressors).

. Chemical reaction (example: homogeneous reactors).

. Combination of any two or more of the above processes(example: heterogeneous reactors).

V. According to the number of phases in the system

. Homogeneous models (formed of one phase).

. Heterogeneous models (more than one phase).VI. According to the number of stages in the system

. Single stage.

. Multistage.VII. According to the mode of operation of the system

. Batch (closed or isolated system).

. Continuous (open system).VIII. According to the system’s thermal relation with the surroundings

. Adiabatic (neither heating nor cooling).

. Nonadiabatic (example: cocurrent or countercurrent cool-ing or heating).

IX. According to the thermal characteristics of the system

. Isothermal.

. Nonisothermal.

All of these classifications are, of course, interactive, and the best approachis to choose a main classification and then have the other classification assubdivisions of this main classification.

1.6 THE STORY OF CHEMICAL ENGINEERING INRELATION TO SYSTEM THEORY ANDMATHEMATICAL MODELING

Chemical engineering as we witness today at the beginning of the 21stcentury has evolved into a very demanding discipline extending to a widespectrum of industries and requiring a wide spectrum of knowledge in manydiversified fields. The easy days of chemical engineers (or chemical technol-ogists), being simply mechanical engineers with some background in indus-trial chemistry, have long gone. Norton, at MIT, started, in 1888, what maybe called the first chemical engineering curriculum. However, it was more ofa chemical technology than a chemical engineering curriculum as we haveknown it for the last three to four decades. It consisted mostly of descriptivecourses in industrial chemistry and chemical technology. In about 1923came the preliminary steps in the direction of classifying chemical engineer-ing equipment and processes, on a higher level of generality. This wasrepresented by the concept of unit operations, where, for example, distilla-tion and extraction are taught as unified courses not necessarily related to aspecific industry.

This conceptual approach spread worldwide and started to give birth tomore of these generalized subjects. Systematic, though simple, methods ofdesign were developed (e.g., McCabe–Thiele diagrams) and the concept ofequilibrium stages became well established. The emphasis at this stage, whichextended to the 1960s, was on the overall behavior of the chemical equipmentwithout real involvement into the details of the microscale processes.

The second milestone in chemical engineering came in 1960 with thepublication of Transport Phenomena, by Bird et al. [2]. Their new approachemphasized the microscale processes and the analogy among mass, heat,and momentum transfer in different processes.

This stage witnessed an explosion of changes in chemical engineeringboth in research and education, giving birth to new well-defined fields. Aprime example is chemical engineering kinetics, which gradually evolvedinto the rich discipline under the title ‘‘Chemical Reaction Engineering,’’which emphasizes the design, analysis, and optimization of different types ofchemical reactors.

The chemical engineering discipline started to ‘‘bifurcate’’ faster andfaster year after year. Process dynamics and control (PDC) became animportant branch of chemical engineering. Plant design, economics, andother specialized disciplines within chemical engineering started to grow ata rapid rate.

These developments were accompanied by a steady growth in pro-ductivity, sophistication, and a high level of competition in the chemical

industry. In order to meet these demands, a typical chemical engineeringcurriculum had to be crowded with many subjects. In addition to basicscience courses (mathematics, physics, chemistry, biology, thermodynamics,etc.), the student had to be taught the new chemical engineering disciplines:mass and heat balance, mass and heat transfer, chemical reaction engineer-ing, multistage operations, process dynamics and control, plant design andeconomics, and so forth, in addition to supplementary subjects from otherengineering disciplines such as mechanical, civil, and electronic engineering.The training of chemical engineers left little room for what James Wei callsthe ‘‘Third Paradigm,’’ which should emphasize a ‘‘global outlook to therelation between Engineering and Society’’ [3].

Despite this crowding of the chemical engineering curriculum, a prac-ticing engineer or an experienced professor can look at any curriculum andfind a lot to be desired. Students are learning about chemical reaction engi-neering principles, which is a must, but they do not learn much about actualindustrial reactors; they learn classical control theory, which is a must, butthey do not know enough about digital control, in a time when most plantsare discarding analog control and installing digital control systems. The listis endless and offers a strong temptation to add more and more courses, butbecause the students are incompressible (as Levenspiel once put it [4]), it isvery difficult to add more courses in response to the legitimate desires ofexperienced industrialists and professors.

The situation becomes even more acute with the expansion of chemicalengineering into new fields, especially biotechnology, the electronic indus-tries, new materials, and composite membranes. In addition to that, chem-ical engineers are qualified and obliged to play a leading role in theenvironmental challenge that is facing the human society in a dangerousand complicated manner.

In a nutshell, we do have a problem that cannot be solved by quanti-tative measures; it is so acute that it needs, actually, a change of concept.This change of concept requires looking at chemical engineering from asystem theory point of view.

1.7 THE PRESENT STATUS OF CHEMICALINDUSTRY AND UNDERGRADUATECHEMICAL ENGINEERING EDUCATION

This book is designed for undergraduate chemical engineering students. Thesystem approach is utilized in order to structure undergraduate chemicalengineering education in a new fashion suitable for the present state ofchemical, biochemical, and related industries.

The chemical industry has revolutionized human life to such an extentthat it invaded every domain of modern life in our society. The industrialdevelopment and the innovative work of many pioneering chemical engi-neering researchers, coupled with the advancement in computer technology,makes the training of chemical engineers for the future quite a challengingtask. A chemical engineer graduating today is expected in his/her career todeal with a wide range of problems that need a sound fundamental basis aswell as an arsenal of practical knowledge. Of course, much of the practicalknowledge is acquired in the industry after graduation; however, under-graduate training is the critical factor that determines the degree of successof the trajectory of the chemical engineers after graduation. The socioeco-nomic, safety, and environmental challenges, together with the fast expan-sion in the use of digital computers, make the task even more difficult. Is itplausible today to produce a chemical engineer who is not fluent in theapplication of computer power to chemical engineering problems? Is it pos-sible to be satisfied with a graduate who knows enough about chemicalengineering, but who is illiterate with regard to some of the basic computersoftware and hardware necessary for computer-controlled experimentationor operation of equipment? Considering the socioeconomic implications ofchemical, biochemical, and related industries, is it wise to produce chemicalengineers who are not sufficiently socially aware of the problems of theirsociety and the connections between these problems and their profession? Ofcourse, we should be aware that the 4 or 5 years of chemical engineeringeducation is not and should not be a substitute for the whole life experiencethat he or she will acquire before university education or after graduation.However, these 4–5 years are crucial for the production of our hoped-for‘‘Chemical Engineer Citizen,’’ a chemical engineer who can grasp the grow-ing fascinating opportunities of our modern society as well as coping with itsdifficulties and problems.

1.8 SYSTEM THEORY AND THEMATHEMATICAL MODELING APPROACHUSED IN THIS BOOK

The challenges facing the chemical engineering profession in its process ofhealthy and fruitful intercourse with society, as well as with first and secondnatures, cannot be advanced to higher levels with one single idea. Certainly,man has learned that life is much more complicated than the deterministicviews of the 18th and 19th centuries. A more complex view of nature,society, and man-made processes is emerging, from simple monotony tocomplex bifurcations which were thought to represent the highest degreeof complexity before the revolutionary discovery of chaos, strange attrac-

tors, and fractals structures. Scientific development in the last few yearsshould teach us that, most probably, nobody will say the last word in any-thing and that the most that anybody can hope for, which is very honorable,is to achieve one’s part in a successful iteration of the continuous human riseto higher levels of civilization and intellect with all its dynamical beauty andits enjoyable new challenges and difficulties. Our argument regarding therole of systems theory and mathematical modeling in undergraduate chem-ical engineering education will, hopefully, represent one small part of acomplex set of changes that are needed to take the chemical engineeringscience forward, through a tortuous route of developments, to match therevolutionary changes the human society is witnessing scientifically, techno-logically, socially, and politically. We argue, in this section of the book, thatthe extraordinary expansion in the domain of chemical engineering requiresa new step in the direction of generalized classification that will open newand faster expanding horizons for this important discipline. This can beachieved through a radical change of the undergraduate chemical engineer-ing syllabus, in order to make it based on system theory and the mathema-tical modeling approach, which is a very effective step forward in organizingknowledge and achieving a much higher level of economy of information.Before getting deeper into the subject, it may be worthwhile to present adescription of system theory and mathematical modeling in chemical engi-neering, which will add further to the simpler understanding we presented inthe previous section.

1.8.1 Systems and Mathematical Models

In this subsection, we discuss very briefly the basic principles of systems andmathematical modeling theories, with special emphasis on chemical biochem-ical engineering problems, in a manner that enriches the very simple explana-tion we presented earlier. System theory is the more general, more abstractpart, whereas mathematical modeling is more applied and less abstract.

As explained earlier, a system is a whole consisting of elements orsubsystems. The concept of systems–subsystems and elements is relativeand depends on the level of analysis. The system has a boundary thatdistinguishes it from the environment. The system may exchange matterand/or energy with the environment depending on the type of systemfrom a thermodynamical point of view. A system (or subsystem) is describedby its elements (or subsystems), the interaction between the elements, and itsrelation with the environment. The elements of the system can be materialelements distributed topologically within the boundaries of the system andgiving the configuration of the system, or they can be processes taking placewithin the boundaries of the system and defining its function. They can also

be both, together with their complex interactions. An important property ofthe system wholeness is related to the principle of the irreducibility of thecomplex to the simple, or of the whole to its elements; the whole system willpossess properties and qualities not found in its constituent elements. Thisdoes not mean that certain information about the behavior of the systemcannot be deduced from the properties of its elements, but, rather, it addssomething to them.

The systems can be classified on different bases as briefly explainedearlier. The most fundamental classification is that based on thermodynamicprinciples, and on this basis, systems can be classified into the followingclasses:

Isolated Systems

Isolated systems are systems that exchange neither energy nor matter withthe environment. The simplest chemical engineering example is the adiabaticbatch reactor. These systems tend toward their thermodynamic equilibriumwith time, which is characterized by maximum entropy (highest degree ofdisorder).

Closed Systems

Closed systems are systems that exchange energy with the environmentthrough their boundaries, but do not exchange matter. The simplestexample is a nonadiabatic batch reactor. These systems also tend towarda thermodynamic equilibrium with time, characterized by maximumentropy (highest degree of disorder).

Open Systems

Open Systems are systems that exchange both energy and matter withthe environment through their boundaries. The most common chemicalengineering example is the continuous-stirred tank reactor (CSTR). Thesesystems do not tend toward their thermodynamic equilibrium with time, butrather toward a ‘‘stationary nonequilibrium state,’’ which is characterizedby minimum entropy production.

It is clear from the above classification that batch processes are usuallyof the isolated or closed type, whereas the continuous processes are usuallyof the open type. As can be seen from the above definitions, the systemtheory is very abstract and, in general, it treats any system regardless ofwhether a mathematical model for this system can be built or not; mathe-matical modeling, on the other hand, is less abstract and more applied thanthe system concept.

For continuous processes, a classification of systems from a mathe-matical point of view is very useful for both model formulation andalgorithms for model solution. According to this basis, systems can beclassified as shown briefly earlier.

Lumped Systems

Lumped systems are systems in which the state variables describing thesystem are lumped in space (invariant in all space dimensions). The simplestexample is the perfectly mixed continuous-stirred tank reactor. These sys-tems are described at steady state by algebraic equations, whereas theunsteady state is described by initial-value ordinary differential equationsfor which time is the independent variable.

Distributed Systems

Distributed systems are systems in which the state variables are varying inone direction or more of the space coordinates. The simplest example is theplug flow reactor. These systems are described at steady state, either byordinary differential equations [where the variation of the state variablesis only in one direction of the space coordinates (i.e., one-dimensional sys-tems), and the independent variable is this space direction] or partial differ-ential equations [where the variation of the state variables is in more thanone direction of space coordinates (i.e., two- or three-dimensional systems)and the independent variables are these space directions]. The ordinarydifferential equations describing the steady state of the one-dimensionaldistributed systems can be either initial-value differential equations (e.g.,plug flow systems) or two-point boundary-value differential equations(e.g., systems with superimposed axial dispersion). The equations describingthe dynamic behavior of distributed systems are invariably partial differen-tial equations.

Another classification of systems, which is important for deciding thealgorithm for model solution, is that of linear and nonlinear systems.The equations of linear systems can usually be solved analytically,whereas the equations of nonlinear systems are almost always solvednumerically. In this respect, it is important to recognize the significantfact that physical systems are almost always nonlinear and that linearsystems are either an approximation that should be justified or areintentionally linearized in the neighborhood of a certain state of thesystem and are strictly valid only in this neighborhood. This locallinearization is usually applied to dynamical system to investigatelocal stability characteristics of the system.

A third classification, which is relevant and important in chemicalengineering, is the classification based on the number of phasesinvolved within the boundaries of the system. According to this clas-sification, the systems are divided as follows:

Homogeneous Systems

Homogeneous systems are systems in which only one phase is involved in theprocesses taking place within the boundaries of the system. In reaction sys-tems, the behavior of these systems is basically governed by the kinetics of thereactions taking place, without the interference of any diffusion processes.

Heterogeneous Systems

Heterogeneous systems are systems in which more than one phase isinvolved in the processes taking place. In reaction systems, the behaviorof these systems is not only governed by the kinetics of the reactions takingplace, but also by the complex interaction between the kinetics and diffusionprocesses. When the system does not involve a chemical reaction, then thesystem behavior is not governed by processes taking place in one phase, but,rather, by the totality of the processes taking place in the different phasesand the interaction between them. The modeling and analysis of these sys-tems is obviously much more complicated than for homogeneous systems.Heterogeneous systems are globally divided into two- and three-phasesystems. In more detail, they are divided as follows:

1. Liquid–liquid systems2. Gas–liquid systems3. Gas–solid systems: catalytic and noncatalytic4. Liquid–solid systems: catalytic and noncatalytic5. Gas–solid–liquid systems: catalytic and noncatalytic

1.8.2 Mathematical Model Building: General Concepts

The process of classification and building of mathematical models (designequations) has been simply discussed earlier. Here, we give more detailsabout this process which add to the buildup of knowledge for the readerin this direction.

Building a mathematical model for a chemical engineering systemdepends to a large extent on the knowledge of the physical and chemicallaws governing the processes taking place within the boundaries of thesystem. This includes the different rates of mass, heat, and momentumtransfer, rates of reactions and rates of adsorption–desorption, and soforth. It also includes the thermodynamic limitations that decide the feasi-bility of the process to start with, as well as heat production and absorption.

Mass and heat transfer rates are both dependent on the proper descriptionof the fluid flow phenomena in the system. The ideal case is when all of theseprocesses are determined separately and then combined into the system’smodel in a rigorous manner. However, very often this is quite difficultin experimental measurement; therefore, special experiments need to bedevised, coupled with the necessary mathematical modeling, in order todecouple the different processes implicit in the measurements.

In the last two decades, there has been a considerable advancement inthe development of mathematical models of different degrees of sophistica-tion for chemical engineering processes. These models are taking their part indirecting design procedures as well as in directing scientific research. It isimportant in this respect to recognize the fact that most mathematical modelsare not completely based on rigorous mathematical formulation of the phy-sical and chemical processes taking place within the boundaries of the sys-tem. Every mathematical model, contains a certain degree of empiricism. Thedegree of empiricism limits the generality of the model, and as our knowledgeof the fundamentals of the processes taking place increases, the degree ofempiricism decreases and the generality of the model increases. The existenceof models at any stage, with their appropriate level of empiricism, helpsgreatly in the advancement of the knowledge of the fundamentals, and, there-fore, helps to decrease the degree of empiricism and increase the level of rigorin the mathematical models. Models will always contain certain simplifyingassumptions believed by the model-builder not to affect the predictive natureof the model in any manner that sabotages the purpose of it. Mathematicalmodels (design equations) are essential for design, optimization, and controlof different equipments for chemical, biochemical, and related industries.

Different models with different degrees of sophistication can be built.Models which are too simplified will not be reliable and will not serve thepurpose, whereas models which are too sophisticated will present anunnecessary and sometimes expensive overburden. In undergraduate chemi-cal engineering education, the concept of model sophistication and its rela-tion to the purpose of the model building should be emphasized. This bookwill explain and discuss this extremely important issue at every stage ofdevelopment, starting from material and energy balance to heterogeneousdistributed models.

1.8.3 Outline of the Procedure for Model Building

This issue has been discussed earlier in a simplified manner. It is useful forthe gradual buildup of the reader’s understanding of modeling and simula-tion to visit the problem again on a higher level of rigor and to distinguishclearly the difference between modeling and simulation.

First, we start with procedure for model building, followed by thedistinction between modeling and simulation, followed by the Amundsonreport and its emphasis on modern chemical engineering education, andfinally ending up by the relation among systems theory, mathematicalmodeling, and modern chemical engineering education.

The procedure for model building can be summarized in the followingsteps:

1. The identification of the system configuration, its environment,and the modes of interaction between the system and its environ-ment.

2. The introduction of the necessary justifiable simplifying assump-tions.

3. The identification of the relevant state variables that describe thesystem.

4. The identification of the processes taking place within the bound-aries of the system.

5. The determination of the quantitative laws governing the rates ofthe processes in terms of the state variables. These quantitativelaws can be obtained from information given in the literatureand/or through an experimental research program coupled withthe mathematical modeling program.

6. The identification of the input variables acting on the system.7. The formulation of the model equations based on the principles

of mass, energy, and momentum balances appropriate to the typeof system.

8. The development of the necessary algorithms for the solution ofthe model equations.

9. The checking of the model against experimental results (labora-tory, pilot plant, or commercial units) to ensure its reliability andcarrying out a re-evaluation of the simplifying assumptions. Thisre-evaluation process may result in imposing new simplifyingassumptions or relaxing some of them.

It is clear that these steps are interactive in nature and the results ofeach step should lead to a reconsideration of the results of all previous ones.In many instances, steps 2 and 3 are interchanged in the sequence, depend-ing on the nature of the system and the degree of knowledge regarding theprocesses taking place.

Of course the modeling of an existing process will differ from themodeling to design a new plant based on known technology. And bothwill be different from using modeling to aid in the development of a noveltechnology.

1.9 MODELING AND SIMULATION INCHEMICAL ENGINEERING

Aris [5], in his 1990 Dankwarts Memorial Lecture entitled ‘‘MannersMakyth Modellers,’’ distinguishes between modeling and simulation in aspecial manner which follows the reasoning of Smith [6] in ecological mod-els. This reasoning is very useful for chemical engineering students to com-prehend early in their career. The reasoning goes as follows (having thebeautiful and very intellectual characteristics of the writings of ProfessorAris, the readers are advised to read the classical papers referred to in thisreasoning):

It is an essential quality in a model that it should be capable ofhaving a life of its own. It may not, in practice, need to be sunderedfrom its physical matrix. It may be a poor thing, an ill-favored thingwhen it is by itself. But it must be capable of having this indepen-dence. Thus Liljenroth (1918) in his seminal paper on multiplicity ofthe steady states can hardly be said to have a mathematical model,unless a graphical representation of the case is a model. He worksout the slope of the heat removal line from the ratio of numericalvalues of a heat of reaction and a heat capacity. Certainly he isdealing with a typical case, and his conclusions are meant to haveapplication beyond this particularity, but the mechanism for doingthis is not there. To say this is not to detract from Liljenroth’spaper, which is a landmark of the chemical engineering literature,it is just to notice a matter of style and the point at which a math-ematical model is born. For in the next papers on the question ofmultiple steady states, those of Wagner (1945), Denbigh (1944,1947), Denbigh et al. (1948) and Van Heerden (1953), we do notfind more general structures. How powerful the life that is instinctin a true mathematical model can be seen from the Fourier’s theoryof heat conduction where the mathematical equations are fecund ofall manner of purely mathematical developments. At the other end

of the scale a model can cease to be a model by becoming too large

and too detailed a simulation of a situation whose natural line of

development is to the particular rather than the general. It ceases to

have a life of its own by becoming dependent for its vitality on its

physical realization [the emphasis is ours]. Maynard Smith (1974)was, I believe, the first to draw the distinction in ecological modelsbetween those that aimed at predicting the population level withgreater and greater accuracy (simulation) and those that seek todisentangle the factors that affect population growth in a more

general way (model). The distinction is not a hard and fast one, butit is useful to discern these alternatives [5].

The basis of the classification given by Aris is very interesting, true,and useful. It is typical of the Minnesota group founded by Amundson some50 years ago. This important research group in the history of chemicalengineering has almost never verified the models (or the variety of interest-ing new phenomena resulting from them) against experiments or industrialunits. Experimental verifications of the new and interesting steady-state anddynamic phenomena discovered by the Minnesota group were carried out atother universities, mostly by graduates from Minnesota. The most interest-ing outcome is the fact that not a single phenomenon, which was discoveredtheoretically using mathematical models by the Minnesota group, was notexperimentally confirmed later. This demonstrates the great power of themathematical modeling discipline as expressed by Aris [5], where the modelis stripped of many of its details in order to investigate the most fundamen-tal characteristics of the system. It also demonstrates the deep insight intophysical systems that can be achieved using mathematics, as Ian Stewartputs it: ‘‘Perhaps mathematics is effective in organizing physical existencebecause it is inspired by physical existence . . . . The pragmatic reality is thatmathematics is the most effective method that we know for understandingwhat we see around us’’ [7]. In this undergraduate book, a different, morepragmatic definition for mathematical modeling and simulation than that ofAris [5] will be adopted. The definition we adopt is that mathematical mod-eling will involve the process of building up the model itself, whereas simu-lation involves simulating the experimental (or pilot plant or industrial)units using the developed model. Thus, simulation in this sense is closelylinked to the verification of the model against experimental, pilot plant, andindustrial units. However, it will also include the use of the verified modelsto simulate a certain practical situation specific to a unit, a part of a produc-tion line, or an entire production line.

However, because universities are responsible also for the preparationof the students for research careers, this book will also include the elementsdiscussed by Aris. In this case, the definition by Aris [5] must come stronglyinto play. Because much of the chemical engineering research work today isinterdisciplinary in nature, especially in the relatively new fields such asbiotechnology and microelectronics, it should be clear that mathematicalmodeling and system theory are the most suitable and efficient means ofcommunication between the different disciplines involved. Therefore,chemical engineering education based on system theory and mathematicalmodeling as portrayed in this book seems to be the best approach to preparethe chemical engineers for this interdisciplinary research. The fact that

Henry Poincare, the undisputed first discoverer of Chaos, was trained first inengineering, following the Napoleonic traditions, helps to emphasize theneed for a broad education of the engineering researchers of tomorrow,which is hard to achieve through the present structure of engineering syllabi.

Another very general definition of models is given by StephenHawking [8], who relates models to theories of the universe:

A theory is just a model of the universe, or a restricted part of it,and a set of rules that relate quantities in the model to observationsthat we make. It exists only in our minds and does not have anyother reality (whatever that might mean). A theory is a good theoryif it satisfies two requirements: it must accurately describe a largeclass of observations on the basis of the model that contains only afew arbitrary elements, and it must make definite predictions aboutthe results of future observations.

One of the major findings of the Minnesota group, using mathematicalmodeling, is the discovery of a wide variety of static and dynamic bifurca-tion phenomena in chemical reactors. Although theoretical studies on thebifurcation behavior have advanced considerably during the last three dec-ades, the industrial appreciation of these phenomena remains very limited. Itis of great importance that practically oriented chemical engineers dealingwith the mathematical modeling of industrial units become aware of them.These phenomena are not only of theoretical and academic importance, butthey have very important practical implications (e.g., for industrial FluidCatalytic Cracking Units [9–12]). Most chemical engineers complete theirundergraduate education without knowing almost anything about thesephenomena. A large percentage of chemical engineers finish their post-graduate education without knowing anything about these phenomena.This book will introduce undergraduate students to these phenomena in avery simple manner early in their education.

1.10 AMUNDSON REPORT AND THE NEED FORMODERN CHEMICAL ENGINEERINGEDUCATION

The report of the committee on ‘‘Chemical Engineering Frontiers: ResearchNeeds and Opportunities’’ [13], known as the Amundson report, outlinesbeautifully the picture of chemical engineering in the next decades. Thereport is quite optimistic about the future of the profession, and the authorsof this book strongly share this optimism. Chemical engineering played animportant part in human development in the last few decades and it isexpected to play an even larger role in the future.

Although the ‘‘Amundson Report’’ emphasizes the challenges facingthe American chemical and related industries, the report should not belooked upon from this narrow point of view. In fact, the report is farmore reaching than that and is to a great extent relevant to the worldwidechemical engineering profession. The report, without ignoring classicalchemical engineering problems, stresses a number of relatively newchemical engineering fields:

1. Biotechnology and Biomedicine2. Electronic, Photonic, and Recording Materials and Devices3. Polymers, Ceramics, and Composites4. Processing of Energy and Natural Resources5. Environmental Protection, Process Safety, and Hazardous Waste

Management6. Computer-Assisted Process and Control Engineering7. Surfaces, Interfaces, and Microstructures

Emphasizing these fields without being able to ignore classical chemicalengineering problems helps only to emphasize the view that chemical engi-neering is entering a new era. In addition, the chemical engineering commu-nity cannot ignore its fundamental scientific responsibilities toward therevolution in scientific knowledge created by the discovery of chaos, strangeattractors, and fractal structures, especially that many manifestations ofthese phenomena are evident in typical chemical engineering systems.

The report is nowmore than 13 years old and the issues addressed in thisreport must reflect themselves on chemical engineering undergraduate educa-tion. This can only be achieved through the adoption of system theory andmathematical modeling in undergraduate chemical engineering education.

These considerations lead to the inevitable conclusion that a higherlevel of organization of thinking and economy of knowledge is needed inundergraduate chemical engineering education, as briefly discussed earlierand will be detailed in the rest of the book.

1.11 SYSTEM THEORY AND MATHEMATICALMODELING AS TOOLS FOR MOREEFFICIENT UNDERGRADUATE CHEMICALENGINEERING EDUCATION

In order to achieve a higher level of organization and economy of knowl-edge, it will be extremely useful for the students to be introduced early intheir undergraduate education to the basic principles of system theory. Thebest and most general classification of systems is based on thermodynamicprinciples. Therefore, it is possible that the students, after passing their basic

science courses, be exposed to a course in thermodynamics that emphasizesthe basic concepts of thermodynamics. The basic principles of system theorycan either be integrated into this course or may be taught in a separatecourse. This course should emphasize not only the basic principles of systemtheory but also its relevance to chemical engineering systems. In fact, such acourse can be used as an elegant and efficient tool for introducing thestudents to chemical engineering systems. The students should learn howto classify chemical engineering systems into their main categories: isolated,closed, and open systems. They should learn how to divide the system intoits subsystems or elements depending on the level of analysis. They shouldlearn not only that distillation is a unit operation regardless of the specifickind of distillation, as the unit operation paradigm teaches us, but to extendtheir mind further and learn that continuous, heterogeneous, multistageprocesses are systems that are open, formed of more than one phase andmore than one stage, that there are certain processes taking place withintheir boundaries that need to be expressed in terms of state variables, andthat input variables should be specified and parameters should be identified(regardless of whether the process is distillation, extraction, drying, or multi-stage catalytic reactors).

Of course, at this early stage, students will not be able yet to developspecific models for specific processes because they would not yet havestudied the laws governing the rates of these processes and their form ofdependence on the state variables. They even may not be able, at this stage,to identify completely the state variables of the system. However, this earlytraining in system theory will orient the students mind and their furthereducation in the framework of the system approach.

The students can then be ready to view their other chemical engineer-ing courses in a new light; those courses should also be changed to empha-size the system approach, where all of the rate processes are treated in aunified fashion, emphasizing the laws governing the rates of these processesand its dependence on state variables, rather than dividing it into masstransfer, heat transfer, momentum transfer, and rates of reactions. Thetransport phenomena paradigm can be easily extended in this direction.This will also allow higher emphasis on the interaction between these pro-cesses and the effects resulting from these interactions (e.g., the interactionbetween mass transfer and chemical reaction and its implication for thebehavior of the system).

It is natural that, at this stage, students be introduced to the formula-tion of some simple mathematical models (design equations) for certainsystems. When the students are familiar with the laws governing the ratesof different processes in terms of the state variables, they will be ready for anintensive applied course on mathematical modeling of chemical engineering

systems. This course should cover the basic principles of mathematical mod-eling theory, the main classifications of mathematical models, the proce-dures for building mathematical models, application to the developmentof mathematical models for a large number of chemical engineeringprocesses in the petrochemical, petroleum refining, biochemical, electronicindustries, as well as some mathematical models for biological systems.

It is important to note that material and energy balances represent thebasis of any rational and useful chemical engineering education. In this book,we use an approach that generalizes the material and energy balance equa-tions in modular forms for the most general cases with multiple inputs,multiple outputs, and multiple reactions. These generalized material andenergy balance equations will represent the basis for all material and energybalance problems, with less complicated cases treated as special cases of thesegeneralized mathematical formulation. These material and energy balanceequations will themselves be used to develop both lumped and distributedmathematical models (design equations). This approach will represent aserious achievement with regard to economy of knowledge, organizationof thinking, and systematic optimization of information for chemicalengineering students. The extension of design equations (steady-statemodels) to dynamic equations (unsteady-state models) will also be shown.

So far, we have emphasized the practical usefulness of the approach. Inthis last sentence, we should also emphasize that the approach is beautifuland elegant: ‘‘He gets full marks who mixes the useful with the beautiful’’[14].

In this preliminary section, we have explained to the reader thatchemical engineering is expanding very quickly and that a new approachbased on system theory and mathematical modeling is needed for a higherlevel of organization of thinking and economy of information. The concepthas been briefly introduced and discussed and the influence of this approachon the structure of chemical engineering teaching has been discussed in alimited fashion, only to show how it can be integrated into a modern andprogressive chemical engineering syllabus and how it does affect theapproach to teaching material and energy balance and rate processes inan integrated manner. This approach is adopted in this book, and theapproach will be clearer as more details and applications are given through-out the book.

1.12 SUMMARY OF THE MAIN TOPICS IN THISCHAPTER

It is useful for the reader to summarize all of the ideas and classifications.Although each of the following subjects will be discussed in more detail, it is

very useful for the reader to have an ‘‘eagle’s-eye view’’ to know where it allfits.

1.12.1 Different Types of Systems and Their MainCharacteristics

What Is a System?

It is a whole, formed of interactive parts.

Isolated Systems

Do not exchange matter or energy with the surrounding. Example isadiabatic batch reactors.

Most Basic Characteristics. Tend toward thermodynamic equili-brium characterized by maximum entropy [it is stationary with time,stationary in the sense of ‘‘death’’ (i.e., nothing happens)].

Closed Systems

Do not exchange matter with the environment or surrounding, but doexchange energy. Example is a non-adiabatic batch reactor.

Most Basic Characteristics. Tend toward thermodynamic equili-brium characterized by maximum entropy [it is stationary with time,stationary in the sense of ‘‘death’’ (i.e., nothing happens)].

Open Systems

Exchange matter and energy with the environment or surrounding (actuallystating that it exchanges matter is sufficient because this implies an exchangeof the energy in the matter transferred). Example is a CSTR.

Most Basic Characteristics. Do not tend toward thermodynamicequilibrium, but rather toward what may be called a ‘‘stationary nonequili-brium state’’; it is stationary with time but not in the sense of being ‘‘dead.’’Things do happen; the stationary nature comes from the balance betweenwhat is happening within the boundaries of the system and the exchange ofmatter with the surrounding.

Important note A stationary nonequilibrium state can also be called a‘‘point attractor.’’ We will see later that other types of attractor are alsopossible.

Relative Nature of Systems, Subsystems, and Elements

. Subsystems are parts of the system, that together with their inter-actions define the system.

. The element is the smallest ‘‘subsystem’’ of the system, accordingto level of analysis.

. Subsystems and elements are relative and depend on the level ofanalysis.

. Subsystems, elements, and ‘‘whole’’ system can be defined ondifferent basis.

Linear and Nonlinear Systems

Linear and nonlinear systems are usually defined on the basis of the form ofthe appearance of the state variables in the process. However, what are‘‘state variables’’?

State Variables and Parameters

State variables are the variables that describe the state of the system (i.e.,temperature of the fluid in a tank, or concentration of a component in areactor or on a tray of a distillation column).

Parameters are constants (or variables) that are imposed (or chosen)for the system and determine the state of the system through their effect onthe processes. We can classify them as follows:

1. Input parameters: feed flow rate, feed concentration, and soforth.

2. Design parameters: diameter, height, and so forth of a vessel.3. Process parameters: mass transfer coefficients, kinetic rate con-

stants, and so forth.4. Physical parameters: density, viscosity, and so forth.

The classical definition of linear/nonlinear systems is related to whenthe system is described by a black box with empirical relation(s) betweeninput(s) (I) and output(s) (O) and the relation between I and O is linear/nonlinear (refer to Fig. 1.1). However, with more models developed on aphysicochemical basis, the usual definition is as follows: The model in linearif all state variables appear to the power 1 and without two state variablesappearing as multiplied with each other. The model is nonlinear otherwise.

For example,

y ¼ ax

is a linear model, where y and x are state variables and a is a parameter.However,

y ¼ ax2

is a nonlinear model. Also,

dx

dt¼ ax

is a linear model and

dx

dt¼ ax2

is a nonlinear model.

1.12.2 What Are Models and What Is the DifferenceBetween Models and Design Equations?

Models in the sense used in this book are equations describing the process; inother words, a model is some kind of mathematical equation(s) (of differentdegrees of rigor and complexity) that is (are) able to predict what we do notknow using the information that we know. In most cases, it is as follows:

. The output is unknown, and the input is known.

. The input is known, the desired output is fixed and it is the designparameters which are to be computed to obtain the desired output.

. It can be that the system design parameters are known, the desiredoutput is fixed, and it is the input parameters which are to becalculated.

As a matter of fact,

Mathematical models=Design equations

When the equations are based on fundamentals of the different processesand are rather rigorous and complex, they are called mathematical models.However, when they are simple and highly empirical, they are called designequations. In the present age of powerful computers, the computer is able tosolve complicated nonlinear equations in a short time using suitable solutiontechniques. Thus, the boundaries between mathematical models and designequations should disappear.

Types of Model

There are different bases on which models are classified; none of them issufficient in itself. The integral of all definitions and classifications gives thecomplete picture.

Examples are as follows:

Homogeneous and heterogeneous modelsLumped and distributed modelsRigorous (relatively rigorous) and empirical modelsLinear and nonlinear models

Steady-state and unsteady-state modelsDeterministic and stochastic models.

1.12.3 Summary of Numerical and Analytical SolutionTechniques for Different Types of Model

The various solution techniques are presented in Figures 1.6–1.10.Most of the efficient solution techniques (whether analytical or numer-

ical) will be presented in this book.

Programming

For most chemical engineering problems, the solution algorithm is quitecomplex and needs to be solved using a computer program/software.Students can use whatever programming language or computational envir-onment they want (a short list of some programming environments andsoftwares is given in Appendix F). Examples are as follows:

. Fortran with IMSL subroutines library

. C and C++ with necessary libraries

. Matlab

. Polymath

. Mathcad

. Mathematica.

Figure 1.6 Solution techniques for linear algebraic equations.

Figure 1.7 Solution techniques for nonlinear algebraic equations.

Figure 1.8 Solution techniques for linear ordinary differential equations.

Figure 1.9 Solution techniques for nonlinear ordinary differential equations.

Figure 1.10 Solution techniques for partial differential equations.

REFERENCES

1. Smith, J.M., Van Hess, H.C., Abbott, M.M., and Van Hess. Introduction to

Chemical Engineering Thermodynamics, 6th ed. McGraw-Hill, New York, 2000.

2. Bird, R.B., Stewart, W.E., and Lightfoot, E.N. Transport Phenomena. Wiley,

New York, 1960.

3. Wei, J. Educating Chemical Engineers for the Future, In Chemical Engineering

in a Changing Environment, Engineering Foundation Conference (S.L. Sandler

and B.A. Finlayson, eds.), 1988, pp. 1–12.

4. Levenspiel, O. Private discussion during AIChE meeting, San Francisco, 1989.

5. Aris, R. Manners makyth modellers. Chem. Eng. Res. Des. 69(A2), 165–174,

1991.

6. Smith, J.M. Models in Ecology. Cambridge University Press, Cambridge, 1974.

7. Stewart, I. Does God Play Dice: The New Mathematics of Chaos, Penguin,

London, 1989.

8. Hawking, S.W. A Brief History of Time: From the Big Bang to Black Holes.

Bantam Press, London, 1989.

9. Iscol, L. The dynamics and stability of a fluid catalytic cracker, Joint Automatic

Control Conference, 1970, pp. 602–607.

10. Elnashaie, S.S.E.H. and El-Hennawi, I.M. Multiplicity of the steady states in

fluidized bed reactors, IV. Fluid catalytic cracking. Chem. Eng. Sci. 34, 1113–

1121, 1979.

11. Edwards, W.M. and Kim, H.N. Multiple steady states in FCC unit operations.

Chem. Eng. Sci., 43, 1825–1830, 1998.

12. Elshishini, S.S. and Elnashaie, S.S.E.H. Digital simulation of industrial fluid

catalytic cracking units, bifurcation and its implications. Chem. Eng. Sci., 45,

553–559, 1990.

13. Amundson, N.R. Frontiers in Chemical Engineering, Research Needs and

Opportunities, National Academy Press, Washington, DC, 1988.

14. Aris, R. A quotation from Horace. Comments made during a discussion on

Chemical Engineering Education, 10th International Symposium on Chemical

Reaction Engineering, (ISCRE 10), 1988.

PROBLEM

Problem 1.1

State three of the systems that you have studied in your chemical engineer-ing courses and show whether they are open, closed, or isolated. Define thefollowing for each of the systems: the state variables, the processes takingplace in them, the laws governing the processes in terms of the state vari-ables, and the system input variables.

2

Material and Energy Balances

2.1 MATERIAL AND ENERGY BALANCES

2.1.1 A Simple, Systematic, and Generalized Approach

Material and energy (M&E) balances are the basis of almost all chemicalengineering calculations. M&E balances, in themselves, are used for theinventory of overall materials, different species and energy for single units,and total process flowsheets. It is essential, as a first step, for the design ofsingle units as well as overall processes formed of a number of units to carryout M&E balance calculations, in order to determine the basic character-istics of unit/processes and is a prelude for the detailed design of these units/processes. The tightly controlled inventory for units/process using M&Ebalances, which is as rigorous as possible, is not only important for design,operation and control but also for tight control over pollution. By defini-tion, the polluting components can be escaping reactants or impurities withthe feed or escaping products or side products. The tight control over thematerial balances and the comparison with continuous measurements afterthe plant is built is one of the best and most reliable means to predictpollution emissions from the plant (specially fugitive emissions).

The rigorous M&E balances are also the basis for the rigorous designand control equations as will be shown in Chapter 5. M&E balances seem sosimple so that it can look intuitive; for example, we gave you six oranges,

you have eaten two of them, then the remaining are certainly four oranges.We gave you $200, you spent $50 and gave a friend $30, so the remainingdollars are certainly $120. This intuitive approach was used in M&E bal-ances for a long time.

However, it is important in our present age of large complicatedprocesses and large effective computers to systematize the process ofM&E balances into generalized equations, with all simpler cases comingout of these equations as special cases. The first textbook using thisapproach is the book by Reklaitis [1]; however he did not explain thisapproach fully. This approach not only allows the systematic computeriza-tion of M&E balance calculations for small as well as large complicatedprocesses but also presents a simple, clear, and very useful link betweenM&E balance equations on the one hand and design/control equations onthe other, as shown in Chapters 3–6.

In order to make the subject exceedingly easy and smooth for thereader, we will develop this generalized and systematic approach andthese generalized equations starting from the intuitive approach (which isvery dear to a number of generations of chemical engineers). Our approachhere will make the subject very easy for the reader and will make the finalgeneralized equation almost obvious.

2.1.2 Development of Material Balance Relations

Semi-Intuitive Approach for Nonreacting Systems

Consider the system in Figure 2.1; it is obviously an open system. So afterstart-up, it reaches the state rigorously called the ‘‘stationary nonequili-brium state’’ (see Chapter 1) and commonly known in chemical engineeringas the ‘‘steady state.’’ It is a separation unit with no reactions taking place.

Note that

Xi (Weight fraction of component iÞ¼ Mass of component i in the stream

Total mass of the stream

This is a completely defined mass balance diagram for this single unit; inother words, there are no unknowns.

Let us now start training the reader on mass balance by hiding some ofthe known variables. How many variables can we hide and refind? (We willanswer this question in stages, coming at the end in a very logical systematicway to the rules of degrees of freedom analysis for single unit and completeflow sheets).

So let us have the following problem (Fig. 2.2) by hiding some of theknown variables in Figure 2.1. Counting the number of unknowns (denoted

by ?) shows that it has seven unknowns. Can we find out those sevenunknowns through the laws of mass balance? Shall we start solving directlyor we can check if it is solvable or not?

Figure 2.1 Complete material balance over a separation unit.

Figure 2.2 Material balance diagram with some missing values.

Let us first check whether the equations are solvable or not. In otherwords, from an inuititive mass balance point of view, do we have as manypossible equations as unknowns or not?

Number of Mass Balance Equations

1. Component balance

FXif ¼ FDXiD þ FW XiW for i ¼ A;B;C; and D

ð2:1Þ--ð2:4ÞBecause we have four components, we have four equations

2. Overall balance

F ¼ FD þ FW ð2:5Þ3. Summation of weight fractions for each stream is equal to 1:X

i

Xif ¼ 1:0 ð2:6ÞXi

XiD ¼ 1:0 ð2:7ÞXi

XiW ¼ 1:0 ð2:8Þ

Are all of the above eight equations independent? No, because the summingof Eqs. (2.1)–(2.4) and then using Eqs. (2.6)–(2.8) results in Eq. (2.5). So,actually, we have only seven independent equations, but we also have sevenunknowns. This leads to that the degrees of freedom are equal to zero andthe problem is solvable. This is one of the simplest examples for the degrees-of-freedom analysis.

Strategy of Solution for This Simple Case

With which mass balance shall we start?

If we start with a component balance on, say, component A (this is a badchoice); then we will have

FXAf ¼ FD XAD þ FWXAW ð2:9ÞIn this equation, we have three unknowns (XAf ; FW , and XAW ). Also, theproduct FWXAW can be considered a type of nonlinearity, which is notdesirable when we solve a sequence of linear equations.

Certainly, it is very clear that Eq. (2.9) is a very bad choice because ithas to be solved simultaneously with at least two other equations.

So, What Is a Good Choice for Starting the Solution?

It is clear that the overall balance is the only starting balance that gives anequation with one unknown. It is clear that this is the best starting step.Thus, we start by using the overall balance Eq. (2.5),

F ¼ FD þ FW

which becomes

100 ¼ 60þ FW

thus giving

FW ¼ 40 min/g

What Is the Next Best Step?

After FW has been calculated, we have other components that will give anequation in one variable. Try component B:

FXBf ¼ FDXBD þ FWXBW

On substituting the corresponding values, we get

100XBf ¼ ð60Þð0:4Þ þ ð40Þð0:15ÞTherefore,

XBf ¼ 0:3

Similarly, try component D:

FXDf ¼ FDXDD þ FWXDW

On substituting the corresponding values, we get

ð100Þð0:1Þ ¼ ð60ÞXDD þ ð40Þð0:19ÞTherefore,

XDD ¼ 0:04

Until now we have used the overall mass balance equation and two of thecomponent mass balance equations (we are left with only one) and we stillhave three

Pi Xi ¼ 1:0 relations.

What Is the Next Best Move?

If we choose to use a component mass balance on component A, we willhave an equation with two unknowns. If we choose to use a componentmass balance on component C, we will have an equation with two

unknowns. For example, the best choice is to start to use theP

i Xi ¼ 1:0relations:

XAf þ XBf þ XCf þ XDf ¼ 1:0

Thus,

XAf þ 0:3þ 0:1þ 0:1 ¼ 1:0

and we get

XAf ¼ 0:5

Now, if we use a component mass balance on component A, we will have anequation with only one unknown:

FXAf ¼ FDXAD þ FWXAW

Substitution of values gives

ð100Þð0:5Þ ¼ ð60Þð0:5Þ þ 40XAW

Calculation gives

XAW ¼ 0:5

We have thus exhausted the component balances; we can then directly usethe remaining two

Pi Xi ¼ 1:0 relations. UsingX

i

XiD ¼ 1:0

we get

XAD þ XBD þ XCD þ XDD ¼ 1:0

Substituting the values gives

0:5þ 0:4þ XCD þ 0:04 ¼ 1:0

Thus,

XCD ¼ 0:06

Also, for the other stream,Xi

XiW ¼ 1:0

we get

XAW þ XBW þ XCW þ XDW ¼ 1:0


0:5þ 0:15þ XCW þ 0:19 ¼ 1:0

Thus,

XCW ¼ 0:16

Note: By the proper sequence of solution, we went through the problemwithout solving any simultaneous equations. Each equation we havewritten in the proper sequence had only one unknown.

Suppose now that we have the following arrangement of the knownand unknown variables (keeping the total number of unknowns the same) asshown in Figure 2.3. Counting the number of unknowns, there are obviouslythe same number of unknowns as earlier (seven unknowns). The unawarechemical engineer who does not use the degrees-of-freedom concept fluentlywill immediately go into trying to solve this problem, will spend a long timeand lots of frustration, and will end up not being able to solve it, whereas thechemical engineer with good awareness regarding the use of the degrees-of-freedom analysis will realize in a few seconds that the problem is notsolvable. Why? Because, actually, with the given structure of unknowns,we have only six relations not seven. Why? This is because for the feedstream the relationX

i

Xif ¼ 1:0

Figure 2.3 Material flow diagram with unknown values.

is redundant, because all of the weight fractions of the four components aregiven in this stream.

Now let us pose the same mass balance problem in a third fashion(Fig. 2.4). Here, we have eight unknowns and seven relations. Thus, thedegree of freedom is 1, so strictly speaking, the problem is not solvable.However, actually, it can be solved for an assumed value of F or FD or FW ,which will be called the basis and we will notice a simple linear relationbetween the solution and the chosen basis. This linear relation makes thesolution quite useful.

1. Assume F¼100 min/g (basis)

Here, we have seven unknowns and seven relations.

Step 1

100 ¼ FD þ FW

which gives

FW ¼ 100� FD

Step 2

UseXi

Xif ¼ 1:0

Figure 2.4 Material flow diagram with unknown values.

To get


which gives

XAf þ XBf ¼ 0:8

Step 3

The mass balance on component A is


which gives

100XAf ¼ FDð0:5Þ þ ð100� FDÞð0:5ÞThus,

XAf ¼ 0:5

Step 4

From the relation in step 2, we get

XBf ¼ 0:8� 0:5 ¼ 0:3

Step 5

The mass balance on component B is



ð100Þð0:3Þ ¼ FDð0:4Þ þ ð100� FDÞð0:15ÞThus,

FD ¼ 60 min/g

Then, using the relation in step 1, we get,

FW ¼ 40 min/g

obtaining the rest of the variables is really straightforward. Thereader is advised to compute the remaining unknown values.

2. Assume a basis of F=50 g/min

Step 1

50 ¼ FD þ FW

which gives

FW ¼ 50� FD

Step 2

Use

Xi

Xif ¼ 1:0

to get


which gives

XAf þ XBf ¼ 0:8

Step 3

The mass balance on component A is


which gives

50XAf ¼ FDð0:15Þ þ ð50� FDÞð0:5ÞThus,

XAf ¼ 0:5

Step 4

From the relation in step 2, we get

XBf ¼ 0:8� 0:5 ¼ 0:3

Step 5

The mass balance on component B is



ð50Þð0:3Þ ¼ FDð0:4Þ þ ð50� FDÞð0:15ÞThus,

FD ¼ 30 min/g

Then, using the relation in step 1, we get

FW ¼ 20 min/g

Thus, the difference in the values by taking different basis can besummarized as in Table 2.1. Because

F(first basis)

F(second basis)¼ 2

we have

FD(first basis)

FD(second basis)¼ 2 and

FW (first basis)

FW (second basis)¼ 2

It should be noted that in both cases the weight fractions of all componentsin all streams remain the same.

This is the essence for the wide and very useful use of basis with regardto mass or molar flow rates.

Cases Where Taking Basis Does Not Solve the Problem ofthe Number of Unknowns Being Larger Than the Numberof Relations

What will be the situation when we have eight unknowns (all of them weightfractions, as shown in Figure 2.5)? In this case, let us proceed a few steps inthe calculations. We can obtain XAf by


Table 2.1 Comparison of Values for

Different Bases

Basis: F1 ¼ 100 g/min Basis: F2 ¼ 50 g/min

FD1 ¼ 60 g/min FD2 ¼ 30 g/min

FW1 ¼ 40 g/min FW2 ¼ 20 g/min

On substituting the values, we get

100XAf ¼ ð60Þð0:5Þ þ ð40Þð0:5Þ

Thus, we get

XAf ¼ 0:5

After that, we cannot go any further with the computations; the remainingequations are six in number and the unknowns are seven in number and nobasis can be taken because F , FD, and FW have specified values.

2.2 SINGLE AND MULTIPLE REACTIONS:CONVERSION, YIELD, AND SELECTIVITY

Now, we move one important step forward toward the mass balance of unitswith chemical reactions (chemical reactors). In order to reach this stage in asystematic way, we start by analyzing the different types of reaction andintroduce the necessary additional information needed for the mass balanceof units with chemical reactions.

Figure 2.5 Mass flow diagram where basis does not work.

2.2.1 Single Reactions

A single reaction is a reaction formed of one stoichiometric relation regard-less of the number of components (species) involved as reactants orproducts.

Example of Single Reactions

A ! B (two components, irreversible)A,B (two components, reversible)Aþ B ! C þD (four components, irreversible)Aþ B,C þD (four components, reversible)5Aþ 8Bþ 9C ! 3Dþ 6F (five components, irreversible)3Aþ 2B, 5C þ 7D (four components, reversible)

A reversible reaction such as

A,B

is strictly depicting two reactions. However, it has many relations to singlereactions because of the special nature of the two reactions involved.However, in this book, for straightforward consistency we will alwaystreat it as two reactions.

The number in front of each component is called the stoichiometricnumber and it is an indication of the proportion (in moles) of the compo-nent entering the reaction in comparison with other components (reactantsor products).

The following stoichiometric relations are strictly equivalent:

5Aþ 10B ! C þ 20D

Aþ 2B ! 15C þ 4D

50Aþ 100B ! 10C þ 200D

Conversion and Yield for a Single Reaction

A reactor with a single reaction can be completely defined (and solved) interms of one variable (conversion of any one of the reactants, or the yield ofany one of the products, or the concentration of one of the components, orthe rate of reaction of one of the components) because all other quantitiescan be computed in terms of this single variable as long as the stoichiometricequation is fully defined. This will be clearly known after we define conver-sion and yield.

The conversion of a reactant component is defined as the number ofmoles reacted of the specific component divided by the original number ofmoles of the same specific component. For continuous operation, this state-ment will read: Conversion of a reactant component is defined as thenumber of moles reacted per unit time of the specific component dividedby the feed numbers of moles per unit time of the same specific component(i.e., molar flow rate).

For a reactor as the one shown in Fig. 2.6, the reaction is

Aþ B ! C þD ðfour components, irreversible)

The conversion of component A is defined as

xA ¼ nAf � nA

nAf

The conversion of component B is defined as

xB ¼ nBf � nB

nBf

The relation between xA and xB depends on the stoichiometric numbersof each component. This will be shown and discussed later. Even if thestoichiometric numbers are different,

2Aþ 5B ! 7C þ 8D

the definitions still remain strictly the same:

xA ¼ nAf � nA

nAf

Figure 2.6 Molar flow diagram across a reactor.

and

xB ¼ nBf � nB

nBf

Only the relation between xA and xB will be different because of thestoichiometric numbers, as will be explained later.Note: It is possible to use the same basis for the two conversions, such as

xA ¼ nAf � nA

nAf

and

xB ¼ nBf � nB

nAf

Of course, the relation between xA and xB will be different because of thedifferent ways we define the basis for each component.

Yield of a Product (e.g., C )

The yield of a product C is the number of moles of C formed (per unit timefor continuous processes) divided by the original (or feed for continuousprocesses) number of moles of reactant A (per unit time for continuousprocesses); that is,

Y ¼ nC � nCf

nAf

or we can reference the yield to reactant B; that is,

Y ¼ nC � nCf

nBf

Also, the yield can be referred to the number of moles of A (or B) thatreacted:

Y 0 ¼ nC � nCf

nAf � nAor Y 00 ¼ nC � nCf

nBf � nB

Mass Balance Calculations for a Reactor with a SingleReaction

As we mentioned earlier, the mass balance of the reactor can be definedcompletely in terms of only one variable (xA; xB;Y;Y 0;Y 00, etc.) when thereis a single reaction.

2.2.2 Degrees-of-Freedom Analysis

Solved Example 2.1

For the reactor shown in Figure 2.7, with known feed conditions ðnAf ;nBf ; nCf ; nDf Þ, determine the relations giving the output variables ðnA; nB;nC; nD) in terms of one variable ðxA or xB or Y) for the followingreaction:

2Aþ 3B ! 3C þ 5D

If the number of unknowns is four (nA; nB; nC; nDÞ plus one extra variable(the rate of reaction, conversion, or yield; the relation among them will begiven later), then the total number of unknowns is five.

The number of mass balance relations for four components is four. Werequire an additional relation in terms of conversion (or yield). Suppose thatthe conversion is given, then this is an extra given ‘‘known variable.’’ Thus,the degrees of freedom is zero and the problem is solvable.

Solution for This Continuous Process

xA ¼ nAf � nA

nAf

where nAf � nA is the number of moles of component A reacted per minuteðRA ¼ nAf xAÞ, where RA is the rate of consumption of component A (molesof A consumed per unit time).

Figure 2.7 Molar flow diagram for solved example 2.1.

From the stoichiometric equation, we have the number of moles ofcomponent B reacted per unit time ðRB ¼ 3

2number of moles of A reacted

per unit time):

RB ¼ 32ðnAf � nAÞ ¼ 3

2nAf xA

where RB is the rate of consumption of component B.

From Stoichiometric Equations

The number of moles of C produced per unit time ðRC ¼ 32number of moles

of A reacted per unit time) is

RC ¼ 32ðnAf � nAÞ ¼ 3

2nAf xA

The number of moles of D produced per unit time ðRD ¼ 52number of moles

of A reacted per unit time) is

RD ¼ 52ðnAf � nAÞ ¼ 5

2nAf xA

Now, the simplest rational thinking will tell us that

nA ¼ nAf � RA

nB ¼ nBf � RB

nC ¼ nCf þ RC

nD ¼ nDf þ RD

Thus,

nA ¼ nAf � nAf xA ¼ nAf ð1� xAÞnB ¼ nBf � 3

2nAf xA

nC ¼ nCf þ 32nAf xA

nD ¼ nDf þ 52nAf xA

Thus, all the exit variables from the reactor are given in terms of the knowninput variables nAf ; ; nBf ; nCf ; nDf , and one conversion (i.e., xA).

Numerical Values for Example 2.1

nAf ¼ 10 mol/min

nBf ¼ 30 mol/min

nCf ¼ 2 mol/min

nDf ¼ 1:5 mol/min

Then, for any given conversion, we can obtain nA; nB; nC, and nD. Say, forexample, the conversion is 80%; then,

xA ¼ 0:8

From the above equations,

nA ¼ 10ð1� 0:8Þ ¼ 10ð0:2Þ ¼ 2 mol/min

nB ¼ 30� 32ð10Þð0:8Þ ¼ 30� 12 ¼ 18 mol/min

nC ¼ 2þ 32ð10Þð0:8Þ ¼ 2þ 12 ¼ 14 mol/min

nD ¼ 1:5þ 52ð10Þð0:8Þ ¼ 1:5þ 20 ¼ 21:5 mol/min

Exactly the same can be done in terms of xB or in terms of Y .The reader should practice obtaining similar relations once in terms of

xB and once in terms of Y .

Rate of Reaction

The rate of reaction in the context of material and energy balance is definedas the number of moles produced of a component per unit time. Of course,produced or consumed in the definition is a matter of convention. We willchoose the convention that R is a rate of production. However, this isarbitrary, and the reader should practice the derivation with R, which isthe rate of consumption.

According to the above sign convention, the rate of reaction (produc-tion) of component A is obviously negative, because actually A is a reactant(it is being consumed):

RA ¼ ðnA � nAf ÞIf we write the rate of consumption of A and call it RA, then it is positiveand defined as

RA ¼ ðnAf � nAÞThis simple fact should be very clear in the mind of the reader. It is reallyvery simple; it is like saying that John lost $100, which is exactly the same assaying that John gained � $100.

However, of course, we have to follow a certain convention. Do wework with the rates of production (and in this case, rates of consumption areonly negative rates of production) or do we work with the rates of consump-tion (and in this case, rates of production are only negative rates of con-sumption)? The choice is arbitrary; however, the chemical engineer should

be very clear about the above meanings because in the complex process ofdesign in the chemical industry, different people (and different books andmanuals) may choose the convention differently, which may cause confusionif the chemical engineer is not completely aware of the above simple andfundamental facts.

We will choose to work with the rates of production. Thus, for thereaction

2Aþ 3B ! 3C þ 5D

we will have the following:

Rate of production of A ¼ RA ¼ nA � nAf (negative)

Rate of production of B ¼ RB ¼ nB � nBf (negative)

Rate of production of C ¼ RC ¼ nC � nCf (positive)

Rate of production of D ¼ RD ¼ nd � nDf (positive)

Therefore, we have four rates of reaction: RA, RB, RC and RD. Does thismean that the single reaction is defined by four different rates of reactions?Of course not! The four rates are related by the stoichiometric numbers.Intuitively, we can easily note that

RB ¼ 32RA

RC ¼ �32RA

RD ¼ �52RA

Thus, one rate of reaction (e.g., RA; it can of course be RB or RC or RD, butonly one rate of reaction) defines automatically, together with the stoichio-metric numbers, the other rates of reactions as long as it is a single reactionsystem.

We can obviously define the rates of reaction in terms of RB or RC orRD (but only one rate) as follows:

1. The above relations are in terms of component A.

2. In terms of RB,

RA ¼ 23RB

RC ¼ �RB

RD ¼ �53RB

3. In terms of RC,

RA ¼ �23RC

RB ¼ �RC

RD ¼ 53RC

4. In terms of RD,

RA ¼ �25RD

RB ¼ �35RD

RC ¼ 35RD

Thus, the rates of reaction of a single reaction can be all expressed in termsof one rate of reaction as long as the stoichiometric numbers for this singlereaction are all known. We will define the generalized rate of reaction ðrÞ in alater section.

2.2.3 Relations Among Rate of Reaction, Conversion,and Yield

We have made it very clear earlier that the single reaction system is fullydefined (and solvable) in terms of any conversion for one of the reactants orany yield of one of the products. We have also made it very clear that allrates of reaction of the components of a single reaction are fully defined interms of one rate of reaction.

Does this mean that we need one conversion (or yield) plus one rateof reaction to define (and solve) the system? Of course not! This willcontradict the strict statement that the system is completely defined (andsolvable) in terms of one variable. In fact, we need only either oneconversion (or yield) or one rate of reaction. This is because rates ofreaction are directly related to conversions (or yields), as will be shownin the next few lines.

It also has to be very clear to the reader that if he/she has a degree offreedom in a problem that is not zero, he/she cannot use a conversion (oryield) and rate of reaction as two given relations. If conversion (or yield)is used, any rate of reaction information is redundant from the degree-of-freedom point of view and vice versa. This simple fact will be made evenclearer after we show the relations between conversion (or yield) and rate ofreaction.

Rate of Reaction, Conversions, and Yields

Consider the reaction

2Aþ 3B ! 3C þ 5D

taking place in the reactor in Figure 2.8. It is straightforward to define RA asfollows:

Rate of production of component A,

RA ¼ nA � nAf

which is negative because A is consumed not produced.The fractional conversion xA is defined as

xA ¼ nAf � nA

nAf¼ Number of moles of A reacted

Number of moles of A fed

Therefore, from the above two relations, it is very clear that

xA ¼ � RA

nAf

or, equivalently,

RA ¼ �nAf xA

Thus, there is a direct relation between RA and xA.

Figure 2.8 Molar flow diagram.

Similar relations can be found between RA and Y (yield) as follows.Because

RA ¼ nA � nAf

YC (yield of component C) can be defined as

Yc ¼nC � nCf

nAf¼ Number of moles of C produced

Number of moles of A fed

However, we know from the stoichiometry of the reaction that the numberof moles (per unit time for continuous process) produced of C is 3

2times the

number of moles of A reacted; that is,

nC � nCf ¼ 32ðnAf � nAÞ

Therefore,

Yc ¼32ð�RAÞnAf

which gives

RA ¼ � 23YCnAf

Similar relations can be developed between any rate of reaction ðRA;RB;RC;or RD) and any reactant conversion (xA or xB) or any yield ðYC or YD). Thereader should practice deriving these relations.

The following should be clear:

1. For a single reaction, the definition of one conversion (or yield)makes any other definition of any other conversion (or yield) orany other rate of reaction redundant.

2. For a single reaction, the definition of one rate of reaction makesany other definition of any other rate of reaction or conversion(or yield) redundant.

Solved Example 2.2

Again consider the reaction

2Aþ 3B ! 3C þ 5D

taking place in the reactor in Figure 2.9 with the shown feed molar flowrates of various components. The number of unknowns for this reactor isfour ðnA; nB; nC; nD), but because there is a single reaction, there is an extraunknown (a rate of reaction or a conversion for the above equation). Thiswill add up to five unknowns.

Relations

We have four independent mass balance relations. Thus, the problem asshown is not solvable. Why? The reason is that if we define the degrees offreedom as the difference between number of unknowns and the number ofrelations, we have

Degrees of freedom ¼ 5 unknowns� 4 mass balance relations ¼ 1

(The reader can obviously define the degree of freedom in the oppositesense; i.e. as the difference between number of relations and the numberof unknowns. This gives the degrees of freedom to be 4� 5 ¼ �1; thisshould not cause any problem to the reader). Now, if this problem is tobe solved, something else should be defined, let us consider the followingcases:

1. Conversion of A defined as 80%

In this case, the problem is solvable because the degree of free-dom now is zero. For

xA ¼ nAF � nAnAf

we get

0:8 ¼ 10� nA10

giving

nA ¼ 2 min/mol

Now, because 8 mol of A was consumed, the number of moles ofB consumed is equal to ð3=2Þð8Þ ¼ 12 mol/min. Thus,


nB ¼ 40� 12 ¼ 28 mol/min

Because, 8 moles of A was consumed, the number of moles of Cproduced is equal to ð3=2Þð8Þ ¼ 12 mol/min. Thus,

nC ¼ 3þ 12 ¼ 15 mol/min

Also, because 8 moles of A was consumed, the number of molesof D produced is ð3=2Þð8Þ ¼ 20 mol/min and

nD ¼ 1:5þ 20 ¼ 21:5 mol/min

Making the Above Calculations More Systematic

As we agreed earlier, we define R as the rate of production; there-fore,

nA ¼ nAf þ RA

nB ¼ nBf þ RB

nC ¼ nCf þ RC

nD ¼ nDf þ RD

However, we also agreed earlier that all the rates of reaction canbe expressed in terms of one rate of reaction and the stoichio-metric numbers; thus,

RB ¼ 32RA

RC ¼ �32RA

RD ¼ �52RA

Therefore, the four mass balance equations can be written interms of RA:

nA ¼ nAf þ RA

nB ¼ nBf þ 32RA

nC ¼ nCf � 32RA

nD ¼ nDf � 52RA

However, we also agreed that the rates of reaction can beexpressed in terms of conversion (or yield) and vice versa.Thus, we have

RA ¼ �nAf xA

This will give RA ¼ �8 mol/min, which can be substituted in theprevious equation (or we write the equations in terms of xA).

Therefore, the mass balance equations can be written as

nA ¼ nAf � nAf xA

nB ¼ nBf � 32nAf xA

nC ¼ nCf þ 32nAf XA

nD ¼ nDf þ 52nAf xA

Substituting the numbers of the specific example in the aboverelations give the same results as we obtained earlier:

nA ¼ 10� ð10Þð0:8Þ ¼ 10� 8 ¼ 2 min/mol

nB ¼ 40� 32ð10Þð0:8Þ ¼ 40� 12 ¼ 28 min/mol

nC ¼ 3þ 32ð10Þð0:8Þ ¼ 3þ 12 ¼ 15 min/mol

nD ¼ 1:5þ 52ð10Þð0:8Þ ¼ 1:5þ 20 ¼ 21:5 min/mol

2. Rate of reaction of A defined as RA ¼ �8 mol/min

In this case, we use the relations

nA ¼ nAf þ RA

nB ¼ nBf þ 32RA

nC ¼ nCf � 32RA

nD ¼ nDf � 52RA

Therefore, we again obtain the same results we obtained earlier:

nA ¼ 10� 8 ¼ 2 min/mol

nB ¼ 40� 32ð8Þ ¼ 28 min/mol

nC ¼ 3� 32ð�8Þ ¼ 15 min/mol

nD ¼ 1:5� 52ð�8Þ ¼ 21:5 min/mol

The reader is advised to solve for cases where other conversionðxBÞ is defined or one of the yields ðYC or YDÞ or one other rate ofreaction ðRB;RC, or RD) is defined.

An Important Question

Can other relations (other than one conversion or one yield or one rate ofreaction) be used to solve the problem?

To answer this question, let us consider the same problem (Fig. 2.10).The reaction is again the same,

2Aþ 3B ! 3C þ 5D

and we have the same reactor with the same feeds and feed flow rates.Instead of giving one conversion (or one yield) or one rate of reaction wegive one relation,

nB ¼ 14nA

Can we solve the problem? Yes, of course, because the degrees of freedomis zero.

We have the following mass balance (with a single chemical reaction)relations:

nA ¼ nAf þ RA ð2:10ÞnB ¼ nBf þ 3

2RA ð2:11Þ

nC ¼ nCf � 32RA ð2:12Þ

nD ¼ nDf � 52RA ð2:13Þ

we also have

nB ¼ 14nA ð2:14ÞOn substituting the feed conditions, Eq. (2.10) becomes

nA ¼ 10þ RA ð2:15Þ


On substituting the feed conditions and using Eq. (2.14), Eq. (2.11) gives

14nA ¼ 40þ 32RA ð2:16Þ

Substituting nA from Eq. (2.15) into Eq. (2.16) gives

14ð10þ RAÞ ¼ 40þ 32RA

which becomes

100 ¼ �12:5RA

Therefore, the rate of reaction of component A is obtained as

RA ¼ �8 min/mol

Then, substituting RA in Eqs. (2.10)–(2.13) gives

nA ¼ 10� 8 ¼ 2 min/mol

nB ¼ 40� 32ð8Þ ¼ 28 min/mol

nC ¼ 3þ 32ð8Þ ¼ 15 min/mol

nD ¼ 1:5þ 52ð8Þ ¼ 21:5 min/mol

Thus, from the given relation ðnB ¼ 14nAÞ, we are able to calculate RA andthen the solution proceeds as before. However, the reader cannot considerthis relation together with a given rate of reaction or conversion as twoindependent relations. To make this point even clearer, let us consider thesame problem but with five flow rates unknown (meaning it is a sixunknowns problem because of the single reaction) (Fig. 2.11).

The reaction is the same as earlier:

2Aþ 3B ! 3C þ 5D


Total number of unknowns ¼ 6 unknowns (5 molar flow ratesþ1 rate of reaction)

Mass balance reactions ¼ 4Degrees of freedom ¼ 2

Thus, this problem is not solvable.If we define any one relation,

xA ¼ 0:8 or RA ¼ �8 or nB ¼ 14nA

we will have degree of freedom ¼ 6� 5 ¼ 1. Still the problem is unsolvable!What if we define any two of the above relations, say,

RA ¼ �8 and nB ¼ 14nA

Now, the absent-minded chemical engineer or the one who is not strictly clearabout the degrees-of-freedom concept (the one who does not know at allabout the degree-of-freedom-concept will be in even worse condition) willcount as follows: I have six unknowns, I have four mass balance relations,and two extra relations (RA ¼ �8 and nB ¼ 14nA), then the degree of freedomis equal to zero and the problem is solvable! This is completely wrong!!! RA ¼�8 and nB ¼ 14nA are not two independent relations and the problem is stillnot solvable. The reader should try to solve it to see why it is unsolvable.

Another Important Question

Can two other independent relations be defined that will make the aboveproblem solvable? The answer is a conditional yes, depending on the trueindependence of these relations.

Case 1: For the above problem, we will define

RA ¼ �8 andnBf

nAf¼ 4

Now, the problem is solvable, because immediately from the second relationwe can find that nBf ¼ 40 and the problem is solvable.

This appears too obvious, let us try something that is not that obvious.

Case 2: For the above problem, we define

RA ¼ �8 and nC ¼ 7:5nA

Is this solvable? Let us try.We have

nA ¼ nAf þ RA ð2:10Þ

nB ¼ nBf þ 32RA ð2:11Þ



Using RA ¼ �8 and nC ¼ 7:5nA, we get

nA ¼ 10� 8 ¼ 2 min/mol

nB ¼ nBf � 32ð8Þ ¼ ðnBf � 12Þ min/mol

nC ¼ 3þ 32ð8Þ ¼ 15 min/mol

nD ¼ 1:5þ 52ð8Þ ¼ 21:5 min/mol

The problem is not solvable, as we still have nB and nBf as unknowns.We notice from the above that automatically by applying

RA ¼ �8

we get

nCnA

¼ 7:5

The relation we gave as nC ¼ 7:5nA is actually redundant and cannot beused to obtain the unknowns nB and nBf .

Conclusion

Because for given feed conditions, the output conditions (for the case of asingle reaction) are completely defined in terms of one and only one variable(conversion of any reactant, or yield of any product, or rate of reaction ofany component), then only one relation related to these variables or onerelation relating the output variables together can be used in the solution ofthe problem (and in the determination of the degrees of freedom).

2.3 GENERALIZED MATERIAL BALANCE

In the previous section, we have seen that, for a single reaction, the ratesof reaction for different components can all be expressed in terms of the rateof reaction of one component (together with the stoichiometric numbers), orthe conversion of one of the reactants, or the yield of one of the products (ofcourse, together with the stoichiometric numbers). These information andrelations for the single reaction are adequate for the solution of any massbalance problem with a single reaction and can be easily extended to multi-ple reaction systems, as will be shown later. However, in this section, we willtry to make the calculations even more systematic. This will require, first,that we introduce the sign convention for the stoichiometric numbers, as we

introduced earlier the sign convention for the rates of reaction (positive forproduction and negative for consumption).

2.3.1 Sign Convention for the StoichiometricNumbers

The sign convention is arbitrary, and the reader should be very clear aboutthis and should not memorize certain relations under a certain sign conven-tion. The equations you use should be always correct under the chosen signconvention. However, it is best to adhere to the sign convention used bymost chemical engineers and textbooks.

Most books use the following sign convention:

Ri ¼ rate of production of component iRate of production ¼ positiveRate of consumption ¼ negativeStoichiometric number of products ¼ positiveStoichiometric number of reactants ¼ negativeStoichiometric numbers for inerts ¼ zero

We will use different sign conventions and show that as long as it is con-sistent it is correct.

I. First Sign Convention (Not Commonly Used)

Ri ¼ rate of production of component i ¼ negativeStoichiometric number of reactants (�R) ¼ positiveStoichiometric number of products ð�PÞ ¼ negative

We use the reaction given earlier,

2Aþ 3B ! 3C þ 5D

Using this uncommon sign convention, we get

RA ¼ rate of production of component A ¼ negative�A ¼ 2�B ¼ 3�C ¼ �3�D ¼ �5

The balance equations developed earlier are

nA ¼ nAf þ RA ð2:10Þ

nB ¼ nBf þ 32RA ð2:11Þ



and the relations between R’s are

RB ¼ 32RA ð2:17Þ

RC ¼ �32RA ð2:18Þ

RD � 52RA ð2:19Þ

Of course, we could have used RB, RC, or RD as the basis rather than RA. Toavoid this multiplicity of choices, we can define a rate of reaction that doesnot depend on the component. It is only related to the reaction; let us callthis rate ‘‘the generalized rate of reaction, r’’:

r ¼ Ri

�i

We can easily show that r does not depend on the choice of i. If i ¼ A, then

rA ¼ RA

�a¼ RA

2ð2:20Þ

If i ¼ B, then

rB ¼ RB

�B¼ RB

3¼ 1

3

3

2RA

� �¼ RA

2ð2:21Þ

If i ¼ C, then

rC ¼ RC

�C¼ RC

�3¼ 1

�3

�3

2RA

� �¼ RA

2ð2:22Þ

If i ¼ D, then

rD ¼ RD

�D¼ RD

�5¼ 1

�5

�5

2RA

� �¼ RA

2ð2:23Þ

Therefore, it is clear that

rA ¼ rB ¼ rC ¼ rD ¼ r ¼ generalized rate of reaction:

Thus, for the given uncommon sign convention (which is usually not used),

r ¼ Ri

�i) always negative ð2:24Þ

Therefore, the mass balance equations can be written as

nA ¼ nAf þ�A�i

Ri ¼ nAf þ �Ar ð2:25Þ

nB ¼ nBf þ�B�i

Ri ¼ nBf þ �Br ð2:26Þ

nC ¼ nCf þ�C�i

Ri ¼ nCf þ �Cr ð2:27Þ

nD ¼ nDf þ�D�i

Ri ¼ nDf þ �Dr ð2:28Þ

where Ri is the rate of production of the component i used as a basis for rand �i is its stoichiometric number according to the sign convention chosen.

Equations (2.25)–(2.28) can be rewritten in general as

ni ¼ nif þ �ir

�i > 0 for reactants

�i < 0 for products

and

r ¼ Ri

�i) always negative

ð2:29Þ

II. Second Sign Convention (Commonly Used)

Ri ¼ rate of production of component i ¼ positiveStoichiometric number of reactants ð�RÞ ¼ negativeStoichiometric number of products ð�PÞ ¼ positive

Relation (2.29) remains the same, but r is now always positive (even when Ri

is negative),

ni ¼ nif þ �ir

�i < 0 for reactants

�i > 0 for products

and

r ¼ Ri

�i) always positive

ð2:30Þ

Relations Among Ri; r; xi, and Limiting Components

RA ¼ rate of production ¼ nA � nAf

and

xA ¼ nAf � nA

nAfð2:31Þ

Thus,

xA ¼ � RA

nAfð2:32Þ

or

RA ¼ �nAf xA ¼ � �A�A

nAf xA

RB ¼ �B�A

RA ¼ � �B�A

nAf xA

RC ¼ �C�A

RA ¼ � �C�A

nAf xA

RD ¼ �D�A

RA ¼ � �D�A

nAf xA

In general, we can write

Ri ¼ � �i�A

nAf xA ð2:33Þ

Therefore, the relation between the rate of reaction of any component andthe conversion for a specific equation does not depend on the stoichiometricnumber sign convention. Of course, we can choose any other componentother than A as a base for conversion (if it is another reactant, then it will beconversion, but if it is a product, then it will be called yield, not conversion).Thus, the general form of Eq. (2.33) is

Ri ¼ � �i�k

nkf xk ð2:34Þ

where nkf is the feed number of moles on which basis x (conversion) isdefined.

Now, the generalized rate of reaction is

r ¼ Ri

�i¼ � nkf xk

�kð2:35Þ

When we substitute into the mass balance relation, we have

ni ¼ nif þ �ir ð2:36Þwhich is actually

ni ¼ nif ��i�A

nAf xA ð2:37Þ

2.3.2 The Limiting Component

This is a very important concept regarding the rational definition of theconversion for a certain feed component; thus, this definition should notviolate the natural laws of the reaction itself. When the stoichiometric num-bers for all reactants are equal, the problem is rather trivial; it is even moretrivial when the stoichiometric numbers of the reactants are equal and thefeed is equimolar. However, when the stoichiometric numbers of the reac-tants are not equal and/or the feed molar flows of the different componentsare not equal, the problem is not trivial, although it is very simple.

Consider, first, the following reaction with equal stoichiometricnumbers for the different components:

Aþ B ! C þD

taking place in the reactor in Figure 2.12 with the shown feed molar flowrates for the four components.

If we specify the conversion of A as xA ¼ 90%, then we find that

nA ¼ nAf � nAf xA ¼ 10� 9 ¼ 1 min/mol


Now,

nB ¼ nBf � nBf xA ¼ 6� 9 ¼ �3 min/mol!!!!

Obviously, this is impossible and it means that the molar flow rate of com-ponent B does not allow 90% conversion of component A; however, if wespecify the conversion of B as xB ¼ 90%. Then, we have

xB ¼ nBf � nB

nBfand nB ¼ nBf � nBf xB ¼ 6� 6ð0:9Þ ¼ 0:6 min/mol

Therefore,

nA ¼ nAf � nBf xB ¼ 10� 6ð0:9Þ ¼ 4:6 min/mol

Thus, B is the limiting component, which means that it is the componentwhich gets exhausted or reacted completely before the other reactant isexhausted. In other words, its conversion can be specified as high as100% without violating the natural laws of the reaction.

Therefore, for reactions in which all of the reactants have the samestoichiometric numbers, the limiting component is the reactant with thelowest number of the moles in feed. For this case of equal stoichiometricnumbers for the reactants, if we additionally have equimolar feed, than anyreactant can be the limiting reactant.

2.3.3 Reactions with Different StoichiometricNumbers for the Reactants

Consider the following chemical reaction with different stoichiometric num-bers for different components:

2Aþ 10B ! 6C þ 3D ð2:38ÞNow, if we take for this reaction that the limiting component is the compo-nent with the lowest number of moles in the feed, we will fall into a seriousmistake. How?

Considering the case shown in Figure 2.13, if we wrongly use theprevious statement (the limiting component is the reactant with the lowestnumber of moles in the feed), we will strongly say that A is the limitingcomponent. Let us check, say, the conversion of component A, xA ¼ 90%,then

nA ¼ nAf � nAf xA ¼ 10� 9 ¼ 1 min/mol

Now,

nB ¼ nBf �10

2nBf xA ¼ 20� ð5Þð10Þð0:9Þ ¼ �25 min/mol!!!!

Obviously this is wrong. A is not the limiting component (or reactant); infact, B is the correct limiting component (despite the fact that nBf > nAf ; thisis because j�Bj >> j�AjÞ. The reader could find that out directly from look-ing at the ratio of the stoichiometric numbers. However, for complexsystems, it is not that obvious.

A Deceptive Wrong Procedure for the Determination of theLimiting Component

Let us refer to the generalized form of the mass balance equations forreactants:

ni ¼ nif þ �ir


ni ¼ nif þ�i�jRj

where Rj ¼ rate of production (negative for reactants).Let us make

ni ¼ nif þ�i�jRjmax

the limit of this. The reactant i with the largest absolute value of ð�i=�jÞRjmax

is the limiting component.However, now, what is Rjmax

? It is Rj at 100% conversion (obvious,isn’t it?).

Therefore may be it is better to write the above relation in terms ofconversion xj as follows. Because

Rj ¼ nj � njf


and

xj ¼njf � nj

njf

therefore

xj ¼ � Rj

njf

From the above relation, we get

Rj ¼ �xjnjf

and, therefore,

Rjmax¼ �ð1:0Þnif (as maximum conversion is 100% or xj ¼ 1:0Þ

Therefore, the mass balance relation will be reduced to

ni ¼ nif ��i�jnjf (at maximum conversion xj ¼ 1:0Þ

Thus, the quantity

~xxi ¼ �i�jnjf

is the measure for the controlling component i. The component i with thelargest value of ~xxi is the rate-limiting component. Here, reactant i with thelargest value of ~xxi is the limiting component. (Be careful, this is a wrongconclusion!! Although, it will be correct in the next deceiving example).

Application to the Previous Problem

Now, for the case shown in Figure 2.13 and for reaction (2.38), we want tofind the reactant i (A or B) which represents the limiting component. Thebase component j is arbitrary, we can actually choose whatever we like.Choose component j to be A. Then,

~xxi ¼�A�A

nAf

which gives

~xxA ¼ �A�A

� �10 ¼ 10

and

~xxB ¼ �B�A

� �10 ¼ 10

2

� �10 ¼ 50

Thus, the limiting components is B (as ~xxB is higher than ~xxAÞ. Therefore, B,again, is the limiting component regardless of the basis j taken (A or B). Isthe procedure correct, or can it cause errors? The reader should think aboutit and check this question before proceeding to the next part, where acounterexample is given.

Counterexample

The previous procedure for the determination of the limiting reactant,although it ‘‘sounds’’ logical and when applied to the previous examplegave correct results, it is actually wrong!! It is important for the reader torealize that he/she cannot develop a procedure and check for one case andfrom the result of this single case conclude that it is correct. Things shouldbe thought about much deeper.

We will now give a counterexample (Fig. 2.14) showing that theprevious procedure is not correct. We will only change the feed conditions;the reaction remains the same as in the Eq. (2.38) case.

Let us apply the previous procedure:

1. Choose component A as j.

~xxi ¼�A�A

nAf

which gives

~xxA ¼ �A�A

� �10 ¼ 10

and

~xxB ¼ �B�A

� �10 ¼ 10

2

� �10 ¼ 50


Thus, B is the limiting component according to this (wrong)procedure.

2. Choose component B as j.

~xxi ¼�B�B

nBf

which gives

~xxB ¼ �B�B

� �nBf ¼

10

10

� �2000 ¼ 2000

and

~xxA ¼ �A�B

� �nBf ¼

2

10

� �2000 ¼ 400

Thus, B is the limiting component according to this procedure(wrong again!!).

In fact, B is certainly not the limiting component. For if B is the limitingcomponent, then we should be able to define xB ¼ 100% without violatingthe natural laws of the reaction.

Let us do that and see. Just take xB ¼ 90% ¼ 0:9. Now, calculate

nA ¼ nAf ��A�B

xBnBf

Thus,

nA ¼ 10� 2

10ð0:9Þð2000Þ

� �¼ �350!!!

Obviously wrong!! So although the procedure gave the correct conclusionwhen nBf ¼ 20, it gave the wrong conclusion when nBf ¼ 2000; therefore, itis not a correct procedure.

The Correct Procedure for Determining the LimitingComponent

From the mass balance,

ni ¼ nif þ �ir

The limiting situation for any component i is when it is exhausted (i.e.,100% conversion and therefore ni ¼ 0Þ. Thus, for this situation, we get

0 ¼ nif þ �ir

which gives

rLi ¼ � nif

�i

If we adhere to the more common sign convention for stoichiometric num-bers, which is �i < 0 for reactants, then rLi will always be positive. It isobvious now that the reactant which will deplete first will be the reactanti with the smallest rLi.

Solved Example 2.3

Let us apply this simple (and hopefully correct) criterion to the previouscases:

2Aþ 10B ! 6C þ 3D

Case 1 (Fig. 2.15)

rLA ¼ 10

2¼ 5

rLB ¼ 20

10¼ 2

The limiting component is B, which is a correct conclusion.

Case 2 (Fig. 2.16)

rLA ¼ 10

2¼ 5

rLB ¼ 2000

10¼ 200

The limiting component is A, which is a correct conclusion.


In the previous examples, we used cases which are simple and almostobvious in order to check and develop a procedure. Now, as we have devel-oped a simple and nice procedure, let us use it to find the limiting reactantfor a slightly complex case:

3Aþ 8Bþ 13C þ 2D ! Products

The feeds are as shown in Figure 2.17.What is the limiting reactant (component)?

rLA ¼ 15

3¼ 5

rLB ¼ 22

8¼ 2:75

rLC ¼ 43

13¼ 3:308

rLD ¼ 8

2¼ 4

Thus, the limiting reactant is B (correct), as it has the lowest rLi value.



2.3.4 Multiple Reactions and the Special Case of aSingle Reaction

Now, we move to the more complex case of multiple reactions. Multiplereactions are defined by more than one variable depending on the number ofindependent reactions. The concept of independent and dependent reactionswill be discussed later, and until we do that, we will deal only with inde-pendent reactions.

Consider the two independent consecutive reactions:

A ! B ! C

It can also be written as two separate reactions:

A ! B and B ! C

If we define the conversion of A in Figure 2.18 as

xA ¼ nAf � nA

nAf

then we can find nA in terms of conversion xA (if it is given) as follows:


However, we cannot find nB or nC. Why?The reason is that we cannot know how much of the converted A

(nAf xA) has changed to B and how much to C. Therefore, we need to defineone more variable. For example, we can define the yield of one of the othercomponents (B or C):

Yield of B ¼ YB ¼ nB � nBf

nAf

Figure 2.18 Molar flow diagram for a multiple-reaction system.

or

Yield of C ¼ YC ¼ nC � nCf

nAf

To completely specify the problem, we need to specify any two of the fol-lowing three variables: xA, YB, or YC. A specification of the third variable isredundant because it can be deducted from the other two variables.

Solved Example 2.4

In this example, we prove that if xA and YB are defined, then YC is auto-matically defined because


nB ¼ nBf þ nAf YB

and

nC ¼ nCf þ nAf xA � nAf YB

which means that if xA and YB are defined, then we can compute nC andthen YC. Also, if xA and YC are defined, then YB is automatically defined,and if YB and YC are defined, then xA is also automatically defined.

Different Definitions of Yield

1. Yield of component i, Yi:

Yi ¼ni � nif

nAf

This is the amount of i produced as related to initial amount(or rate) of reactant fed.

2. Yield of component i, Yi:

Yi ¼ni � nif

nAf xA

This is the amount (or rate) of i produced as related to theamount (or rate) of the reactant that has reacted.

Selectivity

Selectivity has many definitions; one of the most popular one is

Si ¼Number of moles of i produced

Number of moles of all products produced

In the above case, it is defined as

SB ¼ nBnB þ nC

Multiple Reactions in Terms of Rates of Reaction

The simplest case is the above case, where the reactions are independent andalso all of the stoichiometric numbers are equal to unity.

A ! B R1

B ! C R2

R1 is defined as the rate of production of A by reaction 1 (negative, ofcourse). Then the rate of production of B by reaction 1 is R1; this is to say,

RA1 ¼ R1 (negative)

RB1 ¼ �R1 (positive)

The rate of production of B by reaction 2 is RB2 ¼ R2 (negative) and the rateof production of C by reaction 2 is RC2 ¼ �R2 (positive). Therefore, for thereactor shown in Figure 2.19, we can write

nA ¼ nAf þ RA1 ¼ nAf þ R1 where R1 ¼ RA1 is negative ð2:39ÞnB ¼ nBf þ RB1 þ RB2

¼ nBf � R1 þ R2 where R1 and R2 are negative ð2:40ÞnC ¼ nCf þ RC2 ¼ nCf � R2 where R2 is negative ð2:41Þ

Note that for this reaction,Xi

nif ¼Xi

ni; where i�A, B, and C


that is, the reaction is not accompanied by any change in the number ofmoles.

Relations Among Conversions, Yields, and Rates of Reaction

1. As

xA ¼ nAf � nA

nAfand R1 ¼ nA � nAf

Thus,

xA ¼ �R1

nAf

finally giving

R1 ¼ �nAf xA

2. As YB ¼ nB � nBf

nAf

From relation (2.40),

nB � nBf ¼ R2 � R1

Thus,

YB ¼ R2 � R1

nAf¼ R2 þ nAf xA

nAf

Finally giving

R2 ¼ nAf ðYB � xAÞ

3. As

YC ¼ nC � nCf

nAf

We get

YC ¼ � R2

nAf

finally giving

R2 ¼ �nAf YC

Therefore, when R1 and R2 are defined, then xA, YB, and YC can be deducedeasily from them. When xA and YB, or xA and YC, or YB and YC are defined,then R1 and R2 can be easily deduced from them. Thus, the system is fullydefined in terms of xA and YB, or xA and YC, or YB and YC, or R1 and R2.

For N independent reactions, the system is completely defined in termsof N conversions and yields or N rates of reaction.

A Case of Multiple Reactions with Slightly More ComplexStoichiometry

In this case, let us consider the following two reactions:

3Aþ 5B ! 6C þ 8D

4C þ 8K ! 4F þ 7L

These are two independent reactions and we are given two rates of reaction:

RA1 (rate of production of A in reaction 1, negative)

RC2 (rate of production of C in reaction 2, negative)

Consider the reactor in Figure 2.20 with the shown feed molar flow rates ofthe seven components involved in the two reactions.

For the two values for RA1 and RC2,

RA1 ¼ �6 mol/min

RC2 ¼ �2 mol/min


the mass balance relations are

nA ¼ nAf þ RA1 ¼ nAf þ 33RA1

nB ¼ nBf þ 53RA1

nC ¼ nCf þ �63RA1 þ 4

4RC2

nD ¼ nDf þ �83RA1

nK ¼ nKf þ 84RC2

nF ¼ nFf þ �44RC2

nL ¼ nLf þ �74RC2

Therefore, on using the given values of the rates of reactions (RA1 ¼ �6mol/min and RC2 ¼ �2 mol/min), we get

nA ¼ 4 mol/min

nB ¼ 15 mol/min

nC ¼ 12 mol/min

nD ¼ 19 mol/min

nK ¼ 6 mol/min

nF ¼ 2 mol/min

nL ¼ 5 mol/min

Note that this reaction is accompanied by a change in the number of moles:Pnif ¼ 10þ 25þ 2þ 3þ 10þ 0þ 1:5 ¼ 51:5 min/molPni ¼ 4þ 15þ 12þ 19þ 6þ 2þ 5 ¼ 63 min/mol

Let us try to make the relations more systematic by using stoichiometricnumbers (with its sign conventions, of course).The following common sign convention is used:

�i ¼ positive for products

�i ¼ negative for reactants

�i ¼ zero for inerts in the specific reaction

�ij is the stoichiometric number of component i in reaction j, when i is areactant �ij < 0 and when i is a product, then �ij > 0, and, finally, �ij ¼ 0when i is not involved in the reaction.

The mass balance relations become

nA ¼ nAf þ�A1�A1

RA1 þ�A2�C2

RC2

nB ¼ nBf þ�B1�A1

RA1 þ�B2�C2

RC2

nC ¼ nCf þ�C1�A1

RA1 þ�C2�C2

RC2

nD ¼ nDf þ�D1

�A1RA1 þ

�D2

�C2RC2

nK ¼ nKf þ�K1

�A1RA1 þ

�K2

�C2RC2

nF ¼ nFf þ�F1�A1

RA1 þ�F2�C2

RC2

nL ¼ nLf þ�L1�A1

RA1 þ�L2�C2

RC2

Now, according to the sign convention we use for this problem, �A1 ¼ �3,�A2 ¼ 0, �B1 ¼ �5, �B2 ¼ 0, �C1 ¼ 6, �C2 ¼ �4, �D1 ¼ 8, �D2 ¼ 0, �K1 ¼ 0,�K2 ¼ �8, �F1 ¼ 0, �F2 ¼ 4, �L1 ¼ 0, and �L2 ¼ 7. Substituting the values inthe above mass balance relations gives

nA ¼ 4 mol/min

nB ¼ 15 mol/min

nC ¼ 12 mol/min

nD ¼ 19 mol/min

nK ¼ 6 mol/min

nF ¼ 2 mol/min

nL ¼ 5 mol/min

This is the same result as we obtained earlier.From the above relation, we note that we can define generalized rates

for each reaction:

r1 ¼RA1

�A1and r2 ¼

RC2

�C2where r1 is the generalized rate of the first reaction and r2 is the generalizedrate of the second reaction.

Thus, we can write for the above two reactions that

ni ¼ nif þX2j¼1

ð�ijrjÞ;where i � A;B;C;D;K;F; and L:

Generalizing and using summation relation, we can write for N reactions:

ni ¼ nif þXNj¼1

ð�ijrjÞ

(for the balance of any component i in a system of N reactions), where

rj ¼Rij

�ij

where i is the component for which the rate of production in the reaction j isgiven, Rij is the rate of production of component i in reaction j, and �ij is thestoichiometric number of component i in reaction j.

Thus, the generalized mass balance equation for any system withsingle-input, single-output and N reactions (as shown in Fig. 2.21) isgiven by

ni ¼ nif þXNj¼1

�ijrj ð2:42Þ

for any number of components i and any number of reactions N.If we have a single reaction, then the above expression becomes

ni ¼ nif þ �ir

For no reactions (nonreacting systems), it becomes

ni ¼ nif

For a system with multiple-inputs, multiple-outputs and multiple reactions(the most general case), as shown in Figure 2.22, the mass balance equation is

XKk¼1

nik ¼XLl¼1

nifl þXNj¼1

�ijrj; i ¼ 1; 2; . . . ;M ð2:43Þ

This equation is the most general mass balance equation. It applies to allpossible mass balance cases. If there is only one input and one output, then


it reduces to Eq. (2.42). If there are no reactions, then it reduces to thefollowing form:

XKk¼1

nik ¼XLl¼1

nifl

and so on.

2.3.5 The Algebra of Multiple Reactions (LinearDependence and Linear Independence ofMultiple Reactions)

The number of variables (whether conversions and yields or rates of reac-tion) to define a system of reactions is equal to the number of independentreactions. Therefore, for a system of M reactions, we should determine thenumber of linearly independent reactions N.

Because this process requires a little knowledge of linear algebra(determinants and matrices), a limited review is given in Appendix A (matrixalgebra). Here, we introduce the concept through a simple example.

2.3.6 The Most General Mass Balance Equation(Multiple-Input, Multiple-Output, and MultipleReactions)

This is the most general one-phase (homogeneous) chemical engineeringlumped system. It has F input streams (and we use L as the counter forthe input streams ðnif1 ; nif2 ; . . . ; nifl ; . . . ; nifL ÞÞ and K output streams (we use k

Figure 2.22 Molar flow diagram for multiple-inputs, multiple-outputs, and mul-

tiple reactions.

as the counter for output streams ðni1 ; ni2 ;ik ; . . . ; niK ÞÞ with M ði ¼ 1; 2 . . . ;MÞ components and N reactions ðr1; r2; . . . ; rj; . . . ; rNÞ. Therefore the gen-eralized material balance equation is

XKk¼1

nik ¼XLl¼1

nifl þXNj¼1

�ijrj; i ¼ 1; 2; . . . ;M

There are M equations having N reactions ðrj; j ¼ 1; 2; . . . ;NÞ and �ij is thestoichiometric number of component i in reaction j. The generalized rateof reaction for reaction j is given by rj. Note that rj is the overall rate ofreaction for the whole unit (it is not per unit volume or per unit mass ofcatalyst and so on; that is why our equations are mass balance equations,not design equations as shown later).

Introductory Example

Consider the following system of reversible isomerization reactions:

1-butene , cis-2-butenecis-2-butene , trans-2-butenetrans-2-butene , 1-butene

Let us call 1-butene component A, cis-2-butene component B, and trans-2-butene component C. Then the reaction network can be rewritten as

A,B

Rate ¼ r1 ¼R1

�A1

B , C

Rate ¼ r2 ¼R2

�B2

C , A

Rate ¼ r3 ¼R3

�C3

There are apparently three reactions and three components. Let us write themass balance equations in terms of the rates of reaction ðr1; r2, and r3) for afeed of nAf ; nBf , and nCf :

nA ¼ nAf þ �A1r1 þ �A3r3

nB ¼ nBf þ �B1r1 þ �B2r2

nC ¼ nCf þ �C2r2 þ �C3r3

When we introduce the values (with correct signs) for the stoichiometricnumbers, we get

nA ¼ nAf � r1 þ r3

nB ¼ nBf þ r1 � r2

nC ¼ nCf þ r2 � r3

If nAf , nBf , and nCf are known, then in the above three equations we have sixunknowns: nA, nB, nC, r1, r2, and r3. Without thorough examination wewould say that we have to define r1, r2, and r3 (three rates of reaction) inorder to solve for nA, nB, and nC. This is actually not true and the problem issolvable with only two specifications.

Why Is That?

Because the given three reactions are not linearly independent. The systemactually contains only two independent reactions. The mass balance equa-tions could have been written in terms of the two modified rate variables

r01 ¼ r1 � r3 and r02 ¼ r1 � r2

We can rewrite the mass balance relations as

nA ¼ nAf � r01nB ¼ nBf þ r02nC ¼ nCf þ r01 � r02

Therefore, there are only two unknown variables (r01 and r02) other than nA,nB, and nC. How can we tell beforehand that the reactions are linearlyindependent or not? Let us illustrate that first for the previous example.

A , B ð2:44ÞB , C ð2:45ÞC , A ð2:46Þ

We will write them in the following algebraic form:

A� Bþ 0� C ¼ 0 ð2:47Þ0� Aþ B� C ¼ 0 ð2:48Þ

�1� Aþ 0� Bþ C ¼ 0 ð2:49Þ

The matrix of the coefficients is

A ¼1 �1 0

0 1 �1

�1 0 1

0BB@

1CCA

It is a 3� 3 matrix; if its determinant is zero (singular matrix), then the rankof this matrix is less than 3, which means there is linear dependence:

Det½A� ¼ 1ð1Þ þ 1ð�1Þ þ 0 ¼ 1� 1 ¼ 0

This means that the matrix is singular and the above three equations arelinearly dependent. In fact addition of Eqs (2.47) and (2.48) gives Eq. (2.49).

Consider another reaction set; it is the steam reforming of methane. Inthis set of reactions, there are five components and three reactions:

CH4 þH2O , COþ 3H2

COþH2O , CO2 þH2

CH4 þ 2H2O , CO2 þ 4H2

We will write these three reactions in algebraic form. Let us call CH4

component A, H2O component B, CO component C, H2 component D,and CO2 component E. We get

1Aþ 1B� 1C � 3D� 0E ¼ 0 ð2:50Þ0Aþ 1Bþ 1C � 1D� 1E ¼ 0 ð2:51Þ1Aþ 2Bþ 0C � 4D� 1E ¼ 0 ð2:52Þ

The matrix of the coefficients is

1 1 �1 �3 0

0 1 1 �1 �1

1 2 0 �4 �1

0BB@

1CCA

For these three reactions to be linearly independent, we must have at leastone 3� 3 submatrix which is nonsingular. This is not achieved in the abovematrix because all 3� 3 submatrices are singular and therefore the rank isless than 3 and the system is linearly dependent. Therefore, Eqs. (2.50)–(2.52) are linearly dependent.

2.4 SOLVED PROBLEMS FOR MASS BALANCE

Readers are advised to solve the problem on their own before inspecting thegiven solution.

Solved Example 2.5

In the Deacon Process for the manufacture of chlorine, HCl and O2 react toform Cl2 and H2O. Sufficient air (21% O2, 79% N2) is fed to the reactor tosupply 30% excess oxygen, and the fractional conversion of HCl is 70%.Calculate the molar composition of the product stream.

Solution

4HClþO2 ! 2H2Oþ 2Cl2

For simplification, we can write

4Aþ B ! 2C þ 2D

where HCl � A, O2 � B, H2O � C, Cl2 � D, and N2 � E

We can write the generalized mass balance (see Fig. 2.23) as

XKk¼1

nik ¼XLl¼1

nifl þXNj¼1

�ijrj

where L ¼ 1, K ¼ 1, and N ¼ 1, giving

ni ¼ nif þ �ir

The fractional conversion of component A is

xA ¼ nAf� nA

nAf

¼ 0:7

thus giving

nA ¼ 0:3 nAf


We have

nBf(requiredÞ ¼ 1

4nAf

As 30% excess is fed, we get

nBf¼ ð1:3Þ 1

4nAf

thus giving

nBf¼ 0:325 nAf

Also, we have

nEf¼ 79

21nBf


nEf¼ 1:223 nAf

Let RA be the rate of production of component A; then, we get

RA ¼ �ðnAf� nAÞ

which is equal to

RA ¼ �0:7 nAf

Now, we have

r ¼ �0:7 nAf

�A

As �A ¼ �4, we get

r ¼ 0:175 nAf

From the mass balance relations, we get, for component A,

nA ¼ nAfþ �Ar


nA ¼ 0:3 nAfð2:53Þ

For component B,

nB ¼ nBfþ �Br

As �B ¼ �1 and nBf¼ 0:325nAf

, we get

nB ¼ 0:15nAfð2:54Þ

For component C,

nC ¼ nCfþ �Cr

As �C ¼ 2 and nCf¼ 0, we get

nC ¼ 0:35 nAfð2:55Þ

For component D,

nD ¼ nDfþ �Dr

As �D ¼ 2 and nDf¼ 0, we get

nD ¼ 0:35 nAfð2:56Þ

For component E,

nE ¼ nEfþ �Er

As �E ¼ 0 and nEf¼ 1:223nAf

, we get

nE ¼ 1:223 nAfð2:57Þ

Thus, the output mixture is as follows:

nA ¼ 0:3 nAf; nB ¼ 0:15 nAf

; nC ¼ 0:35 nAf; nD ¼ 0:35 nAf

;

nE ¼ 1:223 nAf

The total output mixture is

nTotalout ¼ ð0:3þ 0:15þ 0:35þ 0:35þ 1:223Þ nAf

which is equal to

nTotalout ¼ 2:373 nAf

Now, to calculate the composition of exit stream (mole fraction),

Mole fraction of A ¼ XA ¼ nAnTotalout

¼ 0:3 nAf

2:373 nAf

¼ 0:3

2:373¼ 0:1264

Mole fraction of B ¼ XB ¼ nBnTotalout

¼ 0:15 nAf

2:373 nAf

¼ 0:15

2:373¼ 0:06321

Mole fraction of C ¼ XC ¼ nCnTotalout

¼ 0:35 nAf

2:373 nAf

¼ 0:35

2:373¼ 0:14749

Mole fraction of D ¼ XD ¼ nDnTotalout

¼ 0:35 nAf

2:373 nAf

¼ 0:35

2:373¼ 0:14749

Mole fraction of E ¼ XE ¼ nEnTotalout

¼ 1:223 nAf

2:373 nAf

¼ 1:223

2:373¼ 0:51538

Just as a check, the sum of the mole fractions of exit stream should be equalto 1. Thus,

X5p¼1

Xp ¼ 0:1264þ 0:06321þ 0:14749þ 0:14749þ 0:51538

¼ 0:99997 ffi 1:0

Hence, our calculations are correct.

Solved Example 2.6

Chlorine oxide gas is used in the paper industry to bleach pulp and it isproduced by reacting sodium chlorate, sulfuric acid, and methanol in lead-lined reactors:

6NaClO3 þ 6H2SO4 þ CH3OH ! 6ClO2 þ 6NaHSO4 þ CO2 þ 5H2O

(a) Suppose 14 moles of an equimolar mixture of NaClO3 andH2SO4 are added per 1 mole of CH3OH per hour; determinethe limiting reactant.

(b) For the above given molar ratios, calculate the reactant flowsrequired to produce 10 metric tons per hour of ClO2, assuming90% conversion (of the limiting component) is achieved.

Molecular weights of the components are Cl2 ¼ 70:906, Na ¼ 23:0,S ¼ 32:0, C ¼ 12:0, H ¼ 1:0, and O ¼ 16:0.

Solution

The reaction (Fig. 2.24) can be written as

6Aþ 6Bþ C ! 6Dþ 6F þM þ 5K

Part a: The limiting rates are given by

rLA¼ �nAf

�A¼ �7

�6¼ 1:16666

rLB¼ �nBf

�B¼ �7

�6¼ 1:16666

rLC¼ �nCf

�C¼ �1

�1¼ 1

As rLChas the smallest numeral value, component C is the limiting agent.

It is required to produce 10,000 kg/h of component D. The molecularweight of DðMDÞ is 67 kg/kmol. Therefore,

nD ¼ WD

MD

¼ 10,000

67¼ 149:253 kmol/h ð2:58Þ

Also, it is given in the problem that the conversion of component C is equalto 0.9:

xC ¼ 0:9 ð2:59ÞFrom the stoichiometric relations in the feed stream,

nBf¼ nAf

ð2:60Þ

nCf¼ 1

7nAf

ð2:61Þ

Figure 2.24 Molar flow diagram for example 2.6.

Moreover, we have

nDf¼ 0 ð2:62Þ

nFf¼ 0 ð2:63Þ

nMf¼ 0 ð2:64Þ

nkf ¼ 0 ð2:65ÞDegrees-of-freedom analysis (i.e., the problem is solvable or not)

Total number of unknowns:

nif for i ¼ A;B;C;D;F;M and K : 7 unknowns

ni for i ¼ A;B;C;D;F;M and K : 7 unknowns

Rate of reaction (single reaction) : 1 unknown

Total : 15 Unknowns

Known and specified relations/values:

Number of mass balance relations: 7 for 7 componentsNumber of given relations (equations 2.59--2.65) : 8Total: 15 known relationsDegrees of freedom ¼ 15� 15 ¼ 0

Hence, the problem is correctly specified and solvable.Let us solve it.

Part b: Because we know nD, from the balance for component D we canget r (because we know that nDf

¼ 0Þ as follows:nD ¼ nDf

þ �Dr


nD ¼ 0þ 6r

Finally, we get

149:253 ¼ 6r

thus giving

r ¼ 149:253

6¼ 24:8756 kmol/h ð2:66Þ

Now, we know the conversion of component C, xC ¼ 0:9. We can write

xC ¼ nCf� nC

nCf

¼ nAf=7� nC

nAf=7

¼ 0:9 ð2:67Þ

Also, we have

nC ¼ nCfþ �Cr ð2:68Þ

On substituting the values (�C ¼ �1Þ, we get

nC ¼ nAf

7� r

Finally, we get

nC ¼ nAf

7� 24:8756 ð2:69Þ

Using Eqs. (2.69) and (2.67), we get

nAf=7� ðnAf

=7� 24:8756ÞnAf

=7¼ 0:9

On solving, we get

nAf¼ 193:477 kmol/h ð2:70Þ

From Eq. (2.60), we get

nBf¼ nAf

Hence,

nBf¼ 193:477 kmol/h ð2:71Þ


nCf¼ nAf

7

Thus,

nCf¼ 27:6396 kmol/h ð2:72Þ

Another (easier) route of solution

Instead of trying to obtain nAfand then obtain nBf

and nCf, we can actually

compute nCf, and then from it, nAf

and nBfas follows:

We compute r as we did earlier:

nD ¼ nDfþ �Dr ð2:65Þ

thus giving

r ¼ 24:8756 kmol/h ð2:66Þ

We have

xC ¼ nCf� nC

nCf

¼ nAf=7� nC

nAf=7

¼ 0:9 ð2:67Þ

From the balance equation for component C, we have

nC ¼ nCfþ �Cr

Thus, we get

nCf� nC ¼ r ð2:68Þ

From Eqs. (2.67) and (2.68), we get

xC ¼ 0:9 ¼ r

nCf

¼ 24:8756

nCf

ð2:69Þ

Thus,

nCf¼ 24:8756

0:9¼ 27:63958 kmol/h

Therefore,

nAf¼ 7nCf

¼ 7ð27:63958Þ ¼ 193:477 kmol/h

Finally,

nBf¼ nAf

¼ 193:477 kmol/h

Solved Example 2.7

Acetaldehyde, CH3CHO, can be produced by catalytic dehydrogenation ofethanol, C2H5OH, via the reaction

C2H5OH ! CH3CHOþH2

There is, however, a parallel reaction producing ethyl acetate,CH3COOC2H5:

2C2H5OH ! CH3COOC2H5 þ 2H2

Suppose that, in a given reactor, the conditions are adjusted so that a con-version of 95% ethanol is obtained with a 80% yield of acetaldehyde basedon the consumption of C2H5OH. Calculate the mole fraction of H2 in theproduct stream.

Solution

C2H5OH ! CH3CHOþH2

and

2C2H5OH ! CH3COOC2H5 þ 2H2

For simplification, we can write

A ! Bþ C, and the generalized rate of reaction is r1

2A ! Dþ 2C, and the generalized rate of reaction is r2

where C2H5OH � A, CH3CHO � B, H2 � C, and CH3COOC2H5 � D:We can write the generalized mass balance equation (Fig. 2.25) as

XKk¼1

nik ¼XLl¼1

nifl þXNj¼1

�ijrj

where N ¼ 2, K ¼ 1, and L ¼ 1.Thus, in a more simplified form, we can write

ni ¼ nif þX2j¼1

�ijrj

Fractional conversion of component A ¼ xA ¼ nAf� nA

nAf

¼ 0:95

thus giving

nA � nAf¼ �0:95 nAf

ð2:70Þthus, nA ¼ 0:05 nAf

ð2:71Þ

Yield of B¼ YB ¼ nB � nBf

xAnAf

¼ 0:8

thus giving

nB � nBf¼ ð0:8Þð0:95ÞnAf


Further calculation gives

nB � nBf¼ 0:76 nAf

ð2:72ÞAs nBf

¼ 0, we get

nB ¼ 0:76 nAfð2:73Þ

From the mass balance equation for component B,

nB ¼ nBfþ �B1

r1 þ �B2r2

We get

ðnB � nBfÞ ¼ r1 þ 0

thus giving

nB ¼ r1 ¼ 0:76 nAfð2:74Þ

From the mass balance equation for component A,

nA ¼ nAfþ �A1

r1 þ �A2r2

We get

ðnA � nAfÞ ¼ �r1 � 2r2

Using Eqs. (2.70) and (2.74) in the above relation gives

�0:95 nAf¼ �0:76 nAf

� 2r2

Finally, we get

r2 ¼ 0:095 nAfð2:75Þ

From the simplified generalized mass balance relation, we have the fol-lowing.

For component C,

nC ¼ nCfþ �C1

r1 þ �C2r2

As �C1¼ 1, �C2

¼ 2, and nCf¼ 0, we get

nC ¼ 0:76 nAfþ 0:19 nAf

On solving, we get

nC ¼ 0:95 nAfð2:76Þ

For component D,

nD ¼ nDfþ �D1

r1 þ �D2r2

As �D1¼ 0, �D2

¼ 1, and nDf¼ 0 we get

nD ¼ 0þ 0þ r2

On substituting the value, we get

nD ¼ 0:095 nAfð2:77Þ

Thus, the output mixture is

nA ¼ 0:05 nAf; nB ¼ 0:76 nAf

; nC ¼ 0:95 nAf;

nD ¼ 0:095 nAf

The total output mixture is

nTotalout ¼ ð0:05þ 0:76þ 0:95þ 0:095Þ nAf

which is equal to

nTotalout ¼ 1:855 nAf

Now, to calculate the composition of exit stream (mole fraction),

XA ¼ nAnTotalout

¼ 0:05 nAf

1:855 nAf

¼ 0:027

XB ¼ nBnTotalout

¼ 0:76 nAf

1:855 nAf

¼ 0:410

XC ¼ nCnTotalout

¼ 0:95 nAf

1:855 nAf

¼ 0:512

XD ¼ nDnTotalout

¼ 0:095 nAf

1:855 nAf

¼ 0:051

Just as a check, the sum of the mole fractions of exit stream should be equalto 1. Thus,

X4p¼1

Xp ¼ 0:027þ 0:410þ 0:512þ 0:051 ffi 1:0

Hence, our calculations are correct.

2.5 HEAT EFFECTS

The only difference between simple linear heat balance problems and theheat balance problems, which are rather complex and nonlinear, is theexistence of chemical reactions [i.e., the conversion of one species or more

to other species causing release of heat (exothermic reactions) or absorptionof heat (endothermic reactions)]. If the heat of reaction is negligible, then thecomplex nonlinear heat balance equation reduces to the same simple linearheat balance equations.

2.5.1 Heats of Reactions

It is very important for the reader to understand the concept and the meth-ods of calculating the heat of reaction before he/she starts heat balanceproblems for systems with chemical reactions. This is due to the fact thatthe heat of reaction is the heart of the problem here.

The heat of reaction is usually defined as

Heat of reaction ¼ Enthalpy of products� Enthalpy of reactants

Heats of reaction as defined above are:

Positive for endothermic reactions

Negative for exothermic reactions

The enthalpies of products and reactants here are meant to be in the samemolar proportions given in the stoichiometric equation. For example, forthe hypothetical reaction,

2Aþ 3B ! 7C þ 8D ð2:78Þ

the heat of reaction for this reaction (as written) should be calculated as

�HR ¼ 7HC þ 8HD � 2HA � 3HB ð2:79Þ

This heat of reaction is not per mole of A or B or C or D; actually, it is perthe reaction as stoichiometrically written. Let us suppose that HA is in J/molof A, HB is in J/mol of B, HC is in J/mol of C, and HD is in J/mol of D; then,it is clear that �HR in Eq. (2.79) is not calculated per mole of any of thecomponents. If we insist at this stage to refer to the heat of reaction per moleof A, B, C, or D, then we can just for the sake of clarity write it as follows:

�HR ¼ joules/2 moles of A reacted

¼ joules/3 moles of B reacted

¼ joules/7 moles of C produced

¼ joules/8 moles of D produced

Therefore, if we want the heat of reaction per mole of any of the compo-nents, then we will have to do the following simple division by the absolutevalue of the stoichiometric number of the corresponding component:

�HA ¼ �HR

j�AjJ/mol of A reacted

�HB ¼ �HR

j�BjJ/mol of B reacted

�HC ¼ �HR

j�CjJ/mol of C produced

�HD ¼ �HR

j�DjJ/mol of D produced

Actually, in any problem we always need the rate of heat production orabsorption (J/time); therefore, we should always combine the heat of thereaction with the rate of reaction.

For reaction (2.78), which is a single reaction, we can express thisamount of heat as follows:

Q ¼ �HAjRAj J/time

where RA is the rate of reaction of A. It can also be written as

Q ¼ �HBjRBj J/time

or

Q ¼ �HCjRCj J/time

or

Q ¼ �HDjRDj J/time

Q ¼ Positive for endothermic reactions (heat is absorbed bythe reaction)

Q ¼ Negative for exothermic reactions (heat is produced bythe reaction)

All of these relations give the same answer, of course. This can easily bedemonstrated by inspecting the relation between RA and RB as well as �HA

and �HB; for example,

jRBj ¼ jRAjj�Bjj�Aj

and

�HA ¼ �HR

j�Aj; �HB ¼ �HR

j�BjThus,

�HB ¼ �HA

j�Ajj�bj

Therefore,

�HBjRBj ¼ �HA

j�Ajj�Bj

jRAjj�Bjj�Aj

¼ �HAjRAj

We can also express Q in terms of �HR and the generalized rate of reaction(r) as

Q ¼ �HAjRAj ¼�HR

j�AjjRAj ¼ �HRr

Similarly,

Q ¼ �HBjRBj ¼�HR

j�BjjRBj ¼ �HRr

and so on.

Important Notes

1. Note that for this irreversible single reaction, we always take theabsolute values for rates of reaction and stoichiometric numbers,because the sign of Q is determined by the sign of the heat ofreaction (whether the reaction is exothermic or endothermic) andnot by whether we are referring to a product or a reactant.However, when we use the generalized rate of reaction ðrÞ, thisissue is automatically addressed because r is always positive, as wehave shown in the previous section.

2. Heats of reactions obtained from tables or from the literature willhave the units and the basis given to them in the data source; forexample, if you find �H of a reaction given as J/mol of A reacted,then this �H actually is �HA and is defined as

�HA ¼ �HR

j�Ajwhere �HR is for the reaction as stoichiometrically written.

3. The sign convention of �HR is defined by

�HR ¼X

Hiproducts �X

Hireactants ð2:80Þ

This results in

�HR ¼ Positive for endothermic reaction

�HR ¼ Negative for exothermic reaction

Of course, scientifically there is nothing wrong in reversing thesign convention to be

�HR ¼X

Hireactants �X

Hiproducts ð2:81Þ

In this case, we will have

�HR ¼ Positive for exothermic reaction

�HR ¼ Negative for endothermic reaction

However, the first sign convention (2.80) is the one adoptedworldwide. In some cases, ��HR is used instead of �HR. Inthis case, it is obvious that

��HR ¼ Positive for exothermic reaction

��HR ¼ Negative for endothermic reaction

2.5.2 Effects of Temperature, Pressure, and Phaseson Heat of Reaction

We can now write the heat of reaction as

�HR ¼Xi

½j�ijproductsHiðT;P; iÞ� �Xi

½j�ijreactantsHiðT;P; iÞ�

where i defines the phase of component i.With the sign convention discussed earlier, �i > 0 for products and �i

< 0 for reactants. We can write

�HR ¼Xi

�iHiðT;P; iÞ

where i is the number of all components (reactants and products) involvedin the reaction.

Because the enthalpies are function of T , P, and i then �HR isobviously also a function of T , P, and i.

Important note: T and P are numbers, and i represents definition ofphases (gas or liquid or solid). The change of �HR with T or i (effect ofP is usually negligible except for processes at very high pressure) can beobtained from the change of enthalpies with T or i. Suppose that theheat of reaction of a certain reaction is given at T8, P8, and 8i and is re-presented by �HRðT8;P8; 8i Þ. We want to compute the heat of reactionat T , P8, and i; this can be achieved through the fact that

�HRðT;P8; iÞ ��HRðT8;P8; 8iÞ¼Xi

f�iHiðT;P8; iÞ � �iHiðT8;P8; 8iÞg

The right-hand side is a difference between enthalpies and thus can be easilycalculated for both changes from T8 ! T (sensible heat) and from 8i ! i

(change of phase, latent heat).

For the change of temperature only

�HRðT;P8; 8iÞ ��HRðT8;P8; 8iÞ¼Xi

f�iHiðT;P8; 8iÞ � �iHiðT8;P8; 8iÞg

¼Xi

�i

ðTT0

CPi dT

For the change of phase, if 8i and i represent different phases, then theheats of phase transition and the temperatures of phase transitions must beinvolved as clearly illustrated in the following example.

An Illustrative Example

Given the heat of the reaction

4NH3ðgÞ þ 5O2ðgÞ ! 4NOðgÞ þ 6H2OðlÞ

at 1 atm and 298 K to be �279:33 kcal/mol. Calculate the heat of reaction at9208C and 1 atm with water in the vapor phase.

Solution

Given

�HRðT8;P8; 8iÞ ¼ �279:33 kcal

where

T8 ¼ 298K;P8 ¼ 1 atm

and

8i ¼NH3 gasO2 gasNO gasH2O liquid

8>><>>:

What is required to compute is

�HRðT;P8; iÞ ¼ ?

where

T ¼ 920þ 273 ¼ 1193 K; P8 ¼ 1 atm

and

i ¼NH3 gasO2 gasNO gasH2O vapor

8>><>>:

Now,

�HRðT8; p8; i8Þ ¼Xi

�iHiðT8;P8; i8Þ

and

�HRðT;P8; iÞ ¼Xi

�iHiðT;P8; iÞ

Then,

�HRðT;P8; iÞ ��HRðT8;P8; 8iÞ ¼Pi

f�iHiðT;P8; iÞ � �iHiðT8;P8; 8iÞg¼ �4fHNH3

ð1193; 1; gasÞ �HNH3

ð298; 1; gasÞg � 5fHO2ð1193; 1; gasÞ

�HO2ð298; 1; gasÞg þ 4fHNOð1193; 1;

gasÞ �HNOð298; 1; gasÞg þ 6fHH2O

ð1193; 1; vaporÞ �HH2Oð298; 1;liquid)g

Thus, for the right-hand side, we get

� 4

ð1193298

CpNH3dT � 5

ð1193298

CPO2dT þ 4

ð1193298

CPNO dT

þ 6

ð373298

CPH2OðliqÞ dT þ�HH2O

ðvapÞ þð1193373

CPH2OðvapÞ dT

� �

Finally, we get

�HRðT;P8; iÞ ¼�HRðT8;P8; i8Þ � 4

ð1193298

CPNH3dT � 5

ð1193298

CPO2dT þ 4

ð1193298

CPNOdT

þ 6

ð373298

CPH2OðliqÞ dT þ�HH2O

ðvapÞ�

þð1193373

CPH2OðvapÞ dT

�

where �HH2OðvapÞ is the latent heat of vaporization.

2.5.3 Heats of Formation and Heats of Reaction

What is the heat of formation? The heat of formation of a compound isthe heat of reaction evolved (or absorbed) during the formation of thiscompound from its elementary constituents; for example, for the reaction

Cþ 2H2 ! CH4

the heat of this reaction is actually the heat of formation of CH4 (methane).The heats of formation of large number of compounds are tabulated at

standard conditions (heat of formation tables available in Perry’s ChemicalEngineer’s Handbook, 1997). The standard conditions are usually defined at258C and 1 atm and the phase that is normal for compounds under theseconditions (carbon is solid, H2O is liquid, CO2 is gas, etc.).

The heats of reaction under standard conditions can be computedfrom the heats of formation of the different components in this reaction(reactants and products) under standard conditions through the followingformula, which can be easily proven:

�H8R ¼Xi

�i�H8Fi

Here, i is the numbering for the species (reactants and products) involved inthe reaction: �i > 0 for products and �i < 0 for reactants. �H8R is negativefor exothermic reactions and positive for endothermic reactions.

2.5.4 Heats of Combustion and Heats of Reaction

The heat of combustion is the heat evolved from oxidizing the componentcompletely with a stoichiometric amount of oxygen; for example,

CH4 þ 2O2 ! CO2 þ 2H2O

The heat evolved from the above reaction is the heat of combustion ofmethane.

SþO2 ! SO2

The heat evolved from this reaction is the heat of combustion of sulfur.The heat of reaction can be computed from the heat of combustion

using the formula

�H8R ¼ �X

�i�H8Ci

Any heat of reaction is computed at standard conditions. �H8RðT8;P8; i8Þcan be used to compute the heats of reaction at any other condition �HR

ðT;P8; iÞ using the previously discussed simple procedure.

2.6 OVERALL HEAT BALANCE WITH SINGLEAND MULTIPLE CHEMICAL REACTIONS

The heat balance for a nonadiabatic reactor is given in Figure 2.26. i repre-sents the numbering of all components involved in the reaction (reactantsand products). If any product does not exist in the feed, then we put itsnif ¼ 0; if any reactant does not exist in the output, then we put its ni ¼ 0.Enthalpy balance isX

i

nif HiðTf ;Pf ; if Þ þQ ¼Xi

niHiðT;P; iÞ ð2:82Þ

where Q is the heat added to the system (the reactor)

Figure 2.26 Heat balance for a nonadiabatic reactor.

Note: We did not include any heats of reactions; it is automatically in-cluded, as will be clear in the next few lines.

Define a reference state, and add and subtract the following terms from Eq.(2.82):X

i

nif HirðTr;Pr; irÞ andXi

niHirðTr;Pr; irÞ

This addition of two terms and subtracting of the same two terms togetherwith rearrangement makes the heat balance equation have the followingform: X

i

nif fHif ðTf ;Pf ; if Þ �HirðTr;Pr; irÞg þQ

¼Xi

nifHiðT;P; iÞ �HirðTr;Pr; irÞg þXi

ðni � nif ÞHirðTr;Pr; irÞ

From the mass balance of a single reaction, we have

ni ¼ nif þ �ir ð2:83Þwhere

r ¼ Ri

�i

Equation (2.83) can be rearranged as

ni � nif ¼ �ir

Therefore, the heat balance equation becomesXi


¼Xi

nifHiðT;P; iÞ �HirðTr;Pr; irÞg þXi

�irHirðTr;Pr; irÞ

This equation can be rearranged to giveXi


¼Xi

nifHiðT;P; iÞ �HirðTr;Pr; irÞg þ rXi

�iHirðTr;Pr; irÞ

It is clear that, by definition,Xi

�iHirðTr;Pr; irÞ ¼ ð�HRÞr

where ð�HRÞr is the heat of reaction at the reference conditions.Thus, the heat balance equation becomes

Xi


¼Xi

nifHiðT;P; iÞ �HirðTr;Pr; irÞg þ rð�HRÞr

In other words,

fEnthalpy in (above reference conditions)+Heat addedg

¼ Enthalpy out (above reference conditions)

+Heat absorbed by reaction (at reference conditions)

�

The terms of heat absorbed by reaction (for a single reaction) is rð�HRÞrNote: If r ¼ 0 (i.e., a nonreacting system), then we obtain a simple heatbalance relation for the nonreacting system. The heat of reaction ð�HRÞris evaluated at the reference state ðTr;Pr; rÞ.

The term rð�HRÞr needs some more clarification. For the sake ofbrevity, let us use:

Enthalpy in (at above reference conditions) ¼ IinEnthalpy out (at above reference conditions) ¼ Iout

Then,

Iin þQ ¼ Iout þ rð�HRÞrThe sign convention ensures that r is always positive and this is more con-venient in the heat balance; ð�HRÞr is the heat of reaction for the reactionas stoichiometrically written as discussed earlier and re-explained in thefollowing notes:

Note 1: If the reaction is endothermic, then ð�HRÞr is positive and it isthe heat absorbed or consumed; therefore, it is clear that it must be addedto the right-hand side with Iout as shown above.

If the reaction is exothermic, then ð�HRÞr is negative and the heat isevolved; then, it will be effectively added as positive quantity to the left-hand side.

Note 2: Suppose that the reaction we are dealing with is in the form

2Aþ 4B ! 5C þ 6D

Then, ð�HRÞr is evaluated from the enthalpies as follows:

ð�HRÞr ¼Xi

�iHir

This (�HRÞr is for the reaction stoichiometry as written; if RA is given, thenwe must divide ð�HRÞr by �A or RA by �A, giving

ð�HRÞrRA

�A¼ ð�HRÞrr

Suppose that the heat of reaction is given from experiment or tabulated inthe units of r, �HRA (¼ J/mol of A); then, r to be used should be based onthe above reaction rewritten as

Aþ 2B ! 52C þ 3D

However, never use �HRA (¼ J/mol of A); use r ¼ RA=�A with j�Aj ¼ 2from the first stoichiometric equation! This will mean that you have actuallydivided all of the stoichiometric numbers by the stoichiometric coefficient ofA twice: once when you used �HRA which is equal to ð�HRÞr=�A and oncewhen you computed r to be equal to RA=�A. This mistake is very commonand many books and chemical engineers make this serious mistake. Thereader should understand this point very clearly.

Note 3: The heat generation (or absorption) term will always have thesign of ð�HRÞr or �HRA, because it is always multiplied by the general-ized rate of reaction r, which is always positive.

2.6.1 Heat Balance for Multiple Reactions and theSpecial Case of a Single Reaction

Consider that we have a number of reactions taking place simultaneously.The heat of reaction in the heat balance equation should account for all theheats produced/absorbed by all the reactions.

If the number of components is M, the number of reactions is N; weuse i as the component counter and j as the reaction counter. Therefore, theheat balance equation becomes

XMi¼1


¼XMi¼1

nifHiðT;P; iÞ �HirðTr;Pr; irÞg þXNj¼1

rjð�HRÞrj ð2:84Þ

It is convenient to write these heat balance equations in a shorter form byrealizing that the Hif is the enthalpy at Tf ;Pf ; if without writing themexplicitly between brackets, and so on for other terms. Therefore, Eq.(2.84) can be rewritten as

XMi¼1

nif ðHif �HirÞ þQ ¼XMi¼1

niðHi �HirÞ þXNj¼1

rjð�HRÞrj

An Illustrative Example

Steam reforming of methane includes the following three reactions:

CH4 þH2O , COþ 3H2; generalized rate of reaction is r1

COþH2O , CO2 þH2; generalized rate of reaction is r2

CH4 þ 2H2O , CO2 þ 4H2; generalized rate of reaction is r3

We can label these components as follows: CH4 �A, H2O �B, CO �C,H2 �D, CO2 �E, and Inerts � I . Consider single-input, single-output.The mass balance equations are all obtained from the general equationsdiscussed earlier:

ni ¼ ni;f þX3j¼1

�ijrj; i � A;B;C;D;E; I

Thus, the mass balance equations are simply

nA ¼ nAf þ �A1r1 þ �A2r2 þ �A3r3 ¼ nAf � r1 � r3

nB ¼ nBf þ �B1r1 þ �B2r2 þ �B3r3 ¼ nBf � r1 � r2 � 2r3

nC ¼ nCf þ �C1r1 þ �C2r2 þ �C3r3 ¼ nCf þ r1 � r2

nD ¼ nDf þ �D1r1 þ �D2r2 þ �D3r3 ¼ nDf þ 3r1 þ r2 þ 4r3

nE ¼ nEf þ �E1r1 þ �E2r2 þ �E3r3 ¼ nEf þ r2 þ r3

nI ¼ nIf þ �I1r1 þ �I2r2 þ �I3r3 ¼ nIf

The heat balance general equation is

X6i¼1

nif ðHif �HirÞ þQ ¼X6i¼1

niðHi �HirÞ þX3j¼1

rjð�HRÞj

Thus, the heat balance equation (long and tedious, but very straightfor-ward) is

nAf ðHAf �HArÞ þ nBf ðHBf �HBrÞ þ nCf ðHCf �HCrÞþ nDf ðHDf �HDrÞ þ nEf ðHEf �HErÞ þ nIf ðHIf �HIrÞþQ ¼ nAðHA �HArÞ þ nBðHB �HBrÞ þ nCðHC �HCrÞþ nDðHD �HDrÞ þ nEðHE �HErÞ þ nI ðHI �HIrÞþ ½r1�HR1 þ r2�HR2 þ r3�HR3�

2.6.2 The Most General Heat Balance Equation (forMultiple Reactions and Multiple-Input andMultiple-Output System Reactor with MultipleReactions)

In this case, we just sum up the enthalpies of the input streams and theoutput streams. If we have F input streams (and we use l as the counter forthe input streams) and K output streams (we use k as the counter for outputstreams) with L components and N reactions, then we have the followingmost general heat balance equation:

XLl¼1

XMi¼1

nifl ðHifl�HirÞ

!( )þQ ¼

XKk¼1

XMi¼1

nikðHik�HirÞ

!( )

þXNj¼1

rjð�HRÞrj

This is the most general heat balance equation for a multiple-input,multiple-output (MIMO), multiple reactions (and, of course, multicom-ponents) system.

2.7 SOLVED PROBLEMS FOR ENERGYBALANCE

The reader is advised first to solve each problem before inspecting the givensolution.

Solved Example 2.8

Two liquid streams of carbon tetrachloride are to be evaporated to producea vapor at 2008C and 1 atm. The first feed is at 1 atm and 308C with a flowrate of 1000 kg/h and the second feed stream is at 708C and 1 atm and has aflow rate of 500 kg/h. Calculate the heat that must be supplied to theevaporator.

The data given are as follows:

Carbon tetrachloride � CCl4 � component AMolecular weight of A ¼ MA ¼ 153:84Normal boiling of A ¼ TBPA

¼ 349:7 K ¼ 76:78CLatent heat of vaporization ofA at boiling point��A ¼ 36,882.1 J/molSpecific heat in the vapor phase

CPAV¼ a0 þ b0T þ c0T2 þ d 0T3 þ e0T4 J/mol K

where a0 ¼ 8:976, b0 ¼ 0:42, c0 ¼ �7:516� 10�4, d 0 ¼ 6:237� 10�7,and e0 ¼ �1:998� 10�10

Specific heat in the liquid phase

CPAL¼ aþ bT þ cT2 þ dT3J/mol K

where a ¼ 12:284, b ¼ 1:09475, c ¼ �3:183� 10�3, andd ¼ 3:425 � 10�6.

Solution (see Fig. 2.27)

The heat balance on the evaporator gives

nAf1HAf1þ nAf2HAf2

þQ ¼ nAHA ð2:85Þ

From the mass balance, we have

nA ¼ nAf1 þ nAf2 ð2:86Þ

Substitution from Eq. (2.86) into Eq. (2.85) gives

nAf1HAf1 þ nAf2HAf3 þQ ¼ ðnAf1 þ nAf2 ÞHA

Rearrangement gives

Q ¼ nAf1 ðHA �HAf1Þ þ nAf2ðHA �HAf2Þ ð2:87Þ

where HA is the enthalpy of carbon tetrachloride in the vapor phase at2008C (473 K) and 1 atm, HAf1

is the enthalpy of carbon tetrachloride inthe liquid phase at 308C (303 K) and 1 atm; and HAf2

is the enthalpy ofcarbon tetrachloride in the liquid phase at 708C (343 K) and 1 atm.Therefore,

Figure 2.27 Heat flow diagram for solved example 2.8.

ðHA �HAf1Þ ¼

ð349:7303

CPALdT þ �A þ

ð473349:7

CPAVdT

¼ð349:7303

ðaþ bT þ cT2 þ dT3Þ dT þ �A

þð473349:7

ða0 þ b0T þ c0T2 þ d 0T3 þ e0T4Þ dT

¼ aT þ bT2

2þ c

T3

3þ d

T4

4

" #349:7

303

þ�A

þ a0T þ b0T2

2þ c0

T3

3þ d 0 T

4

4þ e0

T5

5

" #473

349:7

Thus, we get

ðHA�HAf1Þ¼ að349:7�303Þ þ b

2½ð349:7Þ2�ð303Þ2� þ c

3½ð349:7Þ3�ð303Þ3�

þ d

4½ð349:7Þ4 � ð303Þ4� þ 36882:1|fflfflfflffl{zfflfflfflffl}

�A

þa0ð473� 349:7Þ

þ b0

2½ð473Þ2 � ð349:7Þ2� þ c0

3½ð473Þ3 � ð349:7Þ3�

þ d 0

4½ð473Þ4 � ð349:7Þ4� þ e0

5½ð473Þ5 � ð349:7Þ5�

By substituting the values of a, b, c, d a0, b0, c0, d 0, and e0 and performing thestraightforward computation, compute the value of HA �HAf1

:The reader is requested to perform this computation and obtain the

value of HA �HAf1 ; we will label it_

A here, that is,

HA �HAf1� _

A J/mol

Similarly, we can compute HA �HAf2 (the reader is requested to performthis computation), we will label the value as

_

B,

HA �HAf2� _

B J/mol

Now, after computing the values of_

A and_

B, the heat balance equation(2.87) can be rewritten as

Q ¼ nAf1_

Aþ nAf2_

B

nAf1 ¼1000

MA

¼ 1000

153:84¼ 6:5 kmol/h ) nAf1 ¼ 6500 mol/h

and

nAf2 ¼500

MA

¼ 500

153:84¼ 3:25 kmol/h ) nAf2 ¼ 3250 mol/h

Thus,

Q ¼ 6500_

Aþ 3250_

B J/h

or

Q ¼ 6:5_

Aþ 3:25_

B kJ/h

For More Practice

The reader is encouraged to carry out the following tasks to get a betterunderstanding of the concepts used in this section:

1. Solve the above problems by numbers and obtain the numericalvalues of

_

A,_

B, and Q.2. Solve the above problem if the second feed stream is fed at 858C

instead of 708C. Compare the Q obtained with that of first case.

Solved Example 2.9

Compute the heat of reactions for the following reactions:

(a) Oxidation of sulfur at 600 K

SðsÞ þO2ðgÞ ! SO2ðgÞ(b) Decomposition of propane at 500 K

C3H8 ! C2H2 þ CH4 þH2

Solution

Part a: The heat of reaction at 600 K can be calculated as follows:

The reaction is:

SðsÞ þO2ðgÞ ! SO2ðgÞFor simplicity write the reaction as, Aþ B ! C.The heat of reaction can be written in terms of the heat of formation asfollows:

ð�HrÞ298 ¼X3i¼1

�ið�HfiÞ298l ; i ¼ A;B; and C

Thus, we get

ð�HrÞ298 ¼ �Að�HfAÞ298 þ �Bð�HfB

Þ298 þ �Cð�HfCÞ298

(We will no longer write 298 because it is well known and it is the tempera-ture at which the standard tables are tabulated.) The superscript degree signmeans at 298 K and 1 atm here and in Part b. Thus, we can write

�H8r ¼ ��H8fA|fflffl{zfflffl}zero

��H8fB|fflffl{zfflffl}zero

þ�H8fC

Thus, we get

�H8r ¼ �H8fC at 258C ð298 KÞIf we also want it evaluated at 600 K, we derive the relations


�ið�Hfi

Þ298

and


�ið�HfiÞ600

On subtracting the above two relations, we get

ð�HrÞ600 � ð�HrÞ298 ¼X3i¼1

�i �Hfi

� �600

� �Hfi

� �298

n oWe can write

ð�HrÞ600 ¼�H8r þ �A½ð�HfAÞ600 � ð�HfA

Þ298� þ �B½ð�HfBÞ600

� ð�HfBÞ298� þ �C½ð�HfC

Þ600 � ð�HfCÞ298�

which can be rewritten as

ð�HrÞ600 ¼ �H8r �ðTmp

298

CPSAdT|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl}

solid

þ �A|{z}transition

þð600Tmp

CPLAdT|fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl}

liquid

26664

37775

�ð600298

CPBdT þ

ð600298

CpCdT

Part b: The heat of reaction at 500 K can be calculated as:

The reaction is:

C3H8 ! C2H2 þ CH4 þH2

For simplicity write the reaction as, A ! Bþ C þD.The heat of reaction at standard conditions can be written in terms of heatsof formation as follows:


�ið�HfiÞ298; i ¼ A;B;C; and D

Thus we get

�H8r ¼ �A�H8fA þ �B�H8fB þ �C�H8fC þ �D�H8fD


�H8r ¼ ��H8fA þ�H8fB þ�H8fC þ�H8fD

If we want it at 500 K, then we get

ð�HrÞ500 � ð�HrÞ298 ¼X4i¼1

�ifð�HfiÞ500 � ð�Hfi

Þ298g

We can write

ð�HrÞ500 ¼ �H8r þ �A½ð�HfAÞ500 � ð�HfA

Þ298Þ þ �B½ð�HfBÞ500

� ð�HfB Þ298� þ �C½ð�HfC Þ500 � ð�HfC Þ298Þ� þ �D½ð�HfDÞ500� ð�HfD

Þ298�which can be rewritten as

ð�HrÞ600 ¼ �H8r �ð500298

CPAdT þ

ð500298

CPBdT þ

ð500298

CPCdT

þð500298

CPDdT

Solved Example 2.10

Ethylene is made commercially by dehydrogenating ethane:

C2H6ðgÞ , C2H4ðgÞ þH2ðgÞ�HRð2008CÞ ¼ 134:7 kJ/mol

Ethane is fed to a continuous adiabatic reactor at TF ¼ 20008C. Calculatethe exit temperature that would correspond to 100% conversion (xA ¼ 1:0,where A � C2H6). Use the following data for the heat capacities in yourcalculations:

CPA¼ 49:37þ 0:1392T

CPB¼ 40:75þ 0:1147T

CPC¼ 28:04þ 4:167� 10�3T

where A � C2H6ðgÞ, B � C2H4ðgÞ, and C � H2ðgÞ; CPjis in J/mol 8C, and

T is the temperature (in 8C).


For simplicity, let us represent the reaction as

A , Bþ C

We take the basis as 100 mol/min. The generalized heat balance equation isXnif ðHif

�HirÞn o

þQ ¼X

niðHi �HirÞn o

þ ð�HÞRr

Based on the basis and the rate of reaction definition in mass balance,r ¼ 100 mol/min. Take the reference temperature as 2008C; thus, we get

100

ð2000200

CPAdT þ Q|{z}

zero

¼ 100

ðT200

CPBdT þ 100

ðT200

CPCdT

þ ð134:7� 103Þð100ÞOn substituting the values, we get

100

ð2000200

ð49:37þ 0:1392TÞ dT ¼ 100

ðT200

ð40:75þ 0:1147TÞ dT

þ 100

ðT200

ð28:04þ 4:167� 10�3TÞ dT

þ ð134:7� 103Þð100ÞSolving the integrals gives

49:37T þ 0:1392T2

2

" #2000

200

¼ 40:75T þ 0:1147T2

2

" #T

200

þ 28:04T þ 4:167� 10�3T2

2

" #T

200

þ 134:7� 103

Figure 2.28 Energy flow diagram for solved example 2.10.

Solving it further gives

49:37ð2000� 200Þ þ 0:1392ð20002 � 2002Þ2

" #

¼ 40:75ðT � 200Þ þ 0:1147ðT2 � 2002Þ2

" #

þ 28:04ðT � 200Þ þ 4:167� 10�3ðT2 � 2002Þ2

" #þ 134:7� 103

Mathematical calculations give

ð5:93835� 10�5ÞT2 þ 0:06879T � 232:15 ¼ 0

Solution of this quadratic equation gives

T ¼�0:06879�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið0:06879Þ2 þ 4ð5:93835� 10�5Þð232:15Þ

q2ð5:93835� 10�5Þ

Neglecting the negative solution, we get

T ¼ �0:068709þ 0:2446734

1:1876� 104

Thus, T ¼ 1480:9988C (the answer)

Solved Example 2.11

Acetic acid is cracked in a furnace to produce the intermediate ketene via thereaction

CH3COOHðlÞ ! CH2COðgÞ þH2OðgÞ

The reaction

CH3COOHðlÞ ! CH4ðgÞ þ CO2ðgÞ

also occurs to an appreciable extent. It is desired to carry out cracking at7008C with a conversion of 80% and a fractional yield of methane (CH4) of0.0722. Calculate the required furnace heating rate for a furnace feed of 100kmol/h acetic acid. The feed is at 3008C.

The data given are as follows:Heats of formation at standard conditions (258C, 1 atm):

�H8fCH2CO¼ �14:6

kcal

g mol

�H8fH2O¼ �57:8

kcal

g mol

�H8fCH3COOH¼ �103:93

kcal

g mol

�H8fCH4¼ �17:89

kcal

g mol

�H8fCO2¼ �94:05

kcal

g mol

The specific heats are given by the following relations (J/gmol K):

CPH2OðlÞ (liquid water) ¼ 18:2964þ 0:04721T � 0:001338T2

� 1:3� 10�6 T3

CPH2OðgÞ ¼ 34:047þ 9:65� 10�3T � 3:2988� 10�5T2

� 2:044� 10�8T3 þ 4:3� 10�12T4

CPCO2 ðgÞ¼ 19:02þ 7:96� 10�2T � 7:37� 10�5T2

þ 3:745� 10�8T3 � 8:133� 10�12T4

CPCH4 ðgÞ¼ 3:83� 7:366� 10�2T þ 2:909� 10�4T2 � 2:638

� 10�7T3 þ 8� 10�11T4

CPCH3COOHðlÞ ¼ �36:08þ 0:06T � 3:9� 10�5T2 � 5:6� 10�7T3

CPCH3COOHðgÞ ¼ 14:6þ 2:3� 10�2T � 11:02� 10�5T2 þ 2:6

� 10�9T3 � 2:8� 10�13T4

CPCH2COðg;lÞ ¼ 4:11þ 2:966� 10�2T � 1:793� 10�5T2 þ 4:22

� 10�9T3

Boiling point of acetic acid ¼ 118:18CBoiling point of ketene ¼ TK

Latent heat of vaporization of acetic acid ¼ 96:76 cal/gLatent heat of vaporization of ketene ¼ �k cal/g


In order to carry out the necessary heat balance, we must first compute theinput and output streams and their composition. This means that we shouldformulate and solve the mass balance problem first.

Mass Balance The two reactions can be written as

A ! Bþ C with the generalized rate, r1

and

A ! Dþ E with the generalized rate, r2

The feed is pure acetic acid with a feed rate of nAf ¼ 100 kmol/h.

Degrees of Freedom To make sure that the problem is solvable we firstcheck the degree of freedom.

The number of unknowns ¼ 5ðnA; nB; nC; nD; nEÞ þ 2ðr1; r2Þ ¼ 7Number of mass balance equations ¼ 5Number of specifications (xA;YDÞ ¼ 2Total number of equations and specifications available ¼ 5þ 2 ¼ 7Degree of freedom ¼ 7� 7 ¼ 0

Thus, the mass balance problem is solvable. Note that nA can be easilycomputed from nAf and the given conversion:

xA ¼ nAf � nA

nAf


Thus, we get

0:8 ¼ 100� nA100

and the value of nA is equal to

nA ¼ 20kmol

h

nD can be easily computed from the fractional yield of methane, YD (kmolof D produced per kmol of A reacted):

YD ¼ 0:0722 ¼ nD � nDf

nAf xA¼ nD

80

which gives

nD ¼ 80� 0:0722 ¼ 5:776 kmol/h

The rest of the components are calculated from their mass balance equations

nB ¼ nBf þ �B1r1 þ �B2r2

which gives

nB ¼ r1

Also,

nC ¼ r1

and, similarly,

nE ¼ r2

r1 and r2 can be easily obtained from the mass balance equations for com-ponents A and D. The mass balance for A gives

nA ¼ nAf þ �A1r1 þ �A2r2


20 ¼ 100� r1 � r2

Therefore,

r1 þ r2 ¼ 80 kmol/h ð2:88ÞThe mass balance for D gives

nD ¼ nDf þ �D1r1 þ �D2r2


5:776 ¼ 0þ 0þ r2

or we can write

r2 ¼ 5:776 kmol/h

Using this value of r2 in Eq. (2.88), we get

r1 ¼ 80� 5:776 ¼ 74:224 kmol/h

Now, all input streams and output streams are known and we can carry outthe heat balance to find Q.

Heat Balance (see Fig. 2.30) The heat balance equation is given by

Xnif Hif ðTf ;Pf ; f Þ �HirðTr;Pr; rÞ �n o

þQ

¼X

ni½HiðT;P; Þ �HirðTr;Pr;Pr; rÞn io

þX2j¼1

rjð�HRÞrj

Take the reference condition to be 258C and 1 atm (standard conditions).The first term on the left-hand side of the above equation can thus be writtenas X

nif ½Hif �Hir� ¼ nAf ðHAf �HArÞ þ 0þ 0þ 0þ 0

Figure 2.30 Molar flow diagram (with temperature shown) for heat balance for

solved example 2.11.

Which can be rewritten as

ð100� 1000Þð118:1þ273

25þ273

CPCH3COOHðlÞ dT þ �CH3COOH

�

þð300þ273

118:1þ273

CPCH3COOHðgÞ dT � � If

Now, also, the first term on the right-hand side of the heat balance equationcan be written as:

Xni ½Hi �Hir� ¼

nA|{z}20�1000

ð118:1þ273

25þ273

CPCH3COOHðlÞ dT þ �CH3COOH þð700þ273

118:1þ273

CPCH3COOHðgÞ dT

� �

þnB

ð118:1þ273

25þ273

CPCH2COðlÞ dT þ �CH2CO þð700þ273

TK

CPCH2COðgÞ dT

� �

þnC

ð100þ273

25þ273

CPH2OðlÞ dT þ �H2Oþð700þ273

100þ273

CPH2OIðgÞ dT

� �

þnD

ð700þ273

25þ273

CPCH4 ðgÞdT

� �þ nE

ð700þ273

25þ273

CPCO2 ðgÞdT

� �

8>>>>>>>>>>>>>><>>>>>>>>>>>>>>:

9>>>>>>>>>>>>>>=>>>>>>>>>>>>>>;

� I

Let the right-hand side of the above equation be equal to some value I .After computing If and I as shown above (note that reader has to do thenecessary units conversion to keep all the units consistent), the heat balanceequation is

If þQ ¼ I þ r1ð�HR1Þ þ r2ð�HR2Þ�HR1 and �HR2 are to be computed from the heats of formation:

�HR1 ¼ �H8fB þ�H8fC ��H8fA

and

�HR2 ¼ �H8fD þ�H8fE ��H8fAThus, Q can be easily computed:

Q ¼ I � If þ r1ð�HR1Þ þ r2ð�HR2ÞThe reader should perform the above computation to get the numericalvalue of the result.

More Practice for the Reader

Solve the above problem for 90% acetic acid conversion and a fractionalyield of methane of 0.08 and a reactor temperature of 8008C.

Solved Example 2.12

Consider the two-stage ammonia synthesis loop shown in Figure 2.31. In theflowsheet, a cold stoichiometric feed of N2 and H2 is introduced between the

reactor stages. In each reactor, the following synthesis reaction takes placewith specific conversion.

N2 þ 3H2 , 2NH3

or we can represent it by Aþ 3B , C for simplification.The effluent of the second reactor stage is cooled by exchanging heat

with the input stream to the first reactor stage. After further cooling, stream5 is separated to recover a product stream containing all of the NH3 (C) andsome N2ðAÞ and H2ðBÞ. The following specification are given:

1. The feed rate (in stream 1) is 97.8 mol/h with 15% N2 ðAÞ and75% H2 ðBÞ and the feed temperature is 508C.

2. The input stream to the first reactor stage (stream 7) is at 4258C.

Figure 2.31 Flowsheet of solved example 2.12.

3. Input stream to separator (stream 5) is at 508C and the tempera-tures of the outlet streams (streams 6 and 9) are equal to eachother.

4. The recycle stream (stream 6) is to contain no NH3 ðCÞ and0.9945 each of the N2 ðAÞ and H2ðBÞ fed to the separator. Thisstream contains no inerts.

5. The conversion of N2 ðAÞ is 10% in stage 1 and 12.33% in stage2.

6. All of the units are adiabatic except the cooler, which is non-adiabatic.

Assume all streams are gas phase and that pressure effects are neglected.

a. Show that the problem is correctly specified.b. Outline a calculation procedure.c. Solve the problem to obtain the temperatures and molar flow

rates of all components in all streams.d. Construct a flowchart for a computer program for your solution.

Solution

We will start the solution of this problem with the degrees-of-freedomanalysis.

First, let us identify all of the given values and specified relations forthe problem:

1. nTotal1 ¼ nT1¼ 97:8 mol/h

2. YA1¼ 0:15 (molar fraction of component A in stream 1)

3. YB1¼ 0:75 (molar fraction of component B in stream 1)

4. T1 ¼ 50þ 273 ¼ 323 K5. nC1

¼ 06. T7 ¼ 425þ 273 ¼ 698 K7. T5 ¼ 323 K8. T6 ¼ T9

9. nC6¼ 0

10. nA6¼ 0:9945 nA5

11. nB6¼ 0:9945 nB5

12. xA1¼ nA7

� nA8

nA7

¼ 0:1

13. xA2¼ nA2

� nA3

nA2

¼ 0:1233

14. nl6 ¼ 0, where I � inerts

Thus, we have 14 specified relations.

Process Degrees of Freedom Total number of unknowns,

Reactor 1: 11Mixer: 10Reactor 2: 6Exchanger: 10Cooler: 6Separator: 5Total: 48

Mass and Heat Balance Relations

Reactor 1: 5Mixer: 5Reactor 2: 5Separator: 5Cooler: 5Exchanger: 9Total: 34

Note that the mass balance relations for the cooler (four in number) andexchanger (eight in number) are described by the following relations:For one side of the exchanger,

ni3 ¼ ni4 ; i ¼ A;B;C; and I

Thus, giving four different relations.For the other side of the exchanger,

ni6 ¼ ni7 ; i ¼ A;B;C; and I

Thus, giving four different relations.For the cooler,

ni4 ¼ ni5i ¼ A;B;C; and I

Thus, giving four relations.Given specified relations ¼ 14ð9þ 5 as shown above)

So we have

Total known/specified relations ¼ 34þ 14 ¼ 48Total unknowns ¼ 48

Thus,

Degrees of freedom ¼ 48� 48 ¼ 0

As the degrees of freedom is equal to zero, this problem is correctly specifiedand hence is solvable.

Degrees of Freedom (DF) for Each Unit Separately

Reactor 1

Number of unknowns ¼ 10þ 1 ¼ 11Number of specified relations

T7 ¼ 698 K and xA1¼ nA7

� nA8

nA7

¼ 0:1

so specified relations ¼ 2Mass and heat balance equations ¼ 5DF ¼ 11� ð5þ 2Þ ¼ 11� 7 ¼ 4

Mixer

Number of unknowns ¼ 15Number of specified relations

nT1¼ 97:8 mol/h; YA1

¼ 0:15; YB1¼ 0:75; nC1

¼ 0;

T1 ¼ 323 K

so specified relations = 5Mass and heat balance equations ¼ 5DF ¼ 15� ð5þ 5Þ ¼ 15� 10 ¼ 5

Reactor 2


xA2¼ nA2

� nA3

nA2

¼ 0:1233

so total specified relations ¼ 1Mass and heat balance equations ¼ 5DF ¼ 11� ð5þ 1Þ ¼ 11� 6 ¼ 5

Exchanger


T7 ¼ 698 K and nC6¼ 0

so total specified relations ¼ 2

Number of mass balance relations ¼ 8Number of heat balance relations ¼ 1DF ¼ 20� ð2þ 8þ 1Þ ¼ 20� 11 ¼ 9

Cooler


T5 ¼ 323 K

so total specified relations ¼ 1Number of mass and heat balance relations ¼ 5DF ¼ 11� ð1þ 5Þ ¼ 11� 6 ¼ 5

Separator


nC6¼ 0; nA6

¼ 0:9945nA5; nB6

¼ 0:9945nB5; T5 ¼ 323 K;

T6 ¼ T0; nI6 ¼ 0

so total specified relations ¼ 6Number of mass and heat balance relations ¼ 5DF ¼ 15� ð6þ 5Þ ¼ 4

Overall Mass Balance (see Figs. 2.32 and 2.33)

Number of unknowns ¼ 10þ 2þ 1 ¼ 13

Figure 2.32 Energy flow diagram.

Number of specified relations

nT1¼ 97:8 mol/h; YA1

¼ 0:15; YB1¼ 0:75; T1 ¼ 323 K;

nI9 ¼ 0:978

so total specified relations ¼ 5Number of mass and heat balance relations = 5DF ¼ 13� ð5þ 5Þ ¼ 13� 10 ¼ 3

Note that

r1 ¼ rate of reaction in reactor 1r2 ¼ rate of reaction in reactor 2r1 þ r2 ¼ R

The overall balance for component A is

nA9¼ nA1

þ �A1r1 þ �A2

r2


nA9¼ nT1

YA1� r1 � r2


nA9¼ ð97:8Þð0:15Þ � r1 � r2

We finally get

nA9¼ 14:67� R ð2:89Þ

Similarly,

nB9¼ nB1

þ �B1r1 þ �B2

r2

Figure 2.33 Showing solution strategy: where to start and end.

On substituting the values we get,

nB9¼ ð97:8Þð0:75Þ � 3r1 � 3r2

thus giving

nB9¼ 73:35� 3R ð2:90Þ

and

nC9¼ nC1

þ �C1r1 þ �C2

r2

Substituting of values gives

nC9¼ 0þ 2r1 þ 2r2

thus giving

nC9¼ 2R ð2:91Þ

and

nI9 ¼ nI1

Substitution of values gives

nI9 ¼ ð97:8Þð0:1Þand we get

nI9 ¼ 9:78 ð2:92ÞHeat balanceX

nif ðHif �HirÞn o

þQ ¼X

niðHi �HirÞn o

þ r1�Hr þ r2�Hr

Take the reference as 258C ¼ 25þ 273 ¼ 298 KThus, we get

nA1

ð323298

CPAdT þ nB1

ð323298

CPBdT þ nC1

ð323298

CPCdT þ nI1

ð323298

CPIdT

�þQ

¼ nA9

ðT9

298

CPAdT þ nB9

ðT9

298

CPBdT þ nC9

ðT9

298

CPCdT þ nI

ðT9

298

CPIdT

�þ Rð�HrÞ

In the above equation, every value is known except R;Q, and T9. Therefore,the above is a relation between R;Q, and T9. Thus any of them can beexpressed in term of the other two; for example, we can writeT9 ¼ F1ðR;QÞ. Also, all ni9 ðnA9

, nB9, nC9

) are expressed in terms of Ronly. After we have expressed five unknowns in the exit stream number 9

(as well as Q), all in terms of R and Q, we can move to the separator asshown in the following step.

Separator (see Fig. 2.34)

Number of unknowns,

ni5 ; ni6 ;T5;T6;R; and Q ¼ 12 unknowns

Number of specified relations,

nC6¼ 0; nA6

¼ 0:9945nA5; nB6

¼ 0:9945nB5; T5 ¼ 323 K;

T6 ¼ T9; nI6 ¼ 0

so total specified relations ¼ 6Number of mass and heat balance relation ¼ 5DF ¼ 12� ð6þ 5Þ ¼ 1

Mass Balance

nA5¼ nA6

þ nA9


nA5¼ 0:9945nA5

þ 14:67� R

Finally, it gives

nA5¼ 2667:27� 181:8R ð2:93Þ

Figure 2.34 Mass and energy flow diagram for separator.

Also, we have

nA6¼ 0:9945 nA5

Thus, we get

nA6¼ 2652:6� 180:8R ð2:94Þ

For component B, we get

nB5¼ nB6

þ nB9

On substituting the values, we finally get

nB5¼ 13336:36� 545:454R ð2:95Þ

Also, we have

nB6¼ 13263:01� 542:45R ð2:96Þ

For component C, we get

nC5¼ nC6

þ nC9


nC5¼ 2R ð2:97Þ

Also, we have specified

nC6¼ 0 ð2:98Þ

For component I (the inert), we get

nI5 ¼ nI6 þ nI9


nI5 ¼ 9:78 ð2:99ÞAlso, we get

nI6 ¼ 0 ð2:100Þ

Heat BalanceXni5ðHi5 �HirÞ ¼

Xni6 ðHi6 �HirÞ þ

Xni9 ðHi9 �HirÞ

We have

T6 ¼ T9

So, let us take the reference temperature

Tref ¼ T6 ¼ T9

Thus, the right-hand side of the above-mentioned heat balance becomeszero, and we getX

ni5 ðHi5�Hi6

Þ ¼ 0

This can be rewritten as

nA5

ð323T6

CPAdT þ nB5

ð323T6

CPBdT þ nC5

ð323T6

CPCdT þ nI5

ð323T6

CPIdT ¼ 0

On substituting the values of nA5, nB5

, nC5, and nI5 from Eqs. 2.93, 2.95, 2.97,

and 2.99, we get

ð2667:27� 181:8RÞð323T0

CPAdT þ ð13336:36� 545:45RÞ

ð323T6

CPBdT

þ 2R

ð323T6

CPCdT þ 9:78

ð323T6

dT ¼ 0

This gives a direct relation between R and T6 as follows:

T6 ¼ T9 ¼ F2ðRÞAs we had established earlier,

T9 ¼ F1ðR;QÞTherefore, we say

F1ðR;QÞ ¼ F2ðRÞwhich can be written as

Q ¼ F3ðRÞCooler (see Fig. 2.35) Now we can look at the cooler as having only thefive unknowns of stream 4 and all other variables are in terms of R.

Number of unknowns ¼ 5þ R ¼ 6Number of specified relations ¼ 0Number of mass and heat balance relations ¼ 5DF ¼ 6� 5 ¼ 1

Mass Balance

The mass balance for each component can be simply written as

nA4¼ nA5

¼ 2667:27� 181:8 R

nB4¼ nB5

¼ 13336:36� 545:45R

nC4¼ nC5

¼ 2R

nI4 ¼ nI5 ¼ 9:78

Heat Balance

The generalized heat balance equation is

Xni4ðHi4

�HirÞ þQ ¼X

ni5ðHi5�HirÞ

Take the reference temperature as

Tref ¼ T4

We get

Q ¼X

ni5 ðHi5�Hi4

Þ

Calculations give

Q ¼ nA5

ð323T4

CPAdT þ nB5

ð323T4

CPBdT þ nC5

ð323T4

CPCdT þ nI5ð323

T4

CPIdT

On substituting the values of nA5; nB5

; nC5, and nI5 from Eqs. (2.93), (2.95),

(2.97), and (2.99), we get

Figure 2.35 Mass and energy flow diagram for the cooler.

ð2667:27� 181:8RÞð323T4

CPAdT þ ð13336:36� 545:45RÞ

ð323T4

CPBdT þ 2R

ð323T4

CPCdT þ 9:78

ð323T4

CPIdT ¼ Q

As Q ¼ F3ðRÞ and all the above terms involve either R or T4, we can write

T4 ¼ F4ðRÞThus, we have nA5

; nB5; nC3

, and nI5 , all functions of R, and Q ¼ F3ðRÞ,together with T4 ¼ F4ðRÞ. Thus, whenever we compute R, we will get thevalues of Q and T4, and back substitution will give us the values of all thevariables.

Now we move to the other path as shown in Figure 2.33.

Mixer (see Fig. 2.36)

Mass BalanceMass balances for all the components give the following:

For component A:

nA1þ nA8

¼ nA2


nA2¼ 14:67þ nA8

For component B:

nB1þ nB8

¼ nB2

Figure 2.36 Mass and energy flow diagram for the mixer.


nB2¼ 73:35þ nB8

For component C:

nC1þ nC8

¼ nC2


nC2¼ nC8

For component I :

nI1 þ nI8 ¼ nI2


nI2 ¼ 9:78þ nI8

Heat Balance

The generalized heat balance equation isXni1ðHi1

�HirÞ þX

ni8ðHi8�HirÞ ¼

Xni2 ðHi2

�HirÞTake the reference temperature as

Tref ¼ T2

We getXni1ðHi1

�Hi2Þ þ

Xni8 ðHi8

�Hi2Þ ¼ 0

Calculations give

14:67

ð323T2

CPAdT þ 73:35

ð323T2

CPBdT þ 0þ 9:78

ð323T2

CPIdT

þ nA8

ðT8

T2

CPAdT þ nB8

ðT8

T2

CPBdT þ nC8

ðT8

T2

CPCdT þ nI8

ðT8

T2

CPIdT ¼ 0

Thus, we get

T2 ¼ F5ðT8; ni8 Þ; i ¼ A;B;C; and I

so, we can write

T2 ¼ F5ðT8; ni8 ÞAs ni8 and ni2 are related, we can write

T2 ¼ F6ðT8; ni2 Þ

Reactor 2 (see Fig. 2.37)

For component A, we get

nA3¼ nA2

þ �A2r2

On substituting the values (note that �A2¼ �1Þ, we get

nA3¼ 14:67þ nA8

� r2 ð2:101ÞFor component B, we get

nB3¼ nB2

þ �B2r2

On substituting the values (note that �B2¼ �3Þ, we get

nB3¼ 73:35þ nB8

� 3r2 ð2:102ÞFor component C, we get

nC3¼ nC2

þ �C2r2

On substituting the values (note that �C2¼ 2Þ, we get

nC3¼ nC8

þ 2r2 ð2:103ÞFor component I , we get

nI3 ¼ nI2 þ �I2r2

On substituting the values (note that �I2 ¼ 0Þ, we get

nI3 ¼ 9:78þ nI8 ð2:104ÞNow, we make use of the conversion value given in the problem:

xA2¼ 0:1233 ¼ nA2

� nA3

nA3

Figure 2.37 Mass and energy flow diagram for reactor 2.


0:1233 ¼ ð14:67þ nA8Þ � nA3

nA3

Using the mass balance relations derived above, we get

0:1233 ¼ ð14:67þ nA8Þ � ð14:67þ nA8

� r2Þð14:67þ nA8

� r2Þwhich can be simplified to

r2 ¼ 1:61þ 0:11nA8ð2:105Þ

Heat Balance


�HirÞ þ Q|{z}zero

¼X

ni3 ðHi3�HirÞ þ r2ð�HrÞ


Tref ¼ 258C ¼ 298 K

We can write

nA2

ðT2

298

CPAdT þ nB2

ðT2

298

CBdT þ nC2

ðT2

298

CPCdT þ nI2

ðT2

298

CPIdT

¼ nA3

ðT3

298

CPAdT þ nB3

ðT3

298

CPBdT þ nC3

ðT3

298

CPCdT þ nI3ðT3

298

CPIdT þ r2ð�HrÞ

Note that all ni2 can be put in terms of ni8 as shown earlier (wherei ¼ A;B;C, and I). Therefore, the above heat balance equation gives arelation among ni8 ; ni3 ;T2;T3; and r2. Thus, we can write

T3 ¼ Gðni8 ; ni3 ;T2; r2Þ; i ¼ A;B;C; and I

We can further write

T3 ¼ Gðni8 ; ni3 ;F5ðT8; ni8Þ; r2ÞFinally, we can write

T3 ¼ F7ðni8 ; ni3 ;Tg; r2Þ

However, we already know

ni3 ¼ FFðni8 ; r2ÞThus, we get

T3 ¼ F8ðni8 ;T8; r2ÞThen, from Eq. (2.105), we have

r2 ¼ ~FFðnA8Þ

Finally, we get

T3 ¼ F9ðni8 ;T8Þ

Reactor 1 (See Fig. 2.38) For component A, we get

nA8¼ nA7

þ �A1r1

On substituting the values (note that �A1¼ �1Þ, we get

nA8¼ nA7

� r2

This can be rewritten as

nA7¼ nA8

þ r2 ð2:106ÞIn the very same way, we can write the mass balance equations for compo-nents B, C, and I , we get

nB7¼ nB8

þ 3r1 ð2:107ÞnC7

¼ nC8� 2r1 ð2:108Þ

nI7 ¼ nI8 ð2:109ÞNow, we make use of the conversion value given in the problem:

xA1¼ 0:1 ¼ nA7

� nA8

nA7

Figure 2.38 Mass and energy flow diagram for reactor 1.

Using the mass balance relations derived above, we get

0:1 ¼ ðnA8þ r1Þ � nA8

ðnA8þ r1Þ

which can be simplified to

r1 ¼ 0:111nA8ð2:110Þ

As can be seen, r1 is only a function of nA8. Moreover, all ni7 involve both

r1 and ni8 , so we can easily say that

ni7 ¼ GGðni8 Þ i ¼ A;B;C; and I

Heat Balance


�HirÞ þ Q|{z}zero

¼X

ni8 ðHi8�HirÞ þ r1ð�HrÞ


Tref ¼ 258C ¼ 298 K

It is given that

T7 ¼ 4258C ¼ 698 K

We can write

nA7

ð698298

CPAdT þ nB7

ð698298

CPBdT þ nC7

ð698298

CPCdT þ nI7

ð698298

CPIdT

¼ nA8

ðT8

298

CPAdT þ nB8

ðT8

298

CPBdT þ nC8

ðT8

298

CPCdT þ nI8ðT8

298

CPIdT þ r1ð�HrÞ

On substituting the values from the mass balance relations derived above,we get

ðnA8þ 0:111nA8

Þð698298

CPAdT þ ½nB8

þ 3ð0:111nA8Þ�ð698298

CPBdT

þ ½nC8� 2ð0:111nA8

Þ�ð698298

CPCdT þ nI8

ð698298

CPIdT

¼ nA8

ðT8

298

CPAdT þ nB8

ðT8

298

CPBdT þ nC8

ðT8

298

CPCdT

þ nI8

ðT8

298

CPIdT þ ð0:111nA8

Þð�HrÞ

Therefore, the above heat balance equation gives a relation between ni8 andT8. Thus, we can write

T8 ¼ G2ðni8Þ; i ¼ A;B;C; and I

Exchanger (See Fig. 2.39)

Mass Balance

The mass balance on streams 3 and 4 gives the following: For component A,we get

nA3¼ nA4


ð14:67� 1:61Þ þ nA8ð1� 0:11Þ ¼ F 00

AðnA8Þ

Figure 2.39 Mass and energy flow diagram for the exchanger.

As the above relation is in terms of only nA8, we can solve it to get the value

of nA8.Similarly, we obtain the values for nB8

, nC8, and nI8 using the mass

balances for components B, C, and I . Note that ni3 ¼ ni4 , where i ¼ A;B;C,and I . Because the value of nA8

has been calculated, we can get the value ofT4 as T4 ¼ F 00

T ðnA8). Also, because nA8

has been calculated, T3 ¼ F9ðni8 ;T8Þbecomes

T3 ¼ F10ðT8ÞThen from the individual mass balance for the separator, we have

ni6 ¼ F 000ðRÞ ¼ F 000ðnA8Þ:

Thus, it is known. Also, we get

T6 ¼ F000ðRÞ ¼ F

000ðnA8Þ:

Thus, it is also known. Using the above two known values, ni7 and T7 areknown.

Heat Balance

The generalized heat balance equation isXni3 ðHi3

�HirÞ þX

ni6 ðHi6�HirÞ

¼X

ni7ðHi7�HirÞ þ

Xni4 ðHi4

�HirÞBecause

ni3 ¼ ni4 and ni6 ¼ ni7

we getXni3ðHi3

�Hi4Þ ¼

Xni7 ðHi7

�Hi6Þ

On substituting the values, we see that the above expression has only oneunknown (T8), hence, the above expression can be solved to get the value ofT8, and on back substituting, we get all the unknowns in the problem.

REFERENCE

1. Reklaitis, G. V. Introduction to Material and Energy Balances. Wiley, New

York, 1983.

PROBLEMS

Obtain any missing data from the literature (e.g., Perry’s ChemicalEngineering Handbook 1999).

Problem 2.1

Superphosphate is produced by reacting calcium phosphate with sulfuricacid according to the reaction

Ca3ðPO4Þ2 þ 2H2SO4 ! CaH4ðPO4Þ2 þ 2CaSO4

If 20,000 kg/day of raw calcium phosphate containing 14% inert impuritiesis reacted with 15,000 kg/day of H2SO4 of 92% concentration, determine therate of production of superphosphate assuming the reaction is 95% com-plete (95% conversion). Which is the limiting reactant?

Problem 2.2

Carbon disulfide is used in viscose rayon and cellophane manufacture and inthe production of carbontetrachloride. In the generally preferred process,vaporized sulfur is reacted with methane according to the reactions

CH4 þ 4S ! CS2 þ 2H2S

CH4 þ 2S ! CS2 þ 2H2

CH4 þ 2H2S ! CS2 þ 4H2

For a feed containing 4 mol of sulfur per mole methane, calculate thecomposition of the product if 90% conversion of methane and 70% con-version of sulfur are achieved.

Problem 2.3

Formaldehyde can be made by the partial oxidation of natural gas usingpure oxygen made industrially from liquid air.

CH4 þO2 ! CH2OþH2O

The natural gas must be in large excess. The CH4 is heated to 4008C and O2

to 3008C and introduced into the reactor. The products leave the reactor at6008C and have the following analysis: H2O 1.9%, CH2O 11.7%, O2 3.8%,and CH4 82.6%. How much heat is removed from the reactor by cooling perhour for a production rate of 200 kg/h formaldehyde.

Problem 2.4

Methane and oxygen react in the presence of a catalyst to form formalde-hyde. In a parallel side reaction, some of the methane is instead oxidized tocarbon dioxide and water:

CH4 þO2 ! HCHO þH2O

CH4 þ 2O2 ! CO2 þ 2H2O

The feed to the reactor contains equimolar amounts of methane and oxygen.The fractional conversion of methane is 95%, and the fractional yield

of formaldehyde is 90%. Calculate the molar composition of the reactoroutput stream and the selectivity of formaldehyde production relative tocarbon dioxide production.

In another case, the reactor output stream contains 45 mol% formal-dehyde, 1% carbon dioxide, 4% unreacted methane, and the balance oxygenand water. Calculate the fractional conversion of methane, the fractionalyield of formaldehyde and the selectivity of formaldehyde production rela-tive to carbon dioxide production.

Problem 2.5

Ethylene oxide is produced by the catalytic partial oxidation of ethylene:

C2H4 þ 12O2 ! C2H4O

An undesired competing reaction is the combustion of ethylene:

C2H4 þ 3O2 ! 2CO2 þ 2H2O

The feed to the reactor (not the fresh feed to the process) contains 3 mol ofC2H4 per mole of oxygen. The fractional conversion of ethylene in thereactor is 20%, and the yield of ethylene oxide based on ethylene consumedis 80%. A multiple-unit process is used to separate the products ðC2H4O,CO2, H2O) from the output gases leaving the reactor: C2H4 and O2 arerecycled back to the reactor, C2H4O is sold as a product, and CO2 andH2O are discarded.

(a) Construct a degree-of-freedom table. Is the problem correctlyspecified?

(b) Calculate the molar flow of C2H4 and O2 in the fresh feed. Whatis the mass fraction of C2H4 in the feed?

(c) Calculate the overall conversion and the overall yield based onethylene fed.

(d) Calculate the selectively of C2H4O based on reactor output.

Problem 2.6

Oxychlorination of hydrocarbons refers to a chemical reaction in whichoxygen and hydrogen chloride react with a hydrocarbon in the vaporphase over a supported copper chloride catalyst to produce a chlorinatedhydrocarbon and water. The oxychlorination of ethylene to produce 1,2-dichloroethane [commonly called ethylene dichloride (EDC)] is of the great-est commercial importance. EDC is a precursor for poly(vinyl chloride)

(PVC), which is one of the most widely used commercial plastics. The over-all oxychlorination reaction of ethane is given as follows:

2C2H4 þ 4HClþO2 ! 2C2H4Cl2 þ 2H2O

Determine the mass, moles, and weight percent of the reactant and productstreams if 2000 kmol of the limiting reactant HCl is fed with 10% excess airand 5% excess ethane. Ninety-five percent conversion of ethylene occurs inthe reactor.

Problem 2.7

A mixture containing 68.4% H2, 22.6% N2, and 9% CO2 react according tothe following reaction scheme:

N2 þ 3H2 , 2NH3

CO2 þH2 , COþH2O

The reaction proceeds until the mixture contains 15.5% NH3 and 5% H2O.Calculate the mole fraction of N2, H2, CO2, and CO.

Problem 2.8

A gas mixture consisting of NO2 and air is bubbled through a process intowhich water is fed at a rate of 55.5 kmol/h (as shown in Fig. P2.8). Waterabsorbs most of the NO2 and none of the air. The volumetric flow rate ofthe feed gas is determined from the pressure drop across the orifice. Anequation relating the pressure drop and volumetric flow rate is given by

V ðm3=hÞ ¼ 13:2 h0:515

Figure P2.8 Process diagram for Problem 2.8.

where h is in mm Hg. The molar flow rate of the gas mixture is related tovolumetric flow rate by the ideal gas law

PV ¼ nRT

where R ¼ 0:082 L atm/g mol K, V is the volumetric flow rate, and n is themolar flow rate.

An instrument was used to measure the composition of the gasmixture, where the mole fraction x of the mixture is determined from thereading M obtained from the instrument using the relation

x ¼ 5:00� 10�4 e0:06M

The following data are recorded for the inlet gas mixture: T ¼ 248C,P ¼ 11:2 atm, h (orificemeter) ¼ 210 mm Hg, and M ¼ 80:4. For the outletgas, M ¼ 11:6.

Determine the amount of NO2 absorbed by the water.

Problem 2.9

The recycle process, for producing perchloric acid (HClO4) is shown inFigure P2.9. The reaction

BaðClO4Þ2 þH2SO4 ! BaSO4 þ 2HClO4

proceeds in the reactor with 85% conversion of Ba(ClO4Þ2. The ratio of thenumber of moles of Ba(ClO4Þ2 in feed (fed to the mixer in the figure) to thenumber of moles of H2SO4 fed to the reactor is 1.0 : 1.1. The process mainfeed (fed to the mixer in the figure) consists of 88% (by weight) Ba(ClO4)2

Figure P2.9 Flowsheet of the process for Problem 2.9.

and the rest (12% by weight) is HClO4. The recycle contains onlyBa(ClO4)2. Calculate all flow rates and compositions for all the streams inthe process.

Problem 2.10

Refined sugar (sucrose) can be converted to glucose and fructose by theinversion process

C12H22O11|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl}sucrose

þ H2O ! C6H12O6|fflfflfflfflffl{zfflfflfflfflffl}d-glucose

þ C6H12O6|fflfflfflfflffl{zfflfflfflfflffl}d-fructose

The combined quantity glucose/fructose is called inversion sugar. If 90%conversion of sucrose occurs on one pass through the reactor, what wouldbe the recycle stream flow per 100 kg/h of sucrose solution entering theprocess shown in Figure P2.10? What is the concentration of inversionsugar in the recycle stream and in the product stream? The concentrationsof the components in the recycle stream and product streams are the same.Note: All percentages are weight percentages.

Problem 2.11

Product P is produced from reactant R according to the reaction

2R ! PþW

Unfortunately, because both reactant and product decompose to form theby-product B according to the reactions

R ! BþW

P ! 2BþW

only 50% conversion of R is achieved in the reactor and the fresh feedcontains 1 mol of inerts I per 11 mol of R. The unreacted R and inerts I

Figure P2.10 Process flow diagram for Problem 2.10.

are separated from the products and recycled. Some of the unreacted R andinerts I must be purged to limit the inerts I in the reactor feed to 12% on amolar basis (see Fig. P2.11). For a product stream produced at a rate of1000 lb mol/h and analyzing 38% product P on molar basis, calculate thefollowing (see Fig. P2.11):

(a) The composition of the recycle stream on molar basis(b) The fraction of the recycle which is purged [purge/purge+re-

cycle)](c) The fresh feed rate in lb mol/h(d) The composition of the product stream on a molar basis(e) The fraction of reactant R which reacts via the reaction

R ! BþW .

Problem 2.12

In the recycle system shown in Figure P2.12, a feed of 1000 mol/h consistingof one-third A and two-thirds B is mixed with a recycle stream and reactedfollowing the stoichiometry

Aþ B ! D

In the reactor, 20% of the entering A is converted to product. The resultingstream is separated so that the recycle stream contains 80% of A, 90% of B,and 10% of D fed to the separator.


(a) Show that the process is correctly specified (i.e., mass balance issolvable).

(b) Determine a calculation order for manual solution.(c) Solve the problem.

Problem 2.13

Acetic anhydride can be produced from acetic acid by catalytic cracking. Inthe conceptual process shown in Figure P2.13, the acetic acid is reacted, the



products are separated, and the unreacted acetic acid as well as inerts arerecycled. The fresh feed consists of 1 mol of inerts per 50 mol of acetic acid,and the product analyses 46% ðCH3COÞ2O, 50% H2O, and 4% CH2CO(molar basis).

(a) Determine the degree of freedom and a calculation order.(b) Calculate the fraction of the acetic acid which must be purged to

prevent the inerts from accumulating.(c) Calculate the conversion of acetic acid per pass.

Problem 2.14

A stream of saturated water at 10 bar is available to exchange heat with abrine solution at 1 bar and 508C. If the brine flow is twice the water flow andif the water stream can be cooled to 758C, what is the temperature to whichthe brine can be heated? The brine can be assumed to have the properties ofwater.

Problem 2.15

Given that the net heat of combustion of CH4 (g) is �191:76 kcal/g mol,calculate its heat of formation. The heats of formation of CO2 and H2O areas follows

�H8fCO2 ðgÞ¼ �94:0518 kcal/g mol

�H8fH2OðgÞ ¼ �57:7979 kcal/ g mol

Problem 2.16

In the double-pipe heat exchanger shown in Figure P2.16, oxygen at the rateof 100 kmol/h is being heated by condensing saturated steam at 1.5 bar.Oxygen enters the inner pipe at 258C and leaves at 2008C. The exchanger iswell insulated such that no heat is lost to the atmosphere.

Find the amount of steam condensed per hour of operation if themolar heat capacity of oxygen is given by the expression

CP ¼ 29:88� 0:11384� 10�1T þ 0:43378� 10�4T2

where T is in K and CP is in kJ/kmol.

Problem 2.17

The temperature in a CO shift converter can be moderated by the injectionof excess steam. Assuming a feed of 30% CO, 20% H2, and 50% H2O at6008F and assuming 90% of CO will be converted. Determine the additional

5508F steam required (per mole of feed) to maintain the reactor outlettemperature below 8508F. The reaction stoichiometry is:

COðgÞ þH2OðgÞ ! CO2ðgÞ þH2ðgÞ

Problem 2.18

In the production of ethylene oxide by partial oxidation of ethylene, thefollowing reactions take place:

2C2H4 þO2 ! 2C2H4O

C2H4 þ 3O2 ! 2CO2 þ 2H2O

In a given reactor with a feed consisting of 12%C2H4 and the rest air, 26%conversion of C2H4 and an 78% fractional yield of C2H4O from C2H4

converted are attained when the reactor is operated at 2458C. Calculatethe required heat removal rate from the reactor for a feed rate of 1200kmol/h if the reactor feed mixture is at 1058C. All species are in the gasphase.

Problem 2.19

Carbon monoxide is completely burned at a pressure of 1 atm with excessair. If reactants enter at 2008F, the products leave at 18008F and the heatlosses are negligible, what percentage of excess air was used?

Problem 2.20

Methanol is synthesized from CO and H2 at 50 atm and 550 K over acopper-based catalyst. With a feed of 75%H2 and 25% CO at these condi-tions, a product stream consisting of 20% CH3OH, 1% CH4, 20% CO, and

Figure P2.16 Flow diagram for Problem 2.16.

the rest H2O and small amounts of CO2 and H2 are produced. The primaryreaction

COðgÞ þ 2H2ðgÞ ! CH3OHðgÞThe two secondary reactions are

COðgÞ þ 3H2ðgÞ ! CH4ðgÞ þH2OðgÞCOðgÞ þH2OðgÞ ! CO2ðgÞ þH2ðgÞ

If the feed and product streams are maintained at 50 atm and 500 K, what isthe heat transfer rate required to maintain isothermal conditions? Must heatbe removed or added to the reactor?

Problem 2.21

For the reaction

2Aþ B ! C

the heat of reaction (at 300 K) is �10000 cal/g mol. The heat capacities ofthe components are

CPA¼ 16:0� 1:5� 103

T

CPB¼ 11:0� 0:5� 103

T

CPC¼ 25:0� 1000

T

In the above relations, T is in K and CP in cal/g mol K. The heat capacityequations are valid in the range 300 K � T � 1000 K.

(a) Derive an equation for the heat of reaction as a function oftemperature.

(b) Calculate the temperature at which the reaction changes fromexothermic to endothermic.

(c) Calculate the heat of reaction at 500 K assuming that substanceA undergoes a change of phase at 400 K with �HVLA

¼ 928cal/g mol after which its heat capacity becomes constant at10 cal/g mol K.

Problem 2.22

A hot process stream is cooled by heat exchanger with boiler feed water(BFW), thus producing steam. The BFW enters the exchanger at 1008C and100 bar; the stream is saturated at 100 bar. The process flow stream is 1000

kmol/h, its inlet molar enthalpy is equal to 2000 kJ/kmol, and the exitenthalpy is 800 kJ/kmol. Calculate the required BFW flow rate (see Fig.P2.22).

Problem 2.23

Methanol at 6758C and 1 bar is fed to an adiabatic reactor where 25% of itis dehydrogenated to formaldehyde according to the reaction

CH3OHðgÞ ! HCHOðgÞ þH2ðgÞwhich can be represented by the following reaction for simplification:

A ! Bþ C

Calculate the temperature of the gases leaving the reactor. The given dataare as follows:

Average heat capacity of CH3OHðgÞ ¼ CP0A¼ 17 kcal/g mol 8C

Average heat capacity of HCHO(g)=CP0B¼ 12 kcal/g mol 8C

Average heat capacity of H2ðgÞ ¼ CP0C¼ 7 kcal/g mol 8C

Heat of formation of CH3OH(g) at 258C and 1 bar ¼ �48:08 kcal/g mol

Heat of formation of HCHO(g) at 258C and 1 bar ¼ �27:7 kcal/g molHeat of formation of H2ðgÞ at 258C and 1 bar ¼ 0:0 kcal/g mol

Problem 2.24

An evaporator is a special type of heat exchanger in which steam is used toheat a solution to partially boil off some of the solvent. In the evaporatorshown in Figure P2.24, a brine containing 1% (weight) salt in water is fed at


1 bar and 508C. The exit brine contains 2% (weight) salt and is saturatedliquid at 1 bar. The evaporated water is saturated steam at 1 bar. If satu-rated steam at 2 bar is used as heat source and its condensate is assumed tobe saturated liquid at 2 bar, calculate the kilograms of 2-bar steam requiredper kilogram of evaporated water

Problem 2.25

Acetic acid can be produced via the reaction

3C2H5OHþ 2Na2Cr2O7 þ 8H2SO4 ! 3CH3COOH þ 2Cr2ðSO4Þ3þ 2Na2SO4 þ 11H2O

In the recycle stream shown in Figure P2.25, 90% overall conversion ofC2H5OH is obtained with a recycle flow equal to the feed rate of freshC2H5OH.

The feed rates of fresh H2SO4 and Na2Cr2O7 are 20% and 10%,respectively, in excess of the stoichiometric amounts required for the freshC2H5OH feed. If the recycle stream contains 94% H2SO4 and the restC2H5OH, calculate the product flow and the conversion of C2H5OH inthe reactor.

Problem 2.26

The dehydrogenation of ethanol to form acetaldehyde takes place accordingto the stoichiometric relation

Figure P2.24 Setup for Problem 2.24.

C2H5OHðgÞ ! CH3CHOðgÞ þH2ðgÞ�H8R ¼ �HrRjat 258C ¼ 68:95 kJ/mol of ethanol reacted

The reaction is carried out in an adiabatic reactor. Ethanol vapor is fed tothe reactor at 3008C and a conversion of 30% is obtained. Calculate theproduct (outlet) temperature of this adiabatic reactor.

Heat Capacities[Average values � Assumed constant (independent of temperature)]:

C2H5OHðgÞ : CP ¼ 0:11 kJ/mol 8CC3CHOðgÞ : CP ¼ 0:08 kJ=mol 8CH2ðgÞ : CP ¼ 0:029 kJ/mol 8C

Problem 2.27

Ethylene is made commercially by dehydrogenating ethane:

C2H6ðgÞ , C2H4ðgÞ þH2ðgÞ�HRð2008CÞ ¼ 134:7 kJ/mol

Ethane is fed to a continuous adiabatic reactor at Tf ¼ 20008C. Calculatethe exit temperature that would correspond to 95% conversion (xA ¼ 0:95,where A � C2H6Þ. Use the following data for the heat capacities in yourcalculations:


CPA¼ 49:37þ 0:1392T

CPB¼ 40:75þ 0:1147T

CPC¼ 28:04þ 4:167� 10�3T

where A � C2H6ðgÞ; B � C2H4ðgÞ, and C � H2ðgÞ; CPjis in J/mol 8C and

T is the temperature (in 8C)

Problem 2.28

Synthesis gas is upgraded to higher methane content in the recycle systemshown in Figure P2.28. The feed gas contains a small amount of methaneand analyzes 22% CO, 13% CO2, and 65% H2 (mol%) on a methane-freebasis. The product stream (stream 6) analyzes (all mol %) CO 5%, H2 9%,CH4 50%, CO2 27%, and H2O 9%. Both feed and product are gases at2008F. In the reactor, the following two reactions take place:

COþ 3H2 ! CH4 þH2O

COþH2O ! CO2 þH2

The reactor effluent is cooled to 5008F in exchanger 1 and then it is furthercooled in exchanger 2. Part of the effluent is split off as product. The


remainder is sent to a separator in which the stream is cooled to 908F and95% of the H2O is separated as liquid. The remaining gas stream is reheatedin exchanger 2, mixed with the fresh feed, and returned to the reactor.Assume that the mixer, splitter, exchanger 2, and reactor operate adiabati-cally. Use the average CP values given in Table P2.28.

(a) Check if the process is correctly specified (i.e., process degrees offreedom ¼ 0); if not, assume any missing information and givesound engineering justification for your assumption.

(b) Determine a calculation order of the problem.(c) Write a computer program for the solution based on subroutines

for the standard four mass and heat balance modules.(d) Calculate the reactor outlet temperature.

Problem 2.29

A vertical lime kiln (Fig. P2.29) is charged with pure limestone (CaCO3) andpure coke (C), both at 258C. Dry air is blown in at the bottom and providesthe necessary heat for decomposition of the carbonate by burning thecarbon to carbon dioxide. Lime (CaO) leaves the bottom at 9508C andcontains 5% carbon and 1% CaCO3. The kiln gases leave the top at6008C and consist only of CO2 and N2. Assume heat losses are negligible.The two reactions are as follows:

CaCO3ðsÞ ! CaOðsÞ þ CO2ðgÞCðsÞ þO2ðgÞ ! CO2ðgÞ

Heats of formation are given in terms of kcal/gmol:

�H8fCaCO3ðsÞ¼ �289:5; �H8fCaOðsÞ ¼ �151:7;

�H8fCO2 ðgÞ¼ �289:5

Table P2.28 Values of Specific Heats

for Problem 2.28

CPðgÞ(Btu/lb mol 8F)

CPðlÞ(Btu/lb mol 8F)

CO 7.3 —

H2 7.0 —

CH4 12.0 —

CO2 11.4 —

H2O 8.7 18.0

The molar heat capacities (CPÞ are given in units of cal/g mol K:

CPCaCO3ðsÞ¼ 28:6; CPCðsÞ ¼ 4:6 CPCaOðsÞ ¼ 13:7;

CPCO2 ðgÞ¼ 12:2; CPN2 ðgÞ

¼ 7:5; CPO2ðgÞ¼ 8:0

(a) Calculate the heats of reaction at 258C.(b) Analyze the degrees of freedom of this process. Can the material

and energy balance be coupled?(c) Calculate the required feed ratio of CaCO3 to C required.

Problem 2.30

In the manufacture of sulfur dioxide by the direct oxidation of sulfur withpure oxygen,

SþO2 ! SO2

Cool sulfur dioxide must be recycled to the burner to lower the flame tem-perature below 1000 K so as not to damage the burner. The flowsheet andthe temperatures of the different streams are given in Figure P2.30. For perlb mol of SO2 product, calculate the lb mol SO2 recycled and the pounds ofsteam produced.

Given data for sulfur: melting point ¼ 1138C, heat of fusion (at melt-ing point) ¼ 0:3 kcal/mol, and heat capacity of solid ¼ 5:8 cal/mol K.Obtain any other necessary data from appropriate tables.

Figure P2.29 Schematic diagram of the kiln for Problem 2.29.

Problem 2.31

The production of ethylene oxide by partial oxidation of ethylene over asilver catalyst involves the exothermic reaction

2C2H4ðgÞ þO2ðgÞ ! 2C2H4OðgÞand the secondary, even more exothermic, reaction

C2H4ðgÞ þ 3O2ðgÞ ! 2CO2ðgÞ þ 2H2OðgÞTemperature control is essential and is achieved by boiling a hydrocarbonheat transfer fluid on the outside of the reactor tubes. This vaporized fluidis then condensed in a heat exchanger by transferring heat to a stream ofsaturated liquid water at 100 bar to produce saturated stream at 100 bar.The reactor feed is 11% C2H4, 13% O2 and the rest N2 at 3608C at 10bar. The reactor product stream is at 3758C and 10 bar and the conver-sion of C2H4 is 22% with a 83% selectivity for C2H4O. At the operatingpressure used in this system, the heat transfer fluid boils at 3508C with aheat of vaporization of 500 Btu/lb and has a liquid heat capacity of 0.8Btu/lb 8C and a vapor heat capacity of 0.4 Btu/lb 8C. The flow of heattransfer fluid is adjusted so that it enters the reactor as liquid at 3408Cand leaves as a two-phase mixture with vapor fraction of 21% (see Fig.P2.31).

(a) Calculate the mass of steam produced per mole of C2H4O pro-duced.

(b) Calculate the recirculation rate of the heat transfer fluid per moleof C2H4O produced.


(c) Why is a heat transfer fluid used instead of the direct heat transferusing water?

(d) What pressure would the reactor tubes have to withstand if waterwere boiled at 3508C on the outside of the reactor tubes?


3

Mathematical Modeling (I):Homogeneous Lumped Systems

This chapter concentrates on the transformation of material and energybalance equations of homogeneous lumped systems into mathematicalmodels (design equations).

3.1 MATHEMATICAL MODELING OFHOMOGENEOUS LUMPED PROCESSES

3.1.1 Basic Concepts for the Mathematical Modeling ofProcesses

Mathematical modeling of mass transfer, heat transfer, and reaction in thepetrochemical, petroleum refining, biochemical, and electronic systems hasbeen a very strong tool for design, simulation, and control as well asresearch and development (R&D) in these industries. It has led to a morerational approach for design and control in addition to elucidating manyimportant phenomena associated with these systems. It has also led to ahigher level of organization for research in these fields. Rigorous highlysophisticated mathematical models of varying degrees of complexity arebeing developed and used in industrial design and control as well as aca-demic and industrial research (1–3). An important point to be noted withregard to the state of the art in these relatively well-established fields is thatsteady-state modeling is more advanced than unsteady-state modelingbecause of the additional complexities associated with unsteady-state beha-vior and the additional physicochemical information necessary (1,4–8).

Based on these achievements, continuous processing has advanced inmost chemical and related industries over the years and became thedominant processing mode. Computerized reliable design packageshave been developed and computer control of units and whole plantshave been introduced widely (9–13). All of this has been coupled with aconsiderable increase in the productivity of industrial units and plants with ahigh reduction in manpower and tight control over the product quality.

On the industrial and academic levels, these advancements have led toinnovative designs and configurations that open new and exciting avenuesfor developing new compact units with very high productivity (14–17).

In this book, a system approach is adopted which treats any process orprocessing plant as a system consisting of subsystems, with their propertiesand interactions giving the overall system its characteristics. Before gettinginto detail, it is important to give a more applied exposition of system theorythan that given in Chapter 1. To make the exposition more applied, we willgive some details regarding the principles of mathematical models buildingwith some emphasis on fixed-bed catalytic reactors, which represent one ofthe most important units in the above fields.

3.1.2 Systems and Mathematical Models

In this subsection, the basic concepts of system theory and principles formathematical modeling of chemical processes are briefly revisited. The basicprinciples have been discussed in Chapter 1, these basic principles are dis-cussed here once again in a simple and brief manner.

A system is a whole consisting of elements and subsystems. The con-cept of system–subsystems–elements is relative and depends on the level ofanalysis. The system has a boundary that distinguishes it from the environ-ment. The system may exchange matter and/or energy with the environ-ment, depending on the type of system from a thermodynamics point ofview. A system (or subsystem) is described by its elements (or subsystems),the interaction between the elements, and its relation with the environment.The elements of the system can be material elements distributed topologi-cally within the boundaries of the system and giving the configuration of thesystem, or they can be processes taking place within the boundaries of thesystem and defining its functions. They can also be both, together with theircomplex interactions. An important property of the irreducibility of thecomplex to the simple or of the whole to its elements is related to the factthat the whole system will possess properties and qualities not found in itsconstituent elements. This does not mean that certain information aboutthe behavior of the system cannot be deduced from the properties of itselements, but it rather adds to it.

Systems can be classified on different bases. The most fundamentalclassification is that based on thermodynamic principles, and on this basis,they can be classified into isolated systems, closed systems, and open systems(18,19), as discussed in Chapter 1.

3.1.3 What Are Mathematical Models and Why DoWe Need Them?

Because much of the basic information for detailing the answer to thisquestion are discussed in other parts of this book, we present a brief answerto this question here.

For the first part of the question, we present briefly Denn’s answer (20):

A mathematical model of a process is a system of equations whosesolution, given specified input data, is representative of the responseof the process to a corresponding set of inputs.

This brief answer contains the essential components of a true mathematicalmodel. However, for the definition to be more explicitly complete, we have toadd that the term process in the above context means the configuration of theunit and all of the parameters related to its dimension and design. We havealso to add that the answer given by Denn (20) is phrased for the simulationof existing processes. Of course, mathematical models are basically rigorousdesign equations which can and must be used for design purposes. In designproblems, the output variables are specified as required from the processtogether with the specifications of the product and their rates of production.Thus, in most design problems, the output variables are predetermined, aswell as the input variables, and it is the design parameters which we seek tocompute. Therefore, both input and output variables are fed to the model(actually the computer program of the model) in order to compute (usually inan iterative manner) some of the design parameters of the unit itself and alsoto check that the obtained design does not violate the physically imposedconstraints. When the design results violate the physical constraints, themodel is further used to suggest solutions for this crucial problem.

The second part of the question has been explicitly and implicitlyanswered in many places in this book and we can summarize the answerin the following points regarding the advantages of using rigorous mathe-matical models in design:

1. It gives more precise and optimal design of industrial units.2. It allows the investigation of new more effective designs.3. It can be used for simulation of the performance of existing units

to ensure their smooth operation.

4. It is usable for the optimization of the performance of existingunits and compensation for external disturbances in order toalways keep the unit working at its optimum conditions.

5. Dynamic models are used in the design of control loops fordifferent industrial units to compensate for the dynamic effectsassociated with external disturbances.

6. Dynamic models are also used for stabilization of the desirableunstable steady state by designing the necessary stabilizingcontrol.

7. It is utilized in the organization and rationalization of safetyconsiderations.

8. It is extremely useful for training purposes.9. It provides an excellent guide regarding environmental impact

and protection.

3.1.4 Empirical (Black Box) and Physical(Mathematical) Models

Empirical models are based on input–output relations that do not take intoconsideration the description of the processes taking place within the sys-tem. This is why it is sometimes referred to as the ‘‘black-box’’ approach,because the system is considered as a nontransparent box with attentionfocused only on input and output variables, without much concern aboutwhat is going on inside the box. Many procedures exist for determining theappropriate experimental design and the way in which data are converted toequations. It is usually necessary to make some a priori assumptions aboutthe physical structure of the system as well as the mathematical structure ofthe equations. Empirical models are highly unreliable and can hardly beextrapolated outside the region where the experiments were conducted.However, they are relatively easy and some empirical models are sometimesused as parts of an overall physical (or mathematical) model. Thus, empiri-cal models are unreliable as overall process models, but sometimes they areused as parts of overall mathematical models.

On the other hand, physical (or mathematical) models are built on thebasis of understanding the processes taking place within the boundaries ofthe system, the laws governing the rates of these processes, the interactionbetween the processes, as well as the interaction with the input variablesand surrounding environment. These models are sometimes called physicalmodels. This is to distinguish them from empirical models, because theydescribe the physical situation instead of ignoring it. Empirical models, onthe other hand, relate input and output through linear and nonlinearregression. Physical models are also called mathematical models because

of the mathematical description of the processes taking place within theboundaries of the system and in the description of the overall system.

Of course, no mathematical model is completely rigorous. Each math-ematical model will always contain some empirical parts. As the degree ofempiricism decreases, the model reliability increases. As an example, in thedevelopment of mathematical models for packed beds, the mass and heattransfer resistances are usually computed using the empirical j-factor corre-lations for mass and heat transfer between the solid surface and the bulkfluid. Having some of the components of the mathematical models asempirical relations does not make the model empirical; it is only thatsome of the parameters involved in the quantitative description of theprocesses taking place within the system are determined from empiricalrelations obtained experimentally.

To illustrate the above points in a simple manner, let us consider anonisothermal adiabatic fixed-bed reactor packed with nonporous catalystpellets for a simple exothermic reaction

A ! B

Assume that the pellets are of large thermal conductivity and that the rate ofreaction is first order in the concentration of component A:

r ¼ k0e�E=RTCA

kmol

kg catalyst hð3:1Þ

A mathematical model for this system will be a heterogeneous model takinginto consideration the differences in temperature and concentration betweenthe bulk gas phase and the solid catalytic phase. For the catalyst pellet, thesteady state is described by algebraic equations as follows.

Using the symbols in the schematic diagram of the catalyst pellet asshown in Figure 3.1, the mass balance for component A (for a case withnegligible intraparticle mass and heat transfer resistances) is given by

aPkgAðCAB � CAÞ ¼ WPr ð3:2Þ

Figure 3.1 Nonporous catalyst pellet.

where aP is the external surface area of the pellet (in m2),WP is the weight ofthe catalyst pellets (in kg), kgA is the external mass transfer coefficient ofcomponent A (in m=h), r is the rate of reaction per unit mass of the catalyst,CA is the concentration of component A in the catalyst pellet, and CAB

is thebulk concentration around the pellet. The heat balance for the pellet is givenby

aPhðT � TBÞ ¼ WPrð��HRÞ ð3:3Þwhere ��HR is the heat of reaction (in kJ=k mol, h is the external heattransfer coefficient (in kJ=m2 KhÞ, T is the pellet temperature (in K), and TB

is the bulk temperature (in K). The bulk phase balances are shown in Figure3.2. The bulk mass balance equation for the variation of CAB along thelength of the reactor (details of its derivation will be given in a later section)is

qdCAB

dl¼ kgAAtaSrBðCA � CABÞ ð3:4Þ

Similarly, the bulk temperature variation along the length of reactor is givenby

qrmixCPmix

dTB

dl¼ hAtaSrBðT � TBÞ ð3:5Þ

with initial conditions (at l ¼ 0ÞCAB ¼ CAf and TB ¼ Tf ð3:6Þ

where q is the volumetric flow rate (in m3=h), rmix is the density of the gasmixture, rB is the density of the catalyst (in kg/m3, At is the cross-sectionalarea of the catalyst tube (in m2), CPmix

is the specific heat of the mixture (inkJ/kg), and aS is the specific surface area of the catalyst (in m2=kg catalyst).

To obtain the effect of input variables (e.g., q; CAf , and Tf on theoutput variables CB and TB), Eqs. (3.1)–(3.6) must be solved simultaneouslyand numerically to obtain necessary results, as shown in Figure 3.3, where Lis the length of the reactor.

Figure 3.2 Mass and energy flow diagram for the bulk phase.

In the above simple model, the parameters of the system are obtainedfrom independent experiments to evaluate k0, E, and ��HR. The heat andmass transfer coefficients (h and kgA, respectively) are obtained from empiri-cal j-factor correlations. For each value of Tf shown in Figure 3.3, theequations are solved to obtain the output variables CAB and TB. The solu-tion also provides the concentration and temperature profiles along thelength of the catalyst bed. It will also provide the temperature of the catalystpellet T and the concentration CA at every position along the length of thereactor. This is obviously a physical (mathematical) model, although someof the parameters are obtained using empirical relations (e.g., h and kgAfrom j-factor correlations).

If the relations in Figure 3.3 are to be obtained by an empirical model,a number of experiments have to be performed, where, for example, Tf ischanged and CABjat l¼L is measured experimentally, and then the results arefitted to a polynomial (or any other suitable function) by nonlinear regres-sion to obtain an empirical model of the form

CABjat l¼L ¼ f Tf

� �If other input variables need to be incorporated, then a very large number ofexperiments must be performed and multiple regression is used to obtain anempirical model of the form

CABjat l¼L ¼ f Tf ; q;CAf

� �The difference in rigor, reliability, and predictability between the twoapproaches should be obvious from this simple example. It is obviousthat such empirical relations have no physical basis and, therefore, it is

Figure 3.3 Effect of feed temperature on output concentration of reactant A.

highly unreliable and it is not very wise (nor necessary nowadays) to try toextrapolate it.

3.2 MATHEMATICAL MODEL BUILDING:GENERAL CONCEPTS

Building a mathematical model for any chemical engineering systemdepends to a large extent on the level of knowledge regarding the physicaland chemical processes taking place within the boundaries of the system andthe interaction between the system and the environment. It is, of course, alsoessential to know the laws governing the processes taking place within theboundaries of the system as well as the laws governing its interaction withthe surroundings. For example, if we consider the more general catalyticreacting systems, these will include the diffusion mechanism and rates ofdiffusion of reacting species to the neighborhood of active centers of reac-tion, the chemisorption of the reacting species on these active sites, thediffusion of reactants through the pores of the catalyst pellets (intraparticlediffusion), the mechanism and kinetic rates of the reaction of these species,the desorption of products, and the diffusion of products away from thereaction centers. It also includes the thermodynamic limitations that decidethe feasibility of the process to start with, and also includes heat productionand absorption as well as heat transfer rates. Of course, diffusion rates andheat transfer rates are both dependent to a great extent on the properdescription of the fluid flow phenomena in the system. The ideal case iswhen all of these processes are determined separately and then combinedinto the system’s model in a rigorous manner. However, very often this isquite difficult to achieve; therefore, special experiments need to be devised,coupled with the necessary mathematical modeling, in order to decouple thedifferent processes interacting in the measurements.

Mathematical models (design equations) of different degrees of sophis-tication and rigor were built in last three to four decades to take their part indirecting design procedure as well as directing scientific research in all fieldsof chemical engineering and its related disciplines. It is important in thisrespect to stress the fact mentioned a number of times previously that mostmathematical models are not completely based on rigorous mathematicalformulation of the physical and chemical processes taking place within theboundaries of the system and the interaction between the system and itsenvironment. Every mathematical model contains a certain degree ofempiricism. The degree of empiricism, of course, limits the generality ofthe model, and as our knowledge of the fundamentals of the processestaking place increases, the degree of empiricism decreases and the degreeof rigor (generality) of the model increases. The existing models at any stage,

with this stage’s appropriate level of empiricism, helps greatly in theadvancement of the knowledge of the fundamentals and therefore helps todecrease the degrees of empiricism and increase the level of rigor in themathematical models. In addition, any model will contain simplifyingassumptions which are believed, by the model builder, not to affect thepredictive nature of the model in any manner that sabotages the purposeof the model.

With a given degree of fundamental knowledge at a certain stage ofscientific development, one can build different models with different degreesof sophistication depending on the purpose of the model building and thelevel of rigor and accuracy required. The choice of the appropriate level ofmodeling and the degree of sophistication required in the model is an artthat needs a high level of experience. Models that are too simplified will notbe reliable and will not serve the purpose, whereas models that are toosophisticated will present unnecessary and sometimes expensive overburden.Models that are too sophisticated can be tolerated in academia and maysometimes prove to be useful in discovering new phenomena. However,oversophistication in modeling can hardly be tolerated or justified in indus-trial practice.

The procedure for model building differs depending on the processitself and the modeling group. However, a reasonably reliable procedurecan be summarized in the following steps:

1. The identification of the system configuration, its environment,and the modes of interaction between them.

2. The introduction of the necessary justifiable simplifying assump-tions.

3. The identification of the relevant state variables that describe thesystem.

4. The identification of the processes taking place within the bound-aries of the system.

5. The determination of the quantitative laws governing the rates ofthe processes in terms of the state variables. These quantitativelaws can be obtained from the known information in the litera-ture and/or through an experimental research program coupledwith the mathematical modeling program.

6. The identification of the input variables acting on the system.7. The formulation of the model equations based on the principles

of mass, energy, and momentum balance appropriate to the typeof system.

8. The development of the necessary algorithm for the solution ofthe model equations.

9. The checking of the model against experimental results to ensureits reliability and re-evaluation of the simplifying assumptionswhich may result in imposing new simplifying assumptions orrelaxing some of them.

It is clear that these steps are interactive in nature and the results of each stepshould lead to a reconsideration of the results of all previous steps. In manyinstances, steps 2 and 3 are interchanged in the sequence, depending on thenature of the system and the degree of knowledge regarding the processestaking place, within its boundaries, and its interaction with the environment.

3.2.1 Classification of Models

We revisit the different types of models which are essential in model devel-opment for the different chemical/biochemical engineering processes. Batchprocesses are usually of the isolated- or closed-system type, whereas thecontinuous processes are almost always of the open-system type.

For continuous processes, a classification from a mathematical pointof view is very useful for both model formulation and algorithms for modelsolution. According to this basis, systems can be classified as follows:

1. Lumped systems: These are systems in which the state variablesdescribing the system are lumped in space (invariant in all spacedimensions). The simplest chemical engineering example is theperfectly mixed continuous-stirred tank reactor. These systemsare described at steady state by algebraic equations, whereasthe unsteady state is described by initial-value ordinary differen-tial equations for which time is the independent variable.

2. Distributed systems: These are systems in which the state variablesare varying in one or more direction of the space coordinates.The simplest chemical engineering example is the plug flow reac-tor. These systems are described at steady state by either anordinary differential equation [when the variation of the statevariable is only in one direction of the space coordinates, (i.e.,one-dimensional models) and the independent variable is thespace direction] or partial differential equations [when the varia-tion of the state variables is in more than one direction of thespace coordinates (i.e., two-dimensional models or higher), andthe independent variables are these space directions]. The ordin-ary differential equations of the steady state of the one-dimen-sional distributed parameter models can be either initial-valuedifferential equations (e.g., plug flow models) or two-pointboundary-value differential equations (e.g., models with super-

imposed axial dispersion or diffusion through pellets or mem-branes). The equations describing the unsteady state of distribu-ted models are invariably partial differential equations.

Another classification of systems, which is very important for decidingthe algorithm for model solution, is that of linear and nonlinear systems.The equations of linear systems can usually be solved analytically, whereasthe equations of nonlinear systems are almost always solved numerically. Inthis respect, it is important to recognize the important fact that physicalsystems are almost always nonlinear, and linear systems are either anapproximation that should be justified or intentionally linearized in theneighborhood of a certain state of the system and are strictly valid only inthis neighborhood.

A third classification, which is very relevant and important to chemicalengineers, is the classification based on the number of phases involvedwithin the boundaries of the system. According to this classification systemsare divided as follows (21):

1. Homogeneous systems: These are systems in which only one phaseis involved in the process taking place within the boundaries ofthe system. The behavior of these systems (for reacting systems) isbasically governed by the kinetics of the reaction taking placewithout the interference of any diffusion processes betweenphases. Nonreacting homogeneous systems are usually very sim-ple (e.g., mixer, splitter, filling/emptying of a tank, etc.).

2. Heterogeneous systems: These are systems in which more thanone phase is involved in the processes taking place within theboundaries of the system. The behavior of these systems (forreacting systems) is governed not only by the kinetics of thereactions taking place but also by the complex interactionbetween the kinetics and the relevant diffusion processes. Themodeling and analysis of these systems is obviously much morecomplicated than for homogeneous systems. It is clear that thesystems for fixed-bed catalytic reactors fall into the category ofheterogeneous systems and, more specifically into the category ofgas–solid systems; therefore, the behavior of the system is depen-dent on a complex interaction between kinetics and diffusion.Nonreacting heterogeneous systems are not very simple (e.g.,absorption, distillation, adsorption, etc.).

The reader interested in more details about system theory and thegeneral concepts of mathematical modeling will find a large number ofvery good and interesting books (22–24).

3.2.2 Difference Between Modeling and Simulation

Aris (25) in his 1990 Dankwerts Memorial Lecture entitled ‘‘MannersMakyth Modelers’’ distinguishes between modeling and simulation in aspecial manner that follows the reasoning of Smith (26) in ecologicalmodels,

It is essential quality in a model that it should be capable of havinga life of its own. It may not, in practice, need to be sundered fromits physical matrix. It may be a poor, an ill-favored thing when it isby itself. But it must be capable of having this independence.

He also adds in the same lecture:

At the other end of the scale a model can cease to be a model bybecoming too large and too detailed a simulation whose natural lineof development is to the particular rather than the general. It ceasesto have a life of its own by becoming dependent for its vitality on itsphysical realization.

As discussed in Chapter 1, the basis of the classification given by Arisis very interesting, true, and useful. It is typical of the Minnesota school,which actually revolutionized the field of mathematical modeling in chemi-cal engineering with their extensive work in this field since the mid-1940suntil today. Experimental verifications of the arsenal of new and interestingsteady-state and dynamic phenomena discovered by the Minnesota schoolwere carried out in other universities, mostly by graduates from Minnesota.The most interesting outcome is the fact that not a single phenomena whichwas discovered theoretically using mathematical models by the Minnesotagroup was not experimentally confirmed later. This demonstrates the greatpower of mathematical modeling discipline as expressed by Aris (25), wherethe model is stripped of many of its details in order to investigate the mostfundamental characteristics of the system. The Minnesota school using thisapproach has achieved a real revolution in chemical engineering in generaland in chemical engineering mathematical modeling in particular.

In this book, a different definition for mathematical modeling andsimulation than the one of Aris (25) will be adopted. Our definition willbe more pragmatic and will serve our specific purpose in this book. Thedefinition we will adopt is that mathematical modeling will involve theprocess of building up the models themselves, whereas simulation involvessimulating experimental or industrial units using the developed models.Thus, simulation in this sense is actually closely linked to the verificationof the model against experimental and industrial units. However, it will alsoinclude the use of the verified models to simulate a certain practical situation

specific to a unit, a part of a production line, or an entire production line.The main purpose of developing and verifying these models is to use them asrigorous design equations.

3.2.3 Design Equations and Mathematical Models

From the process design point of view, design means deciding on the con-figuration of the unit, its dimensions, and the optimal input variables. Ofcourse, the guidelines for such a design are the imposed constraints, such asthe amount of raw material to be processed in the unit, the properties of theraw material and the products, chemical and physical paths between the rawmaterial and the products, the rate of these paths, the thermochemicalproperties of the materials and processes (or paths), the maximum tempera-ture that the material of construction and catalyst can withstand, the qualityof the products and their rate of production, and so forth.

Design can be carried out based on previous experience with very fewcalculations and that was the case at the very early stages of the chemicalindustry before even chemical engineering was established. Certainly, someelements of this highly empirical approach still exists in today’s advanceddesign procedure. A relatively higher degree of rationalization as comparedwith the highly empirical approach will be based on scaling up from benchscale to pilot plant, then to commercial scale directly or through a semi-commercial-scale unit. This procedure includes extensive experimental workwith a limited amount of computation. Another procedure is to use designequations that are of a certain degree of rigor coupled with design guidelinesthat accumulated over the years.

What are usually called design equations are usually expressed interms of nonlinear sets of algebraic, differential, or integral equations andcan be called mathematical models.

Actually, there is no difference in principle between design equationsand mathematical models. It is just that when the process is describedby a highly empirical and simplified set of equations, we call them designequations, whereas when the equations are more rigorous and thereforemore reliable, they are called a mathematical model.

3.2.4 Simplified Pseudohomogeneous ModelsVersus Rigorous Heterogeneous Models

We will use the industrially important fixed bed-catalytic reactor as a toolto illustrate some basic modeling principles. The main steps for reactionstaking place in fixed-bed catalytic reactors are as follows:

1. External mass transfer from the bulk gas phase to the externalsurface of the pellet

2. Intraparticle diffusion of reactants through the pores of thecatalyst pellets toward the center of the pellet

3. Chemisorption of reaction molecules onto the surface of thepellet

4. Surface reaction of the adsorbed species5. Desorption of the product molecules6. Intraparticle diffusion of product molecules toward the surface of

the pellet7. External mass transfer of the product molecules from the surface

of the pellet to the bulk gas phase

This sequence of events makes the concentrations under which the reactionsactually take place quite different form those at the measurable bulk con-ditions. In addition to that, when the heats of reactions are appreciable,there will also be a temperature difference between the bulk gas phase andthe solid catalyst phase.

Heterogeneous models (to be discussed in Chapter 6) take into con-sideration these differences between the bulk gas phase and the catalyst solidphase, whereas pseudohomogeneous models ignore these differences. Thesedifferences between the two phases are best expressed through a coefficientcalled the effectiveness factor Z (or sometimes called the efficiency factor)which is defined as the ratio of the actual rate of reaction to the rate ofreaction when mass and heat transfer resistances are neglected.

In other words,

Z ¼ Correct rate of reaction computed from heterogeneous models

Rate of reaction computed from homogeneous models

The effectiveness factor computation involves the solution of nonlinear two-point boundary-value differential equations of varying degrees of complex-ity depending on the specific reactions taking place in the reactor. Theeffectiveness factors vary with the change of the bulk-gas-phase conditionsand therefore must be computed at every point along the length of thereactor. Efficient numerical techniques have been developed for solvingthis problem.

From the above simple discussion, it is clear that the pseudo-homogeneous model is simply a heterogeneous model but with Z ¼ 1:0[or at least Z ¼ constant (i.e., it is not changing along the length of thereactor)]. Therefore, when Z approaches 1.0, the pseudohomogeneousmodels are valid for design, operation, and optimization of catalytic

reactors. However, when Z is far from unity and is varying along the lengthof the reactor, the use of the heterogeneous model is a must.

In industrial fixed-bed catalytic reactors, the catalysts are usually rela-tively large to avoid a pressure drop, and, therefore, the Z’s are usually farfrom unity and heterogeneous models are necessary for the accurate design,operation, and optimization of these reactors. For example, ammonia con-verters have Z’s is in the range 0.3–0.7, and an error of 30–70% results fromthe use of the pseudohomogeneous models. For steam-reforming Z is in therange 10�2�10�3; therefore, the use of pseudohomogeneous models will giveextremely wrong predictions (10,000–100,000% error).

In some cases of highly exothermic reactions and reactions with non-monotonic kinetics, Z can be much larger than unity and, therefore, the useof pseudohomogeneous models can give very erroneous results.

It is clear that the use of pseudohomogeneous models is not suitablefor most industrial fixed-bed catalytic reactors, and heterogeneous modelsshould almost always be used. To the contrary, for fluidized-bed catalyticreactors, the catalyst particles are in the form of fine powder and, therefore,Z’s are very close to unity. However, in fluidized-bed reactors, the hydro-dynamics of the fluidized bed should be taken into consideration especiallythe bubbles of gas rising through the fluidized powder. Thus, for fluidizedbeds, although the difference in conditions between the gas surrounding thesolid particles and the solid particles in the emulsion phase can be neglected,the difference in concentration and temperature between the emulsion phaseand the bubble phase cannot be neglected.

In the petrochemical industry, fluidized beds are used in the pro-duction of polyethylene and polypropylene, and in the petroleum refiningindustry, they are used in fluid catalytic cracking (FCC).

In this chapter, we will concentrate on the development of homo-geneous models; and Chapter 6 is devoted to the heterogeneous systems.

3.2.5 Steady-State Models Versus Dynamic Models

Steady-state models are those sets of equations which are time invariant anddescribe the conditions of the system at rest (i.e., when the states of thesystem are not changing with time). This will automatically presupposethat the system parameters are also time invariant (i.e., input variables,heat transfer coefficients, catalyst activity, and so forth are not changingwith time). Of course, this is a theoretical concept, for no real system canfulfill these requirements perfectly. However, this theoretical concept rep-resents the basis for the design and optimization of almost all chemical/biochemical engineering equipment. The philosophy is that we assumethat the system can attain such a time-invariant state and design the system

on that basis. Then, we design and implement the control system that always‘‘pushes’’ the system back to its optimally designed steady state.

For design and optimization purposes, we use steady-state models,whereas for start-up and control of units, we must use dynamic models(unsteady-state models). To illustrate these concepts we use a very simpleexample of a consecutive reaction,

A �!k1 B �!k2 C

taking place in an isothermal continuous-stirred tank reactor (CSTR), withB being the desired product. For simplicity, we assume that the feed is pureA (contains no B or C):

CBf ¼ CCf ¼ 0

Now, the unsteady-state mass balance equation for component A is:

VdCA

dt¼ qCAf � qCA � Vk1CA ð3:7Þ

and for component B is:

VdCB

dt¼ �qCB þ Vðk1CA � k2CBÞ ð3:8Þ

with the initial conditions (t ¼ 0)

CA ¼ CA0 and CB ¼ CB0

We want to find the size of the reactor V that gives maximum concentrationof the desired product B for given q, CAf , k1, and k2. To achieve this firsttask, we use the steady-state equations, which can be simply obtained bysetting the time derivatives in Eqs. (3.7) and (3.8) equal to zero, thus giving

qðCAf � CAÞ ¼ Vk1CA ð3:9Þand

qCB ¼ Vðk1CA � k2CBÞ ð3:10ÞSome simple manipulations of Eqs. (3.9) and (3.10) give

CB ¼ qk1CAfV

ðqþ Vk1Þðqþ Vk2Þð3:11Þ

To obtain Voptimum, we differentiate Eq. (3.11) with respect to V to get

dCB

dV¼ qk1CAf

q2 � k1k2V2

ðqþ Vk1Þ2ðqþ Vk2Þ2" #

ð3:12Þ

Then, putting dCB=dV ¼ 0 gives Voptimum (denoted as VoptÞ,

Vopt ¼qffiffiffiffiffiffiffiffiffik1k2

p ð3:13Þ

Therefore,

CBmax¼ qk1CAfVopt

ðqþ Voptk1Þðqþ Voptk2Þð3:14Þ

If the reactor is operating at this output concentration, disturbances willcause it to deviate from it. The deviation can be dynamic, (varying withtime) or static (very slow disturbance) so that the system settles down to anew steady state.

If we want to follow the change of the state variables with time due toa disturbance, say a change in q, we have to solve the unsteady-state equa-tions. Suppose that the system is at its optimum steady-state conditions;then, the initial condition ðt ¼ 0Þ is given by

CA ¼ CAoptand CB ¼ CBopt

where

CAopt¼ qCAf

qþ Voptk1ð3:15Þ

and CBoptis equal to CBmax

in Eq. (3.14).If q changes to q0, then we insert q0 in the dynamic equations instead of

q and solve the differential equations from t ¼ 0 to higher values of time inorder to follow the change with time. At large values of t, the system willsettle to new steady-state values corresponding to the new q0. These newvalues of CA and CB will be C 0

A and C 0B:

C 0A ¼ q0CAf

ðq0 þ Voptk1Þð3:16Þ

C 0B ¼ q0k1CAfVopt

ðq0 þ Voptk1Þðq0 þ Voptk2Þð3:17Þ

Vopt is used here as a symbol to indicate the value of V chosen to give CBmax

for q. However, for q0, the volume Vopt is no longer optimum.This steady-state deviation from CBmax

can be compensated for by thechange of another variable such as CAf and/or the operating temperaturethat will change k1 and k2. The requirement for this compensation (say, infeed concentration) will be that

q0C 0Af

q0 þ Voptk01

¼ qCAf

qþ Voptk1ð3:18Þ

and

q0k01C0Af Vopt

ðq0 þ Voptk01Þðq0 þ Voptk

02Þ

¼ qk1CAfVopt

ðqþ Voptk1Þðqþ Voptk2Þð3:19Þ

These are two equations in three unknowns ðC 0Af , k

01 and k02). However, note

that both k01 and k02 are functions of the new temperature T 0. Equations(3.18) and (3.19) can be solved for C 0

Af and T 0 (gives k01 and k02) to find thenew feed concentration and temperature needed to maintain the system atits original optimum state despite the change of q to q0.

However, the question arises with regard to the dynamic variation dueto change in input. What is to be done when the variation in input para-meters is continuous (i.e., when the system does not have enough time tosettle to any new steady state)? In this case, the dynamic model equationsmust be used to design a controller that introduces compensation which ischanging with time. It is also important in this respect to make clear thatwhen the disturbances are very slow (e.g., slowly deactivating catalyst), thenthe quasi-steady-state approximation can be used, where the change withtime is considered as a series of steady states each corresponding to the valueof the changing variable at the sequence of time intervals.

3.2.6 A Simple Feedback Control Example

A simple example can be used to illustrate the concept of the use of dynamicmodels in simulation and control. Consider the water tank shown in Figure3.4, where the valve at the bottom discharges water at a rate proportional tothe head h. It is well known that the discharge is proportional to

ffiffiffih

p; how-

ever, we use the assumption that it is proportional to h in order to make theequations linear and, therefore, illustrate the ideas in a simple manner.

The mass balance on water gives

qin ¼ q0 þ Adh

dt¼ Chþ A

dh

dtð3:20Þ

where C is the valve coefficient and A is the cross-sectional area of thecylindrical tank. This is a simple linear equation. Suppose that the tankwas originally empty; then, at t ¼ 0, h ¼ 0. Also suppose that qin is constant;then, Eq. (3.20) can be solved analytically and we get the change of h withtime:

hðtÞ ¼ qinAa

ð1� e�atÞ ð3:21Þ

where

a ¼ C

A

The steady-state height can be obtained from the steady-state equation

qin ¼ ChSS ð3:22ÞThus,

hSS ¼qinC

or from the solution of Eq. (3.21) by putting t ! 1,

hSS ¼qinC

ð3:23Þ

The change of h with time from h ¼ 0 to h ¼ hSS follows Eq. (3.21).

Figure 3.4 Schematic diagram for the controlled water tank (PC ¼ proportional

controller).

If there is a disturbance in qin while the height is at hSS, then h willchange with time until it reaches the new steady state. If we want to keep theheight at hSS, then we have to install a control system which will get thesystem back to hSS every time h changes from hSS.

In such situation, we can have a level controller with the set pointadjusted at hSS; therefore, any deviation of the measured h from hSS willcreate a signal proportional to h� hSS. This signal will actuate the valve tochange its position in order to compensate for the deviation from hSSthrough the proportional law:

C ¼ CSS þ Kðh� hSSÞ ð3:24ÞThus, when h > hSS, C increases, allowing an increase in q0 to bring h downto hSS. When h < hSS, C decreases, allowing a decrease in q0 to bring h up tohSS.

The dynamic model equation for this closed-loop controller systembecomes

qin ¼ CSS þ Kðh� hSSÞ½ �hþ Adh

dtð3:25Þ

and the response of this closed-loop system can be investigated using Eq.(3.25) in order to find the value of K that gives the best system response toexternal disturbances.

After this simple discussion, we can emphasize that, in general,‘‘steady-state models are used for design and optimization, whereas dynamicmodels are used for start-up, shutdown, and process control.’’

3.3 GENERIC AND CUSTOMIZED MODELS

Generic models is a term usually used to describe models which are notdeveloped for a specific unit but are simple models which have the qualita-tive behavior of the unit rather than the exact quantitative behavior. It givestrends for the behavior of the unit and is therefore not verified againstspecific units. They are usually simple and formed from a collection ofsemiempirical relations that are supposed to give the qualitative behaviorof the unit. These models are quite useful in many cases, especially fortraining purposes. However, it is our experience that in some situations,these generic models do not even give the correct qualitative behavior ofthe unit.

In this section, we will always refer to heterogeneous catalytic chemicalreactors as our industrial example. This is because many other importantchemical engineering processes can be treated as special cases of thesecomplex units, as will be shown later. The mathematical modeling ofhomogeneous reactors is a special case of heterogeneous catalytic reactors

when the diffusional resistances are neglected; the modeling of a nonreactingsystem is a special case when the rates of reactions are set equal to zero.

For catalytic reactors (which represent some of the most complex che-mical engineering units), generic models are less widespread than in separa-tion and heat transfer processes. This is because catalytic systems have theirown specific characteristics that depend on the intrinsic kinetics, the diffu-sional processes, as well as the reactor’s configuration and thermal character-istics. The term generic in the field of chemical reactors is usually and looselyapplied to reactors models based on thermodynamic equilibrium (Gibbs free-energy reactor models) or at their highest level of ‘‘sophistication’’, they referto plug flow and perfectly mixed homogeneous steady-state models. Thesemodels hardly represent actual catalytic reactors. Thermodynamic equili-brium models do not represent even homogeneous reactors because theyneglect the kinetics of reactions altogether, whereas pseudohomogeneousplug flow and perfectly mixed models, although they do not neglect thekinetics of the reaction, they actually completely neglect diffusional pro-cesses. In heterogeneous reactors, the concept of the effectiveness factor Zis extremely important. It expresses the effect of diffusional resistance by onenumber. The pseudohomogeneous models, which neglect diffusional resis-tances, assume that the effectiveness factors Z’s are equal to unity (or anyother constant number). Taking into consideration the fact that diffusionalprocesses expressed in terms of effectiveness factors can be very different fromunity (or constancy) and depend on intrinsic kinetics, operating variables,particle size, and so forth, one can easily realize that such models cannotseriously be a representation of industrial catalytic reactors. For example, Zfor steam reforming of natural gas is in the range 10�2�10�3(and variesstrongly along the length of the reactor), whereas for some highly exothermicreactions, Z may reach values of several hundreds and sometimes severalthousands. These simple and clear facts show very clearly that generic modelsare actually not suitable for serious consideration when we are dealing withindustrial catalytic reactors as well as other industrial chemical engineeringprocesses involving diffusional limitations (e.g., adsorption column).

On the other hand, customized models for catalytic reactors include allof the main processes taking place inside the catalytic reactors. The mostimportant of these processes for catalytic reactors are those associated withthe catalyst pellets, namely intrinsic kinetics (which includes chemisorptionand surface reaction), intraparticle diffusion of mass and heat, external massand heat transfer resistances between the catalyst surface and the bulk of thefluid, as well as all the heat production and heat consumption accompanyingthe catalytic reaction.

For fluidized-bed reactors, the diffusional processes associated withthe catalyst pellets are not very important because the particles are quite

small; however, the intrinsic kinetics is still of prime importance, as well asthe hydrodynamics of the fluidization process, especially those associatedwith the bubbles and the mass and heat transfer between the bubble anddense phases, as will be shown in Chapter 6.

Such customized models for industrial catalytic reactors (or any otherchemical engineering units) should be verified against the performance ofindustrial units. After some verification work and identification of the basicphysicochemical parameters of the industrial units, the model should be ableto predict the behavior of the unit without adjustable parameters. Theintroduction of too many adjustable parameters reduces the predictivepower of the model.

Such kinetic-diffusion highly rigorous models should not be verifiedagainst inefficiently designed units operating very close to thermodynamicequilibrium for in this case, the verification will not be critical enoughbecause many models without the required degree of precision will predictcorrect results. Units operating near thermodynamic equilibrium are notgood for testing models based on rate processes and which are built forthe correct prediction of the behavior of these units away from thermo-dynamic equilibrium.

3.3.1 Practical Uses of Different Types of Models

Mathematical models are very useful and efficient tools for the design, opti-mization, and control of different chemical engineering units. Steady-statemodels are suitable for design and optimization, whereas unsteady-statemodels are necessary for dynamic analysis, start-up, shutdown, and control.

Rigorous and reliable models are excellent tools during the designstage and also during the entire operational life period of units and plants.When correctly developed, these models can eliminate the stage of pilotplant and semicommercial units and, therefore, achieve a great economicalsaving. When a pilot plant is still needed, rigorous mathematical modelshelp to exploit the pilot plant to the maximum level and will also makethe pilot plant stage shorter, less expensive, and more fruitful.

We will still use heterogeneous catalytic reactors as our illustrationexample, for the reasons described earlier (i.e., mathematical models ofmost other important chemical engineering units will be shown to be specialcases of the heterogeneous catalytic reactors). Our descriptive discussion herewill include steady states as well as dynamic models and their applications.

3.3.2 Steady-State Models

Models for industrial catalytic reactors, like all other units, can be built withdifferent degrees of sophistication and rigor. The level of sophistication and

rigor is strongly linked to the intended use of the model. Obviously, themore sophisticated and rigorous the model, the more expensive it is to buyor develop because highly sophisticated models require a higher level ofexpertise and more time and effort for their development.

The most beneficial practical use of these models is in the design andoptimization of these industrial units. In fact, most industrial unit’s designand optimization is based on steady-state design, with the dynamic modelsdeveloped in a later stage for the design of the proper control loops in orderto keep the reactor dynamically operating near its optimum steady-statedesign in the face of external disturbances.

Models intended for design, optimization, and on-line steady-stateadjustment of variables to keep the reactor operating at its optimum con-ditions should be sophisticated, rigorous, high-fidelity models. However, ifthe models are intended for operator training or rough design calculations,then models of lower level of rigor and fidelity can be used.

For high-fidelity models, there is also an optimum degree of sophisti-cation to be used, for if one tries to describe all of the processes taking placein the unit to the highest degree of rigor, the model can become too com-plicated and too expensive. The expert developer of the model shouldemphasize the most influential processes in the model’s development andtry to formulate it to the highest degree of rigor. However, less influentialprocesses should be treated in a more approximate and less rigorous man-ner. There is always a trade-off between the accuracy of the model and itscost; a critical point exists beyond which added accuracy is too small tojustify the extra cost involved in the extra development of the model. Thiscritical point is not easily distinguished and depends to a high degree on theexperience of the model developers. We can call this side of the problem ‘‘theoptimum degree of model sophistication’’.

The practical advantages gained from the use of steady-state models indesign, optimization, and operation of catalytic reactors are tremendous. Itis estimated that about 80�85% of the success of the process depends on thesteady-state design and the remaining 15�20% depends on the successfuldynamic control of the optimum steady state. These estimates are, of course,made for a process operating smoothly with conventional control which isnot model based. However, in certain cases, inefficient dynamic control maycause temperature runaway or a complete shutdown of the process.

3.3.3 Dynamic Models

For most continuous processes operating under steady-state conditions,dynamic models are used to design appropriate control loops that minimizethe deviation of the process dynamically from the optimum steady-state

operation. This is the classical control objective whether we use analog ordigital computer control. However, a more crucial but less common objec-tive is when the process is intrinsically unstable and the control objective isto stabilize this intrinsically unstable process. This objective does not cancelthe first one, but rather adds to it, for either the stabilizing controller is alsoable to achieve the first objective or extra control loops are added to achievethat.

Dynamic models are essential for discovering intrinsic instability andfor choosing the proper control action to stabilize the unstable process.With regard to the classical control objective, although it may be achievedwithout the need of a model for the process, this, in fact, can be quitedangerous because the closed-loop dynamics of a process can be unstable,even when the process is intrinsically stable. Also, the model-based control isalmost always more efficient and robust than control not based on reliabledynamic models.

In industrial catalytic reactors with their heterogeneous and dis-tributed nature (variation of the state variables with respect to the spacecoordinates), dynamic temperature runaways may occur, especially forhighly exothermic reactions. A reliable dynamic model is one of the bestways to discover and monitor these temperature runaways, which may causeexplosions or, at the least, emergency shutdowns, which are quite expensive,especially with today’s large-capacity production lines.

Both steady-state and dynamic models are essential tools for the fullexploitation of modern digital computers used in the design, optimization,and control of petrochemical and petroleum refining plants, as well as bio-chemical systems. They allow the on-line computation of optimum operat-ing conditions and optimum actions to be taken in order to keep the plant atits maximum production capacity. These models are also essential for theoptimum performance of adaptive control, where the set points obtainedfrom steady-state design need to be changed in the case of long-termchanges in feed conditions (especially changes in feedstock composition).

3.3.4 Measures for the Reliability of Models andModel Verification

‘‘A theory is just a model of the universe, or a restricted part of it,and a set of rules that relate quantities in the model to observationsthat we make. It exists only in our minds and does not have anyother reality (whatever that might mean). A theory is a good theoryif it satisfies two requirements: it must accurately describe a largeclass of observations on the basis of a model that contains only a

few arbitrary elements, and it must make definitive predictionsabout the results of future observations.’’ (27)

The reliability of a mathematical model for any chemical engineeringprocess depends on many integrated factors which are inherent in the struc-ture of the model itself. Although the final test of the model is through itscomparison with experimental or industrial results, this, in fact, is notenough; more scientific and intellectual logical insight into the model isnecessary. Blind reliance upon comparison with experimental and industrialdata, although of great value, can sometimes be extremely misleading.Therefore, before the model is put to the test through a comparison withexperimental data, physical and chemical information should be put into themodel. The model should stand the test of logical and scientific theoreticalanalysis before it is put to the test of a comparison with experimental data.Before we proceed further, it is important to make the above statementclearer, for it may sound strange to many practitioners who put greatfaith on the comparison of the model with experimental results. To thatend, we recall that ‘‘a mathematical model of a process is a system ofequations whose solution, given specific input data, is representative ofthe response to a corresponding set of data.’’ We also recall a simple theo-retical fact: ‘‘One never proves that a theory is correct (validate a model), forthis would require an infinite number of experiments.’’ In other words, themodel must have a certain logic in order to be reliable without the need touse a large number of experimental runs for comparison.

In this sense, the situation changes drastically; it becomes the follow-ing. The model should contain, in a rational and scientifically sound basis,the important processes taking place within the system. These processes areincluded in the model through the use of scientifically sound laws andthrough the correct estimation of the parameters. Such a model should bea good representation of the real system; it should match the results of thereal system. A few experiments need to be compared with the model pre-dictions to confirm these facts and/or fine-tune the model to represent realitymore accurately.

To illustrate the above argument by a simple industrial example, con-sider the industrial steam reformer, where three reversible reactions aretaking place catalytically in the reactor. If the industrial or experimentalreactor is operating under conditions such that the exit conditions are closeto the thermodynamic equilibrium of the mixture and the comparisonbetween model predictions and industrial performance is based on exit con-ditions (which is actually the case, because it is quite difficult to measureprofiles of variables along the length of an industrial steam reformer), thenthis comparison is not a valid check for the accuracy of the model with

regard to its correct prediction of rates of reaction and mass transfer. This isbecause the results do not depend on the rates because the reactions areclose to thermodynamic equilibrium. Therefore, if a diffusion-reactionmodel which is not rigorously built is compared with a large number ofindustrial cases, all close to thermodynamic equilibrium, it does not meanthat the model is reliable, because a thermodynamic equilibrium model(which is much simpler) will also give good results. Such a model, althoughchecked successfully against a large number of industrial cases, may proveto be a failure when compared with a case operating far from thermo-dynamic equilibrium.

Another example, it is our experience that models using the simplifiedFickian diffusion for the catalyst pellet give good results for cases not veryfar from thermodynamic equilibrium. However, for cases far from thermo-dynamic equilibrium, such models fail and the more rigorous dusty gasmodel must be used for the modeling of the catalyst pellets.

Denn (20) discusses a similar case for coal gasifiers, where the kinetic-free model (thermodynamic equilibrium model) fails in certain regions ofparameters, necessitating the use of a kinetic model.

In a nutshell, the reliability of the model depends on the following:

1. The structure of the model as a combination of accurate descrip-tion of the processes taking place within the system, together withthe interaction between the processes themselves and betweenthem and the environment

2. The verification of the model against experimental and industrialunits, provided that these tests are for critical cases that actuallytest the rates of the processes in the system

Verification of the model does not mean the use of a number of adjus-table parameters. The main role of the model is prediction, and adjustableparameters reduce the predictive nature of the model; when overused, theycan reduce it to zero. It is best when the model is verified without anyadjustable parameters. When this is too difficult, one (or more) set(s) ofdata can be used to adjust some parameters; however, there should beenough data to test the model successfully against other sets of data withoutany adjustable parameters.

3.4 ECONOMIC BENEFITS OF USING HIGH-FIDELITY CUSTOMIZED MODELS

These benefits can be classified into the following two broad categories:design and operation, and control.

3.4.1 Design and Operation

Chemical reactors (as well as many other chemical/biochemical processunits) design differs fundamentally from other engineering design in thefact that ‘‘Design Safety Factors’’ which are widely and comfortably usedin other engineering designs are not applicable in the case of many chemical/biochemical engineering processes. This is due to the complexity of theseunits, in which a large number of processes are taking place. However, thecore of the matter is the fact that, for chemical reactors, reaction networksare usually formed of complex consecutive and parallel steps which give riseto selectivity and yield problems. For a large class of reactions, the depen-dence of the yield of desired product(s) on design variables is nonmono-tonic. This nonmonotonic dependence implies that the use of ‘‘Design SafetyFactors’’ can be rather catastrophic for the process.

To illustrate this point in the simplest possible fashion, consider theconsecutive reaction

A ! B ! C

taking place in an isothermal CSTR with component B being the desiredproduct. For this simple system, the yield of the desired component B, YB,will depend on the reactor volume VR or the volumetric flow rate to thereactor qR in the manner shown schematically in Figure 3.5. It is clear from

Figure 3.5 Schematic representation of YB versus VR (for certain values of qR and

other parameters), or YB versus 1�qR (for certain values of VR and other para-

meters).

Figure 3.5 that there is an optimum value of VR (or 1�qR), VRO, which gives

the maximum yield of B, YBmax. It is obvious that the celebrated engineering

‘‘Design Safety Factors’’ used in many engineering designs (includingseparation processes when the number of trays are increased over the designcalculated value by multiplying it by a safety factor or dividing by what iscalled tray efficiency) can be disastrous in the case of certain chemical/biochemical units.

Although the above simple illustration of the concept for a homo-geneous isothermal lumped system is applicable to other more complicatedsystems, the situation for catalytic reactors is much more involved becauseof the complexity of the intrinsic kinetics as well as the complex inter-action among reactions, heat release (or absorption), and mass and heatdiffusion inside the reactor. For example, many catalytic and biocatalyticreactions show nonmonotonic dependence of the rate of reaction on reac-tant concentrations. For example, the hydrogenation of benzene to cyclo-hexane over different types of nickel catalyst has an intrinsic rate ofreaction of the form

r ¼ kCACB

ð1þ KACA þ KBCBÞ2

where A is hydrogen and B is benzene. This rate equation has nonmonotonicdependence on reactant concentrations, and, therefore, in some regions ofparameters, the rate of reaction may decrease when the concentrationincreases!

Furthermore, the interaction between diffusion and reaction in porouscatalyst pellets gives rise to an arsenal of complicated behaviors. The sim-plest is the reduction of the rate of reaction due to diffusional resistances,giving rise to effectiveness factors less that 1.0. However, effectivenessfactors greater than 1.0 are also widespread in catalytic and biocatalyticsystems for nonmonotonic kinetics and/or nonisothermal pellets. In addi-tion, negative effectiveness factors (for reversible reactions) indicating thatthe reactions reverse their direction due to diffusion are also possible forimportant industrial catalytic reactors with multiple reactions [e.g., steamreforming of natural gas, partial oxidation of o-xylene to phthalic anhy-dride, methanation, etc. (28)]. Nonmonotonic change of the effectivenessfactor Z along the length of the reactor is also possible for reversible singlereactions (29, 30). Fluctuation in the value of the effectiveness factor alongthe reactor length is possible in many industrial reactors (e.g., steam reform-ing of natural gas and methanol synthesis).

From the above, it is clear that the widespread engineering designprocedure of using simple design equations coupled with a ‘‘Design SafetyFactor’’ is not generally applicable in the design of catalytic and non-

catalytic chemical and biochemical reactors. Therefore, rigorous high-fide-lity, steady-state models (design equations) should be used for the precisedesign of these units. This does not mean that it is not possible to designsuch units without such models. This false statement is contradictory tothe history of the chemical and biochemical industries because differentunits were designed and operated before chemistry and chemical engineer-ing principles were even discovered. Units can be designed and operatedusing empirical techniques and accumulation of practical experience andeven using trial and error and sequential scaling up. However, such designswill have a very high cost and will not be the most efficient and safedesigns. With today’s large continuous production lines, design cost, pro-ductivity, and safety are of the highest concern.

The first step in the design of any unit is the sizing of the unit for agiven configuration and operating conditions. The high-fidelity mathemati-cal models allow optimizing the design not only with respect to operatingand design parameters but also with regard to the configuration of the unititself. Design and optimization on this basis can have a very strong impacton the process. To take a very well-known example from an importantchemical industry, consider an ammonia production line of 1200 MTPDcapacity. For this line, an improvement of 1% in the production line willmean about 1.5–2 million $/year in extra revenues. Optimization of anexisting plant using high-fidelity models can lead to up to 10% improvementin the ammonia production (equivalent to 15–20 million $/year in addedrevenues). The opportunities for improvement are much higher when thehigh-fidelity models and optimization algorithms are used in the designstage.

Despite the great importance of efficient control of processes andproduction lines, it is now well established in industry that the overallvalue of the project is not dominated by controller’s action but by the abilityto predict the best operating point in terms of process variables. For exam-ple, 0.03–0.05 $/barrel might be achieved by holding an FCC plant undertight control near a desired point or constraint. Complicated control loopswill help to achieve smaller excursions and faster responses to set-pointchanges and load disturbances by moving the operation to the desiredpoint faster and with less variation in product yields. However, 0.3–0.5$/barrel might be realized by moving the steady state of the plant from asteady state which was not properly optimized to new and near-optimalsteady-state operating conditions.

The above brief discussion gives an elementary idea about the eco-nomic benefits of using rigorous high-fidelity, steady-state models in thedesign and operation of industrial catalytic reactors. The same principleapplies to other industrial chemical and biochemical units also.

The follow-up of the operation of the plant using such modelsallows an instant and deep insight into what is going on inside theunits. This allows smooth operation and early warning of troubles thatmay cause an expensive shutdown or an even more expensive and catas-trophic accident.

3.4.2 Control

Steady-state models are invaluable in what we may call steady-state control.The term steady-state control refers to situations in which the operator is nothighly concerned about the dynamic behavior of the system, but is mostlyconcerned about operating the unit at its optimum steady state in the face oflong-term external disturbances. This is a situation in which one of theoperating parameters changes (usually feedstock composition), and thesteady-state control question is: What are the input variables that need tobe changed in order to keep the unit at the same steady state and producingthe desired yield and production rates? Both off-line and on-line use of high-fidelity mathematical models can be utilized here. Of course, the on-lineoption is more efficient and reliable, however, it has a higher cost withregard to fixed investment (computer, measurement, and manipulationhardware).

On the other hand, dynamic control is when it is desired to avoidthe losses associated with dynamic excursions of the process from steady-state optimum operation resulting from transient disturbances and thecorresponding process responses.

Economic incentives for advanced control (which is usually modelbased) can be considered to fall into three categories:

1. The major benefit results from moving the steady-state operationto a better operating point, which has been discussed earlier andwhich mainly depends on high-fidelity, steady-state models.

2. Improved control allows operation closer to any limiting con-straint, thus resulting in additional benefits. This relies uponsteady-state as well as dynamic models.

3. Smaller excursions and faster responses to setpoint changes andload disturbances move the operation back to the desired steadystate faster and with less variation in the yield and productivity(and, of course, the product quality). This relies mainly onsteady-state and dynamic models.

Of course, all of the above is for intrinsically stable processes. For intrinsi-cally unstable processes, stabilization of the unstable process must precedeall of the above-discussed points. We should note that in some cases,

unstable operation may give higher yield and productivity than stable states.However, this is well beyond the scope of this undergraduate book.

Dynamic models are also essential for the smooth and fast start-up ofthe process and for shutdown procedures. These two crucial stages in theplant operation are intrinsically unsteady state and therefore dynamicmodels are the ones applicable to these stages. In the start-up and shutdownphases, the controller settings are very different from the settings at steadystate operation because the objectives are completely different.

It was estimated that the close control over an ammonia productionline based on objective 3 given above, can achieve a 2–3% improvement inammonia productivity. For an ammonia plant with a capacity of 1200MTPD, this amounts to about 2.4–6.0 million $/year.

The use of rigorous, high-fidelity, steady-state mathematical models inthe design stage allows integration between steady-state design and controlconsiderations. This is due to the fact that changes in the process designcould influence profoundly the process dynamics. Thus, for the optimaloverall design of the process, control configurations should be consideredduring the design of the process itself. This crucial objective is best achievedusing rigorous, high-fidelity, steady-state and dynamic models.

3.5 INCORPORATION OF RIGOROUS MODELSINTO FLOWSHEET SIMULATORS ANDPUTTING MATHEMATICAL MODELS INTOUSER-FRIENDLY SOFTWARE PACKAGES

Commercial steady-state flowsheet simulators are usually computerpackages equipped with an extensive physical properties database and mod-ular steady-state mass and heat balance calculations facilities. These simu-lators usually also contain modules for the process design of multistageoperations such as distillation, absorption, extraction, and so forth. Mostof these simulators are based on equilibrium stage calculations coupled withempirical formulas for calculation of stage efficiencies. Very few packagescontain design procedures based on rates of mass and heat transfer.Therefore, most of these packages are not completely suitable for the designof continuous contact separation processes such as packed and spray col-umns. However, most of them contain some highly empirical correlationsfor the rough design of such equipment.

With regard to chemical reactors, the most sophisticated simulatorsusually contain modules for idealized plug flow and perfectly mixed contin-uous-stirred tank reactors. The less sophisticated simulators usually contain

only procedures for reactor design based on thermodynamic equilibrium(Gibbs free energy).

However, with regard to catalytic reactors, these simulators usuallycontain nothing. This state of affairs is due to the fact that each catalyticreactor has its own characteristics. The development of catalytic reactormodels needs a deep understanding of catalysis, kinetics, kinetic modeling,mass and heat transfer processes, reactor modeling, and advanced numericaltechniques. The development of models for these catalytic reactors needs amodeler with long experience in a number of fields related to catalyticprocesses and it is also a time-consuming process.

High-fidelity models for industrial catalytic reactors can be incorpo-rated into these flowsheet simulators via a number of techniques dependingon the characteristics of the flowsheet simulator. In some steady-state flow-sheet simulators, there is a built-in facility for the user to add his ownmodule. In such cases, the high-fidelity reactor model can be added asone of these special modules. The reactor model in this case can make fulluse of the physical properties database available in the simulator. However,in most flowsheet simulators, this facility is not available; in such cases, themodule can be used with the flowsheet simulator on the basis that the outputvariables from the last unit preceding the catalytic reactor is used as an inputfile to the catalytic reactor model, and when the calculations for the reactorare complete, the output file from the reactor model is used as an input fileto the unit next to the reactor in the process flowsheet. Clearly, this lasttechnique is less efficient and takes more effort to implement than in the firstcase.

In many cases, the catalytic reactor model is used as a stand-alone unitin the design, simulation, and optimization of catalytic reactors. There aresome typical cases in the petrochemical industry where the catalytic reactorsdominate the production lines (e.g., the ammonia production line usuallycontains about six catalytic reactors representing almost 90% of the produc-tion line). In these cases, an evaluation has to be made in order to decidewhether the reactor modules should be added to the flowsheet simulatoror the few noncatalytic processes should be borrowed from the flowsheetsimulator and added to a specially formulated flowsheet simulator ofcatalytic reactors forming the production line.

With regard to dynamic simulation, very few simulators exist and theyusually give the trends of the dynamic behavior rather than the precisesimulation of the dynamics. High-fidelity dynamic models for catalytic reac-tors are quite rare and they are usually developed through special orders andagreements. The incorporation of these models into commercial dynamicsimulators follows basically the same general principles outlined for thesteady-state models.

Putting Mathematical Models into User-Friendly SoftwarePackages

Great efforts are expended in the development and verification of rigorous,high-fidelity mathematical models for any chemical and biochemical engi-neering unit(s). Therefore, it is quite important to maximize the benefitsfrom these models and to facilitate their use to a wider spectrum of users.This aim is achieved through putting the models into user-friendly softwarepackages with advanced graphic capabilities. This facilitates the use of themodel by many people in the plant or design office without much back-ground in mathematical modeling, numerical analysis, or computer pro-gramming. Also, the results are usually presented in both graphical andtabulated forms. The user can have a summary screen with all of the mostimportant input–output variables or the user can zoom to any set of vari-ables and get more details about them. The user can also obtain a hardcopyof the results independently or use an on-line insertion of the tabulated orgraphical results into a report. In addition to these facilities, the user canfollow the progress of the solution through interesting visual means.

Of course, it is beyond the scope of this undergraduate book to aim atteaching and training the reader on how to put the models into these soft-ware packages. However, we can give a general outline for the procedures ofdeveloping such computer packages. An overall introduction for undergrad-uate chemical and biochemical engineers can be summarized in the follow-ing very general and brief points:

I. What is a software package?

A software package is a collection of modules interlinked toachieve certain purpose(s).

II. Components of software packages

1. Input section for acquiring necessary information2. Database for retrieving basic data necessary for computations3. Calculation section for creating new data from input, a data-

base, and a mathematical model (a numerical algorithm is, ofcourse, developed and incorporated into the software pack-age for the accurate solution of the model equations)

4. Output section for presenting computed informationIII. Basic decisions for software design

1. Language selection: Fortran, C, C++, and so forth.2. Operating system selection: DOS, Unix, VMS, windows, and

so forthIV. Design steps for software packages

A. Preliminary steps:

1. Select (or develop) the proper mathematical model withthe appropriate state variables

2. Identify input parameters3. Identify calculation procedures4. Identify output variables (the state variables at the exit of

the unit)5. Test run with I/O from files

B. Input and Output:1. Design input screen layout2. Test run readings from screen3. Select output variables4. Design output selection screen5. Decide output variable format

C. Alert the user:1. Select indicative intermediate variables2. Decide on intermediate displays3. Test displays of intermediate results4. Add as many variables and indicators as necessary to

keep the user alertD. Error trapping (input):

1. Include limits on input parameters2. Check input parameters for type and value3. Check other related values

Warning: Take nothing for granted, you have to check everythingand test run it.

E. Error trapping (calculations):1. Check for overflow and underflow2. Give user a proper warning when there is a problem in

calculations3. Advise user to possible action4. Always make an option to quit gracefully

F. Error trapping (files):1. Check for existence2. Never write over a file3. Never erase a file4. Avoid end of file trap

G. Fine-tuning:1. Avoid redundant inputs2. Keep a consistent use of keys3. Keep messages in one area4. Use short indicative messages

V. Ultimate package

1. General in nature2. Based on sound mathematical model and an efficient

solution algorithm3. Easy to use by nonspecialists4. User friendly with error trapping5. Flexibility in data entry6. Provides intermediate, summary, and detailed results7. Industrially verified against industrial units

3.6 FROM MATERIAL AND ENERGY BALANCESTO STEADY-STATE DESIGN EQUATIONS(STEADY-STATE MATHEMATICAL MODELS)

Let us first review the most general material and heat balance equations andall of the special cases which can be easily obtained from these equations.This will be followed by the basic idea of how to transform these materialand energy balance equations into design equations, first for lumped systemsand then followed by the same for distributed systems. We will use homo-geneous chemical reactors with multiple inputs, multiple outputs, andmultiple reactions. It will be shown in Chapter 6 how to apply the sameprinciples to heterogeneous system and how other rates (e.g., rates of masstransfer) can systematically replace (or is added to) the rates of reactions.

This approach will put the mathematical modeling of all chemical/biochemical engineering systems into one unified and very easy-to-use fra-mework. We will also extend this unified framework to dynamic models.

3.6.1 Generalized Mass Balance Equation (Fig. 3.6)

This is the most general one-phase (homogeneous) chemical/biochemicalengineering lumped system. It has L input streams nif1; nif2 ; . . .;

�nifl ; . . .; nifL) and K output streams ni1 ; ni2 ; . . . ; nik ; . . . ; niK Þ

�with N reactions

r1; r2; . . . ;ð rj; . . . ; rNÞ and the number of species is M i ¼ 1; 2; . . . ;Mð Þ.The generalized material balance equation is

XKk¼1

nik ¼XLl¼1

nifl þXNj¼1

�ijrj; i ¼ 1; 2; . . . ;M ð3:26Þ

These are M equations having N reactions rj; j ¼ 1; 2; . . . ;N� �

and �ij isthe stoichiometric number of component i in reaction j. The generalizedrate of reaction for reaction j is given by rj. Note that rj is the overallrate of reaction for the whole unit (it is not per unit volume, or per unit

mass of catalyst, and so forth, and that is why our equations are massbalance equations, not design equations, as shown later).

This general system can also be represented as three systems: a mixer, asingle input–single output reactor, and a splitter, as shown in Figure 3.7.

Equation (3.26) can be put in an equivalent form consisting of threesets of equations:

ni ¼XLl¼1

nifl and ni ¼ nif for i ¼ 1; 2; . . . ;M ð3:27Þ

ni ¼ ni þXNj¼1

�ijrj and ni ¼ nif for i ¼ 1; 2; . . . ;M ð3:28Þ

XKk¼1

nik ¼ nif ¼ ni for i ¼ 1; 2; . . . ;M ð3:29Þ

Figure 3.6 Mass flow diagram.

Figure 3.7 Another representation of the system.

Note also that Eq. (3.26) is the general form for all kinds of special cases:

1. For a single reaction N ¼ 1ð Þ, the equation becomes

XKk¼1

nik ¼XLl¼1

nifl þ �ir for i ¼ 1; 2; . . . ;M

2. For a single input L ¼ 1ð Þ, the equation becomes

XKk¼1

nik ¼ nif þXNj¼1

�ijrj for i ¼ 1; 2; . . . ;M

3. For a single output K ¼ 1ð Þ, the equation becomes

ni ¼XLl¼1

nifl þXNj¼1

�ijrj for i ¼ 1; 2; . . . ;M

4. For no reactions N ¼ 0ð Þ, the equation becomes

XKk¼1

nik ¼XLl¼1

nifl for i ¼ 1; 2; . . . ;M

5. For N ¼ 0 and L ¼ 1, we have thus the equation for a singleinput splitter:

XKk¼1

nik ¼ nif

6. For N ¼ 0 and K ¼ 1, we have the equation of a mixer:

ni ¼XLl¼1

nifl

and so on.Now, can we change this general mass balance equation to a design

equation? Very simply, we use r0 instead of r, where r0 is the rate of reactionper unit volume of the reactor, and the equation becomes

XKk¼1

nik ¼XLl¼1

nifl þXNj¼1

�ijr0jVR

Now, this is not a mass balance equation; it is actually a design equationwhich can be used for the design of a lumped system (e.g., an isothermalCSTR). Let us illustrate this for a very simple case of a first-order irrever-sible liquid-phase reaction, where

A ! B

and the rate of reaction is given by r0 ¼ kCA, where k is the reaction rateconstant (in s�1) and CA is the concentration of component A (in gmol=LÞ.Later in this book, we will show how the same principles can be applied todistributed system and also for other rates like the rate of mass transfer forheterogeneous systems (Chapter 6).

Let us consider the single–input, single–output case, thus,

ni ¼ nif þ �ir0VR

which, when applied to component A, gives

nA ¼ nAf � kCAVR ð3:30ÞWe have a simple problem; that is, the existence of CA and nA in the sameequation. For liquid-phase systems when there is no change in volume, thisproblem is trivial (for a gas-phase system with a change in number of molesaccompanying the reaction, although it is simple, it is not as trivial, as willbe shown later).

For this case of the liquid-phase system, we can easily write

nA ¼ qCA and nAf ¼ qCAf

Thus, Eq. (3) becomes

qCA ¼ qCAf � kCAVR ð3:31ÞThus, for a specific volumetric flow rate q and the rate of reaction constantk, we can obtain the volume of the reactor that is needed to achieve a certainconversion.

Rearranging Eq. (3.31) gives

VR ¼ qCAf

qþ kð ÞCA

Thus, for example, if we want 90% conversion, then CA ¼ 0:1CAf and VR

can be obtained as

VR ¼ q

0:1 qþ kð ÞThis is how simply the mass balance equation has been turned into a designequation (a mathematical model for the lumped isothermal system).

The same very simple principles apply to the heat balance equationsfor a nonisothermal system and also for distributed systems as shown in thefollowing section. It also applies to the heterogeneous system, as shown inChapter 6.

Now, we will apply these simple principles to batch reactors, CSTRs,and tubular reactors (distributed systems), starting with isothermal systemsfollowed by nonisothermal systems.

3.6.2 Isothermal Reactors (Temperature Is Constant)

We will illustrate the design problem using the consecutive reactionsnetwork problem for CSTR, batch and tubular reactors. We consider thefollowing reaction network:

3A �!k1 2B �!k2 5C

The Rates of Reaction

. This network is a three-component, two-reaction network. Thetwo reactions are obviously independent.

. Accordingly, the system can be described in terms of two statevariables, for example, nA and nB or xA and YB. [nA and nB arethe molar flow rates (or number of moles of A and B respectively,and xA ¼ ðnAf � nAÞ=nAf and YB ¼ ðnB � nBf Þ=nAf are the conver-sion of A and yield of B, respectively. nAf and nBf are the molarflow rates at the entrance (feed point) of the reactor for thecontinuous-flow reactors (CSTR and tubular) or the initial numberof moles in a batch reactor].

The rates of reaction can be given for the production or consumption of thecomponents or they can be given for the reactions that lead to the general-ized rates of reaction studied earlier.

The network can be written as

3A �!k1 2B ð3:32Þ2B �!k2 5C ð3:33Þ

Define all rates for components as production rates:

RA ¼ Rate of production of A ¼ �K1CA

moles of A converted

L s

RB ¼ Rate of production of B ¼ 23K1CA � K2CB

moles of B produced

L s

RC ¼ Rate of production of C ¼ 52K2CB

moles of C produced

L s

In a more detailed fashion, we write

RA1 ¼ Rate of production of A in reaction ð3:32Þ¼ �K1CA

moles of A converted

L s

RB1 ¼ Rate of production of B in reaction ð3:32Þ

¼ �B1�A1

ð�K1CAÞ ¼2

�3

� �ð�K1CAÞ ¼

2

3K1CA

� �moles of B produced

L s

RB2 ¼ Rate of production of B in reaction ð3:33Þ¼ �K2CB

moles of B produced

L s

RC2 ¼ Rate of production of C in reaction ð3:33Þ

¼ �C2�B2

ð�K2CBÞ ¼5

�2

� �ð�K2CBÞ ¼

5

2K2CB

� �moles of C produced

L s

As explained earlier, it is much more convenient to define the rates for thereaction rather than the components:

3A �!k1 2B r01 is the generalized rate of this first reaction

2B �!k2 5C r02 is the generalized rate of this second reaction

The generalized rates of reactions r01 and r02 (moles per unit volume per unittime) can be defined as explained in the following very simple steps.

The generalized rate r0j for any reaction j is defined as the rate ofproduction of any component in reaction j divided by the stoichiometricnumber of the said component; therefore,

r01 ¼RA1

�A1¼ �K1CA

�3¼ K1

3CA ¼ k1CA

gmol

L s

where

k1 ¼K1

3

or (giving exactly the same result),

r01 ¼RB1

�B1¼ 2=3K1CA

2¼ K1

3CA ¼ k1CA

gmol

L s

where

k1 ¼K1

3

and

r02 ¼RB2

�B2¼ �K2CB

�2¼ K2

2CB ¼ k2CB

gmol

L s

where

k2 ¼K2

2

or (giving exactly the same result)

r02 ¼RC2

�C2

¼ 5=2K2CB

5¼ K2

2CB ¼ k2CB

gmol

L s

where

k2 ¼k22

Note: Notice that we called the generalized rates used above r01 and r02 asper unit volume of the reactor, whereas earlier we called them r1 and r2,as the overall rates. In other words, r01 and r02 are in moles per unitvolume per unit time, whereas r1 and r2 were in moles per unit time forthe whole unit. As explained in the previous few pages, this is the onlyreal difference between the mass balance (inventory) equations and thedesign equations; for example,

1. For a CSTR,

rj ¼ VCSTRr0j

2. For a tubular reactor,

rj ¼ �Vtr0j

where �Vt is a difference element of the volume of the tubularreactor.

3. For a batch reactor,

rj ¼ V�tr0j

where �t is a difference element of time.

With this simple and obvious background material, we can tackle thedesign problem for different configurations of isothermal reactors.

The general mass balance equation for the reacting system with singleinput–single output (SISO) and N multiple reactions is given by

ni ¼ nif þXNj¼1

�ijrj i ¼ 1; 2; . . . ;M

where M is the number of components, N is the number of reactions,rj ¼ mol/time ¼ Vr0j and r0j ¼ mol/(volume)(time)

I. The Continuous-Stirred Tank Reactor (CSTR)

The design equation for component A is

nA ¼ nAf þ �A1Vr01 þ �A2Vr

02

It can be rearranged in the form

nA ¼ nAf � 3Vk1CA ð3:34ÞFor component B,

nB ¼ nBf þ �B1Vr01 þ �B2Vr

02

On rearrangement, we get

nB ¼ nBf þ 2Vk1CA � 2Vk2CB ð3:35ÞFor component C,

nC ¼ nCf þ �C1Vr01 þ �C2Vr

02

On rearrangement,

nC ¼ nCf þ 5Vk2CB ð3:36ÞFrom Eqs. (3.34)–(3.36), by addition we get

nA3

þ nB2þ nC

5¼ nAf

3þ nBf

2þ nCf

5ð3:37Þ

Equation (3.37) is essentially the same overall mass balance relationobtained for any kind of reactor:

XMi¼1

ni�i�� ¼XM

i¼1

nif

�i��

It is obvious that we need only Eqs (3.34) and (3.35) to solve the system,because nC can be obtained from Eq. (3.37) when nA and nB are computed.Design equations (3.34) and (3.35) can be formulated in term of nA and nBor in terms of CA and CB or in terms of xA (conversion of A) and YB (yieldof B) as discussed earlier.

We will illustrate this for the more difficult gas-phase case (isothermaland constant pressure). Formulation of the design equations in terms of nAand nB can be done as follows:

CA ¼ nAq

and CB ¼ nBq

where q is the volumetric flow rate.Also, we know that

Pq ¼ nTRT

ðfor nonideal gases we have to use the compressibility factor ZÞ:where nT is the total molar flow rate. Thus, the relation between thevolumetric flow rate and nT (the total molar flow rate), T (temperature),and P (pressure) will be

q ¼ nTRT

P


q ¼ anT

where

nT ¼ nA þ nB þ nC and a ¼ RT

P

From Eq. (3.37),

nC ¼ 5

Pnif

�i�� nA

3� nB

2

!¼ f1ðnA; nBÞ

and

q ¼ a nA þ nB þ f1ðnA; nBÞ½ � ¼ f2ðnA; nBÞwhere f1 and f2 are functions of nA and nB only. Thus,

CA ¼ nAq

¼ nAf ðnA; nBÞ

¼ gAðnA; nBÞ ð3:38Þ

CB ¼ nBq

¼ nBf ðnA; nBÞ

¼ gBðnA; nBÞ ð3:39Þ

where gA and gB are functions of nA and nB only.Using Eqs. (3.38) and (3.39) in Eqs. (3.34) and (3.35), we get

nA ¼ nAf � 3Vk1gAðnA; nBÞ ð3:40Þ

nB ¼ nBf þ 2Vk1gAðnA; nBÞ � 2Vk2gBðnA; nBÞ ð3:41Þ

Equations (3.40) and (3.41) are two equations in two variables nA and nB;they can be solved to obtain nA and nB. When nA and nB are computed, wecan easily compute q (is a function of nA and nB), CA, CB, xA, and YB, where

xA ¼ nAf � nA

nAfand YB ¼ nB � nBf

nAf

and CA and CB in terms of nA, nB, and q as shown above.

Exercise

Formulate the design equations for the above-discussed CSTR in terms ofCA and CB and in terms of xA and YB.

II. Batch Reactors

This is a system which is ‘‘distributed’’ with respect to time.For the batch reactor with variable volume (constant pressure), the

gas-phase reaction is

3A �!k1 2B

2B �!k2 5C

The design equation for component A (carried out over an element of time,�t) is

nA tþ�tð Þ ¼ nA tð Þ þ �A1Vr01�t

This difference form can be written as

nA þ�nA ¼ nA þ �A1Vr01�t

which, after rearrangement and taking the limit as �t ! 0, gives

dnAdt

¼ �A1Vr01 ¼ �A1Vk1CA ¼ �3Vk1CA ð3:42Þ

Similarly,

dnBdt

¼ �B1Vr01 þ �B2Vr

02 ¼ Vð�B1r01 þ �B2r

02Þ

Thus,

dnBdt

¼ V 2k1CA � 2k2CB½ � ¼ 2V k1CA � k2CBð Þ ð3:43Þ

Similarly,

dnCdt

¼ �C1Vr02 ¼ 5Vk2CB ð3:44Þ

Here, we will demonstrate some of the possible simple manipulations thathelp to reduce the dimensionality of many problems.

We can easily obtain a relation among nA, nB, and nC, and thereforeeliminate one of the three differential equations. Write Eqs. (3.42)–(3.44) asfollows:

1

3

dnAdt

¼ �Vk1CA ð3:45Þ

1

2

dnBdt

¼ Vðk1CA � k2CBÞ ð3:46Þ

1

5

dnCdt

¼ Vk2CB ð3:47Þ

By addition of equations (3.45)–(3.47), we get

d

dt

nA3

þ nB2þ nC

5

� �¼ 0

Thus, by integration, we get

nA3

þ nB2þ nC

5¼ C1

where C1 is a constant and is evaluated from the initial conditions, at t ¼ 0,nA ¼ nAf , nB ¼ nBf , and nC ¼ nCf . Therefore,

nA3

þ nB2þ nC

5¼ nAf

3þ nBf

2þ nCf

5ð3:48Þ

Equation (3.48) can be rewritten in the generalized form

XMi¼1

ni�i�� ¼XM

i¼1

nif

�i�� ð3:49Þ

where M is the number of components. Therefore, it is clear that Eq. (3.48)can be used to compute nC without the need to solve the differentialequation (3.47).

Now, we are left with Eqs. (3.45) and (3.46), which have the statevariables nA, nB, CA, and CB; however, it is very easy to note that (for thecase of constant reactor volume)

CjV ¼ nj ð3:50ÞTherefore, Eqs. (3.45) and (3.46) can be rewritten as

dnAdt

¼ �3k1nA ð3:51Þ

and

dnBdt

¼ 2ðk1nA � k2nBÞ ð3:52Þ

with the initial conditions, at t ¼ 0, nA ¼ nAf and nB ¼ nBf . The two equa-tions are linear and can be solved analytically to obtain the change of nA andnB with time as follows. Note that Eq. (3.51) does not include nB and canthus be solved independently of Eq. (3.52) as follows:

dnAnA

¼ �3k1dt

Integration gives

ln nA ¼ �3k1tþ C1

Because, at t ¼ 0, nA ¼ nAf , we have,

ln nAf ¼ C1

and

lnnAnAf

� �¼ �3k1t

Thus,

nA ¼ nAf e�3k1t ð3:53Þ

Substituting the value of nA from Eq. (3.53) into Eq. (3.52) gives

dnBdt

¼ 2k1nAf e�3k1t � 2k2nB

which can be arranged in the following form in order to use the integrationfactor method to solve it:

dnBdt

þ 2k2nB ¼ 2k1nAf e�3k1t

Multiplying by the integration factor (e2k2t), we get

e2k2tdnBdt

þ 2e2k2tk2nB ¼ 2k1nAf e�3k1te2k2t

We can put this in the following form:

d nBe2k2t

� �dt

¼ 2k1nAf eð2k2�3k1Þt

By integration we obtain

e2k2tnB ¼ 2k1nAf eð2k2�3k1Þt

2k2 � 3k1þ C1 ð3:54Þ

At t ¼ 0, nB ¼ nBf ; thus we get:

nBf ¼2k1nAf

2k2 � 3k1þ C1

Therefore,

C1 ¼ nBf �2k1nAf

2k2 � 3k1ð3:55Þ

Substituting the value of C1 from Eq. (3.55) into Eq. (3.54) gives

e2k2tnB ¼ 2k1nAf eð2k2�3k1Þt

2k2 � 3k1þ nBf �

2k1nAf

2k2 � 3k1

which can be rearranged as

e2k2tnB ¼2k1nAf eð2k2�3k1Þt � 1

� �2k2 � 3k1

þ nBf

Thus,

nB ¼2k1nAf e�3k1t � e�2k2t

� �2k2 � 3k1

þ nBf e�2k2t ð3:56Þ

Equations (3.53) and (3.56) give the change of nA and nB with time; if nC isrequired, it can be easily computed using Eq. (3.48), as the values of nA andnB are known from Eqs. (3.53) and (3.56).

If the volume change is also required, then from the following relationfor ideal gases

PV ¼ nTRT ðfor nonideal gases we have use compressibility factor ZÞ

we get

V ¼ nA þ nB þ nCð ÞRTP

¼ a nA þ nB þ nCð Þ

Thus,

V ¼ a

nAf e

�3k1t þ2k1nAf e�3k1t � e�2k2t

� �2k2 � 3k1

þ nBf e�2k2t

þXMi¼1

nif

�i�� nAf e

�3k1t

3� 1

2

2k1nAf e�3k1t � e�2k2t� �2k2 � 3k1

� 1

2nBf e

�2k2t

!

which simplifies to give

V ¼ aXMi¼1

nif

�i�� þ 2

3nAf e

�3k1t þk1nAf e�3k1t � e�2k2t

� �2k2 � 3k1

þ 1

2nBf e

�2k2t

0@

1Að3:57Þ

If the conversion and yield are required, then we obtain the following:

Conversion of A,

xA ¼ nAf � nA

nAf

Using Eq. (3.53), we get

xA ¼ 1� nAnAf

� �) 1� e�3k1t

Yield of B,

YB ¼ nB � nBf

nAf

Using Eq. (3.56), we get

YB ¼ nBnAf

� nBf

nAf

� �¼

2k1 e�3k1t � e�2k2t� �2k2 � 3k1

þ nBf

nAf

� �e�2k2t � nBf

nAf

� �

This can be further simplified to get

YB ¼2k1 e�3k1t � e�2k2t

� �2k2 � 3k1

þ nBf

nAf

� �e�2k2t � 1� �

From this relation, we can obtain the optimum time for obtaining themaximum yield of B ðYBmax

Þ as depicted in Figure 3.8:

dYB

dt¼ 2k1

2k2 � 3k1�3k1e

�3k1t þ 2k2e�2k2t

� �þ nBf

nAf

� ��2k2e

�2k2t� �

Now, we obtain the optimum time topt� �

by equating dYB=dt to zero:

0 ¼ 2k12k2 � 3k1

�3k1e�3k1t þ 2k2e

�2k2t� �

þ nBf

nAf

� ��2k2e

�2k2t� �

Solving the above equation to obtain topt, gives YBmax.

For the special case of nBf ¼ 0, we get

2k2e�2k2topt ¼ 3k1e

�3k1topt

This gives

e 3k1�2k2ð Þtopt ¼ 3k12k2

Thus,

topt ¼ln 3k1=2k2ð Þ3k1 � 2k2

Thus, for such a reaction and for nBf ¼ 0, the optimum time topt� �

forobtaining the maximum yield of B ðYBmax

Þ, which is only a function of k1and k2 as shown above.

III. Tubular Reactors

For the homogeneous tubular reactor (the simplest model is the plug flow),the design equations are again obtained from the mass balance equations

Figure 3.8 Maximum yield.

after turning them into design equations as shown earlier for the batchreactor by taking the balance over an element of length or volume insteadof time. Note that in the batch case, ni is the number of moles of componenti, whereas, here, for the tubular reactor, ni is the molar flow rate of compo-nent i. Thus the balance over an element �l will be

nA l þ�lð Þ ¼ nA lð Þ þ �AAt�lr0

Again considering the same reaction network,

3A �!k1 2B �!k2 5C

Rearranging and taking the limit as �l ! 0 gives the following differentialequation for the change of the molar flow rate of component A along thelength of the tubular reactor:

dnAdl

¼ �3Atk1CA ð3:58Þ

The above equation can also be written in terms of dV :

dnAdV

¼ �3k1CA

where

dV ¼ At dl

and At is the cross-sectional area of the tubular reactor.With the initial condition at

l ¼ 0 (or V ¼ 0Þ;nA ¼ nAf ðor CA ¼ CAf Þ

Important notes: Note that in the batch reactor, nA is the number ofmoles of component A, whereas, here, in the continuous reactor, nA is themolar flow rate of A (mol/time). Also in the batch case, CA is the concen-tration of the component A inside the batch reactor ðnA=V ¼ number ofmoles in reactor divided by the volume of reactor). In the present contin-uous case, this in not the case; in fact, CA ¼ nA=q is molar flow rate ofcomponent A divided by the total volumetric flow rate. These facts willslightly complicate the formulation of the design equation as will beshown in this section.

In this continuous-flow system (gas flow system), the change in num-ber of moles causes a change in the volumetric flow rate (q) and no change inpressure. The balance on component B gives

dnBdl

¼ Atð2k1CA � 2k2CBÞ ð3:59Þ

Similarly,

dnBdV

¼ ð2k1CA � 2k2CBÞ

For component C,

dnCdl

¼ 5Atk2CB ð3:60Þ

From Eqs. (3.58)–(3.60), it is clear that the same relation among nA, nB, andnC obtained earlier for the CSTR and batch reactors can be obtained:

nA3

þ nB2þ nC

5¼ nAf

3þ nBf

2þ nCf

5ð3:61Þ

Therefore, only Eqs. (3.58) and (3.59) need to be solved [because nC can becomputed from Eq. (3.61) when nA and nB have been computed). However,these equations contain the expression for components A and B in twodifferent forms; we have nA and nB as well as CA and CB. We can handlethis situation by writing everything in terms of nA and nB, or CA and CB, orxA and YB, as discussed earlier.

Formulation in terms of nA and nB yields

CA ¼ nAq

and CB ¼ nBq

Also,

Pq ¼ nTRT

q ¼ nTRT

P

Therefore,

q ¼ anT

where

nT ¼ nA þ nB þ nC

From Eq. (3.61),

nC ¼ 5

Pnif

�i�� nA

3� nB

2

!

Therefore, we can write

nT ¼ nA þ nB þ 5

Pnif

�i�� nA

3� nB

2

!

which gives

nT ¼ 5

Pnif

�i��

!� 2

3nA � 3

2nB

This can be rewritten simply as

nT ¼ 5nAf

3þ 5

nBf

2þ nCf �

2

3nA � 3

2nB

On reorganizing, we have

nT ¼ nCf þ 135nAf � 2nA� �þ 1

25nBf � 3nB� �

Thus, nT can now be written in terms of nA and nB:

nT ¼ f1 nA; nBð ÞThus,

q ¼ af1 nA; nBð ÞSo we get

CA ¼ nAaf1ðnA; nBÞ

¼ gAðnA; nBÞ ð3:62Þ

and

CB ¼ nBaf1ðnA; nBÞ

¼ gBðnA; nBÞ ð3:63Þ

Substituting Eqs. (3.62) and (3.63) into Eqs. (3.58) and (3.59) gives

dnAdl

¼ �3Atk1gAðnA; nBÞ ð3:64Þ

dnBdl

¼ At 2k1gAðnA; nBÞ � 2k2gBðnA; nBÞ½ � ð3:65Þ

Equations (3.64) and (3.65) are two differential equations that can be solvedsimultaneously using one of the standard subroutines and the initial condi-tions at l ¼ 0 (or V ¼ 0), nA ¼ nAf and nB ¼ nBf . Of course, by computingnA and nB at any position along the length of the reactor, all other variablescan be computed, including xA, YB, CA, CB, q, and so forth.

Formulation in terms of CA and CB (instead of in terms of nA and nB) is asfollows: For this case, we write Eq. (3.58) as

d qCAð Þdl

¼ �3Atk1CA

Thus,

qdCA

dlþ CA

dq

dl¼ �3Atk1CA ð3:66Þ

From the formulation in terms of nA and nB we have

q ¼ a nA þ nB þ nCð ÞThus,

dq

dl¼ a

dnAdl

þ dnBdl

þ dnCdl

� �ð3:67Þ

Substituting Eqs. (3.58)–(3.60) into Eq. (3.67), we get

dq

dl¼ aAtð�3k1CA þ 2k1CA � 2k2CB þ 5k2CBÞ

which reduces to

dq

dl¼ aAt �k1CA þ 3k2CB½ � ð3:68Þ

Substituting the value of dq=dl from Eq. (3.68) into Eq. (3.66), we get

qdCA

dlþ CAaAt �k1CA þ 3k2CBð Þ ¼ �3Atk1CA


dCA

dl¼ At

q�aCAð�k1CA þ 3k2CBÞ � 3k1CA½ � ð3:69Þ

Equation (3.59) can also be rearranged in term of CB and q, and the result-ing equation is solved simultaneously with Eqs. (3.68) and (3.69).

It is clear that the formulation in term of nA and nB is more straight-forward than the formulation in terms of CA and CB.

Exercise

Formulate the design equations in terms of xA and YB, as defined earlier.

The Simple Special Case of No Change in Number of Moles (or Liquid

Phase) Systems

The simple procedure given here is applicable to gas phase systems with nochange in the number of moles and also to liquid phase in general, where thechange of number of moles accompanying the reaction has negligible effecton volumes and volumetric flow rates.

Consider the reaction

A �!k1 B �!k2 C

This is a much simpler case and the design equations can be formulatedeasily in terms of CA and CB (or any other form: nA and nB or xA and YB).

CSTR

The balance for component A (design equation) is given by

nA ¼ nAf � Vk1CA

In this case,

nA ¼ qCA

where q is a constant. So we can also write

nAf ¼ qCAf

Thus, the above equation can be written as

qCA ¼ qCAf � Vk1CA

Similarly,

qCB ¼ qCBf þ Vk1CA � Vk2CB

qCC ¼ qCCf þ Vk2CB

and

CA þ CB þ CC ¼ CAf þ CBf þ CCf

Clearly, the resulting design equations are much simpler than the gas-phasecase, which is accompanied by a change in the number of moles.

Batch Reactor

The balance for component A is

dnAdt

¼ �Vk1CA

where

nA ¼ VCA and V ¼ constant

Thus,

d VCAð Þdt

¼ �Vk1CA

Because V is constant, we can write,

VdCA

dt¼ �Vk1CA

which gives

dCA

dt¼ �k1CA

Similarly,

dCB

dt¼ k1CA � k2CB

and

dCC

dt¼ k2CB

with initial conditions at t ¼ 0, CA ¼ CAf , CB ¼ CBf , and CC ¼ CCf . Also,


Again, the design equations are much simpler than the gas-phase case, which isaccompanied by change in the number of moles.

Plug Flow (Tubular)

The balance for component A is

dnAdl

¼ �Atk1CA

Here,

nA ¼ qCA and q ¼ constant

Therefore,

qdCA

dl¼ �Atk1CA

Similarly,

qdCB

dl¼ At k1CA � k2CBð Þ

and,

qdCC

dl¼ Atk2CB

With initial conditions at l ¼ 0 ðor V ¼ 0Þ, CA ¼ CAf , CB ¼ CBf , andCC ¼ CCf and


As seen for the previous two cases, these design equations are much simplerthan the gas-phase system accompanied by a change in the number of moles.

3.6.3 Nonisothermal Reactors

The generalized heat balance equation for L inputs, K outputs, M com-ponents, and N reactions is the following single equation:

XLl¼1

XMi¼1

nifl ðHifl�HirÞ þQ ¼

XKk¼1

XMi¼1

nik Hik �Hirð Þ þXNj¼1

rj�Hj

Similar to what we have shown earlier for the mass balance, the aboveequation is the very general heat balance equation for homogeneoussystems. All other cases are special cases of this generalized form:

1. For no reactions, we just put all rj’s equal to zero.2. For adiabatic operation, we simply put Q ¼ 0.3. For single-input, single-output, and single reaction, the general

equation becomes

XMi¼1

nif Hif �Hir

� �þQ ¼XMi¼1

ni Hi �Hirð Þ þ r�H

and so on.

Now, we start with some specific examples combining the mass and heatbalance equations. In order to turn this heat balance equation into a part ofthe design equations, we simply replace rj with Vr0j, as we did earlier for massbalance.

I. Nonisothermal CSTR

Mass Balance

For a case of single-input, single-output, and N reactions, the mass balanceequations for any component is given by

ni ¼ nif þXNj¼1

�ijrj ð3:70Þ

where i ¼ 1; 2; 3; . . . ;M, M is the number of components, N is the numberof reactions, and rj is the generalized overall rate of reaction for reaction j[in mol/time (rate of reaction for the entire unit)].

As we have discussed earlier, in order to turn the mass balance [Eq.(3.70)] into a design equation, we write it as

ni ¼ nif þ VXNj¼1

�ijr0j ð3:71Þ

Here, r0j is the generalized rate of reaction for reaction j [in mol/(time �volume)] (it is equal to rj for a unit volume of reactor). For a single reaction,the design equation simply becomes

ni ¼ nif þ V�ir0 ð3:72Þ

Heat Balance

The heat balance for a single-input, single-output and N reactions can bewritten as

XMi¼1

nif Hif �Hir

� �þQ ¼XMi¼1

ni Hi �Hirð Þ þXNj¼1

ð�HjÞrrj ð3:73Þ

where ð�HjÞr is the heat of reaction for reaction j and Q is the heat added tothe system.

Derivation of the Heat Balance Equation (3.73) for a Single Reaction

To remind the reader of the derivation of this equation, which is quiteimportant, we derive it again here. The heat balance equation is as follows:

XMi¼1

nif Hif þQ ¼XMi¼1

niHi ð3:74Þ

However, now we have to put it in the form of enthalpy difference, this canbe done as follows:

1. SubtractPM

i¼1 nif Hir from the left-hand side (therefore, subtract itfrom the right-hand side so as not to change the equation).

2. SubtractPM

i¼1 niHir from the right-hand side (therefore, subtract

it from the left-hand side so as not to change the equation).

Thus, we get

XMi¼1

nifHif �XMi¼1

nif Hir �XMi¼1

niHir þQ ¼XMi¼1

niHi �XMi¼1

nif Hir �XMi¼1

niHir

ð3:75ÞRearranging gives

XMi¼1

nif Hif �Hir

� �þQ ¼XMi¼1

ni Hi �Hirð Þ þXMi¼1

ni � nif� �

Hir ð3:76Þ

However, from the mass balance for a single reaction, we have

ni ¼ nif þ �ir ð3:77ÞUsing Eq. (3.77) in Eq. (3.76), we get

XMi¼1

nif Hif �Hir

� �þQ ¼XMi¼1

ni Hi �Hirð Þ þXMi¼1

�irHir ð3:78Þ

From the definition of the heat of reaction, we have

XMi¼1

�iHir ¼ �Hr

Therefore, Eq. (3.78) becomes

XMi¼1

nif Hif �Hir

� �þQ ¼XMi¼1

ni Hi �Hirð Þ þ r �Hrð Þ ð3:79Þ

where �Hr is the heat of reaction at reference conditions r for the reactionas stoichiometrically written.

What do we mean by ‘‘the heat of reaction for the reaction as stoichio-metrically written’’? Suppose we have

3Aþ 5B ! 6C þ 8D

Because we define the heat of reaction as

�Hr ¼ 8HD þ 6HC � 3HA � 5HB

which is

�Hr ¼XMi¼1

�iHi

this �Hr is not per mole of A, B, C, or D, but it is actually per 3 mol of Areacted, 5 mol of B reacted, 6 mol of C produced, or 8 mol of D produced. If

we want it per mole of any component, we should divide by �i�� , where i is

that particular component for which we want to make the heat of reactionper mole of it.

In most cases, �Hr is computed from the heat of formation or heats ofcombustion as follows:

�Hr ¼X

�i�Hfi

or

�Hr ¼ �X

�i�Hci

These heats of reaction are the correct ones to be used in Eq. (46), becausethey are for the reaction as stoichiometrically written.

However, if �Hr is obtained experimentally and is defined as �Hr,which is in calories per mole of A, then the correct �Hr to be used in Eq.(3.79) is �Hr �A

�� ; in general,

�Hr ¼ ð�Hri �i�� Þ

For reactor design purposes, of course r should be written as Vr0 (forCSTR), or �Vr0 (for tubular), or V�tr0(for batch). Thus, Eq. (3.79) for aCSTR should be written as

XMi¼1

nif Hif �Hir

� �þQ ¼XMi¼1

ni Hi �Hirð Þ þ Vr0 �Hrð Þ

For multiple reactions, it will be

XMi¼1

nif ðHif �HirÞ þQ ¼XMi¼1

niðHi �HirÞ þ VXNj¼1

r0jð�HrjÞ ð3:80Þ

whereM is total number of components involved (i.e., reactants + products+ inerts) and N is the number of reactions.

For multiple inputs and multiple outputs, we have

XLl¼1

XMi¼1

nifl Hifl�Hir

� �þQ ¼XKk¼1

XMi¼1

nik Hik �Hirð Þ þ VXNj¼1

r 0j�Hj

ð3:81ÞBecause our entire book prior to heterogeneous systems deals with

homogeneous systems, in this part of the book no change of phase isinvolved and therefore the change of enthalpies are changes in sensibleheats only. Therefore, Eq. (3.81) can be rewritten as

XMi¼1

nif

ðTf

Tr

Cpi dT þQ ¼XMi¼1

ni

ðTTr

Cpi dT þ VXNj¼1

r0jð�HrjÞ ð3:82Þ

Equations (3.71) and (3.82) are the basic design equations for the design ofnonisothermal CSTR. For tubular and batch reactors, the same equationsare slightly modified to put them in a suitable differential equation form.

II. Nonisothermal Tubular Reactors

This case, of course, belongs to the distributed system type, which will becovered in more detail in Chapter 4. For this distributed case, Eq. (3.71)becomes

niðlþ�lÞ ¼ niðlÞ þ At�lXNj¼1

�ijr0j


ni þ�ni ¼ ni þ At�lXNj¼1

�ijr0j

After rearrangement and taking the limit as �l ! 0, we get the differentialequation

dnidl

¼ At

XNj¼1

�ijr0j ð3:83Þ

For a single reaction, the above equation becomes

dnidl

¼ At�ir0 ð3:84Þ

To get the heat balance design equation, an enthalpy balance over an ele-ment �l for a single reaction gives

XMi¼1

ðniHiÞl þQ0�l ¼XMi¼1

ðniHiÞlþ�l

where M is the total number of components involved (reactants + products+ inerts), including components not in the feed but created during thereaction (they will have nif ¼ 0Þ and Q0 is the rate of heat removed oradded per unit length of reactor. Rearranging gives

XMi¼1

�niHi

�l�Q0 ¼ 0

Taking the limit as �l ! 0, we obtain the differential equation

XMi¼1

d niHið Þdl

�Q0 ¼ 0

which gives

XMi¼1

nidHi

dlþHi

dnidl

� ��Q0 ¼ 0 ð3:85Þ


XMi¼1

nidHi

dlþHiAt�ir

0� �

�Q0 ¼ 0

Rearrangement gives

XMi¼1

nidHi

dlþ Atr

0XMi¼1

Hi�i

!�Q0 ¼ 0

Because

XMi¼1

�iHi ¼ ð�HÞ � heat of reaction

we get

XMi¼1

nidHi

dlþ Atr

0ð�HÞ �Q0 ¼ 0

Finally the above equation can be written as

XMi¼1

nidHi

dl¼ Atr

0ð��HÞ þQ0 ð3:86Þ

For multiple reactions, it will be

XMi¼1

nidHi

dl¼

XNj¼1

Atr0jð��HjÞ

!þQ0 ð3:87Þ

The design equations for the multiple reactions case are Eqs. (3.83) and(3.87), whereas for the single-reaction case, the design equations are Eqs.(3.84) and (3.86).

III. Nonisothermal Batch Reactor

For this case, mass balance design equations are obtained as follows(balance over a time element �t):

ni tþ�tð Þ ¼ ni tð Þ þ V�tXNj¼1

�ijr0j

Rearranging and taking the limit as �t ! 0 gives

dnidt

¼ VXNj¼1

�ijr0j ð3:88Þ

For single reaction, it will be

dnidt

¼ V�ir0 ð3:89Þ

The heat balance (for a single reaction) is

XMi¼1

ðniHiÞt þQ0�t ¼XMi¼1

ðniHiÞtþ�t

After rearranging and taking the limit as �t ! 0, we obtain the followingdifferential equation:

XMi¼1

d niHið Þdt

�Q0 ¼ 0

After differentiation, we obtain

XMi¼1

nidHi

dtþHi

dnidt

� �" #�Q0 ¼ 0 ð3:90Þ


XMi¼1

nidHi

dtþHiV�ir

0� �" #

�Q0 ¼ 0

Rearranging gives

XMi¼1

nidHi

dtþ Vr0

XMi¼1

Hi�i

!�Q0 ¼ 0

BecauseX�iHi ¼ ð�HÞ � heat of reaction

we obtain

XMi¼1

nidHi

dtþ Vr0ð�HÞ �Q0 ¼ 0

The above equation can be rearranged and written as

XMi¼1

nidHi

dt¼ Vr0ð��HÞ þQ0 ð3:91Þ

For multiple reactions,

XMi¼1

nidHi

dt¼ V

XNj¼1

r0jð��HjÞ !

þQ0 ð3:92Þ

Equations (3.88) and (3.92) are the design equations for the multiple-reactions case and equations (3.89) and (3.91) are the design equations forthe single-reaction case.

3.7 SIMPLE EXAMPLES FOR THE GENERALEQUATIONS

In this section, we show the design (model) equations for the three types ofreactors. We consider only the one-phase (homogeneous) case.

I. Nonisothermal CSTR

Consider the simplest case of a single reaction with no change in the numberof moles:

A �!k B; with r0 ¼ kCA

Equation (3.72) becomes

nA ¼ nAf þ Vð�1ÞkCA ð3:93ÞBecause there is no change in the number of moles, q (the volumetric flowrate) is constant; therefore,

nA ¼ qCA and nAf ¼ qCAf

Thus, Eq. (3.93) becomes

qCA ¼ qCAf � VkCA

and k depends on the temperature according to the Arrhenius relation

k ¼ k0e�E=RTð Þ

The mass balance design equation is thus

qCA ¼ qCAf � Vk0e�E=RTð ÞCA ð3:94Þ

Equation (3.94) has two variables, CA and T ; therefore, we must use theheat balance equations also.

We assume that there is no change in phase and use Eq. (3.82) andafter putting N ¼ 1 (single reaction), we get

XMi¼1

nif

ðTf

Tr

Cpi dT

!þQ ¼

XMi¼1

ni

ðTTr

Cpi dT

!þ VkCAð�HrÞ

We further assume adiabatic operation Q ¼ 0ð Þ and also assume that all ofthe Cpi’s are constant (or using constant average values) to obtain

XMi¼1

nif CpiðTf � TrÞ !

¼XMi¼1

niCpiðT � TrÞ !

þ VkCAð�HrÞ

We further assume an average Cpmix, which is the same for feed and

products, to obtain

CpmixðTf � TrÞ

XMi¼1

nif

!¼ Cpmix

ðT � TrÞXMi¼1

ni

!þ VkCAð�HrÞ

We further assume that nTf ¼ nT (actually, there is no change in the numberof moles for this simple reaction; therefore, for this specific, case this is aphysical fact not an assumption) and therefore the equation reduces to

nTC0pTf ¼ nTC

0pT þ VkCA �Hrð Þ

C 0p ¼ Cpmix

, where C 0p is the molar specific heat and nT is the molar flow rate.

We can also write it in terms of specific heat Cp(per unit mass as):

qrCpTf ¼ qrCpT þ Vk0e�E=RTð ÞCA �Hrð Þ ð3:95Þ

where q is the volumetric flow rate and r is the average constant density ofthe mixture. Equations (3.94) and (3.95) are the design equations for thisnonisothermal case; we can put them in a dimensionless form as follows:

XA ¼ XAf � a0e �g=Yð ÞXA ð3:96Þwhere

XA ¼ CA

CAref

; XAf ¼CAf

CAref

; a0 ¼ Vk0q

; Y ¼ T

Tref

and g ¼ E

RTref

Similarly, the heat balance equation can be written in a dimensionless formas

Y � Yf ¼ a0e� g=Yð ÞXAb ð3:97Þwhere

b ¼ ð��HrÞCref

rCpTref

When analyzed, eqs. (3.96) and (3.97) show the complex behavior of thissystem and will be discussed later.

If we take

CAref¼ CAf and Tref ¼ Tf

Then the dimensionless heat balance equation becomes

Y � 1 ¼ a0e� g=Yð ÞbXA ð3:98Þand the dimensionless mass balance equation becomes

XA � 1ð Þ ¼ �a0e� g=Yð ÞXA ð3:99ÞThe two nonlinear algebraic equations [Eqs. (3.98) and (3.99)] have twovariables (XA and Y) and can be solved simultaneously for given valuesof a0, b, and g.

Note: The nine physical parameters (also design and operating para-meters) are q; r;C0

P;Tf ;V; k0;E;CAf ; and��Hr. They all are lumped intothree parameters: a0, b, and g

Can we get some insight into the characteristics of this system and itsdependence on the parameters before we solve its design equations (3.98)and (3.99)? Let us use some chemical engineering intelligence together withEqs. (3.98) and (3.99):

Y � 1ð Þ ¼ a0e� g=Yð Þ bXA ð3:98Þ

XA � 1ð Þ ¼ �a0e� g=Yð ÞXA ð3:99ÞMultiplying Eq. (3.99) by b and adding it to Eq. (3.98) gives

Y � 1þ b XA � 1ð Þ ¼ 0


Y ¼ 1þ b 1� XAð Þ ð3:100Þ

What is the maximum possible value of Y? It is when XA ! 0 (i.e., at100% conversion):

Ymax ¼ 1þ b

For the exothermic reaction, b is called as the maximum dimensionlessadiabatic temperature rise.

Can we reduce the two nonlinear coupled equations to one equation inone variable and the other is a simple explicit linear equation? Yes, we can dothat through the two alternative equivalent route shown in Table 3.1. Fromeither of the two routes, we get one equation in terms of Y (one variable); wecan solve it and get the value of Y . Then, the value of xA can be obtainedeasily from Eq. (3.101) or (3.102).

Take, for example,

1

a0Y � 1ð Þ ¼ e� g=Yð Þð1þ b� YÞ

Table 3.1 Alternate Routes to Solve Eqs. (3.98) and (3.99)

Route 1 Route 2

Equation (3.99) can be rearranged as

XA þ a0e�ðg=YÞXA ¼ 1


XA ¼ 1

1þ a0e�ð�=YÞÞ ð3:101Þ

Substitute Eq. (3.101) into Eq. (3.98)

to eliminate XA and obtain

Y � 1 ¼ a0e�ð�=YÞb1þ a0e�ð�=YÞ

We can rearrange it in the following form:

1

a0ðY � 1Þ|fflfflfflfflfflffl{zfflfflfflfflfflffl}RðYÞ

¼ be�ð�=YÞ

1þ a0e� ð�=YÞ|fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl}GðYÞ

which gives

RðYÞ ¼ GðYÞwhere RðYÞ is a heat removal function and

GðYÞ is a heat generation function.

Equation (3.100) can be rearranged

to give

XA ¼ 1þ b� Y

bð3:102Þ

Use Eq. (3.102) in Eq. (3.98) to

eliminate XA and obtain

Y � 1 ¼ a0e�ð�=YÞbð1þ b� YÞ

bwhich can be written as

1

a0ðY � 1Þ|fflfflfflfflfflffl{zfflfflfflfflfflffl}RðYÞ

¼ e�ð�=YÞð1þ b� YÞ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}GðYÞ

where RðYÞ is a heat removal

function and GðYÞ is a heat

generation function

It can be solved numerically (using Newton–Raphson, bisectional methods,etc.; see Appendix B), but for the sake of illustration, we solve it graphically(as shown in Fig. 3.9).

1

a0Y � 1ð Þ|fflfflfflfflfflffl{zfflfflfflfflfflffl}R Yð Þ

¼ e� g=Yð Þð1þ b� YÞ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}G Yð Þ

where, G Yð Þ is the heat generation function and R Yð Þ is the heat removalfunction.

It is clear from Figure 3.9 that for an exothermic reaction, there is apossibility of three steady states (stationery nonequilibrium states), whereasfor an endothermic reaction, there is only one steady state. For moreinformation regarding this very important phenomenon (multiplicityphenomenon or bifurcation behavior), see Chapter 7.

II. Nonisothermal Tubular Reactor

Consider the simple reaction

A �!k B

Figure 3.9 Plot of heat generation and heat removal functions (points of inter-

section are the steady states).

where r0 ¼ kCA. The mass balance design equation [Eq. (3.84)] for com-ponent A is

dnAdl

¼ �AtkCA

Because q is constant, nA ¼ qCA and dnA=dl ¼ qdCA=dl; thus,

qdCA

dl¼ �Atk0e

� E=RTð ÞCA ð3:103Þ

The heat balance design equation is

XMi¼1

nidHi

dl¼ Atk0e

� E=RTð ÞCAð��HÞ þQ

For adiabatic operation, Q ¼ 0, and for no change in phase, it becomes

XMi¼1

nid CpiT� �dl

¼ Atk0e� E=RTð ÞCAð��HÞ

For constant Cpi, we can write

XMi¼1

niCpi

� � dTdl

¼ Atk0e� E=RTð ÞCAð��HÞ

Using constant molar Cpmixgives,

nTCpmix

dT

dl¼ Atk0e

� E=RTð ÞCAð��HÞ

If we use Cpmixper unit mass instead of per unit mole and call it Cp, we can

write

qrCp

dT

dl¼ Atk0e

� E=RTð ÞCAð��HÞ ð3:104Þ

Equations (3.103) and (3.104) are the design equations for this case. Theycan be put in dimensionless form as we did for the CSTR. On the samebasis, the reader can develop these equation for the batch reactor case.

Note: �H here is not at reference condition as in the CSTR case, it canbe assumed constant or its variation with temperature can be taken intoconsideration.

Questions

1. Derive the simplified design equations for the batch nonisothermalreactor.

2. Write the tubular reactor and the batch reactor equations indimensionless form.

3. Compare your dimensionless groups for the batch and the tubularreactors with those for the CSTR and comment.

3.8 MODELING OF BIOCHEMICAL SYSTEMS

Biochemical systems can be classified into two main categories: The firstdeals with enzyme systems and the second deals with processes catalyzed bywhole-cell micro-organisms.

3.8.1 Modeling of Enzyme Systems

Biochemical systems can be handled in a very similar manner as was done sofar for chemical systems in the previous chapters and sections. The fewsimple differences will be clarified in this subsection through simple andillustrative examples. Also, some of the most important terminology willbe introduced.

Microbial systems. These are systems in which there is a biochemicalreaction which is catalyzed by micro-organisms.

Micro-organisms. They contain large number of enzymes and theyexist in a tissue structure. They not only catalyze the biochemicalreaction, but they themselves grow also.

Micro-organism operation (Fig. 3.10):1. Micro-organisms behave as a catalyst when fed with the substrate

(reactant).2. They grow and accumulate.3. They release product(s)4. They are responsive to surroundings (pH, temperature, etc.).

Figure 3.10 Micro-organism operation.

Microbial reactions

Substrates+Micro-organisms ! Products+More micro-organisms

The process is autocatalytic, because the micro-organism is not onlycatalyzing the reactions, but it also grows and reproduces during thereaction.

The parameter used to describe the fraction of substrate (reactant)converted to micro-organism is called the ‘‘yield factor’’ YSð Þ:

YS ¼ gram of micro-organism produced

gram of substrate consumed

Characteristics of microbial systems

1. Micro-organisms grow and accumulate.2. The reaction network is quite complex.3. Historical experience from many processes such as fermentation

are important for process development.

Enzyme Systems

There are more than 2000 different kinds of known enzyme (e.g., trypsin,pepsin, etc.).

Example

Glucose��!Glucose isomeraseFructose

Enzymes also act as catalysts, but they differ from micro-organisms, asshown in the few simple points given below.

Enzymes

1. They do not grow like microorganisms.2. Very selective in nature.3. Soluble in some solvent to form a solution.

Micro-organisms, in contradiction, are characterized by the following:

1. They grow and accumulate in the presence of substrate andnutrients.

2. They contain a large number of enzymes in their network tissues.3. They are not soluble in solution, they remain suspended.

Both enzymes and micro-organisms are expensive and they have to be pro-tected from washing out with the product. Thus, enzyme and micro-organismimmobilization is used to prevent their washout.

Immobilization

There are two main types of immobilization techniques:

1. Attachment to solid surface: This technique is usually used formicroorganisms through attaching the micro-organism onto asolid surface either by physical or chemical bonding in order toprevent its mobility. The advantage of this technique is that thereis very little or no mass transfer resistance. The disadvantage ofthis method is that if micro-organisms are used in a reactor wherethe flow rate is high, then the shear stress will remove the micro-organisms from the solid’s surface.

2. Microencapsulation: This technique is suitable for both enzymesand micro-organisms where they are entrapped in capsules madeof materials which are permeable only to substrates and products,but are impermeable to enzyme or micro-organism. The advan-tage of this technique is that it keeps the enzymes or micro-organ-isms inside the capsules even under high-flow-rate turbulentconditions. The main disadvantage of this method is that it intro-duces additional mass transfer resistances.

Enzyme Kinetics

A simple enzyme kinetic can be expresssed as,

S �!E P

where S is the substrate and P is the product. Concentration and pH are themost important variables affecting the rates of enzyme reactions, and that iswhy they are the main state variables for enzyme systems.

Michaelis-Menten Kinetics

This is the simplest type of enzyme reaction giving rise to monotonic enzymekinetics. The mechanism can be simply expressed by the following steps:

E þ S,k1k�1

ES ð3:105Þ

where E is an enzyme, S is a substrate, and ES is a enzyme–substratecomplex. The equilibrium constant for this reversible reaction is KS,which is given by

KS ¼ E½ � S½ �ES½ � ¼ k�1

k1

The enzyme–substrate complex gives the product P and reproduces theenzyme E according to the reaction

ES �!K Pþ E ð3:106ÞThe rate of product production through this irreversible reaction is given by

r ¼ K ES½ � ð3:107ÞWe cannot measure the intermediate concentration of ES ES½ �ð Þ , so we tryto describe the rate of reaction in terms of the substrate concentration S½ �.E reacts with S and at steady state; thus, we have the relation

K ES½ � ¼ k1 S½ � E½ � � k�1 ES½ � ð3:108Þwhere E½ � is the amount of enzyme available for reaction at any time. This isequal to the difference between the total amount of enzyme and the amountof enzyme used for the formation of ES. Mathematically, it can be written as

E½ � ¼ ET½ � � ES½ �where ET½ � is the total amount of enzyme. Substituting this relation of E½ � inEq. (3.108) gives

K ES½ � ¼ k1 S½ � ET½ � � ES½ �� k�1 ES½ �which can be rearranged to give

ES½ � K þ k1 S½ � þ k�1

� � ¼ k1 ET½ � S½ �Thus, we can obtain ES½ � as a function of ET½ � and S½ � as follows:

ES½ � ¼ k1 ET½ � S½ �K þ k1 S½ � þ k�1

ð3:109Þ

Equation (3.109) can be manipulated into the following form:

ES½ � ¼ k1k1

� �ET½ � S½ �

�K

k1þ k�1

k1þ k1k1

S½ ��

Because

k�1

k1¼ KS

we can rewrite the equation as

ES½ � ¼ ET½ � S½ �ðK=k1Þ þ KS þ S½ � ð3:110Þ

From Eq. (3.107), the reaction rate is given as

r ¼ K ES½ �

Substituting the expression for ES½ � from Eq. (3.110) into Eq. (3.106) gives

r ¼ K ET½ � S½ �ðK=k1Þ þ KS þ S½ � ð3:111Þ

Let

K ET½ � ¼ Vmax


r ¼ Vmax S½ �ðK=k1Þ þ KS þ S½ � ð3:112Þ

If k1 is large, then the term K=k1 is negligible, but KS ¼ k�1=k1ð Þ is notnegligible because it is a division of two large numbers. If we assume thatK=k1 ¼ 0, it means that the reaction given by Eq. (3.105) is very fast ascompared to the reaction given by Eq. (3.107). Equation (3.112) becomes

r ¼ Vmax S½ �KS þ S½ � ð3:113Þ

Equation (3.113) is called the Michaelis–Menten equation, and the follow-ing can be deduced from the equation:

1. If S½ � is very small as compared with KS, a first-order reaction rater is dominating, which can be described graphically by a straightline with slope Vmax=KS and the rate is r ¼ Vmax S½ �=KS.

2. If KS is small as compared to S½ � (at high concentration), then thereaction rate r is a horizontal line having one value, r ¼ Vmax, andthat will be described graphically by a straight line parallel to the xaxis. The Michaelis–Menten kinetics is graphically shown inFigure 3.11.

Important Notes

1. The actual differential equations which describe the reaction andthe relation between S and ES are

d S½ �dt

¼ �k1 S½ � E½ � þ k�1 ES½ � ð3:114Þ

and

d ES½ �dt

¼ k1 S½ � E½ � � k�1 ES½ � � K ES½ � ð3:115Þ

where

E½ � ¼ ET½ � � ES½ �

2. If d S½ �=dt and d ES½ �=dt in Eqs. (3.114) and (3.115) are set equal tozero, then Eq. (3.112) is satisfied and that means that the dynamicsof the system is fast (equilibrium).

3. When using Eq. (3.113) to develop unsteady-state models, weshould notice that there are many assumptions, including thatthe operation is quasi-steady-state. Also, we assume that ES pro-duced from reaction (3.105) is the same as that reacted in reaction(3.107). Thus, Michaelis–Menten Eq. (3.113) is not rigorousenough for unsteady-state models.

Substrate-Inhibited Enzyme Kinetics

For noncompetitive substrate inhibition, the overall reaction is given by

S ! P ð3:116ÞThe formation of the active enzyme complex can be shown as

S þ E,ES ð3:117ÞThe equilibrium constant for reaction (3.117) can be written as

KS ¼ S½ � E½ �ES½ � ¼ k�1

k1

The formation of inactive enzyme complex follows the pathway

ES þ S,ES2 ð3:118Þ

Figure 3.11 Graphical representation of Michaelis–Menten kinetics.

The equilibrium constant for reaction (3.118) is

K 0S ¼ ES½ � S½ �

ES2½ �The active enzyme complex produces the product and enzyme again accord-ing to the pathway

ES �!K Pþ E

The overall rate can be written as

r ¼ K ES½ � ¼ K S½ � E½ �KS

Also,

ET½ � ¼ E½ � þ ES½ � þ ES2½ �By using the equilibrium relations of reactions (3.116)–(3.118) in the defini-tion of ET½ �, we get

ET½ � ¼ E½ � þ S½ � E½ �KS

þ ES½ � S½ �K 0

S

By taking E½ � as a common part, where ES½ � ¼ S½ � E½ �=KS, we get

ET½ � ¼ E½ � 1þ S½ �KS

þ S½ �2KSK

0S

!

from which we obtain

E½ � ¼ ET½ ��

1þ S½ �KS

þ S½ �2KSK

0S

!

From the definition of the reaction rate,

r ¼ K S½ � E½ �KS

We substitute the relation for E½ � to get

r ¼ K S½ �KS

� �ET½ �

�1þ S½ �

KS

þ S½ �2KSK

0S

!¼ K S½ � ET½ �

KS þ S½ � þ S½ �2=K 0S

!

or we can write

r ¼ Vmax S½ �KS þ S½ � þ S½ �2=K 0

S

ð3:119Þ

The difference between Eqs. (3.113) (Michaelis–Menten) and (3.119) is thatin Eq. (3.119) when S½ �2 is large, it will dominate the equation leading to anonmonotonic behavior.

We can see from the graph on the right-hand side of Figure 3.12 theexistence of a negative slope region (after the dashed line), where the rate ofreaction decreases with the increase in concentration of substrate (inhibi-tion). The negative slope region means that as the substrate concentrationincreases, the reaction rate decreases r ¼ K= S½ �ð Þ.

Equation (3.119) represents what we call nonmonotonic kinetics. Thishas a very important implication on the behavior of the bioreactors whensuch bioreactions are conducted.

Explanation of non-monotonic kinetics (Fig. 3.13)

Positive reaction order (region I). Itmeans that the rate of ES½ �producedfrom reaction (3.117) is higher than that consumed in reaction (3.118).

Zero reaction order (region II). It means that the rate of ES½ � producedfrom reaction (3.117) is equal to that consumed by reaction (3.118).

Negative reaction order (region III). It means that the rate of ES½ �produced from reaction (3.117) is less than that consumed by reac-tion (3.118).

Competitive Substrate Inhibition

The above substrate inhibition mechanism is called consecutive substrateinhibition. Another mechanism is the competitive substrate inhibition asshown in this subsection.

Figure 3.12 Plots showing the difference between the behavior of rates expressed

by Eqs. (3.113) and (3.119).

E þ S,ES Equilibrium constant ¼ KS ð3:120Þ2S þ E,ES2 Equilibrium constant ¼ K 0

S ð3:121ÞES �!K Pþ E Rate of reaction, r ¼ K ES½ �

We know that

KS ¼ E½ � S½ �ES½ � and K 0

S ¼ S½ �2 E½ �ES2½ �

The reaction rate can be written as

r ¼ KS½ � E½ �KS

ð3:122Þ

Now the total enzyme balance gives

ET½ � ¼ E½ � þ ES½ � þ ES2½ �By substituting the relations for ES½ � and ES2½ � in terms of KS and K 0

S, weget

ET½ � ¼ E½ � þ E½ � S½ �KS

þ E½ � S½ �2K 0

S

which can be rearranged to give

ET½ � ¼ E½ � 1þ S½ �KS

þ S½ �2K 0

S

!

We can obtain E½ � as

Figure 3.13 Different kinetics regions.

E½ � ¼ ET½ ��

1þ S½ �KS

þ S½ �2K 0

S

!

By substituting this expression of E½ � in the rate equation (3.123), we get

r ¼ K S½ �KS

ET½ �1þ S½ �=KS þ S½ �2=K 0

S

!

or we can write

r ¼ Vmax S½ ��

KS þ S½ � þ KS S½ �2K 0

S

" #ð3:124Þ

We can generalize the reaction rate equation of competitive and non-competitive substrate inhibitions by using the following formula:

r ¼ Vmax S½ �KS þ S½ � þ aI S½ �2 � ð3:125Þ

If aI ¼ 1=K 0S, we have noncompetitive (consecutive) inhibition; if

aI ¼ KS=K0S we have competitive inhibition.

Continuous-Stirred Tank Enzyme Reactor (or Enzyme CSTR)(Fig. 3.14)

For simplicity, assume that the enzymes are immobilized inside of thereactor (i.e., no washout) but, at the same time, there is no mass transferresistance, then the equation is simply

qSf ¼ qS þ Vr ð3:126ÞNote: If r is per unit mass, then multiply by the enzyme concentration CE .

By dividing Eq. (3.126) by V , we get

Figure 3.14 Enzyme CSTR.

q

V

� �Sf ¼

q

V

� �S þ r ð3:127Þ

where q=Vð Þ is the dilution rate D time�1� �

, which is the inverse of theresidence time, t ¼ V=q Therefore,

D Sf � S� � ¼ r ð3:128Þ

Equation (3.128) can be easily solved; the nature of the solution(s) willdepend on the rate of the bioreaction function r. We will call it the consump-tion function C Sð Þ½ �; the left-hand side D Sf � S

� �� will be called the supply

function S Sð Þ½ �. Let us solve them graphically as shown in Figure 3.15.It is clear that for Michaelis–Menten kinetics, we have only one steady

state for the whole range of D, whereas for the substrate-inhibited kinetics,we can have more than one steady state over a certain range of D values(this can be easily visualized by Fig. 3.16).

More details regarding multiplicity of the steady states are shown inFigure 3.16. The stability of the different steady states is stated below.

The simplest definition of the stability of steady states is as follows:

Stable steady state. If disturbance is made and then removed, thesystem will return back to its initial steady state.

Figure 3.15 Graphical solution of Eq. (3.128).

Unstable steady state. If a disturbance is made and then removed, thesystem will not return to its initial steady state.

Corresponding to Figure 3.16, we can describe the stability behavior of thethree steady states (denoted by points A, B, and C in Fig. 3.16).

Steady-state point A. From a static point of view, it is stable becauseslightly to the right of it, the rate of consumption of the substrate isgreater than the rate of supply. This will give rise to a decrease in theconcentration and the reactor goes back to its steady state at A. Also,for concentrations slightly to the left of steady state A, the rate ofconsumption is smaller than the rate of supply. Therefore, the con-centration will increase and the reactor will go back to steady state A.

Steady-state point B. From the static point of view, it is unstablebecause for any concentration change slightly to the right of thispoint, the rate of substrate consumption is smaller than the rate ofsubstrate supply. Therefore, the concentration continues to increaseand the system does not go back to steady state B. Also, as theconcentration is changed slightly to the left of steady state pointB, the rate of substrate supply is lower than the rate of consumptionand, therefore, the concentration continues to decrease and neverreturns to steady state B.

Figure 3.16 Multiple steady states and their stability characteristics.

Steady-state point C. This steady state is stable for the same reasonsdescribed for steady-state point A. (The reader is advised to performthe tests described for steady state points A and B to verify thatsteady-state point C is indeed stable).

At which stable steady state will the reactor operate? This is not knownand the behavior of the reactor is described by its previous history (non-physical, nonchemical condition).

Sources of multiplicity

1. Nonmonotonic dependence of a rate process (or more) on at leastone state variable

2. Feedback of information (a recycle in a tubular reactor or axialdispersion, which will be discussed later) (see Fig. 3.17).

More details about multiplicity of steady states and bifurcationbehavior are given in Chapter 7.

3.8.2 Modeling of Microbial Systems

Microbial bioreactors differ from enzyme reactors with regard to the factthat the biocatalyst (micro-organism) grows with the progress of the reac-tion (i.e., it is an autocatalytic process).

Micro-organism concentration ¼ Xg

L

� �and the reaction is as follows:

S þ X ! Pþ X

The following biochemical yield factors replace the stoichiometric numbersused in reactions:

YS ¼ Grams of micro-organism produced

Grams of substrate consumed

Figure 3.17 Recycle is a source of multiplicity.

YP ¼ Grams of product produced

Grams of substrate consumed

or

Y 0P ¼ YP

YS

¼ Grams of product produced

Grams of micro-organism produced

Monod Equation and Substrate Inhibition

This is the most common reaction rate equation used. It resembles theenzyme rate equation, but it is more empirical:

m ¼ Grams of micro-organism produced

Grams of micro-organisms existing� Time¼ time�1

Mathematically, it can be written as

m ¼ mmax S½ �KS þ S½ �

For the case of substrate inhibition,

m ¼ mmax S½ �KS þ S½ � þ K S½ �2 ð3:129Þ

. There can be product inhibition also.

. There can be cell inhibition also, where, after a time, the cells stopthe reaction.

If the reaction is taking place in a batch reactor, then

d VXð Þdt

¼ m VXð Þ

where m is the rate of production of the micro-organism and VX is the totalweight of the micro-organism at any time t.

Simplifying Assumptions

1. Assume V is constant.2. Assume dilute solution where the micro-organism is small and is

accumulating slowly.

With these simplifying assumptions, the equation can be modified to give

VdS

dt¼ � mVX

YS

(consumption rate of substrate)

This gives

dS

dt¼ � mX

YS

ð3:130Þ

Also, we get

dX

dt¼ mX ð3:131Þ

with initial condition at t ¼ 0, X ¼ Xf and S ¼ Sf . By rearranging Eqs.(3.130) and (3.131) and adding them, we get

YS

dS

dtþ dX

dt¼ �mX þ mX ¼ 0


d YSS þ Xð Þdt

¼ 0

To solve this differential equation, we integrate with the initial conditions att ¼ 0, X ¼ Xf and S ¼ Sf :ð

d YSS þ Xð Þdt

¼ð0ð

d YSS þ Xð Þ ¼ C ¼ YSSf þ Xf

We get

YSS þ X ¼ YSSf þ Xf

On rearranging, we get

X ¼ Xf þ YS Sf � S� �

By substituting the value of X in Eq. (3.130) and after some rearrangement,we get the design equation in terms of one state variable S:

YS

dS

dt¼ � mmax S½ �

KS þ S½ � Xf þ YS Sf � S� � �

where

m ¼ mmax S½ �KS þ S½ �

For other cases, another mechanism can also decrease (eat) the micro-organism through ‘‘death’’. Therefore, we introduce a term kd to incor-porate this decrease in micro-organism number,

dX

dt¼ mX � kdX ¼ X m� kdð Þ ð3:132Þ

and for the substrate, we can write

dS

dt¼ � mX

YS

ð3:133Þ

The two differential equations (3.132 and 3.133) can to be solved simulta-neously with suitable initial conditions.

Continuous-Stirred Tank Fermentor (Without Micro-organismDeath) (Fig. 3.18)

For this case and with the assumption of constant volume, the unsteady-state equations are given as follows:

For micro-organisms

FXf þ mXV ¼ FX þ VdX

dt

For substrate

FSf ¼ FS þ mXVYS

þ VdS

dt

With initial conditions at t ¼ 0;S ¼ Sf and X ¼ Xf . The steady-stateequations can be obtained by simply setting

dX

dt¼ dS

dt¼ 0

Figure 3.18 Continuous-stirred tank fermentor.

A much more detailed and involved heterogeneous modeling of microbialreaction systems is discussed in Chapter 6.

Solved Example

The continuous anaerobic digester used for treating municipal wastesludge is essentially like any other CSTR except that it involves multiphasebiological reactions. Material essentially consisting of organisms of con-centration CXO mol/m3 and a substrate of concentration CSO mol/m3 is fedto a certain anaerobic digester at F m3/day. Use the following additionalassumptions:

1. The reactions of interest take place only in the liquid phase and thereactor (liquid) volume V (m3) is constant.

2. The organism concentration in the digester CX is uniform, as is thesubstrate concentration CS.

Obtain a model for this digester process by carrying out organism andsubstrate balances and using the following constitutive relationships:

1. The organism growth rate (in the reactor) is given by

rX ¼ dCX

dt¼ mCX

where m, the specific growth rate, is itself given by the Monodfunction

m ¼ m0CS

KS þ CS

� �

Here, m0 is the maximum specific growth rate and KS is the satu-ration constant.

2. The yield (the rate of growth of organism ratioed to the rate ofconsumption of substrate) in the reactor is given by

rX ¼ �YXS rS

Solution

The process can be shown schematically as in Figure 3.19. The microbialreaction is given by

S!X P

The rate of production of the micro-organism is given by

rX ¼ mCX ¼ m0CS

KS þ CS

� �CX

The rate of consumption of the substrate is given by (note the negative signin the rate as the substrate concentration decreases due to its consumptionby the micro-organisms)

rS ¼ dCS

dt¼ � 1

YXS

m0CS

KS þ CS

� �CX

An appropriate model may be obtained from a micro-organism materialbalance and a substrate mass balance:

Substrate balance

VdCS

dt¼ FCSO � FCS � 1

YXS

m0CS

KS þ CS

� �CX ð3:134Þ

Micro-organism balance

VdCX

dt¼ FCXO � FCX þ m0

CS

KS þ CS

� �CX ð3:135Þ

With initial conditions at t ¼ 0;CS ¼ CS 0ð Þ and CX ¼ CX 0ð Þ.Equation (3.134) and (3.135) with initial conditions formulate the dynamicmodel for this microbial system.

Note: Volume V is assumed to be constant.

Figure 3.19 Schematic diagram of the process.

The steady state is obtained by setting the differential terms equal to zero inEqs. (3.134) and (3.135):

dCS

dt¼ dCX

dt¼ 0

The steady-state equations are

FCSO � FCSSS� 1

YXS

m0CSSS

KS þ CSSS

� �CXSS

¼ 0

and

FCXO � FCXSSþ m0

CSSS

KS þ CSSS

� �CXSS

¼ 0

where CSSSis the steady-state concentration of the substrate and CXSS

is thesteady-state concentration of the micro-organism.

REFERENCES

1. Eigenberger, G. Modelling and Simulation in Industrial Chemical Reaction

Engineering. Springer Series in Chemistry and Physics Vol. 18. Springer-

Verlag, New York, 1981, pp. 284–304.

2. Salmi, T. Modeling and simulation of transient states of ideal heterogeneous

catalytic reactors. Chem. Eng. Sci, 43(3), 503–511, 1988.

3. Aube, F., and Sapoundjiev, H. Mathematical model and numerical simulations

of catalytic flow reversal reactors for industrial applications. Computers Chem.

Eng., 24(12), 2623–2632, 2000.

4. Elnashaie, S. S., Gaber, A. H., and El-Rifai, M. A. Dynamic analysis of enzyme

reactors exhibiting substrate Iinhibition multiplicity. Chem. Eng. Sci. 32(5),

557–559, 1977.

5. Baiker, A., and Bergougnan, M. Investigation of a fixed-bed pilot plant reactor

by dynamic experimentation. Part 1. Apparatus and experimental results. Can.

J. Chem. Eng. 63(1), 138–145, 1985.

6. Baiker, A., and Bergougnan, M. Investigation of a fixed-bed pilot plant reactor

by dynamic experimentation. Part 2. Simulation of reactor behavior. Can. J.

Chem. Eng. 63(1), 146–154, 1985.

7. Elnashaie, S. S. E. H., Moustafa, T., Alsoudani, T., and Elshishini, S. S.

Modeling and basic characteristics of novel integrated dehydrogenation-hydro-

genation membrane catalytic reactors. Computers Chem. Eng., 24(2–7), 1293–

1300, 2000.

8. Ajbar, A., Alhumazi, K., and Elnashaie, S. S. E. H. Classification of static and

dynamic behavior in a fluidized-bed catalytic reactor. Chem. Eng. J., 84(3), 503–

516, 2001.

9. Elnashaie, S. S. E. H., Alhabdan, F. M., and Adris, A. M. A Software

Computer Package for the Simulation of Industrial Steam Reformers for

Natural Gas. Institute of Chemical Engineering Symposium Series No. 114.

1989, pp. 59–67.

10. Hoelzer, D., and Hahn, H. H. Development of a software package for the

dimensioning and operation of activated sludge plants. Korresp. Abwasser

42(10), 1836–1838, 1841–1846, 1995.

11. Li, Ke, Crittenden, J. C., and Hand, D. W. A software package for modeling of

the H2O2/UV advanced oxidation process. Proceedings of the Annual

Conference of the American Water Works Association, 2001, pp. 1421–1424.

12. Schnelle, P. D., Jr., and Richards, J. R. A review of industrial reactor control:

Difficult problems and workable solutions. Chemical Process Control (CPC3),

Proceedings of the 3rd International Conference on Chemical Process Control,

1986, pp. 749–802.

13. Richards, J. R., and Schnelle, P. D., Jr. Perspectives on industrial reactor

control. Chem. Eng. Prog. 84(10), 32–36, 1988.

14. Boreskov, G. K., and Matros, Yu. Sh. Fixed catalyst bed reactors operated in

steady- and unsteady-state conditions. Recent Adv. Eng. Anal. Chem. React.

Syst., 142–155, Wiley, New York, 1984.

15. Adris, A., Grace, J., Lim, C., and Elnashaie, S. S. E. H. Fluidized bed reaction

system for steam/hydrocarbon gas reforming to produce hydrogen, U.S. Patent

5,326,550, June 5, 1994.

16. Makel, D. Low cost microchannel reformer for hydrogen production from

natural gas. California Energy Commission (CEG), Energy Innovations

Small Grant (EISG) Program, 1999.

17. Sammells, A. F., Barton, A. F., Peterson, D. R., Harford, S. T., Mackey, R.,

Van Calcar, P. H., Mundshau, M. V., and Schultz, J. B. Methane conversion to

syngas in mixed conducting membrane reactors. 4th International Conference

on Catalysis in Membrane Reactors, 2000.

18. Prigogine, I., George, C., Henin, F., and Rosenfeld, L. Unified formulation

of dynamics and thermodynamics. With special reference to non-equilibrium

statistical thermodynamics. Chem. Scr., 4(1), 5–32, 1973.

19. Nicolis, G., and Prigogine, I. Self-organization in nonequilibrium systems: From

dissipative structures to order through fluctuations. Wiley, New York, 1977.

20. Denn, M. M. Process Modeling. Longman Science and Technology, London,

1986.

21. Froment, G. F., and Bischoff, K. B. Chemical Reactor Analysis and Design.

Wiley, New York, 1990.

22. Dransfield, P. Engineering Systems and Automatic Control, Prentice-Hall,

Englewood Cliffs, NJ, 1968.

23. Von Bertalanffy, L. General Systems Theory. George Braziller, New York, 1968.

24. Robert Shaw, J. E., Mecca, S. J., and Rerick, M. N. Problem Solving. A System

Approach. Academic Press, New York, 1964.

25. Aris, R. Manners makyth modellers. Chem. Eng. Res. Des., 69(A2), 165–174,

1991.

26. Smith, J. M. Models in Ecology. Cambridge University Press, Cambridge, 1974.

27. Hawking, S. W. A Brief History of Time: From the Big Bang to Black Holes.

Bantam Press, London, 1989.

28. Soliman, M. A. and Elnashaie, S. S. E. H. Negative effectiveness factors for

cyclic reversible reactions, Chem. Eng. Sci. 45(7), 1925–1928, 1990.

29. Elnashaie, S. S. E. H., Abashar, M. E., and Alubaid, A. S. Simulation and

optimization of an industrial ammonia reactor, Ind. Eng. Chem. Res. 27(11),

2015–2022, 1988.

30. Alhabdan, F. M. and Elnashaie, S. S. E. H. Simulation of an ammonia plant

accident using rigorous heterogeneous models—Effect of shift converters

disturbances on methanator. Math. Computer Model. 21(4), 85–106, 1995.

PROBLEMS

Problem 3.1

Derive the steady-state mass balance equations for an isothermal contin-uous-stirred tank reactor in which a consecutive homogeneous reaction

A �!k1 B �!k2 C

is taking place. Put the resulting equation in a matrix form and suggestthe sequence of solution procedures to obtain the output of the threecomponents.

Problem 3.2

A semibatch reactor is run at constant temperature by varying the rate ofaddition of one of the reactants A. The irreversible exothermic reactiontaking place in the reactor is first order in reactants A and B:

Aþ B ! C

The tank is initially filled to its 40% level with the pure reactant B at aconcentration CBO. Cooling water is passed through a cooling coil immersedin the reactor and reactant A is slowly added to the perfectly stirred vessel.Write the equations describing the system. Without solving the equations,try to sketch the profiles of FA (feed flow rate) and CA (concentration of A inthe reactor), and CB (concentration of B in the reactor) with time.

Problem 3.3

A perfectly mixed nonisothermal adiabatic reactor carries out a simple first-order exothermic reaction in the liquid phase:

A ! B

The product from the reactor is cooled from the output temperature T toa temperature TC and is then introduced into a separation unit where theunreacted A is separated from the product B. The feed to the separationunit is split into two equal parts: top product and bottom product; the

bottom product from the separation unit contains 95% of the unreacted A inthe effluent of the reactor and 1% of the B in the same stream. This bottomproduct, which is at temperature TC (i.e., the separation unit is isothermal), isrecycled and mixed with the fresh feed to the reactor and the mixed stream isheated to the reactor feed temperature Tf before being introduced into thereactor. Write the steady-state mass and heat balance equations for thissystem assuming constant physical properties and heat of reaction. (Note:concentrate your modeling attention on the nonisothermal, adiabatic reac-tor, and for the rest of units, carry out a simple mass and heat balance inorder to define the feed conditions to the reactor.)

Problem 3.4

Consider a system that initially consists of 1 mol of CO and 3 mol of H2 at1000 K. The system pressure is 25 atm. The following reactions are to beconsidered:

2CO þ 2H2 , CH4 þ CO2 ðAÞCOþ 3H2 , CH4 þH2O ðBÞCO2 þH2 , H2Oþ CO ðCÞ

When the equilibrium constants for reactions A and B are expressed interms of the partial pressures of the various species (in atm), the equilibriumconstants for these reactions have the values KPA

¼ 0:046 and KPB¼ 0:034.

Determine the number of independent reactions, and then determine theequilibrium composition of the mixture.

Problem 3.5

A perfect gas with molecular weight M flows at a rate W0 (kg/h) into acylinder through a restriction. The flow rate is proportional to the squareroot of the pressure drop over the restriction:

W0 ¼ K0

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiP0 � P

pwhere P is the pressure (N/m2 absolute) in the cylinder, P0 is constantupstream pressure. The system is isothermal. Inside the cylinder a pistonis forced to the right as the pressure P builds up. A spring resists the move-ment of the piston with a force that is proportional to axial displacement Xof the piston:

FS ¼ KS X ðNÞThe pressure on the spring side is atmospheric and is constant. The pistonis initially at X ¼ 0 when the pressure in the cylinder is zero. The cross-

sectional area of the cylinder is A (m2). Assume that the piston has negligiblemass and friction.

(a) Derive the equations describing the system, (i.e., the dynamicbehavior of the system).

(b) What will the steady-state piston displacement be?(c) What is the effect of relaxing the assumption of negligible mass

of the piston on both the dynamic behavior and steady-stateposition of the piston?

(d) What is the effect of relaxing the assumption of constant atmo-spheric pressure in the spring side (behind the piston)?

Problem 3.6

The reaction of ethylene and chlorine in liquid ethylene dichloride solutionis taking place in a CSTR. The stoichiometry of the reaction is,

C2H4 þ Cl2 ! C2H4Cl2

Equimolar flow rates of ethylene and chlorine are used in the followingexperiment, which is carried out at 368C. The results of the experimentare tabulated in Table P3.6.

(a) Determine the overall order of the reaction and the reaction rateconstant.

(b) Determine the space time (space time is the residence time whichis the ratio between the volume of the reactor and the volumetricflow rate) necessary for 65% conversion in a CSTR.

(c) What would be the conversion in a PFR having the space timedetermined in part (b)?

Table P3.6 Experimental Data

Space time (s)

Effluent chlorine concentration

(mol/cm3)

0 0.0116

300 0.0094

600 0.0081

900 0.0071

1200 0.0064

1500 0.0060

1800 0.0058

In parts (b) and (c), assume that the operating temperature and the initialconcentrations are the same as in part (a).

Problem 3.7

Some of the condensation reactions that take place when formaldehyde Fð Þis added to sodium paraphenolsulfonate Mð Þ in an alkaline-aqueous solu-tion have been studied. It was found that the reactions could be representedby the following equations:

FþM ! MA; k1 ¼ 0:15 L/gmol minFþMA ! MDA; k2 ¼ 0:49MAþMDA ! DDA; k3 ¼ 0:14MþMDA ! DA; k4 ¼ 0:14MAþMA ! DA; k5 ¼ 0:04MAþM ! D; k6 ¼ 0:056FþD ! DA; k7 ¼ 0:50FþDA ! DDA; k8 ¼ 0:50

where M, MA, and MDA are monomers and D, DA, and DDA are dimers.The process continues to form trimers. The rate constants were evaluatedusing the assumption that the molecularity of each reaction was identical toits stoichiometry.

Derive a dynamic model for these reactions taking place in a single,isothermal CSTR. Carefully define your terms and list your assumptions.

Problem 3.8

The conversion of glucose to gluconic acid is a relatively simple oxidation ofthe aldehyde group of the sugar to a carboxyl group. A micro-organism in abatch or continuous fermentation process can achieve this transformation.The enzyme (a biocatalyst) glucose oxidase, present in the micro-organism,converts glucose to gluconolactone. In turn, the gluconolactone hydrolyzesto form the gluconic acid. The overall mechanism of the fermentationprocess that performs this transformation can be described as follows:

Cell growth

Glucose þ Cells ! Cells

Rate of cell growth

RX ¼ k1CX 1� CX

K1

� �

where, CX is the concentration of cells and k1 and K1 are rate con-stants.

Glucose oxidation

GlucoseþO2 ��!Glucose oxidase

GluconolactoneþH2O2

and

H2O2 ! 12O2 þH2O

Rate of gluconolactone formation

RGl ¼k2CXCS

K2 þ CS

where CS is the substrate (glucose) concentration and k2 and K2 arerate constants.

Gluconolactone hydrolysis

GluconolactoneþH2O ! Gluconic acid

Rate of gluconolactone hydrolysis

RGH ¼ 0:9k3CGl

Rate of gluconic acid formation

RG ¼ k3CGl

where CGl is the gluconolactone concentration and k3 is a rate con-stant.

Overall rate of substrate (glucose) consumption

RS ¼ 1:011k2CXCS

K2 þ CS

� �

At operating conditions of 328C and pH ¼ 6:8, the values of the parametersare as follows:

k1 ¼ 0:949

K1 ¼ 3:439

k2 ¼ 18:72

K2 ¼ 37:51

k3 ¼ 1:169

(a) Develop a model for the batch fermentor and calculate thechange of the concentration of the species with time for thefollowing initial conditions (demonstrate the calculations for afew steps using the simple Euler’s method):

CX 0ð Þ ¼ 0:05 mg/mL

CGl 0ð Þ ¼ 0:0 mg/mL

CGA 0ð Þ ¼ 0:05 mg/mL (gluconic acid concentration)

CS 0ð Þ ¼ 50:0 mg/mL

(b) Develop the unsteady-state model for the continuous fermentor.Find the feed flow rate and fermentor active volume, which gives60% substrate (glucose) conversion at steady state and 50 kg/h ofgluconic acid. Show, using the unsteady-state model, the dynamicresponse of the fermentor to disturbances in the input feed flowrate.

Problem 3.9

A perfectly mixed, isothermal CSTR has an outlet weir as shown in FigureP3.9. The flow rate over the weir is proportional to the height of liquid overthe weir, how, to the 1.5 power. The weir height is hw. The cross-sectionalarea of the tank is A. Assume constant density.

A first-order chemical reaction takes place in the tank:

A �!k B

Derive the equations describing the system.

Figure P3.9 Schematic diagram of Problem 3.9.

Problem 3.10

The liquid in a jacketed, nonisothermal CSTR is stirred by an agitatorwhose mass is significant compared with the reaction mass. The mass ofthe reactor wall and the mass of the jacket wall are also significant. Write theenergy equations for the system. Neglect radial temperature gradients in theagitator, reactor wall, and jacket wall.

Problem 3.11

The reaction

3A ! 2Bþ C

is carried out in an isothermal semibatch reactor as shown in Figure P3.11.Product B is the desired product. Product C is a very volatile by-productthat must be vented off to prevent a pressure buildup in the reactor. GaseousC is vented off through a condenser to force any A and B back into thereactor to prevent loss of reactant and product.

Assume FV is pure C. The reaction is first-order in CA. The relativevolatilities of A and C to B are aAB ¼ 1:2 and aCB ¼ 10. Assume perfectgases and constant pressure. Write the equations describing the system.Carefully list all assumptions.

Figure P3.11 Schematic diagram for Problem 3.11.

Problem 3.12

A vertical, cylindrical tank is filled with well water at 658F. The tank isinsulated at the top and bottom but is exposed on its vertical sides to coldnight air at 108F. The diameter of the tank is 2 ft and its height is 3 ft. Theoverall heat transfer coefficient is 20 Btu/h 8F ft2. Neglect the metal wall ofthe tank and assume that the water in the tank is perfectly mixed.

(a) Calculate how many minutes it will be until the first crystal of iceis formed.

(b) How long will it take to completely freeze the water in the tank?The heat of fusion of water is 144 Btu/lbm.

Problem 3.13

An isothermal, first-order, liquid-phase, irreversible reaction is conducted ina constant volume batch reactor:

A �!k B

The initial concentration of reactant A at the beginning of the batch is CAO.The specific reaction rate decreases with time because of catalyst deactiva-tion as k ¼ k0e

�bt.

(a) Solve for CA tð Þ(b) Show that in the limit as b ! 0, CA tð Þ ¼ CAOe

�k0t

(c) Show that in the limit as b ! 1, CA tð Þ ¼ CAO

Problem 3.14

There are 3580 lbs of water in the jacket of a reactor that are initially at1458F. At time equal to zero, 708F cooling water is added to the jacket at aconstant rate of 418 lbs/min. The holdup of water in the jacket is constantbecause the jacket is completely filled with water and excess water isremoved from the system on pressure control as cold water is added.Water in the jacket can be assumed to be perfectly mixed.

(a) How many minutes does it take the jacket water to reach 998F ifno heat is transferred to the jacket?

(b) Suppose a constant 358,000 Btu/h of heat is transferred into thejacket from the reactor, starting at time equal zero when thejacket is at 1458F. How long will it take the jacket water toreach 998F if the cold water addition rate is constant at 416lbs/min?

Problem 3.15

Consider an enzymatic reaction being carried out in a CSTR. The enzyme issubject to substrate inhibition, and the rate equation is given by Eq. (3.119):

r ¼ Vmax S½ �KS þ S½ � þ S½ �2=K 0

S

(a) Sketch a plot of reaction rate versus the substrate concentrationfor this system. Use this plot to show that for sufficiently highinlet substrate concentration, there is a range of dilution rates forwhich multiple steady states are possible.

(b) Explain how would you calculate the highest and lowest dilutionrates for which multiple steady states are possible.

4

Mathematical Modeling (II):Homogeneous Distributed Systemsand Unsteady-State Behavior

In Chapter 3, we covered the mathematical modeling of lumped systems aswell as some preliminary examples of distributed systems using a systematic,generalized approach. The examples for distributed systems were preliminaryand they were not sufficiently generalized. In this chapter, we introduce suffi-cient generalization for distributed systems and give more fundamentally andpractically important examples, such as the axial dispersion model resultingin two-point boundary-value differential equations. These types of modelequations are much more difficult to solve than models described by initial-value differential equations, specially for nonlinear cases, which are solvednumerically and iteratively. Also, examples of diffusion (with and withoutchemical reaction) in porous structures of different shapes will be presented,explained, and solved for both linear and nonlinear cases in Chapter 6.

In this chapter, the unsteady-state term(s) to be introduced intothe models will be explained and discussed for homogeneous as well asheterogeneous systems, approximated using pseudohomogeneous model.Rigorous steady-state and dynamic heterogeneous models will be coveredin Chapter 6. The approach adopted in this chapter regarding dynamicmodeling, like the rest of the book, is a very general approach which canbe applied in a very systematic and easy-to-understand manner. It consists ofexplaining how to formulate the unsteady-state term(s) in any situation andwhere to add it to the equations of the steady-state model in order to make

the model an unsteady state (dynamic) model. These dynamic models aresuitable for start-up, shutdown, process control, and investigation of thedynamic stability characteristics of the system, as will be shown in Chapter 5.

4.1 MODELING OF DISTRIBUTED SYSTEMS

A distributed system is characterized by variation of the state variable(s)along one or more of the space coordinates. When the state variable(s) arevarying along only one of the space coordinates, then the distributed modelis called a one-dimensional model, and is described by ordinary differentialequations (ODEs). When the variation is along two of the space coordi-nates, the model is called a two-dimensional model, and is described bypartial differential equations (PDEs). When the variation is along threespace coordinates, it is called a three-dimensional model (PDEs). In thisbook, only one-dimensional distributed models will be covered. The reasonfor that is that one-dimensional distributed models are sufficient for mostpractical purposes, in addition to the fact that the mathematical formulationand solution of higher-dimensional models is beyond the scope of thisundergraduate book.

4.1.1 Isothermal Distributed Systems

Here, we present the basic idea for the formulation of one-dimensionaldistributed models for a single-input, single-output, and single-reactiontubular reactor. Axial dispersion is neglected in this preliminary stage;then, it will be introduced and discussed later in this chapter.

Figure 4.1 illustrates a single reaction in a tubular reactor. The overallsteady-state mass balance is

ni ¼ nif þ sir

where r is the overall rate of reaction for the whole reactor. This means thatit is not a design (model) equation; it is a mass balance equation. Thetechnique used for lumped system [i.e., replacing r by r0 (rate per unit

Figure 4.1 Mass flow along a tubular reactor.

volume) multiplied by the reactor volume Vð Þ� is not directly applicablehere. Why? Because r0 changes along the length of the reactor.

The solution is to carry out the balance over an element of length �l(or volume �V ¼ At�l, where At is the cross-sectional area of the tubularreactor). Thus the mass balance over this element becomes

ni þ�nið Þ ¼ ni þ sir0At�l

Rearranging and dividing the equation by �l gives

�ni�l

¼ siAtr0

Taking the limit as �l ! 0 gives the differential mass balance (design)equation for this simple distributed model:

dnidl

¼ siAtr0

or (because �V ¼ At�l)

dnidV

¼ sir0

where, i ¼ 1; 2; 3; . . . ;M and r0 is the rate of reaction per unit volume of thereactor.

For multiple reactions, it becomes

dnidV

¼XNj¼1

sijr0j

where M is the number of components and N is the number of reactionswith the initial condition at l ¼ 0 (or V ¼ 0Þ; ni ¼ nif .

For a simple case having no change in the number of moles with thereaction (or liquid-phase system), we can write

ni ¼ qCi

giving

qdCi

dV¼XNj¼1

sijr0j

with initial condition at V ¼ 0;Ci ¼ Cif .For multiple-input, multiple-output, the differential equation will not

change; we will have only the equivalent of a mixer at the feed conditionsand a splitter at the exit. The simplest way to handle multiple-input, multi-ple-output situations for distributed systems is to use the modular approachdiscussed earlier and as shown in Figure 4.2.

4.1.2 Nonisothermal Distributed Systems

Similarly, the heat balance for the distributed system is to be handled bycarrying out the heat balance over an element (see Fig. 4.3).

For the nonadiabatic case, Q0 6¼ 0; Q0 is the heat flux per unit lengthJ=minmð Þ.

A heat balance over the �l element gives

XMi¼1

niHi þQ0�l ¼XMi¼1

niHi þXMi¼1

� niHið Þ

which, on rearrangement, gives

XMi¼1

� niHið Þ�l

¼ Q0

Taking the limiting case as �l ! 0, we get

XMi¼1

d niHið Þdl

¼ Q0

Figure 4.2 Modules and modular approach.

On differentiating the left-hand side of the above equation, we get,

XMi¼1

nidHi

dlþHi

dnidl

� �¼ Q0 ð4:1Þ

From mass balance for the distributed system with a single reaction, we have

dnidl

¼ Atsir0 ð4:2Þ

Substituting Eq. (4.2) into Eq. (4.1), we get

XMi¼1

nidHi

dlþHiAtsir

0� �

¼ Q0

Rearrangement gives

XMi¼1

nidHi

dl

� �þXMi¼1

HiAtsir0� � ¼ Q0

which can be written as,

XMi¼1

nidHi

dl

� �þ Atr

0XMi¼1

Hisið Þ|fflfflfflfflfflffl{zfflfflfflfflfflffl}

�H

¼ Q0

Defining �H as the heat of reaction (not at reference conditions), the aboveequation becomes

XMi¼1

nidHi

dl

� �¼ Atr

0 ��Hð Þ þQ0

This derivation can be easily extended to multiple reactions by replacing thefirst term of the right-hand side (RHS) by

Figure 4.3 The heat balance setup for a tubular reactor.

At

XNj¼1

r0j ��Hj

� �

where N is the number of reactions.

Simplified Case

If there is no change in phase and only one reaction takes place, then we canwrite the heat balance design equation as

XMi¼1

nid CpiT� �dl

� �¼ Atr

0 ��Hð Þ þQ0

If Cpi is taken as an average constant value for each i (not varying withtemperature), then we get

XMi¼1

niCpi

dT

dl

� �¼ Atr

0 ��Hð Þ þQ0

Taking an average constant C 0pmix gives

XMi¼1

niC0pmix

dT

dl

� �¼ Atr

0 ��Hð Þ þQ0


XMi¼1

ni

!|fflfflfflfflffl{zfflfflfflfflffl}

nt

C 0pmix

dT

dl¼ Atr

0 ��Hð Þ þQ0

The above equation gives

ntC0pmix

dT

dl¼ Atr

0 ��Hð Þ þQ0

where, nt is the total molar flow rate and C 0pmix is the molar specific heat.

For constant volumetric flow rate qð Þ and specific heat per unit massCp

� �, this equation can be written in the more ‘‘popular’’ but ‘‘simplified’’

form

qrmixCp

dT

dl¼ Atr

0 ��Hð Þ þQ0

with initial condition at l ¼ 0;T ¼ Tf .

4.1.3 Batch Systems (Distributed in Time)

Batch reactors can be considered unsteady-state lumped systems with no inputand no output, or it can be considered a distributed system in time (not inspace) with the initial conditions as the feed conditions and the final conditionsas the exit conditions.

In either case, the batch reactor design equations can be developed asfollows. With the regard to the heat balance (see Fig. 4.4), we get

XMi¼1

niHi þQ�t ¼XMi¼1

niHi þXMi¼1

� niHið Þ

After the same kind of manipulation we did for the tubular reactor, we get

XMi¼1

nidHi

dt

� �¼ Vr0 ��Hð Þ þQ

For multiple reactions,

XMi¼1

nidHi

dt

� �¼XNj¼1

Vr0j ��Hð Þj þQ

whereM is the number of components and N is the number of reactions. Wecan simplify the above equation to (as we did earlier for the tubular reactor)

VrmixCp

dT

dt¼XMj¼1

Vr0j ��Hð Þj þQ

where, Cp is an average constant specific heat per unit mass of the mixture.The initial condition is at t ¼ 0;T ¼ Tf .The mass and heat balance design (model) equations must be solved

simultaneously (see Fig. 4.5, as a reminder of how we derived the mass

Figure 4.4 Heat balance for a batch reactor.

balance design equation for the batch reactor). The mass balance design(model) equation is easily obtained (for a single reaction):

ni þ�ni ¼ ni þ siVr0�t

which, after some simple manipulations as discussed earlier, also gives

dnidt

¼ siVr0

For multiple reactions, the equation becomes

dnidt

¼ VXNj¼1

sijr0j

� �The initial condition at t ¼ 0 is given by ni ¼ nif

For the special case of no change in the number of moles accompany-ing the reaction (or liquid phase) giving constant V , we can write

ni ¼ VCi

Thus, the mass balance design equation becomes

dCi

dt¼XNj¼1

sijr0j

� �The initial condition at t ¼ 0 is given by Ci ¼ Cif .

4.2 THE UNSTEADY-STATE TERMS INHOMOGENEOUS AND HETEROGENEOUSSYSTEMS

The heterogeneous system here is described by a pseudohomogeneousmodel. For complete heterogeneous systems, see Chapter 6.

The unsteady-state (dynamic) models differ from the steady-statemodel through the addition of the dynamic (accumulation/depletion)term. This term is always added to the side of the equation having the

Figure 4.5 Mass balance for a batch reactor.

output stream. This term has many names, dynamic or unsteady state, oraccumulation/depletion, or, sometimes, capacitance.

4.2.1 Lumped Systems

Isothermal Homogeneous Lumped Systems

We will start with the simplest isothermal lumped homogeneous case, alumped system, as shown in Figure 4.6. The steady-state equation (for asingle reaction) is given by

ni ¼ nif þ Vsir0

For the unsteady state, all that we will do is add the unsteady-state(dynamic) term to the side of equation having the output stream. In thiscase, the equation becomes

dn0idt|{z}

dynamic term

þ ni ¼ nif þ Vsir0 ð4:3Þ

For the homogeneous system, we can write n0i (the molar content of thetank) as

n0i ¼ VCi

and if V is constant, then

dn0idt

¼ VdCi

dt


Now, Eq. (4.3) becomes

VdCi

dtþ ni ¼ nif þ Vsir

0

Isothermal Heterogeneous Lumped Systems

Suppose that there are catalyst particles in the reactor (see Fig. 4.7). Thetotal volume of the catalyst is V 1� eð Þ and the gas or liquid volume is Ve.The concentration of the component in the liquid or gas is Ci (in mol=LÞ,whereas on the catalyst surface, the concentration is Cis in ðmol=g(or cm3 or cm2Þ of catalystÞ. For illustration, we will use Cis with units ofgmol=cm2(surface) of catalystÞ. Therefore,

n0i ¼ VeCi þ V 1� eð ÞSvCis

where Sv is the specific surface area of the catalyst (in cm2/cm3). Forconstant V and e,

dn0idt

¼ VedCi

dtþ V 1� eð ÞSv

dCis

dtð4:4Þ

We can relate Ci and Cis through a simple linear adsorption isotherm, asfollows:

Cis ¼ KiCi

If we assume Ki to be constant, on differentiation we get

dCis

dt¼ Ki

dCi

dtð4:5Þ


Using Eq. (4.5) in Eq. (4.4), we get

dn0idt

¼ Ve|{z}a1

þV 1� eð ÞSvKi|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl}a2

0@

1A dCi

dt

Usually, a2 is very large for fine or porous particles.In most practical cases,

a2 >>> a1

(a2 is thousands of times greater than a1), so a1 can be neglected as com-pared to a2:

a1|{z}neglected

þa2

0@

1A dCi


0

and the suitable dynamic (unsteady state) model equation for this hetero-geneous system (described by pseudohomogeneous model, because masstransfer resistances are negligible) becomes

a2dCi


0

4.2.2 Distributed Systems

The same principles apply to distributed systems. However, all the balancesare over differential elements, giving rise to partial differential equation asshown in Figure 4.8.

Homogeneous Isothermal Systems

The dynamic term is carried out over the element and is added to the term ofthe exit stream on the left-hand side of the equation, giving

ni þ�ni þ At�l@Ci

@t¼ ni þ siAt�lr0

Figure 4.8 Mass flow diagram for a distributed system.

which can be put into the following partial differential equation form:

@ni@l

þ At

@Ci

@t¼ siAtr

0

For constant volumetric flow rate qð Þ, the above equation can be written as

q@Ci

@lþ At

@Ci

@t¼ siAtr

0

Dividing both sides by the cross-sectional area of the tubular reactor Atð Þand recognizing that q=At ¼ v, which is velocity of the flow, we get

v@Ci

@lþ @Ci

@t¼ sir

0 ð4:6Þ

The initial conditions at l ¼ 0 and t ¼ 0 are, Ci ¼ Cif , and Ci ¼ CiðlÞjt¼0,respectively. Obviously for multiple reactions, Eq. (4.6) becomes

v@Ci

@lþ @Ci

@t¼XNj¼1

sijr0j

� �

Heterogeneous Isothermal Systems

This is a case with negligible mass transfer resistances, as described by thepseudohomogeneous model. For a full heterogeneous system, see Chapter 6.This situation is a bit more complicated compared to the lumped system. Wewill consider a two-phase system with no mass transfer resistance betweenthe phases and the voidage is equal to e (see Fig. 4.9). The mass balancedesign equation over the element �l is:

ni þ�ni þ At�le@Ci

@tþ At�lð1� eÞSv

@Cis

@t¼ ni þ siAt�lr0

Figure 4.9 Mass flow diagram for a two-phase distributed system.

which can be put into the following partial differential equation form:

@ni@l

þ Ate@Ci

@tþ Atð1� eÞSv

@Cis

@t¼ siAtr

0 ð4:7Þ

Here, Sv is the solid specific surface area (per unit solid volume).We can relate Ci and Cis through a simple linear adsorption isotherm:

Cis ¼ KiCi

If we assume Ki to be constant, on differentiation we get

dCis

dt¼ Ki

dCi

dtð4:8Þ


@ni@l

þ Ate@Ci

@tþ Atð1� eÞSvKi

@Ci

@t¼ siAtr

0

For a constant flow rate q, we get

q@Ci

@lþ Ateþ Atð1� eÞSvKið Þ @Ci

@t¼ siAtr

0

On dividing the equation by At, we obtain

v@Ci

@lþ e|{z}

a1

þ ð1� eÞSvKi|fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl}a2

0@

1A @Ci

@t¼ sir

0

As discussed earlier, a2 >>> a1. So we can write

a1|{z}negligible

þa2

0@

1A @Ci

@tþ v

@Ci

@l¼ sir

0 ð4:9Þ

With the initial conditions at l ¼ 0 and t ¼ 0, Ci ¼ Cif and Ci ¼ Ci lð Þ��t¼0

,respectively. Needless to mention, for multiple reactions equation (4.9) willbecome

a1|{z}negligible

þ a2

0@

1A @Ci

@tþ v

@Ci

@l¼XNj¼1

sijr0j

� �

The reader can now handle any dynamic (capacitance) case through acorrect physical understanding, together with rational mathematicalexpression.

4.2.3 Nonisothermal Systems

Lumped Homogeneous Systems

For the steady state, the heat balance design equation for a single reaction is

qrCp T � Tf

� � ¼ Vr0 ��Hð ÞFor the unsteady state, after introducing the earlier assumption of nochange in phase, average constant specific heats, constant volume and volu-metric flow rate, and then adding the dynamic term with the output on theleft-hand side of the equation, we get

VrCp

dT

dtþ qrCp T � Tf

� � ¼ Vr0 ��Hð Þ

where, before the approximation, the first term on the left-hand side (thedynamic term) is d

PniHi=dtÞ and the second term is

Pni Hi �Hirð Þ

�P nif Hif �Hir

� �.

Lumped Heterogeneous Systems

Described by a pseudohomogeneous TS ¼ Tð Þ system and solid is stationaryinside the system with no solid input or output:

VerCp þ V 1� eð ÞrsCps

� dTdt

þ qrCp T � Tf

� � ¼ Vr0 ��Hð Þ

For gas–solid systems,

rCp << rsCps

Distributed Homogeneous Systems

The steady state is given by

qrCp

dT

dl¼ Atr

0 ��Hð Þ

The unsteady state is given by

AtrCp

@T

@tþ qrCp

@T

@l¼ Atr

0 ��Hð Þ

at l ¼ 0;T ¼ Tf , and at t ¼ 0;T ¼ TðlÞ��t¼0

.

Distributed Heterogeneous Systems

The distributed heterogeneous system (described by pseudohomogeneousT ¼ TS) is given by (with no input or output solids)

AterCp þ At 1� eð ÞrsCps

� � @T@t

þ qrCp

@T

@l¼ Atr

0 ��Hð Þ

at l ¼ 0;T ¼ Tf , and at t ¼ 0;T ¼ TðlÞ��t¼0

.As an exercise the reader can develop equations for the case of

continuous solid flow in and out of the reactor.

4.3 THE AXIAL DISPERSION MODEL

The assumption of plug flow is not always correct. The plug flow assumesthat the convective flow (flow by velocity q=At ¼ v, caused by a compressoror pump) is dominating over any other transport mode. In fact, this is notalways correct, and it is sometimes important to include the dispersion ofmass and heat driven by concentration and temperature gradients.However, the plug flow assumption is valid for most industrial units becauseof the high Peclet number. We will discuss this model in some detail, notonly because of its importance but also because the techniques used tohandle these two-point boundary-value differential equations are similarto that used for other diffusion–reaction problems (e.g., catalyst pellets) aswell as countercurrent processes and processes with recycle. The analyticalanalysis as well as the numerical techniques for these systems are verysimilar to this axial dispersion model for tubular reactors.

4.3.1 Formulation and Solution Strategy for the AxialDispersion Model

The axial dispersion of mass is described by Fick’s law:

Ni ¼ �Di

dCi

dl

where, Ni is the mass flux of component i (in mol=cm2 s), Di is the diffusivityin component i (in cm2=sÞ, Ci is the concentration of component i (inmol=cm3), and l is the length (in cm).

The axial dispersion of heat is described by Fourier’s law:

q ¼ ��dT

dl

where q is the heat flux (in J=cm2 sÞ, � is the thermal conductivity (inJ=cm s K), and T is the temperature (in K).

We will demonstrate the introduction of the axial dispersion of massand heat for a single reaction in the tubular reactor operating at steady statewith the generalized rate of reaction per unit volume r0

� �.

In Figure 4.10, the mass balance for the convective and diffusion flowsare shown. It should be noted that ni is the convective flow, Ni is the diffu-

sion flow, and A is the cross-sectional area of the reactor tube. The steady-state mass balance with axial dispersion gives

ni þ�ni þ At Ni þ�Nið Þ ¼ ni þ AtNi þ siAt�lr0

Dividing the equation by �l and taking the limit �l ! 0, we get

dnidl

þ At

dNi

dl¼ siAtr

0 ð4:10Þ

This equation can be put in a dimensionless form after introducing simpleassumptions (constant volumetric flow rate q). With this assumption we canwrite

qdCi

dlþ At

dNi

dl¼ siAtr

0

Now, we use Fick’s law,

Ni ¼ �Di

dCi

dl

Assuming Di to be constant, we get

dNi

dl¼ �Di

d2Ci

dl2

and the mass balance design equation becomes

qdCi

dl� AtDi

d2Ci

dl2¼ siAtr

0

Dividing by At gives

vdCi

dl�Di

d2Ci

dl2¼ sir

0

where, v is the velocity of flow ¼ q=Atð Þ.Defining a dimensionless length

o ¼ l

L

Figure 4.10 Convective and diffusion flows.

where L is the total length of the tubular reactor, we get

v

L

dCi

do� Di

L2

d2Ci

do2¼ sir

0

On rearranging, we get,

1

PeM

d2Ci

do2� dCi

doþ si

L

vr0 ¼ 0

where PeM is the Peclet number for mass ¼ Lv=Dið Þ.Applying the same procedure for heat balance with axial dispersion of

heat, we getXniHi þ AtqþQ0�l ¼

XniHi þ�ðniHiÞ½ � þ Atðqþ�qÞ

The above equation can be rearranged to giveX d niHið Þdl

þ At

dq

dl¼ Q0

Differentiating the product niHi in the first term givesXnidHi

dlþX

Hi

dnidl

þ At

dq

dl¼ Q0 ð4:11Þ

Equation (4.10) can be rewritten as

dnidl

¼ siAtr0 � At

dNi

dl

Substitute this equation in the heat balance design equation (4.11), weobtainX

nidHi

dlþX

Hi siAtr0 � At

dNi

dl

� �þ At

dq

dl¼ Q0

which can be rearranged to the following form:

XnidHi

dlþ Atr

0XsiHi � At

XHi

dNi

dl

� �� At�

d2T

dl2¼ Q0

After some approximations (for one-phase systems), as was used in previouschapters, we can approximate the above equation to the following form:

ntC0pmix

dT

dlþ Atr

0 �Hð Þ � At

XHi

dNi

dl� At�

d2T

dl2¼ Q0

where C 0pmix

is the average molar specific heat of the mixture and ni is theaverage total molar flow rate. This equation can also be written in thefollowing form:

At�d2T

dl2� qrCpmix

dT

dlþ Atr

0 ��Hð Þ þ At

XHi

dNi

dl� ¼ �Q0 ð4:12Þ

where Cpmixis the average mass specific heat for the mixture, q is the average

volumetric flow rate, and r is the average density.

A Problematic Term Usually Neglected in All Books Without EverMentioning Anything About It

In Eq. (4.12), the term At

PHidNi=dlÞ is the most problematic. It represents

the enthalpy carried with mass axial dispersion. We make the followingapproximations in order to handle this term (without blindly neglecting it):

The First and Simplest Consider that the term At

PHidNi=dl, accounts

for the heat transferred with the axial dispersion of mass to be accountedfor through a small empirical correction in � ! �e.

Second and Less Simple The term At

PHidNi=dl can be written as

follows:

�At

XHiDi

d2Ci

dl2

which gives approximately

�AtDav

XHi

d

dl

dCi

dl

� �ð4:13Þ

From the mass balance equation for plug flow (an assumption),

qdCi

dl¼ siAtr

0


dCi

dl¼ At

q

� �sir

0 ð4:14Þ

Using Eq. (4.14) in Eq. (4.13) gives

�AtDav

XHi

d

dl

At

q

� �sir

0� �


�AtDav

XHi

At

q

� �si

dr0

dl

It reduces to

�AtDav

At

q

� �dr0

dl

XsiHi

ReplacingP

siHi with �H gives

�A2t Dav

q

dr0

dl�Hð Þ ð4:15Þ

From Eq. (4.15), we reach at the following expression for this problematicterm:

At

XHi

dNi

dl¼ A2

t Dav

q��Hð Þ dr

0

dl

Using the above expression in the heat balance equation gives

At�d2T

dl2� qrC 0

pmix

dT

dlþ Atr

0 ��Hð Þ þ A2t Dav

q��Hð Þ dr

0

dl¼ �Q0


At�d2T

dl2� qrC 0

pmix

dT

dlþ At ��Hð Þ r0 þ At

q

� �|fflffl{zfflffl}1=v

Dav

dr0

dl

26664

37775 ¼ �Q0 ð4:16Þ

Dividing the equation by At gives

�d2T

dl2� vrC 0

pmix

dT

dlþ ��Hð Þ r0 þDav

v

dr0

dl

� �¼ �Q0

For most systems,

r0 >>>Dav

v

dr0

dl

So for this case, we can rewrite the above equation as

�d2T

dl2� vrC 0

pmix

dT

dlþ r0 ��Hð Þ ¼ �Q0 ð4:17Þ

Note: For cases where r0 is not much larger than ðDav=vÞðdr0=dlÞ, thisterm can not be neglected and thus must be included in the equation.

Using the dimensionless form o ¼ l=L in Eq. (4.17), we get

�

L2

d2T

d l=Lð Þ2 �vrC 0

pmix

L

dT

d l=Lð Þ þ r0 ��Hð Þ ¼ �Q0

Multiplying both sides by L=vrC 0pmix

gives

�

L2

L

vrC 0pmix

d2T

do2� vrC 0

pmix

L

L

vrC 0pmix

dT

doþ r0 ��Hð Þ L

vrC 0pmix

¼ �Q0 L

vrC 0pmix

which can be reorganized into the following form:

�

LvrC 0pmix

d2T

do2� dT

doþ r0 ��Hð Þ L

vrC 0pmix

¼ �Q0 L

vrC 0pmix

Let

�

LvrC 0pmix

¼ 1

PeH

where PeH is the dimensionless Peclet number for heat transfer, and we get

1

PeH

d2T

do2� dT

doþ ��Hð ÞL

rC 0pmix

1

vr0 ¼ �Q

where

Q ¼ Q0 L

vrC 0pmix

Now, let us consider that the rate of reaction is for a simple first-orderirreversible reaction is

r0 ¼ k0e� E=RTð ÞCA

and let us define dimensionless temperature and concentration as follows:

y ¼ T

Tf

(dimensionless temperature)

and

xA ¼ CA

CAf

(dimensionless concentration)

Using the rate of reaction and the dimensionless temperature and concen-tration, we get

1

PeH

d2y

do2� dy

doþ ��Hð ÞCAf

rC 0pmix

Tf|fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl}

Dae�ð�=yÞxA ¼ � Q

Tf

which can be simplified as

1

PeH

d2y

do2� dy

doþ Dae

�ð�=yÞxA ¼ �QQ ð4:18Þ

Here,

¼ ��Hð ÞCAf

rC 0pmix

Tf

(thermicity factor)

Da ¼k0L

v(Damkohler number)

and

QQ ¼ Q

Tf

Similarly, the mass balance design equation for this case is

1

PeM

d2xAdo2

� dxAdo

�Dae�ð�=yÞxA ¼ 0 ð4:19Þ

Note that both mass and heat balance design equations (4.18) and (4.19) aretwo-point boundary-value differential equations. Thus, each one of themrequires two boundary conditions. These boundary conditions are derivedas shown in this solved example (Example 4.1). In Example 4.1, the bound-ary conditions are developed for the mass balance design equation. For thenonisothermal case, the boundary conditions for the heat balance designequations are left as an exercise for the reader.

Solved Example 4.1 Axial Dispersion Model

In this solved example, we present the development of the isothermal modeland its boundary conditions for a case where the equation is linear and canbe solved analytically. Also presented is another case, where the model isnonlinear and we describe its numerical solution using Fox’s iterativemethod.

For a steady-state, isothermal, homogeneous tubular reactor considerthe simple reaction

A ! B

taking place in a tubular reactor; axial dispersion is not negligible (Pecletnumber Pe ¼ 15:0). Answer the following design questions:

A. If the reaction is first order, what is the value of Da (Damkohlernumber) to achieve a conversion of 0.75 at the exit of the reactor?

B. If the reaction is second order and the numerical value of Da

(Damkohler number) is the same as in part 1 (although its defini-tion is slightly different), find the exit conversion using the Fox’siterative method (explain your formulation of adjoint equationsfor the iterative solution of the nonlinear two-point boundary-value differential equation).

Solution (Fig. 4.11)

The mass balance design equation is

qCA þ AtðNAÞ ¼ qðCA þ�CAÞ þ AtðNA þ�NAÞ þ At�lkCnA

where n is the order of reaction. Canceling similar terms from both sides ofthe equation gives

0 ¼ q�CA þ At�NA þ At�lkCnA

Dividing the equation by �l and taking the limit as �l ! 0 gives

0 ¼ qdCA

dlþ At

dNA

dlþ AtkC

nA ð4:20Þ

From Fick’s law for diffusion we have

NA ¼ �DA

dCA

dl

which, upon differentiation, gives

dNA

dl¼ �DA

d2CA

dl2ð4:21Þ

Figure 4.11 Axial dispersion mass flow.


0 ¼ qdCA

dl� AtDA

d2CA

dl2þ AtkC

nA

Divide both sides by At and note that v, the velocity of flow, equals q=At, weobtain

0 ¼ vdCA

dl�DA

d2CA

dl2þ kCn

A

We reorganize the equation to get

DA

d2CA

dl2� v

dCA

dl� kCn

A ¼ 0

We define the following dimensionless variables:

o ¼ l

L(dimensionless length, where L is the total length of the

reactor tube)

x ¼ CA

CAf

(dimensionless concentration, where CAf is the feed

concentration of AÞ

Using these dimensionless variables, we get

DA

L2

d2 CA

�CAf

� �d l=Lð Þ2 � v

L

d CA

�CAf

� �d l=Lð Þ � kCn�1

Af

CnA

CnAf

¼ 0

Multiplying the equation with L and dividing it by v we obtain

DA

Lv

d2x

do2� dx

do� kLCn�1

Af

v

!xn ¼ 0

Define Pe ¼ Lv=DA and Da ¼ kLCn�1Af =v to get

1

Pe

d2x

do2� dx

do�Dax

n ¼ 0 ð4:22Þ

where

for the first-order reaction

Da ¼kL

v

for the second-order reaction

Da ¼kLCAf

v

Boundary Conditions

At the exit,

o ¼ 1;dx

do¼ 0 ð4:23Þ

At the entrance,

o ¼ 0;1

Pe

dx

do¼ x� 1 ð4:24Þ

Boundary condition (4.24) can be obtained as shown in Figure 4.12.A mass balance at the entrance at l ¼ 0 gives

qCAf ¼ qCA þ AtNA

Dividing by At and using Fick’s law, we get

vCAf ¼ vCA �DA

dCA

dl

Rearranging in dimensionless form gives

1 ¼ x�DA

Lv

dx

d!

which can be written in its more popular form

1

Pe

dx

do¼ x� 1

Thus, the model equations are as follows

A. For the First-Order Reaction

1

Pe

d2x

do2� dx

do�Dax ¼ 0 ð4:22Þ

Figure 4.12 Mass flow at the entrance.

The boundary conditions are as follows:

At o ¼ 1;

dx

do¼ 0 ð4:23Þ

At o ¼ 0;

1

Pe

dx

do¼ x� 1 ð4:24Þ

The present problem is a design problem because the required conversion isgiven and it is required to find the value of Da. However, for educationalpurposes, before we present the solution of this design problem, we will firstpresent the simulation problem when Da is given and the conversion isunknown. For illustration, we will take Da ¼ 1:4. The solution steps forthis simulation problem is as follows:

1. Solution of the second-order differential equation (4.22)Equation (4.22) can be rewritten as:

d2x

do2� Peð Þ dx

do� PeDað Þx ¼ 0

This is a linear second-order ordinary differential equation whichcan be solved using the operator method (see Appendices B andC).

Thus, the characteristic equation is

�2 � Pe�� PeDað Þ ¼ 0

The roots of this characteristic equation is obtained as:

�1;2 ¼ Pe�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiPe2 þ 4PeDa

p2

Thus, for Pe ¼ 15:0 and Da ¼ 1:4, the characteristic roots (eigen-values) �1 and �2 are given by

�1;2 ¼15�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi152 þ 4 15ð Þ 1:4ð Þ

p2

which simplifies to

�1;2 ¼15� 17:58

2

We get �1 ¼ 16:29 and �2 ¼ �1:29:

The solution of this second-order differential equation has thefollowing general form:

x ¼ C1e�1o þ C2e

�2o

For the present specific case, the dimensionless concentration x isgiven by

x ¼ C1e16:29o þ C2e

�1:29o

The constants C1 and C2 can be calculated using the boundaryconditions.

The differential of x is given by

dx

do¼ C1�1e

�1o þ C2�2e�2o

First, the boundary condition at o ¼ 1:0 is

dx

do¼ C1�1e

�1 þ C2�2e�2 ¼ 0

It gives

C1 16:29ð Þe16:29 ¼ �C2 �1:29ð Þe�1:29

Further simplification yields

C2 ¼ 5:4488� 108� �

C1

The second boundary condition at o ¼ 0 is

1

Pe

dx

do¼ x� 1: Therefore,

1

PeðC1�1 þ C2�2½ Þ ¼ ðC1 þ C2Þ � 1

which gives

1

1516:29ð ÞC1 þ �1:29ð Þ 5:448� 108C1

� � �¼ C1 þ 5:448� 108C1

� � �� 1

Thus we can obtain C1 as:

C1 ¼ 16:9� 10�10

and because

C2 ¼ 5:4488� 108� �

C1

C2 is obtained as

C2 ¼ 0:9208

Thus, the solution is

x ¼ 16:9� 10�10� �

e16:29o þ ð0:9208Þe�1:29o

At the exit, o ¼ 1:0,

xðo ¼ 1:0Þ|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl}at exit

¼ 16:9� 10�10� �

e16:29 þ ð0:9208Þe�1:29

giving


¼ 0:27248

As x ¼ CA=CAf , the conversion is

CAf �CA

CAf

¼ 1� x

so the value of conversion at the exit for Da ¼ 1:4 and Pe ¼ 15:0is

Conversion ¼ 1� 0:27348 ¼ 0:7265

Note that we were just lucky that we took the value of Da ¼ 1:4,which gave us the conversion close to 0:75 in the design problem.This is just luck (do you know why?).

2. The design problemDa is unknown, but xðo ¼ 1:0Þ at the exit is 0.25 (because thegiven conversion is 0.75). In this case, the same procedure isfollowed, but it is slightly more difficult and lengthier. Why?

All the steps are the same as above until we reach the calcula-tion of �1 and �2 with Da unknown; thus, we have

�1;2 ¼15�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi152 þ 4 15ð ÞDa

p2

giving

�1;2 ¼ 7:5�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi56:25þ 15Da

pThus, �1 and �2 are given by

�1 ¼ 7:5þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi56:25þ 15Da

pand �2 ¼ 7:5�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi56:25þ 15Da

p

Now, we have


�2o

and

dx

do¼ C1�1e

�1o þ C2�2e�2o

The first boundary condition (at o ¼ 1:0) is

dx

do¼ C1�1e

�1 þ C2�2e�2 ¼ 0

Therefore,

C2 ¼ �C1�1e�1

�2e�2

ð4:25Þ

The other boundary condition is at ðo ¼ 0Þ1

Pe

dx

do¼ x� 1


1

15C1�1 þ C2�2ð Þ ¼ C1 þ C2ð Þ � 1 ð4:26Þ

Substituting Eq. (4.25) in Eq. (4.26), we get

1

15C1�1 �

C1�1e�1

�2e�2

� ��2

� �¼ C1 �

C1�1e�1

�2e�2

� �� 1

which gives

C1

�115

� �115

� �e�1��2

� �¼ C1 1� �1

�2e�1��2

� �� 1

Further rearrangement gives

C1

�115

� 1

� �þ �1

�2e�1��2

� �1� �2

15

� �� ¼ �1

So we can write

C1 ¼ � �115

� 1

� �þ �1

�2e�1��2

� �1� �2

15

� �� 1

Because

�1 � �2 ¼ 7:5þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi56:25þ 15Da

p� �� 7:5�


p� �

we have

�1 � �2 ¼ 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi56:25þ 15Da

p¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi225þ 60Da

pThus, C1 is given as

C1 ¼� 7:5þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi56:25þ 15Da

p15

� 1

!"

þ 7:5þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi56:25þ 15Da

p7:5�


peð225þ60DaÞ

12

0@

1A

� 1� 7:5�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi56:25þ 15Da

p15

!#�1

ð4:27Þ

We can write Eq. (4.27) as

C1 ¼ F1 Dað Þand from Eq. (4.25), we have

C2 ¼ �C1

�1�2

e�1��2

Therefore,

C2 ¼ �F1 Dað Þ 7:5þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi56:25þ 15Da

p7:5�


p effiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi225þ60Da

p


C2 ¼ �F2 Dað Þwe return to the dimensionless concentration equation


�2o

which at exit ðo ¼ 1:0Þ can be written as


¼ C1e�1 þ C2e

�2

Giving


¼ F1 Dað Þ e7:5þð56:25þ15DaÞ12|fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl}

f1ðDaÞþF2 Dað Þ e7:5�ð56:25þ15DaÞ

12|fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl}

f2ðDaÞ

Now, for 75% conversion, the equation becomes

0:25 ¼ F1 Dað Þf1 Dað Þ þ F2 Dað Þf2 Dað Þwhich can be written as

F1 Dað Þf1 Dað Þ þ F2 Dað Þf2 Dað Þ � 0:25|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}FðDaÞ

¼ 0

Finally, we have this single equation in terms of Da:

F Dað Þ ¼ 0

This last equation is a nonlinear algebraic equation in Da; it canbe solved using the Newton–Raphson method [if the differentia-tion is very lengthy and cumbersome, which is the case here, thenin the Newton–Raphson method, you can use the modifiedNewton–Raphson by approximating @F=@Dað Þn by Fn�1 � FnÞ=�ðDn�1

a �DnaÞ or, easier and sure to converge, use the bisectional

method, which is straightforward and sure to converge andis applicable to this case, which is a single equation F insingle variable Da]. For Newton–Raphson, modified Newton–Raphson, and Bisectional methods, see Appendix B.

Hint: If all these very elementary techniques are too painful for thereader, then just plot FðDaÞ versus Da for different values of Da and findthe value of Da for which FðDaÞ ¼ 0.

B. For the Second-Order Reaction and Fox’s Iterative Method for Nonlinear

Two-Point Boundary-Value Differential Equation

For the second-order reaction, the equation is

1

Pe

d2x

do2� dx

do�Dax

2 ¼ 0

with

Da ¼kLCAf

v

where Da is the value obtained from the previous part (it will be close to orgreater than 1.4) and the Peclet number is

Pe ¼ 15:0

The boundary conditions are as follows:

At o ¼ 1

dx

do¼ 0

At o ¼ 0

1

Pe

dx

do¼ x� 1

For the numerical solution of this nonlinear equation, the first step is to putthe second-order differential equation in the form of two first-order differ-ential equations. Let

x ¼ x1

so

dx

do¼ dx1

do

and let

dx1do

¼ x2

so

d2x

do2¼ d2x1

do2¼ dx2

do

The second-order differential equation can now be rewritten as the set of thefollowing two first-order differential equations:

1

Pe

dx2do

� x2 �Dax21 ¼ 0

and

dx1do

¼ x2

Thus, the equivalent two first-order differential equations are

dx1do

¼ x2

dx2do

¼ Pex2 þ PeDax21

with the boundary conditions:

At o ¼ 1

x2ð1Þ ¼ 0

At o ¼ 0

x2ð0Þ ¼ Pe x1ð0Þ � 1½ �:

If we want to start the integration using the marching technique(Euler, Runge–Kutta, or using any subroutine from Polymath, Matlab, orIMSL libraries for the solution of initial-value differential equations, seeAppendix B), then we need to assume x1ð0Þ or x2ð0Þ. Suppose we chooseto assume x1ð0Þ ¼ xn1ð0Þ, where n refers to the nth iteration (for the initialguess, it will be n ¼ 1). Then, xn2ð0Þ can be easily and directly computed fromthe relation

xn2ð0Þ ¼ Pe xn1ð0Þ � 1½ Þ�Now, knowing xn1ð0Þ and xn2ð0Þ, we can integrate forward using any of thesubroutines until we reach o ¼ 1:0, but we will find that xn2ð1Þ 6¼ 0. So, ourinitial guess or choice is not correct and we have to choose another value,xnþ11 ð0Þ, and then try again and so on until we find a guess xnþ1

1 ð0Þ, which,after integration until o ¼ 1:0, will give us xnþ1

2 ð1Þ ¼ 0.

How can we accelerate this process? We can use Fox’s method (which is amodified Newton–Raphson method for two-point boundary-value differentialequations).

For a certain iteration xn1ð0Þ, the next iteration xnþ11 ð0Þ can be obtained

using the following process.Our objective is to have

x2ð1Þ ¼ f x1ð0Þð Þ ¼ 0

However, for any guess of xn1ð0Þ, we will not have this satisfied. Actually, wewill have f n xn1ð0Þð Þ 6¼ 0. Then we expand the function f , which is a functionof x 0ð Þ using Taylor series expansion, as follows:

f nþ1 xnþ11 ð0Þ� � ¼ f n xn1ð0Þð Þ þ @f xn1ð0Þð Þ

@x1ð0Þ� �n

xnþ11 ð0Þ � xn1ð0Þ

� �þ � � �

Note: n and nþ 1 are numbers of the iteration (not exponentials!)

Our objective is that

f nþ1 xnþ11 ð0Þ� � ¼ 0

and f n xn1ð0Þð Þ is calculated from the integration at o ¼ 1:0 as xn2ð1Þ.Therefore, from the Taylor series expansion, we get

0 ¼ xn2ð1Þ þ@x2ð1Þ@x1ð0Þ� �n


�� þNeglect higher-order terms

Rearrangement gives

xnþ11 ð0Þ ¼ xn1ð0Þ �

xn2ð1Þ@x2ð1Þ=@x1ð0Þð Þn

In the above relation, if we calculate @x2ð1Þ=@x1ð0Þð Þn, then the iteration cango on. How do we calculate this differential @x2ð1Þ=@x1ð0Þð Þn?We will do that through the following steps:Define a variable (called adjoint variable)

y1ðoÞ ¼@x2ðoÞ@x1ð0Þ

� �

which leads to

@x2ð1Þ@x1ð0Þ� �

¼ y1ð1:0Þ

we also define

y2ðoÞ ¼@x1ðoÞ@x1ð0Þ

� �

How can we compute y1ðoÞ so that we can get xnþ11 ð0Þ? This can be achieved

by formulating adjoint equations for y1ðoÞ that are solved simultaneouslywith the x1 and x2 differential equations. However, the adjoint equationsmust be initial-value differential equations. The method of doing thisfollows:Because

dx1ðoÞdo

¼ x2ðoÞ

if we differentiate both sides of the above equation partially with respect tox1ð0Þ, we get

d @x1ðoÞ�@x1ð0Þ

�do

¼ @x2ðoÞ@x1ð0Þ

By the definition of y1ðoÞ and y2ðoÞ given above, we get

dy2ðoÞdo

¼ y1ðoÞ

which is the first adjoint equation. To get the second adjoint equation, welook at the second differential equation of the state variable x2. Because

dx2ðoÞdo

¼ Pex2ðoÞ þ PeDax21ðoÞ

differentiating the above equation with respect to x1ð0Þ givesd @x2ðoÞ

�@x1ð0Þ

�do

¼ Pe@x2ðoÞ@x1ð0Þ

þ PeDa

@x21ðoÞ@x1ð0Þ


dy1ðoÞdo

¼ Pe y1ðoÞð Þ þ PeDa

@x21ðoÞ@x1ðoÞ

@x1ðoÞ@x1ð0Þ

giving

dy1ðoÞdo

¼ Pey1ðoÞ þ 2PeDax1ðoÞy2ðoÞ

which is the second adjoint equation.The initial conditions at o ¼ 0 for the adjoint variables are

y1ð0Þ ¼@x2ð0Þ@x1ð0Þ

From the boundary condition at o ¼ 0, we have

x2ð0Þ ¼ Pe x1ð0Þ � 1½ �Differentiation gives

@x2ð0Þ@x1ð0Þ

¼ Pe ¼ y1ð0Þ

and


¼ 1

Now the problem is ready for the iterative solution.The system equations and the boundary conditions are

dx1do

¼ x2

and

dx2do

¼ Pex2 þ PeDax21

With the following boundary conditions:At o ¼ 1

x2ð1Þ ¼ 0

At o ¼ 0

x2ð0Þ ¼ Pe x1ð0Þ � 1ð Þ

The adjoint equations and their initial conditions are

dy2ðoÞdo

¼ y1ðoÞ

and

dy1ðoÞdo

¼ Pey1ðoÞ þ 2PeDax1ðoÞy2ðoÞ

with the initial conditions at o ¼ 0 of

y1ð0Þ ¼ Pe and y2ð0Þ ¼ 1:0

The iterative formula is

xnþ11 ð0Þ ¼ xn1ð0Þ �

xn2ð1Þyn1ð1Þ

Important Note: The iterative solution for this two-point boundary-valuedifferential equation utilizing the adjoint equations is shown in Figure4.13. Notice that this procedure may be unstable numerically and we mayneed to reverse the direction of the integration from the exit of the reactortoward the inlet as explained in the following subsection.

4.3.2 Solution of the Two-Point Boundary-Value DifferentialEquations and Numerical Instability Problems

Consider the axial dispersion model

1

Pe

d2x

do2� dx

do�Dax

2 ¼ 0 ð4:28Þ

with the following boundary conditions:

At o ¼ 1

dx

do¼ 0

At o ¼ 0

1

Pe

dx

do¼ x� 1

The iterative procedure as explained earlier is to put the equation in theform of two first-order differential equations:

dx1do

¼ x2

Figure 4.13 Solution scheme for the two-point boundary-value differential equa-

tions of the isothermal nonlinear axial dispersion model.

dx2do

¼ Pex2 þ PeDax21


At o ¼ 1

x2ð1Þ ¼ 0

At o ¼ 0

x2ð0Þ ¼ Pe x1ð0Þ � 1ð Þ ð4:29ÞIf we employ forward integration and iteration, then following twoproblems may arise:

1. You guess x1ð0Þ with its errors; then, x2ð0Þ is defined automati-cally from the boundary condition (4.29). For a large value ofPe(>5.0) the error in x1ð0Þ propagates to x2ð0Þ amplified by Pe.Thus, the x1ð0Þ error is clearly amplified.

2. Consider the linear version of this equation and inspect thecharacteristic roots (eigenvalues) of the differential equation.The characteristic equation for this linear case is

�2 � Pe��DaPe ¼ 0

giving the characteristic roots (eigenvalues)

�1;2 ¼Pe�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiPe2 þ 4DaPe

p2

This leads to one negative eigenvalue and one positive eigenvalue:

�1 ¼Peþ


p2

which is positive

�2 ¼Pe�


p2

which is negative

This is illustrated as follows:

Case 1

For Pe ¼ 15.0 and Da ¼ 1:5, the eigenvalues are

�1 ¼15þ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi152 þ 4 1:5ð Þ 15ð Þ

p2

¼ 16:37

�2 ¼15�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi152 þ 4 1:5ð Þ 15ð Þ

p2

¼ �1:374

The solution of the differential equation is

x1 ¼ C1e16:37o þ C2e

�1:374o

Note that �1�� >> �2

�� ; e16:37o indicates strong instability in the o direction

and e�1:374o indicates stabilization in the o direction. For such a case, it isvery likely to have numerical instability.

Case 2

For Pe ¼ 0.1 and Da ¼ 1:5, the eigenvalues are

�1 ¼ 0:88|fflfflfflfflfflffl{zfflfflfflfflfflffl}Positive

and �2 ¼ �0:681|fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl}Negative

Note that �1�� and �2

�� are very comparable; therefore, numerical instabilityis less likely. Thus, it is clear that the integration in the positive directionfrom o ¼ 0 to o ¼ 1:0 is not very wise. The following backward integrationis much more stable and avoids the complexities due to the above tworeasons.

Backward Integration. Consider again the differential equations

dx1do

¼ x2

dx2do

¼ Pex2 þ PeDax21


At o ¼ 1

x2ð1Þ ¼ 0

At o ¼ 0;

x2ð0Þ ¼ Pe x1ð0Þ � 1ð ÞWe use the following simple transformation from o to o0 in order to achievebackward integration:

o0 ¼ 1� o

Thus, at o ¼ 1:0;o0 ¼ 0, and at o ¼ 0;o0 ¼ 1. The differential equationsbecome

� dx1do0 ¼ x2

� dx2do0 ¼ Pex2 þ PeDax

21

Now, the set of differential equations by its boundary conditions can bewritten as

dx1do0 ¼ �x2 ð4:30Þ

dx2do0 ¼ �Pex2 � PeDax

21 ð4:31Þ

with the following boundary conditions

At o0 ¼ 0

x2ð0Þ ¼ 0 ð4:32ÞAt o0 ¼ 1

x2ð1Þ ¼ Pe x1ð1Þ � 1½ � ð4:33ÞNow, the iteration proceeds by assuming that x1ð0Þ ðx1 at o0 ¼ 0,which is at the exitÞ and the function F x1ð0Þð Þ should be zero at o0 ¼ 1,which means that

x2ð1Þ � Pe x1ð1Þ � 1ð Þ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}F x1ð0Þð Þ

¼ 0 ð4:34Þ

Thus, the Taylor series expansion of F x1ð0Þð Þ can be written as

Fnþ1 x1ð0Þð Þ|fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl}equal to zero

¼ Fn x1ð0Þð Þ þ @F

@ x1ð0Þð Þ� �n


� �

From Eq. (4.34), differentiating with respect to x1 0ð Þ gives@F

@ x1ð0Þð Þ� �n

¼ @x2ð1Þ@x1ð0Þ� �

� Pe@x1ð1Þ@x1ð0Þ� �

We define the following adjoint variables:

y1ðo0Þ ¼ @x1ðo0Þ@x1ð0Þ

� �

y2ðo0Þ ¼ @x2ðo0Þ@x1ð0Þ

� �

Differentiate the differential equation (4.30) with respect to x1ð0Þ to get

d @x1ðo0Þ�@x1ð0Þ �do0 ¼ � @x2ðo0Þ

@x1ð0Þ

which can be written in terms of the adjoint variables as follows:

dy1ðo0Þdo0 ¼ �y2ðo0Þ

Similarly, differentiating the differential equation (4.31) gives

d @x2ðo0Þ�@x1ð0Þ �do0 ¼ �Pe

@x2ðo0Þ@x1ð0Þ

� PeDa

@x21ðo0Þ@x1ð0Þ

which gives

dy2ðo0Þdo0 ¼ �Pe y2ðo0Þ �� PeDa

@x21ðo0Þ@x1ðo0Þ

@x1ðo0Þ@x1ð0Þ

giving

dy2ðo0Þdo0 ¼ �Pe y2ðo0Þ �� PeDa 2x1ðo0Þ� � @x1ðo0Þ

@x1ð0ÞFurther simplification gives

dy2ðo0Þdo0 ¼ �Pe y2ðo0Þ� �� PeDa 2x1ðo0Þ� �

y1ðo0Þ

The final form is

dy2ðo0Þdo0 ¼ �Pe y2ðo0Þ �� 2PeDax1ðo0Þy1ðo0Þ

The two adjoint equations are thus given as

dy1ðo0Þdo0 ¼ �y2ðo0Þ ð4:35Þ

dy2ðo0Þdo0 ¼ �Pe y2ðo0Þ� �� 2PeDax1ðo0Þy1ðo0Þ ð4:36Þ

The boundary conditions at o0 ¼ 0 for these adjoint equations are


¼ 1 ð4:37Þ


¼ 0 ð4:38Þ

The reader can use model equations (4.30)–(4.33) and these adjointequations (4.35)–(4.38) to establish the iteration procedure as discussedearlier.

Solved Example 4.2

Two CSTRs in series are operated isothermally. The reaction is first orderand the two tanks have equal volumes.

(a) Formulate a steady-state mathematical model for this process.(b) Choose volumes of the two reactors and temperatures (isother-

mal and equal for both CSTRs) so that the conversion from thesecond CSTR is 0.78. The feed concentration is given as 0.45 mol/m3, and the first-order reaction rate constant k is given byk ¼ k0 e

�E=RT , where k0 ¼ 2:9� 107 h�1 and E=R ¼ 5300 K.(c) Formulate a steady-state mathematical model for the process

when the second reactor is a distributed system (tubular, plugflow unit).

(d) Find the process exit concentration for part (c), with all para-meters being the same as part (b).

(e) Derive an unsteady-state model for case (a) (two CSTRs).


Consider the reaction as

A ! B

with the generalized rate of reaction as

r0j ¼ k0 e�E=RTj CAj

where j ¼ 1 and 2 (the two reactors).

Part (a)

The steady-state mathematical model is

nA 1¼ nAf

þ sA V1 r01 ð4:39Þ

nA 2¼ nA 1

þ sA V2 r02 ð4:40Þ

Figure 4.14 Mass flow diagram for CSTRs in series.

Part (b)

Given

V1 ¼ V2 ¼ V and T1 ¼ T2 ¼ T

and

r01 ¼ k0 e�E=RT1 CA 1

and r02 ¼ k0 e�E=RT2 CA 2

From Eqs (4.39) and (4.40), we get

qCA 1¼ qCAf

� V k0 e�E=RT CA 1

ð4:41ÞqCA 2

¼ qCA 1� V k0 e

�E=RT CA 2ð4:42Þ

Given k0 ¼ 2:9� 107 h�1 and E=R ¼ 5300K, the overall conversion is

xA ¼ CAf� CA 2

CAf

On substituting the given values, we get

0:78 ¼ 0:45� CA 2

0:45

Thus,

CA 2¼ 0:099 mol=m3

Equations (4.41) and (4.42) can be rewritten respectively as

qCA 1¼ q 0:45ð Þ � V 2:9� 107

� �e� 5300=T CA 1

ð4:43Þq 0:099ð Þ ¼ qCA 1

� V 2:9� 107� �

e� 5300=T 0:099ð Þ ð4:44ÞNote that these are two equations and the number of unknowns is fourq;CA 1

;V;T� �

; thus, the degrees of freedom are 2. It means that we canchoose two of the four unknowns and use the two equations to get theother two unknowns. The choice of the two unknowns depends on ourengineering judgment. As an example, choose T ¼ 300 K and V ¼ 2 m3.Then, on solving Eqs. (4.43) and (4.44) simultaneously, we get q ¼ 1:093 m3

and CA1¼ 0:2115 mol=m3:

Part (c) (see Fig. 4.15)

The mass balance equation for the CSTR at steady state is,

nA 1¼ nAf

þ sA V1 r0

and for the plug flow reactor (PFR) it is

dnAdV

¼ sA r0

The above two equations can be rewritten as

qCA 1¼ qCAf

� V k0 e�E=RT CA 1

ð4:45Þ

qdCA

dV¼ � k0 e

�E=RT CA ð4:46Þ

with the initial condition at V ¼ 0

CA ¼ CA 1

[note that the value of CA 1here is the same as CA 1

in part (b)].

Part (d)

CA is the same as CA 1in part (b). Now, we have to simply integrate Eq.

(4.46) with the initial conditions to get the value of concentration at the exitof the PFR (volume of the PFR is equal to 2 m3):

dCA

dV¼ � k0 e

�E=RT=q|fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl}a¼constant

CA

with initial condition at V ¼ 0

CAjt¼0 ¼ 0:2115

It becomes

dCA

dV¼ � �CA

On integration, we get

lnCA ¼ ��V þM

Figure 4.15 Mass flow diagram for CSTR followed by a plug flow reactor.

where M is the constant of integration and its value can be found using theinitial condition. We get M ¼ �1:5535; thus, we get

lnCA ¼ ��V � 1:5535

To get the exit concentration after the PFR, we substitute V ¼ 2 into theabove relation to get

lnCA2¼ � 0:56395ð Þ 2ð Þ � 1:5535

where � ¼ 0:56395. On calculation, we get

CA2¼ 0:068 mol/m3

Part (e)

The unsteady-state model for the arrangement is shown in Figure 4.14. Forthe first CSTR,

dnA 1

dt|ffl{zffl}Unsteady-state term

þ nA 1¼ nAf

þ sA V1 r01


VdCA 1

dtþ qCA 1

¼ qCAf� V k0 e


Similarly for the second CSTR,

VdCA 2

dtþ qCA 2

¼ qCA 1� V k0 e


The initial conditions at t ¼ 0 are

CA1¼ CA1

0ð Þ and CA2¼ CA1

0ð Þ:

PROBLEMS

Problem 4.1

The hydrogenolysis of low-molecular-weight paraffins takes place in a tub-ular flow reactor. The kinetics of the propane reaction may be assumed to befirst order in propane in the regime of interest. From the data given in TableP4.1, determine the reaction rate constants at the indicated temperaturesand the activation energy of the reaction.

Feed ratio H2=C3H8 ¼ 2:0 in all casesReactor pressure ¼ 7.0 MPa in all cases

For this problem, the stoichiometry of the main reaction may be consideredto be of the form

H2 þ C3H8 ! CH4 þ C2H6

Problem 4.2

Consider the following homogeneous gas-phase reaction:

Aþ B ! C þD

The reaction is essentially ‘‘irreversible’’ and the rate of production of com-ponent A in a constant-volume batch reactor is given by

RA ¼ �kCACB

At the temperature of interest, k ¼ 100 m3/mol s. Compounds A and B areavailable in the off-gas stream from an absorption column at concentrationsof 21 mol/m3 each. 15 m3/s of this fluid is to be processed in a long iso-thermal tubular reactor. If the reactor is assumed to approximate a plugflow reactor, what volume of pipe is required to obtain 78% conversion ofspecies A?

Problem 4.3

Many of the techniques used by chemical engineers are also helpful in thefood processing industry. For example, consider the problem of sterilizingfood after it has been placed in cylindrical cans. Normally, it is assumed thatall harmful bacteria will be killed if the food temperature is raised to somevalue T1. The heating process is accomplished by placing the can in asterilization bath that is maintained at a high temperature T0. Develop anequation for the sterilization time. Also, by selecting various values for the

Table P4.1 Experimental Data

Temperature,

Tð8CÞSpace time

(s)

Fraction propane

converted

538

593

0

42

98

171

40

81

147

0

0.018

0.037

0.110

0.260

0.427

0.635

system parameters, see if you can determine whether or not it is necessary toconsider the resistance to heat transfer in a stagnant film surrounding thecan.

Problem 4.4

It would be possible to use a double-pipe heat exchanger as a reactor. For asingle, irreversible, exothermic reaction, this unit would have the greatadvantage that the heat generated by the reaction could be used to raisethe temperature of the reacting material, which would eliminate the need forany heating fluid. Derive a dynamic model for the system.

Problem 4.5

A simple first-order reaction

A ! B

takes place in a homogeneous tubular reactor. The reactor is isothermal andaxial dispersion is important, with a Peclet number Pe ¼ 20:0.

(a) Find the value of the Damkohler number Dað Þ so that the exitconversion is 0.8. Note that the equation and its two-pointboundary-value conditions are linear and therefore can be solvedanalytically (to get the characteristic equation, eigenvalues, etc.).You should be careful to note that this is a design equation, not asimulation one. Therefore, although the equations are linear, youwill need some nonlinear iterations to obtain the solution at theend.

(b) Construct the numerical solution procedure using Fox’s methodwhen the reaction is second order for the same numerical value ofthe Damkohler number you obtained in part (a).

(c) Repeat the same you did in part (a) but for the Peclet numbergoing once to zero and once to infinity. Compare and discussboth results.

Problem 4.6

Under appropriate conditions, compound A decomposes as follows:

A ! B ! C

The desired product is B and the rate constant for the first and secondreactions are equal to the following values:

k1 ¼ 0:1 min–1

k2 ¼ 0:05 min–1

The feed flow rate is 1000 L/h, and the feed concentration of the reactant Ais equal to CAf ¼ 1:4 mol/L, whereas CBf ¼ CCf ¼ 0.

(a) What size of PFR will maximize the yield of B, and what is theconcentration of B in the effluent stream from this optimumreactor? What will be the effect of axial dispersion if the Pecletnumber is equal to 10.0, and what is the effect when the Pecletnumber is equal to 500.0?

(b) What size of CSTR will maximize the yield of B, and what is theconcentration of B in the effluent stream from this optimizedCSTR?

(c) Compare and comment on the results of previous two parts.(d) What will be the effect if the reactors are nonisothermal, adia-

batic, and both the reactions are exothermic.

Problem 4.7

A consecutive reaction takes place in different reactor configurations. Bothreactions are first order. The rate constant for the first reaction is equal to 5min–1, and for the second reaction, it is equal to 2 min–1. The feed is purereactant. Calculate the following:

(a) For a batch reactor, the optimum operating time and maximumyield of the intermediate desired product

(b) For a CSTR, the optimum residence time and the maximum yieldof the intermediate desired product

(c) For a tubular reactor, the optimum residence time and maximumyield of the intermediate desired product. What is the effect ofaxial dispersion when the Peclet number is equal to 0.1 and whenit is equal to 1000.0?

(d) Compare the results obtained in previous parts and comment.(e) Redo parts (a)–(d) if both the reactions follow second-order

kinetics.

Problem 4.8

One approximate way to obtain the dynamics of a distributed system is tolump them into a number of perfectly mixed sections. Prove that a series ofN mixed tanks is equivalent to a distributed system as N goes to infinity.

Problem 4.9

In an isothermal batch reactor, 70% of a liquid reactant is converted in 13min. What space time and space velocity are needed to effect this conversionin a plug flow reactor?

Problem 4.10

Assuming a stoichiometry A ! R for a first-order gas reaction, we calculatethe size of plug flow reactor needed to achieve a required conversion of 99%of pure A feed to be 32 liters. In fact, however, the reaction stoichiometry isA ! 3R. With this corrected stoichiometry, what is the required reactorvolume?

Problem 4.11

Consider the packed-bed tubular reactor whose schematic diagram is shownin Figure P4.11. It is a hollow cylindrical tube of uniform cross-sectionalarea A, packed with solid catalyst pellets, in which the exothermic reactionA ! B is taking place. The packing is such that the ratio of void space to thetotal reactor volume—the void fraction—is known; let its value be repre-sented by e. The reactant flows in at one end at constant velocity v, and thereaction takes place within the reactor. Obtain a theoretical model that willrepresent the variation in the reactant concentration C and reactor tempera-ture T as a function of time and spatial position z. Consider that the tem-perature on the surface of the catalyst pellets TS is different from thetemperature of the reacting fluid and that its variation with time and posi-tion is also to be modeled.

Use the following assumptions:

1. The reacting fluid flows through the reactor with a flat velocityprofile (implying that there are no temperature, concentration, orvelocity variations in the radial direction).

Figure P4.11 Schematic diagram of problem 4.11.

2. Transfer of material and energy in the axial (i.e., z) direction is byconvective forces only; there is no diffusion in this direction.

3. The rate of reaction is first order with respect to the reactantconcentration, with a rate constant k. It does not depend expli-citly on position along the reactor; it only depends on reactantconcentration and temperature. The heat of reaction may betaken as ��H.

4. The heat transfer from the solid catalyst to the fluid is assumed tofollow Newton’s law of cooling; the area over which this heattransfer takes place is assumed to be a known constant AS.

5. The fluid density r and specific heat capacity Cp are constant; thesolid catalyst particles are all assumed to be identical, with iden-tical density rS and identical specific heat capacity CpS.

6. The catalyst packing is assumed to be uniform, so that across anycross section of the reactor, the number and arrangement ofparticles are identical. (This allows the use of arbitrarily locatedmicroscopic element for developing the model).

7. There are no heat losses to the atmosphere.

If you need to make any additional assumptions, state them clearly.

Problem 4.12

A plug flow reactor is to be designed to produce the product D from Aaccording to the following reaction:

A ! D; rate RD ¼ 60CA mol/L s

In the operating condition of this reactor, the following undesired reactionalso takes place:

A ! U; rate RU ¼ 0:003CA

1þ 105CA

mol/L s

The undesired product U is a pollutant and it costs $10/mol U to dispose ofit, whereas the desired product D has a value of $35/mole D. What sizereactor should be chosen in order to obtain an effluent stream at its max-imum value?

Pure reactant A with volumetric flow rate of 15 L/s and molar flowrate of 0.1 mol/s enters the reactor. Value of pure A is $5/mole A.

Problem 4.13

The vapor-phase cracking of acetone, described by the endothermic reaction

CH3COCH3 ! CH2COþ CH4

takes place in a jacketed tubular reactor. Pure acetone enters the reactor at atemperature of T0 ¼ 1030 K and a pressure of P0 ¼ 160 kPa, and the tem-perature of the external gas in the heat exchanger is constant atTe ¼ 1200 K. Other data are as follows:

Volumetric flow rate: q ¼ 0:003 m3=sVolume of the reactor: VR ¼ 1:0 m3

Overall heat transfercoefficient: U ¼ 110 W=m2K

Total heat transfer area: A ¼ 160 m2=m3reactorReaction constant: k ¼ 3:56 e 34200 1=1030�1=Tð Þ½ � s�1

Heat of reaction: �HR ¼ 80700þ 6:7 T � 298ð Þ�5:7� 10�3 T2 � 2982

� ��1:27� 10�6 T3 � 2983

� �J=mol

Heat capacity of acetone: CPA¼ 26:65þ 0:182T � 45:82

� 10�6T2 J=mol KHeat capacity of ketene: CPK

¼ 20:05þ 0:095T � 31:01� 10�6T2J=mol K

Heat capacity of methane: CPM¼ 13:59þ 0:076T � 18:82� 10�6T2J=mol K

Determine the temperature profile of the gas along the length of the reactor.Assume constant pressure throughout the reactor.

Problem 4.14

Formulate steady-state and dynamic models for the tubular heat exchangershown in Figure P4.14 (Hint: We will get a partial differential equation forthe dynamic model).

Figure P4.14 Tubular heat exchanger.

The following nomenclature is to be used:

Q ¼ amount of heat transferred from the steam to the liquid per unittime and per unit of heat transfer area

A ¼ cross-sectional area of the inner tubev ¼ average velocity of the liquid (assumed to be constant)D ¼ external diameter of the inner tube

List all of the assumptions made to derive the model and suggest analgorithm to solve the problem numerically.

5

Process Dynamics and Control

5.1 VARIOUS FORMS OF PROCESS DYNAMIC MODELS

The more logical manner to formulate dynamic mathematical models is toformulate it in terms of real time. However, sometimes it is convenient toexpress these dynamic models in a domain other than the real physical timedomain. These situation arise either due to ease of solution and equationmanipulation in another domain (the Laplace domain) or due to onlinemeasurement and control, as is shown in this chapter.

Therefore, the different dynamic models can be classified on a differentbasis than the previous classification.

A useful and interesting classification is the following:

1. State-space models

The main characteristics of these physical/rigorous models are asfollows:

. The state variables occur explicitly along with the input andoutput variables.

. The models are physically formulated from the process firstprinciples.

2. Input–output models

. They strictly relate only the input and output variables.

. They can occur in the Laplace or z-transform domain, or inthe frequency domain as well as in the time domain.

. In the Laplace or z-transform domain, these input–outputmodels usually occur in what is known as the ‘‘transformdomain transfer function’’ form; in the frequency domain,they occur in the ‘‘frequency-response (or complex vari-able)’’ form; in the time domain, they occur in the‘‘impulse-response (or convolution)’’ form.

. Input–output models can be obtained from the transforma-tions of the state-space models, but they can also beobtained directly from input–output data correlations.

The above two main classifications can be further detailed into the following:

1. The state-space (differential equations or difference equations)form

2. The transform-domain (Laplace or z-transform) form3. The frequency-response (complex variable) form4. The impulse-response (convolution) form

The four forms are related to each other as shown in the following sections.

5.2 FORMULATION OF PROCESS DYNAMIC MODELS

5.2.1 The General Conservation Principles

In the previous chapters, we presented the general material and heat balancedesign equations in full detail and have shown the assumptions that reducethese equations into a simple form. Here, we will use the simplified formdirectly.

We can express the balance equation as follows:

‘‘What remains accumulated within the boundaries of a system isthe difference between. What was added (input). What was taken out (output). What was generated by internal production, or disappeared by

internal consumption’’

In other, shorter terms, it is

Accumulation ¼ Input�Outputþ Internal production ðor �consumptionÞ

We can also express this in terms of rate:

‘‘The rate of accumulation of a conserved quantity q within theboundaries of a system is the difference between the rate at which

this quantity is being added to the system and the rate at which it isbeing taken out (removed) plus the rate of internal production, orminus the rate of consumption.’’

Rate of accumulation of q ¼Rate of input of q�Rate ofoutput of q þRate ofproduction of q

8<:

5.2.2 Conservation of Mass, Momentum, and Energy

Such balances could be made over the entire system, to give ‘‘overall’’ or‘‘macroscopic’’ balances, or they could be applied to portions of the systemof differential size, giving ‘‘differential’’ or ‘‘microscopic’’ balances.

Mass Balance

We can write the conservation of mass equation as

Rate of accumulation of mass ¼Rate input of mass�Rate outputof massþRate generation ofmass�Rate depletion of mass|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

net rate of generation of mass

If there are n components, one can formulate a mass balance for eachcomponent (n equations) plus one overall mass balance equation, butfrom the nþ 1 equations, only n equations are independent.

Note: The total mass balance equations will not have any generation/de-pletion terms, these will always be zero (as long as we do not have a nu-clear reaction changing mass into energy). Of course, the same does notapply to the molar balance equations when there are chemical reactionsassociated with a change in the number of moles, as shown earlier.

Momentum Balance

Momentum balance is based on Newton’s second law equation:

Rate of accumulationor momentum

¼Rate input of momentum� Rate outputof momentumþRate of forces acting onthe volume element|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

net rate of generation of momentum

‘‘Generation’’ of momentum must be due to forces acting on the volumeelement (over which the balance is carried out), and this is ‘‘done’’ at a rateequal to the total sum of these forces.

Energy Balance

Energy balance is based on the first law of thermodynamics:

Rate of accumulation of energy ¼

Rate input of energy �Rateoutput of energy þRate ofgeneration of energy �Rate ofexpenditure of energy|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

net rate of generation of energy

5.2.3 Constitutive Equations

The next step after the conservation equations is the introduction of explicitexpressions for the rates that appear in the balance equations. The followingare the most widely used:

1. Equations of the properties of matter. Basic definitions of mass,momentum, and energy in terms of physical properties such as r,CP, T , and so forth.

2. Transport rate equations

. Newton’s law of viscosity (for momentum transfer)

. Fourier’s heat conduction law (for heat transfer)

. Fick’s law of diffusion (for mass transfer)

3. Chemical kinetic rate equations

. Law of mass action

. Arrhenius expression for temperature dependence of reac-tion rate constants

4. Thermodynamic relations

. Equations of state (e.g., ideal gas law, van der Waal’s equa-tion)

. Equations of chemical and phase equilibria

Example of Mathematical Model

Consider the heating (with no chemical reaction) tank shown in Figure 5.1.The unsteady mass balance equation is given by

ni þdnidt

¼ nif

with no chemical reaction. The general heat balance equation is given by

Xnif Hif �Hir

� �þQ ¼X

ni Hi �Hirð Þ þ dP

Hi

� �dt

We can introduce the following assumptions:

1. The tank is well mixed T 0 ¼ T� �

.2. r (fluid density), Cp (fluid specific heat), and � (heat of vaporiza-

tion of the steam in the heating coil) are constant.3. All of the heat of condensation of steam is given to the liquid (no

heat accumulation in coils, stirrers, walls of tank, etc.).4. There are negligible heat losses to the atmosphere.5. There is no change of phase inside the tank.

Mass balance

Accumulation of mass is given by

d

dtrVð Þ

Which can be simplified as

rdV

dt(for constant density)

Mass input is given by

Fi ¼ rqi

Figure 5.1 Continuous heating tank.

where qi is the input volumetric flow rate. Mass output is given by

F ¼ rq

where q is the output volumetric flow rate. Thus,

rdV

dt¼ rqi � rq

which can be reduced to give

dV

dt¼ qi � q

For qi ¼ q, we get dV=dt ¼ 0, which means that V is constant.

Energy balance

. Change in kinetic and potential energies is negligible.

. T is the reference temperature.

The rate of accumulation of energy is given by

rCp

d

dtV T � Tð Þ½ �

The rate of heat input (feed) is equal to

rqiCp Ti � Tð ÞThe rate of heat output (outlet stream) is given by

rqCp T � Tð ÞThe rate of heat input (heating coil) is equal to

�rsqs

where qs is the steam volumetric flow rate in the steam heating coil and rs isthe steam density and � is the latent heat of steam. Thus, the overall energybalance will be given by

rCp

d

dtV T � Tð Þ½ � ¼ rqiCp Ti � Tð Þ þ �rsqs � rqCp T � Tð Þ

For qi ¼ q, and V ¼ constant, the above equation becomes

rVCp

dT

dt¼ rqCp Ti � Tð Þ þ �rsqs

This equation can be put in the following form:

dT

dt¼ rqCp

rVCp

Ti � Tð Þ þ �rsrVCp

qs

Now, we define the following normalized parameters:

y ¼ V

q¼ Residence time and b ¼ �rs

rVCp

Thus, the equation can be put in the following form:

dT

dt¼ � 1

yT þ bqs þ

1

yTi ð5:1Þ

The steady-state relation is given by

0 ¼ � 1

yTS þ bqsS þ 1

yTiS

Exercise

1. Solve the above equation for certain values of TiS and qsS toobtain TS; the subscript S denotes steady state.

2. Calculate the effect of the change in b, y, TiS, and qsS on thesteady-state temperature TS.

Deviation Variables

Equation (5.1) can be written in terms of the deviation variables (deviationfrom the steady state equations) in the following form

y ¼ T � TS

u ¼ qs � qsS

d ¼ Ti � TiS

Thus,

dy

dt¼ � 1

yyþ buþ 1

yd ð5:2Þ

If the process is initially at steady state, then y 0ð Þ ¼ 0.

5.2.4 The Laplace Transform Domain Models

Laplace transformation is a simple mathematical technique that transformsdifferential equations from the time domain tð Þ to the Laplace domain sð Þ.Some detail about Laplace transformation is given later in this chapter. Oneof the main characteristics of Laplace transformation is to transform deri-vatives in the time domain tð Þ to algebraic form in the Laplace domain [i.e.,dy tð Þ=dt ) s y sð Þ�. Using this property, we find that the Laplace transforma-tion of Eq. (5.2) gives

sy sð Þ ¼ � 1

yy sð Þ þ bu sð Þ þ 1

yd sð Þ

Rearranging to put the output y sð Þ on one side of the equation gives

sþ 1

y

� �y sð Þ ¼ bu sð Þ þ 1

yd sð Þ

Again, on rearranging, we obtain the following input–output relation:

y sð Þ ¼ by1þ ys

� �u sð Þ þ 1

1þ ys

� �d sð Þ

Now, we introduce the following definition of what we call ‘‘transfer func-tions’’ [gu sð Þ and gd sð Þ:

by1þ ys

� �¼ gu sð Þ

and

1

1þ ys

� �¼ gd sð Þ

Finally, we get,

y sð Þ ¼ gu sð Þu sð Þ þ gd sð Þd sð Þ

This is called the ‘‘transfer-function model,’’ where the output y sð Þ is relatedto the inputs u sð Þ and d sð Þ through a relation involving the transfer functionsgu sð Þ and gd sð Þ. This is the basis of block diagram algebra, which will bediscussed in a later section.

5.2.5 The Frequency-Response Models

This is still a transfer-function model, but in the frequency domain, wherewe introduce

s ¼ jo

where o is the frequency and j ¼ ffiffiffiffiffiffiffi�1p

. If we put jo instead of s in gu sð Þ, theLaplace transform becomes a Fourier transform gu joð Þ and we can convertthe transform domain model to a frequency-response model as shown inTable 5.1 for both gu sð Þ and gd sð Þ. Therefore, the frequency-domain modelbecomes

yð joÞ ¼ guð joÞuð joÞ þ gdð joÞdð joÞ

5.2.6 Discrete Time Models

The linear discrete single-input, single-output (SISO) processes are describedby xðkÞ, uðkÞ, dðkÞ, and yðkÞ, which are the state, control, disturbance, and

output variables, respectively, at the discrete time instant tk ¼ k�t, where�t is the sampling interval. Thus, the model is given by

xðkþ 1Þ ¼ axðkÞ þ buðkÞ þ gdðkÞand

yðkÞ ¼ cxðkÞThe multiple-input, multiple-output (MIMO) version is

Xðkþ 1Þ ¼ A XðkÞ þ B uðkÞ þ � dðkÞand

YðkÞ ¼ C XðkÞ

Table 5.1 The Transfer Function in the Frequency Domain

The gu sð Þ transfer function The gd sð Þ transfer function

Simply substitute s by jo to get

guð ðoÞ ¼y

jyoþ 1

It can be arranged in the form

guð ðoÞ ¼yð1� jyoÞ

jyoþ 1ð Þð1� jyoÞwhich gives

guð ðoÞ ¼yð1� jyoÞ

1� jyoþ jyo� j2y2o2


guð ðoÞ ¼y� jy2o

1þ y2o2

and, finally, can be divided into

the real part (Re) and the Imaginary

part (Im) to give

Re guðoÞ½ � ¼ y

1þ yoð Þ2

Im guðoÞ½ � ¼ �oy2

1þ yoð Þ2

Simply substitute s by jo to get

gd ððoÞ ¼1

jyoþ 1

It can be arranged in the form

gd ððoÞ ¼ð1� jyoÞ

jyoþ 1ð Þð1� jyoÞwhich can be rewritten as

gd ððoÞ ¼1� jyo

1þ y2o2

and, finally, can be divided into the

real part ðReÞ and the Imaginary part

ðImÞ to give

Re gd ð ðoÞ½ � ¼ 1

1þ yoð Þ2

Im gd ð ðoÞ½ � ¼ �oy

1þ yoð Þ2

The nonlinear discrete model is

XðkÞ ¼ f XðkÞ; uðkÞ; dðkÞð Þ

and

YðkÞ ¼ h XðkÞð ÞThe state-space models are used most frequently in dynamic analysis

because of the following:

. They give real-time behavior, suitable for computer simulation ofprocess behavior.

. They are used almost exclusively for the analysis of nonlinearsystems behavior.

. Another important use of state-space models is in representingprocesses in which some states are not measured.

All the other model forms are based on input/output relationships so thatany aspect of the process not manifested as either input or measured outputis not likely to be represented.

5.2.7 SISO and MIMO State-Space Models

The state-space models (in terms of real time) can be classified as follows:

Linear Lumped SISO Processes

One input variable u tð Þ, one disturbance d tð Þ, and one state variable x tð Þ aremodeled by

dxðtÞdt

¼ axðtÞ þ buðtÞ þ gdðtÞ

Note: The output is not necessarily x tð Þ [it can be yðtÞ], but is related tox tð Þ through the linear relation

yðtÞ ¼ cxðtÞ

Linear Lumped MIMO Processes

The MIMO processes can be two dimensional (the simplest MIMO case), asin the following simple example:

dx1ðtÞdt

¼ a11x1ðtÞ þ a12x2ðtÞ þ b11u1ðtÞ þ b12u2ðtÞ þ g11d1ðtÞ þ g12d2ðtÞ

dx2ðtÞdt

¼ a21x1ðtÞ þ a22x2ðtÞ þ b21u1ðtÞ þ b22u2ðtÞ þ g21d1ðtÞ þ g22d2ðtÞ

and the single output is,

yðtÞ ¼ c1x1ðtÞ þ c2x2ðtÞIn the matrix form, this can be rewritten as

d

dt

x1

x2

� �¼ a11 a12

a21 a22

� �x1

x2

� �þ b11 b12

b21 b22

� �u1

u2

� �þ g11 g12

g21 g22

� �d1

d2

� �

and

yðtÞ ¼ c1 c2� � x1

x2

� �

or in more general form for any n-dimensional system, we can write

dX

dt¼ A X þ B uþ � d

and

YðtÞ ¼ C X

where

X ðn�1Þ ¼

x1

x2

..

.

xn

0BBBB@

1CCCCA; A n�nð Þ ¼

a11 a12 � � � a1n

a21 a22 � � � a2n

..

. ... . .

. ...

an1 an2 � � � ann

0BBBB@

1CCCCA

B n�nð Þ ¼

b11 b12 � � � b1n

b21 b22 � � � b2n

..

. ... . .

. ...

bn1 bn2 � � � bnn

0BBBB@

1CCCCA; � n�nð Þ ¼

g11 g12 � � � g1ng21 g22 � � � g2n

..

. ... . .

. ...

gn1 gn2 � � � gnn

0BBBB@

1CCCCA

C m�nð Þ ¼

c11 c12 � � � c1n

c21 c22 � � � c2n

..

. ... . .

. ...

cm1 cm2 � � � cmn

0BBBB@

1CCCCA; uðn�1Þ ¼

u1

u2

..

.

un

0BBBB@

1CCCCA

d ðn�1Þ ¼

d1

d2

..

.

dn

0BBBB@

1CCCCA; Y ðm�1Þ ¼

y1

y2

..

.

ym

0BBBB@

1CCCCA

Nonlinear Lumped Systems

dX

dt¼ f X; u; dð Þ

where f X; u; dð Þ are nonlinear functions, and the output vector is given by

YðtÞ ¼ h XðtÞð Þ

Distributed Systems

These are systems where the state variables vary with one or more of thestate coordinates.

One-dimensional model

@T

@tþ v

@T

@z¼ UAS

rACP

TS � Tð Þ

This partial differential equation describes one side of the steam-heated shell and tube heat exchanger.

High-dimensional models

@X

@t¼ f X; u;rX;r2X; . . .

� �r is the Laplacian operator representing the vector of partial deriva-

tive operators in the spatial directions.

5.2.8 SISO and MIMO Transform Domain Models

SISO

The SISO linear model will be represented after applying Laplace transfor-mation as

yðsÞ ¼ guðsÞuðsÞ þ gdðsÞdðsÞwhere yðsÞ is the process output, guðsÞ is the process transfer function (in thetransform domain), uðsÞ is the process input, gdðsÞ is the disturbance transferfunction (in the transfer domain), and dðsÞ is the process disturbance.

MIMO

The MIMO linear model will be represented after applying the Laplacetransformation as

YðsÞ ¼ GuðsÞuðsÞ þ GdðsÞdðsÞwhere

YðsÞðn�1Þ ¼

y1

y2

..

.

yn

0BBBB@

1CCCCA GuðsÞ n�mð Þ ¼

gu11 gu12 � � � gu1mgu21 gu22 � � � gu2m

..

. ... ..

. ...

gun1 gun2 � � � gunm

0BBBB@

1CCCCA

GdðsÞ n�lð Þ ¼

gd11 gd12 � � � gd1lgd21 gd22 � � � gd2l

..

. ... ..

. ...

gdn1 gdn2 � � � gdnl

0BBBBB@

1CCCCCA

uðsÞðm�1Þ ¼

u1

u2

..

.

um

0BBBB@

1CCCCA; dðsÞðl�1Þ ¼

d1

d2

..

.

dl

0BBBB@

1CCCCA

Here, GuðsÞ and Gd ðsÞ are the matrix transfer functions.

5.2.9 SISO and MIMO Frequency-Response Models

For continuous time systems, we have the following.

SISO

yð joÞ ¼ guð joÞuð joÞ þ gdð joÞdð joÞwhere guð joÞ and gd ð joÞ are frequency-response transfer functions. Also

gð joÞ ¼ Re gðoÞ½ � þ jIm gðoÞ½ �

MIMO

Yð joÞ ¼ Guð joÞuð joÞ þ Gdð joÞdð joÞ

5.2.10 SISO and MIMO Discrete Time Models

SISO

It relates the z-transform of the sampled point signal to that of the outputsignal by the following relation:

yyðzÞ ¼ gguðzÞuuðzÞ þ ggdðzÞddðzÞwhere gguðzÞ and ggdðzÞ are the z-transform functions of the discrete timesystem.

MIMO

YYðzÞ ¼ GGuðzÞuuðzÞ þ GGdðzÞddðzÞwhere GGuðzÞ and GGdðzÞ are the transfer-function matrices. Transform-domain transfer functions are used extensively in process dynamics andcontrol.

Figure 5.2 shows the interrelationship between the discussed processmodel forms in a nice, concise manner. Uses of the three different types ofmodels are as follows

1. State space: Real-time simulation of process behavior, nonlineardynamic analysis.

2. Transform domain: Linear dynamic analysis involving well-characterized input functions; control system design.

3. Frequency response: Linear nonparametric models for processesof arbitrary mathematical structure; control system design.

Continuous and Discrete Time Models

The conversion of continuous-time models to discrete-time form is achievedby discretizing the continuous process model. This will be discussed in somedetail later.

5.3 STATE-SPACE AND TRANSFER DOMAINMODELS

These are the most important types of model. Most process models formu-lated from first principles occur in the state-space form, but the preferredform for dynamic analysis and controller design is the transfer domain form.However, after dynamic analysis and controller design, the task of controlsystem simulation and real-time controller implementation are carried out inthe time domain, requiring the state-space form again. The interrelation

between state-space and transfer domain models is shown in Figure 5.3. TheSISO process is given by

dxðtÞdt

¼ axðtÞ þ buðtÞ þ gdðtÞ

and

yðtÞ ¼ cxðtÞ

Figure 5.2 Interrelationship between process model forms.

Figure 5.3 Interrelation between state-space and transform domain models.

Laplace transformation gives

sxðsÞ ¼ axðsÞ þ buðsÞ þ gdðsÞ ð5:3Þand

yðsÞ ¼ cxðsÞ ð5:4ÞEquation (5.3) can be rewritten as

ðs� aÞxðsÞ ¼ buðsÞ þ gdðsÞOn rearrangement, it gives

xðsÞ ¼ b

s� auðsÞ þ g

dðsÞs� a

Consider the following more general input–output Laplace transformmodel:

yðsÞ ¼ guðsÞuðsÞ þ gdðsÞdðsÞwhere

guðsÞ ¼Ku

tusþ 1and gdðsÞ ¼

Kd

tdsþ 1

The solution (realization) is given by

yðsÞ ¼ Ku

tusþ 1

� �uðsÞ þ Kd

tdsþ 1

� �dðsÞ ð5:5Þ

We can write yðsÞ in terms of x1ðsÞ and x2ðsÞ in a simple form as follows:

yðsÞ ¼ x1ðsÞ þ x2ðsÞ ð5:6ÞNote:

yðsÞ ¼ CTxðsÞwhere

CT ¼ 1

1

� �

Now, x1ðsÞ is given by

x1ðsÞ ¼Ku

tusþ 1

� �uðsÞ

The above relation, on rearrangement, gives

tusx1ðsÞ þ x1ðsÞ ¼ KuuðsÞ

x2ðsÞ is given by

x2ðsÞ ¼Kd

tdsþ 1

� �dðsÞ


tdsx2ðsÞ þ x2ðsÞ ¼ Kd dðsÞThus, realization in the time domain gives

tudx1dt

þ x1 ¼ KuuðtÞ

and

tddx2dt

þ x2 ¼ Kd dðtÞ

together with

yðtÞ ¼ x1ðtÞ þ x2ðtÞat t ¼ 0, x1ð0Þ ¼ 0, and x2ð0Þ ¼ 0. This is one possible realization, and it isnot unique.

Realization is one of the more popular names for this process of transferringfrom Laplace domain to the time domain. It actually uses inverse transfor-mation from the Laplace domain to the time domain. An even simplerexample will make this process very clear. Consider the following verysimple input–output relation in the Laplace domain:

yðsÞ ¼ gðsÞuðsÞwhere

gðsÞ ¼ K

tsþ 1

Thus, we can write

yðsÞ ¼ K

tsþ 1

� �uðsÞ


tsþ 1ð ÞyðsÞ ¼ KuðsÞOn rearrangement, we get

tsð ÞyðsÞ þ yðsÞ ¼ KuðsÞ

Inverse Laplace transformation of the above relation gives

tdy

dtþ yðtÞ ¼ KuðtÞ

The nonuniqueness (it is a linear nonuniqueness in contrast to nonlinearnonuniqueness discussed in Chapter 7) can be shown in a more generalform:

y sð Þ ¼ gu sð Þu sð Þ þ gd sð Þd sð Þwhere

guðsÞ ¼Ku

tusþ 1and gdðsÞ ¼

Kd

tdsþ 1

Now, we can write

y sð Þ ¼ c1x1 sð Þ þ c2x2 sð Þ ð5:7Þand

x1ðsÞ ¼a1u sð Þtusþ 1

and x2ðsÞ ¼a2d sð Þtdsþ 1

On substituting the values of x1ðsÞ and x2ðsÞ in Eq. (5.7), we get

y sð Þ ¼ a1c1tusþ 1

� �u sð Þ þ a2c2

tdsþ 1

� �d sð Þ

Now,

a1c1tusþ 1

¼ Ku

tusþ 1and

a2c2tdsþ 1

¼ Kd

tdsþ 1

which gives

Ku ¼ a1c1 and Kd ¼ a2c2

This shows that the degree of freedom is equal to 2.Thus, there is an infinite combination of a1c1 and a2c2 that satisfy the

above equations. This is the linear nonuniqueness.

A special case: For the input–output Laplace transform model,

yðsÞ ¼ Ku

tusþ 1

� �uðsÞ þ Kd

tdsþ 1

� �dðsÞ

If we consider the special case of tu ¼ td ¼ t, we get

yðsÞ ¼ 1

tsþ 1KuuðsÞ þ Kd dðsÞ½ �


tsþ 1ð ÞyðsÞ ¼ KuuðsÞ þ Kd dðsÞand inverse transformation to the time domain gives

tdyðtÞdt

þ yðtÞ ¼ KuuðtÞ þ Kd dðtÞ

We can set yðtÞ ¼ xðtÞ and write

tdxðtÞdt

þ xðtÞ ¼ KuuðtÞ þ Kd dðtÞ

and

yðtÞ ¼ xðtÞ

5.4 INTRODUCTORY PROCESS CONTROL CONCEPTS

In this part of the chapter, we present in a very condensed/simple mannerthe basic process control concepts.

5.4.1 Definitions

The process control system is the entity that is charged with the responsi-bility for the following important tasks: monitoring outputs, making deci-sions about how best to manipulate inputs so as to obtain desired outputbehavior, and effectively implement such decisions on the process. Thus, itdemands that three different tasks be done:

1. Monitoring process output variables by measurements2. Making rational decisions regarding what corrective action is

needed on the basis of the information about the current (outputvariables) and the desired (set point) state of the process

3. Effectively implementing these decisions on the process

It can have different levels of sophistication as follows:

. The simplest are, of course, the manual control systems.

. More advanced are the automatic control systems.

. The most advanced are the digital control systems.

The control hardware elements are as follows:

1. Sensors. Sensors are also called measuring devices or primaryelements. Examples are the following:

. Thermocouples

. Differential pressure cells (for liquid level measurements)

. Gas/liquid chromatographs (for analysis of gas samples)

2. Controllers. The controller is the decision-maker or the heart ofthe control system. It requires some form of intelligence.Controllers can be of the following types:

. Pneumatic (now almost obsolete)

. Electronic (more modern)

. Digital (for more complex operations, they are some kindof a special-purpose small digital computer)

Currently, the first two types of controller (pneumatic and elec-tronic) are limited to simple operations.

3. Transmitters. Examples are pneumatic (air pressure), electricalsignals, and digital signals.

4. Final control elements

. They implement the control command issued by the con-troller.

. They are mostly ‘‘control valves’’ [usually pneumatic (i.e.,air driven)]

. Other types include variable-speed fans, pumps, compres-sors, conveyors, relay switches, and so forth.

5. Other hardware elements

. Transducers: to convert an electric signal from an electroniccontroller to a pneumatic signal needed for a control valve.

. A/D and D/A converters (for computer devices): A/D isanalog to digital and D/A is for digital to analog. Theyare needed simply because the control system operates onanalog signals (electric voltage or pneumatic pressures),whereas the computer operates digitally (giving out andreceiving only binary numbers).

Figure 5.4 shows a typical sequential setup for the control system compo-nents.

Some additional process control terminologies are as follows:

Set point: The value at which the output is to be maintained, fed to thecontroller to compare with the measurement, and make decisions.

Regulatory control: This is used when it is required to keep the processat a specific fixed set point.

Servo control: This is used when it is required to make the outputtrack a changing set point.

5.4.2 Introductory Concepts of Process Control

We will illustrate the basic concepts of process control using a typical gen-eric chemical process as shown in Figure 5.5. A typical chemical process canbe a petrochemical plant, a refinery (or a section of it), a biochemical plant,and so forth. Let us consider it to be a part of a refinery (the fractionatingpart). It will typically consist of the following:

1. Processing units. Typically these are the following:

. The storage tanks

. The furnaces

. The fractionation towers

and their auxiliary equipments.

Figure 5.4 Logical representation of the control system components.

Figure 5.5 A typical chemical process.

2. Raw materials. These usually include the following:

. Crude oil

. Air

. Fuel gas

3. Finished products. A fractionating unit will produce a wide rangeof products such as the following:

. Naphtha

. Light gas oil

. Heavy gas oil

. High-boiling residues

4. Heat input. It is mainly the furnace generated heat.

5. Heat output. It is usually the heat removed from the condenser.

Basic Principles Guiding the Operation of Processing Units

1. It is desirable to operate the processing unit safely (includingenvironmental safety).

2. Specific production rates must be maintained.3. Product quality specifications must be maintained.

Process dynamics and control is that aspect of chemical engineeringwhich is concerned with the analysis, design, and implementation of controlsystems that facilitate the achievement of specified objectives of processsafety, production rate, and product quality.

Here, we examine a typical industrial process control problem.Consider the furnace shown in Figure 5.6, where crude oil is preheatedto be fed to the fractionating column. The crude oil feed flow rate fluc-tuates considerably, however we need to supply it to the fractionator at aconstant temperature T. Consider Tm to be the highest safe temperaturefor the type of metals used for the heating tubes and Tt is the tubetemperature.

Control Problem Statement

‘‘Deliver crude oil feed to the fractionator at a constant temperature Tð Þand flow rate F0ð Þ regardless of all factors potentially capable of causing thefurnace outlet temperature Tð Þ to deviate from this desired value, makingsure that the temperature of the tube surfaces within the furnace does notexceed the value Tmð Þ at any time.’’

In short, keep T ¼ T, F ¼ F0, and Tt < Tm (where Tt is the tubetemperature).

A dialogue between the plant engineer (PE) and the control engineer(CE) regarding this problem involves the following phases:

Phase 1. Keep the outlet temperature T at the desired value T. This isto be achieved through a controller (e.g., PID¼ proportional integralderivative, as discussed in the next part of this chapter) which measuresthe output temperature using a thermocouple and compares it with thedesired temperature (set point, T) and then the deviation is used to con-trol the furnace heating fuel gas. If T > T, then we decrease the fuel gasflow rate, whereas if T < T, then we increase the fuel gas flow rate, and,finally, if T ¼ T then we keep the fuel gas flow rate constant. This iswhat we call a feedback control system. Its main advantage is that it cancompensate for the deviation of T from T regardless of the cause ofthis deviation. Its disadvantage is that it takes action only after thedisturbance(s) has (have) caused measurable deviation of T from T. Thisfeedback control system is shown in Figure 5.7.

Phase 2. Suppose that we cannot tolerate the time lag associated withthe effect of the input disturbance(s) appearing as deviation in T . In thiscase, we can use what we call the feed-forward control system. This isbased on measuring the input flow rate, and as soon as this flow rate devi-ates from the design value F0, we take control action to manipulate thefuel gas flow rate before the effect of disturbance in F appears as a devia-tion in T . The advantage of feed-forward control is that the control ac-

Figure 5.6 Preheater furnace for crude oil.

tion is performed before the disturbance starts to cause the deviation inthe output variables. However, the main disadvantages are as follows:

1. There has to be some relation to relate feed flow rate to fuel gasfeed rate.

2. The control system will not respond to input disturbances otherthan F .

The feed-forward control arrangement for this process is shown in Figure5.8.

Phase 3. The two types of feedback and feed-forward can be combinedas shown in Figure 5.9 (the feed-forward controller sets the setpoint forthe feedback controller) or the cascade system as shown in Figure 5.10.

5.4.3 Variables of a Process

It is important to distinguish the different types of variables of a process.There are many classifications for these variables.

Process variables. Process variables include variables such as tempera-ture, flow rate in/out, pressure, concentration, and so forth. Thesecan be divided into two types:

1. Input variables: These are capable of influencing the process con-ditions.

Figure 5.7 Feedback control system.

Figure 5.8 Feed-forward control system.

Figure 5.9 The feed-forward controller sets the set point for the feedback

controller.

2. Output variables: These provide information about the processconditions.

State variables versus output variables

State variables. They are the minimum set of variables essential forcompletely describing the internal state (or condition) of a process.In other words, they are the variables which describe the state of thesystem (e.g., temperature, concentration, pH, etc.)

Output variables. They are some measurements, either of a single statevariable or a combination of state variables at the outlet of theprocess.

Input variables. They can be divided into two main types:

. Manipulated or control variables (we can have control overthem)

. Disturbance variables (we do not have control over them)

Measured and unmeasured variables. Output variables which are char-acterized by the fact that we need to take samples for analysis arenot called measured variables because they are not measured on-line.

Figure 5.10 Cascade control.

Figure 5.11 shows a typical process, where the input is divided intotwo types: manipulated variables and input disturbances. Input disturbancescan be divided further, into measured and unmeasured disturbances.The output variables to be controlled are also divided into measured andunmeasured variables.

Examples

Stirred heating tank process. The input variables are as shown in Figure5.12 include the feed flow rate Fi and feed temperature Ti, whereas theoutput variables are F and T . The steam flow rate Q is the input manipu-lated variable.

A furnace. The input disturbances for this furnace are the input flow rateFi, the input temperature Ti, the fuel supply pressure PF , and the feedheat content �F . The manipulated variables are airflow rate QA and fuelflow rate QF , whereas F and T are the output variables. (see Fig. 5.13).

5.4.4 Control Systems and Their Possible Configurations

In this subsection, the reader is introduced to a number of the more typicalcontrol system configurations (briefly discussed earlier for a specific exam-ple).

1. Feedback control. For this configuration, the control action istaken after the effect; that is, for any disturbance entering theprocess, no control action is taken except after the effect of thedisturbance appears in the output (see Fig. 5.14).

Figure 5.11 Types of input–output variables.

Figure 5.12 Variables of a stirred heating tank.

Figure 5.13 Variables of a furnace.

2. Feed-forward control. For this configuration action is takenbefore the effect (i.e., before the process is affected). The mainadvantage is that the decision of the controller is taken using amodel for the process. The model relates the input disturbanceand output variables in order to calculate the input changes thatcompensates for the disturbances to keep the output at its desiredvalue.

The disadvantages are as follows:

. The negative effects associated with inaccuracy of modelsused.

. The choice of the disturbances to be measured and thepossible effect of the unmeasured disturbances. It doesnot act to compensate for any unmeasured disturbances.

. The controller has no information about the conditionsexisting at the process output.

A typical feed-forward control configuration is shown in Figure 5.15.

3. Open-loop control. Open-loop control is where the set point tothe controller is pre-programmed to follow a certain path (servocontrol) as shown in Figure 5.16.

Other control configurations, such as cascade control have been shownearlier in the book.

Figure 5.14 Typical feedback control system.

5.4.5 Overview of Control Systems Design

Here, we will present a preliminary introduction to the design of controlsystems using a very simple example.

General Principles Involved in Designing a Control System

Step 1. Assess the process and define the control objectives to answerthe following questions:

1. Why is there a need for control?

Figure 5.15 Typical feed-forward control arrangement.

Figure 5.16 Open-loop control arrangement.

2. Can the problem be solved only by control, or there is anotheralternative (such as redesigning part of the process)?

3. What is the control of the process expected to achieve?

Step 2. Selection of the process variables (output and manipulatedvariables as well as disturbances in the case of feed-forward control)in order to answer the following questions:

1. What is the crucial output variables to be measured (for feedbackcontrol)?

2. What are the most serious disturbances, and which ones aremeasurable (for the feed-forward control)?

3. Which inputs to be chosen as the input manipulated variables(for both feed-forward and feed-backward controls).

Step 3. Selection of the control structure, whether open-loop, feed-back, feed-forward, or a combination of feed-backward and feed-forward control is required.

Step 4. Design of the controller(s) may have different degrees of com-plexity, but it basically involves the formulation of a control lawwhich utilizes as much information as possible from the process.This control law is used to produce a control decision to be usedto adjust the manipulated variables.

Process control engineers must have a good understanding ofthe process and its dynamics. They must also understand the steady-statecharacteristics of the process.

Example

We will follow the above four steps and apply them to a very simple exam-ple. This example is the filling/emptying of a liquid tank with valves at boththe inlet and outlet, as shown in Figure 5.17.

For the process model without control, it is very easy to write thedifferential equations (the mathematical model) describing the dynamicbehavior of the tank. It will be

AC

dh

dt¼ Fi � F ð5:8Þ

with the initial conditions at t ¼ 0, h ¼ h0. F and Fi should be certain con-stant values or functions of h. For example, F ¼ C

ffiffiffih

p, where C is called the

valve coefficient, and Fi ¼ B, where B is some constant value. Thus, Eq.(5.8) can be rewritten

AC

dh

dt¼ B� C

ffiffiffih

pð5:9Þ

Let us consider the following steady-state situation: At steady state,dh=dt ¼ 0 (there is no height change with time). So,

B ¼ CffiffiffiffiffiffiffihSS

pwhich givesffiffiffiffiffiffiffi

hSSp

¼ B

C

Therefore, the steady-state height hSS is given by

hSS ¼ B

C

� �2

The dynamic behavior is described by Eq. (5.9).Equation (5.9) is a nonlinear differential equation. It can be solved

numerically by using any of the well-known subroutine packages such asPolymath, MathCAD, IMSL libraries, and so forth. However, can we usesome transformations to solve this equation analytically? Yes, it can be doneas follows. Let

B� Cffiffiffih

p¼ y

Thus, we can writeffiffiffih

p¼ B� y

C

Figure 5.17 Continuously filling and emptying tank.

On differentiation of y with respect to h, we get

dy

dh¼ �C

1

2h�1=2

� �ð5:10Þ

We can also write

dy

dt¼ dy

dh

dh

dt

Therefore,

dy

dt¼ �C

2h�1=2

� �dh

dtð5:11Þ

Rearranging gives,

dh

dt¼ � 2

ffiffiffih

p

C

!dy

dt

Substituting the value offfiffiffih

pgives,

dh

dt¼ � 2

C2ðB� yÞ dy

dtð5:12Þ

Substitution from Eq. (5.12) into Eq. (5.9), gives

AC � 2

C2ðB� yÞ

� �dy

dt¼ B� C

B� yð ÞC

� �

Rearranging gives

2AC

C2|ffl{zffl}a

y� Bð Þ dydt

¼ y


y� B

ydy ¼ 1

adt ð5:13Þ

where

a ¼ 2AC

C2(a constant value)

Rearrangement of Eq. (5.13) gives

1� B

y

� �dy ¼ 1

adt


y� B ln y ¼ 1

atþ C1

Boundary condition is at t ¼ 0; y ¼ y0, where y0 ¼ B� Cffiffiffiffiffih0

p. So we get the

value of constant of integration as

C1 ¼ y0 � B ln y0

Using this value of the constant of integration, we obtain the solution

y� B ln y ¼ 1

atþ y0 � B ln y0

which can be rearranged in the following form:

y� y0ð Þ � B lny

y0

� �¼ 1

at ð5:14Þ

We can easily find the change of y with t using this simple equation.Obviously, we can transfer y back to h very easily. This completes theexample.

Let us now look at the effect of different control-loop configurationson the model equations and the behavior of the system.

First configuration

The first configuration we look at is the simple feedback-level control usingthe tank level h as the measured variable and the output flow rate F as thecontrol (or manipulated) variable as shown in Figure 5.18, where Fi is theinput variable (disturbance), F is the output variable (control or manipu-lated variable), and h is the output variable (liquid-level measurement). Themodel equation is given by

AC

dh

dt¼ Fi � F ð5:15Þ

At steady state, we can write

FiSS � FSS ¼ 0 ð5:16ÞWe can write Eq. (5.15) at steady state as

AC

dhSSdt

¼ FiSS � FSS ¼ 0 ð5:17Þ

To convert the variables into deviation variables, subtract Eq. (5.17)from Eq. (5.15) to get

AC

d h� hSSð Þdt

¼ Fi � FiSSð Þ � F � FSSð Þ ð5:18Þ

Define the deviation variables as

y ¼ h� hSSð Þ

d ¼ Fi � FiSSð Þ

u ¼ F � FSSð ÞThus Eq. (5.18) can be rewritten in terms of deviation variables as

AC

dy

dt¼ d � u

The above equation can be rearranged to give

dy

dt¼ 1

AC

d � 1

AC

u ð5:19Þ

The simple control law is a proportional control on F ; that is

F ¼ FSS þ K h� hSSð Þ ð5:20Þwhere K is the proportional controller gain. Equation (5.20) can be re-arranged as

F � FSS ¼ K h� hSSð Þ

Note: Notice that this linear formulation is restricted by the fact thatwhen K ¼ 0, then F ¼ FSS, regardless of the height inside the tank, mean-ing that h increases to infinity for any positive disturbance.

Figure 5.18 Feedback-level control.

The above control law equation can be written in terms of deviationvariables as

u ¼ Ky

Let us suppose that Fi increases from FiSS to Fi ¼ FiSS þ x, then we have

d ¼ FiSS þ x� FiSS ¼ x


dy

dt¼ 1

AC

x� 1

AC

Kyð Þ ð5:21Þ

with the initial condition at t ¼ 0, y ¼ 0 (which means that at t ¼ 0, thesystem is at its steady state). If x is a constant, then we can write

xAC

¼ a1 andK

AC

¼ a2


dy

dt¼ a1 � a2y ð5:22Þ


dy

a1 � a2y¼ dt

On integrating the above equation, we get

� 1

a2ln a1 � a2yð Þ ¼ tþ C1

We obtain the value of constant C1 from the initial condition at t ¼ 0, y ¼ 0,Therefore the value of integration constant is equal to

C1 ¼ � 1

a2ln a1ð Þ

and we get

� 1

a2ln a1 � a2yð Þ ¼ t� 1

a2ln a1ð Þ


1

a2ln a1ð Þ � ln a1 � a2yð Þ½ � ¼ t


lna1

a1 � a2y

� �¼ a2t

Alternatively, we can rearrange it as

lna1 � a2y

a1

� �¼ �a2t

After some simple manipulations, we get

y ¼ a1a2

1� e�a2t� � ð5:23Þ

With Eq. (5.23), we can compute the value of y for any value of t.Substituting the values of a1 and a2 in Eq. (5.23) gives

y tð Þ ¼ xAC

AC

K1� e�ðK=ACÞt� �


y tð Þ ¼ xK

1� e�ðK=ACÞt� �

ð5:24Þ

Note: We cannot put K ¼ 0 in this solution, as it will give us the 0=0form, which is indeterminate. Thus, if K ¼ 0 a2 ¼ 0ð Þ, then there is nocontrol. For this case, Eq. (5.22) reduces to

dy

dt¼ a1

which actually is

dy

dt¼ x

AC

The above equation can be integrated to give

y ¼ xAC

tþ C1

with initial condition at t ¼ 0; y ¼ 0. The constant of integration C1 ¼ 0 forthis initial condition. So we get the solution as

y ¼ xAC

t (for no control)

Figure 5.19 shows the behavior of system with and without the con-trol. Note that the ‘‘no control’’ (by putting K ¼ 0) means that the valve is

not affected by the change in height of liquid in the tank as explained earlier.Of course, this is not the real situation when there is no control. When thereis no control, the flow output will increase (proportional to the square rootof the height of liquid in the tank) until the system reaches a new steadystate.

Second configuration

The second configuration we look at is the simple feed-forward-level controlwith the input flow rate Fi as the measured variable and the output flow rateF as the control variable, as shown in Figure 5.20. If inflow is equal tooutflow, then whatever happens to Fi, F follows and, therefore, h remainsconstant. Thus, the control law is

F ¼ Fi

Thus,

AC

dh

dt¼ 0

where h ¼ constant, depending on the accuracy of measurement, regulators,and different time lags.

Third configuration

The third configuration (Fig. 5.21) is feedback control strategy, but it isdifferent from the first feedback configuration. The inflow Fi is used asthe control variable rather than the outlet flow, but still measuring thelevel ðhÞ, which is an output variable. This is the reason it is still a feedbackstrategy.

Figure 5.19 Behavior of system with and without control.

Here, Fi lost its status as a disturbance variable and became a control(or manipulated) variable, whereas F was solely dependent upon h (which,in turn, depends on the C value of the valve coefficient) and lost its identityas a control (or manipulated) variable. Thus, the output variable h is themeasured variable and the input variable Fi is the control (or manipulated)variable.

Figure 5.20 Feed-forward control.

Figure 5.21 Special type of feedback configuration.

For simplicity of illustration, let us consider

F ¼ Ch

(instead of F ¼ Cffiffiffih

p). This is done just for the simplicity of handling a

linear differential equation rather than a nonlinear differential equation.Thus, the dynamic model equation for the controlled tank system is

AC

dh

dt¼ Fi � Ch ð5:25Þ

Using the deviation variables

y ¼ h� hSS

u ¼ Fi � FiSS

gives the following simple deviation equation for Eq. (5.25):

AC

dy

dt¼ u� Cy

The above differential equation can be rearranged as

dy

dt¼ � C

AC

� �yþ 1

AC

u ð5:26Þ

If this system is originally at steady state, then the initial condition will be att ¼ 0, h ¼ hSS or, in terms of deviation variable, y ¼ 0. Now, if we useproportional control, the control law equation will be

Fi ¼ FiSS þ K h� hSSð ÞConsider the following physical implications for K > 0:

. If h > hSS, then Fi > FiSS and, therefore, h continues to increase.

. If h < hSS, then Fi < FiSS and, therefore, h continues to decrease.

This is obviously the opposite of control; thus, we either must have K < 0,or a much better choice is the control law

Fi ¼ FiSS þ K hSS � hð ÞThen,

Fi � FiSS ¼ �K h� hSSð Þwhich can be rewritten in a more simplified form as deviation variables:

u ¼ �Ky

Differential Eq. (5.26) becomes

dy

dt¼ � C

AC

� �yþ 1

AC

�Kyð Þ ð5:27Þ


dy

dt¼ � C þ K

AC

� �|fflfflfflfflfflffl{zfflfflfflfflfflffl}

a

y

and can be thus written in the very simple form

dy

dt¼ �ay

With the initial condition at t ¼ 0; y ¼ 0. On integration of the differentialequation, we obtain

ln y ¼ �atþ lnB

where B is the constant of integration. On rearrangement we get,

lny

B

� �¼ �at


y ¼ Be�at

To get the value of integration constant B, the initial condition is used to get

B ¼ 0

Thus, the solution of the differential equation is thus

y ¼ 0

This means that the system is always at the steady state (except for the effectof delays in the control loop).

Because of this integration constant problem (it is not really a pro-blem, it is actually correct), we go back to the original Eq. (5.25):

AC

dh

dt¼ Fi � Ch

At steady state, the equation is

AC

dhSSdt

¼ FiSS � ChSS ¼ 0

This gives the steady-state height as

hSS ¼FiSS

C

Introduction of the proportional controller gives

Fi ¼ FiSS þ K hSS � hð Þ or Fi ¼ FiSS � K h� hSSð ÞThus, we get

AC

dh

dt¼ FiSS � K h� hSSð Þ � Ch ð5:28Þ

At steady state,

AC

dhSSdt

¼ FiSS � K hSS � hSSð Þ � ChSS ¼ 0

From the dynamics point of view for this special feedback control loop, wehave

dh

dt¼ FiSS

AC

� 1

AC

Kh� KhSS þ Chð Þ


dh

dt¼ FiSS þ KhSS

AC

� �|fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl}

a1

� K þ Cð ÞAC|fflfflfflffl{zfflfflfflffl}a2

h

The above equation can be written in a simplified form as

dh

dt¼ a1 � a2h ð5:29Þ

where

a1 ¼FiSS þ KhSS

AC

� �

a2 ¼K þ C

AC

with the initial condition at t ¼ 0; h ¼ hSS. We can write the differential Eq.(5.29) as

dh

a1 � a2h¼ dt

Integration gives

� 1

a2ln a1 � a2hð Þ ¼ tþ C1 ð5:30Þ

where C1 is the constant of integration and the initial condition is att ¼ 0; h ¼ hSS. Thus, we obtain the integration constant C1 as

� 1

a2ln a1 � a2hSSð Þ ¼ C1

On substituting the value of C1, the solution of differential equationbecomes

� 1

a2ln a1 � a2hð Þ ¼ t� 1

a2ln a1 � a2hSSð Þ


1

a2ln a1 � a2hSSð Þ � ln a1 � a2hð Þ½ � ¼ t

Further rearrangement yields

lna1 � a2h

a1 � a2hSS

� �¼ �a2t

After some manipulations, we get

h ¼ a1a2

� a1 � a2hSSa2

� �e�a2t ð5:31Þ

Now, we calculate the following and substitute these values in Eq. (5.31):

a1a2

¼ FiSS þ KhSSAC

AC

K þ C¼ FiSS þ KhSS

K þ C

anda1 � a2hSS

a2¼ FiSS þ KhSS

K þ C� hSS ¼ FiSS � ChSS ¼ 0

Thus, the solution (5.31) can be written in following form:

h ¼ FiSS þ KhSSK þ C

� FiSS � ChSSð Þ|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl}equal to zero

e�a2t

which gives

h ¼ FiSS þ KhSSK þ C

� �

and this can be rearranged to give

h ¼ ChSS þ KhSSK þ C

¼ K þ C

K þ C

� �hSS

which basically gives

h ¼ hSS

Thus, with this feed-forward control strategy and with F linearly dependenton h together with the assumption that all loop delays are neglected, then hremains at hSS all the time.

Some Concluding Remarks

1. Note that we used a simplifying assumption, F ¼ Ch, and notF ¼ C

ffiffiffih

p, just for the simplicity of using linear differential equa-

tion.

2. Modeling errors. Intrinsic (inherent) degree of errors in mathe-matical models are unavoidable and, of course, they affect thecontrol policy.

3. Implementation problems. These include the following:

. Time delays

. Imperfect measurements

. Inaccurate transmission

. Control valve inertia (leading to inaccurate valve actuation)

4. Complicated process control structure. Consider the exampleshown in Figure 5.22. A number of elementary questions arisewith regard to this relatively complicated control structure. Forexample, how did we choose that the hot stream is not the effec-tive control (manipulated) variable for the control of the level(rather than the control of the temperature)? In more general

Figure 5.22 A relatively complicated control structure.

terms, which control variable should be manipulated (regulated/controlled) to control which output variable (T , level) to achievemaximum effectiveness? Assuming that the shown input/outputpairing is the proper one, the question still remains, can the coldstream regulate the liquid level without upsetting the hot stream’stask of regulating the liquid temperature, and vice versa? Thismutual interference is called interaction and it is very essential forMIMO design.

5.5 PROCESS DYNAMICS AND MATHEMATICALTOOLS

In this section, we discuss some basic ideas associated with processdynamics. Consider the process shown in Figure 5.23. A number of possibleresponses [output variable change with time y tð Þ] to a step change in inputvariable u tð Þ depending on the characteristic of the system or the process areshown in Figure 5.24.

5.5.1 Tools of Dynamic Models

Process Model and Ideal Forcing Functions

The input–output relation of the process can be obtained by disturbing theinput and recording the corresponding output. However this is expensive,time-consuming, and, most importantly, assumes that the unit does existand is not in the design stage. To avoid these main three limitations, weshould use mathematical models. This is a task which can be stated in aclearer form as follows: Given some form of mathematical representation ofthe process, investigate the process response to various input changes; thatis, given a process model, find yðtÞ in response to inputs uðtÞ and dðtÞ. Thistask needs (1) a process model and (2) well-characterized input functions(forcing functions).

Figure 5.23 Input-output process (response to step input disturbance).

Mathematical Tools

The mathematical tools available for this task include the following:

. Dynamic mathematical models: usually in the form of differentialequations (ordinary and partial, linear, and nonlinear).

. Laplace transform: used in the development of transfer functions,which are the most widely used model form in process controlstudies. The Laplace transform converts an ordinary differentialequation (ODE) to an algebraic equation and, likewise, converts apartial differential equation (PDE) into an ordinary differentialequation (ODE).

. For a nonlinear differential equation, either numerical solutions orlinearization in the neighborhood of a certain state is used.

Figure 5.24 Some of the different forms of response.

. The z-transform may be used to convert a difference equation toan algebraic equation (it is used in digital control, which is thedominant form of control for last three to four decades).

. Multivariable systems utilize the short-hand methods of matrices.

Digital computers are used to obtain numerical solutions for systemswhich are described by equations that cannot be solved analytically.

5.6 THE LAPLACE TRANSFORMATION

As we mentioned earlier, some transformations are useful in handling pro-cess dynamics and control problems. The most popular for linear (orlinearized) systems is the Laplace transformation. The basic definition ofLaplace transformation is

f ðsÞ ¼ð10

e�stf ðtÞ dt ¼ L f ðtÞ� �where s is a complex variable. Laplace transformation is a mapping from thet-domain (time domain) to the s-domain (Laplace domain). Later, f ðsÞ isused instead of f ðsÞ for easier representation.

5.6.1 Some Typical Laplace Transforms

If f ðtÞ ¼ 1, then

f ðsÞ ¼ L f ðtÞ� � ¼ L 1f g ¼ð10

e�stð1Þ dt ¼ �1

se�st

� �10

¼ � 1

se�1 � e�0 �

¼ � 1

s0� 1½ �

giving

L 1f g ¼ 1

s

If f ðtÞ ¼ e�at

then

f ðsÞ ¼ L f ðtÞ� �¼L e�at� �¼ð1

0

e�ste�at dt ¼ð10

e� sþað Þt dt ¼ �1

sþ ae� sþað Þt

� �10

¼ �1

sþ ae�1 � e�0 � ¼ �1

sþ a0� 1½ �

finally giving

L e�at� � ¼ 1

sþ a

We usually use tables and some properties to get the Laplace transforms. Atable of the most important Laplace transforms is given in Appendix D. Inorder to cover a wider range of functions using a limited table like the one inAppendix D, we must utilize some important properties of the Laplacetransformation. The main properties of Laplace transform are as follows:

1. All functions necessary in the study of process dynamics andcontrol have Laplace transforms.

2. f ðsÞ has no information for t < 0, because the integral starts att ¼ 0. However, this fact practically represents no problem,because t is always time and we are interested in t > t0, wheret0 is the initial time that can arbitrarily be set to zero (for t < 0,we can put f ðtÞ ¼ 0Þ:

3. It is not possible for two different functions f ðtÞ and gðtÞ to havethe same Laplace transform. f ðtÞ and f ðsÞ are called the transferpair and this transfer pair is unique.

4. The Laplace transform operation is linear:

L C1 f1ðtÞ þ C2 f2ðtÞ� � ¼ C1L f1ðtÞ

� �þ C2L f2ðtÞ� �

5.6.2 The Inverse Laplace Transform

The inverse Laplace formula is

f ðtÞ ¼ L�1 f ðsÞ� � ¼ 1

2j

þC

estf ðsÞ ds

which is a complex contour integral over the path represented by C (calledthe Bromwich path). Tables of the inverse transform are also available (seeAppendix D).

5.6.3 The Transform of Derivatives

Because

L f ðtÞ� � ¼ ð10

e�stf ðtÞ dt ð5:32Þ

we have

Ldf ðtÞdt

�¼ð10

e�st df ðtÞdt

� �dt ¼ sf ðsÞ � f ð0Þ|{z}

at t¼0

Simple proof. Let

e�stf ðtÞ ¼ y

Then,

dy

dt¼ �se�stf ðtÞ þ e�st df ðtÞ

dt

Rearranging gives

df ðtÞdt

¼ 1

e�st

dy

dtþ se�stf ðtÞ

� �ð5:33Þ

Substituting Eq. (5.33) into Eq. (5.32) givesð10

e�st df ðtÞdt

� �dt ¼

ð10

e�st 1

e�st

dy

dtþ se�stf ðtÞ

� �dt

¼ð10

dy

dt

� �dtþ s

ð10

e�stf ðtÞ dt

¼ð10

dyþ s

ð10

e�stf ðtÞ dt

¼ yð1Þ � yð0Þ þ sf ðsÞ

Because

yð1Þ ¼ e�1f ð1Þ ¼ 0

yð0Þ ¼ e�0f ð0Þ ¼ f ð0Þ

we have

Ldf ðtÞdt

�¼ sf ðsÞ � f ð0Þ|{z}

at t¼0

Similarly, we can obtain the Laplace transformation for higher derivatives:

Ld2f ðtÞdt2

( )¼ s2f ðsÞ � sf ð0Þ � f 1ð0Þ

where

f 1ð0Þ ¼ df ðtÞdt

��t¼0

The general form is

Ldnf ðtÞdtn

�¼ snf ðsÞ � sn�1f ð0Þ � sn�2f 1ð0Þ � sn�3f 2ð0Þ � � � � � f n�1ð0Þ

ð5:34Þwhere

f n�1ð0Þ ¼ dn�1f ðtÞdtn�1

��t¼0

Equation (5.34) can also be written in the form

Ldnf ðtÞdtn

�¼ snf ðsÞ �

Xn�1

k¼0

skf ðn�1�kÞð0Þ

where f ið0Þ is the ith derivative of f ðtÞ at t ¼ 0 and f 0ð0Þ is the function f ðtÞat t ¼ 0.

If all the initial conditions of f ðtÞ and its derivatives are zero (distur-bance variables), then

Ldnf ðtÞdtn

�¼ snf ðsÞ �

Xn�1

k¼0

skf ðn�1�kÞð0Þ|fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl}each term equal to zero

And, therefore, the Laplace transformation of the derivatives is given by

Ldnf ðtÞdtn

�¼ snf ðsÞ

The transforms of integrals is

L

ðt0

f ðt0Þdt 0 �

¼ 1

sf ðsÞ

Proof: Reader is advised to see the proof in any elementary mathematicstext.

5.6.4 Shift Properties of the Laplace Transform

If

L f ðtÞ� � ¼ f ðsÞ

then

L eatf ðtÞ� � ¼ ð10

eate�stf ðtÞ dt

¼ð10

e�ðs�aÞtf ðtÞ dt

¼ f ðs0Þ � f ðs� aÞwhere

s0 ¼ s� a

that is to say,

L eatf ðtÞ� � ¼ f ðs� aÞThus, the s variable in the transform has been shifted by a units.

A more important result is the shift with regard to the inverse of theLaplace transform:

L�1 f ðs� aÞ� � ¼ eatf ðtÞLet us use a very simple example:

f ðtÞ ¼ 1

then

f ðsÞ ¼ L 1f g ¼ 1

s

Find

L�1 1

sþ a

�The solution is

L�1 1

s� �að Þ �

¼ f ðtÞ ¼ e�at

The shift in time is given by

L f ðt� aÞ� � ¼ e�asf ðsÞThe proof is as follows:

L f ðt� aÞ� � ¼ ð10

e�stf ðt� aÞ dt

Define

t� að Þ ¼ t0

which gives dt ¼ dt0. For the limit t ¼ 0, we get t0 ¼ �a; for t ¼ 1, we gett0 ¼ 1. Thus, we can write,

L f ðt� aÞ� � ¼ ð1�a

e�s t 0það Þf ðt0Þ dt 0

¼ð1�a

e�st0e�asf ðt0Þ dt0

¼ e�as

ð0�a

e�st0f ðt0Þ dt 0|fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl}equal to zero

þð10

e�st0f ðt0Þ dt0|fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl}f ðsÞ

26664

37775

It is always assumed that f ðt0Þ for t0 < 0 is equal to zero. Thus

L f ðt� aÞ� � ¼ e�asf ðsÞFigure 5.25 shows the change in the behavior of a function due to the shift intime.

5.6.5 The Initial- and Final-Value Theorems

These theorems relate the limits of t ! 0 and t ! 1 in the time domain tos ! 1 and s ! 0, respectively, in the Laplace domain.

Initial-value theorem

limt!0

f ðtÞ½ � ¼ lims!1 sf ðsÞ �

Figure 5.25 Behavior due to shift in time.

Final-value theorem

limt!1 f ðtÞ½ � ¼ lim

s!0sf ðsÞ �

5.6.6 Use of Laplace Transformation for the Solution ofDifferential Equations

The condensed information regarding the Laplace transformation in theprevious sections is sufficient to utilize it in solving linear differential equa-tions, as shown in the following subsection.

Consider the following differential equation:

tdy

dtþ y ¼ KuðtÞ ð5:35Þ

For uðtÞ ¼ 1, yð0Þ ¼ 0; we get, by Laplace transformation,

tsyðsÞ þ yðsÞ ¼ K

s


tsþ 1ð ÞyðsÞ ¼ K

s


yðsÞ ¼ K

s tsþ 1ð Þ ð5:36Þ

Using partial fractions, we can write the right-hand side of Eq. (5.36) in theform

K

s tsþ 1ð Þ ¼a

sþ b

tsþ 1

We can start to obtain the constants a and b by putting the equation in theform

K

s tsþ 1ð Þ ¼a tsþ 1ð Þ þ bs

s tsþ 1ð Þwhich gives

K ¼ atþ bð Þsþ a

On equating the coefficients, we get

a ¼ K

b ¼ �Kt

Thus, we can write

yðsÞ ¼ K

s� Kt

tsþ 1ð ÞIn a more simplified form,

yðsÞ ¼ K

s� K

s� �1=tð ÞOn taking the inverse Laplace transform of the two parts, we get

yðtÞ ¼ K � Ke�t=t


yðtÞ ¼ K 1� e�t=t� � ð5:37ÞThis is the solution of our differential equation.

The reader is advised to solve the same differential equation using theoperator method (see Appendix C).

Solution of Higher-Order ODE by Laplace Transformation

Consider the second-order ODE for which y00 represents the second deriva-tive of y and y0 represents the first derivative of y:

y00ðtÞ þ 5y0ðtÞ þ 6yðtÞ ¼ f ðtÞ ð5:38Þwith

f ðtÞ ¼ 1; yð0Þ ¼ 1:0; y0ð0Þ ¼ 0

The Laplace transformation of the differential Eq. (5.38) gives

s2yðsÞ � syð0Þ � y0ð0Þ �þ 5 syðsÞ � yð0Þð Þ½ � þ 6yðsÞ ¼ 1

s

For yð0Þ ¼ 1:0 and y0ð0Þ ¼ 0, this equation becomes,

s2yðsÞ � s� 0þ 5 ½syðsÞ � 1�ð Þ þ 6yðsÞ ¼ 1

s

Further simplification yields

s2yðsÞ þ 5syðsÞ þ 6yðsÞ ¼ 1

sþ sþ 5ð Þ

Rearrangement gives

s2 þ 5sþ 6� �

yðsÞ ¼ 1

sþ sþ 5ð Þ

From the above equation, we can obtain the expression for yðsÞ as

yðsÞ ¼ ð1=sÞ þ sþ 5ð Þs2 þ 5sþ 6

which finally simplifies to,

yðsÞ ¼ 1þ s2 þ 5s

s s2 þ 5sþ 6� � ð5:39Þ

Using partial fractions and employing the fact that

s2 þ 5sþ 6 ¼ sþ 3ð Þ sþ 2ð Þwe can write the right-hand side of Eq. (5.39) as

1þ s2 þ 5s

s s2 þ 5sþ 6� � ¼ A

sþ B

sþ 3þ C

sþ 2ð5:40Þ

Rearrangement gives

1þ s2 þ 5s

s sþ 3ð Þ sþ 2ð Þ ¼A sþ 3ð Þ sþ 2ð Þ þ Bs sþ 2ð Þ þ Cs sþ 3ð Þ

s sþ 3ð Þ sþ 2ð ÞFrom the above equation, we get

1þ s2 þ 5s ¼ A s2 þ 5sþ 6� �þ B s2 þ 2s

� �þ C s2 þ 3s� �


1þ s2 þ 5s ¼ s2 Aþ Bþ Cð Þ þ s 5Aþ 2Bþ 3Cð Þ þ 6A

Equating the coefficients give

6A ¼ 1

Aþ Bþ C ¼ 1

5Aþ 2Bþ 3C ¼ 5

On solving the above three linear algebraic equations in terms of A, B, andC, we get

A ¼ 1

6; B ¼ � 5

3; C ¼ 5

2

Using the inverse Laplace transformation on Eq. (5.40), we get the solutionof the differential equation as

y tð Þ ¼ 1

6� 5

3e�3t þ 5

2e�2t

5.6.7 Main Process Control Applications of Laplace andInverse Transformations

Figure 5.26 shows a flow diagram for the Laplace transformation andinverse transformation. It is clear that the main function of the Laplacetransformation is to put the differential equation (in the time domain)into an algebraic form (in the s-domain). These s-domain algebraic equa-tions can be easily manipulated as input–output relations.

5.7 CHARACTERISTICS OF IDEAL FORCINGFUNCTIONS

The ideal forcing functions are the following:

1. Ideal step function (Fig. 5.27)

uðtÞ ¼ 0 for t < 0A for t > 0

or

uðtÞ ¼ AHðtÞ

Figure 5.26 Laplace transformation and the inverse transformation.

where HðtÞ is the Heaviside function defined as

HðtÞ ¼ 0 for t < 01 for t > 0

The Laplace transformation of the step function is

uðsÞ ¼ A

s

2. Dirac delta function (Fig. 5.30). It is the limit of the rectangularpulse function (Fig. 5.28). Let

A ¼ 1

b

Thus,

A b ¼ 1

bb ¼ 1

So, as b ! 0;A ! 1.The Dirac delta function is a rectangular-pulse function of zero

width and unit area:

L dðtÞ� � ¼ ð10

e�std tð Þ dt

¼ðe�e

e�std tð Þ dt

¼ e�0

Thus,

L dðtÞ� � ¼ 1

Figure 5.27 Ideal step function.

For any function f ðtÞðt0þe

t0�ed t� t0ð Þ f tð Þ dt ¼ f t0ð Þ

Here, d t� t0ð Þ is the general shifted delta function (zero every-where except at t ¼ t0). For t0 ¼ 0, we getðe

�ed tð Þ f tð Þ dt ¼ f 0ð Þ

For f ðtÞ ¼ Aðe�e

d tð ÞAdt ¼ A

3. The ideal rectangular-pulse function (Fig. 5.29)

u tð Þ ¼0 for t < 0A for 0 < t < b0 for t > b

8<:

Hðt� bÞ ¼ 0 for t < b1 for t > b

So

u tð Þ ¼ A H tð Þ �H t� bð Þ½ �Thus

u sð Þ ¼ A1

s� e�bs 1

s

� �¼ A

s1� e�bs� �

Figure 5.28 Ideal rectangular pulse function which becomes Dirac as b ! 1.

4. Ideal impulse function (Dirac delta function) (Fig. 5.30)

u tð Þ ¼ Ad tð Þ

It is infinite at the point t ¼ 0 and zero elsewhere,

dðtÞ ¼ 1 for t ¼ 00 elsewhere

andð1�1

d tð Þdt ¼ðe�e

d tð Þdt ¼ 1

u sð Þ ¼ L u tð Þ� � ¼ A

Figure 5.29 Ideal rectangular-pulse function.

Figure 5.30 Ideal impulse function.

The Dirac delta function is the derivative of the unit step (orHeaviside function) (Fig. 5.31). H tð Þ is the unit step function:

d H tð Þ½ �dt

¼ d tð Þ

It is not possible to implement it exactly experimentally.

5. The ideal ramp function (Fig. 5.32)

uðtÞ ¼ 0 for t < 0At for t > 0

The Laplace transformation is

u sð Þ ¼ L u tð Þ� � ¼ A

s2

6. The ideal sinusoidal function (Fig. 5.33)

uðtÞ ¼ 0 for t < 0A sin otð Þ for t > 0

The Laplace transformation is

u sð Þ ¼ L u tð Þ� � ¼ Aos2 þ o2

It is difficult to implement exactly experimentally. However, theresponse to a theoretical sinusoidal function is very useful inprocess dynamics and design of controllers.

Figure 5.31 Unit step function.

Realization of Ideal Forcing Functions

If the input of interest is the steam flow rate (Figure 5.34), then,

1. Step input. Open the steam valve a given percentage at t ¼ 0 suchthat Q changes by A units.

2. Pulse input. Open the steam valve at t ¼ 0, hold at the new valuefor a duration of b time units, and then return to the old value.

3. Impulse (impossible to realize perfectly). Open the steam valve(wide open) at t ¼ 0 and instantaneously (or as soon as possiblethereafter) return to the initial position.

4. Ramp input. Gradually open the steam valve such that Qincreases linearly. Ramp ends when the steam valve is fully open.

5. Sinusoidal input. The only practical way to achieve this is to con-nect a sine-wave generator to the steam valve. Realizing high-frequency sinusoidal input may be limited by the valve dynamics.

Figure 5.32 Ideal ramp function.

Figure 5.33 Ideal sinusoidal function.

Important Highlights

Process Model

. The idea of utilizing a collection of mathematical equation as a‘‘surrogate’’ for a physical process is at once ingenious as well asexpedient. More importantly, however, this approach has nowbecome indispensable to proper analysis and design of processcontrol systems (and also design of equipment).

. Classical control theory concentrates on the transfer-functionmodel in the s-domain (whether it is obtained from Laplace trans-formation of rigorous models, or from empirical fitting to the Bodediagram of the rigorous model, or from fitting to experimentalresponses).

Mathematical Description of Chemical Process

. Process control requires understanding of process dynamics.

. The main use of the mathematical model is as a convenient ‘‘sur-rogate’’ for the physical system, making it possible to investigatesystem response under various input conditions, both rapidly andinexpensively without necessarily tampering with the actual entity.

. It is usually not the ‘‘exact’’ equivalent of the process (but we canmake it so, and we should use ‘‘optimum degree of sophistication’’).

Figure 5.34 Variables of a stirred heating tank.

. Note the title ‘‘lumped/distributed parameter systems (models)’’ isnot correct because it is not the ‘‘parameter’’ that is distributed, itis the state variable in fact.

Process Characteristics and Process Models

. Dependent variables: all inputs, output, disturbances, and so forth.

. Independent variables: time and space

. Linear and nonlinear systems.

. Lumped (ODEs) and distributed (PDEs) systems

. Discrete-time system

Even though the tacit assumption is that the variables of a process donot ordinarily change in ‘‘jumps,’’ that they normally behave as smoothcontinuous functions of time (and position), there are situations in whichoutput variables are deliberately sampled and control action implementedonly at discrete points at time. The process variables then appear to changein a piecewise constant fashion with respect to the now discretized time.Such processes are modeled by difference equations and are referred to asdiscrete-time systems.

5.8 BASIC PRINCIPLES OF BLOCK DIAGRAMS,CONTROL LOOPS, AND TYPES OF CLASSICALCONTROL

Laplace transformation is the best way to handle classical linear controlproblems using P (proportional), I (integral), and D (derivative) controllers.The controller is activated by the error e tð Þ (the difference between themeasured variable and the set point) to give its output signal y tð Þ. Therelation between e tð Þ and y tð Þ depends on the type of controller used. It ismuch easier to handle this relation in the Laplace domain. In the Laplacedomain, e tð Þ becomes ~EE sð Þ and y tð Þ becomes ~YY sð Þ and they are related toeach other through the transfer function G sð Þ.

Types of Classical Control (Fig. 5.35)

Proportional Controller

For this type of control, the time domain relation between the input andoutput is

y tð Þ ¼ KCe tð ÞIn the Laplace domain, the relation becomes

~YY sð Þ ¼ KC~EE sð Þ

Thus, the transfer function GC is given by GC ¼ KC for a proportionalcontroller.

Integral Controller

For this type of controller, the input–output relation in the time domain is

y tð Þ ¼ KI

ðt0

e tð Þ dt

In the Laplace domain, it becomes

~YY sð Þ ¼ KI

s~EE sð Þ

Thus, GC ¼ KI=s for an integral controller.Note that the integral relation in the time domain has changed into an

algebraic relation in the Laplace domain.

Derivative Controller

In the time domain, the input–output relation is

y tð Þ ¼ KD

de tð Þdt

In the Laplace domain, it is

~YY sð Þ ¼ KDs ~EE sð ÞThus, GC ¼ KDs for a derivative controller.

Note that the differential relation in the time domain has changed intoan algebraic relation in the Laplace domain.

Three-Mode Controller (PID: Proportional, Integral, and Derivative)

The input–output proportional-integral-derivative relation in the timedomain is

y tð Þ ¼ KCe tð Þ þ KI

ðt0

e tð Þ dtþ KD

de tð Þdt

In the Laplace domain, the above relation becomes an algebraic relation:

Figure 5.35 Input–output relation.

~YY sð Þ ¼ KC~EE sð Þ þ KI

s~EE sð Þ þ KDs ~EE sð Þ

which can be rearranged in the simpler form

~YY sð Þ ¼ KC þ KI

sþ KDs

� �~EE sð Þ

Thus,

GC ¼ KC þ KI

sþ KDs

� �

for a PID controller.

A Very Simple Example

Consider the tank shown in Figure 5.36. The unsteady-state mass balanceequation is given by the differential equation

Adh

dt¼ qi � q0 ð5:41Þ

where A is the cross-sectional area of the tank and h is the liquid heightinside the tank.

Figure 5.36 Filling tank with feedback proportional controller.

With the initial condition (at steady state) given by t ¼ 0, h ¼ hSS; qi ¼qiSS; q0 ¼ q0SS which means that at t ¼ 0, the system is at steady state.

We define deviation (or hat) variables as follows:

hh ¼ h� hSS

qqi ¼ qi � qiSS

qq0 ¼ q0 � q0SS

Thus, the differential Eq. (5.41) becomes

Adhh

dt¼ qqi þ qiSSð Þ � qq0 þ q0SSð Þ


Adhh

dt¼ qqi � qq0 þ qiSS � q0SSð Þ|fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl}

zero

Thus, the above equation simplifies to give

Adhh

dt¼ qqi � qq0 ð5:42Þ

with the initial condition at t ¼ 0; hh ¼ 0; qqi ¼ 0, and qq0 ¼ 0, which meansthat all deviation (hat) variables are equal to zero at initial time.

The Laplace transform of Eq. (5.42) gives

As ~hh sð Þ ¼ ~qqi sð Þ � ~qq0 sð Þwhich can be written as

~hh sð Þ ¼ 1

As~qqi sð Þ �

1

As~qq0 sð Þ ð5:43Þ

This is the open-loop equation (without the effect of feedback control)in the Laplace domain.

We can put the equation in the form of the shown open-loop blockdiagram in Figure 5.37,

Closed Loop (Including the Effect of the Feedback Control)

Consider a proportional controller relating ~hh sð Þ to ~qq0 sð Þ. In time scale, weconsider a very simple proportional controller making the flow rate out ofthe tank proportional to the height in the tank (this is an approximation toexplain the steps without the lengthy linearization steps):

q0 ¼ KCh

Note: We consider q0 / h rather than q0 /ffiffiffih

pfor simplicity of illustra-

tion. We will discuss the nonlinear case and the associated linearizationprocess later.

Using deviation variables, we can write

q0 � q0SS ¼ KCh� KChSS

In terms of deviation (hat) variables,

qq0 ¼ KChh

Taking the Laplace transform of the above equation gives

~qq0 sð Þ ¼ KC~hh sð Þ:

Because qq0 0ð Þ ¼ 0 and hh 0ð Þ ¼ 0 (5.44)

We can represent this in a block diagram as shown in Figure 5.38.From equations (5.43) and (5.44), we get

~hh sð Þ ¼ 1

As~qqi sð Þ �

1

AsKC

~hh sð Þ ð5:45Þ

Figure 5.37 Block diagram for the open-loop system.

Figure 5.38 Block diagram for the closed-loop control.

Closed-Loop Block Diagram

From Eq. (5.45), we can construct the following closed-loop (feedback)block diagram as shown in Figure 5.39. We can also rearrange Eq. (5.45)into the following form:

~hh sð Þ 1þ KC

As

� �¼ 1

As~qqi sð Þ

The above equation can be reduced to

~hh sð Þ ¼ 1=As

1þ KC=As~qqi sð Þ


~hh sð Þ ¼ GC sð Þ ~qqi sð Þ ð5:46Þwhere

GC sð Þ ¼ 1=As

1þ KC=As

Equation (5.46) can be drawn as shown in Figure 5.40.A more general form for the closed loop (feedback) controlled process

can be drawn as in Figure 5.41.

Figure 5.39 Alternate block diagram for the closed-loop control.

The transfer functions for the different parts can be written as follows:

Process, Open-Loop Equation

X sð Þ ¼ Gd sð ÞD sð Þ � GP sð ÞC sð Þ ð5:47Þ

Controller Equation

C sð Þ ¼ GC sð ÞX sð Þ ð5:48ÞFrom relations (5.47) and (5.48), we obtain

X sð Þ ¼ Gd sð ÞD sð Þ � GP sð ÞGC sð ÞX sð Þwhich can be rearranged in the form

X sð Þ 1þ GP sð ÞGC sð Þ½ � ¼ Gd sð ÞD sð Þ

Figure 5.40 Block diagram for Eq. (5.46).

Figure 5.41 A more general block diagram.

Finally, we get the relation between the output X sð Þ and the disturbanceD sð Þ:

X sð Þ ¼ Gd sð Þ1þ GP sð ÞGC sð ÞD sð Þ

Of course, this relation can be written as

X sð Þ ¼ G sð ÞD sð Þwhere G sð Þ is the overall transfer function given by

G sð Þ ¼ Gd sð Þ1þ GP sð ÞGC sð Þ

For the Tank Problem

Gd sð Þ ¼ 1

As

and

GP sð Þ ¼ 1

As

For the proportional controller,

GC sð Þ ¼ KC ð¼ CSS þ KC; where CSS is the valve coefficient and

KC is the proportional gain of the controller)

so we get

X sð Þ|ffl{zffl}~hh sð Þ

¼ 1=As

1þ KCð1=AsÞD sð Þ|ffl{zffl}~qqi sð Þ

Response to a Step Input in the Feed

Figure 5.42 shows the step change in input flow rate qi to the tank and itseffect on the deviation variable qqi ¼ qi � qiSS. For this step function, we havein term of the deviation variable qqi:

For t < 0, qqi ¼ 0.

For t 0; qqi ¼ A

The Laplace transformation for ~qqi sð Þ � D sð Þ½ � is given by

D sð Þ ¼ A

s

Therefore, X sð Þ½� ~hh sð Þ� is given by

X sð Þ ¼ 1=As

1þ KCð1=AsÞA

s¼ 1

As

1

1þ KCð1=AsÞA

s

¼ A

A

1

s

1

s

1

ðAsþ KCÞ=As¼ A

A

1

s2As

Asþ KC

Thus, after some simple manipulations, X sð Þ is given by

X sð Þ ¼ A1

s KC þ As� � ð5:49Þ

Using partial fractions, we get

1

s KC þ As� � ¼ a1

sþ a2

KC þ As

¼ a1 KC þ As� �þ a2s

s KC þ As� �

¼ a1KC þ Aa1 þ a2ð Þss KC þ As� �

From the above relation, we get

a1KC ¼ 1

giving

a1 ¼1

KC

Figure 5.42 Step change in input flow rate and corresponding change in deviation

variable (not to scale).

and

a1Aþ a2 ¼ 0; leading to

a2 ¼ � A

KC

Thus, we can write

1

s KC þ As� � ¼ 1=KC

sþ �A=KC

KC þ As

Therefore, relation (5.49) becomes

X sð Þ ¼ 1

KC

1

s� 0ð Þ �1

KC

1

s� �KC=A� �

!A ð5:50Þ

Taking the inverse Laplace transformation of Eq. (5.50), we get

X tð Þ ¼ hh tð Þ ¼ A

KC

e0t � e�ðKC=AÞt� �

ð5:51Þ

Let us say

KC

A¼ 1

t

We get

hh tð Þ ¼ A

KC

1� e�t=t� � ð5:52Þ

Does hh tð Þ go to zero as t ! 1 (i.e., h ! hSSÞ?The answer is ‘‘no’’; there is an offset which decreases as KC increases,

hh 1ð Þ ¼ A

KC

offsetð Þ

Thus, as KC increases, the offset decreases, as can be seen in Figure 5.43.Note that at steady state, the deviation variables (after the step increase in qqiand without controller) will be given by the simple relation

qqi ¼ A ¼ CSS hSSnew � hSS� �

Thus, the new steady state due to the input increase (without controller) willbe

hSSnew � hSS ¼ hhSSnew ¼ A

CSS

Now, when we have a proportional controller with a proportionalgain ¼ KC, we get

q0 ¼ CSShþ KC h� hSSð ÞHere KC is the controller proportional gain. When h ¼ hSS (at steady state),we have

q0SS ¼ CSShSS

Using the above two equations, we get the deviation variable

qq0 ¼ CSS þ KCð Þhh ¼ KChh

Thus, as KC increases, KC increases over CSS. The response becomes fasterand the offset smaller, as seen in Figure 5.44.

If KC ¼ 0, then KC ¼ CSS. Note that in all of the above calculations,we have used the relation q0 ¼ Ch for simplicity to illustrate the main ideas

(in fact, we should have used q0 ¼ Cffiffiffih

p). For KC > 0, it goes toward

hh 1ð Þ ¼ A=KC offsetð Þ. As KC increases, the offset decreases. WhenKC � 1:0, then the system tends to hh ¼ 0, which is the original steadystate or the set point.

What if the Controller Is PI?

In this case, the controller transfer function is given by

GC sð Þ ¼ KC þ KI

s

� �

Therefore,

X sð Þ|ffl{zffl}~hh sð Þ

¼ 1=Aszffl}|ffl{Gd sð Þ

1þ ð1=AsÞ|fflffl{zfflffl}GP sð Þ

KC þ KI=s� �|fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl}

GC sð Þ

D sð Þ|ffl{zffl}~qqi sð Þ

ð5:53Þ

Figure 5.43 Effect on offset.

For a step input, ~qqi sð Þ ¼ A=s

Thus, from Eq. (5.53), we get

~hh sð Þ ¼ 1=As

1þ ð1=AsÞ KC þ KI=s� � A

sð5:54Þ

On simplification, Eq. (5.54) gives

~hh sð Þ ¼ A

As2As2 þ KCsþ KI

As2

!�1

¼ A

As2 þ KCsþ KI

ð5:55Þ

Using partial fractions for Eq. (5.55), we get

~hh sð Þ ¼ A

s� �1ð Þ s� �2ð Þ ¼a1

s� �1þ a2s� �2

¼ a1s� a1�2 þ a2s� a2�1s� �1ð Þ s� �2ð Þ


a1 þ a2 ¼ 0

thus giving a1 ¼ �a2, and

�a1�2 � a2�1 ¼ A

which gives a1 ¼ A=ð�2 � �1Þ. From the above two equations, we obtain

a1 ¼A

�2 � �1and a2 ¼

A

�1 � �2

Now, to calculate �1 and �2,

�1;2 ¼�KC �

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiK

2C � 4AKI

q2A

Figure 5.44 Effect of controller.

So,

�1 � �2 ¼1

2A�KC þ


2C � 4AKI

qþ KC þ


2C � 4AKI

q� �

giving

�1 � �2 ¼1

A


2C � 4AKI

q� �

To have real values, choose 4AKI < K2C. Thus,

�1 � �2 ¼1

A


2C � 4AKI

q� �and �2 � �1 ¼ � 1

A


2C � 4AKI

q� �

On taking the inverse Laplace transformation of

~hh sð Þ ¼ a1s� �1ð Þ þ

a2s� �2ð Þ

we get

hh tð Þ ¼ a1e�1t þ a2e

�2t

Substitution of the values of a1 and a2 gives

hh tð Þ ¼ A

�2 � �1e�1t þ A

�1 � �2e�2t


hh tð Þ ¼ A

�1 � �2e�2t � e�1t� � ð5:56Þ

If both �1 and �2 are negative, then the closed-loop control system is stable,and as t ! 1, the value of hh tð Þ ! 0, as can be observed from Eq. (5.56).

What is the condition that �1 and �2 are negative? It is that KC > 0and 4AKI > 0 (and the condition of 4AKI < K

2C is also satisfied) which is, of

course, always satisfied. Thus, it is clear that the integral control actionalways prevents the offset, since �1 and �2 are always negative.

Another Example: Feedback Control of a CSTR

Consider the first-order irreversible chemical reaction

A ! B

with rate of reaction equal to r ¼ kCA taking place in the continuous-stirredtank reactor (CSTR) shown in Figure 5.45. The CSTR is assumed to beisothermal.

Open Loop, Uncontrolled

We start by looking at the open-loop (uncontrolled) case. The first step is todevelop the dynamic model. The dynamic mass balance equation is given bythe relation

ni þdnidt

¼ nif þ V�ir

Considering a constant flow rate and constant volume, we get

ni ¼ qCi; nif ¼ qCif ;dnidt

¼ VdCi

dt

Writing the balance for component A gives

qCAf ¼ VkCA þ qCA þ VdCA

dt

The above equation can be rewritten as

V

q

� �|ffl{zffl}

t

zffl}|ffl{residencetime

dCA

dt¼ CAf � CA

� �� Vk

q

� �|fflffl{zfflffl}

a

CA

Figure 5.45 Continuous-stirred tank reactor.

where

t ¼ V

q

a ¼ Vk

q

In terms of normalized parameters, it can be written as

tdCA

dt¼ CAf � CA

� �� aCA ð5:57Þ

For the control problem the initial condition is at t ¼ 0 and CA ¼ CASS. We

obtain the steady state CASSfrom the steady-state equation

CAf ¼ 1þ að ÞCASS


CASS¼ CAf

1þ að Þ ¼CAf

1þ Vk=qð Þ ¼qCAf

qþ Vkð ÞSubtracting

VdCA

dt¼ q CAf � CA

� �� VkCA

from

VdCASS

dt¼ q CAfSS � CASS

� �� VkCASS

to get the unsteady-state equation in terms of deviation variables, we get thefollowing equation in terms of deviation variables:

VdCCA

dt¼ q CCAf � CCA

� �� VkCCA ð5:58Þ

where

CCA ¼ CA � CASS

CCAf ¼ CAf � CAfSS

with initial condition at t ¼ 0, CCA ¼ 0 and CCAf ¼ 0:Laplace transformation of Eq. (5.58) gives

Vs ~CCA sð Þ ¼ q ~CCAf sð Þ � qþ Vkð Þ ~CCA sð ÞThe above equation can be rearranged as

~CCA sð Þ Vsþ qþ Vkð Þ ¼ q ~CCAf sð Þ

Finally, the relation between the output ~CCA sð Þ and the input ~CCAf sð Þ is given by

~CCA sð Þ ¼ q

ðVsþ qþ VkÞ~CCAf sð Þ ð5:59Þ

Equation (5.59) can be put in the form of a block diagram, as shown inFigure 5.46.

Now, if

qþ Vk

q¼ a0

we get

~CCA sð Þ ¼ V

q|{z}t

sþ qþ Vk

q|fflfflffl{zfflfflffl}a0

0BB@

1CCA

�1

~CCAf sð Þ

Thus, the above equation becomes

~CCA sð Þ ¼ 1

tsþ a0~CCAf sð Þ ð5:60Þ

Therefore, the input–output relation in the Laplace domain is given by

~CCA sð Þ ¼ 1

a0 þ ts~CCAf sð Þ ð5:61Þ

Consider a step function in the feed concentration; thus

~CCAf sð Þ ¼ A

s

where A is the size of the step. So, for this input, we get the followingresponse (output) using Eq. (5.61):

~CCA sð Þ ¼ 1

a0 þ tsA

s



~CCA sð Þ ¼ A

t1

sþ a0=tð Þ s� 0ð Þ� �

Using partial fractions, we get

1

sþ a0=t|{z}g

0@

1A s� 0ð Þ

¼ a1sþ g

þ a2s¼ a1sþ a2sþ a2g

sþ gð Þs

where

g ¼ a0

t


a2g ¼ 1

therefore,

a2 ¼1

g

and

a1 þ a2 ¼ 0

so it gives

a1 ¼ � 1

g

Thus, we get the response (still in the Laplace domain but rearranged for theease of inversion)

~CCA sð Þ ¼ A

t�1=g

s� �gð Þ þ1=gs� 0

� �ð5:62Þ

On taking the inverse Laplace transformation of Eq. (5.62), we get

~CCA tð Þ ¼ A

a01� e�ða0=tÞt� �

ð5:63Þ

Old Steady State (for CCAf=0Þ and New Steady State (for CCAf=A)

The old steady state CCASSold

� �is given for the following conditions:

CCAf ¼ 0 and CCASS¼ 0

The new steady state CCASSnew

� �(Fig. 5.47) can be obtained from the

steady state equation

0 ¼ q CCAf|{z}A

�CCASSnew

0B@

1CA� VkCCASSnew

which gives

qþ Vkð ÞCCASSnew¼ Aq

thus giving the new steady state as

CCASSnew¼ Aq

qþ Vk

Of course, the same can be obtained by setting t ! 1 in Eq. (5.63).

What if we apply a feedback control?

. Deviations of CA from CASSare used to manipulate q (assume

that the flow rate response is constant, with no time delays, i.e.,qin ¼ qout ¼ q always and the volume V is also constant).

. Note that as the flow rate increases, CA increases and vice versa. Inother words, when we want to compensate for an increase in CA,we should decrease q.

Thus, for proportional control,

q ¼ qSS þ K CASS� CA

� �because when CA is higher than CASS

, we want to decrease CA, of course, andin order to decrease it, we should decrease q and vice versa. Thus, thedynamic equation with the feedback controller becomes,

Figure 5.47 Attainment of new steady state.

qSS þ K CASS�CA

� � �|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}qc

CAf ¼VkCA þ VdCA

dtþ CA qSS þ K CASS

� CA

� � �|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}qc

ð5:64ÞNote the weak nonlinearity resulting from the control in the left-hand sideas well as the last term of the equation on the right-hand side.

The Steady-State Equation

The open-loop steady state is obtained by putting CASS� CA ¼ 0 in Eq.

(5.64):

qSSCAfSS¼ VkCASS

þ VdCASS

dtþ qSSCASS

ð5:65Þ

The closed-loop (controlled) steady-state equation is given by

qSSCAf þ K CASS� CA

� �CAf ¼VkCA þ V

dCA

dtþ qSSCA

þ K CASS� CA

� �CA

ð5:66Þ

Subtract Eq. (5.65) from Eq. (5.66) and define the following deviation vari-ables:

CCA ¼ CA � CASS

CCAf ¼ CAf � CAfSS

Equation (5.66) in terms of deviation (or hat) variables becomes

qSSCCAf þ K CAf CCA|fflfflffl{zfflfflffl}a

¼ VkCCA þ VdCCA

dtþ qSSCCA þ K CACCA|fflffl{zfflffl}

b

with initial condition at t ¼ 0, CCA ¼ 0 and CCAf ¼ 0. Note that we have twononlinear terms a and bð Þ. For linear analysis, we need to linearize them.For nonlinear analysis, we can solve numerically and we do not need tolinearize them.

5.9 LINEARIZATION

For the term labeled a in the above equation,

CAf CCA ¼ CAf CASS� CA

� � ¼ CAfCASS|fflfflfflfflffl{zfflfflfflfflffl}linear in CAf

�CAfCA|fflfflffl{zfflfflffl}nonlinear

To linearize the nonlinear term (last term of the above relation), we use theTaylor’s series expansion (and neglecting the higher-order terms), we get

CAfCA¼CAfSSCASS

þ @ CAfCA

� �@CA

��SS

CASS� CA

� �þ @ CAfCA

� �@CAf

��SS

CAfSS�CAf

� �

Thus, the nonlinear term a is given in the following linearized form:

CAfCA ¼ CAfSSCASS

þ CAfSSCCA þ CASS

CCAf|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}linearized form of CAf CA

ð5:67Þ

Therefore, using Eq. (5.67), we get the expression of CAf CCA ¼ a as follows,

CAf CCA ¼ CAfCASS|fflfflfflfflffl{zfflfflfflfflffl}linear in CAf

� CAfCA|fflfflffl{zfflfflffl}non-linear

¼ CAfCASS� CAfSSCASS|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

�CCAf CASS

�CAfSSCCA � CASSCCAf

¼ �CCAf CASS� CAfSS

CCA � CASSCCAf

¼ �2CCAf CASS� CAfSS

CCA

Thus, the nonlinear term a has been linearized.The reader is encouraged to do the same for the other nonlinear term

(b) and analyze the resulting linear differential equation.The reader should show the conditions for the stability of the system

also show the effect of the proportional controller gain on stability, speed ofresponse, and offset.

5.10 SECOND-ORDER SYSTEMS

Second-order models are usually used to empirically represent some pro-cesses (D.J. Cooper, Control Station for Windows—A Software for ProcessControl Analysis, Tuning and Training. http://www.ControlStation.com).

Consider a typical second-order system described by the followingsecond-order differential equation:

a2d2y

dt2þ a1

dy

dtþ a0y ¼ bf tð Þ ð5:68Þ

Here, y is the state variable, t is an independent variable (time), f tð Þ is theforcing function (nonhomogeneous term), and a0, a1, a2 and b are the para-meters. Usually, the typical differential Eq. (5.68) for second-order systemsis written in a different form:

t2d2y

dt2þ 2xt

dy

dtþ y ¼ KPf tð Þ ð5:69Þ

where

t ¼ffiffiffiffiffia2a0

r

x ¼ a12ffiffiffiffiffiffiffiffiffia0a2

p

KP ¼ b

a0

The physical significance of the above parameters are as follows: t is thenatural period of oscillation of the system [the autonomous system whenf tð Þ ¼ 0], x is damping factor, and KP is static (steady state) gain of thesystem

If Eq. (5.69) is in terms of deviation variables (or the hat variables) and

yy 0ð Þ ¼ 0;dyy 0ð Þdt

¼ 0; andd2yy 0ð Þdt2

¼ 0

then the Laplace transformation of Eq. (5.69) gives

t2s2 ~yy sð Þ þ 2xts ~yy sð Þ þ ~yy sð Þ ¼ KP~ff sð Þ

which can be arranged in the following input ~ff sð Þ and output ~yy sð Þ form:

~yy sð Þ ¼ KP

t2s2 þ 2xtsþ 1

� �|fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}transfer function¼GðSÞ

~ff sð Þ ð5:70Þ

The transfer function is given by

G sð Þ ¼ KP

t2s2 þ 2xtsþ 1

� �ð5:71Þ

Response to Unit Step Input

For a unit step input, ~ff sð Þ is given by

~ff sð Þ ¼ 1

s

So, using Eq. (5.70), we get the output ~yy sð Þ in the form

~yy sð Þ ¼ KP

t2s2 þ 2xtsþ 1

� �1

s

Using the partial fractions (or Heaviside theorem), we can obtain the inverseLaplace transform.

Suppose the roots of t2s2 þ 2xtsþ 1 are �1 and �2; we can write

~yy sð Þ ¼ KP

s� 0ð Þ s� �1ð Þ s� �2ð Þand

�1; �2 ¼ � 2xt�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi4x2t2 � 4t2

p2t2

which gives

�1 ¼ � xtþ

ffiffiffiffiffiffiffiffiffiffiffiffiffix2 � 1

pt

and �2 ¼ � xt�

ffiffiffiffiffiffiffiffiffiffiffiffiffix2 � 1

pt

It is clear that the character of �1 and �2 will depend on the value of x:

1. For x > 1:0, both �1 and �2 are real (the system is called over-damped).

2. For x ¼ 1:0, �1 and �2 are equal to �x=t (the system is calledcritically damped). These are repeated roots, so the reader isadvised to be careful while taking the inverse transform.

3. For x < 1:0, �1 and �2 are complex in nature (the system is calledunderdamped or oscillatory).

5.10.1 Overdamped, Critically Damped, and UnderdampedResponses

The reader is advised to do the complete analytical manipulation for thethree cases.

. x > 1:0, distinct real roots

. x ¼ 1:0, repeated real roots

. x < 1:0, complex roots

The typical responses for the above-mentioned cases are shown in Figure5.48.

5.10.2 Some Details Regarding the Underdamped Response

As an example for the reader, some analysis of the underdamped system isgiven here (refer to Figure 5.49). Note the following:

. o ¼ radian frequency (radians per unit time)

. f ¼ cyclic frequency (cycles per unit time)

. T ¼ period of one cycle ¼ 1=f

. o ¼ 2f ¼ 2=T

The main characteristics of the response are as follows:

1. Overshoot

A

B¼ exp � xffiffiffiffiffiffiffiffiffiffiffiffiffi

1� x2p

!

2. Decay ratio

C

A¼ exp � 2xffiffiffiffiffiffiffiffiffiffiffiffiffi

1� x2p

!¼ overshootð Þ2

Figure 5.48 Typical overdamped, critically damped, and underdamped systems.

Figure 5.49 Typical underdamped system.

3. Period of oscillation

T ¼ 2tffiffiffiffiffiffiffiffiffiffiffiffiffi1� x2

p ¼ 2

o;

where o ¼ffiffiffiffiffiffiffiffiffiffiffiffiffi1� x2

p=t

4. Natural period of oscillation

Tnatural ¼ T when x ¼ 0 i.e., Tnatural ¼ 2tð Þ5. Response time: When the system approaches �5% of its final

value, it is considered that the system has reached practicallythe steady state and the time elapsed is the response time.

6. Rise time: It is the time when the response curve cuts the hori-zontal steady-state line for the first time.

5.11 COMPONENTS OF FEEDBACK CONTROL LOOPS

Figure 5.50 shows the typical components found in a control loop. It can beseen that there are four main components involved: process, measuringdevice, controller, and the final control element. The controller mechanismgets the value of the set point and directs the final control element to carryout the actions. Disturbances enter the system through the process, whichaffect the process variables.

Figure 5.50 Typical components of a control loop.

Example 1: Pressure control

Figure 5.51 shows an example of a tank pressure control mechanism. It canbe seen that the measured variable is the pressure inside the tank. Thecontroller mechanism receives the pressure value and its set point andaccordingly directs the control valve to either open or close to keep thepressure constant inside the tank.

Example 2: Temperature control

Figure 5.52 shows the control loop for a heat exchanger. The outlet tem-perature of the fluid to be heated is the measured variable and it is sent tothe controller mechanism to be compared with the temperature set point.Based on the error, the controller takes action to either increase or decreasethe steam flow rate into the heat exchanger.

5.12 BLOCK DIAGRAM ALGEBRA

5.12.1 Typical Feedback Control Loop and the TransferFunctions

In this subsection, we present the typical control loop (as shown in Fig. 5.53)with its transfer functions, inputs–outputs, and their relations. It is also asimple introduction to block diagram algebra.

Figure 5.51 Typical tank pressure control loop.

The process: The input variables [m sð � and d s½ Þ] enter the process andthe transfer functions are related through the relation

y sð Þ ¼ GP sð Þm sð Þ þ Gd sð Þ d sð Þ

The measuring device: The input, output, and transfer function of themeasuring instrument are related through the relation

ym sð Þ ¼ Gm sð Þy sð Þ

Figure 5.52 Typical temperature control loop for heat exchanger.

Figure 5.53 Typical control loop with transfer functions and input–output

variables.

Controller mechanism: The controller is formed of the following twoparts:

Comparator: This part has the following relation producing the errorbetween the measured variable and the set point:

e sð Þ ¼ ysp sð Þ � ym sð ÞController: This part produces the controller action C sð Þ from the errore sð Þ, where Gc sð Þ is the transfer function of the controller dependingon the mode of the controller, as shown earlier:

C sð Þ ¼ Gc sð Þe sð ÞFinal control element: The relation for the final control element is givenby (usually the final control element is a valve)

m sð Þ ¼ Gf sð ÞC sð Þ

5.12.2 Algebraic Manipulation of the Loop TransferFunctions

Backward substitution into the process equation gives

m sð Þ ¼ Gf sð ÞGc sð Þe sð ÞSubstitution of the e sð Þ value gives

m sð Þ ¼ Gf sð ÞGc sð Þ ysp sð Þ � ym sð Þ� �Substitution of ym sð Þ gives

m sð Þ ¼ Gf sð ÞGc sð Þbysp sð Þ � Gm sð Þy sð ÞcSubstituting into the process equation gives the relation

y sð Þ ¼ GP sð Þ Gf sð ÞGc sð Þ ysp sð Þ � Gm sð Þy sð Þ �� þ Gd sð Þ d sð Þwhich is further expanded to give

y sð Þ ¼ GP sð ÞGf sð ÞGc sð Þ �

ysp sð Þ � GP sð ÞGf sð ÞGc sð ÞGm sð Þ �y sð Þ þ Gd sð Þ d sð Þ

Rearrangement of the above equation gives

y sð Þ 1þ GP sð ÞGf sð ÞGc sð ÞGm sð Þ � ¼ GP sð ÞGf sð ÞGc sð Þ �|fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

G sð Þ

ysp sð Þ þ Gd sð Þ d sð Þ

ð5:72ÞThe forward path overall transfer function is equal to

G sð Þ ¼ GP sð ÞGf sð ÞGc sð Þ

Thus, Eq. (5.72) can be written as

y sð Þ 1þ G sð ÞGm sð Þ½ � ¼ G sð Þysp sð Þ þ Gd sð Þ d sð Þ


y sð Þ ¼ G sð Þ1þ G sð ÞGm sð Þ� �|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl}

Gsp sð Þ

ysp sð Þ þ Gd sð Þ1þ G sð ÞGm sð Þ� �|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl}

Gload sð Þ

d sð Þ

Now, we define

Gsp sð Þ ¼ G sð Þ1þ G sð ÞGm sð Þ and Gload sð Þ ¼ Gd sð Þ

1þ G sð ÞGm sð ÞThus, we get

y sð Þ ¼ Gsp sð Þysp sð Þ þ Gload sð Þ d sð Þ ð5:73Þ

Equation (5.73) can be illustrated as a block diagram, as shown in Figure5.54.

Two types of closed-loop controls use the following:

1. The disturbance does not change, it is the set point which changesd sð Þ ¼ 0 �

. This is called the ‘‘servo problem’’ and the processequation becomes

y sð Þ ¼ Gsp sð Þysp sð Þ

2. The set point is constant ysp sð Þ ¼ 0 �

, but the disturbancechanges. This is called the ‘‘regulatory problem’’ and the processequation becomes

y sð Þ ¼ Gload sð Þ d sð Þ


Example: Control of the Liquid Level in a Tank

In this example, the liquid level inside a tank is controlled using a DPC(differential pressure cell). The DPC sends out the measured variable to thecomparator, where it is compared with the set point. Based on the error, thecontrol action is transmitted to the final control element (a control valve) inorder to either increase or decrease the outflow rate of liquid from the tank,as shown in Figure 5.55.

5.12.3 Block Diagram and Transfer Functions

These are related to the model equations shown earlier and are repeatedagain here (see Fig. 5.56).

The transfer functions in the process part of the loop ðGp and GdÞ: The dy-namic equation can be simply written as

Adh

dt¼ qi � q0 ð5:74Þ

In terms of deviation variables

Adhh

dt¼ qqi � qq0 ð5:75Þ

with initial condition, at t ¼ 0, hh ¼ 0, qqi ¼ 0, and qq0 ¼ 0. Taking the Laplacetransformation of Eq. (5.75) yields

As ~hh sð Þ ¼ ~qqi sð Þ � ~qq0 sð Þ

Figure 5.55 Liquid-level control in a tank.

and rearrangement gives

~hh sð Þ ¼ 1

As~qqi sð Þ �

1

As~qq0 sð Þ ð5:76Þ

The transfer function for the measuring instrument Gm sð Þ½ �: The measuringdevice is a DPC. It will be described in most cases by a second-ordersystem as follows:

t2d2hmdt2

þ 2xtdhmdt

þ hm ¼ KP�P ¼ KPah ð5:77Þ

After Laplace transformation, Eq. (5.77) can be put in the following form:

~hhm sð Þ~hh sð Þ

¼ aKP

t2s2 þ 2xtsþ 1¼ Gm sð Þ ð5:78Þ

Comparator:

~ee sð Þ ¼ ~hhm sð Þ � ~hhsp sð Þ ð5:79Þ

The transfer function for the controller (PI controller):

~CC sð Þ~ee sð Þ ¼ KC 1þ 1

tI s

� �¼ GC sð Þ ð5:80Þ

where

tIKC

¼ 1

KI

givesKC

tI¼ KI

Figure 5.56 Block diagrams and transfer functions.

Control valve (final control element) for a first-order system

~qq0 sð Þ~CC sð Þ ¼

KV

1þ tI s¼ Gf sð Þ ð5:81Þ

Algebraic manipulations: Equation (5.76) can be rewritten as

~hh sð Þ ¼ 1

As|{z}Gd sð Þ

~qqi sð Þ|ffl{zffl}di sð Þ

� 1

As|{z}Gp sð Þ

~qq0 sð Þ

Making use of relations (5.78)–(5.81), we get

~hh sð Þ ¼ Gd sð Þ ~ddi sð Þ � Gp sð ÞGf sð ÞC sð Þ¼ Gd sð Þ ~ddi sð Þ � Gp sð ÞGf sð ÞGc sð Þ~ee sð Þ¼ Gd sð Þ ~ddi sð Þ � Gp sð ÞGf sð ÞGc sð Þ½ ~hhm sð Þ � ~hhsp sð Þ�¼ Gd sð Þ ~ddi sð Þ � Gp sð ÞGf sð ÞGc sð ÞGm sð Þ ~hh sð Þ þ Gp sð ÞGf sð ÞGc sð Þ ~hhsp sð Þ

Rearrangement gives

1þ Gp sð ÞGf sð ÞGc sð ÞGm sð Þ �~hh sð Þ ¼ Gp sð ÞGf sð ÞGc sð Þ|fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl}

G sð Þ

~hhsp sð Þ þ Gd sð Þ ~ddi sð Þ

Set

G sð Þ ¼ Gp sð ÞGf sð ÞGc sð Þto get

1þ G sð ÞGm sð Þ½ � ~hh sð Þ ¼ G sð Þ ~hhsp sð Þ þ Gd sð Þ ~ddi sð ÞRearranging gives

~hh sð Þ ¼ G sð Þ1þ G sð ÞGm sð Þ|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl}

Gsp sð Þ

~hhsp sð Þ þ Gd sð Þ1þ G sð ÞGm sð Þ|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl}

Gload sð Þ

~ddi sð Þ ð5:82Þ

In shorter notation

~hh sð Þ ¼ Gsp sð Þ ~hhsp sð Þ þ Gload sð Þ ~ddi sð Þ ð5:83ÞThe above relation can be drawn as shown in Figure 5.57.

Servo problem:

~ddi sð Þ ¼ 0

Regulatory control:

~yysp sð Þ ¼ 0

For the regulatory problem

~hh sð Þ~ddi sð Þ

¼ Gd sð Þ1þ G sð ÞGm sð Þ

Substitution of the values of Gd sð Þ, G sð Þ, and Gm sð Þ gives~hh sð Þ~ddi sð Þ

¼ 1

As

� �1þ 1

AsKC 1þ 1

tI s

� �KV

1þ tVs

� �aKp

t2ps2 þ 2xtpsþ 1

!" #�1

ð5:84Þ

5.13 SOME TECHNIQUES FOR CHOOSING THECONTROLLER SETTINGS

There are a wide range of techniques for finding the optimal settings ofcontrollers. With the advent of digital control, most of these techniquesare now obsolete; however, some of them are still used as guidelines. Wegive a very brief idea regarding these techniques in this section.

5.13.1 Choosing the Controller Settings

Many criteria can be used to find the optimal controller settings. Some ofthem are based on the time response and others on the frequency response.

We will give an example of the time response criteria. The differentperformance criteria that can be used include the following:

1. Keep the maximum deviation (error) as small as possible.2. Achieve short settling times.

Figure 5.57 Block diagrams and transfer functions for Eq. (5.83).

3. Minimize the integral of the errors until the process settles downto the desired set point.

4. And others depending on the process and its requirements.

How does one achieve that? It may be through the following:

1. Certain limit on overshoot.2. Criterion for rise time and/or settling time.3. Choosing an optimal value for the decay ratio.4. Some characteristics for the frequency response.5. Other techniques not covered here.

Sometimes, a combination of a number of different criteria is used.Figure 5.58 shows the response of system to a unit step change in load

for different controllers, namely proportional (P), integral (I) and derivative(D).

Example

We will illustrate one of the techniques using the tank problem. For the tankproblem explained earlier, take Gm ¼ Gf ¼ 1:0 and look at the servoproblem di sð Þ ¼ 0

�; we substitute the values of G sð Þ and Gm sð Þ to get

y sð Þ ¼ tI sþ 1

t2s2 þ 2xtsþ 1ysp sð Þ ð5:85Þ

where

t ¼ffiffiffiffiffiffiffiffiffiffiffiffiffitItpKPKC

rand x ¼ 1

2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffitI

tpKPKC

r1þ KPKCð Þ

Figure 5.58 Response to unit step change in load.

A step response of y sð Þ for a unit step input in ysp sð Þ gives that the decayratio is equal to

Decay ratio ¼ C

A¼ exp � 2xffiffiffiffiffiffiffiffiffiffiffiffiffi

1� x2p

!


C

A¼ exp

�2 12

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffitI=tpKPKC 1þ KPKCð Þp

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� 1

4tI=tpKPKC

� �1þ KPKCð Þ2

q0B@

1CA ð5:86Þ

One of the most widely used criterion is that

C

A¼ 1

4ð5:87Þ

Using criterion (5.87) together with Eq. (5.86) and after some manipula-tions, we get

�2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffitI

4tpKPKC � t 1þ KPKCð Þ2s0@

1A 1þ KPKCð Þ ¼ ln

1

4

� �ð5:88Þ

For a given process, generally tp and KP are known (e.g., take tp ¼ 10 andKP ¼ 0:1). Now, we have an equation in tI and KC, and for each value of KC,we will get a corresponding value of tI , as shown in Table 5.2. The final choicedepends upon good chemical engineering understanding of the process.

5.13.2 Criteria for Choosing the Controller Settings fromthe Time Response

Mainly time integral performance criteria are used:

e tð Þ ¼ ysp tð Þ � y tð Þ or y tð Þ � ysp tð ÞThe following are typical examples:

1. Minimize the integral of the square error (ISE)

ISE ¼ð10

e2 tð Þ dt

Table 5.2 Different Values of �I and KC

KC 1.0 10.0 30.0 50.0 100.0

�I 0.127 0.12 0.022 0.007 0.001

2. Minimize the integral of the absolute value of the error (IEA)

IEA ¼ð10

e tð Þ�� dt3. Minimize the integral of the time-weighted absolute error (ITAE)

ITAE ¼ð10

t e tð Þ�� dtWhich one of the above criterion has to be chosen? Again, it depends on theprocess characteristics.

5.13.3 Cohen and Coon Process Reaction Curve Method

This is one of popular techniques for choosing controller settings. FromFigure 5.59,

~yym sð Þ~CC sð Þ ¼ GfGpGm|fflfflfflfflffl{zfflfflfflfflffl}

GPRC

where

GPRC ¼ GfGpGm

Thus

~yym sð Þ ¼ GPRC sð Þ ~CC sð ÞFor a step input in ~CC sð Þ ¼ A=s (where A is the amplitude of the step input),we get a response like the one shown in Figure 5.60.

Figure 5.59 Block diagram.

GPRC sð Þ can be approximated by

GPRC sð Þ ¼ ~yym sð Þ~CC sð Þ ffi Ke�td s

1þ ts

where

K ¼ B

Aand t ¼ B

S

Choosing the Controller Setting According to COHEN and COON

1. For proportional (P),

GPRC sð Þ ¼ KC ¼ 1

K

ttd

1þ td3t

� �

2. For proportional-integral (PI),

where,

GPRC sð Þ ¼ KC 1þ 1

tI s

� �

KC ¼ 1

K

ttd

0:9þ td12t

� �

Figure 5.60 Response to a step input.

And,

tI ¼ td30þ 3td=t9þ 20td=t

� �

3. For proportional–integral–differential PIDð Þ;

GPRC sð Þ ¼ KC 1þ 1

tI sþ tDs

� �

and we get,

KC ¼ 1

K

ttd

4

3þ td4t

� �

where,

tI ¼ td32þ 6td=t13þ 8td=t

� �for Gc sð Þ ¼ KC þ KI

sþ KDs

and

tD ¼ td4

11þ 2td=t

� �

This empirical technique can also be used without development ofa model, through obtaining the response curve experimentally.

SOLVED EXAMPLES

Solved Example 5.1

Figure 5.61 shows an underground tank used for storing gasoline for sale tothe public. Recently, engineers believe that a leak has developed, which

Figure 5.61 Schematic diagram for Solved Example 5.1.

threatens the environment. Your supervisor has assigned you the task ofmodeling the height in the tank as a function of the supply flow F1, the salesflow F2, and the unmeasured leakage F3. It is assumed that this leakage isproportional to the height of liquid in the tank (i.e., F3 ¼ bh). The tank is acylinder with constant cross-sectional area A. The density of gasoline is rand assumed to be constant. Find the transfer-function model for the levelin the tank as a function of the supply flow and the sales flow. Which ofthese input variables would you describe as a control variable and which youdescribe as a disturbance variable?

Solution

. Wehave to get the relation between h, F1, and F2 [i.e., h ¼ f F1;F2Þð �:

. Unmeasured leakage F3 ¼ bh; usually, it is F3 ¼ bffiffiffih

p. However, to

keep equation linear we use this linear relation.. Density and area of the cross section of the tank are constant.

We write the generalized mass balance as

ni þdnidt

¼ nif þReaction|fflfflfflfflffl{zfflfflfflfflffl}zero

Thus, we get

n2 þ n3 þdn

dt¼ n1

Because the density and molecular weight of gasoline coming in and goingout are same, we get

F2 þ F3 þdV

dt¼ F1

and we can write

V ¼ Ah

where A ¼ constant. On differentiation, we get

dV

dt¼ A

dh

dt

Also,

F3 ¼ bh

Thus, we get

Adh

dt¼ F1 � F2 � bh

with initial condition at t ¼ 0, h ¼ h0. The disturbance variable will be F1

and the control variable is h.

Note 1: You can have the control variable as F2 also, because you canhave control over the pumping out of gasoline. Also, if you have a mea-surement of h, you can use F2 as the manipulated variable to keep h closeto the value you want.

Note 2: If we fix the leakage, then F3 ¼ 0, and the model equation be-comes

Adh

dt¼ F1 � F2

with initial condition, at t ¼ 0, h ¼ h0.The steady state of the tank is when the differential term is equal to

zero; hence, we get

F1SS � F2SS � bhSS ¼ 0

Let us define the following deviation variables:

hh ¼ h� hSS

FF1 ¼ F1 � F1SS

FF2 ¼ F2 � F2SS

Thus, the unsteady-state equation can be written in terms of deviationvariables as follows:

Adhh

dt¼ FF1 � FF2 � bhh

with initial condition at t ¼ 0, hh ¼ 0.Taking the Laplace transforms of the above equation and rearrange-

ment gives the required transfer-function model:

~hh sð Þ ¼~FF1 sð Þ

Asþ b�

~FF2 sð ÞAsþ b

Or we can write

~hh sð Þ ¼ ð1=bÞ ~FF1 sð ÞA=bÞsþ 1ð � ð1=bÞ ~FF2 sð Þ

ðA=bÞsþ 1

Solved Example 5.2

(a) Figure 5.62 shows a system for heating a continuous-flow waterkettle using a hot plate. Assuming that the hot-plate temperature

can be changed instantaneously by adjusting the hot-plate rate ofheat input and assuming uniform water kettle temperature, showthat the following is a reasonable model for the process:

rVCp

dT1

dt¼ c T2 � T1ð Þ þ FrCpT0 � FrCpT1

where r is the density of water (a constant value), Cp is the heatcapacity of water (a constant value), c is a constant, and F thevolumetric flow rate in and out of the kettle.

(b) If the heat capacity of the hot-plate material is assumed uniformand given as Cp2 and its mass and effective lateral area for heattransfer to the atmosphere are given as m and Ac, respectively,obtain a second modeling equation that adequately describes thedynamics of T2, the hot-plate temperature, in response to changesin Q, the rate of heat input. It may be assumed that the atmo-spheric temperature is Ta, a constant; the heat transfer coefficientmay also be taken as a constant value equal to h.

Solution

Part a. The assumptions are as follows:

. r and Cp are constant over the operating range.

. Twater ¼ T1 (ideal mixing inside the kettle).

Figure 5.62 Schematic diagram for Solved Example 5.2.

. The instantaneous change in hot-plate temperature and the energyinput from plate to kettle is directly proportional to T2 � T1.

. Heat losses to the atmosphere are negligible.

. The heat capacity of heating plate is neglected.

The generalized heat balance equation is

Xnif Hif �Hir

� �þQ ¼X

ni Hi �Hirð Þ þ 0|{z}no reaction

þX d niHi

� �dt

Because we have only one component (water), the above equation becomes

nf Hf �Hr

� �þQ ¼ n H �Hirð Þ þ d nH� �dt

Now, if there is no change in phase

Hf �Hr ¼ðT0

Tr

CpM dT

Note: Here, CpM is the molar heat capacity, (i.e., the heat capacity permole of water).

As it has been assumed that the value of Cp is constant within theoperating range, we get

Hf �Hr ¼ CpM T0 � Trð ÞSimilarly,

H �Hr ¼ CpM T1 � Trð ÞThe heater duty (the heating plate rate of heat supply, where the heatingplate is at temperature T2) is given by

Q ¼ c T2 � T1ð ÞThe last term on the right-hand side of the heat balance equation can bechanged in terms of temperature as follows:

d nH� �dt

¼ ndH

dt¼ n

d H �Hr

� �dt

and

H �Hr

� � ¼ ðT1

Tr

CpM dT ¼ CpM T1 � Trð Þ

Thus, we get

nd H �Hr

� �dt

¼ CpM

d T1 � Trð Þdt

¼ CpM

dT1

dt

Thus, our heat balance equation becomes

nf CpM T0 � Trð Þ þ c T2 � T1ð Þ ¼ nCpM T1 � Trð Þ þ nCpM

dT1

dt

Because there is no chemical reaction,

n ¼ nf

Thus, we get

nCpM T0 � Trð Þ þ c T2 � T1ð Þ ¼ nCpM T1 � Trð Þ þ nCpM

dT1

dt


nCpM T0 � T1ð Þ þ c T2 � T1ð Þ ¼ nCpM

dT1

dt

Let the mass flow rate of water be m; then,

nCpM ¼ mCp

Similarly, for the contents (if the mass content of the tank is m), we can write

nCpM ¼ mCp ðCp is per unit mass)

Thus, the heat balance equation becomes

mCp T0 � T1ð Þ þ c T2 � T1ð Þ ¼ mCp

dT1

dt

Obviously

m ¼ Fr and m ¼ Vr

where F is the volumetric flow rate, r is the density of water, and V is thevolume of the tank contents.

Then, the heat balance equation becomes

FrCp T0 � T1ð Þ þ c T2 � T1ð Þ ¼ VrCp

dT1

dt

which on rearrangement, gives

VrCp

dT1

dt¼ c T2 � T1ð Þ þ FrCpT0 � FrCpT1

with initial condition, at t ¼ 0, T1 ¼ T1 0ð Þ.

Part b. Take into consideration the heating capacity of the heating plate.There is no flow of material; hence,X

nif Hif ¼X

niHi ¼ 0

and we get

Q ¼ Q� c T2 � T1ð Þ � Ach T2 � Tað Þwhere Q is the heat coming into the plate, c T2 � T1ð Þ is the heat lost to thewater, and Ach T2 � Tað Þ is the heat lost to the surrounding air, We thenhave

QdðnHÞdt

¼ mCp2

dT2

dt(as shown in part a)

Thus, the equation for the plate becomes

mCp2

dT2

dt¼ Q� c T2 � T1ð Þ � Ach T2 � Tað Þ

with initial condition at t ¼ 0, T2 ¼ T2 0ð Þ.

Solved Example 5.3

The water heater as shown in Figure 5.63 is a well-mixed, constant-volumeVlð Þ tank through which fluid flows at a constant mass flow rate w (or rlFwhereas F is the volumetric flow rate and rl is the liquid density); the specificheat capacity of the fluid is Cpl . Because the incoming fluid temperature Ti issubject to fluctuations, an electric coil to which a simple proportional con-

Figure 5.63 Schematic representation of the Solved Example 5.3.

troller is used to regulate the temperature of the liquid in the heater. Tofacilitate the design of this controller, it is desired to obtain a theoreticalprocess model.

The electrical coil is made of a metal whose specific heat capacity is Cp,with a surface area A and mass mc; the overall heat transfer coefficient isgiven as U. The temperature controller is provided with information aboutthe tank temperature T , through a thermocouple probe; it is designed tosupply energy to the coil at a rate of

Qc ¼ Kc Td � Tð Þ

where Kc is a predetermined constant and Td is the desired tank tempera-ture.

Develop a theoretical model for the water heater. You may wish tomake use of the following assumptions:

1. The agitation is assumed perfect, so that the temperature withinthe heater may be considered uniform.

2. The coil temperature Tc is also assumed uniform at any instant,but different from the boiler liquid temperature T .

3. The physical properties of all the components of the process areassumed constant.

4. There are no heat losses to the atmosphere.

Cast your model in terms of deviation variables, and present in the state-space form.

Solution

A reasonable theoretical model for the water heater based on the givenassumptions is obtained from the heat balances for the tank and for the coil.

The overall tank heat balance (using similar principles as in Example 5.2) is

VlrlCpl

dT

dt¼ UA Tc � Tð Þ þ wCpl Ti � Tð Þ � wCpl T � Tð Þ

where T is some reference temperature. The above equation simplifies togive

VlrlCpl

dT

dt¼ UA Tc � Tð Þ þ wCpl Ti � Tð Þ

with initial condition at t ¼ 0, T ¼ T 0ð Þ.

The overall heat balance for the coil is as follows. Using principles similar tothose used in Example 5.2, we get

mcCp

dTc

dt¼ KcðTd � TÞ �UAðTc � TÞ

with initial condition at t ¼ 0, Tc ¼ Tc 0ð Þ. The two differential equations arecoupled and must be solved simultaneously.

Let us define the following deviation (or hat variables):

TT ¼ T � TSS

TTc ¼ Tc � TcSS

TTi ¼ Ti � TiSS

TTd ¼ Td � TdSS

Also, let us define the following parameters:

a1 ¼UA

rlVlCpl

a2 ¼w

rlVl

b1 ¼UA

mcCpc

b2 ¼Kc

mcCpc

Then, using the above-defined deviation variables and parameters, thedifferential equations for the tank and coil become

dTT

dt¼ � a1 þ a2ð ÞTT þ a1TTc þ a2TTi

dTTc

dt¼ b1 � b2� �

TT � b1TTc þ b2TTd

with initial conditions at t ¼ 0, TT ¼ 0 and TTc ¼ 0. Because the tank tem-perature is the sole measured variable, we have the complete water heatermodel in terms of deviation variables in state-space form.

Solved Example 5.4

The process shown in Figure 5.64 is a continuous stirred mixing tank used toproduce FB L/min of brine solution of mass concentration CB g/liter. Theraw materials are fresh water, supplied at a flow rate of Fw L/min, and ahighly concentrated brine solution (mass concentration CBf g/L) supplied ata flow rate of FBf L/min. The volume of the material in the tank is V L, theliquid level in the tank is h m, and the tank’s cross-sectional area A m2

� �is

assumed to be constant.

(a) Assuming that the tank is well mixed, so that the brine concen-tration in the tank is uniformly equal to CB, and that the flow rateout of the tank is proportional to the square root of the liquidlevel, obtain a mathematical model for this process.

(b) If the process is redesigned to operate at constant volume, obtaina new process model and compare it with the one obtained in part(a).

Solution

Part a. An appropriate mathematical model may be obtained for thisprocess by carrying out material balances on salt and on water as follows.

The salt material balance is

d

dtVCBð Þ ¼ FBf CBf � FBCB

Figure 5.64 Schematic diagram for Solved Example 5.4

and because,

V ¼ Ah; A ¼ constant

FB ¼ kffiffiffih

p

we get

Ad

dthCBð Þ ¼ FBf CBf � k

ffiffiffih

pCB

The above differential equation can be rewritten as

A CB

dh

dtþ h

dCB

dt

� �¼ FBf CBf � CBk

ffiffiffih

p

The water material balance is obtained as follows. Assume that mixingx g of salt with � L of water does not alter appreciably the volume of theresulting mixture. This implies that FBf L of brine contains FBf L of water:

d

dtrWV� � ¼ rW FBf þ Fw

� �� rWFB

where rW is the constant density of water. Thus, we get

Adh

dt¼ FBf þ Fw � k

ffiffiffih

pð5:89Þ

Using the above differential equation to substitute the expression for dh=dtin the salt material balance, we get

A CB

FBf þ Fw � kffiffiffih

p

A

!þ h

dCB

dt

" #¼ FBf CBf � CBk

ffiffiffih

p


dCB

dt¼ 1

Ah�CB FBf þ Fw

� �þ FBf CBf

� ð5:90Þ

with initial conditions at t ¼ 0, CB ¼ CB 0ð Þ and h ¼ h 0ð Þ (5.91)

Thus, Eq. (5.89) along with Eq. (5.90) and initial conditions (5.91) give themathematical model of this system.

Notes about this model:

. It is a set of two coupled nonlinear ODEs, even though thecoupling is ‘‘oneway,’’ CB is affected by changed in h, but h doesnot depend on CB.

. At steady state, Eq. (5.90) indicates that

CBSS¼ FBf CBf

FBf þ Fw

which is consistent with well-known mixing rules.

Part b. By redesigning the process to operate at constant volume, h be-comes constant and the process model simplifies to only one equation:

dCB

dt¼ a �CB FBf þ Fw

� �þ FBf CBf

� ð5:92Þ

where

a ¼ 1

Ah¼ constant

If the constant volume is achieved by fixing FBf and Fw, and manipulatingonly CBf , then observe that Eq. (5.92) will now be a linear ODE. If, onthe other hand, constant volume is achieved by allowing both FBf and Fw

to vary but in such a way that FBf þ Fw ¼ FB, then Eq. (5.92) will be non-linear.

PROBLEMS

Problem 5.1

Linearize the following nonlinear functions:

(a) f xð Þ ¼ ax1þ a� 1ð Þx, where a is a constant.

(b) f Tð Þ ¼ eA=TþB, where A and B are constants.

(c) f vð Þ ¼ K vð Þ0:8, where K is a constant.

(d) f hð Þ ¼ K hð Þ32, where K is a constant.

Problem 5.2

Linearize the following ODEs, which describe a nonisothermal CSTR withconstant volume. The input variables are Tf , TJ , CAf , and F .

VdCA

dt¼ F CAf � CA

� �� VkCA

VrCp

dT

dt¼ FrCp Tf � T

� �� VkCA �UA T � TJð Þ

where

k ¼ k0e�E=RT

Problem 5.3

Solve the following differential equations using Laplace transforms:

(a)d2x

dt2þ 3

dx

dtþ x ¼ 1 with x 0ð Þ ¼ dx

dt

��t¼0

¼ 0

(b)d2q

dt2þ dq

dt¼ t2 þ 2t with q 0ð Þ ¼ 0 and

dq

dt

��t¼0

¼ �2

Problem 5.4

Obtain y tð Þ for the following,

(a) y sð Þ ¼ s2 þ 2s

s4

(b) y sð Þ ¼ 2s

s� 1ð Þ3

Problem 5.5

The function f tð Þ has the Laplace transformation

ff sð Þ ¼ 1� 2e�s þ e�2s

s2

Obtain the function f tð Þ and plot its graph (its variation with t).

Problem 5.6

Given a system with the transfer function

y sð Þx sð Þ ¼

T1sþ 1

T2sþ 1

find y tð Þ if x tð Þ is a unit-step function. If T1=T2 ¼ 5, sketch y tð Þ versus t=T2.Show the numerical minimum, maximum, and ultimate values that mayoccur during the transient.

Problem 5.7

Simplified equations can describe a first-order, irreversible, exothermic reac-tion in a CSTR containing a heating coil to illustrate many of the principlesof process dynamics. However, these equations neglect the possibility of

dynamic effects in the heating coil. Assuming plug flow in the coil and nowall capacitance or resistance of the wall, derive an appropriate set ofequations that includes the coil dynamics. Linearize the equations andobtain the transfer functions relating changes in the hot fluid inlet tempera-ture and velocity to the reactor composition. Discuss the effect of this mod-ification on the nature of characteristic equation and the stability of thesystem.

Problem 5.8

Derive a transfer function relating the tube outlet temperature to the shellinlet temperature for a two-tube-pass, single-shell-pass heat exchanger.

Problem 5.9

When chemists undertake a laboratory study of a new reaction, they oftentake data in a batch reactor at conditions corresponding to complete con-version or thermodynamic equilibrium. If the process economics appearpromising, a pilot plant might be constructed to study the reaction in acontinuous system. Normally, data are gathered at several steady-state oper-ating conditions in an attempt to ascertain the most profitable operatingregion.

Are any of these basic laboratory data useful if we are interested inestablishing the dynamic characteristics of a process? If you were in chargeof the whole project, what kind of experiments would you recommend? Howwould you try to sell your approach to top management? How would youexpect the cost of your experimental program to compare to the conven-tional approach?

Problem 5.10

A simple model for a pair of exothermic parallel reactions, A ! B andA ! C, in a CSTR containing a cooling coil might be written as

VdCA

dt¼ qðCAf � CAÞ � k1VC

2A � k2VCA

VdCB

dt¼ �qCB þ k1VC

2A

VCprdT

dt¼ qCprðTf � TÞ þ ��H1ð Þk1VC2

A þ ��H2ð Þk2VCA

�UAcKqc1þ Kqc

T � Tcð Þ

(a) See if you can list the assumptions implied by these equations.(b) Describe a procedure for calculating the steady-state composi-

tions and temperature in the reactor.(c) Linearize these equations around the steady-state operating

point.(d) Calculate the characteristic roots of the linearized equations.

Problem 5.11

You have measured the frequency response of an industrial furnace andfound that the transfer function relating the temperature of the effluentstream to the pressure supplied to a pneumatic motor valve on the fuelline could be represented by

~TT

~PP¼ 40e�20s

900sþ 1ð Þ 25sþ 1ð Þ8Cpsi

Also, the response of the outlet temperature to feed temperature change is

~TT

~TTf

¼ e�600s

60sþ 1ð Þ

where the time constants are given in seconds.If a fast-acting temperature-measuring device is available, which has a

gain of 0.2 psi=8C, select the gains for various kinds of pneumatic controllersthat use air pressure for the input and output signals. Then, calculate theresponse of the closed-loop systems to a 208C step change in the feed tem-perature.

Problem 5.12

A step change of magnitude 3 is introduced into the following transferfunction

y sð Þx sð Þ ¼

10

2s2 þ 0:3sþ 0:5

Determine the overshoot and frequency of the oscillation.

Problem 5.13

The flow rate F of a manipulated stream through a control valve with equalpercentage trim is given by

F ¼ CVax�1

where F is flow in gallons per minute and CV and a are constants set by thevalve size and type. The control valve stem position x (fraction wide open) isset by the output signal of an analog electronic feedback controller whosesignal range is 4–20 mA. The valve cannot be moved instantaneously. It isapproximately a first-order system:

tvdx

dtþ x ¼ � 4

16

The effect of the flow of the manipulated variable on the process tempera-ture T is given by

tpdT

dtþ T ¼ KpF

Derive one linear ordinary differential equation that gives the dynamicdependence of process temperature on controller output signal .

Problem 5.14

An isothermal, first-order, liquid-phase, reversible reaction is carried out ina constant-volume, perfectly mixed continuous reactor:

A,k1k2

B

The concentration of product B is zero in the feed, and in the reactor, it isCB. The feed rate is equal to F .

(a) Derive a mathematical model describing the dynamic behavior ofthe system.

(b) Derive the steady-state relationship between CA and CAf . Showthat the conversion of A and the yield of B decrease as k2increases.

(c) Assuming that the reactor is at this steady-state concentrationand that a step change is made in CAf to CAf þ�CAf , find theanalytical solution that gives the dynamic response of CA tð Þ:

Problem 5.15

Process liquid is continuously fed into a perfectly mixed tank in which it isheated by a steam coil. The feed rate F is 50,000 lbm/h of material with aconstant density r of 50 lbm/ft

3 and heat capacity Cp of 0.5 Btu/lbmoF.

Holdup in the tank V is constant at 4000 lbm. The inlet feed temperatureTf is 80oF.

Steam is added at a rate S lbm/h that heats the process liquid up totemperature T . At the initial steady state, T is 190oF. The latent heat ofvaporization �S of the steam is 900 Btu/lbm.

(a) Derive a mathematical model of the system and prove thatprocess temperature is described dynamically by the ODE

tdT

dtþ T ¼ K1Tf þ K2S

where

t ¼ V

F

K1 ¼ 1

K2 ¼�SCpF

(b) Solve for the steady-state value of steam flow SSS.(c) Suppose a proportional feedback controller is used to adjust the

steam flow rate,

S ¼ SSS þ Kc 190� Tð ÞSolve analytically for the dynamic change in T tð Þ for a stepchange in inlet feed temperature from 80oF down to 50oF.What will be the final values of T and S at the new steady statefor a Kc of 100 lbm/h/8F?

Problem 5.16

(a) Design of a water supply tank: You are required to design acylindrical tank to supply a chemical plant with a constantwater feed of 50 m3/h.

(b) Design of feedback control loop: The feed to this tank is sufferingfrom some fluctuations with time, whereas the feed to the plantneeds to be constant at 50 m3/h. Introduce the necessary feedbackcontrol loop using a PI (proportional–integral) controller. Derivethe unsteady-state model equations for the closed-loop system(including the dynamics of controller, measuring instrument(DPC¼ differential pressure cell) and final control element (pneu-matic control valve). Obtain the transfer functions (and theirconstants) for the DPC and control valve from the literature.Put the equations in linear form using Taylor series expansion,then use Laplace transformation to put the equations in the form

of transfer functions. Find the optimal controller settings (gains)using the following techniques:

(a) The 14decay ratio criterion

(b) Cohen and Coon reaction curve method

(c) Evaluation of controller performance: Using the nonlinear modelfor the closed loop and solving the equations numerically, com-pare the performance of the nonlinear system with the controllersettings obtained using the above two methods.

Problem 5.17

You are required to design two cylindrical tanks in series to supply achemical plant with a constant water feed of 35 m3/h. The feed to the firsttank keeps fluctuating, whereas the feed to the plant needs to be constant at35 m3/hr. Design a suitable control loop (PI control action) for this system.Draw appropriate block diagrams for both the open- (uncontrolled) andclosed- (controlled) loop cases. Find the optimal control settings (gains)using the 1

4decay ratio criterion, and the Cohen and Coon criterion and

compare the results of both methods.

6

Heterogeneous Systems

Most real chemical and biochemical systems are heterogeneous. What do wemean by heterogeneous? We mean that the system is formed of more thanone phase, with strong interaction between the various phases.

. Distillation is a heterogeneous system formed of at least a gas anda liquid. Distillation is not possible except when at least two phasesare present (packed-bed distillation columns are formed of threephases, taking into consideration the solid packing phase).

. Absorption is another gas–liquid system (two phases). Packed-bedabsorbers have three phases when taking into consideration thesolid packing of the column.

. Adsorption is a solid–gas (or solid–liquid) two-phase system.

. Fermentation is at least a two-phase system (the fermentationliquid mixture and the solid micro-organisms catalyzing the fer-mentation process). Fermentation can also be a three-phase systembecause aerobic fermentation involves gaseous oxygen which isbubbled through the fermentor. Immobilized packed-bed aerobicfermentors are formed of four phases; two of them are solid (themicro-organism and the solid carrier), one is liquid, and one is agaseous phase.

. Gas–solid catalytic systems are two-phase systems involving asolid catalyst with reactants and products in the gas phase.

433

It is easily noticed that, in the earlier parts of this book, when dealing withone-phase systems, the emphasis was mainly on reacting systems. Thereason was simply that for one-phase systems, the nonreacting cases arealmost trivial (e.g., mixers and splitters).

However, in the present chapter dealing with heterogeneous systems,the nonreacting systems are as nontrivial as the reacting systems.

6.1 MATERIAL BALANCE FORHETEROGENEOUS SYSTEMS

6.1.1 Generalized Mass Balance Equations

Let us first remember the generalized mass balance equations for a onephase system. (see Fig. 6.1). The generalized mass balance relation isgiven by

XKk¼1

nik ¼XLl¼1

nif l þXNj¼1

sijrj ð6:1Þ

where rj is the overall generalized rate of reaction for reaction j and sij is thestoichiometric number of component i in reaction j.

The design equation for a lumped system is given by,

XKk¼1

nik ¼XLl¼1

nif l þ VXNj¼1

sijr0j ð6:2Þ

where r0j is the generalized rate of the jth reaction per unit volume ofreactor.

Figure 6.1 Material balance for homogeneous systems.

Let us, for simplicity, start by the simple single input–single output,single reaction mass balance which can be represented as shown in Figure6.2. Now, we will move from the homogeneous one-phase system (as shownin Fig. 6.2) to the corresponding heterogeneous system.

6.1.2 Two-Phase Systems

If the system is formed of two phases, we can draw the mass balanceschematic diagram as in Figure 6.3. The interaction between the two phasesis usually some kind of mass transfer from phase I to phase II or vice-versa(also, it can be in one direction for one component and in the oppositedirection for another component). The easiest way is to make the balancein one direction, as shown in Figure 6.4, and the sign of the driving force foreach component determines the direction of mass transfer for the specificcomponent. Therefore, the mass balance equations can be written as

ni þRMi ¼ nif þ siIrI ð6:3Þand

ni �RMi ¼ nif þ siIIrII ð6:4ÞNote that Eqs. (6.3) and (6.4) are coupled through the term RMi, the overallrate of mass transfer of component i.

Let us take a simpler case with no reaction for which Eqs. (6.3) and(6.4) will become

ni þRMi ¼ nif

and

ni �RMi ¼ nif

For constant flow rates, we can write

qICi þRMi ¼ qICif

Figure 6.2 Mass balance for single-input, single-output, and single reaction

system.

and,

qIICi �RMi ¼ qIICif

where qI and qII are the constant volumetric flow rates in phases I and II,respectively.

Then, RMi should be expressed in terms of Ci and Ci as follows:

RMi ¼ amKgiCi � Ci

� �where am is the total area for mass transfer between the two phases and Kgi

is the mass transfer coefficient (for component i).Therefore, we have

qICi þ amKgiCi � Ci

� � ¼ qICif ð6:5Þ

Figure 6.3 Interaction between two phases.

Figure 6.4 Interaction between different phases with the interaction in one direc-

tion.

and

qIICi � amKgiCi � Ci

� � ¼ qIICif ð6:6ÞTherefore, for given values of qI; qII; am; kgi ;Cif ; and Cif , we can com-

pute the values of Ci and Ci from the Eqs. (6.5) and (6.6).

The countercurrent case: For countercurrent flow streams, we have thesituation as shown in Figure 6.6. The two equations remain the same:

qICi þ amKgiCi � Ci

� � ¼ qICif ð6:5Þand

qIICi � amKgiCi � Ci

� � ¼ qIICif ð6:6ÞThe difference between cocurrent and countercurrent flows does not appearin a single lumped stage. It appears in a sequence of stages (or in a distrib-uted system) as shown later.

6.1.3 The Equilibrium Case

For a system in which the contact between the two phases is long and/or therate of mass transfer is very high, the concentrations of the different com-ponents in the two phases reach a state of equilibrium (a state where nofurther mass transfer is possible). More details regarding equilibrium stateshave been covered in thermodynamics courses. The equilibrium relationsrelate the concentrations in the two phases, such that for example,

Ci ¼ fi Cið Þ ð6:7ÞIn certain narrow regions of concentration, we can use linear relations likethe following:

Ci ¼ KiCi ð6:8Þwhere Ki is the equilibrium constant for component i. In this case, weneither know nor need the rates of mass transfer. The simple and systematicapproach is to add Eqs. (6.5) and (6.6) for both the cocurrent case and the

Figure 6.5 Cocurrent flow of streams.

countercurrent case (as the mass balance equations are same for both flowcases).

Cocurrent case (Fig. 6.5): Adding Eqs. (6.5) and (6.6) gives

qICi þ qIICi ¼ qICif þ qIICif ð6:9ÞThis is a single algebraic equation with two unknowns (Ci and Ci), but theequilibrium relation (6.7) can be used to make the equation solvable:

qICi þ qII fi Cið Þ ¼ qICif þ qIICif ð6:10Þand as soon as Eq. (6.10) is solved and the value of Ci is obtained, it isstraightforward to obtain the value of Ci from the equilibrium relation (6.7).

Countercurrent case (Fig. 6.6): Adding Eqs. (6.5) and (6.6) gives

qICi þ qIICi ¼ qICif þ qIICif ð6:11ÞAgain, we can replace Ci using relation (6.7) to get

qICi þ qII fi Cið Þ ¼ qICif þ qIICif ð6:12ÞAs evident, there is no difference in the equations [compare Eqs. (6.10) and(6.12)] for the cocurrent and countercurrent cases. There will be evidentdifferences when we have more than one stage of cocurrent or countercur-rent operations, or a distributed system.

6.1.4 Stage Efficiency

For many mass transfer processes (e.g., distillation, absorption, extraction,etc.) which are multistage (in contradistinction to continuous, e.g., packed-bed columns), each stage (whether a cocurrent or countercurrent process)can be considered at equilibrium. These stages are usually called ‘‘the idealstages.’’ The deviation of the system from this ‘‘ideality’’ or ‘‘equilibrium’’ iscompensated for by using what is called ‘‘stage efficiency’’, which variesbetween 1.0 (for an ideal stage) and 0 (for a ‘‘useless stage’’). Usually,stage efficiencies are in the range 0.6–0.8, but they differ widely for different

Figure 6.6 Counter-current flow of streams.

processes, components, and designs (specially retention time) for a particu-lar stage.

However, in some cases, the concept of ‘‘stage efficiency’’ can causeproblems and unacceptable inaccuracies may arise. The rate of mass transferbetween the phases should be used in both the mass balance and the designequations, except when we are sure that equilibrium is established betweenthe two phases.

It is clear from the above that the main concepts used in the massbalance for the one-phase system can be simply extended to heterogeneoussystem by writing mass balance equations for each phase and taking theinteraction (mass transfer) between the phases into account. The sameapplies when there are reactions in both phases. Only the rate of reactionterms must be included in the mass balances of each phase, as shownearlier for homogeneous (one phase) systems.

6.1.5 Generalized Mass Balance for Two-PhaseSystems

For the sake of generality, let us write the most general mass balance equa-tions for a two-phase system, where each phase has multiple inputs andmultiple outputs, with each phase undergoing multiple reactions within itsboundaries.

Figure 6.7 presents the generalized mass balance for a heterogeneoussystem; i represents all the components in the two phases (reactants, pro-ducts and inerts).

Now the mass balance equations can be written as follows:

Phase I

XKk¼1

nik þRMi ¼XLl¼1

nifl þXNj¼1

sijrj ð6:13Þ

where

RMi ¼ overall rate of mass transfer of component i from phase I tophase II

rj ¼ overall generalized rate of reaction for reaction j in phase Isij ¼ stoichiometric number of component i in reaction j in phase IK ¼ total number of output streams for phase IL¼ total number of input streams for phase IN¼ total number of reactions in phase I

Phase II

XKk¼1

nik�RMi ¼

XLl¼1

niflþXNj¼1

sijrj ð6:14Þ

where

rj ¼ overall generalized rate of reaction for reaction j in phase IIsij ¼ stoichiometric number of component i in reaction j in phase II

K ¼ total number of output streams for phase IIL¼ total number of input streams for phase IIN ¼ total number of reactions in phase II

To change the above mass balance equations into design equations, we justreplace rj by Vr0j and rj by Vr 0j , where r

0j and r 0j are the rates of reaction per

unit volume of reaction mixture (or per unit mass of catalyst for catalyticreactions, etc.). More details about the design equations is given in thefollowing sections. Note that these terms can also be VI r0j (where VI isthe volume of phase I and r0j is the rate of reaction per unit volume ofphase I) and VII r

0j (where VII is the volume of phase II and r0j is the rate

of reaction per unit volume of phase II).

Figure 6.7 A two-phase heterogeneous system with multiple inputs–multiple out-

put, and multiple reactions in each phase and mass transfer between the two phases.

6.2 DESIGN EQUATIONS (STEADY-STATEMODELS) FOR ISOTHERMAL,HETEROGENEOUS LUMPED SYSTEMS

For simplicity and clarity, let us consider a two-phase system. Each phasehas single input and single output with a single reaction taking place in eachphase (see Fig. 6.8.)

The mass balance equations can be written as

ni þRMi ¼ nif þ sir (for phase I) ð6:15Þand

ni �RMi ¼ ni þ sir (for phase II) ð6:16Þwhere

ni ¼ molar flow rate of component i out of phase Ini ¼ molar flow rate of component i out of phase IInif ¼ molar flow rate of component i fed to phase Inif ¼ molar flow rate of component i fed to phase IIr ¼ generalized rate of the single reaction in phase Isi ¼ stoichiometric number of component i in the reaction in phase Ir ¼ generalized rate of the single reaction in phase IIsi ¼ stoichiometric number of component i in the reaction in phase IIRMi ¼ overall mass transfer rate of component i from phase I tophase II

Now, in order to turn these mass balance equations into design equations,we just turn all of the rate processes (r, r, and RMi) into rates per unitvolume of the process unit (or the specific phase). For example, let us con-sider the following definitions for the rates:

Figure 6.8 A two-phase heterogeneous system with single input–single output and

single reaction in each phase.

r0 ¼ rate of reaction in phase I per unit volume of the process unitr 0 ¼ rate of reaction in phase II per unit volume of the process unitRM0

i ¼ is the rate of mass transfer of component i from phase I tophase II per unit volume of the process unit

V ¼ volume of the process unit (¼ VI þ VII )

Thus, Eqs. (6.15) and (6.16) can be rewritten as

ni þ VRM0i ¼ nif þ Vsir

0 (for phase I) ð6:17Þ

and

ni � VRM0i ¼ ni þ Vsir

0 (for phase II) ð6:18ÞFor example, consider the case of first-order irreversible reactions in bothphases with constant flow rates (qI in phase I and qII in phase II) andconcentrations in phase I are Ci0s and in phase II are Ci0s. The rate ofreactions are given by

r0 ¼ kCA; ðA ! B in phase I)

and

r 0 ¼ kCB; ðB ! C in phase II)

Note that i ¼ 1; 2; 3; . . .A; . . .B; . . . . Thus, Eqs. (6.17) and (6.18) become

qICi þ Va0mKgi Ci � Ci

� � ¼ qICif þ VsikCA ð6:19Þ

qIICi � Va0mKgi Ci � Ci

� � ¼ qIICif þ Vsik CB ð6:20Þ

where a0m is the area of mass transfer per unit volume of the process unit andKgi

is the coefficient of mass transfer of component i between the twophases.

Simple Illustrative Example

Let us consider an example where the flow rates and rates of reactions are asshown in Figure 6.9. Note that CAf

is fed to phase I while CBfand CCf

are

equal to zero. For phase II, CBfis fed while CAf

and CCfare equal to zero.

The design equations are as follows:

Phase I: For component A,

qICA þ Va0mKgACA � CA

� � ¼ qICAf� VkCA ð6:21Þ

For component B,

qICB þ Va0mKgBCB � CB

� � ¼ qICBf|fflffl{zfflffl}equal to zero

þVkCA ð6:22Þ

For component C,

qICC þ Va0mKgCCC � CC

� � ¼ qICCf|fflffl{zfflffl}equal to zero

ð6:23Þ

Phase II: For component A,

qIICA � Va0mKgACA � CA

� � ¼ qIICAf|fflfflffl{zfflfflffl}equal to zero

ð6:24Þ

For component B;

qIICB � Va0mKgBCB � CB

� � ¼ qIICBf� VkCB ð6:25Þ

For component C,

qIICC � Va0mKgCCC � CC

� � ¼ qIICCf|fflfflffl{zfflfflffl}equal to zero

þVkCB ð6:26Þ

The above six equations are the design equations for this two-phase isother-mal system when both phases are lumped systems.

Figure 6.9 Mass flow diagram for a two-phase system.

6.3 DESIGN EQUATIONS (STEADY-STATEMODELS) FOR ISOTHERMAL, DISTRIBUTEDHETEROGENEOUS SYSTEMS

As explained in detail earlier for the homogeneous system, the distributedsystem includes variation in the space direction, and, therefore for the designequations, we cannot use the overall rate of reaction (rate of reaction perunit volume multiplied by the total volume) and the overall rate of masstransfer (the mass transfer per unit area of mass transfer multiplied by thearea of mass transfer, where the area of mass transfer is treated as themultiple of the area per unit volume of the process unit multiplied by thevolume of the process unit).

In the case of distributed systems, we need to take the balance on anelement; then, we take the limit when the size of element goes to zero, and soforth, as was detailed earlier for a distributed homogeneous system. As anillustration, let us consider a case with a single reaction

A ! B

taking place in phase I (no reaction is taking place in phase II) and we con-sider that mass transfer between the two phases takes place (see Fig. 6.10).

Thus, the balance for phase I (the design equation) is

ni þ�ni þ�VRM0i ¼ ni þ�Vsir

0 ð6:27Þand for phase II, it will be (for the cocurrent case),

ni þ�ni ��VRM0i ¼ ni þ 0 ð6:28Þ

Figure 6.10 A distributed heterogeneous system.

because there is no reaction in this phase.Arranging Eq. (6.27) gives

dnidV

þRM0i ¼ sir

0 ð6:29Þ

and rearranging Eq. (6.28) gives

dnidV

�RM0i ¼ 0 ð6:30Þ

Both are differential equations with the initial conditions (for the cocurrentcase) at V ¼ 0, ni ¼ nif and ni ¼ nif .

If the flow rate is constant and the reaction is first order (r0 ¼ kCA), wecan write Eqs. (6.29) and (6.30) in the following form:

qdCi

dVþ a0mKgi

Ci � Ci

� � ¼ sikCA ð6:31Þ

and,

qdCi

dV� a0mKgi Ci � Ci

� � ¼ 0 ð6:32Þ

and the initial conditions (for the cocurrent case) are at V ¼ 0;Ci ¼ Cif andCi ¼ Cif , where, i ¼ A and B. Specifically, for the two componentsA and B, we can write the following:

Phase I

qdCA

dV¼ �a0mKgA

CA � CA

� �� kCA ð6:33Þ

and

qdCB

dV¼ �a0mKgB

CB � CB

� �þ kCA ð6:34Þ

Phase II

qdCA

dV¼ a0mKgA CA � CA

� � ð6:35Þ

and

qdCB

dV¼ a0mKgB

CB � CB

� � ð6:36Þ

with the initial conditions (for the cocurrent case) at V ¼ 0;CA ¼ CAf, CB ¼

CBf;CA ¼ CAf

and CB ¼ CBf:

Generalized Form

Now, we develop the very general equation, when there are N (the counter isj) reactions taking place in phase I and N (the counter is j) reactions takingplace in phase II. (Refer to Fig. 6.11.)

Then, the design equations for the cocurrent case are as follows:

Phase I

ni þ�ni þ�VRM0i ¼ ni þ�V

XNj¼1

sijrj0

which, after the usual manipulation, becomes

dnidV

þRM0i ¼

XNj¼1

sijrj0 ð6:37Þ

Phase II

Similarly for phase II, the design equation is,

dnidV

�RM0i ¼

XNj¼1

sijrj0 ð6:38Þ

The initial conditions are (for cocurrent case), at V ¼ 0; ni ¼ nif andni ¼ nif .

Note that for distributed systems, the multiple-input problem corre-sponds to adding an idle-stage (with no reaction or mass transfer) massbalance at the input for each phase, as shown in Figure 6.12, which is avery simple problem. The same applies to multiple outputs.

Figure 6.11 Mass flow for the heterogeneous distributed system.

There are cases in which one of the phases is distributed and the otherphase is lumped. In such cases, the distributed phase will contribute its massand/or heat transfer term to the lumped phase through an integral, as shownlater with regard to the modeling of a bubbling fluidized bed.

For the countercurrent case, the equation of phase II will simply be

� dnidV

�RM0i ¼

XNj¼1

sijrj0 ð6:39Þ

with the boundary conditions at V ¼ Vt; ni ¼ nif , where Vt is the totalvolume of the process unit.

6.4 NONISOTHERMAL HETEROGENEOUSSYSTEMS

Based on the previous chapters, it should be quite easy and straightforwardfor the reader to use the heat balance and heat balance design equationsdeveloped earlier for homogeneous systems and the principles we used for

Figure 6.12 Multiple inputs can be combined together as a single feed.

developing the mass balance and mass balance design equations for hetero-geneous systems. We will, as usual, start with the lumped system.

6.4.1 Lumped Heterogeneous Systems

Let us first remember how we rigorously developed the heat balance equa-tions and nonisothermal design equations for homogeneous systems. Thelumped homogeneous system is schematically shown in Figure 6.13.

We have the following heat balance equation (developed earlier):XnifHif þQ ¼

XniHi ð6:40Þ

and then we remember that we transformed this enthalpy equation into anenthalpy difference equation as follows (using enthalpy of each componentat reference condition Hir

):Xðnif Hif � nif Hir

Þ þX

nifHirþQ ¼

XðniHi � niHir

Þ þX

niHir

From this equation, we can rearrange to obtainXnif ðHif �Hir Þ þQ ¼

XniðHi �HirÞ þ

XniHir � nif Hir Þ� ð6:41Þ

As we did earlier in Chapter 2, from the mass balance we get

ni ¼ nif þ sir

which can be rewritten as,

ni � nif ¼ sir

Now, we can write,Xni � nif� �

Hir ¼X

sirHir ð6:42ÞThus, relation (6.42) can be rewritten as r

PsiHir

, and this can be finallyrewritten as r�Hr. Thus the heat balance equation becomes, as shown inChapter 2,X

nif ðHif �Hir Þ þQ ¼X

ni Hi �HirÞ þ r�Hr

� ð6:43Þ

Figure 6.13 Heat balance for a homogeneous system.

In order to convert the equation into nonisothermal (heat balance) designequation, we just do the single step we did several times before: We replace rwith r0V , where r0 is the rate of reaction per unit volume and V is the volumeof the reactor. Thus, Eq. (6.43) becomesX

nif ðHif �Hir Þ þQ ¼X

ni Hi �Hir Þ þ Vr0�Hr

� ð6:44ÞNote that the basis of r0 andV depends on the type of process. For example, ifwe are dealing with a gas–solid catalytic system, we will usually define r0 as perunit mass of the catalyst and replace V with WS (the weight of the catalyst).

We do not need to repeat the above for multiple reactions (N reac-tions); Eq. (6.44) will become

Xnif ðHif �Hir Þ þQ ¼

XniðHi �Hir Þ þ V

XNj¼1

r0jð�HrjÞ

where ð�HrjÞ is the heat of reaction for reaction j.

The Heterogeneous System

For the heterogeneous system, the problem is very simple. We just write theabove equation for each phase, taking into account Q as the heat transferbetween the two phases. For a nonadiabatic system, the heat added fromoutside, Qexternal (which is the Q in Eq. (6.44)], will be added to the phasereceiving it, or distributed between the two phases if it is added to bothphases (this will depend very much on the configuration and knowledgeof the physical system as will be shown with the nonadiabatic bubblingfluidized-bed catalytic reactor example).

Now, let us consider an adiabatic two-phase system with each phasehaving one reaction, as shown schematically in Figure 6.14.Here,Q is the heattransfer between the two phases. For phase I, the heat balance equation isX

nif ðHif �Hir Þ �Q ¼X

ni Hi �Hir Þ þ rI�HrI

� ð6:45Þ

Figure 6.14 Heat balance for a two-phase lumped system.

and for phase II, it isXnif ðHif �Hir

Þ þQ ¼X

ni Hi �HirÞ þ rII�HrII

� ð6:46ÞTo turn these heat balance equations into the nonisothermal heat balancedesign equations, we just define the rate of reaction per unit volume (or perunit mass of the catalyst depending on the system) and the heat transfer isdefined per unit volume of the process unit (or per unit length, whichever ismore convenient). Thus, Eqs. (6.45) and (6.46) become as follows:

For phase I,Xnif ðHif �Hir Þ � VQ0 ¼

Xni Hi �Hir Þ þ Vr0�HrI

� ð6:47Þand for phase II, it isX

nif ðHif �HirÞ þ VQ0 ¼

Xni Hi �Hir

Þ þ Vr 0�HrI

� ð6:48ÞThe rate of heat transfer per unit volume of the process unit Q0 is given by

Q0 ¼ a0hh T � T� � ð6:49Þ

where a0h is the area of heat transfer per unit volume of the system, h is theheat transfer coefficient between the two phases, T is the temperature ofphase II, and T is the temperature of phase I.

Obviously, for multiple reactions in each phase, the change in theequations is straightforward, as earlier. For N reactions (counter j) inphase I and N reactions (counter j) in phase II, the equations become asfollows:

For phase I,

Xnif Hif �Hir

� �� Va0hh T � T� � ¼X

ni Hi �Hir

� �þ VXNj¼1

rj0�Hrj

ð6:50ÞFor phase II, it is

Xnif Hif �Hir

� �þ Va0hh T � T� � ¼X

ni Hi �Hir

� �þ VXNj¼1

rj0�Hr

j

ð6:51ÞNow, the whole picture is almost complete; what is remaining is the heatbalance design equation for the distributed two-phase system and thedynamic terms in heterogeneous systems.

6.4.2 Distributed Systems (see Fig. 6.15)

For phase I with one reaction,XniHi ��V

XRM0

iHi ��VQ0 ¼X

niHi þ�ðniHið ÞÞ ð6:52Þ

Equation (6.52) becomes

X d niHið ÞdV

¼�Q0 �X

RM0iHi


XnidHi

dVþX

Hi

dnidV

¼ �Q0 �X

RM0iHi ð6:53Þ

From the mass balance design equation [Eq. 6.37], when written for a singlereaction we get

dnidV

þRM0i ¼ sir

0 ð6:54Þ

Substituting for dni=dV from Eq. (6.54) into Eq. (6.53) gives

XnidHi

dVþX

Hi sir0 �RM0

i

� ¼ �Q0 �X

RM0iHi

which gives

XnidHi

dVþ r0�H ¼ �Q0 ð6:55Þ

Figure 6.15 Heat balance for a heterogeneous distributed system.

For phase II, it will be (for cocurrent operation)

XnidHi

dVþ r0�H ¼ Q0 ð6:56Þ

If the change is only sensible heat, we can use an average constant CP for themixture. Then, as shown earlier for the homogeneous system, we can writeEqs. (6.55) and (6.56) as,

qrCP

dT

dVþ r0�H ¼ �a0hh T � T

� �for Phase I ð6:57Þ

and

q rCP

dT

dVþ r0�H ¼ a0hh T � T

� �for Phase II ð6:58Þ

with the initial conditions (for the cocurrent case) at V ¼ 0, T ¼ Tf andT ¼ Tf .

For the countercurrent case,

qrCP

dT

dVþ r0�H ¼ �a0hh T � T

� �for Phase I ð6:59Þ

Note that the equation for phase I for countercurrent case is the same as forcocurrent case. For phase II the equation is:

�q rCP

dT

dVþ r0�H ¼ a0hh T � T

� �for Phase II ð6:60Þ

with the following two-point split boundary conditions:

At V ¼ 0;T ¼ Tf :

At V ¼ Vt;T ¼ Tf .

6.4.3 Dynamic Terms for Heterogeneous Systems

The dynamic terms for heterogeneous systems will be exactly the same as forthe homogeneous systems (refer to Chapters 2 and 3), but repeated for bothphases. The reader should take this as an exercise by just repeating the sameprinciples of formulating the dynamic terms for both mass and heat and forboth lumped and distributed systems. For illustration, see the dynamicexamples later in this chapter.

6.5 EXAMPLES OF HETEROGENEOUSSYSTEMS

6.5.1 Absorption Column (High-DimensionalLumped, Steady-State, and Equilibrium StagesSystem)

Let us consider the absorption column (tower), multistage (multitrays), andall stages as equilibrium stages (ideal stages). Component A is absorbedfrom the gas phase to the liquid phase.

Yj ¼ mole fraction of component A in the gas phase leaving the jthtray

nj ¼ molar flow rate of component A in the gas phase leaving the jthtray

Xj ¼ mole fraction of component A in the liquid phase leaving the jthtray

nj ¼molar flow rate of component A in the liquid phase leaving the jthtray

where,

nj ¼ ntjYj

where ntj is the total molar flow rate of the gas phase leaving the jth tray

nj ¼ ntjXj

where ntj is the total molar flow rate of the liquid phase leaving the jth trayThe molar flow streams at tray j are shown in Figure 6.16. The molar

flow balance on the jth tray gives

ntj�1Xj�1 þ ntjþ1

Yjþ1 ¼ ntjXj þ ntjYj ð6:61Þ

Figure 6.16 Molar flow across the jth tray.

For simplicity, we assume that the total molar flow rates of liquid and gasphases remain constant; that is,

ntj ¼ L and ntj ¼ V

Based on the above-stated assumption, Eq. (6.61) becomes

LXj�1 þ V Yjþ1 ¼ LXj þ V Yj ð6:62ÞFor the equilibrium stage, we can write, in general,

Yj ¼ F Xj

� �where F Xj

� �is a function in terms of Xj only. For a linear case, we can write

Yj ¼ aXj þ b

On using the above-mentioned linear relation in Eq. (6.62), we get

LXj�1 þ V aXjþ1 þ b� � ¼ LXj þ V aXj þ b

� �which gives

LXj�1 þ V að ÞXjþ1 ¼ L þ V að ÞXj


LXj�1 � L þ V að Þ|fflfflfflfflfflffl{zfflfflfflfflfflffl}a

Xj þ V að Þ|ffl{zffl}b

Xjþ1 ¼ 0

Thus, we get

LXj�1 � aXj þ bXjþ1 ¼ 0 ð6:63ÞFigure 6.17 is the schematic diagram of the absorption column with themolar flow rates shown for each tray.

Note: The numbering can be reversed in order; it is up to the reader.

Looking at Eq. (6.63), for the first tray j ¼ 1ð Þ we get

LX0 � aX1 þ bX2 ¼ 0


�aX1 þ bX2 þ 0X3 þ 0X4 þ � � � þ 0XN ¼ �LX0

For j ¼ 2,

LX1 � aX2 þ bX3 þ 0X4 þ 0X5 þ � � � þ 0XN ¼ 0

For j ¼ 3,

0X1 þ LX2 � aX3 þ bX4 þ 0X5 þ � � � þ 0XN ¼ 0

Figure 6.17 Absorption column.

Similarly, for j ¼ N � 1,

0X1 þ 0X2 þ 0X3 þ � � � þ LXN�2 � aXN�1 þ bXN ¼ 0

and for j ¼ N,

0X1 þ 0X2 þ 0X3 þ � � � þ 0XN�2 þ LXN�1 � aXN þ bXNþ1 ¼ 0


0X1 þ 0X2 þ 0X3 þ � � � þ 0XN�2 þ LXN�1 � aXN ¼ �bXNþ1

In order to write these equations in a matrix form, define the vector ofthe state variables

X ¼

X1

X2

..

.

XN�1

XN

0BBBBBBB@

1CCCCCCCA

and the matrix of coefficients as

A ¼

�a b 0 0 0 0

L �a b 0 0 0

0 L �a b 0 0

..

. ... . .

. . .. . .

. ...

0 0 0 L �a b

0 0 0 0 L �a

0BBBBBBBBBB@

1CCCCCCCCCCA

The above matrix is a tridiagonal matrix of the following form:

A ¼�a b 0

L . ..

b

0 L �a

0BB@

1CCA

Note: X0 and XNþ1 are not state variables, they are input variables.

Define the input vector matrix as (input vector I is usually denoted by m), sowe get

I ¼ X0

XNþ1

� �¼ m

The input coefficient matrix is given by

B ¼

�L 0

0 0

0 0

..

. ...

0 �b

0BBBBBB@

1CCCCCCA

Thus, the mass balance equations for the absorption column in matrix formcan be written as,

AX ¼ Bm ð6:64Þ

and its solution is given by (see Appendix A)

X ¼ A�1Bm ð6:65Þ

Degrees of Freedom

Example: If all parameters and inputs are defined, then A, B, and m andm are defined. In such a case, we have N unknowns X1;X2;X3; . . . ;XNð Þand N equations; thus, the degrees of freedom is equal to zero, and theproblem is solvable.

If the equilibrium relation is not linear,

LXj�1 þ V F Xjþ1

� � ¼ LXj þ V F Xj

� �where F Xj

� �is a function in terms of Xj only. In this case, the formulation

gives a set of nonlinear equations to be solved numerically (iteratively,using, for example, the multidimensional Newton–Raphson or other mod-ified gradient techniques; see Appendix B).

Nonequilibrium Stages

The overall molar balance we used will not be adequate because the equili-brium relation between the two phases cannot be used. In the present case ofnonequilibrium stages, we must introduce the mass transfer rate between thetwo phases, as shown in Figure 6.18. In the figure, the term RMT is the rateof mass transfer of the component A from the vapor phase. It can beexpressed as,

RMT ¼ Kgjamj

� driving forceð Þ

where Kgj is the mass transfer coefficient and amjis the area of mass transfer

for tray j. Assuming the tray to be perfectly mixed with respect to bothliquid and gas phases, we get

RMT ¼ Kgjamj

Yj � aXj þ b� ��

where

aXj þ b� � ¼ Y

j

which is the gas-phase mole fraction at equilibrium with Xj.The molar balance for the gas phase is given by

V Yjþ1 ¼ V Yj þ Kgj amjYj � aXj þ b

� �� On rearrangement, we get

V þ Kgj amj

� �|fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl}

aj

Yj � V Yjþ1 � Kgj amja

� �|fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl}

bj

Xj � Kgj amjb|fflfflfflfflffl{zfflfflfflfflffl}

gj

¼ 0

thus giving

ajYj � V Yjþ1 � bj Xj � gj ¼ 0 ð6:66Þwhere

aj ¼ V þ Kgjamj

bj ¼ Kgjamja

gj ¼ Kgjamjb

Using Eq. (6.66), we have the following:

For j ¼ 1,

a1Y1 � V Y2 � b1 X1 � g1 ¼ 0

Figure 6.18 Mass transfer on a nonequilibrium stage.

For j ¼ 2,

0Y1 þ a2Y2 � V Y3 � b2 X2 � g2 ¼ 0

For j ¼ N � 1,

0Y1 þ 0Y2 þ 0Y3 þ � � � þ 0YN�2 þ aN�1YN�1 � V YN � bN�1 XN�1

� gN�1 ¼ 0

For j ¼ N,

0Y1 þ 0Y2 þ 0Y3 þ � � � þ 0YN�1 þ aN YN � V YNþ1 � bN XN

� gN ¼ 0

The above equation can be rearranged as

0Y1 þ 0Y2 þ 0Y3 þ � � � þ 0YN�1 þ aN YN � bN XN � gN ¼ V YNþ1

In order to write these equations in a matrix form, we define the matrix ofcoefficients as

AY ¼

a1 �V 0 0 0

0 a2 �V 0 0

..

. ... . .

. . .. ..

.

0 0 0 aN�1 �V

0 0 0 0 aN

0BBBBBB@

1CCCCCCA

The above matrix is a bidiagonal matrix of the form,

AY ¼a1 �V 0

. .. �V

0 an

0B@

1CA

and vectors of the state variables are,

X ¼

X1

X2

..

.

XN

0BBBB@

1CCCCA and Y ¼

Y1

Y2

..

.

YN

0BBBB@

1CCCCA

Also, we get

BX ¼�b1 0

�b2. ..

0 �bN

0BBB@

1CCCA and g ¼

g1g2...

gN

0BBB@

1CCCA

with

BY ¼

0 0

0 0

..

. ...

0 V

0BBB@

1CCCA and mY ¼ 0

YNþ1

� �

Thus, we get

AYY þ bXX � g

Y¼ BYmY ð6:67Þ

Similarly, for the molar balance on liquid phase, we get

LXj�1 þ Kgj amjYj � aXj þ b

� �� ¼ LXj j ¼ 1; 2; . . . ;N ð6:68ÞPutting Eq. (6.68) in matrix form gives

AXX þ bYY � g

X¼ BXmX ð6:69Þ

The reader should define AX ;BX ; gX , and mX .

The matrix Eqs. (6.67) and (6.69) can be solved simultaneously to getthe solution, through the following simple matrix manipulation:

Matrix manipulation


X ¼ A�1X ½g

Xþ BXmX � b

YY � ð6:70Þ


AYY þ bX

A�1X gX þ BX mX � b

YY

h in o� g

Y¼ BY mY

Some simple manipulation gives

AY � bXA�1

X bY

h i|fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

C

Y ¼ BY mY þ gY� b

XA�1

X gX þ BX mX

h in o|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

M

where

C ¼ AY � bXA�1

X bY

M ¼ BYmY þ gY� b

XfA�1

X ½gXþ BXmX �g

Thus, we get

CY ¼ M ð6:71ÞMatrix Eq. (6.71) can be easily solved for vector Y as

Y ¼ C�1M ð6:72Þ

Thus, we obtain the vector Y from matrix Eq. (6.72), and using the solvedvector Y with matrix Eq. (6.70), we can compute the vector X .

6.5.2 Packed-Bed Absorption Tower

Packed-bed absorption towers are distributed systems and Figure 6.19shows the molar flow rates across a small element �l of one of these towers.We consider the absorption of component A from the gas (vapor) phase tothe liquid phase. The rate of mass transfer from the vapor phase to theliquid phase is given by the relation

RMT ¼ am Kg Y � aX þ bð Þð Þ ð6:73Þwhere am is the specific area (mass transfer area per unit volume of column)and aX þ bð Þ ¼ Y is the gas-phase mole fraction at equilibrium with X .

Molar balance on gas phase over a small element �l gives

V Y ¼ V Y þ�Yð Þ þ A�l am Kg Y � aX þ bð Þð Þwhere V is the total molar flow rate of the gas phase (assumed constant), Yis the mole fraction of component A in the gas phase, and X is the mole

Figure 6.19 Schematic diagram of a packed-bed absorption tower.

fraction of component A in the liquid phase. By simple manipulation andrearrangement, we get

V�Y

�l¼ �Aam Kg Y � aX þ bð Þð Þ

Taking the limit value as �l ! 0, we get

VdY

dl¼ �Aam Kg Y � aX þ bð Þð Þ ð6:74Þ

with initial condition, at l ¼ 0; Y ¼ Yf .Molar balance on the liquid phase over a small element �l gives

LX ¼ L X þ�Xð Þ þ A�l am Kg Y � aX þ bð Þð Þwhere L is the total molar flow rate of the liquid phase (assumed constant).By simple manipulation and rearrangement, we get

L�X

�l¼ �Aam Kg Y � aX þ bð Þð Þ

Taking the limit value as �l ! 0, we get

LdX

dl¼ �Aam Kg Y � aX þ bð Þð Þ ð6:75Þ

with boundary condition at l ¼ Ht; X ¼ Xf , where, Ht is the total height ofthe absorption column.

Note that the set of differential equations

VdY

dl¼ �Aam Kg ðY � ðaX þ bÞÞ with initial condition l ¼ 0;

Y ¼ Yf

LdX

dl¼ �Aam Kg Y � aX þ bð Þð Þ with boundary condition l ¼ Ht;

X ¼ Xf

form a set of two-point boundary-value differential equations.

However because of the simplicity of this illustrative problem we can actuallyreduce it to one equation. How? Is it at all possible?

Subtracting Eq. (6.74) and (6.75), we get

VdY

dl� L

dX

dl¼ 0

It can be rewritten as

d V Y � LXð Þdl

¼ 0


V Y � LX ¼ C1 ð6:76Þat l ¼ 0; Y ¼ Yf and X ¼ Xe, where Xe is the mole fraction of componentA in the liquid phase at the exit (at the bottom). Thus, using the initialcondition in Eq. (6.76), we get

V Yf � LXe ¼ C1

So, we have obtained the value of the constant of integration C1.Now, on substituting the value of C1 back in Eq. (6.76), we get

V Y � LX ¼ V Yf � LXe

Thus,

V Y ¼ L X � Xeð Þ þ V Yf


Y ¼ L

VX � Xeð Þ þ Yf ð6:77Þ

On substituting the value of Y from Eq. (6.77) into Eq. (6.75), we get

LdX

dl¼ �Aam Kg

L

VX � Xeð Þ þ Yf � aX þ bð Þ

� �ð6:78Þ

with boundary condition at l ¼ L; X ¼ Xf .Equation (6.78) is a single equation which, when solved for every l,

gives the value of X ; we can use this value of X to obtain the correspondingY by simple substitution into the algebraic Eq. (6.77). However, the pro-blem is still some kind of a two-point boundary-value differential equationbecause Xe at l ¼ 0 (unknown) is on the right-hand side of Eq. (6.78).

The same relation [Eq. (6.77)] can be obtained by mass balance. Massbalance over the shown boundary in Figure 6.20 gives the following relation:

LX þ VYf ¼ LXe þ VY

This gives

V Y ¼ L X � Xeð Þ þ V Yf

thus giving

Y ¼ L

VX � Xeð Þ þ Yf

The above equation is the same as the one obtained by subtracting the twodifferential equations [Eq. (6.77)].

6.5.3 Diffusion and Reaction in a Porous Structure(Porous Catalyst Pellet)

Consider a spherical particle and a simple reaction

A ! B

with rate of reaction equal to

rA ¼ kCA

mol

g catalyst�min

where k is the reaction rate constant and CA is the concentration of com-ponent A (in mol=cm3Þ.

As the considered particle is spherical in shape, it is symmetricalaround the center. The concentration profile inside the catalyst pellet isshown in Figure 6.21. Molar balance on component A over the element�r (here, r is the radial position from the center of the pellet [in cm]) gives

Figure 6.20 Mass balance over the region shown by the dashed lines.

4pr2NA ¼ 4p rþ�rð Þ2 NA þ�NAð Þ þ 4pr2�r� �

rCkCA

where NA is the diffusion flux of component A and rC is the catalyst density,which can be written as

r2NA ¼ r2 þ 2r�rþ �rð Þ2� �NA þ�NAð Þ þ r2�rrCkCA

Further simplification and neglecting higher powers of �r and �r�NA gives

0 ¼ r2�NA þ 2rNA�rþ r2�rrCkCA

On dividing the equation with �r and taking the limits as �r ! 0 and�NA ! 0, we get

0 ¼ r2dNA

drþ 2rNA þ r2rCkCA ð6:79Þ

On using the simple Fick’s law,

NA ¼ �DeA

dCA

drð6:80Þ

Figure 6.21 Elemental molar balance inside a catalyst pellet.

For diffusion in porous structures,

DeA¼ DAe

t

where DA is the molecular diffusivity of component A, e is the pellet’sporosity, and, t is the tortuosity factor.

On assuming DeAto be constant, we get

dNA

dr¼ �DeA

d2CA

dr2ð6:81Þ

Substituting the expressions of NA from Eq. (6.80) and dNA=dr from Eq.(6.81) into Eq. (6.79) gives

0 ¼ �r2DeA

d2CA

dr2� 2rDeA

dCA

drþ r2rCkCA

Rearrangement gives

DeA

d2CA

dr2þ 2

r

dCA

dr

!¼ rCkCA ð6:82Þ

Dividing and multiplying the left-hand side of Eq. (6.82) by R2P (where RP is

the pellet’s radius) gives

DeA

R2P

d2CA

d r2=R2P

� �þ 2

r=RPð ÞdCA

d r=RPð Þ

!¼ rCkCA ð6:83Þ

Defining the dimensionless terms

o ¼ r

RP

and

xA ¼ CA

CAref

Eq. (6.83) can be written as

DeA

R2P

d2xAdo2

þ 2

odxAdo

!¼ rCkxA ð6:84Þ

which is

r2xA ¼ f2xA ð6:85Þ

where

r2 ¼ d

do2þ a

od

do

f2 ¼ rCkR2P

DeA

(Thiele modulus)

The value of a depends on the shape of the particle as follows:

a ¼ 2 for the sphere

a ¼ 1 for the cylinder

a ¼ 0 for the slab

Boundary Conditions

At the center, due to symmetry, we have that at o ¼ 0,

dxAdo

¼ 0

At the surface r ¼ RPÞð , the mass transfer at the surface of catalyst pellet isshown in Figure 6.22. Balance at this boundary condition gives

4pR2PkgA CAB

� CA

��r¼RP

� �þ 4pR2

PNA

��r¼RP

¼ 0


kgA CAB� CA

��r¼RP

� �¼ �NA

��r¼RP

Since from Fick’s law,

NA ¼ �DeA

dCA

dr

Figure 6.22 Mass transfer across the boundary at the surface of a catalyst pellet.

we obtain

kgA CAB� CA

��r¼RP

� �¼ DeA

RP

� �dCA

d r=RPð Þ��r¼RP

We define

xAB¼ CAB

Cref

and get the equation

dxAdo

��o¼1:0

¼ RPkgADeA


ShA

xAB� xA

��o¼1:0

� �

where

RPkgADeA

� �¼ ShA

the Sherwood number for component A. Thus, the second boundary con-dition at o ¼ 1:0 is

dxAdo

¼ ShA xAB� xA

� �

The Limiting Case

When the external mass transfer resistance is negligible (kgA is large, leadingto ShA ! 1), we can write the boundary condition at o ¼ 1:0 as

1

ShA

dxAdo

¼ xAB� xA

As ShA ! 1 we get

xAjo¼1:0 ¼ xAB(corresponding to negligible external mass transfer

resistance)

For both cases of limited ShA and ShA ! 1, we get a two-point boundary-value differential equation. For the nonlinear cases, it has to be solvediteratively (we can use Fox’s method, as explained for the axial dispersionmodel in Chapter 4 or orthogonal collocation techniques as explained inAppendix E).

As the above discussed case is linear, it can be solved analytically. How?

Analytical solution

The second-order differential equation for the catalyst pellet is given by

d2xAdo2

þ 2

odxAdo

¼ f2xA

It can be rewritten as

od2xAdo2

þ 2dxAdo

¼ f2xAo ð6:86Þ

We define the following new variable:

y ¼ xAo ð6:87ÞDifferentiation of Eq. (6.87) gives

dy

do¼ o

dxAdo

þ xA ð6:88Þ

The second differential of Eq. (6.88) gives

d2y

do2¼ o

d2xAdo2

þ dxAdo

þ dxAdo

¼ od2xAdo2

þ 2dxAdo

ð6:89Þ


d2y

do2¼ f2y ð6:90Þ

We can write Eq. (6.90) in the operator form as

D2y� f2y ¼ 0

The characteristic equation for the above equation is

�2 � f2 ¼ 0

which gives the eigenvalues as

�1 ¼ f and �2 ¼ �f

Thus, the solution of Eq. (6.90) is given by

y ¼ C1efo þ C2e

�fo ð6:91ÞC1 and C2 can be calculated using the boundary conditions and the com-plete analytical solution can be obtained.

The reader is advised to carry out the necessary calculations to get thecomplete analytical solution as an exercise.

For nonlinear cases, the nonlinear two-point boundary-value differen-tial equation(s) of the catalyst pellet can be solved using the Fox’s iterativemethod explained for the axial dispersion model in Chapter 4 or the ortho-gonal collocation technique as explained in Appendix E.

6.6 DYNAMIC CASES

6.6.1 The Multitray Absorption Tower

Consider the equilibrium tray (jth tray) as shown schematically in Figure6.23. The holdup Vtð Þ on the tray will have two parts: VL and VG. So we canwrite

Vt ¼ VL þ VG

Thus, we can write the dynamic molar balance equation as

VL

dXj

dtþ VG

dYj

dt¼ LXj�1 þ VYjþ1

� �� LXj þ VYj

� � ð6:92Þ

For equilibrium,

Yj ¼ aXj þ b


VL

dXj

dtþ VGað Þ dXj

dt¼ LXj�1 þ V aXjþ1 þ b

� �� LXj þ V aXj þ b� ��

It can be further simplified to get

VL þ VGað Þ|fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl}a

dXj

dt¼ LXj�1 þ V aXjþ1 þ b

� �� LXj þ V aXj þ b� ��

Figure 6.23 Molar flow on the tray.

The reader can arrange the above equation into matrix form, as we didearlier for the steady-state case; it becomes a matrix differential equation:

adX

dt¼ AX þ Bm ð6:93Þ

with initial condition at t ¼ 0, X ¼ X0, where X0 is the initial conditionsvector.

If the equilibrium relation is not used, then we carry out an unsteadymolar balance for each phase with mass transfer between the two phases.The reader should do that as an exercise.

6.6.2 Dynamic Model for the Catalyst Pellet

For the same parameters and reaction as the steady-state case discussedearlier, unsteady-state molar balance on component A gives

4pr2NA ¼ 4p rþ�rð Þ2 NA þ�NAð Þ þ 4pr2�r� �

rCkCA

þ 4pr2�re� � @CA

@tþ 4pr2�r 1� eð Þ� �

rS@CAS

@t

where, CASis the concentration of the adsorbed (chemisorbed) A on the

surface of the catalyst and is expressed in terms of gmol/g catalyst.On some rearrangement, we get

r2NA ¼ r2 þ 2r�rþ �rð Þ2� �NA þ�NAð Þ þ r2�rrCkCA

þ r2�re� � @CA

@tþ r2�r 1� eð Þ� �

rS@CAS

@t

Further simplification and neglecting higher powers of �r and �r�NA gives

0 ¼ r2�NA þ 2rNA�rþ r2�rrCkCA þ r2�re� � @CA

@t

þ r2�r 1� eð Þ� �rS

@CAS

@t

On dividing the equation by �r and taking the limits as �r ! 0 and�NA ! 0, we get

0 ¼ r2@NA

@rþ 2rNA þ r2rCkCA þ r2e

� � @CA

@tþ r2 1� eð Þ� �

rS@CAS

@t

Dividing the entire equation by r2 gives

0 ¼ @NA

@rþ 2

rNA þ rCkCA þ e

@CA

@tþ 1� eð ÞrS

@CAS

@tð6:94Þ

On using the simple Fick’s law,

NA ¼ �DeA

@CA

@rð6:95Þ

Assuming DeAto be constant, we get

@NA

@r¼ �DeA

@2CA

@r2ð6:96Þ

Substituting Eqs. (6.95) and (6.96) into Eq. (6.94) and after some manipula-tions, we obtain

DeA

@2CA

@r2þ 2

r

@CA

@r

!� rCkCA ¼ e

@CA

@tþ 1� eð ÞrS

@CAS

@tð6:97Þ

If the chemisorption is at equilibrium and the chemisorption isothermis assumed to be linear, then the relation between CA and CAS is given by

CAS ¼ KACm

� �CA ð6:98Þ

where KA is the equilibrium chemisorption constant and Cm is the concen-tration of active sites on the catalyst surface. Thus,

DeA

@2CA

@r2þ 2

r

@CA

@r

!� rCkCA ¼ e

@CA

@tþ 1� eð ÞrSKACm

@CA

@t

Defining the dimensionless variables

o ¼ r

RP

and

xA ¼ CA

CAref

we get

1� eð ÞrSKACm|fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl}aS

þ e|{z}ag

0@

1A @CA

@t¼ DeA

R2P

@2CA

@ r=RPð Þ2 þ2rRP

@CA

@ r=RPð Þ

!� rCkCA

where, 1� eð ÞrSKACm ¼ aS is the catalyst solid surface capacitance ande ¼ ag is the catalyst pore capacitance (usually negligible). Thus, we obtain

R2P

DeA

aS þ ag� � @xA

@t¼ @2xA

@o2þ 2

o@xA@o

!� rCkR

2P

DeA

!xA ð6:99Þ


LeS þ Leg� � @xA

@t¼ r2xA � f2xA ð6:100Þ

where

r2 ¼ d

do2þ a

od

do; f2 ¼ rCkR

2P

DeA

(Thiele modulus)

and

LeS ¼ aSR2P

DeA

is the Lewis number for the solid catalyst surface and

Leg ¼agR

2P

DeA

is the Lewis number for the catalyst gas voidage. The boundary conditionsand initial condition are as discussed earlier: at t ¼ 0, xA oð Þ ¼ xAo

oð Þ. At thecenter, due to symmetry, we have, at o ¼ 0, dxA=do ¼ 0. At the surfacer ¼ RPð Þ, at o ¼ 1:0;

dxAdo

¼ ShA xAB� xA

� �For the limiting case, when LeS >>> Leg, we can neglect Leg and call

LeS ¼ Le; Eq. (6.100) becomes

Le@xA@t

¼ r2xA � f2xA ð6:101Þ

with the same initial and boundary conditions as given earlier.

6.7 MATHEMATICAL MODELING ANDSIMULATION OF FLUIDIZED-BED REACTORS

When a flow of gas stream is passed through a bed of fine powder put in atube (as shown in Fig. 6.24), we can observe the following:

1. In a certain range of flow, we will have a fixed bed (i.e., flowthrough porous medium).

2. The minimum fluidization condition is a condition at which �Pacross the bed is equal to the bed weight; a slight expansionoccurs in this case (Fig. 6.24). The bed properties will be verysimilar to liquid properties (e.g., it can be transferred betweentwo containers).

3. Freely bubbling fluidized bed: With higher flow rate of gas thanthe minimum fluidization limit, we will reach a case in whichthere will be three phases: solid, gas in contact with solid, andgas in bubbles.

The two-phase theory of fluidization is applied: ‘‘Almost all the gas in excessof that necessary for minimum fluidization will appear as gas bubbles’’ (Fig.6.25).

6.7.1 Advantages of Freely Bubbling Fluidized Beds

1. Perfect mixing of solids due to the presence of bubbles (isotherm-ality).

2. Good heat transfer characteristics (high heat transfer coefficient).3. Intraparticle as well as external mass and heat transfer resistances

are negligible (so it can be neglected in modeling). (See Fig. 6.26.)4. All of the advantages of minimum fluidization conditions.

Figure 6.24 Fine powder entrained in a tube starting to fluidize due to gas flow.

Figure 6.25 Fluidized bed.

6.7.2 Disadvantages of Fluidized Beds

1. Bypassing of bubbles through the bed.2. Bubble explosion on surface causes entrainment.3. Difficult mechanical design [e.g., industrial fluid catalytic crack-

ing (FCC) unit has to handle 7000 tons of solid].

Notes:

. Bypassing of bubbles is compensated partially by diffusionbetween dense and bubble phases

. There is a solid exchange among the wake, cloud, and the densephase. It is accounted for in three-phase models.

. Although the dense phase is perfectly mixed, the bubble phase isalmost in plug flow condition.

. It is possible to break the bubbles using baffles or redistributors.Stirrers are not recommended because of vortex formation.

6.7.3 Mathematical Formulation (Steady State)

For illustration, consider a simple reaction

A ! D

Refer to Figure 6.27, where CA 6¼ f1 hð Þ, which means that it is perfectlymixed condition, and CAB

¼ f2 hð Þ, which means that it is plug flowcondition. Here, CA is the concentration of component A in the densephase and CAB

is the concentration of component A in the bubblephase.

Figure 6.26 Diagram showing the main mass and heat transfer regions around the

bubble.

Modeling of the Dense Phase

Molar balance on the dense phase gives

nAf+ Exchange with bubble phase ¼ nA � Vr

For constant volumetric flow rate GI , we can write the above equation as

GICAf þðH0

KgAa CAB � CAð ÞAC dh ¼ GICA þ AIHrbkCA ð6:102Þ

where

rb ¼ bulk density of solid at minimum fluidization conditions

a ¼ cm2 external surface area of bubbles

cm3 volume of bubble phase

KgA ¼ mass transfer coefficient between bubble and dense phase

Figure 6.27 Schematic diagram of a fluidized bed.

Modeling of the Bubble Phase

Molar balance on an element of the bubble phase gives (it is the plug flowmode and assumed to have negligible rate of reaction; see Fig. 6.28),

GCCAB ¼ GC CAB þ�CABð Þ þ KgAa CAB � CAð ÞAC�h

With some manipulation and taking the limits as �CAB! 0 and �h ! 0,

we get

GC

dCAB

dh¼ �KgAa|ffl{zffl}

QE

AC CAB � CAð Þ ð6:103Þ

with initial condition at h ¼ 0; CAB ¼ CAf: Thus, the balance gives the

following two equations:

Dense phase (an integral equation)

GI CAf� CA

� �þQE AC

ðH0

CAB � CAð Þ dh ¼ AI H rb kCA ð6:102Þ

Bubble phase (a differential equation which can be solved analytically)

GC

dCAB

dh¼ �QE AC CAB � CAð Þ ð6:103Þ

at h ¼ 0; CAB ¼ CAf.

Figure 6.28 Mass flow in bubble phase.

Analytical Solution of Differential Equation (6.103)

By separation of variables, we get

dCAB

CAB � CA|{z}constant

0@

1A

¼ � QE AC

GC

� �dh


ln CAB � CAð Þ ¼ � QE AC

GC

� �hþ C1

To get the value of constant of integration C1, we use the initial condition ath ¼ 0,

C1 ¼ ln CAf� CA

� �Thus, we get

lnCAB � CA

CAf� CA

!¼ � QE AC

GC


a

h

Rearrangement gives

CAB � CA

CAf� CA

¼ e�a h

Finally, we get the solution of the differential equation as

CAB � CA ¼ CAf� CA

� �e�a h ð6:104Þ

Substitution of this value (6.104) in the integral Eq. (6.102) of the densephase gives

GI CAf� CA

� �þQE AC CAf

� CA

� � ðH0

e�a h dh ¼ AI H rb kCA

BecauseðH0

e�a h dh ¼ 1

a1� e�aH� �

and

a ¼ QE AC

GC

we get

GI CAf � CA

� �þ GC CAf � CA

� �1� e�aH� � ¼ AIHrbkCA

In a more simplified and organized form

GI þ GC 1� e�aH� �� |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}G

CAf � CA

� � ¼ AIHrbkCA ð6:105Þ

where

G ¼ GI þ GC 1� e�aH� �the modified volumetric flow rate.

Notes

1. Analogy between the above equations and the continuous-stirredtank reactor (CSTR) model can be easily realized, and from thisanalogy, we can define the modified flow rate with the followingphysical significance:

. At very high QE , G ¼ GI þ GCð Þ ¼ G; thus, we approachCSTR.

. At QE ffi 0, G ¼ GI ; thus, we have complete segregation.

2. Output concentration of A is calculated by the relation GI CA þGC CAB

��h¼H

¼ GCAout

Heat Balance Design Equations for the Fluidized-Bed Reactors (seeFig. 6.29)

For a reaction

A ! D

where r ¼ kCA and k ¼ k0 e�E=RT

Heat balance in the dense phase gives

GI T � Tf

� �rCP ¼ AIHrbkCA ��Hð Þ

þðH0

hBaAC TB � Tð Þ dh�UAJ T � TJð Þð6:106Þ

Note that the heat supply/removal coil is considered to be with the dense-phase balance, where

TB ¼ bubble phase temperature (varies with height)TJ ¼ jacket temperatureT ¼ dense phase temperature (constant with height)

U ¼ heat transfer coefficient between the jacket and dense phaseAJ ¼ jacket area available for heat transfer

Heat balance in the bubble phase gives (it is assumed to be in plug flowmode and with negligible rate of reaction), (see Fig. 6.30)

GC rCP TB ¼ GC rCP TB þ�TBð Þ þ hB aAC �h TB � Tð Þwhich, after some rearrangements and taking the limits as �TB ! 0 and�h ! 0, gives

GC rCP

dTB

dh¼ hB aAC T � TBð Þ ð6:107Þ

Analytical Solution of Eq. (6.107)

On separating the variables of differential Eq. (6.107), we get

dTB

T � TB

¼ b dh

where

b ¼ hB aAC

GC rCP

Figure 6.29 Heat transfer in the fluidized bed.


� ln T � TBð Þ ¼ b hþ C1

The integration constant C1 can be calculated using the initial condition ath ¼ 0, TB ¼ Tf . Thus, we get

C1 ¼ � ln T � Tf

� �On substituting the value of constant of integration, we get

lnT � TB

T � Tf

� �¼ �b h


T � TB

T � Tf

¼ e�b h

We can use the above relation in the heat balance for the dense phase (6.106)in order to calculate the heat transfer integral; we get

Figure 6.30 Heat flow in the bubble phase.

GIrCP T � Tf

� � ¼ AIHrbk0e�E=RTCA ��Hð Þ þ hBaAC T � Tf

� ��ðH0

e�bhdh�UAJ T � TJð Þ

BecauseðH0

e�b h dh ¼ 1

b1� e�bH� �

¼ GC rCP

hB aAC

1� e�bH� �

Thus, we can write

GIrCP þ GCrCP 1� e�bH� �h i

T � Tf

� � ¼AIHrbkoe�E=RTCA ��Hð Þ

�UAJ T � TJð ÞFor a fluidized bed without chemical reaction, the heat transfer equationwill be

rCP GI þ GC 1� e�bH� �h i

T � Tf

� � ¼ �UAJ T � TJð Þ ð6:108Þ

6.8 UNSTEADY-STATE BEHAVIOR OFHETEROGENEOUS SYSTEMS: APPLICATIONTO FLUIDIZED-BED CATALYTIC REACTORS

In the mass balance equation, an accumulation term will be added asfollows:

GI þ GC 1� e�ðQEAC=GCÞH� ��

CAf � CA

� � ¼ AIHrbkCA

þAccumulation or depletion with timeð6:109Þ

where

Accumulation or depletion with time ¼ The unsteady state

dynamic term

and we can write the holdup of any component (e.g., component A) as

nA ¼ AIHeCA þ AIHrbCAS ð6:110Þwhere

CA is the gas-phase concentration of component ACAS ¼ solid phase (chemisorbed) concentration of component AnA ¼ number of moles of component A in the fluidized bed reactor

(component A holdup)

rb ¼g catalyst

cm3 bed¼ rS 1� eð Þ

rS ¼ g catalyst

cm3 catalyst

On differentiation of nA [in Eq. (6.110)], we get

dnAdt

¼AIH edCA

dtþ rb

dCAS

dt

� �¼ Accumulation or depletion with time

ð6:111ÞSubstituting Eq. (6.111) into Eq. (6.109) gives

GI þ GC 1� e�ðQEAC=GCÞH� ��

CAf � CA

� � ¼ AIHrbkCA

þ AIH edCA

dtþ rb

dCAS

dt

� � ð6:112Þ

Thus, we get a differential equation in which two variables ðCA and CASÞ arefunctions of time. How do we handle putting CAS (the chemisorbed con-centration of A) in terms of CA (the gas phase concentration of A)? Thesimplest approach is the one given as follows:

Chemisorption mechanism

We give some detail regarding the relation between CA and CAS.

Reactions:

A ! DAþ S , AS at equilibriumAS ! Dþ S

where S is the active site

Thus, we can write

KA ¼ CAS

CAC0S

where,

C 0S ¼ Cm � CAS; where C0

S is the concentration of free active sites.

and, Cm is the total concentration of active sites. Thus, so we get,

KACA Cm � CAS

� � ¼ CAS

Thus,

CAS ¼ KACmCA

1þ KACA

This is a Langmuir isotherm. Considering the low concentration of CA (i.e.,linear relation between CA and CAS), we get

CAS ¼ KACm

� �CA (as earlier) ð6:113Þ

On differentiation of Eq. (6.113) and assuming that KA and Cm are constant,we get

dCAS

dt¼ KACm

� � dCA

dtð6:114Þ

Thus, the final mass balance equation derived earlier [Eq. (6.112)] becomes

GI þ GC 1� e�ðQEAC=GCHÞ� ��

CAf � CA

� � ¼ AIHrbkCA

þ AIH eþ KACmrb� � dCA

dt

ð6:115Þ

Note: It is always difficult to find the values of KA and Cm, except forvery common processes.

Utilizing the nonlinear isotherm will complicate this dynamic term asfollows. Because

CAS ¼ KACmCA

1þ KACA

differentiating gives

dCAS

dt¼ KACm 1þ KACAð Þ�1 dCA

dtþ CA �1ð Þ 1þ KACAð Þ�2KA

dCA

dt

� �

Thus, we get

dCAS

dt¼ KACm

1

1þ KACA

� KACA

1þ KACAð Þ2� �

dCA

dtð6:116Þ

For the nonisothermal case, the heat balance equation is

g T � Tf

� � ¼ AIHk0e�E=RTrbCA ��Hð Þ �UAJ T � TJð Þ � dQ

dt

ð6:117Þwhere

Q ¼ AIHergCPgTg þ AIHrbCPSTS ð6:118Þ

is the heat content of the fluidized bed reactor and

g ¼ GIrCP þ GCrCP 1� e�bH� ��

Assuming negligible heat transfer resistances between the gas and solid, weget

Tg ¼ TS ¼ T

On differentiating Eq. (6.118), we get

dQ

dt¼ AIH ergCPg|fflfflffl{zfflfflffl}

negligible

þrbCPS

0B@

1CA dT

dt

Thus, we get

dQ

dt¼ AIHrbCPS

dT

dtð6:119Þ

The expression for dQ=dt from Eq. (6.119) can be substituted in Eq. (6.117)to get the final unsteady heat balance equation

gðT � Tf Þ ¼ AIHk0e�E=RTrbCAð��HÞ �UAJðT � TJÞ

� AIHrbCPS

dT

dt

ð6:120Þ

Note: For the case of the linear isotherm, we can define a Lewis numberfor the process as follows:

Heat capacitance term

Mass capacitance term¼ AIHrbCPS

AIH eþ KACmrb� � ¼ rbCPS

eþ KACmrb� �

¼ Lewis number Leð Þ

6.9 EXAMPLE: SIMULATION OF A BUBBLINGFLUIDIZED-BED CATALYTIC REACTOR

A consecutive reaction

A ! B ! C

is taking place in a bubbling fluidized-bed reactor with both reactions beingexothermic and of first order. The intermediate product B is the desiredproduct.

Figure 6.31 shows the a schematic representation of this two-phasefluidized-bed reactor with a simple proportional control. It should be notedthat the proportional control is based on the exit temperature (the averagebetween the dense-phase and the bubble-phase temperatures), which is themeasured variable, and the steam flow to the feed heater is the manipulatedvariable.

As we have shown in the previous sections, the dynamics of the densephase are described by nonlinear ordinary differential equations (ODEs) ontime having an integral term for the mass and heat transfer between thedense and the bubble phases. The bubble phase is described by pseudo-steady state linear ODEs on height. The linear ODEs of the bubble phaseare solved analytically and the solution is used to evaluate the integrals indense-phase equations. The reader should do this as practice to reach thedimensionless model equations given next.

The dynamic equations describing this fluidized bed after the above-described manipulations are the following nonlinear ODEs in terms ofdimensionless variables and parameters:

1

LeA

dXA

dt¼ B XAf � XA

� �� a1e� g1=Yð ÞXA

1

LeB

dXB

dt¼ B XBf � XB

� �þ a1e� g1=Yð ÞXA � a2e

� g2=Yð ÞXB

dY

dt¼ B Yf � Y

� �þ a1b1e� g1=Yð ÞXA þ a2b2e

� g2=Yð ÞXB

where Xj is the dimensionless dense-phase concentration of the componentj j ¼ A;B;Cð Þ and Y is the dimensionless dense-phase temperature. The

Figure 6.31 Simulation model for two-phase fluidized-bed reactor with propor-

tional control.

dimensionless feed temperature (after neglecting the dynamics of the feedheater) is given by

Yf ¼ Yf þ K Ym � Yð ÞThe base values for Yf is Yf 0 and B is the reciprocal of the effective residencetime of the bed given by

B ¼ GI þ GC 1� e�að ÞAIH

The yield of the reactor is given by

y ¼ GIXB þ GCXBH

GI þ GCð ÞXAf

where

XBH ¼ XB þ XBf � XB

� �e�a

and

a ¼ QEHAC

GC

The dimensionless parameters are defined as follows:

ai ¼ dimensionless pre-exponential factor for the reaction i¼ rS 1� eð Þkoi

bi ¼ dimensionless exothermicity factor for the reaction i

¼ ��Hið ÞCref

rf CPf Tref

gi ¼ dimensionless activation energy for the reaction i

¼ Ei

RTref

t ¼ normalized time (time/heat capacity of the system)Lej ¼ Lewis number of component j

=CPS

KjCmCPf rf

whererS ¼ solid catalyst density

CPS ¼ specific heat of the catalyst

e ¼ voidage occupied by the gas in the dense phase

��Hi ¼ heat of reaction for reaction i

k0i ¼ pre-exponential factor for reaction i

Cref ¼ reference concentration

Tref ¼ reference temperature

rf ¼ density of the gas

CPf ¼ specific heat of the gas

Ei ¼ activation energy for reaction i

Kj = chemisorption equilibrium constant for component j

Cm = concentration of active sites in the catalyst

The values of parameters used in this simulation are tabulated in Table6.1.

Figure 6.32 shows that the desired steady state (corresponding to thehighest yield of intermediate desired product B) corresponds to a middleunstable steady state. In order to stabilize this desired steady state, the use ofa simple proportional controller is proposed and the behavior is studied fordifferent values of the proportional gain Kð Þ, as shown in Fig. 6.32B. It can

Table 6.1 Data Used for the Simulation

Bed cross-section area, A 3000 cm2

Superficial gas velocity (based on unit cross section of the

whole bed), U2

10.0 cm/s

Minimum fluidization velocity (based on unit cross section

of the whole bed), Umf

0.875 cm/s

Voidage of the dense phase 0.4

Bed height, H 100 cm

Normalized pre-exponential factor for the reaction A ! B 108

Normalized pre-exponential factor for the reaction B ! C 1011

Dimensionless overall thermicity factor for the reaction

A ! B

0.4

Dimensionless overall thermicity factor for the reaction

B ! C

0.6

Dimensionless activation energy for the reaction A ! B 18.0

Dimensionless activation energy for the reaction B ! C 27.0

Lewis number of component A, LeA 1.0

Lewis number of component B, LeB 2.2

Feed concentration of component A, XAf 1.0

Feed concentration of component B, XBf 0.0

Feed concentration of component C, XCf 0.0

Dimensionless feed temperature to the reactor (base value),

Yf

0.553420765035

Set point for the controller, Ym 0.929551298359

be observed that a K value of at least 3.5 is necessary to stabilize thisunstable saddle-type steady state.

The dynamic behavior of this three-dimensional system is quite richand complicated. The reader can find the detailed study in the paper pub-lished by one of the authors of this book (1). A sample of the results isshown in Fig. 6.33, in which the complex oscillations of the state variablescan be observed.

6.10 A DISTRIBUTED PARAMETER DIFFUSION–REACTION MODEL FOR THE ALCOHOLICFERMENTATION PROCESS

The following is an example for the development of a distributed parametermodel for the yeast floc in the alcoholic fermentation process. The modeltakes into consideration the external mass transfer resistances, the masstransfer resistance through the cellular membrane, and the diffusion resis-tances inside the floc. The two-point boundary-value differential equationsfor the membrane are manipulated analytically, whereas the nonlinear two-point boundary-value differential equations of diffusion and reaction inside

Figure 6.32 Yield and van Heerden diagrams for the data in Table 6.1. (A) Yield

yð Þ of B versus dimensionless dense-phase temperature Yð Þ, where Ym is Y at the

maximum yield of B; (B) van Heerden diagram showing the heat generation function

of the system G Yð Þ and heat removal line R Yð Þ for the different values of K versus

the dimensionless dense-phase temperature Y .

the floc are approximated using the orthogonal collocation technique(Appendix E).

The evaluation of the necessary diffusion coefficients have involved arelatively large number of assumptions because of the present limitedknowledge regarding the complex process of diffusion and biochemical reac-tions in these systems.

A comparison between the model and an experimental laboratorybatch fermentor as well as an industrial fed-batch fermentor is also presentedto the reader. The model is shown to simulate reasonably well the experi-mental results, with the largest deviation being for the concentration of yeast.

A heterogeneous model is being developed by taking three mass trans-fer resistances in series for both ethanol and sugar; these are as follows:

1. External mass transfer resistances that depend on the physicalproperties of the bulk fluid (extracellular fluid) and the degreeof mixing as well as temperature.

2. Diffusion through the cell membrane: A process that is not fullyunderstood and which is affected by many factors in a complexway; for example, it is well known that ethanol increases thefluidity of biological membranes; in addition, ethanol causes a

Figure 6.33 Classical dynamic characteristics (time traces) for a case with

K ¼ 3:083691.

change in phospholipid composition and a decrease of the lipid toprotein ratio of the membrane. At steady state, the transport ofsugar through biological membranes is considered a carrier-mediated process, which consists of diffusion and biochemicalreactions in the membrane. Glucose diffusion versus concentra-tion gradient has been beautifully simulated by Kernevez for theglucose pump formed of artificial membranes (2). However, forthe present unsteady-state model, all bioreactions are assumed totake place inside the cell, with the assumption that no reaction istaking place in the membrane. Therefore, Fickian diffusion ofsugar is considered through the membrane with the appropriatediffusion coefficient.

3. Intracellular diffusion, which is dependent on the intracellularconditions. A reasonably rigorous description of this diffusionprocess requires a distributed diffusion–reaction model. A morerigorous description, which is not considered in this undergrad-uate book, requires a more structured diffusion–reaction model.

6.10.1 Background on the Problems Associated withthe Heterogeneous Modeling of AlcoholicFermentation Processes

The rigorous heterogeneous modeling of alcoholic fermentation processrequires a large amount of fundamental information which is not completelyavailable in the literature now. However, rational mathematical modelsbased on the available knowledge regarding this process represent consider-able improvement over the usual pseudohomogeneous models. Such modelsalso help to direct experimental investigations in a more organized manner.

The alcoholic fermentation process is quite a complex process andinvolves the following main processes.

Product Inhibition

The ethanol inhibition effect is of the non-competitive type, where ethanolconcentration affects only the maximum specific growth rate. This depen-dence of the maximum specific rate of growth of ethanol concentrationinvolves two main types of dependence: linear dependence and nonlineardependence. Jobses and Roels [3] have shown that inhibition kinetics can beapproximated by a linear relation between the specific growth rate and theethanol concentration up to 50 g/L ethanol. Above this level, deviation fromlinearity is observed.

As discussed by Elnashaie and Ibrahim [4], many investigatorsreported that produced ethanol (intracellular ethanol) is more toxic than

added ethanol (extracellular ethanol) and an intermediate value of Kp ¼ 35has been observed, which lies between Kp ¼ 105:2 g/L for added ethanol andKp ¼ 3:04 g/L for produced ethanol. This value is believed to represent Kp

for ethanol inhibition when using the heterogeneous model, which distin-guishes between intracellular and extracellular fluids.

Sugar Inhibition

It is well known that a high concentration of sugar inhibits the growth ofyeasts. However, the inhibitory effect of sugar is negligible below a sugarconcentration of 100 g/L [5]. The inhibitory effect of sugar is not included inthe present model; however, it can easily be added to the kinetic rate equa-tion without any extra complications.

Cell Inhibition

In alcoholic fermentation, most kinetic models express cell growth rate as alinear function of cell concentration. However, the experimental results ofCysewski and Wilke [6] showed that at a high cell concentration, the linearrelationship is incorrect.

Ethanol Diffusion

The dispute in the literature regarding the diffusion of ethanol has beendiscussed by Elnashaie and Ibrahim [4]. The slightly more detailed diffusionreaction model for the alcoholic fermentation process presented in this sec-tion of the book will help to throw more light on this dispute than the modelused by Elnashaie and Ibrahim [4]. However, we are certainly still quite farfrom a completely rigorous representation of this rather complex diffusion–reaction problem.

Flocculation of the Yeast

The flocculation of the yeast affects the mass transfer rates and, thus, therate of fermentation considerably through the change of the exposed surfacearea for mass transfer per unit mass of the micro-organism. The flocculationprocess has been investigated experimentally by many researchers; however,the theoretical basis for understanding the flocculation process is severelylacking.

6.10.2 Development of the Model

The model equations are developed for the flocs with three mass transferresistances in series for sugar and ethanol. The present floc model is more

sophisticated than the one developed earlier by Elnahsaie and Ibrahim [4],whereas for the extracellular fluid, the model does not differ from the thatused by Elnashaie and Ibrahim [4].

Single-Floc Model

The floc is formed of a number of cells. This number depends on manyfactors, including the liquid-phase composition and the flocculating ten-dency of the specific strain of the yeast. The structure of the floc is quitecomplex and branched, giving rise to what may be called a fractal structure.In this undergraduate book, we use the more classical approach of reducingthe complex structure of the floc to an equivalent sphere. The geometricdimensions of the floc fractal structure can be used instead. However, thereare certain difficulties associated with the determination of the fractaldimension f and the constant of proportionality a (needed for the calcula-tion of the geometric dimension of the fractal structure), which are beyondthe scope of this undergraduate book. Figure 6.34 shows the idealizedequivalent sphere used in this model, where the internal floc (the equivalentsphere) is considered a substrate sink where sugar reacts to produce ethanol.

Diffusion through the cell membrane (the membrane of the floc equiva-lent sphere) is assumed to obey Fick’s law with membrane diffusion coeffi-cients that take into account the complex nature of the membrane. Carrier-

Figure 6.34 Schematic presentation of the proposed model of an equivalent sphere

for the microbial floc.

mediated transport of sugar through the membrane is not considered becauseof the unsteady-state nature of the model and the fact that the model assumesthat all of the bioreactions take place inside the cell, with no reactions takingplace in the membrane. The external mass transfer resistances are assumed tobe lumped in one hypothetical external mass transfer film, with the masstransfer coefficient depending on the extracellular conditions.

Single-Floc Equations

A schematic diagram showing the details of the proposed model for theequivalent sphere representing the microbial floc is presented in Figure6.34. A number of simplifying assumptions are implicit in the lengthymanipulation of the model equations.

Mass Balance of Ethanol Within the Membrane

If we assume that no reaction is taking place in the membrane, then thediffusion equation becomes

dNp

do¼ d

do�Dpmo

2dP

R1do

" #¼ 0 ð6:121Þ

where

Np ¼ flux of intracellular ethanol (g/cm2 s)Dpm ¼ diffusion coefficient of intracellular ethanol in membrane

(cm2/s)P ¼ intracellular ethanol concentration (g/cm3)o ¼ r=R1 (see Fig. 6.34 for R1)

The diffusivity of ethanol through the membrane is estimated as a functionof ethanol mole fraction in the membrane x1 by the following relation [7]:

Dpm ¼ a22

x1m1=31 þ 1� x1ð Þm1=32

h i3 ð6:122Þ

where a22 is a constant and mi is the viscosity of component i. Also,

x1 ¼P

rwRM þ P 1� RMð Þ � ð6:123Þ

where rw is the density of the floc (g wet weight/cm3 wet volume) ¼ 1.23g/cm3 and RM is the ratio of the molecular weight of ethanol and themembrane lipid.

Substitution of Eqs. (6.122) and (6.123) into Eq. (6.121) and integra-tion gives

f Pð Þ ¼ a22 APþ B

dln cþ dPð Þ � A1

d cþ dPð Þ �A2

2d cþ dPð Þ2� �

¼ C1 �C2

o

ð6:124Þwhere,

A ¼ b3

d3

B ¼ 3b2

d3

!ad � cbð Þ

A1 ¼3b

d3

� �ad � cbð Þ2

A2 ¼1

d3

� �ad � cbð Þ3

a ¼ RMrwb ¼ 1� RM

c ¼ mlipid� �1=3

RMrw

d ¼ methanol� �1=3�RM mlipid

� �1=3a22 ¼ a22=R1

and C1 and C2 are constants of integration to be fitted to boundary condi-tions.

Boundary Condition at the External Surface of the Membrane

The external surface of the membrane corresponds to r ¼ R2 (i.e.,o2 ¼ R2=R1).

The boundary condition at this point is given by

dP

do

��o2

¼ KgpR1

Dpm o2ð Þ

� �Pb � Po2

� � ð6:125Þ

where Kgi is the external mass transfer coefficient for component i (cm/s) andPb is the extracellular ethanol concentration (g/cm3). From Eq. (6.121), weobtain

dP

do

��o2

¼ C2

Dpm o2ð Þo22

ð6:126Þ

Thus, from Eqs. (6.125) and (6.126), we can express the constant of integra-tion C2 in terms of the concentration at the point o2 (i.e., R2=R1):

C2 ¼R2

2

R1

!Kgpðpb � p!2

Þ ð6:127Þ

Equation (6.124) at point o2 can be written as

f Po2

� � ¼ a22 APo2þ B

dln cþ dPo2

� �� A1

d cþ dPo2

� �� A2

2d cþ dPo2

� �2" #

¼ C1 �C2

o2

¼ C1 � R2Kgp Pb � Po2

� �ð6:128Þ

Therefore, C1 can also be written in terms of Po2, using the above relation,

as follows:

C1 ¼ f Po2

� �þ R2Kgp Pb � Po2

� � ð6:129ÞAt the point oþ

1 Rþ1 =R1 ¼ 1:0þ

� �(i.e., Rþ

1 is R1 at the membrane side asdistinguished from R1, which is R1 at the floc side). Equation (6.124) canbe rewritten using Eqs. (6.127) and (6.129) in the following form:

f Poþ1

� �¼ C1 �

C2

o1

¼ f Po2

� �þ R2Kgp Pb � Po2

� �� R22

R1

!Kgp Pb � Po2

� �Thus,

f Poþ1

� �¼ f Po2

� �þ R2Kgp 1� R2

R1

� �Pb � Po2

� � ð6:130Þ

At the interface between the membrane and the floc (the point o�1 ), we have,

DpfdP

do

��o�1

¼ DpmdP

do

��oþ1

ð6:131Þ

Similar to Eq. (6.126), we have

dPm

do

��oþ1

¼ C2

oþ1

� �¼ C2 ð6:132Þ

Thus, from Eqs. (6.131), (6.132), and (6.127), we get

DpfdP

do

��o�1

¼ R22

R1

!Kgp Pb � Po2

� � ð6:133Þ

If we use the one internal collocation point approximation for the diffusionreaction equation inside the floc as shown in the next section, then the left-hand side of Eq. (6.133) can be approximated using the orthogonal colloca-tion formula (refer to Appendix E for orthogonal collocation method), and,thus, Eq. (6.133) becomes

Dpfo�1

A21Pc þ A22Po�1

� �¼ R2

2

R1

!Kgp Pb � Po2

� � ð6:134Þ

where A11 and A22 are collocation constants for the spherical shapeA21 ¼ �3:5;A22 ¼ 3:5ð Þ and c indicates the internal collocation point forthe floc.

Ethanol Mass Balance Inside the Floc

For unsteady-state mass balance inside the floc, considering that all of thereactions producing ethanol and consuming sugar is taking place inside thefloc, we obtain the following parabolic partial differential equations:

@P

@t¼ mmrdYPm

0

YC

þ 1

r2

� �@

@rDpf r

2 @P

@r

� �� ð6:135Þ

whereYP ¼ yield factor for ethanol (g ethanol produced/g sugar consumed)YC ¼ yield factor for yeast (g yeast produced/g sugar consumed)rd ¼ dry density of floc (g dry wt/cm3 wet vol.) ¼ 0.2 g/cm3

mm ¼ maximum specific growth rate (s–1)

Putting the above equation in dimensionless coordinate, o 2 0; 1ð Þ,o ¼ r=R1, we obtain the following equation:

R21

@P

@t¼ mmrdYPm

0R21

YC

þ 1

o2

� �@

@oDpfo

2 @P

@o

� �� ð6:136Þ

Applying the orthogonal collocation technique (refer to Appendix E fororthogonal collocation method) using one internal collocation point at oc,at the internal collocation point we obtain the following differential equation:

R21

dPc

dt¼ B11 DpfcPc

�þ B12 Dpfo�1

Po�1

h iþ�2

SYPm0 ScPc½ � ð6:137Þ

where B11 ¼ �10:5, B12 ¼ 10:5, Sc is the sugar concentration at the internalcollocation point (g/cm3), and

�2S ¼ mmrdR

21

Dsf YC

Equations (6.130), (6.134), and (6.137) are two algebraic equations and onedifferential equation in four variables, namely Poþ

1, Po�

1, Po

1, and Pc; thus,

one equation is missing. This extra equation can be furnished by the equili-brium relation between Poþ

1and Po�

1, which has the following form:

Po�1¼ K1Poþ

1ð6:138Þ

where K is the equilibrium partition constant for the floc side of the mem-brane

Equations (6.130), (6.134), (6.137), and (6.138) can now be solved toobtain four variables at every time during the fermentation process, pro-vided that at these times, the concentration of sugar at the collocation pointis known (the necessary equations are derived in the next section) and thebulk (extracellular) concentrations are known (the extracellular equationsare exactly the same as in Ref. 4; however, for the benefit of the reader, wewill present those equations again in a later section). We will also discuss thesolution algorithm to be employed to solve this complex set of equations.

Mass Balance for Sugar Inside the Membrane

We assume that no reaction is taking place inside of the membrane and alsoassume a pseudo-steady-state for the membrane. Furthermore, we assumeconstant sugar diffusivity through the membrane. Based on these assump-tions, the sugar material balance inside the membrane becomes

r2S ¼ 0 ð6:139Þwhere

r2 ¼ d2

do2þ 2

o

� �d

do

� �and o ¼ r

R1

The boundary condition at the outer surface of the membrane o2 (i.e.,R2=R1) is given by

dS

do

��o2

¼ ShS Sb � So2

� � ð6:140Þ

where

ShS ¼ KgsR1

Dsm

and Kgs is the external mass transfer coefficient for sugar (cm/s) and Dsm isthe diffusion coefficient of sugar in membrane phase (cm2/s).

The boundary condition at the inner surface of the membrane o1 (i.e.,R1=R1 ¼ 1:0) is given by

DsfdS

do

��o�1

¼ DsmdS

do

��oþ1

ð6:141Þ

The general equation of Eq. (6.139) is given by

S ¼ C 01 �

C 02

oð6:142Þ

After some manipulation, we obtain So2in terms of Sþ

o1as follows,

So2¼

Soþ1þ ShSmSb 1� R1=R2ð Þ

1þ ShSm 1� R1=R2ð Þ ð6:143Þ

where

ShSm ¼ ShSR2

R1

� �2

The boundary condition in Eq. (6.141) coupled with the floc differentialequation after simplifying it using the one-internal orthogonal collocationpoint approximation, gives the following relation among S�

o1, Sþ

o1, and the

concentration of sugar at the internal collocation point Sc:

So�1¼

ShSf Sb � Soþ1

� �A22 1þ ShSm 1� R1=R2ð Þ½ � �

A21

A22

� �Sc ð6:144Þ

where

ShSf ¼ ShSmDsmDsf

� �

The change of sugar concentration at the internal collocation point is givenby the following nonlinear differential equation:

R21

Dsf

!dSc

dt¼ B11Sc þ B12So�

1��2

Sm0 ScPc½ � ð6:145Þ

where

�2S ¼ mmrdR

21

DsfYC

m0 ¼ KpS 1� X=Xmð ÞNKp þ P� �

KS þ Sð Þ þK 0

p

K 0p þ P

Kp is the inhibition constant (g/cm3), K 0p is the rate constant (g/cm

3), X is thebiomass concentration (g dry wt/cm3), and Xm is the maximum biomassconcentration (g dry wt/cm3). In addition, we assume the validity of theequilibrium relation between So�

1and Soþ

1written as

So�1¼ K2Soþ

1ð6:146Þ

where K2 is the equilibrium partition constant for the floc side of the mem-brane.

The Extracellular Balance Equation

For the batch fermentor case, the extracellular mass balance equations arethe same as those in Ref. 4; these are as follows:

For sugar,

dSb

dt¼ �3XKgS

rd � X� �

R2

" #Sb � So2

� � ð6:147Þ

For ethanol,

dPb

dt¼ �3XKgp

rd � X� �

R2

" #Po2

� Pb

� � ð6:148Þ

For micro-organisms,

dX

dt¼ mmm

0X ð6:149Þ

where, Kgi is the external mass transfer coefficient for component i(cm/s).

Equations (6.130), (6.134), (6.137), (6.138), and (6.143)–(6.149) are themodel equations that are to be solved together with the physical properties,external mass transfer, and diffusivities correlations to obtain the change ofsugar, ethanol, and micro-organism concentrations with time.

Note: The physical properties, external mass transfer, and diffusivity cor-relations can be obtained from Ref. 4.

6.10.3 Solution Algorithm

For the given initial bulk conditions at t ¼ 0,Pb0, Sb0, and the initial con-ditions for Pc and Sc the solution algorithm is as follows:

1. Equations (6.130), (6.134), and (6.138) together with the relationsfor physical properties and diffusivities [7] are solved to obtainthe values of Po2

, Poþ1, and Po�

1.

2. Equations (6.143), (6.144), and (6.146) together with the relationsfor physical properties and diffusivities [7] are solved to obtainthe values of So2

, Soþ1, and So�

1.

3. The differential Eqs. (6.137) and (6.145) for the concentration atthe internal collocation point are integrated using a fourth-order

Runge–Kutta routine with automatic step size (to ensure accu-racy) to obtain the values of Pc and Sc for the next time step.

4. The differential Eqs. (6.147), (6.148), and (6.149) are integratedusing the same routine to obtain the values of Pb, Sb, and X forthe next time step.

5. Repeat steps 1–4 until you reach the final time of fermentation.

6.10.4 Comparison Between the Model andExperimental/Industrial Data

The model has been solved using the algorithm discussed in Section 6.10.3,the physical parameters estimations are given in Ref. 7, and the kinetic andphysical parameters are given in Ref. 4. The thickness of the membrane istaken as 10�6 cm for the two simulated cases. The results of the presentmodel are compared with the experimental and industrial results.

Figure 6.35 shows an example of the results of the experimentalintracellular ethanol concentration profile versus time together with theprofile predicted by the above-discussed distributed parameter model(DPM) at the internal collocation point. The results of the present DPMmodel give the nonmonotonic shape of the intracellular ethanol concentra-tion and the profile as a whole follows the same experimental trend.

Figure 6.36 shows the profile of the extracellular sugar concentration.The results of the DPM are very close to the experimental results.

Figure 6.35 Intracellular ethanol concentration versus time (*: experimental; —:

DPM for an experimental batch fermentor).

Figure 6.37 shows the extracellular ethanol concentration profileswhere the DPM predicts ethanol concentrations that are very close to theexperimental results.

This discussion of the development of a DPM of the fermentationprocess gives the reader a good overview and the necessary understanding

Figure 6.36 Extracellular sugar concentration versus time (*: experimental; —:


Figure 6.37 Extracellular ethanol concentration versus time (*: experimental; —:


for the development of mathematical models for industrial biochemicalprocesses.

REFERENCES

1. Elnashaie, S. S. E. H., Harraz, H. M., and Abashar, M. E. Homoclinical chaos

and the period-adding route to complex non-chaotic attractors in fluidized bed

catalytic reactors. Chaos Solitons Fractals 12, 1761–1792, 2001.

2. Kernevez, J. P. Control, optimization and parameter identification in immobi-

lized enzyme systems. In Proceedings of the International Symposium on Analysis

and Control of Immobilized Enzyme Systems. (Thomas, D. and Kernevez, J. P,

eds.). North Holland/American Elsevier, 1976, pp. 199–225.

3. Jobses, I. M. L and Roels, J. A. The inhibition of the maximum specific growth

and fermentation rate of Zymomonas mobilis by ethanol. Biotechnol. Bioeng.

28(4), 554–563, 1986.

4. Elnashaie, S. S. E. H. and Ibrahim, G. Heterogeneous modeling for the alcoholic

fermentation process. Appl. Biochem. Biotechnol. 19(1), 71–101, 1988.

5. Ciftci, T., Constantinides, A., and Wang, S. S. Optimization of conditions and

cell feeding procedures for alcohol fermentation. Biotechnol. Bioeng. 25(8),

2007–2023, 1983.

6. Cysewski, G. R. and Wilke, C. R. Process design and economic studies of

alternative fermentation methods for the production of ethanol. Biotechnol.

Bioeng., 20(9), 1421–1444, 1978.

7. Elnashaie, S. S. E. H. and Ibrahim, G. A distributed parameter diffusion–

reaction model for the alcoholic fermentation process. Appl. Biochemi.

Biotechnol. 30(3), 339–358, 1991.

8. Garhyan, P. and Elnashaie, S. S. E. H. Exploitation of static/dynamic bifurca-

tion and chaotic behavior of fermentor for higher productivity of fuel ethanol.

AIChE Annual Meeting, 2001.

PROBLEMS

Problem 6.1

Derive the steady-state mass balance equations for a one-component two-plate absorption tower with ideal (equilibrium) stages; that is, the concen-trations of the transferable component in the liquid and gas leaving eachplate are at equilibrium according to the linearized equilibrium relation

Yj ¼ aXj þ b

where a ¼ 0:72 and b ¼ 0:0, and Yj and Xj are weight fractions of the solute(the transferable component) to inerts in the gas and liquid phases. The massflow rate of the gas and liquid (based on inerts) are G ¼ 66:7 lbs inerts/minand L ¼ 40:8 inerts/min. Solve for the steady-state plate composition whenthe liquid feed is pure and the gas feed has a concentration of 0.3 lb solute/lbinerts.

What is the effect of relaxing the assumption of equilibrium stages.Derive the steady-state mass balance equations for this case of nonidealstages and express the solution vectors in terms of the different coefficientmatrices and input vectors.

Problem 6.2

Flooded condensers are sometimes used in distillation columns. The liquidlevel runs up in the condenser, covering some of the tubes. Thus, a variableamount of heat transfer area is available to condense the vapor. The columnpressure can be controlled by changing the distillate (or reflux) draw-offrate.

Write equations describing the dynamics of the condenser.

Problem 6.3

Derive the unsteady-state mass balance equations for a countercurrentabsorption column with N plates, where a single component is absorbedfrom the vapor phase into the liquid phase as shown in Figure P6.3. Assumeideal stages and that the liquid and gas on each plate are perfectly mixed andthat the liquid and gas molar flow rate from one plate to the other areconstant (dilute system). Assume linear equilibrium relations. Put the result-ing equation in a matrix form. Deduce the steady-state matrix equationfrom the unsteady-state matrix equation.

Problem 6.4

A mixture of two immiscible liquids is fed to a decanter. The heavier liquidA settles to the bottom of the tank. The lighter liquid B forms a layer on thetop. The two interfaces are detected by floats and are controlled by manip-ulating the two flows FA and FB (m3/h) according to the relations

FA ¼ KA hA and FB ¼ KB hA þ hBð Þ

Figure P6.3 Schematic diagram for Problem 6.3.

where hA is the height of the interface above the bottom of the tank and hB isthe height of the light liquid Bð Þ above the interface. The controllers increaseor decrease the flows as the level rise or fall.

The total feed rate is Wo (kg/h). The weight fraction of liquid A inthe feed is XA, and the two densities rA and rB (kg/m3) are constant. Writethe equations describing the dynamic behavior of the system; also expressthe steady-state height of both phases in terms of other parameters.

Problem 6.5

For the partial oxidation of ethylene to ethylene oxide,

C2H4 þ 12O2 ! C2H4O

the catalyst consists of silver supported on alumina, and although it isreasonably specific, appreciable amounts of CO2 and H2O are also formed.Over the range of interest, the yield of ethylene oxide is relatively constant,so that for present purposes, we may regard the reaction stoichiometry as

C2H4 þ 1:5O2 ! 0:6C2H4Oþ 0:8CO2 þ 0:8H2O

The rate of reaction may be expressed as

r ¼ 1:23� 106e�9684=TP0:323C2H4

P0:658O2

where the partial pressures are expressed in atmospheres, temperature isexpressed in degrees Kelvin, and the rate of reaction r is expressed inpound moles per pound of catalyst per hour.

If 18-in. catalyst pellets are packed in 1-in.-inner-diameter tubes, which,

in turn, are immersed in a liquid bath that maintains the tube walls at 2408F,consider the effects of varying the feed temperature and of diluting the feedwith N2 to moderate the thermal effects accompanying the reaction.Consider inlet temperatures from 3508F to 4808F and N2=C2H4 ratiosfrom 0 to 5.0.

Your analysis will be governed by the following constraints:

(a) If the temperature at any point in the reactor exceeds 5508F, theconditions will be inappropriate in that explosions may occur inthis regime.

(b) If the temperature decreases as the mixture moves down thereactor, the reaction must be regarded as self-extinguishing.

(c) If the pressure drop exceeds 14 atm, the calculations must beterminated.

Determine the range of satisfactory performance and the resultant yields forvarious reactor lengths for the following operating specifications:

1. Inlet gas pressure ¼ 15 atm2. Inlet C2H4=O2 ratio (moles) ¼ 4:1.3. Superficial mass velocity ¼ 9000 lb/h ft2.4. Bulk density of catalyst ¼ 81.9 lb/ft3.

External heat and mass transfer effects are to be neglected in your analysis,but you should estimate the potential magnitudes of these effects.

Problem 6.6

Before ethylene feedstocks produced by thermal cracking can be usedchemically for most applications, it is necessary to remove the traces ofacetylene present in such streams. This purification can be accomplishedby selective hydrogenation of acetylene to ethylene. The process involvesadding sufficient hydrogen to the feedstock so that the mole ratio of hydro-gen to acetylene exceeds unity. Using a palladium-on-alumina catalystunder typical reaction conditions (25 atm, 50–2008C), it is possible toachieve extremely high selectivity for the acetylene hydrogenation reaction.As long as acetylene is present, it is selectively adsorbed and hydrogenated.However, once it disappears, hydrogenation of ethylene takes place. Thecompetitive reactions may be written as

C2H2 þH2 ! C2H4 (desired)

C2H4 þH2 ! C2H6 (undesired)

The intrinsic rate expressions for these reactions are both first order inhydrogen and zero order in acetylene or ethylene. If there are diffusionallimitations on the acetylene hydrogenation reaction, the acetylene concen-tration will go to zero at some point within the core of the catalyst pellet.Beyond this point within the central core of the catalyst, the undesiredhydrogenation of ethylene takes place to the exclusion of the acetylenehydrogenation reaction.

(a) In the light of the above facts, what do the principles enunciatedin this chapter have to say about the manner in which the reactorshould be operated and the manner in which the catalyst shouldbe fabricated?

(b) It is often observed that the catalysts used for this purpose incommercial installations do not achieve maximum selectivityuntil they have been on stream for several days. How do youexplain this observation?

Problem 6.7

Show that the equilibrium relationship y ¼ mxþ b reduces to the equationY ¼ mX if we let Y ¼ y� b= 1�mð Þ and X ¼ x� b= 1�mð Þ.

Problem 6.8

A dynamic model for a plate absorber was developed in the chapter for thecase of unit-plate efficiency (ideal stages). In order to make the model morerealistic, we could assume that the plate efficiency is constant throughout thecolumn and is given by the relation,

E ¼ yn � yn�1

yn � yn�1

In this expression yn is the vapor composition in equilibrium with liquidmixture of composition xn, so that the equilibrium relationship becomes

yn ¼ mxn þ b

Develop a dynamic model of a plate absorption column using these assump-tions.

Problem 6.9

Consider the dynamics of an isothermal CSTR followed by a simple (singlestage) separating unit (e.g., an extractor, crystallizer, or a settler). The reac-tion is reversible, A , B, and the effluent stream from the separator, whichis rich in unreacted material, is recycled. It is assumed that the reaction rateis first order, the equilibrium relationship for the separator is linear, and therate of mass transfer between the phases in the separator could be written interms of mass transfer coefficient and a linear driving force. Making certainthat you define all your terms carefully, show that the dynamic model forthe plant can be put into the following form:

Reactor

VdCA

dt¼ qCAf þQCA2 � qþQð ÞCA � V k1CA � k2CBð Þ

VdCB

dt¼ QCB2 � qþQð ÞCB þ V k1CA � k2CBð Þ

Lean phase in separator

HdCA2

dt¼ qþQð ÞCA �QCA2 � kgAav

� �CA2 �m1CA1ð Þ

HdCB2

dt¼ qþQð ÞCB �QCB2 � kgBav

� �CB2 �m2CB1ð Þ

Rich phase in separator

hdCA1

dt¼ kgAav� �

CA2 �m1CA1ð Þ � qCA1

hdCB1

dt¼ kgBav� �

CB2 �m2CB1ð Þ � qCB1

Problem 6.10

If we consider a plate gas absorption unit containing only two trays for acase where the equilibrium relationship is linear, we can write a dynamicmodel for this unit as

Hdx1dt

¼ L1 þ Vmð Þx1 þ L2x2 � Vmx0

Hdx2dt

¼ Vmx1 � L2 þ Vmð Þx2

where x0 is the liquid composition in equilibrium with the vapor streamentering the bottom of the column. For the simplest case, we might hopethat the liquid flow rate was constant, so that L1 ¼ L2 ¼ L. However, thereare situations in which the liquid rate is observed to be time dependent, andwe often model the tray hydraulics with the simple equations

tdL1

dt¼ �L1 þ L2

tdL2

dt¼ �L2 þ L3

where L3 is the liquid rate entering the top plate.First consider the case of constant flow rates:

(a) Find the steady-state solutions of the material balance expres-sions.

(b) Linearize the equations around this steady-state operating point.(c) Determine the characteristic roots of the linearized equations.

Next, try to reproduce parts (a)–(c) for the case of variable flow rates.At what point does the procedure fail. Why?

Problem 6.11

Derive a dynamic model that can be used to describe an adsorption or ion-exchange column. If you desire, consider a particular case where benzene is

being removed from an airstream by passing through a bed packed withsilica gel. Carefully define your terms and list your assumptions.

What is the meaning of steady-state operation in this kind of unit?How does the model change as you change one or more of the assumptions?

Problem 6.12

Show that the dynamic model for a double-pipe, countercurrent heatexchanger can have the same form as the model of a packed absorber.Discuss the assumptions inherent in both the heat exchanger and absorbermodels which might lead to significant differences in the kinds of equationsused to describe each system.

Problem 6.13

Derive a dynamic model for a packed-bed extraction unit that includes anaxial dispersion term for the dispersed phase. Give an appropriate set ofboundary conditions for the model.

Problem 6.14

In the contact process for manufacturing sulfuric acid, sulfur is burned toSO2, the SO2 is reacted with oxygen over a catalyst to produce SO3, and theSO3 gas is absorbed in a packed tower by concentrated sulfuric acid. Theparticular concentration of acid fed to the absorber is a critical design vari-able, for the heat effects can be so large that there might be significantamount of mist formation. Derive a dynamic model describing an SO3

absorber. Carefully define your terms and list your assumptions. Howwould you decide whether you could neglect the accumulation of energyin the packing?

Problem 6.15

Heat regenerators are encountered in a number of large-scale industrialprocesses, such as open-hearth furnaces, liquefaction of a vapor, and theseparation of its components in the liquid state. A hot gas, possibly leav-ing a reactor, is passed through a checkwork of bricks, or even a bed ofstones, so that the heat is removed from the gas and stored in the solid.Then, a cold gas is passed through this bed, normally in the oppositedirection, so that the gas is preheated before it enters the reactor.Derive a dynamic model for the system. Carefully define your terms andlist your assumptions.

Problem 6.16

The catalytic dehydrogenation of butane is important both in the manufac-ture of butadiene and as one step in the synthetic manufacture of gasoline. Ithas been reported that the principal reaction and important secondary reac-tion are as follows:

C4H10 ! C4H8 þH2 (principal reaction)

C4H8 ! C4H6 þH2 (secondary reaction)

In the high-temperature range of interest, an appreciable amount of thereactions occurred by pyrolysis in the homogeneous phase, as well as bythe catalytic path. Also, it was observed that a large number of other reac-tions were taking place, including the following:

(a) The dealkylation or cracking of butane to form methane, ethane,ethylene and propylene

(b) The dealkylation of butanes to form methane, ethane, propane,ethylene, propylene and coke

(c) The dimerization of butadiene to form 4-vinyl cyclohexane-1(d) The decomposition of butadiene to form hydrogen, methane,

ethylene, acetylene, and coke

Obviously, a complete determination of the reaction kinetics, both homo-geneous and catalytic, of this process would be extremely difficult. Hence,for gasoline manufacture, it was suggested to use the following simplifiedreaction model:

C4H10 ! C4H8 þH2

C4H10 ! 0:1C4H8 þ 0:1H2 þ 1:8 (dealkylation products)

C4H8 ! 0:1H2 þ 1:8 (dealkylation products)

with the rate equations

rA ¼ C PA � PRPS=Kð Þ1þ KAPA þ KRsPRsð Þ2 (main dehydrogenation reaction)

rB ¼ kPA

rC ¼ kPR

where,

rA ¼ rate of reaction (moles/kg catalyst h)

C ¼ overall rate constant (moles/kg catalyst h atm)

K ¼ overall gas phase equilibrium constant (atm)

KA ¼ effective adsorption equilibrium constant of butane (1/atm)

KRs ¼ effective average adsorption constant of hydrogen and butene

PRs ¼ 12ðPR þ PsÞ ¼ average partial pressure of hydrogen and butene

rB; rC ¼ rate of cracking reactions (moles/kg catalyst h)

PA;PR ¼ partial pressures of butane and butene

PS ¼ partial pressure of hydrogen

k ¼ cracking reaction rate constant

Using this simplified kinetic model, develop a dynamic model for anonisothermal fixed-bed, catalytic reactor. Include the possibility of feedingsteam to the bed to act as a dilutent. Carefully define your terms and listyour assumptions.

Problem 6.17

The reaction-rate expressions for heterogeneous, catalytic reactions arebased on the assumption that the rates of the individual steps taking placein the overall reaction are equal; that is, an equality exists among the rate ofdiffusion of reactants through a stagnant film of gas surrounding the cata-lyst particle, the rate of adsorption of reactants on the surface of the cata-lyst, the rate of surface reaction; the rate of desorption of the products fromthe surface, and the rate of diffusion of products across the stagnant filmback into the bulk of the gas stream flowing through the catalyst bed. Afurther modification must be made if the catalyst particle is porous and ifthe rate of diffusion into the pores is important.

Normally, the equations we obtain based on this assumption are socomplicated that they are unmanageable. Hence, we often attempt to sim-plify the approach by assuming that one of the steps is rate controlling andthat the others are at equilibrium. The final form of the rate expression, aswell as the values of the unknown constants, is usually obtained by compar-ing the model predictions with experimental data from isothermal reactorsoperated at steady-state conditions.

Discuss, in detail, the applicability of these rate expressions in dynamicmodels.

Problem 6.18

Fluid catalytic cracking is one of the most important processes in the pet-roleum refining industry. The process cracks gas oil to produce high-octane-number gasoline. The Type IV industrial design consists of two bubblingfluidized beds with continuous catalyst circulation between the two vessels;

the reactor and the regenerator. In the reactor gas oil cracks to producehigh-octane-number gasoline; overcracking also occurs, giving rise to lightgases and deposition of carbon on the catalyst that causes catalyst deactiva-tion. The simplest representation of the reaction scheme is the followingconsecutive reaction network using pseudocomponents,

A ! B ! C þ Light gases

where, A is the gas oil, B is the gasoline, and C is the carbon.The cracking reactions are endothermic. In the regenerator, the carbon

deposited on the catalyst is burned off using air and, thus, the catalyst isregenerated. This regeneration reaction is highly exothermic. The catalyst isrecirculating continuously between the reactor and the regenerator.

(a) Develop a dynamic model for this process.(b) Is multiplicity of the steady states possible for this process?

Explain your answer.(c) Discuss the possible sources of instability for this process and

suggest/discuss suitable control loops to stabilize the process.(d) Suggest a simple PI feedback control loop to ‘‘improve’’ the

performance of your unit. Explain the procedure for the designof this control loop.

Problem 6.19

A first-order reversible reaction

A , B

takes place in a nonisothermal porous spherical catalyst pellet. The forwardreaction rate is given by

rf ¼ kf CA and kf ¼ kf 0e� Ef =RTð Þ

The backward reaction rate is given by

rb ¼ kbCA and kb ¼ kb0e� Eb=RTð Þ

The equilibrium constant of the reaction is independent of temperature andis equal to 16.5. The dimensionless activation energy for the forward reac-tion is equal to 18, the net dimensionless exothermicity factor is equal to 0.7,and the external mass and heat transfer resistances are negligible.

Use bulk phase gas concentration and temperature as the referencevalues, such that your XAB ¼ 1:0, XBB ¼ 0:0 and YB ¼ 1:0.

(a) Use a value of the Thiele modulus for the forward reaction thatgives multiple steady states (if possible).

(b) Compute the effectiveness factor(s) for the steady state(s) usingthe orthogonal collocation technique.

(c) Use both integral and derivative formulas for the calculation ofthe effectiveness factor(s) and comment on your results.

Problem 6.20

Jobses and Roels (3) proposed a four-dimensional model to simulate theoscillatory behavior of ethanol fermentation using Zymomonas mobilis.Four state variables in the model are X (micro-organisms), e (an internalkey component related to the rate of growth of micro-organism), S (sub-strate; i.e., sugar), and P (product; i.e., ethanol). The rates of formation ofthese variables obtained in batch experiments are

re ¼ k1 � k2CP þ k3C2P

� � CSCe

KS þ CS

� �

rX ¼ CSCe

KS þ CS

rS ¼ �1

YSX

� �CSCe

KS þ CS

� ��mSCX

rP ¼ 1

YPX

� �CSCe

KS þ CS

� �þmPCX

Parameter values are as follows:

k1ðh�1Þ ¼ 16:0; k2ðm3=kg hÞ ¼ 4:97� 10�1; k3ðm6=kg2 hÞ ¼ 3:83� 10�3;

mSðkg=kg hÞ ¼ 2:16; mPðkg=kg hÞ ¼ 1:1; YSX ðkg=kgÞ ¼ 2:44498� 10�2;

YPX ðkg=kgÞ ¼ 5:26315� 10�2; KSðkg=m3Þ ¼ 0:5

Initial conditions are as follows:

Ce 0ð Þ ¼ 0:06098658 kg=m3; CX 0ð Þ ¼ 1:1340651 kg=m3;CP 0ð Þ ¼ 58:24388 kg=m3 CS 0ð Þ ¼ 45:89339 kg=m3

(a) Formulate a dynamic model for a continuous-stirred tankfermentor (CSTF) of an active volume VF . Assume that theinlet and outlet flow rates remain constant and are equal to q.

(b) Using the given values of the parameters, solve the dynamicmodel and comment on the dependence of dynamic behavioron the dilution rate (in the range 0.0–0.1 h–1) and inlet substratefeed concentration (in the range 140–200 kg/m3). Assume that thefeed to CSTF is pure sugar.

Problem 6.21

As the ethanol produced in fermentation acts as an inhibitor for the process,continuous ethanol removal is generally used to improve the productivity/yield of the fermentation process. Garhyan and Elnashaie (8) modeled acontinuous-membrane fermentor with in situ removal of ethanol producedusing a sweep liquid as shown in Figure P6.21. The rate of ethanol removalis considered to be proportional to the ethanol concentration gradientacross the membrane and area of permeation.

Extend the four-dimensional model of Problem 6.20 to model thiscontinuous-membrane fermentor. List all of the assumptions you makeand justify them.

Figure P6.21 Schematic diagram for Problem 6.21

7

Practical Relevance of Bifurcation,Instability, and Chaos in Chemicaland Biochemical Systems

Multiplicity phenomenon in chemically reactive systems was first observedin 1918 (1) and in the Russian literature in 1940 (2). However, it was notuntil the 1950s that the great interest in the investigation of this phenom-enon started, inspired by the Minnesota school of Amundson and Aris (3–5)and their students (6–12). The Prague school also had a notable contribu-tion to the field (13–16). Since then, this phenomenon has been high on theagenda of chemical engineering research, in general, and chemical reactionengineering research, in particular. The fascination with this phenomenonhas also spread widely to the biological and biochemical engineering litera-ture where it is referred to as ‘‘short-term memory’’ (17,18). The mathema-tical literature has also caught up with the fascination of this phenomenon,where it is treated in a more general and abstract terms under ‘‘bifurcationtheory’’ (19).

A major breakthrough with regard to the understanding of this phe-nomenon in the field of chemical reaction engineering was achieved by Rayand co-workers (20,21) when, in one stroke, they uncovered a large varietyof possible bifurcation behaviours in nonadiabatic continuous-stirred tankreactors (CSTRs). In addition to the usual hysteresis-type bifurcation,Uppal et al. (21) uncovered different types of bifurcation diagrams, themost important of which is the ‘‘isola,’’ which is a closed curve disconnectedfrom the rest of the continuum of steady states.

515

Isolas were also found by Elnashaie et al. (22–24) for enzyme systemswhere the rate of reaction depends nonmonotonically on two of the system’sstate variables (substrate concentration and pH), a situation which wasshown to be applicable to the acetylcholinesterase enzyme system.

Later development in the singularity theory, especially the pioneeringwork of Golubitsky and Schaeffer (19), provided a powerful tool for theanalysis of bifurcation behavior of chemically reactive systems. These tech-niques have been used extensively, elegantly, and successfully by Luss andhis co-workers (7–11) to uncover a large number of possible types of bifur-cation. They were also able to apply the technique successfully to complexreaction networks as well as distributed parameter systems.

Many laboratory experiments were also designed to confirm the exis-tence of bifurcation behavior in chemically reactive systems (25–29) as wellas in enzyme systems (18).

Multiplicity (or bifurcation) behavior was found to occur in othersystems such as distillation (30), absorption with chemical reaction (31),polymerization of olefins in fluidized beds (32), char combustion (33,34),heating of wires (35), and, recently, in a number of processes used for themanufacturing and processing of electronic components (36,37).

Although the literature is rich in theoretical investigations of the bifur-cation behavior and laboratory experimental work for the verification of theexistence of the bifurcation behavior, it is extremely poor with regard to theinvestigation of this phenomenon for industrial systems. In fact, very fewarticles have been published that address the question of whether this phe-nomenon is important industrially or is only of theoretical and intellectualinterest. Research in this field has certainly raised the intellectual level ofchemical engineering and helped tremendously in the development of a moreadvanced and rigorous approach to the modeling of chemical engineeringprocesses. Moreover, in addition to the intellectual benefits of bifurcationresearch to the engineering level of thinking and simulation of difficultprocesses, it seems that the phenomenon is also of great importance andrelevance to certain industrial units. Elnashaie et al. (38) found, using het-erogeneous models, that an industrial TVA ammonia converter is operatingat the multiplicity region. The source of multiplicity in this case is thecountercurrent flow of reactants in the catalyst bed and the cooling tubesand heat exchanger. Multiplicity of the steady states in TVA ammoniaconverters had been observed much earlier by using a simple pseudohomo-geneous model (39,40). The fluid catalytic cracking (FCC) units for theconversion of heavy gas oil to gasoline and light hydrocarbons not onlyexhibit a complex bifurcation behavior but also operate mostly in the middleunstable steady state (41–44). For these systems, the stable high- and low-temperature steady states both give very low gasoline yield.

Despite the extensive interest in the bifurcation behavior of chemicalreactors manifested in the chemical engineering literature, the industrial inter-est in this phenomenon is extremely limited. It seems that the industrial phi-losophy is to avoid these troublesome regions of operating and designparameters which are characterized by instabilities, quenching, ignition,and so forth, where design and control can be quite difficult. This conservativephilosophy was quite justifiable before the great advances achieved during thelast decades in the development of rigorous models for chemical reactors andthe revolutionary advancement achieved in computer power and digital com-puter control. The present state of affairs indicates that industrial philosophyshould change more and more into the direction of exploring the possibleopportunities available with regard to higher conversion, higher yield, andselectivity in this region of operating and design parameters. In fact, in the lasttwo decades, many outstanding academic researchers have demonstrated thepossibility of exploiting these instabilities for higher conversions, selectivities,and yields (45–56). In the Russian literature, Matros and co-workers (57–59)demonstrated the advantage of operating catalytic reactors deliberately underunsteady-state conditions to achieve higher performance. Industrial enter-prises in the West are showing great interest in the concept. Although thisdeliberate unsteady-state operation is not directly related to the bifurcationphenomenon and its associated instabilities, it, nevertheless, demonstrates adefinite change in the conservative industrial attitude.

7.1 SOURCES OF MULTIPLICITY

Catalytic reactors are exceptionally rich in bifurcation and instability pro-blems. These can come from many sources, as summarized in the followingsubsections.

7.1.1 Isothermal Multiplicity (or Concentration Multiplicity)

This is probably the most intrinsic source of multiplicity in catalytic reactorsbecause it results from the nonmonotonic dependence of the intrinsic rate ofreaction on the concentration of reactants and products. Although a decadeago nonmonotonic kinetics of catalytic reactions were considered the excep-tional cases, nowadays it is clear that nonmonotonic kinetics in catalyticreactions is much more widespread than previously thought. Readers canlearn more about various examples from a long list of catalytic reactionsexhibiting nonmonotonic kinetics (60–65). However, nonmonotonic kineticsalone will not produce multiplicity. It has to be coupled with some diffusionprocess, either a mass transfer resistance between the catalyst pellet surfaceand the bulk gas or within the pores of the pellets. For if the flow conditionsand the catalyst pellets size are such that diffusional resistances between the

bulk gas phase and the catalytic active centers are negligible (the system inthis case is described by a pseudohomogeneous model) and the bulk gasphase is in plug flow, then multiplicity is not possible. However, for such apseudohomogeneous reactor, if the bulk phase flow conditions are not inplug flow, then multiplicity of the steady states is possible (66,67). The rangeof deviations from plug flow which gives multiplicity of the steady statescorresponds to shallow beds with small gas flow rates, a situation not applic-able to most industrial reactors.

A word of caution is necessary. Isothermal multiplicity resultingfrom nonmonotonic kinetics occurs only when the nonmonotonic kineticdependence of the rate of reaction on species concentration is sharpenough. For flat nonmonotonic behavior, multiplicity can occur only forbulk-phase concentrations, which are too high to be considered of muchpractical relevance.

7.1.2 Thermal Multiplicity

This is the most widespread and extensively investigated type of multiplicity.It is associated with exothermic reactions. In fact, this type of multiplicityresults from some sort of nonmonotonic behavior associated with thechange of the rate of reaction under the simultaneous variation of reactantsconcentration and temperature accompanying the reactions taking placewithin the boundaries of the system (the reactor). For the case of exothermicreactions, as the reaction proceeds, the reactants deplete, which tends tocause a decrease in the rate of reaction, while heat release increases thetemperature and thus causes the rate of reaction to increase through theArrhenius dependence of the rate of reaction upon temperature. These twoconflicting effects on the rate of reaction associated with the conversion ofreactants lead to a nonmonotonic dependence of the rate of reaction on thereactant concentration. This, in turn, leads to the possibility of multiplesteady states. However, similar to the isothermal case, multiplicity will notoccur in fixed-bed catalytic reactors due to this effect alone; there has to besome diffusional mechanism coupled with the reaction in order to give riseto multiplicity of the steady states. Therefore, for a fixed-bed catalytic reac-tor where the flow conditions are such that external mass and heat transferresistance between the surface of the catalyst pellets and the bulk gas phaseare negligible, the catalyst pellet size, pore structure, and conductivity aresuch that intraparticle heat and mass transfer resistances are also negligible(a system that can be described by a pseudohomogeneous model), the bulkgas phase is in plug flow, and the system is adiabatic or cooled cocurrently,then multiplicity of the steady states is usually not possible. If the bulk flowconditions is not in plug flow, then multiplicity is possible. However, as

indicated earlier, this situation is very unlikely in industrial fixed-bed cata-lytic reactors. Therefore, the main source of multiplicity in fixed-bed cata-lytic reactors is through the coupling between the exothermic reaction andthe catalyst pellet mass and heat transfer resistances.

Isothermal (concentration) multiplicity and thermal multiplicity maycoexist in certain systems, when the kinetics are nonmonotonic, the reactionis exothermic, and the reactor system is nonisothermal.

7.1.3 Multiplicity Due to the Reactor Configuration

Industrial fixed-bed catalytic reactors have a wide range of different config-urations. The configuration of the reactor may give rise to multiplicity of thesteady states when other sources are not sufficient to produce the phenom-enon. The most well known is the case of catalytic reactors where the gasphase is in plug flow and all diffusional resistances are negligible; however,the reaction is exothermic and is counter-currently cooled (68). One of thetypical examples for this case is the TVA-type ammonia converter.

7.2 SIMPLE QUANTITATIVE DISCUSSION OF THEMULTIPLICITY PHENOMENON

Consider three very simple lumped systems, each one described by similaralgebraic equations:

1. CSTR (continuous-stirred tank reactor)2. Nonporous catalyst pellet3. Cell with permeable membrane containing an enzyme

A reaction is taking place in each system with a rate of reaction r ¼ f ðCAÞ,where f ðCAÞ is a nonmonotonic function.

Figure 7.1 shows that for a certain combination of parameters, thesupply line �ðCAf � CAÞ intersects the consumption curve f ðCAÞ in threepoints, giving rise to multiple steady states.

For all the three cases in Table 7.1 [see Eqs. (7.1)–(7.3)]

�ðCAf � CaÞ ¼ f ðCAÞwhere

� ¼ q

V(for CSTR)

� ¼ apKg

Wp

(for catalyst pellet)

� ¼ aCPA

EVC

(for cell)

7.3 BIFURCATION AND STABILITY

In the multiplicity region of operating and design parameters, the stabilitycharacteristics of the system are quite different from those in the uniquenessregion of parameters for the same system. It is also important to note thatfor systems showing multiplicity behavior over a certain region of para-meters, even when the system is operating in the region of unique steadystate, the existence of the multiplicity region will have its implications on thebehavior of the system in the face of external disturbances that may movethe system into this multiplicity region.

7.3.1 Steady-State Analysis

The full appreciation of the stability characteristics of any system requiresdynamic modeling and analysis of the system. Detailed dynamic modeling

Figure 7.1 Hysteresis curve for concentration multiplicity.

Table 7.1 Multiplicity of the Steady States for Three Chemical/Biochemical Systems

CSTR Nonporous catalyst pellet cell containing an enzyme

qðCAf � CAÞ ¼ Vf ðCAÞ ð7:1Þ apKgðCAf � CAÞ ¼ Wpf ðCAÞ ð7:2Þ aCPaðCAf � CAÞ ¼ �EEVCf ðCAÞ ð7:3Þwhere where where

q ¼ volumetric flow rate ap ¼ surface area of pellet aC ¼ surface area of cell

V ¼ reactor volume Wp ¼ mass of pellet VC ¼ volume of cell

CAf ¼ feed concentration Kg ¼ mass transfer coefficient PA ¼ permeability of cell membrane

CA ¼ effluent concentration of reactant CAf ¼ concentration of reactant in bulk

fluid

�EE ¼ weight of enzyme per unit

volume cellf ðCAÞ ¼ reaction rate based on unit

volume of reaction mixture CA ¼ concentration of reactant on pellet

surface

CAf ¼ concentration of reactant in

bulk fluid

f ðCAÞ ¼ reaction rate based on unit

volume of reaction mixture

CA ¼ concentration of reactant

inside the cell

f ðCAÞ ¼ reaction rate based on unit

mass of enzyme

and analysis is beyond the scope of this book; however, a considerableamount of insight into the stability and dynamic characteristics of the sys-tem can be extracted from steady-state analysis as shown in this section. InSection 7.3.2, a simple and brief introduction to the dynamical side of thepicture is given.

A simple and almost obvious illustration of this point is that when it isfound from steady-state analysis that the system is operating in the multi-plicity region, then global stability for any of the steady states is not possi-ble. Every steady state will have its region of asymptotic stability (RAS) or,in the more modern terminology of dynamical systems theory, a basin ofattraction. This fact has very important implications for the dynamic beha-vior, stability, and control of the system. Also, it will be shown that theinitial direction of the dynamic trajectory can be predicted from steady-statearguments. Of course, there remains qualitative and quantitative dynamicquestions that need to be answered and which can only be answered throughdynamic modeling and analysis of the system.

To illustrate the above points, let us consider a simple homogeneousCSTR, where a consecutive exothermic reaction

A �!k1 B �!k2 C

is taking place, and the reactor is at steady-state conditions, as shown inFigure 7.2.

The mass and heat balance equations for the model are given indimensionless form as follows:

XAf ¼ XA þ �1e�y1=yXA ð7:4Þ

XBf ¼ XB þ �2e�y2=yXB � �1e

�y1=yXA ð7:5Þy� yf ¼ �11e

�y1=yXA þ �22e�y2=yXB þ �KKcðyc � yÞ ð7:6Þ

Figure 7.2 Schematic diagram of a CSTR.

where the following dimensionless variables are used:

XA ¼ CA

Cref

XAf ¼CAf

Cref

XB ¼ CB

Cref

XBf ¼CBf

Cref

y ¼ T

Tref

yc ¼Tc

Tref

The following dimensionless parameters are also used:

�i ¼Vki0q

�i ¼Ei

RGTref

i ¼ð�HiÞCref

�CpTref

�KKc ¼UAH

q�Cp

Note:

i ¼ 1; 2

For the special case of adiabatic operation, we set

�KKc ¼ 0:0 ð7:7Þin Eq. (7.6).

The equations of the adiabatic case also represent the case of non-porous catalyst pellets with external mass and heat transfer resistances andnegligible intraparticle heat transfer resistance but with different meaningsto the parameters.

Equations (7.4)–(7.6) can be solved simultaneously for a certain set ofparameters in order to obtain the values XA, XB, and y (the concentrations

and temperature at the exit of the reactor for the CSTR case (and the surfaceof the catalyst pellet for the nonporous catalyst pellet case). However, it ispossible to reduce Eqs. (7.4)–(7.6) to a single nonlinear equation in y,together with two explicit linear equations for the computation of XA andXB once y has been determined. The single nonlinear equation (for XBf ¼ 0Þcan be written as

RðyÞ ¼ ðy� yf Þ � �KKcðyc � yÞ

¼ �11e�y1=yXAf

1þ �1e�y1=y

þ �22e�y2=y�1e

�y1=yXAf

1þ �1e�y1=y

� �1þ �2e

�y2=y� � � GðyÞ ð7:8Þ

The right-hand side is proportional to the heat generation and will betermed the heat generation function, GðyÞ, whereas the left-hand side isproportional to the heat removal due to the flow and the cooling jacketand will be termed the heat removal function:

RðyÞ ¼ 1þ �KKc

� �y� yf þ �KKcyc

� � ð7:9Þ

Equation (7.8) can be solved using any of the standard methods (bisectional,Newton–Raphson, etc.); however, it is more instructive to solve it graphi-cally by plotting GðyÞ and RðyÞ versus y, as shown in Figure 7.3 for a casewith parameters corresponding to three steady states.

The slope of the heat removal line is

b ¼ 1þ �KKc

Figure 7.3 Schematic diagram for the heat generation function GðyÞ and the heat

removal function RðyÞ, for a case with maximum number of steady states equal to 3.

and the intersection with the horizontal axis is at

a ¼ yf þ �KKcyc

1þ �KKc

ð7:10Þ

For the adiabatic case with �KKc ¼ 0, we get

bad ¼ 1 ð7:11Þand

aad ¼ yf ð7:12ÞFor the case shown in Figure 7.3, three steady states are possible: y1, y2, andy3. Some information regarding the stability of the three steady states andthe dynamic behavior can be obtained from this static diagram. The steady-state temperatures y1, y2, and y3 correspond to points where the heat gen-eration and heat removal are equal (steady states). If the reactor is disturbedto a point �yy4 that is not a steady state, then it is easy from this static diagramto determine the direction of temperature change, because at this point, it isclear that the heat generation is higher than the heat removal and, therefore,the temperature will increase. Of course, the full behavior of the tempera-ture–time trajectory can only be determined through the dynamic model ofthe system.

With regard to this limited steady-state test of stability, let us examiney3 using the above argument. Suppose that we disturb y3 by an infinitesimalamount �y, which is positive; in this case, we notice from Figure 7.3 that theheat removal is higher than the heat generation and the system cools downtoward y3. If �y is negative, the heat generation is higher than the heatremoval and, therefore, the system will heat up toward y3. This indicatesthat y3 is stable. However, this is only a necessary condition for stability, butit is not sufficient. Other stability conditions should be checked through thecomputation of the eigenvalues of the dynamic model (69).

If we do the same experiment for y1, we find that it also satisfies thenecessary condition for stability. However, if we check the intermediatetemperature y2, we find that if �y is positive, then the heat generation is higherthan the heat removal and, therefore, the system will heat up away from y2.On the other hand, if �y is negative, the heat removal is higher than the heatgeneration and, therefore, the system will cool down away from y2, comput-ing the eigenvalues of the dynamic model is not necessary because violationof the necessary condition for stability is a sufficient condition for instability.

The shape of the bifurcation diagram can be easily concluded fromthis simple heat generation–heat removal diagram. Let us take, for simpli-city, the adiabatic case with yf as a bifurcation parameter. If we plot y versusthe bifurcation parameter yf , we obtain the S-shaped hysteresis curve shown

in Figure 7.4. The points yfL and yfL are the two limit points which are theboundaries for the multiplicity region; thus, multiplicity of the steady statesexists for values of yf lying in the region

yfL � yf � yfL ð7:13ÞThe curves shown in Figures 7.3 and 7.4 are not the only possible

cases; another case with another set of parameters is shown in Figure 7.5.In this case, five steady states are possible. With regard to static stability, y1is stable, y2 is unstable, y3 is stable, y4 is unstable, and y5 is stable. Thebifurcation diagram for this case is shown in Figure 7.6. In this case, thediagram can be divided into the following regions:

Unique low-temperature steady state for the region

yf � yfL1 ð7:14ÞThree steady states for the region

yfL1 � yf � yfL2

Five steady states in the region


Three steady states in the region


Unique high-temperature steady state in the region

yf yfL4

Figure 7.4 Bifurcation diagram for a case with three steady sates (two limit points).

Figure 7.5 Heat generation and heat removal functions for a case with five steady

states.

Figure 7.6 Schematic bifurcation diagram with five steady states (four limit points).

For some cases, the maximum yield of the desired product corre-sponds to the middle steady state, which is unstable, as shown in Figure7.7. In these cases, efficient adiabatic operation is not possible and nonadia-batic operation is mandatory. However, the choice of the heat transfercoefficient ðUÞ, the area of heat transfer ðAHÞ, and the cooling jackettemperature are critical for the stable operation of the system. The valueof the dimensionless heat transfer coefficient ð �KKcÞ should exceed a criticalvalue ð �KKc;critÞ in order to stabilize the unstable middle steady state. The valueof ð �KKc;critÞ corresponds to line 4 in Figure 7.7.

The increase in �KKc increases the slope of the removal line because theslope is given by ð1þ �KKcÞ. If a bifurcation diagram is drawn for this non-adiabatic case with �KKc as the bifurcation parameter and the jacket coolingtemperature is the temperature of the middle steady state ym, we get thepitchfork-type bifurcation diagram shown in Figure 7.8.

In Figure 7.8, for �KKc ¼ 0 the three steady states yH , ym, and yL arethose of the adiabatic case. For �KKc < �KKc;crit, yH and yL change, but ymremains the same because it is the jacket temperature. However, ym in this

Figure 7.7 Schematic diagram for the case where the maximum yield of B corre-

sponds to a middle unstable steady state.

region is a middle unstable steady state. At �KKc;crit, the multiplicity disap-pears, giving rise to a unique steady state ym. From a static point of view,this steady state is stable. However, dynamically it may be unstable for acertain range of �KKc > �KKc;crit, giving rise to limit cycle behavior in this region(69). If the cooling jacket is not ym, we get ‘‘imperfect pitchfork’’(43,44,77,78).

7.3.2 Dynamic Analysis

The steady states which are unstable using the above-discussed static ana-lysis are always unstable. However, steady states that are stable from a staticpoint of view may prove to be unstable when the full dynamic analysis isperformed. In other words branch 2 in Figure 7.8 is always unstable,whereas branches 1, 3, and 4 can be stable or unstable depending on thedynamic stability analysis of the system. As mentioned earlier, the analysisfor the CSTR presented here is mathematically equivalent to that of acatalyst pellet using lumped models or a distributed model made discreteby a technique such as the orthogonal collocation technique (see AppendixE). However, in the latter case, the system dimensionality will increase con-siderably, with n dimensions for each state variable, where n is the numberof internal collocation points.

For simplicity of presentation, we consider a two-dimensional systemwith one bifurcation parameter:

dX1

dt¼ f1ðX1;X2; �Þ ð7:15Þ

Figure 7.8 Pitchfork bifurcation diagram ðy� �KKc).

and

dX2

dt¼ f2ðX1;X2; �Þ ð7:16Þ

The steady state of this system is given by

f1ðX1;X2; �Þ ¼ 0 and f2ðX1;X2; �Þ ¼ 0 ð7:17ÞWe will consider that Eq. (7.17) has a simple hysteresis-type static bifurca-tion as shown by the solid lines in Figures 7.9A–7.9C. The intermediatestatic branch is always unstable (saddle points), whereas the upper andlower branches can be stable or unstable depending on the eigenvalues ofthe linearized forms of Eqs. (7.15) and (7.16). The static bifurcation dia-grams in Figures 7.9A–7.9C have two static limit points each; these areusually called saddle-node bifurcation points

The stability characteristics of the steady-state points can be deter-mined from the eigenvalue analysis of the linearized version of Eqs. (7.15)and (7.16) which will have the following form:

dxx1dt

¼ g11xx1 þ g12xx2 ð7:18Þ

Figure 7.9 Bifurcation diagram for Eqs. (7.15)–(7.17). (A) A case with two Hopf

bifurcation points; (B) a case with two Hopf bifurcation points and one periodic limit

point; (C) a case with one Hopf bifurcation point, two periodic limit points, and one

homoclinical orbit (infinite period bifurcation point). Solid curve: stable branch of

the bifurcation diagram; dashed curve: saddle points; filled circles: stable limit cycles;

open circles: unstable limit cycles; HB ¼ Hopf bifurcation point; PLP ¼ periodic

limit point.

dxx2dt

¼ g21xx1 þ g22xx2 ð7:19Þ

where

xxi ¼ xi � xiss

and

gif ¼@fi@xj

��ss

Figure 7.9 Continued

where xiss ¼ xi at steady state. The eigenvalues of Eqs. (7.18) and (7.19) arethe roots of the characteristic equation

�2 � trAÞ þ ðdet AÞ ¼ 0ð ð7:20Þwhere

A ¼g11 g12

g21 g22

!

ðtrAÞ ¼ g11 þ g22

ðdet AÞ ¼ g11g22 � g12g21

ð7:21Þ

The eigenvalues for this two-dimensional system are given by

�1;2 ¼ðtrAÞ �

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðtrAÞ2 � 4ðdet AÞ

q2

ð7:22Þ

The most important dynamic bifurcation is the Hopf bifurcation point,when �1 and �2 cross the imaginary axis into positive parts of �1 and �2.This is the point where both roots are purely imaginary and at whichtrA ¼ 0, giving

�1;2 ¼ �iffiffiffiffiffiffiffiffiffiffiffiffiffiðdet A

pÞ ð7:23Þ

At this point, periodic solutions (stable limit cycles) come into exis-tence, as shown in Figure 7.9A–C. This point is called Hopf bifurcation.The case in Figure 7.9A shows two Hopf bifurcation points with a branchof stable limit cycles connecting them. Figure 7.10 is a schematic diagramof the phase plane for this case with � ¼ �1. In this case, a stable limitcycle surrounds an unstable focus and the behavior of the typical trajec-tories are as shown. The case in Figure 7.9B has two Hopf bifurcationpoints in addition to a periodic limit point (PLP) and a branch of unstablelimit cycles in addition to the stable limit cycles branch.

Figure 7.11 shows the phase plane for this case when � ¼ �2. In thiscase, there is an unstable limit cycle surrounding a stable focus and theunstable limit cycle is surrounded by a stable limit cycle. The behavior ofthe typical trajectories is as shown in Figure 7.11.

The case of Figure 7.9C has one Hopf bifurcation point and oneperiodic limit point, and the stable limit cycle terminates at a homoclinicalorbit (infinite period bifurcation). For � ¼ �3, we get a case of an unstablesteady state surrounded by a stable limit cycle similar to the case in Figure7.10. However, in this case, as � decreases below �3, the limit cycle growsuntil we reach a limit cycle that passes through the static saddle point, as

Figure 7.10 Phase plane for � ¼ �1 in Figure 7.9A. Solid curve: stable limit cycle

trajectories; �: unstable saddle; *: stable steady state (node or focus); *: unstable

steady state (node or focus); dashed curve: separatix.

Figure 7.11 Phase plane for � ¼ �2 in Figure 7.9B. Solid curve: stable limit cycle

trajectories; dashed curve: unstable limit cycle; �: unstable saddle; *: stable steady

state (node or focus).

shown in Figure 7.12. This limit cycle represents a trajectory that starts atthe static saddle point and ends after ‘‘one period’’ at the same saddle point.This trajectory is called the homoclinical orbit and will occur at some criticalvalue �HC. It has an infinite period and, therefore, this bifurcation point iscalled ‘‘infinite period bifurcation.’’ For � < �HC, the limit cycle disappears.This is the second most important type of dynamic bifurcation after theHopf bifurcation.

7.3.3 Chaotic Behavior

Limit cycles (periodic solutions) emerging from the Hopf bifurcation pointand terminating at another Hopf bifurcation point or at a homoclinicalorbit (infinite period bifurcation point) represent the highest degree of com-plexity in almost all two-dimensional autonomous systems. However,for higher-dimensional autonomous systems such as the three-dimensionalsystem,

dX1

dt¼ f1ðX1;X2;X3; �Þ ð7:24Þ

dX2

dt¼ f2ðX1;X2;X3; �Þ ð7:25Þ

dX3

dt¼ f3ðX1;X2;X3; �Þ ð7:26Þ

Figure 7.12 Homoclinical orbit.

or the nonautonomous two-dimensional system such as the sinusoidalforced two-dimensional system,

dX1

dt¼ f1ðX1;X2; �Þ þ A sinð!tÞ ð7:27Þ

dX2

dt¼ f2ðX1;X2; �Þ ð7:28Þ

Higher degrees of dynamic complexity are possible, including perioddoubling, quasiperiodicity (torus), and chaos. Phase-plane plots are not thebest means of investigating these complex dynamics for in such cases (whichare at least three-dimensional); the three (or more)-dimensional phase planecan be quite complex, as shown in Figures 7.13 and 7.14 for two of the mostwell-known attractors, the Lorenz strange attractor (70) and the Rosslerstrange attractor (71–73). Instead, stroboscopic maps for forced systems(nonautonomous) and Poincare maps for autonomous systems are bettersuited for the investigation of these types of complex dynamic behavior.

The equations for the Lorenz model are

dX

dt¼ �ðY � XÞ

dY

dt¼ rX � Y � XZ

dZ

dt¼ �bZ þ XY

ð7:29Þ

Figure 7.13 Lorenz strange attractor projected on the two-dimensional xz plane

(� ¼ 10, b ¼ 8=3, r ¼ 28).

The stroboscopic and Poincare maps are different from the phase plane inthat they plot the variables on the trajectory at specific chosen and repeatedtime intervals. For example, for the forced two-dimensional system, thesepoints are taken at every forcing period. For the Poincare map, the intervalof strobing is not as obvious as in the case of the forced system and manytechniques can be applied. Different planes can be used in order to get adeeper insight into the nature of strange attractors in these cases. A periodicsolution (limit cycle) on the phase plane will appear as one point on thestroboscopic (or Poincare) map. When period doubling takes place, period 2

Figure 7.14 Final trajectories of the Rossler attractor [73] for different values of

the parameter a. Left row, top to bottom: limit cycle, a ¼ 0:3; period 2 limit cycle

a ¼ 0:35; period 4, a ¼ 0:375; four-band chaotic attractor, a ¼ 0:386; Right row, top

to bottom: period 6, a ¼ 0:3904; single-band chaos, a ¼ 0:398; period 5, a ¼ 0:4;period 3, a ¼ 0:411. In all cases b ¼ 2 and c ¼ 4. (From Ref. 71.)

will appear as two points on the map, period 4 will appear as four points,and so on. Quasiperiodicity (torus), which looks complicated on the phaseplane, will appear as an invariant close circle with discrete points on themap. When chaos takes place, a complicated collection of points appear onthe stroboscopic map. The shapes formed have fractal dimensions and areusually called strange attractors.

The equations for the Rossler model are

dX

dt¼ �ðY � ZÞ

dY

dt¼ X þ aY

dZ

dt¼ bþ ZðX � cÞ

ð7:30Þ

A strange attractor resulting from such a deterministic model is called‘‘deterministic chaos’’ to emphasize the fact that it is not a random orstochastic variation.

This is the minimum information necessary for the interested reader toget an appreciation of the complex bifurcation, instability, and chaos asso-ciated with chemical and biochemical processes. Although industrial prac-tice in petrochemical and petroleum refining and other chemical andbiochemical industries does not appreciate the importance of these phenom-ena and their implications on the design, optimization, and control of cat-alytic and biocatalytic processes, it is easy to recognize that thesephenomena are important because bifurcation, instability, and chaos inthese systems are generally due to nonlinearity and, specifically, nonmono-tonicity, which is widespread in catalytic reactors either as a result ofexothermicity or as a result of the nonmonotonic dependence of the rateof reaction on the concentration of the reactant species. It is expected that inthe near future and through healthy scientific interaction between industryand academia, these important phenomena will be better appreciated byindustry and that more academicians will turn their attention to the inves-tigation of these phenomena in industrial systems.

For the readers interested in this field, it is useful to recommend thefollowing further reading in bifurcation, instability and chaos:

1. The work of Uppal et al. (20,21) and that of Ray (46) on thebifurcation behavior of continuous-stirred tank reactors.

2. The work of Teymour and Ray (74–76) on bifurcation, instabil-ity, and chaos of polymerization reactors.

3. The work of Elnashaie and coworkers on the bifurcation andinstability of industrial fluid catalytic cracking (FCC) units (42–44,77).

4. The book by Elnashaie and Elshishni on the chaotic behavior ofgas–solid catalytic systems (78).

5. The interesting review article by Razon and Schmitz (79) on thebifurcation, instability, and chaos for the oxidation of carbonmonoxide on platinum.

6. The two articles by Wicke and Onken (80) and Razon et al. (81)give different views on the almost similar dynamics observedduring the oxidation of carbon monoxide on platinum. Wickeand Onken (80) analyze it as statistical fluctuations, whereasRazon et al. (81) analyze it as chaos. Later, Wicke changed hisviews and analyzed the phenomenon as chaos (82).

7. The work of the Minnesota group on the chaotic behavior ofsinusoidally forced two-dimensional reacting systems is usefulfrom the points of view of the analysis of the system as well asthe development of suitable efficient numerical techniques for theinvestigation of these systems (82–85).

8. The book by Marek and Schreiber (86) is important for chemicalengineers who want to get into this exciting field. The book cov-ers a wide spectrum of subjects, including the following: differ-ential equations, maps, and asymptotic behavior; transition fromorder to chaos; numerical methods for studies of parametricdependences, bifurcations, and chaos; chaotic dynamics in experi-ments; forced and coupled chemical oscillators; and chaos indistributed systems. It also contains two useful appendices, thefirst dealing with normal forms and their bifurcation diagrams,and the second, a computer program for construction of solutionand bifurcation diagrams.

9. The software package ‘‘Auto 97’’ (87) is useful in computingbifurcation diagrams using the efficient continuation techniquefor both static and periodic branches of the bifurcation diagram.

10. The software package ‘‘Dynamics’’ (88) is useful in computingFloquette multipliers and Lyapunov exponents.

11. Elnashaie et al. (77) presented a detailed investigation of bifurca-tion, instability, and chaos in fluidized-bed catalytic reactor forboth the unforced (autonomous) and forced (nonautonomous)cases.

The compilation of articles on chaos in different fields published byCvitanovic (89) is also useful as an introduction to the field.

The dynamic behavior of fixed-bed reactors has not been extensivelyinvestigated in the literature. The only reaction which received close atten-tion was the CO oxidation over platinum catalysts. The investigationsrevealed interesting and complex dynamic behavior and have shown thepossibility of oscillatory behavior as well as chaotic behavior (79–81). It iseasy to speculate that more emphasis on the study of the dynamic behaviorof catalytic reactions will reveal complex dynamics like those discovered forthe CO oxidation over a Pt catalyst, for most phenomena are due to thenonmonotonicity of the rate process, which is widespread in catalyticsystems.

There are many other interesting and complex dynamic phenomena inaddition to oscillation and chaos, which have been observed but not fol-lowed in depth both theoretically and experimentally; an example is thewrong directional behavior of catalytic fixed-bed reactors, for which thedynamic response to input disturbances is opposite to that suggested bythe steady-state response (90,91). This behavior is most probably connectedto the instability problems in these catalytic reactors, as shown crudely byElnashaie and Cresswell (90).

It is needless to say that the above briefly discussed phenomena (bifur-cation, instability, chaos, and wrong directional responses) may lead tounexpected pitfalls in the design, operation, and control of industrialfixed-bed catalytic reactors and that they merit extensive theoretical andexperimental research, as well as research directed toward industrial units.

REFERENCES

1. Liljenroth, F. G. Starting and stability phenomena of ammonia-oxidation and

similar reactions. Chem. Met. Eng. 19, 287–293, 1918.

2. Frank-Kamenetskii, D. A. Diffusion and Heat Transfer in Chemical Kinetics.

Plenum Press, New York, 1969.

3. Aris, R, Amundson, and N. R. An analysis of chemical-reactor stability and

control. I. The possibility of local control, with perfect or imperfect control

mechanisms. Chem. Eng. Sci. 7, 121–131, 1958.

4. Aris, R. Chemical reactors and some bifurcation phenomena. In Bifurcation

Theory and Applications in Scientific Disciplines (Gurrel, O. and Rossler, O. E.,

eds.) New York Academy Press, New York, 1979, pp. 314–331.

5. Aris, R., and Varma, A. (eds). The Mathematical Understanding of Chemical

Engineering Systems, Selected Papers of N. R. Amundson. Pergamon Press,

Oxford, 1980.

6. Balakotaiah, V., and Luss, D. Analysis of the multiplicity patterns of a CSTR.

Chem. Eng. Commun. 13, 111–132, 1981.

7. Balakotaiah, V., and Luss D. Structure of the steady-state solutions of lumped-

parameter chemically reacting systems. Chem. Eng. Sci. 37, 1611–1623, 1982.

8. Balakotaiah, V., and Luss, D. Exact steady-state multiplicity criteria for two

consecutive or parallel reactions in lumped-parameter systems. Chem. Eng. Sci.

37, 433–445, 1982.

9. Balakotaiah, V., and Luss, D. Multiplicity features of reacting systems.

Dependence of the steady-state of a CSTR on the residence time. Chem. Eng.

Sci. 38, 1709–1721, 1983.

10. Balakotaiah, V., and Luss, D. Multiplicity criteria for multireaction networks.

AIChE J. 29, 552–560, 1983.

11. Balakotaiah, V., and Luss, D. Global analysis of the multiplicity features of

multireaction lumped-parameter systems. Chem. Eng. Sci. 39, 865–881, 1984.

12. Lee, C. K., Morbidelli, M., and Varma, A. Steady state multiplicity behavior of

an isothermal axial dispersion fixed-bed reactor with nonuniformly active

catalyst. Chem. Eng. Sci. 42, 1595–1608, 1987.

13. Hlavacek, V., Kubicek, M., and Marek, M. Modelling of chemical reactors.

XIV. Analysis of nonstationary heat and mass transfer in a porous catalyst

particle. 1. J. Catal. 15, 17-30, 1969.

14. Hlavacek, V., Kubicek, M., and Marek, M. Modelling of chemical reactors.

XV. Analysis of nonstationary heat and mass transfer in a porous catalyst

particle. 2. J. Catal. 15, 31–42, 1969.

15. Hlavacek, V., Kubicek, M., and Visnak, K. Modeling of chemical reactors

XXVI. Multiplicity and stability analysis of a continuous stirred tank reactor

with exothermic consecutive reactions A ! B ! C. Chem. Eng. Sci. 27, 719–

742, 1972.

16. Hlavacek, V., and Van Rompay, P. On the birth and death of isolas. Chem.

Eng. Sci. 36(10), 1730–1731, 1981.

17. Thomas, D. Proceedings of International Symposium on Analysis and Control of

Immobilized Enzyme Systems, Compiegne (Thomas, D. and Kernevez, J., eds.).

Elsevier, Amsterdam, 1975.

18. Friboulet, A., David, A., and Thomas, D. Excitability, memory, and oscilla-

tions in artificial acetylcholinesterase membranes. J. Membr. Sci. 8, 33–39,

1981.

19. Golubitsky, M., and Schaeffer, D. G. Singularities and Groups in Bifurcation

Theory, Volume I. Applied Mathematical Science, Vol. 51, Springer-Verlag,

New York, 1985.

20. Uppal, A., Ray, W. H., and Poore, A. B. Dynamic behavior of continuous

stirred tank reactors. Chem. Eng. Sci 29(4), 967–985, 1974.

21. Uppal, A., Ray, W. H., and Poore, A. B. The classification of the dynamic

behavior of continuous stirred tank reactors. Influence of reactor residence

time. Chem. Eng. Sci. 31(3), 205–214, 1976.

22. Elnashaie, S. S. E. H., Elrifaie, M. A., Ibrahim, G., and Badra, G. The effect of

hydrogen ion on the steady-state multiplicity of substrate-inhibited enzymic

reactions. II. Transient behavior. Appl. Biochem. Biotechnol. 8(6), 467–479,

1983.

23. Elnashaie, S. S. E. H., El-Rifai, M. A., and Ibrahim, G. The effect of

hydrogen ion on the steady-state multiplicity of substrate-inhibited enzymic

reactions. I. Steady-state considerations. Appl. Biochem. Biotechnol. 8(4), 275–

288, 1983.

24. Elnashaie, S. S. E. H., Ibrahim, G., and Elshishini, S. S. The effect of hydrogen

ions on the steady state multiplicity of substrate-inhibited enzymic reactions.

III. Asymmetrical steady states in enzyme membranes. Appl. Biochem.

Biotechnol. 9(5–6), 455–474, 1984.

25. Root, R. B., and Schmitz, R. A. Experimental study of steady state multiplicity

in a loop reactor. AIChE J. 15(5), 670–679, 1969.

26. Chang, M., and Schmitz, R. A. Oscillatory states in a stirred reactor. Chem.

Eng. Sci 30(1), 21–34, 1975.

27. Hlavacek, V., Mikus, O., Jira, E., and Pour, V. Experimental investigation of

temperature profiles, parametric sensitivity, and multiple steady state in de-

activated fixed bed reactors. Chem. Eng. Sci 35(1–2), 258–263, 1980.

28. Harold, M. P., and Luss, D. An experimental study of steady-state multiplicity

features of two parallel catalytic reactions. Chem. Eng. Sci. 40(1), 39–52, 1985.

29. Marb, C. M., and Vortmeyer, D. Multiple steady states of a crossflow moving

bed reactor: theory and experiment. Chem. Eng. Sci. 43(4), 811–819, 1988.

30. Widagdo, S., Seider, W. D., and Sebastian, D. H. Bifurcation analysis in het-

erogeneous azeotropic distillation. AIChE J. 35(9), 1457–1464, 1989.

31. White, D., Jr., and Johns, L. E. Temperature rise multiplicity in the absorption

of chemically reactive gases. Chem. Eng. Sci. 40(8), 1598–1601, 1985.

32. Choi, K. Y. and Ray, W. The dynamic behavior of fluidized bed reactors for

solid catalyzed gas phase olefin polymerization. Chem. Eng. Sci 40(12), 2261–

2279, 1985.

33. Hsuen, H. K. D., and Sotirchos, S. V. Multiplicity analysis of intraparticle char

combustion. Chem. Eng. Sci 44(11), 2639–2651, 1989.

34. Hsuen, H. K. D., and Sotirchos, S. V. Multiplicity analysis of char combustion

with homogeneous carbon monoxide oxidation. Chem. Eng. Sci. 44(11), 2653–

2665, 1989.

35. Luss, D. A note on steady state multiplicity of an electrically heated wire.

Chem. Eng. Sci. 33(3), 409, 1978.

36. Kushner, M. J. A kinetic study of plasma-etching process. I. A model for the

etching of silicon and silicon dioxide in carbon fluoride (CnFm)/hydrogen and

carbon fluoride (CnFm)/oxygen plasmas. J. Appl. Phys. 53(4), 2923–2938, 1982.

37. Jensen, K. F. Chemical vapor deposition. In Chemical Engineering Materials

Process. Advances in Chemistry Series No. 221, 1989.

38. Elnashaie, S. S. E. H., Mahfouz, A. T., and Elshishini, S. S. Digital simulation

of an industrial ammonia reactor. Chem. Eng. Process 23(3), 165–177, 1988.

39. Baddour, R. F., Brian, P. L. T., Logcais, B. A., and Eymery, J. P. Steady-state

simulation of an ammonia synthesis reactor. Chem. Eng. Sci 20(4), 281–292,

1965.

40. Brian, P. L. T., Baddour, R. F., and Eymery, J. P. Transient behavior of an

ammonia synthesis reactor. Chem. Eng. Sci 20(4), 297–310, 1965.

41. Iscol, L. Automatic Control Conference, 1970, p. 602.

42. Elnashaie, S. S. E. H., and El-Hennawi, I. M. Multiplicity of the steady state in

fluidized bed reactors. IV. Fluid catalytic cracking. Chem. Eng. Sci. 34(9), 1113–

1121, 1979.

43. Elshishini, S. S., and Elnashaie, S. S. E. H. Digital simulation of industrial fluid

catalytic cracking units: bifurcation and its implications. Chem. Eng. Sci. 45(2),

553–559, 1990.

44. Elshishini, S. S., and Elnashaie, S. S. E. H. Digital simulations of industrial

fluid catalytic cracking units—II. Effect of charge stock composition on bifur-

cation and gasoline yield. Chem. Eng. Sci. 45(9), 2959–2964, 1990.

45. Douglas, J. M., and Rippin, D. W. T. Unsteady-state process operation. Chem.

Eng. Sci. 21(4), 305–315, 1966.

46. Ray, W. H. Periodic operation of polymerization reactors. Ind. Eng. Chem.,

Process Des. Dev., 7(3), 422–426, 1968.

47. Gaitonde, N. Y., and Douglas, J. M. Use of positive feedback control systems

to improve reactor performance. AIChE J. 15(6), 902–910, 1969.

48. Bailey, J. E., Horn, F. J. M., and Lin, R. C. Cycle operation of reaction

systems. Effects of heat and mass transfer resistance. AIChE J. 17(4), 819–

825, 1971.

49. Miertschin, G. N., and Jackson, R. Optimal switching policies for fixed bed

catalytic reactors in series. Can. J. Chem. Eng. 49(4), 514–521, 1971.

50. Lindberg, R. C., and Schmitz, R. A. Multiplicity of states with surface reaction

on a blunt object in a convective system. Chem. Eng. Sci. 25(5), 901–904, 1970.

51. Dorawala, T. G., and Douglas, J. M. Complex reactions in oscillating reactors.

AIChE J. 17(4), 974–981, 1971.

52. Renken, A. Use of periodic operation to improve the performance of contin-

uous stirred tank reactors. Chem. Eng. Sci. 27(11), 1925–1932, 1972.

53. Bailey, J. E. Periodic operation of chemical reactors. Review. Chem. Eng.

Commun. 1(3), 111–124, 1974.

54. Bruns, D. D., Bailey, J. E., and Luss, D. Steady-state multiplicity and stability

of enzymic reaction systems. Biotechnol. Bioeng. 15(6), 1131–1145, 1973.

55. Lee, C. K., and Bailey, J. E. Diffusion waves and selectivity modifications in

cyclic operation of a porous catalyst. Chem. Eng. Sci. 29(5), 1157–1163, 1974.

56. Cinar, A., Deng, J., Meerkov, S. M., and Shu, X. Vibrational control of an

exothermic reaction in a CSTR: Theory and experiments. AIChE J. 33(3), 353–

365, 1987.

57. Matros, Yu. Sh. Reactors with fixed catalyst beds. Kinet. Katal. 22(2), 505–512,

1981.

58. Matros, Yu. Sh. Unsteady State Processes in Catalytic Reactors, Naukya,

Novosibirsk, USSR 1982 (in Russian).

59. Boreskov, G. K., and Matros, Yu. Sh. Fixed catalyst bed reactors operated in

steady- and unsteady-state conditions. Recent Adv. Eng. Anal. Chem. React.

Syst. 142–155, 1984.

60. Satterfield, C. N., and Roberts, G. W. Kinetics of thiophene hydrogenolysis on

a cobalt molybdate catalyst. AIChE J. 14(1), 159–164, 1968.

61. Nicholas, D. M., and Shah, Y. T. Carbon monoxide oxidation over a platinum-

porous fiber glass supported catalyst. Ind. Eng. Chem. Prod. Res. Dev. 15(1),

35–40, 1976.

62. Peloso, A., Moresi, M., Mustachi, C., and Soracco, B. Kinetics of the dehy-

drogenation of ethanol to acetaldehyde on unsupported catalysts. Can. J.

Chem. Eng. 57(2), 159–164, 1979.

63. Yue, P. L., and Olaofe, O. Kinetic analysis of the catalytic dehydration of

alcohols over zeolites. Chem. Eng. Res. Deve 62(2), 81–91, 1984.

64. Yue, P. L., and Birk, R. H. Catalytic Dehydration of Ethanol in a Fluidized Bed

Reactor. Institute of the Engineers Symposium Series No. 87. 1984, pp. 487–

494.

65. Das, G., and Biswas, A. K. Initial rate study of the vapor-phase condensation

of aniline to diphenylamine. Can. J. Chem. Eng. 64(3), 473–477, 1986.

66. Elnashaie, S., Marek, M., and Ray, W. H. Effect of catalyst deactivation on the

performance of dispersed catalytic reactors. Chem. React. Eng. Proc. Eur.

Symp. B5, 39–48, 1972.

67. Ray, W. H., Marek, M., and Elnashaie, S. S. E. H. Effect of heat and mass

dispersion on tubular reactor performance. Chem. Eng. Sci. 27(8), 1527–1536,

1972.

68. Luss, D., and Medellin, P. Steady state multiplicity and stability in a counter-

currently cooled tubular reactor. Chem. React. Eng., Proc. Eur. Symp., B4, 47–

56, 1972.

69. Elnashaie, S. S. E. H. Multiplicity of the steady state in fluidized bed reactors.

III. Yield of the consecutive reaction A ! B ! C. Chem. Eng. Sci. 32(3), 295–

301, 1977.

70. Sparrow, C. Bifurcations in the Lorenz Equations. Lecture Notes in Applied

Mathematics, Springer-Verlag, Berlin, 1982.

71. Thompson, J. M. T., and Stewart, H. B. Non-linear Dynamics and Chaos.

Geometrical Methods for Engineers and Scientists. Wiley, New York, 1986.

72. Rossler, O. E. Phys. Lett. 57A, 397, 1976.

73. Rossler, O. E. Phys. Lett. 60A, 392, 1977.

74. Teymour, F., and Ray, W. H. The dynamic behavior of continuous solution

polymerization reactors—IV. Dynamic stability and bifurcation analysis of an

experimental reactor. Chem. Eng. Sci. 44(9), 1967–1982, 1989.

75. Teymour, F., and Ray, W. H. The dynamic behavior of continuous polymer-

ization reactors. V. Experimental investigation of limit-cycle behavior for vinyl

acetate polymerization. Chem. Eng. Sci. 47(15–16), 4121–4132, 1992.

76. Teymour, F., and Ray, W. H. The dynamic behavior of continuous polymer-

ization reactors. VI. Complex dynamics in full-scale reactors. Chem. Eng. Sci.

47(15–16), 4133–4140, 1992.

77. Elnashaie, S. S., Abashar, M. E., and Teymour, F. A. Bifurcation, instability

and chaos in fluidized bed catalytic reactors with consecutive exothermic che-

mical reactions. Chaos Solitons Fractals 3(1), 1–33, 1993.

78. Elnashaie, S. S. E. H., and Elshishni, S. S. Dynamic Modelling, Bifurcation and

Chaotic Behavior of Gas–Solid Catalytic Reactors. Gordon and Breach

Publishers London, 1996.

79. Razon, L. F., and Schmitz, R. A. Intrinsically unstable behavior during the

oxidation of carbon monoxide on platinum. Catal. Rev. Sci. Eng. 28(1), 89–164,

1986.

80. Wicke, E., and Onken, H. U. Statistical fluctuations of conversion and tem-

perature in an adiabatic fixed-bed reactor for carbon monoxide oxidation.

Chem. Eng. Sci. 41(6), 1681–1687, 1986.

81. Razon, L. F., Chang, S. M., and Schmitz, R. A. Chaos during the oxidation of

carbon monoxide on platinum-experiments and analysis. Chem. Eng. Sci. 41(6),

1561–1576, 1986.

82. Wicke, E., and Onken, H. U. Periodicity and chaos in a catalytic packed bed

reactor for carbon monoxide oxidation. Chem. Eng. Sci. 43(8), 2289–2294,

1988.

83. Kevrekidis, I. G., Schmidt, L. D., and Aris, R. On the dynamics of periodically

forced chemical reactors. Chem. Eng. Commun. 30(6), 323–330, 1984.

84. Kevrekidis, I. G., Aris, R., and Schmidt, L. D. The stirred tank forced. Chem.

Eng. Sci. 41(6), 1549–1560, 1986.

85. Cordonier, G. A., Schmidt, L. D., and Aris, R. Forced oscillations of chemical

reactors with multiple steady states. Chem. Eng. Sci. 45(7), 1659–1675, 1990.

86. Marek, M., and Schreiber, I. Chaotic Behavior of Deterministic Dissipative

Systems. Cambridge University Press, Cambridge, 1991.

87. Doedel, E. J., Champneys, A. R., Fairgrieve, T. F., Kuznetsov, Y. A.,

Sandstede, B., and Wang, X. J. AUTO97: Continuation and Bifurcation

Software for Ordinary Differential Equations, 1997.

88. Yorke, J. A. Dynamics: Program for IBM PC Clones. Institute for Science and

Technology, University of Maryland, College Park, 1988.

89. Cvitanovic, P. Universality in Chaos, 2nd ed. Adam Hilger, New York, 1989.

90. Elnashaie, S. S. E., and Cresswell, D. L. Dynamic behavior and stability of

adiabatic fixed bed reactors. Chem. Eng. Sci 29(9), 1889–1900, 1974.

91. Pinjala, V., Chen, Y. C., and Luss, D. Wrong-way behavior of packed-bed

reactors: II. Impact of thermal dispersion. AIChE J. 34(10), 1663–1672, 1988.

8

Novel Designs for IndustrialChemical/Biochemical Systems

The material explained in this book is necessary for the development ofnovel configurations that can revolutionize certain critical processes, espe-cially in the field of clean fuel production.

The rigorous mathematical modeling approach explained in full detailin previous chapters can be used to examine alternative novel configurationsbefore (or without) the expensive pilot-plant and semicommercial unitsstages. Two examples are given in this concluding chapter of the bookand the student, under the supervision of the course instructor, should gettraining in using the mathematical modeling and computer simulationapproach to develop the novel configurations.

8.1 NOVEL REFORMING PROCESS FOR THEEFFICIENT PRODUCTION OF THE ULTRACLEANFUEL HYDROGEN FROM HYDROCARBONS ANDWASTE MATERIALS

8.1.1 Introduction

In the United States, solid waste materials are disposed off in landfills at astaggering annual rate of 200 million tons of municipal solid waste, 300million scrap tires, and 15 million tons of plastic waste. Another major

environmental problem results from the use of hydrocarbons as fuel andfor industrial applications, generating greenhouse gases and other pollu-tants. There is a strategic need for the United States and other countries todevelop alternative sources of clean energy in order to mitigate dependenceon foreign crude oil importation. Chemical engineering students shouldrecognize that recently there has been strong recognition that hydrogenoffers significant advantages as the ultraclean fuel of the future. The mainprocess for hydrogen production is the fixed-bed catalytic steam reformingof natural gas, which is inefficient, highly polluting, and suffers from cat-alyst deactivation, especially when using higher hydrocarbons as feedstock.The objective of this exercise for the student is to use the knowledgelearned in this book to develop a new process which converts a widerange of hydrocarbons (e.g., natural gas, diesel, gasoline, etc.) as well assolid-waste materials into hydrogen and other useful chemicals. The stu-dents are assisted by providing them with the proposed novel configura-tion, which is basically a fast circulating membrane-fluidized bedintegrated to a novel dry reforming process. This process addresses mostof the limitations of the conventional steam reforming process for theproduction of hydrogen/syngas, resulting in an efficient/compact novelreformer for a wide range of feedstocks. This proposed process will offerthe following environmental benefits:

1. Efficient production of an ultraclean fuel, H2, from a wide rangeof feedstocks

2. Mitigation of the solid-waste problem by conversion into cleanfuel

3. Conservation of natural resources4. A novel processing scheme which has minimum environmental

impact

Furthermore, the proposed process will provide several economic and stra-tegic merits, as it will produce hydrogen at prices that are comparable to oreven less expensive than hydrocarbon sources. The reader is requested to usea computer-modeling approach to develop this ultraclean and efficient pro-cess for the conversion of solid wastes and a wide range of hydrocarbonsinto hydrogen and other useful chemicals. The reader should check otherprocesses in the literature and should become aware of the fact that thisproposed novel process is much more efficient, flexible, and cleaner than themodern integrated bubbling fluidized-bed membrane steam reformer(BFBMST) for natural gas (e.g., Ref. 1).

8.1.2 Literature Review

A preliminary literature review is given here as a starter for the reader.However, the reader should update this literature review.

Waste Materials Pretreatment

Waste materials such as municipal solid waste, scrap tires, and waste plasticshave traditionally been placed in sanitary landfills. However, with landfillspace rapidly decreasing in the United States and worldwide, an alternativedisposal method for these waste materials becomes imperative. The recy-cling of solid wastes is a challenging problem, with both economic andenvironmental constraints. Recently, two broad approaches have beenattempted to reclaim solid wastes. The first approach relies on thermal orcatalytic conversion of waste materials into fuel and valuable chemical feed-stocks. Examples of this approach include gasification, pyrolysis, depoly-merization, and liquefaction. The second approach relies on the physicalrecovery of valuable ingredients in the waste materials.

In spite of the experimental viability of conversion methods such asdepolymerization and liquefaction, severe technical and economic limita-tions that prevented their commercial feasibility have been encountered(2,3). Feedstock characteristics and variability constituted a major chal-lenge.

Catalytic Steam Reforming of Hydrocarbons

Nickel-supported catalysts are very efficient for steam reforming of hydro-carbons to hydrogen/syngas. The main reforming reactions are reversibleand highly endothermic as follows:

CH4 þH2O , COþ 3H2 ð�H1 ¼ 206 kJ=molÞCOþH2O , CO2 þH2 ð�H2 ¼ �40 kJ=molÞ

CH4 þ 2H2O , CO2 þ 4H2 ð�H3 ¼ 196 kJ=molÞ

Elnashaie and Elshishini (4) have shown that the rate of steam reforming isnonmonotonic with respect to steam. Steam reforming reactions have astrong tendency to carbon formation, causing deactivation of the catalystas follows:

CH4 ! Cþ 2H2 ð�H ¼ 75 kJ=molÞ2CO ! Cþ CO2 ð�H ¼ �171 kJ=molÞ:

This tendency increases with the increase in temperature and the percentageof higher hydrocarbons in the feed, and it decreases with the increase insteam/hydrocarbon ratio.

Fixed-Bed Reformers: The Present Generation of Reformers

Main Characteristics

Catalyst Tubes: Fixed-bed catalytic reactors with hundreds of tubes(e.g., 100–900) having diameters in the range 7–12 cm and length inthe range 10–15 m. Large catalyst particles to avoid excessive pressuredrop, typically Rashig rings with dimensions 1:6� 0:6� 1:6 cm. Highsteam/methane (S/M) ratio 3 : 6, to avoid carbon formation andcatalyst deactivation. S/M ratio needed increases as the percentage ofhigher hydrocarbons in the feed increases. Typical feed flow rate/catalyst tube: 3.5–4.5 kmol/h/catalyst tube. Typical natural gas feedcomposition: 70–80% CH4, 20–25% higher hydrocarbonsþH2 þ CO2

þN2. Typical feed temperature to catalyst tubes: 730–770 K (850–930 F); typical feed pressure: 2400–2900 kPa (24–29 bar).

Furnace: Huge furnace (top/side fired) to supply the endothermicheat needed for the steam reforming reactions. Typical dimensions:21:8� 35:5� 13:7m.

Typical Furnace Fuel Feed

1. Flow rate (in kmol fuel/kmol natural gas) to catalyst tubes: 0.4–0.9

2. Excess air above stoichiometry: 10–15%3. Fuel/air feed temperature: 320–580 K4. Fuel composition: 10–20% CH4, 65–75% H2 (note the large con-

sumption of hydrogen as fuel in furnace); the rest is CO2 and N2.

Main Previous Attempts to Develop an Efficient Compact SteamReformer

Over the past decade, significant progress has been made toward overcom-ing some of the limitations of conventional fixed-bed reforming systems.These trials include the following: fixed-bed with hydrogen permselectivemembranes (5–7); fixed bed with/without hydrogen permselective mem-branes using methanol feed (8); microchannel reformer for hydrogen pro-duction from natural gas (9); catalytic oxidative steam reforming (10,11);and bubbling fluidized-bed reformer with/without hydrogen permselectivemembranes (1,12,13). The most successful milestone to date involves thebubbling fluidized-bed membrane steam reformer (BFBMSR) for natural

gas (1), which is more efficient than the classical fixed-bed configuration.The present proposed configuration presents a much more efficient andflexible process than the BFBMSR.

The Basic Characteristics of BFBMSRs

1. Using a powdered catalyst with � ¼ 1:0 (100–1000 times higherrate of reaction than conventional fixed-bed steam reformers).

2. Hydrogen-selective membranes ‘‘break’’ the thermodynamicbarrier, giving higher conversions at lower temperatures.

3. In situ burning of natural gas, achieving efficient heat transferwith no large furnace. The hydrogen productivity (moles hydro-gen/h/cm3 of catalyst bed) for these BFBMSRs is 0.43–0.48 ascompared with 0.06–0.12 for the fixed bed, which is more than a300–800% increase in hydrogen productivity.

4. The operating temperature: For the fixed-bed configuration, it isin the range 980–1100 K (1305–1520 F), whereas for theBFBMSR, it is in the range 780–850 K (944–1070 F), about20% reduction in operating temperature.

Hydrogen Fuel (14,15)

Hydrogen is rightly often called the perfect fuel. Its major reserve on earth(water) is inexhaustible. Steam reforming not only extracts the hydrogenfrom the hydrocarbon but also extracts the hydrogen from water. It can beused directly by fuel cells to produce electricity very efficiently (> 50%Þ andwith zero emissions. Ultralow emissions are also achievable when hydrogenis combusted with air to power an engine. Operating fuel cells on natural gas(or other hydrocarbon source) will require a clean/efficient fuel processor toprovide hydrogen to the fuel cell. Total U.S. hydrogen consumption in 1997was approximately 8.5 billion standard cubic feet per day (bscfpd) (¼ 20:53million kg H2/day), with 40% dedicated to petroleum refining and 59% tochemical manufacturing. The U.S. hydrogen production is increasing at arate of 5–10% per year. In most industries, hydrogen is produced using theclassical inefficient fixed-bed reformers.

Hydrogen Production and CO2 Sequestration Costs (16,17)

The proposed novel configuration will not only have an appreciable posi-tive impact on the environment but will also achieve considerable eco-nomic advantages by making clean hydrogen energy economicallycompetitive. The delivered cost of H2 transportation fuel in the NewYork City/New Jersey areas varies considerably from different datasources. It is reported to be in the wide range of $15–40/GJ (¼ $0:054–

0.144/kWh ¼ $1:96–5.23/equivalent gallon gasoline). In some other regionsof the United States (and using other sources of data), the cost ofhydrogen production is $10=GJ (¼ $0:036=kWh ¼ $1:306/equivalentgallon gasoline). The following is hydrogen cost (including reformer capi-tal costþ operating cost, purification, etc.) from another data source:Production cost ¼ $10=GJ ð¼ $0:036=kWh ¼ $1:306/equivalent gallongasoline) and delivered cost ¼ $14=GJ (¼ $0:05=kWh ¼ $1:83/equivalentgallon gasoline). The strategic objective should be to develop processesto produce hydrogen from fossil fuels, waste materials, and renewableenergy sources to meet both central and distributed production goals of$6–8 per million BTUs ($0.82–1.09/kg H2, $0.9–1.2/gallon equivalent gaso-line). The CO2 removal from the exit gases during the production ofhydrogen or from other sources is an important step in decreasing theemission of this greenhouse gas which contributes strongly to globalwarming. The added cost of CO2 separation is approximately $0.65/kgH2 ð¼ $00–0.17/kWh). The cost of removing CO2 from power-plant fluegases is most typically $30–60/ton CO2. The suggested novel CO2 reform-ing which is an integral part of the proposed novel configuration willincrease the H2 productivity, improve the economics of the process, andreduce the CO2 emission considerably.

8.1.3 Limitations of Current Reforming Technologies

Catalytic Steam Reforming Diffusional Limitations

The effectiveness factor � of the catalyst pellets, expressing the fraction ofthe intrinsic rate of reaction being exploited in the fixed-bed configuration, isextremely small (� ¼ 10�2–10�3). This is due to the large catalyst pellet sizesused to avoid excessive pressure drop along the length of the reactor (18–20m). This severe limitation can be ‘‘broken’’ by using a fluidized-bed reac-tor with fine catalyst particles (effectiveness factor ¼ 1:0Þ; thus the fullintrinsic activity of the catalyst is utilized.

The Thermodynamic Equilibrium Barrier for Reversible SteamReforming

The reversible, endothermic steam reforming reactions are thermodynami-cally limited, dictating a very high temperature for high conversion. Thisthermodynamic equilibrium barrier can be ‘‘broken’’ through the continu-ous removal of one (or more) of the products. The most promising techni-que is the hydrogen removal using hydrogen permselective membranes.However, it is also possible to ‘‘break’’ this thermodynamic equilibrium

barrier by the continuous removal of CO2 (21,22) using CaO as a CO2

acceptor according to the reaction

CaOðsÞ þ CO2ðgÞ , CaCO3ðsÞ ð7258CÞThis is an exothermic reaction; therefore, it supplies part of the endothermicheat necessary for reforming. The CaO acceptor can be recovered throughthe reaction

CaCO3ðsÞ , CaOðsÞ þ CO2ðgÞThe above reaction is endothermic and needs high temperature of about975 C.

Either of the two techniques can be used or a third hybrid techniqueutilizing both of them.

Carbon Formation and Catalyst Deactivation (23,24)

The reforming reactions (specially at high temperatures and for higherhydrocarbons) have a strong tendency to deposit carbon on the catalyst,which deactivates it. The usual technique for the solution of this problem isto increase the steam/hydrocarbon ratio in the feed. For the proposed con-figuration, the use of membranes for the removal of hydrogen and/or CO2

‘‘breaks’’ the thermodynamic barrier of the reversible reforming reactions,achieving high equilibrium conversion at low temperatures. The circulatingnature of the proposed fast fluidized bed allows the continuous regenerationof the catalyst in its ‘‘way back’’ to the bed. It is also interesting to note thatthe endothermic carbon formation and the exothermic burning of carbon toregenerate the catalyst results in a net heat supply which is higher thanoxidative reforming. For methane, this heat is accompanied by the produc-tion of two moles of hydrogen per mole of methane, and the excess heat canbe used to carry out steam reforming simultaneously. A design challengeusing this technology is to maximize hydrogen production under autother-mic conditions.

Supply of Heat Necessary for the Endothermic Steam ReformingReactions

The need to use high temperatures and the tendency for carbon formation,which deactivates the catalyst, limits the range of hydrocarbon to be used asfeedstocks in the present generation of reformers. The heat transfer from thefurnace to the catalyst in the tubes is not very efficient, causing an excessiveamount of heat dissipation and environmental pollution. This problem isaddressed in the suggested novel configuration through simultaneous oxida-tive reforming (and more efficiently through carbon formation and carbon

burning as described above). CaO acceptor, and continuous regeneration/circulation of the catalyst.

Materials (Membrane) Limitations

The idea of breaking the thermodynamic equilibrium limitations throughthe use of a selective membrane in not completely new. However, it becamepractically feasible in the last decade due to the impressive advancement inmaterial science and the development of a new generation of highly selectiveinorganic membranes (5,6,25). The reader is requested to review theadvancement in this field to choose the most suitable membranes.

Hydrodynamic Limitations

Whenever the diffusional limitation is ‘‘broken’’ through the use of finecatalyst powder in a bubbling fluidized bed, a new limitation arises relatedto the hydrodynamics of the system. In the bubbling fluidized bed, it is notpossible to fully exploit the very intrinsic kinetics of the powdered catalyst.Fast fluidization (transport) reactor configuration offers excellent potentialto ‘‘break’’ this limitation.

Catalyst Attrition and Entrainment

Catalytic attrition and entrainment represent a problem in bubbling flui-dized-bed reactor, limiting the range of flow rate and dictates the use ofcyclones. In the suggested fast fluidization (transport) reformer, the solidmovement is exploited to the maximum limit in an integrated circulatingconfiguration.

8.1.4 Main Characteristics of the Suggested NovelUltraclean/Efficient Reforming Process Configuration

The proposed novel reformer will utilize five main catalytic processes,namely steam reforming, dry reforming, oxidative steam reforming, andcarbon formation/burning in addition to a number of other noncatalyticprocesses.

Catalytic Processes

1. Steam reforming: Supported nickel catalysts (e.g., 32 mol% Ni/�-alumina, 3 mol% Ni/MgO or Mg-Ca oxide) are very efficient forsteam reforming of hydrocarbons to hydrogen/syngas.

2. Dry reforming: The main reaction is endothermic:

CO2 þ CH4 , 2COþ 2H2 ð�H ¼ 247 kJ=molÞ

Suitable catalysts include Ni-, Rh-, and Ru-supported catalystsand ZrO2-supported Pt catalysts promoted with cerium.

3. Oxidative steam reforming: The main reaction is highly exother-mic:

CH4 þ 12O2 ! COþ 2H2 ð�H ¼ �208 kJ=mol; exothermicÞ

Main suitable catalysts are nickel-based catalysts, titanates-basedperovskite oxides, cobalt-containing catalysts.

4. Carbon formation:

CH4 ! Cþ 2H2 ð�H ¼ 74:8 kJ/mol, endothermic)

5. Carbon burning:

CþO2 ! CO2 ð�H ¼ �393:5 kJ/mol, exothermic)

Other Noncatalytic Processes

The main other noncatalytic reaction is homogeneous combustion ofhydrocarbons. In addition, when CaO is utilized as a CO2 acceptor, twoadditional noncatalytic reactions are involved, as explained earlier.

8.1.5 Components of the Suggested Novel UltracleanProcess for the Production of the Ultraclean FuelHydrogen

The exercise involves the optimization of the process not only with regard tothe design parameters and operating conditions, but also with regard to theconfiguration itself. A simplified diagram of the suggested integrated novelconfiguration is shown in Figure 8.1. A brief description and discussion ofthe different components and their functions and interactions follows. Theywill be the basis for the modeling exercise for the reader.

The suggested novel configuration consists of the following (see Fig.8.1):

1. Fast-circulating fluidized bed (transport reactor). The circulatingfluidized bed (transport reactor) achieves a number of advantagesover the bubbling fluidized bed:

(a) ‘‘Breaks’’ the hydrodynamic limitation ‘‘born’’ after thebreak of the diffusional limitations by using fine catalyst par-ticles, as discussed earlier

(b) Gives near plug flow conditions which are beneficial for con-version and hydrogen selectivity

(c) Allows distribution of the oxidizing agent, as detailed later(d) Allows the flexibility of using different configuration for the

sweep gas (26) in the membrane side(e) Allows the continuous regeneration of catalyst by burning the

reformed carbon.(f) Allows the continuous use and regeneration of CaO for the

continuous removal of CO2

2. Hydrogen permselective membranes. Hydrogen permselectivemembranes are used in the reformer to break the thermodynamicequilibrium barrier of the reforming reactions and to producepure hydrogen in the membrane side.

3. Carbon dioxide acceptors. Addition of CaO to the catalyst cir-culating bed to remove CO2, as described earlier.

4. Oxygen permselective membranes. Oxygen selective membranesare used to supply the oxygen along the height of the reformer forthe oxidative reforming reaction.

5. The catalyst mixture in the reformer. A mixture of mainly asteam reforming catalyst with an optimum percentage of oxida-tive reforming catalyst will be used in the reformer in order to

Figure 8.1 Preliminary suggestion for the novel hydrogen production configura-

tion.

minimize the homogeneous combustion and maximize hydrogenproduction and heat integration.

6. Gas–solid separation unit. The catalyst from the reformer will beseparated from the gases in a gas–solid separator and the regen-erated solid catalyst recycled to the reformer. The hydrocarbon/steam feed will be fed in the downer pipe used for recirculatingthe regenerated catalyst to make use of the residence time in thispipe for the reforming reactions.

7. The novel reactor–regenerator dry reformer. The output gas fromthe steam reformer will be very poor in hydrogen and quite richin CO2 due to hydrogen removal by membranes, regeneration ofcatalyst, and CaCO3. It can be used together with a suitablehydrocarbon in a dry reforming process (27,28). A promisingapproach suggested in the present exercise is to use a novel flui-dized-bed reactor-regenerator system that utilizes the carbonformed on the catalyst in the reactor to supply the necessaryheat for the endothermic dry reforming reaction through theregeneration of the catalyst in the regenerator and recyclingthe hot regenerated catalyst to the reactor [Similar in a sense tothe fluid catalytic cracking (FCC) process]. This part of the unitcan also be used separately as a stand-alone unit for the seques-tration of CO2 produced from different sources (e.g., powerplants), thus contributing to the national/international effortsto control global warming.

8. Feedstocks. To simplify the exercise, natural gas is to be used as thefeedstock. However, other higher hydrocarbons can also be used.

8.1.6 Main Tasks for the Exercise

A number of extensive literature surveys will be carried out by the reader toproduce the necessary data.

Task 1: Kinetics, Catalysis, and Hydrodynamics

1. Collection of steam reforming kinetics. The kinetics of the steamreforming of natural gas (as well as higher hydrocarbons) on anickel catalyst should be obtained from the literature (29,30).

2. Collection of oxidative steam reforming kinetic data. Rate equa-tions should be obtained from the literature for the most promis-ing oxidative reforming catalysts, including Ni-based catalysts,Titanates-based perovskite oxides, and cobalt-containing cata-lysts. Careful consideration will be given to the calorimetry of

the reactions under oxidative conditions. Mixtures of reformingand oxidative reforming catalysts will also be tested.

3. Collection of dry reforming kinetic data. Rate equations shouldbe obtained from the literature (31,32).

4. Carbon formation and catalyst regeneration. Rate equationsshould be obtained from the literature (33).

5. CO2 acceptor and regeneration. Rate equations should beobtained from the literature (33).

6. Hydrodynamics of gas–solid flow and separation. Necessary datashould be obtained from the literature and hydrodynamic modelsof different degree of sophistication/rigor be developed.

Task 2: Separation

The permeation and selectivity of different commercial hydrogen and oxy-gen selective membranes as well as membranes developed by the studentsshould (if possible) be determined using permeation cells and the rate ofpermeation equations will be developed.

Task 3: Mathematical Modeling, Computer Simulation, andOptimization of the Process

Rigorous static/dynamic models should be developed for the differentparts of the novel configuration. The models of the different parts shouldbe integrated for the entire system. The overall model should then beutilized for the prediction of system performance, optimization of thedesign and operating conditions, as well as the development and designof optimal control strategies. This task will utilize all of the data and rateequations collected. Because a number of reactions are exothermic andrecirculation is widely used in this novel configuration, the steady stateand dynamic models should be used to investigate the possible instabilityproblems associated with this configuration. The ultimate goal should beto develop a comprehensive model of the process to be used as a CAD/CAM tool for analysis, design, and optimization of this novel process(34).

Important note: It is stated everywhere that all kinetics and permea-tion parameters as well as other parameters and rate equations are tobe obtained from the literature. However, whenever the reader hasthe experimental facilities to determine any of these parameters, he/she is encouraged to obtain these parameters. This will be useful andadd to the educational benefits of this mathematical modeling andcomputer simulation exercise.

8.2 A NOVEL FERMENTOR

This exercise involves not only novel configuration but, more essentially,also a novel mode of operation of a fermentor.

8.2.1 Introduction

Both ‘‘end-of-pipe’’ as well as ‘‘in-process’’ modifications are not sufficientmeans for achieving long-term economic growth while sustaining a cleanenvironment. It is essential to develop a new generation of technologiesthat achieve pollution avoidance/prevention and produce cleaner fuels uti-lizing abundant waste. Cellulosic biomass is an attractive feedstock for theproduction of fuel ethanol by fermentation, because it is renewable andavailable domestically in great abundance at low cost (in some cases, it hasa negative price). Cellulosic biomass includes tree leaves, forest products(particularly those fast-growing energy trees), agricultural residues (e.g.,corn cobs, corn stalks, rice straws, etc.), waste streams from agriculturalprocessing, municipal waste, yard and wood waste, waste from papermills, and so forth. It is very important for the United States and othercountries to decrease their dependence on hydrocarbons and introducecleaner fuels to the energy and transportation sectors. The main aim ofthis exercise is to use mathematical and computer modeling to develop anefficient and environmentally friendly process for the production of thecleaner fuel ethanol through fermentation of the sugars resulting fromthe hydrolysis of biomass and cellulosic waste and energy crops (e.g.,switch grass). The main bottleneck in this process is associated with thefermentation step, because the sugars produced from the hydrolysis stepare difficult to ferment.

In order to ‘‘break’’ this bottleneck an integrated multidisciplinarysystem as explained in this book should be adopted (35). It consists of thefollowing nonlinearly interacting means:

1. Use of genetically engineered mutated micro-organisms capableof efficiently fermenting these difficult sugars (36,37).

2. The use of advanced fermentor configuration (immobilizedpacked bed) with ethanol-selective membranes for the continuousremoval of ethanol from the fermentation mixture.

3. The exploitation of the oscillatory and chaotic behavior of thesystem at high sugar concentrations, for higher sugar conversionand ethanol productivity (38).

4. Control of chaos for the generation of new attractors with higherethanol productivity.

5. Extensive utilization of reliable mathematical and computermodeling to minimize the cost of development of novel efficientand environmentally friendly technologies capable of achievingMaximum Production–Minimum Pollution (MPMP).

The synergy of the above means to radically improve the performance ofthis process is expected to give considerable increases in ethanol productiv-ity.

Mathematical modeling should be exploited in order to reach theoptimal designs and operating mode/conditions.

Both the continuous-stirred tank configuration with/without ethanolpervaporation membrane separation as well as the packed-bed immobilizedfermentors with their different configurations for the membrane case (cocur-rent, countercurrent, countercurrent with partial blinding and mixed flow)are parts of this exercise for the reader. The kinetics of genetically engi-neered mutated micro-organisms should be used to simulate the fermenta-tion of sugars resulting from the hydrolysis of different types of biomassand/or cellulosic waste. The biokinetic data should be obtained for thedifferent strains of micro-organisms and different feedstocks through exten-sive literature survey by the reader. The effect of the membrane removal ofethanol on the biokinetics should also be investigated. The permeation anddiffusion parameters should be obtained from the literature. Reliable bio-reaction, permeation, and diffusion rate equations for use in the mathema-tical models should be obtained from the literature. Reliable fermentormodels should be developed and used to find the optimal design and oper-ating condition.

8.2.2 Basic Research Description

The development and use of innovative technologies is a must in order todevelop new generation of environmentally benign technologies. Thisimportant strategic aim requires an orchestrated effort utilizing fundamentalknowledge, maximum utilization of mathematical and computer modeling(as discussed in this book), and extensive use of computing resourcescoupled to optimally planned experimentation. This approach will be ableto minimize the cost and systemize the approach and procedure for thedevelopment of efficient and environmentally friendly novel processes andproducts.

This exercise will, of course, concentrate on the mathematical andcomputer modeling part of this strategy.

Continuous Ethanol Removal

It is important for efficient production to use a suitable technique for con-tinuous removal of this inhibitory product, ethanol. There are many tech-niques for the continuous removal of ethanol from the fermentation process.

Membranes

Membranes have been used in various configurations with integrated reac-tion/recovery schemes. Pervaporation is most probably the most promisingtechnique for the efficient continuous removal of ethanol from the fermen-tation mixture for the efficient breaking of the ethanol inhibition barrier(39,40).

Perstraction is a similar technique; however, in this case, a solvent isused to remove product away from the membrane surface. In situations inwhich the solvent is highly toxic, even in dissolved amounts (41), the mem-brane can act to prevent the solubilization of solvent into the aqueous phasecontaining the biocatalyst. Alternatively, in situations in which contactrather than dissolved toxicity is a concern (42), the membrane can be usedto prevent direct contact between the solvent an the biocatalyst. Supportedliquid membranes can also be used in perstraction (43).

Hollow fibers

The economics of Hollow Fiber Ethanol Fermentation (HFEF) was exam-ined (44) and the process was found to be as competitive as conventionalfermentation processes even at the high cost ($4/ft2) of hollow fibers at thetime of that study.

Continuous extractive fermentation of high-strength glucose feeds to ethanol

Extractive fermentation using olel alcohol was used successfully to produce90–95% conversion for higher sugar concentrations (400–450 g/L) (45).

Optimal Fermentor Configuration

Different fermentor configurations can be used, including batch, fed-batch,continuous-stirred tank, packed bed, and so forth. Immobilization of themicro-organism allows the use of high flow rates without facing the problemof micro-organisms washout. However, immobilization adds an additionalmass transfer resistance. The choice of the best configuration being a con-ceptual problem is rather difficult; in this exercise, different configurationsshould be examined using different mutated micro-organisms and differentfeedstocks. Earlier studies by Elnashaie and co-workers (46,47) suggest thatin many cases, the optimal configuration is the countercurrent configurationwith partial blinding for a membrane packed-bed immobilized fermentor.

Immobilization of the micro-organisms also increases the yield of ethanolper gram of sugar converted (48).

Exploitation of Oscillatory and Chaotic Behavior at High SugarConcentrations

It is interesting to note that in the fermentation process oscillations areevident (49,50). Other investigators observed and reported these continuousoscillations (51,52). Elnashaie and co-workers (35,38,46,47) analyzed someof these oscillations and found that in certain regions of operating anddesign parameters, the behavior is not periodic, but chaotic. They also dis-covered that the route to chaotic behavior is through the period-doublingmechanism. Control of chaos through external disturbances (53,54) can beused to control the fermentor in usually unattainable regions with higherethanol yield and productivity.

Bruce and co-workers (55) assumed that the inhibitory effect of etha-nol is the cause of oscillations and argued that ethanol removal using extrac-tive fermentation stops the oscillations. Same preliminary results carried outby the authors of this book, shown in Figures 8.2 and 8.3, support thisconclusion of Bruce and co-workers (55).

It is interesting also to note that periodic attractors may give higherethanol yield than the corresponding steady states. The higher productivity

Figure 8.2 Effect of membrane permeation area on stability and ethanol produc-

tion rate. A case with periodic attractor when area of permeation is 0m2.

of the periodic attractor over the corresponding steady state is shown inFigure 8.2, whereas the higher productivity of the chaotic attractor over thecorresponding steady state is shown in Figure 8.3. This effect will be ofconsiderable importance for the above-discussed unclassical approach tooptimization and control of these bioreactors.

8.2.3 Tasks for the Exercise

Task A. Kinetics of Sugars Fermentation to Ethanol and EthanolMembrane Removal

The objective of this part is to collect the data regarding the kinetics of sugarfermentation to ethanol using different sugars (resulting from the hydrolysisof biomass and cellulosic waste) using different strains of micro-organismsand genetically engineered mutated micro-organisms. The feedstocks aremainly the difficult-to-ferment sugars resulting from the hydrolysis of cellu-losic waste and energy crops. Unstructured and structured kinetic modelsshould be used.

Figure 8.3 Effect of membrane permeation area on stability and ethanol produc-

tion rate. A case with chaotic attractor when area of permeation is 0m2.

Task B. The Continuous-Stirred Tank Fermentor

This part includes modeling (46,53) for the continuous-stirred tank fermen-tor with/without ethanol-selective membranes. It includes modeling for eachof the different feedstocks and each of the different strains of micro-organ-isms suggested. The biokinetic models and parameters obtained from theliterature should be used in the modeling of the continuous-stirred tankfermentor proposed in this part. The membrane flux equations and para-meters obtained from the literature should also be used for the membranecontinuous-stirred tank fermentor.

The modeling and analysis should be carried out over a wide range ofparameters (especially feed concentrations) for the different feedstocks andmicro-organisms. With the basic parameters of the system obtained fromthe literature, it will be wise to build the model, develop an efficient com-puter simulation program for its solution, and investigate the expectedbehavior of this continuous system over a wide range of parameters.

The reader is also advised to refer to the section on distributed para-meter heterogeneous modeling of fermentation (Section 6.10) to obtain avery rigorous model.

Task C. Immobilized Packed-Bed Fermentors With/WithoutMembranes

In this part, the more advanced configuration of immobilized packed-bedfermentor with/without membranes is modeled and simulated (Figure 8.4).The micro-organisms in this configuration is immobilized using different

Figure 8.4 Schematic simplified diagram of the immobilized packed-bed mem-

brane fermentor (cocurrent and countercurrent configurations).

types of carrier (e.g., calcium alginate) with different diffusivities. The diffu-sion coefficients of the beads will be obtained from the literature.

Rigorous heterogeneous steady-state and dynamic models for thisimmobilized packed-bed fermentor with/without membranes should bedeveloped and used to analyze and optimize this novel configuration(refer to the section in Chapter 6 regarding distributed modeling of fermen-tation).

For more senior readers, the dynamic model should be used not onlyto simulate the dynamic behavior of this heterogeneous distributed systembut also to investigate the possible spatio-temporal chaos associated with it.

Important note: As for the previous exercise, it is stated everywherethat all biokinetics and permeation parameters as well as other para-meters are to be obtained from the literature. However, whenever thereader has the experimental facilities to determine any of these para-meters, he/she is encouraged to obtain these values. This will be usefuland add to the educational benefits of this mathematical modelingand computer simulation exercise.

REFERENCES

1. Adris, A., Grace, J., Lim, C., and Elnashaie, S. S. E. H., Fluidized bed reaction

system for system/hydrocarbon gas reforming to produce hydrogen. U.S.

Patent 5,326,550, June 5, 1994.

2. Shelley, M. D., and El-Halwagi, M. M. Techno-economic feasibility and flow

sheet synthesis of scrap tire/plastic waste liquefaction. J. Elastomers Plastics

31(3), 232–254, 1999.

3. Warren, A., and El-Halwagi, M. M. An economic study on the co-generation of

liquid fuel and hydrogen from coal and municipal solid waste. Fuel Process.

Technol. 49, 157–166, 1996.

4. Elnashaie, S. S. E. H., and Elshishini, S. S. Modeling, Simulation and

Optimization of Industrial Fixed Bed Catalytic Reactors. Gordon & Breach

Science, London, 1993.

5. Sammells, A. F., Barton, A. F., Peterson, D. R., Harford, S. T., Mackey, R.,

Van Calcar, P. H., Mundshau, M. V., and Schultz, J. B. Methane conversion to

syngas in mixed conducting membrane reactors. 4th International Conference

on Catalysis in Membrane Reactors, 2000.

6. Sammels, A. F., Schwartz, M., Mackay, R. A., Barton T. F., and Peterson, D.

R. Catalytic membrane reactors for spontaneous synthesis gas production.

Catal. Today 56, 325, 2000.

7. Dyer, P. N., and Chen, C. M. Engineering development of ceramic membrane

reactor system for converting natural gas to H2 and syngas for liquid transpor-

tation fuel. Proceedings of the 2000 Hydrogen Program Review, 2000.

8. Amphlett, J. C., Holland, R. M., Mann, R. F., and Pepply, B. A. Modeling a

steady state catalytic methanol steam reformer for reactor development and

design. AMSE Symposia Series: Emerging Energy Technology, ASME, 1996,

pp. 118–124.

9. Makel, D. Low cost microchannel reformer for hydrogen production from

natural gas. California Energy Commission (CEG), Energy Innovations

Small Grant (EISG) Program, 1999.

10. Hayakawa, T., Andersen, A. G., Shimizu, M., Suzuki, K., and Takehira, K.

Partial oxidation of methane to synthesis gas over some tibanates based per-

ovskite oxides. Catal. Lett. 22, 307–317, 1993.

11. Theron, J. N., Dry, M. E., Steen, E. V., and Fletcher, J. C. G. Internal and

external transport effects during the oxidative reforming of methane on a com-

mercial steam reforming catalyst. Studies Surface Sci. Catal. 107, 455–460,

1997.

12. Elnashaie, S. S. E. H., and Adris, A. Fluidized beds steam reformer for

methane. Proceedings of the IV International Fluidization Conference, 1989.

13. Adris, A., Elnashaie, S. S. E. H., and Hughes, R. Fluidized bed membrane

steam reforming of methane. Can. J. Chem. Eng. 69, 1061–1070, 1991.

14. Ogden, J. Hydrogen: A low polluting energy carrier for the next century. Pollut.

Preven. Rev. 8, 4, 1998.

15. Ogden, J. M. Hydrogen energy systems studies. Proceedings of the 1999 US

DOE Hydrogen Program Review, 1999.

16. Ogden, J. A technical and economical assessment of hydrogen energy systems

with CO2 sequestration. 12th World Hydrogen Energy Conference, 1998.

17. Ogden, J. Hydrogen systems and CO2 sequestration. 9th National Hydrogen

Association Meeting, 1998.

18. Soliman, M. A., Elnashaie, S. S. E. H., Al-Ubaid, A. S., and Adris, A.

Simulation of steam reformers for methane. Chem. Eng. Sci. 43, 1801–1806,

1988.

19. Elnashaie, S. S. E. H., Adris, A., Soliman, M. A., and Al-Ubaid, A. S. Digital

simulation of industrial steam reformers. Can. J. Chem. Eng. 70, 786–793, 1992.

20. Alhabdan, F. M., Abashar, M. A., and Elnashaie, S. S. E. H. A flexible soft-

ware package for industrial steam reformers and methanators based on rigor-

ous heterogeneous models. Math. Computer Model. 16, 77–86, 1992.

21. Han, C., and Harrison, D. P. Simultaneous shift and carbon dioxide separation

for the direct production of hydrogen. Chem. Eng. Sci. 49. 5875–5883, 1994.

22. Brun-Tsekhovoi, A. R., Zadorin, A. R., Katsobashvilli, Y. R., and

Kourdyumov, S. S. The process of catalytic steam reforming of hydrocarbons

in the presence of carbon dioxide acceptor. Hydrogen Energy Progress VII,

Proceedings of the 7th World Hydrogen Energy Conference, 1988, pp. 885–900.

23. Demicheli, M. C., Duprez, D., Barbier, J., Ferretti, O., and Ponzi, E.

Deactivation of steam-reforming catalysts by coke formation. J. Cataly. 145,

437–449, 1994.

24. Barbier, J. Coking of reforming catalysts. Studies Surface Sci. Cataly. 34, 1–19,

1987.

25. Collins, J. P., and Way, J. D. Preparation and characterization of a composite

palladium–ceramic membrane. Ind. Eng. Chem. Res. 32, 3006, 1993.

26. Elnashaie, S. S. E. H., Moustafa, T. M., Alsoudani, T., and Elshishini, S. S.

Modeling and basic characteristics of novel integrated dehydrogenation–hydro-

genation membrane catalytic reactors. Computers Chem. Eng. 24, 1293–1300,

2000.

27. Gadalla, A. M., and Sommer, M. E. Carbon dioxide reforming of methane on

nickel catalysts. Chem. Eng. Sci. 44, 2825, 1989.

28. Bradford, M. C. J., and Vannice, M. A. CO2 reforming of CH4. Catal. Rev. Sci.

Eng. 41, 1–42, 1999.

29. Chen, D., Lodeng, R., Omdahl, K., Anundskas, A., Olsvik, O., and Holmen, A.

A model for reforming on Ni catalyst with carbon formation and deactivation.

Studies Surface Sci. Cataly. 139, 93–100, 2001.

30. Caton, N., Villacampa, J. I., Royo, C., Romeo, E., and Monzon, A. Hydrogen

production by catalytic cracking of methane using Ni–Al2O3 catalysts.

Influence of the operating conditions. Studies Surface Sci. Catal. 139, 391–

398, 2001.

31. Wurzel, T., Olsbye, U., and Mleczko, L. Kinetic and reaction engineering

studies of dry reforming of methane over a Ni/La/Al2O3 catalyst. Ind. Eng.

Chem. Res. 36(12), 5180–5188, 1997.

32. Ferreira-Aparicio, P., Marquez-Alvarez, C., Rodriguez-Ramos, I., Schuurman,

Y., Guerrero-Ruiz, A., and Mirodatos, C. A transient kinetic study of the

carbon dioxide reforming of methane over supported Ru catalysts. J. Catal.

184(10), 202–212, 1999.

33. Arandes, J. M., Abajo, I., Fernandez, I., Lopez, D., and Bilbao, J. Kinetics of

gaseous product formation in the coke combustion of a fluidized catalytic

cracking catalyst. Ind. Eng. Chem. Res. 38(9), 3255–3260, 1999.

34. Natarajan, T. T., and El-Kaddah, N. Finite element analysis of electromagne-

tically driven flow in sub-mold stirring of steel billets and slabs. ISIJ Int. 38(7),

680–689, 1998.

35. Garhyan, P., and Elnashaie, S. S. E. H. Integrated multidisciplinary approach

for efficient production of fuel ethanol from cellulosic waste, International

Conference on Fiber Industry, 2002.

36. Ingram, L. O., Conway, T., and Alterhum, F. U.S. Patent 5,000,000, 1991.

37. Ho, N. W. Y. and Tsao, G. T. U.S. Patent 5,789,210, 1998.

38. Garhyan, P. and Elnashaie, S. S. E. H. Exploration and exploitation of bifur-

cation chaotic behavior for efficient ethanol fermentation. AIChE Annual

Conference, 2002.

39. Yamasaki, A., Ogasawara, K., and Mizoguchi, K. Pervaporation of water/

alcohol mixtures through the PVA membranes containing cyclodextrin oligo-

mer. J. Appl. Polym. Sci. 54(7), 867–872, 1994.

40. Ruckenstein, E., and Liang, L. Pervaporation of ethanol–water mixtures

through polyvinyl alcohol–polyacrylamide interpenetrating polymer network

membranes unsupported and supported on polyethersulfone ultrafiltration

membranes: A comparison. J. Membr. Sci. 110(1), 99–107, 1996.

41. Shukla, R., Kang, W., and Sirkar, K. K. Novel hollow fiber immobilization

techniques for whole cells and advanced bireactors. Appl. Biochem. Biotechnol.

20(2), 571–586, 1989.

42. Steinmeyer, D. E., and Shuler, M. L. Mathematical modeling and simulations

of membrane bioreactor extractive fermentations. Biotechnol. Progr. 6(5), 362–

369, 1990.

43. Deblay, P., Minier, M., and Renon, H. Separation of l-valine from fermenta-

tion broths using a supported liquid membrane. Biotechnol. Bioeng. 35(2), 123–

131, 1990.

44. Naser, S. F., and Fournier, R. L. A numerical evaluation of a hollow fiber

extractive fermentor process for the production of ethanol. Biotechnol. Bioeng.

32(5), 628–638, 1988.

45. Daugulis, A. J., Axford, D. B., Ciszek, B., and Malinowski, J. J. Continuous

fermentation of high-strength glucose feeds to ethanol. Biotechnol. Lett. 16(6),

637–642, 1994.

46. Abasaed, A. E., and Elnashaie, S. S.E .H. A novel configuration for packed bed

membrane fermentors for maximizing ethanol productivity and concentration.

J. Membr. Sci. 82(1–2), 75–82, 1993.

47. Elnashaie, S. S. E. H., Garhyan, P., Haddad, S., Ibrahim, G., Abasaeed, A.,

and Elshishini, S. S. Integrated multidisciplinary approach for efficient produc-

tion of fuel alcohol. AIChE Annual Meeting, 2000.

48. Doran, P. M., and Bailey, J. E. Effects of immobilization on growth, fermenta-

tion properties, and macromolecular composition of Saccharomyces cerevisiae

attached to gelatin. Biotechnol. Bioeng. 28(1), 73–87, 1986.

49. Jobses, I. M. L., Egberts, G. T. C., Luyben, K. C. A. M., and Roels, J. A.

Fermentation kinetics of Zymomonas mobilis at high ethanol concentrations:

oscillations in continuous cultures. Biotechnol. Bioeng. 28, 868–877, 1986.

50. Daugulis, A. J., McLellan, P. J., and Li, J. Experimental investigation and

modeling of oscillatory behavior in the continuous culture of Zymomonas mobi-

lis. Biotechnol. Bioeng. 56, 99–105, 1997.

51. Ghommidh, C., Vaija, J., Bolarinwa, S., and Navarro, J. M. Oscillatory beha-

vior of Zymomonas in continuous cultures: A simple stochastic model.

Biotechnol. Lett. 9, 659–664, 1989.

52. McLellan, P. J., Daugulis, A. J., and Li, J. The incidence of oscillatory behavior

in the continuous fermentation of Zymomonas mobilis. Biotechnol. Progr. 15,

667–680, 1999.

53. Haddad, Sherif, M.Sc. thesis, Salford University, U.K., 1996.

54. Elnashaie, S. S. E. H., and Abashar, M. E. Chaotic behavior of periodically

forced fluidized-bed catalytic reactors with consecutive exothermic chemical

reactions. Chem. Eng. Sci. 49(15), 2483–2498, 1994.

55. Bruce, L. J., Axford, D. B., Ciszek,B., and Daugulis, A. J. Extractive fermenta-

tion by Zymomonas mobilis and the control of oscillatory behavior. Biotechnol.

Lett. 13(4), 291–296, 1991.

Appendix A

Matrices and Matrix Algebra

The following facts make a working knowledge of matrix methods virtuallyindispensable for an engineer dealing with multivariable systems:

1. The convenience of representing systems of several equations incompact, vector-matrix form

2. How such representations facilitate further analysis

It is perhaps useful to briefly review the essential concepts here because mostreaders will have encountered matrix methods previously.

A.1 DEFINITION, ADDITION, SUBTRACTION, ANDMULTIPLICATION

A matrix is an array of numbers arranged in rows and columns, written inthe following notation:

A ¼ a11 a12 a13a21 a22 a23

� �ðA:1Þ

Because the above matrix A contains two rows and three columns, it iscalled a 2� 3 matrix. The elements a11; a12; . . . have the notation aij ; thatis, the element of the ith row and jth column of A.

In general, a matrix with m rows and n columns is said to have dimen-sion m� n. In the special case when n ¼ m, the matrix is said to be square;otherwise, the matrix is said to be rectangular or nonsquare.

As in the case of scalar quantities, matrices can also be added, sub-tracted, multiplied and so forth, but special rules of matrix algebra mustbe followed.

A.1.1 Addition/Subtraction

Only matrices having the same dimensions (i.e. the same number of rowsand columns) can be added or subtracted. For example, given the matricesA, B, and C,

A ¼ a11 a12 a13

a21 a22 a23

� �; B ¼

b11 b12 b13

b21 b22 b23

b31 b32 b33

0B@

1CA

C ¼ c11 c12 c13

c21 c22 c23

� � ðA:2Þ

only A and C can be added or subtracted. B cannot be added or subtractedfrom either A or C because its dimensions are different from both.

The sum of Aþ C ¼ D can be written as

D ¼ Aþ C ¼ a11 a12 a13

a21 a22 a23

� �þ c11 c12 c13

c21 c22 c33

� �

¼ a11 þ c11 a12 þ c12 a13 þ c13

a21 þ c21 a22 þ c22 a23 þ c23

� � ðA:3Þ

Thus, addition (or subtraction) of two matrices is accomplished by adding(or subtracting) the corresponding elements of the two matrices. Theseoperations may, of course, be extended to any number of matrices of thesame order. In particular, observe that adding k identical matrices A

(equivalent to scalar multiplication of A by k) results in a matrix in whicheach element is merely the corresponding element of A multiplied by thescalar k. Thus, when a matrix is to be multiplied by a scalar, the requiredoperation is done by multiplying each element of the matrix by the scalar.For example, for A given in relation (A.2),

kA ¼ ka11 a12 a13a21 a22 a23

� �¼ k a11 k a12 k a13

k a21 k a22 k a23

� �ðA:4Þ

A.1.2 Multiplication

Matrix multiplication can be performed on two matrices A and B to formthe product AB only if the number of columns of A is equal to the number ofrows in B. If this condition is satisfied, the matrices A and B are said to beconformable in the order AB. Let us consider that P is an m� n matrix (mrows and n columns) whose elements pij are determined from the elements ofA and B according to the following rule:

pij ¼Xnk¼1

aikbkj i ¼ 1; 2; . . . ;m; j ¼ 1; 2; . . . ; r ðA:5Þ

For example, if A and B are defined by Eq. (A.2), then A is a 2� 3 matrix, Ba 3� 3 matrix, and the product P ¼ AB, a 2� 3 matrix, is obtained as

P ¼ a11 a12 a13a21 a22 a23

� � b11 b12 b13b21 b22 b23b31 b32 b33

0@

1A

Thus we get

P ¼ a11b11 þ a12b21 þ a13b31ð Þ a11b12 þ a12b22 þ a13b32ð Þ a11b13 þ a12b23 þ a13b33ð Þa21b11 þ a22b21 þ a23b31ð Þ a21b12 þ a22b22 þ a23b32ð Þ a21b13 þ a22b23 þ a23b33ð Þ

� �

ðA:6ÞUnlike in scalar algebra, the fact that AB exists does not imply that BA

exists. So, in general,

AB 6¼ BA ðA:7Þ

Note: In matrix algebra, therefore, unlike in scalar algebra, the order ofmultiplication is very important.

Those special matrices for which AB ¼ BA are said to commute or bepermutable.

A.2 TRANSPOSE OF A MATRIX

The transpose of the matrix A, denoted by AT , is the matrix formed by

interchanging the rows with columns. For example, if

A ¼ a11 a12 a13a21 a22 a23

� �

Algebra

then

AT ¼

a11 a21a12 a22a13 a23

0@

1A ðA:8Þ

Thus, if A is an m� n matrix, then AT is an n�m matrix. Observe, there-

fore, that both products ATA and AA

T can be formed and that each will bea square matrix. However, we note, again, that, in general,

AAT 6¼ A

TA ðA:9Þ

The transpose of a matrix made up of the product of the other matrices maybe expressed by the relation

ðABCDÞT ¼ DTC

TBTA

T ðA:10Þwhich is valid for any number of matrices, A, B, C, D, and so on.

A.3 SOME SCALAR PROPERTIES OF MATRICES

Associated with each matrix are several scalar quantities that are quiteuseful for characterizing matrices in general; the following are some of themost important.

A.3.1 Trace of a Matrix

If A is a square, n� n matrix, the trace of A, sometimes denoted as Tr(A), isdefined as the sum of the elements on the main diagonal; that is,

TrðAÞ ¼Xni¼1

aii ðA:11Þ

A.3.2 Determinants and Cofactors

Of all the scalar properties of a matrix, none is perhaps more important thanthe determinant. By way of definition, the ‘‘absolute value’’ of a squarematrix A, denoted by jAj, is called its determinant. For a simple 2� 2 matrix

A ¼ a11 a12a21 a22

� �ðA:12Þ

the determinant is defined by

jAj ¼ a11a22 � a21a12 ðA:13ÞFor higher-order matrices, the determinant is more conveniently defined interms of cofactors.

A.3.3 Cofactors

The cofactor of the element aij is defined as the determinant of the matrix leftwhen the ith row and the jth column are removed, multiplied by the factorð�1Þiþj. For example, let us consider the 3� 3 matrix

A ¼a11 a12 a13a21 a22 a23a31 a32 a33

0@

1A ðA:14Þ

The cofactor, C12, of element a12 is

C12 ¼ ð�1Þ3 a21 a23a31 a33

�� ¼ � a21a33 � a31a23ð Þ ðA:15Þ

Similarly, the cofactor C23 is

C23 ¼ ð�1Þ5 a11 a12a31 a32

�� ¼ � a11a32 � a31a12ð Þ ðA:16Þ

A.3.4 The General n� n Determinant

For the general n� n matrix A whose elements aij have correspondingcofactors Cij, the determinant jAj is defined as

jAj ¼Xni¼1

aijCij (for any column jÞ ðA:17Þ

or

jAj ¼Xnj¼1

aijCij (for any row iÞ ðA:18Þ

Equations (A.17) and (A.18) are known as the Laplace expansion formulas;they imply that to obtain jAj, one must do the following:

1. Choose any row or column of A.2. Evaluate the n cofactors corresponding to the elements in the

chosen row or column.3. The sum of the n products of elements and cofactors will give the

required determinant. (It is important to note that the sameanswer is obtained regardless of the row or column chosen.)

Algebra

To illustrate, let us evaluate the determinant of the 3� 3 matrix givenby Eq. (A.14) by expanding in the first row. In this case, we have that

jAj ¼ a11C11 þ a12C12 þ a13

that is

a11 a12 a13

a21 a22 a23

a31 a32 a33

�� ¼ a11

a22 a23

a32 a33

�� a12

a21 a23

a31 a33

��þ a13

a21 a22

a31 a32

��

¼ a11a22a33 � a11a32a23 þ a12a31a23 � a12a21a33

þ a13a21a32 � a13a31a22 ðA:19Þ

It is easily shown that the expansion by any other row or column yields thesame result. The reader should try to prove this fact.

A.3.5 Properties of Determinants

The following are some important properties of determinants frequentlyused to simplify determinant evaluation.

1. Interchanging the rows with columns leaves the value of a deter-minant unchanged. The main implications of this property aretwofold:

(a) jAj ¼ jAT j.(b) Because rows and columns of a determinant are interchange-

able, any statement that holds true for rows is also true forcolumns, and vice versa.

2. If any two rows (or columns) are interchanged, the sign of thedeterminant is reversed.

3. If any two rows (or columns) are identical, the determinant iszero.

4. Multiplying the elements of any one row (or column) by the sameconstant results in the value of determinant being multiplied bythis constant factor; if, for example,

D1 ¼a11 a12 a13a21 a22 a23a31 a32 a33

�� and D2 ¼

ka11 a12 a13ka21 a22 a23ka31 a32 a33

��

Then, D2 ¼ kD1:

5. If for any i and j (i 6¼ j) the elements of row (or column) i arerelated to the elements of row (or column) j by a constant multi-plicative factor, the determinant is zero; for example,

D1 ¼ka12 a12 a13ka22 a22 a23ka32 a32 a33

��

where we observe that column 1 is k times column 2; then,D1 ¼ 0.

6. Adding a constant multiple of the elements of one row (or col-umn) to corresponding elements of another row (or column)leaves the value of the determinant unchanged; for example,

D1 ¼a11 a12 a13a21 a22 a23a31 a32 a33

�� and D2 ¼

a11 þ ka12 a12 a13a21 þ ka22 a22 a23a31 þ ka32 a32 a33

��

Then, D2 ¼ D1.7. The value of a determinant that contains a whole row (or col-

umn) of zeros is zero.8. The determinant of a diagonal matrix (a matrix with nonzero

elements only on the main diagonal) is the product of theelements on the main diagonal. The determinant of a triangularmatrix (a matrix for which all elements below the main diagonalor above the main diagonal are zero) is also equal to the productof the elements on the main diagonal.

A.3.6 Minors of a Matrix

Let A be a general m� n matrix with m < n. Many m�m determinantsmay be formed from the m rows and any m of the n columns of A. Thesem�m determinants are called minors (of order m) of the matrix A. Wemay similarly obtain several ðm� 1Þ � ðm� 1Þ determinants from the var-ious combinations of ðm� 1Þ rows and ðm� 1Þ columns chosen from them rows and n columns of A. These determinants are also minors, but oforder ðm� 1Þ.

Minors may also be categorized as first minor, the next highest minoris the second minor, and so forth, until we get to the lowest-order minor, theone containing a single element. In the special case of a square n� n matrix,there is, of course, one and only one minor of order n, and there is no minorof higher order than this. Observe, therefore, that the first minor of a squarematrix is what we have referred to as the determinant of the matrix. The

Algebra

second minor will be of order ðn� 1Þ. Note, therefore, that the cofactors ofthe elements of a square matrix are intimately related to the second minorsof the matrix.

A.3.7 Rank of a Matrix

The rank of a matrix is defined as the order of the highest nonvanishingminor of that matrix (or, equivalently, the order of the highest square sub-matrix having a nonzero determinant). A square n� nmatrix is said to be offull rank if its rank is equal to n. The matrix is said to be rank deficient if therank is less than n.

In several practical applications of matrix methods, the rank of thematrix involved provides valuable information about the nature of the pro-blem at hand. For example, in the solution of the system of linear algebraicequations by matrix methods, the number of independent solutions that canbe found is directly related to the rank of the matrix involved.

A.4 SOME SPECIAL MATRICES

A.4.1 The Diagonal Matrix

If all the elements of a square matrix are zero except the main diagonal fromthe top left-hand corner to the bottom right-hand corner, the matrix is saidto be diagonal. Some examples are

A ¼ a11 0

0 a22

� �and B ¼

b11 0 0

0 b22 0

0 0 b33

0B@

1CA

A.4.2 The Triangular Matrix

A square matrix in which all the elements below the main diagonal are zerois called an upper-triangular matrix. By the same token, a lower-triangularmatrix is one for which all the elements above the main diagonal are zero.Thus,

b11 b12 b13

0 b22 b23

0 0 b33

0B@

1CA and

b11 0 0

b21 b22 0

b31 b32 b33

0B@

1CA

are, respectively, examples of upper- and lower-triangular matrices.

A.4.3 The Identity Matrix

A diagonal matrix in which the nonzero elements on the main diagonal areall unity is called the identity matrix, it is frequently denoted by I:

I ¼

1 0 0 . . . 00 1 0 . . . 00 0 1 . . . 0... ..

. ... . .

.0

0 0 0 0 1

0BBBB@

1CCCCA

The significance of the identify matrix will be obvious later, but, for now, wenote that it has the unique property that

IQ ¼ Q ¼ QI ðA:20Þfor any general matrix Q of conformable order with the multiplying identitymatrix.

A.4.4 Orthogonal Matrix

If ATA ¼ I, the identity matrix, then A is an orthogonal (or orthonormal, or

unitary) matrix. For example, as the reader may easily verify,

A

35

� �45

� �45

� � � 35

� � !

is an orthogonal matrix.

A.4.5 Symmetric and Skew-Symmetric Matrices

When every pair of (real) elements that are symmetrically placed in a matrixwith respect to the main diagonal are equal, the matrix is said to be a (real)symmetric matrix; that is, for a (real) symmetric matrix,

aij ¼ aji and i 6¼ j ðA:21ÞSome examples of symmetric matrices are

A ¼ 1 33 4

� �and B ¼

2 1 51 0 �25 �2 4

0@

1A

Thus, a symmetric matrix and its transpose are identical; that is

A ¼ AT ðA:22Þ

Algebra

A skew-symmetric matrix is one for which

aij ¼ �aji ðA:23ÞThe following are some examples of such matrices:

A ¼ 0 �33 0

� �and B ¼

0 1 5�1 0 �2�5 2 4

0@

1A

For skew-symmetric matrices,

A ¼ �AT ðA:24Þ

By noting that aij can be expressed as

aij ¼ 12aij þ aij� �þ 1

2aij � aijÞ� ðA:25Þ

it is easily established that every square matrix A can be resolved into thesum of a symmetric matrix A

0 and a skew-symmetric matrix A0; that is,

A ¼ A0 þ A

0 ðA:26Þ

A.4.6 Hermitian Matrices

When symmetrically situated elements in a matrix of complex numbers arecomplex conjugates, that is,

aji ¼ �aaij ðA:27Þthe matrix is said to be Hermitian. For example,

3 ð2� 3jÞð2þ 3jÞ 5

� �

is a Hermitian matrix. Note that the (real) symmetric matrix is a specialform of a Hermitian matrix in which the coefficients of j are absent.

A.4.7 Singular Matrices

A matrix whose determinant is zero is said to be singular. A matrix whosedeterminant is not zero is said to be nonsingular.

A.5 INVERSE OF A MATRIX

The matrix counterpart of division is the inverse operation. Thus, corre-sponding to the reciprocal of a number in scalar algebra is the inverse of the

matrix A, denoted by A�1 (read as ‘‘A inverse’’). A matrix for which both

AA�1 and A

�1A give I, the identity matrix, i.e.,

AA�1 ¼ A

�1A ¼ I ðA:28Þ

The inverse matrix A�1 does not exist for all matrices, it exists only if the

following hold:

1. A is square.2. Its determinant is not zero (i.e., the matrix is nonsingular).

The inverse of A is defined as

A�1 ¼ C

T

jAj ¼adjðAÞjAj ðA:29Þ

where C is the matrix of cofactors; that is,

C ¼C11 C12 � � � C1n

C21 C22 � � � C2n

..

. ... . .

. ...

Cn1 Cn2 � � � Cnn

0BBB@

1CCCA

as defined previously, Cij is the cofactor of the element aij of the matrix A.C

T , the transpose of this matrix of cofactors, is termed the adjoint of A andoften abbreviated as adj(A).

Clearly because jAj ¼ 0 for singular matrices, the matrix inverse asexpressed in Eq. (A.29) does not exist for such matrices.

The following should be noted:

1. The inverse of a matrix composed of a product of other matricesmay be written in the form

ðABCDÞ�1 ¼ D�1C

�1B�1A

�1

2. For the simple 2� 2 matrix, the general definition in Eq. (A.29)simplifies to a form easy to remember. Thus, given

A ¼ a11 a12a21 a22

� �

we have that jAj ¼ a11a22 � a12a21, and it is easily shown that

A�1 ¼ 1

jAja22 �a12�a21 a11

� �ðA:30Þ

Algebra

A.6 LINEAR ALGEBRAIC EQUATIONS

Much of the motivation behind developing matrix operations is due to theirparticular usefulness in handling sets of linear algebraic equations; forexample, the following three linear equations:

a11x1 þ a12x2 þ a13x3 ¼ b1

a21x1 þ a22x2 þ a23x3 ¼ b2

a31x1 þ a32x2 þ a33x3 ¼ b3

ðA:31Þ

The above equations contain the unknowns x1, x2, and x3. They can berepresented by the single matrix equation.

Ax ¼ b ðA:32Þwhere

A ¼a11 a12 a13a21 a22 a23a31 a32 a33

0@

1A; x ¼

x1x2x3

0@

1A; b ¼

b1b2b3

0@

1A ðA:33Þ

The solution to these equations can be obtained very simply by premultiply-ing Eq. (A.32) by A

�1 to yield

A�1Ax ¼ A

�1b

which becomes

x ¼ A�1b ðA:34Þ

upon recalling the property of a matrix inverse that A�1A ¼ I and recalling

Eq. (A.20) for the property of the identity matrix.There are several standard, computer-oriented numerical procedures

for performing the operation A�1 to produce the solution given in Eq.

(A.34). Indeed, the availability of these computer subroutines makes theuse of matrix techniques very attractive from a practical viewpoint.

It should be noted that the solution equation (A.34) exists only if A isnonsingular, as only then will A�1 exist. When A is singular, A�1 ceases toexist and Eq. (A.34) will be meaningless. The singularity of A, of course,implies that jAj ¼ 0, and from property 5 given for determinants, this indi-cates that some of the rows and columns of A are linearly dependent. Thus,the singularity of A means that only a subset of the equations we are tryingto solve are truly linearly independent. As such, linearly independent solu-tions can be found only for this subset. It can be shown that the actualnumber of such linearly independent solutions is equal to the rank of A.

Solved Example A.1

To illustrate the use of matrices in the solution of linear algebraic equations,let us consider the following set of equations:

2x1 þ 2x2 þ 3x3 ¼ 1

4x1 þ 5x2 þ 6x3 ¼ 3

7x1 þ 8x2 þ 9x3 ¼ 5

which in matrix notation is given by

Ax ¼ b

where A, b, and x are

A ¼2 2 34 5 67 8 9

0@

1A; b ¼

135

0@

1A; x ¼

x1x2x3

0@

1A

Notice that A is simply a (3� 3) matrix whose inverse can be computed as

A�1 ¼

1 �2 1�2 1 01 2

3 � 23

0@

1A

Thus, from Eq. (A.34), the solution to this set of algebraic equations isobtained as

x ¼ A�1b ¼

1 �2 1

�2 1 0

1 23

� 23

0BB@

1CCA

1

3

5

0BB@

1CCA ¼

1� 6þ 5

�2þ 3þ 0

1þ 63� 10

3

0BB@

1CCA ¼

0

1

� 13

0BB@

1CCA

Thus, x1 ¼ 0, x2 ¼ 1, and x3 ¼ � 13is the required solution.

A.7 EIGENVALUES AND EIGENVECTORS

The eigenvalue/eigenvector problem arises in the determination of the valuesof a constant � for which the following set of n linear algebraic equationshas nontrivial solutions:

a11x1 þ a12x2 þ � � � þ a1nxn ¼ �x1a21x1 þ a22x2 þ � � � þ a2nxn ¼ �x2... ..

. ... ..

. ...

an1x1 þ an2x2 þ � � � þ annxn ¼ �xn

ðA:35Þ

Algebra

This can be expressed as

Ax ¼ �x

or we can write

ðA� �IÞx ¼ 0 ðA:36ÞBeing a system of linear homogeneous equations, the solution to Eq. (A.35)or, equivalently, Eq. (A.36) will be nontrivial (i.e., x 6¼ 0Þ if any only if

jA� �Ij ¼ 0 ðA:37Þthat is, jA� �Ij is a singular matrix

Expanding this determinant results in a polynomial of order n in �:

a0�n þ a1�

n�1 þ a2�n�2 þ � � � þ an�1�þ an ¼ 0 ðA:38Þ

This equation [or its determinant from Eq. (A.37)] is called the characteristicequation of the n� n matrix A, its n roots ð�1; �2; �3; . . . ; �nÞ, which may bereal or imaginary and may not be distinct, are the eigenvalues of the matrix.

Solved Example A.2

Let us illustrate the computation of eigenvalues by considering the matrix

A ¼ 1 23 �4

� �

In this case, the matrix (A� �IÞ is obtained as

ðA� �IÞ ¼ 1 23 �4

� �� 0

0 �

� �¼ ð1� �Þ 2

3 ð�4� �Þ� �

from which the characteristic equation is obtained as

jA� �Ij ¼ �ð1� �Þð4þ �Þ � 6 ¼ �2 þ 3�� 10 ¼ 0

which is a second-order polynomial (quadratic) with the following solutions:

�1; �2 ¼ �5; 2

Thus, in this case, the eigenvalues are both real. Note that �1 þ �2 ¼ �3 andthat �1�2 ¼ �10; also note that Tr(AÞ ¼ 1� 4 ¼ �3 and jAj ¼ �10. Thus,we see that �1 þ �2 ¼ TrðAÞ and �1�2 ¼ jAj.

If we change A slightly to

A ¼ 1 2�3 4

� �

then the characteristic equation becomes

jA� �Ij ¼ 1� � 2�3 4� �

� �¼ �2 � 5�þ 10 ¼ 0

with roots

�1; �2 ¼ 125� ffiffiffiffiffi

15p� �

j �

;where j ¼ ffiffiffiffiffiffiffi�1p

which are complex conjugates. Note, again, that Tr(AÞ ¼ 5 ¼ �1 þ �2 andjAj ¼ 10 ¼ �1�2.

Finally, if we make another small change in A, that is,

A ¼ 1 2�3 �1

� �

then the characteristic equation is

jA� �Ij ¼ �2 � 1þ 6 ¼ 0

which has the solutions

�1; �2 ¼ �ffiffiffi5

p� �j

so the eigenvalues are purely imaginary in this case. Once again, Tr(AÞ ¼ 0 ¼�1 þ �2 and jAj ¼ 5 ¼ �1�2.

The following are some general properties of eigenvalues worth not-ing:

1. The sum of the eigenvalues of a matrix is equal to the trace of thatmatrix:

TrðAÞ ¼Xnj¼1

�j

2. The product of the eigenvalues of a matrix is equal to the deter-minant of that matrix:

jAj ¼Ynj¼1

�j

3. A singular matrix has at least one zero eigenvalue.4. If the eigenvalues of A are �1; �2; �3; . . . ; �n, then the eigenvalues

of A�1 are

1=�1; 1=�2; 1=�3; . . . ; 1=�n; respectively.

Algebra

5. The eigenvalues of a diagonal or a triangular matrix are identicalto the elements of the main diagonal.

6. Given any nonsingular matrix T, the matrices A and �AA ¼ TAT�1

have identical eigenvalues. In this case, such matrices A and �AA arecalled as similar matrices.

Eigenvalues are essential for the study of the stability of different systems.

Solved Example A.3

Let us illustrate the computation of eigenvectors by considering the follow-ing 2� 2 matrix:

A ¼ 1 23 �4

� �for which eigenvalues �1 ¼ �5 and �2 ¼ 2 were calculated in the previousexample.

For �1,

ðA� �1IÞ ¼ 6 23 1

� �

and its adjoint is given by

adjðA� �1IÞ ¼ 1 �2�3 6

� �

We may now choose either column of the adjoint as the eigenvector x1,because one is a scalar multiple of the other. Let us choose

x1 ¼ 1�3

� �as the eigenvector for �1 ¼ �5. Because the norm of this vector isffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi12 þ ð�3Þ2

q¼ ffiffiffiffiffi

10p

, the normalized eigenvector corresponding to �1 ¼ �5will be

�xx1 ¼1ffiffiffiffi10

p

�3ffiffiffiffi10

p

!

Similarly, for �2 ¼ 2,

ðA� �2IÞ ¼ �1 23 �6

� �and its adjoint is given by

adjðA� �2IÞ ¼ �6 �2�3 �1

� �

Thus,

x2 ¼ 21

� �

can be chosen as the eigenvector for eigenvalue �2 ¼ 2. The normalizedeigenvector in this case is

�xx2 ¼2ffiffi5

p

1ffiffi5

p

!

The following are some useful properties of eigenvectors:

1. Eigenvectors associated with separate, distinct eigenvalues arelinearly independent.

2. The eigenvectors xj are solutions of a set of homogeneous linearequations and are thus determined only up to a scalar multiplier;that is, xj and kxj will both be a solution to the equation if k is ascalar constant. We obtain a normalized eigenvector �xxj when theeigenvector xj is scaled by the vector norm kxjk; that is

�xxj ¼xj

kxjk3. The normalized eigenvector �xxj is a valid eigenvector because it is

a scalar multiple of xj . It enjoys the sometimes desirable propertythat it has unity norm; that is,

k �xxjk ¼ 1

4. The eigenvectors xj may be calculated by resorting to methods ofsolution of homogeneous linear equations.

It can be shown that the eigenvector xj associated with the eigenvalue�j is given by any nonzero column of adjðA� �jIÞ. It can also be shown thatthere will always be one and only one independent nonzero column ofadjðA� �jIÞ.

A.8 SOLUTION OF LINEAR DIFFERENTIALEQUATIONS AND THE MATRIX EXPONENTIAL

It is possible to use the results of matrix algebra to provide general solutionsto sets of linear ordinary differential equations. Let us recall a dynamicsystem with control and disturbances neglected:

Algebra

dx

dt¼ Ax with xð0Þ ¼ x0 ðA:39Þ

Here, A is a constant n� n matrix with distinct eigenvalues, and xðtÞ is atime-varying solution vector. Let us define a new set of n variables, zðtÞ, as

zðtÞ ¼ M�1xðtÞ ðA:40Þ

so that

xðtÞ ¼ MzðtÞ ðA:41ÞHere, M is the modal matrix (having the property that matrix M

�1AM is

diagonal) formed of the eigenvectors of matrix A. Substituting Eq. (A.41)into Eq. (A.39) gives

MdzðtÞdt

¼ AMzðtÞ with zð0Þ ¼ M�1x0 ðA:42Þ

Premultiplying both sides by M�1 gives

M�1M

dzðtÞdt

¼ M�1AMzðtÞ

or

dzðtÞdt

¼ �zðtÞ with zð0Þ ¼ z0 ¼ M�1x0 ðA:43Þ

where , ¼ M�1AM

Because � is diagonal, these equations may now be written as

dz1ðtÞdt

¼ �1z1ðtÞ with z1ð0Þ ¼ z10

dz2ðtÞdt

¼ �2z2ðtÞ with z2ð0Þ ¼ z20

..

.

dznðtÞdt

¼ �nznðtÞ with znð0Þ ¼ zn0

Thus, the transformation equation (A.41) has converted the original equa-tion (A.39) into n completely decoupled equations, each of which has asolution of the form

zjðtÞ ¼ e�1tzj0; j ¼ 1; 2; 3; . . . ; n

This may be expressed in matrix notation as

zðtÞ ¼ e�tz0 ðA:44Þ

where

e�t ¼

e�1t 0 0 � � � 0

0 e�2t 0 � � � 0

0 0 e�2t � � � 0

..

. ... ..

. . .. ..

.

0 0 0 0 e�nt

0BBBBBBB@

1CCCCCCCA

ðA:45Þ

It is the matrix exponential for the diagonal matrix �. Equation (A.44) maybe put back in terms of the original variables xðtÞ by using Eq. (A.40):

M�1xðtÞ ¼ e�t

M�1x0

or

xðtÞ ¼ Me�tM

�1x0 ðA:46Þ

This is often written as

xðtÞ ¼ eAtx0 ðA:47Þwhere the matrix eAt is defined as

eAt ¼ MeAtM�1 ðA:48ÞIt is called the matrix exponential of At. It is very important to note that Eqs.(A.47) and (A.48) provide the general solution to any set of homogeneousequations (A.39) provided A has distinct eigenvalues.

Solved Example A.4

To illustrate the solution of linear constant-coefficient, homogeneous differ-ential equations, let us consider the following systems:

dx

dt¼ Ax with xð0Þ ¼ x0 ðA:49Þ

where

A ¼ 1 23 �4

� �and x0 ¼ 1

2

� �

Observe that A is the same matrix we have analyzed in previous examples,so we already know its eigenvalues (�1 and �2) and modal matrix M. Thus,we have that

� ¼ �5 00 2

� �; M ¼ 1 2

�3 1

� �; M

�1 ¼17

� 27

37

17

!

Algebra

Using Eq. (A.48), we get

eAt ¼ Me�tM

�1 ¼ 1 2�3 1

� �e�5t 00 e2t

� � 17

� 27

37

17

!

Thus, we get

eAt ¼e�5t þ 6e2t

7

�2e�5t þ 2e2t

7

�3e�5t þ 3e2t

7

6e�5t þ e2t

7

0BBB@

1CCCA

Finally, we get

xðtÞ ¼ x1ðtÞx2ðtÞ

� �¼

�3e�5t þ 10e2t

7

9e�5t þ 5e2t

7

0BBB@

1CCCA

The extension of these ideas provides solutions for the case of nonhomoge-neous linear differential equations

dxðtÞdt

¼ AxðtÞ þ BuðtÞ þ !dðtÞ with xðt0Þ ¼ x0 ðA:50Þ

The method of obtaining the solution is straightforward. Here, uðtÞ and dðtÞrepresent the control variables and disturbances, respectively. When thematrices A, B, and ! are constant matrices, a derivation completely similarto the one just given leads to the solution

xðtÞ ¼ eAðt�t0Þx0 þðtt0

eAðt��ÞBuð�Þ þ !dð�Þ½ � d� ðA:51Þ

where the matrix exponentials eAðt�t0Þ and eAðt��Þ can be evaluated using Eq.(A.48).

Solved Example A.5

To illustrate the solution of nonhomogeneous equations using Eq. (A.51),let us suppose we have a system equation (A.50) with A the same as theprevious example,

A ¼ 1 23 �4

� �

and

B ¼ 1 0

0 1

� �; ! ¼ 1

0

� �; x0 ¼

1

2

� �;

u ¼ u1

u2

� �; d ¼ d1 (a scalar)

In addition, assume that u1 ¼ u2 ¼ 0, d1 ¼ 1 (a constant disturbance), andt0 ¼ 0. In this case, Eq. (A.51) takes the form

xðtÞ ¼ eAtx0 þðt0

eAðt��Þ!dð�Þ d�

Substituting for eAt and eAtx0 calculated in the previous example, we obtainthe solution as follows:

xðtÞ ¼�3e�5t þ 10e2t

7

9e�5t þ 5e2t

7

0BBB@

1CCCAþ

ðt0

e�5ðt��Þ þ 6e2ðt��Þ

7

�3e�5ðt��Þ þ 3e2ðt��Þ

7

0BBB@

1CCCAd�

or we can write

xðtÞ ¼ x1ðtÞx2ðtÞ

� �¼

� 2

5� 16

35e�5t þ 13

7e2t

� 21

70þ 48

35e�5t þ 13

14e2t

0BB@

1CCA

For the situation where the matrices A, B, and ! can themselves be func-tions of time, the solution requires a bit more computation. In this case,

xðtÞ ¼ (ðt; t0Þx0 þ(ðt; t0Þðtt0

(�1ð�; t0Þ Buð�Þ þ !dð�Þ½ � d� ðA:52Þ

Here, (ðt; t0Þ is an n� n time-varying matrix known as the fundamentalmatrix solution, which may be found from

d(ðt; t0Þdt

¼ AðtÞ(ðt; t0Þ with (ðt0; t0Þ ¼ I ðA:53Þ

Because AðtÞ, BðtÞ, and !ðtÞ can have arbitrary time dependence, the funda-mental matrix solution is usually obtained by numerical methods.Nevertheless, once (ðt; t0Þ is determined, the solution xðtÞ may be readilycalculated for a variety of control and disturbance inputs uðtÞ and dðtÞ usingEq. (A.52).

Algebra

REFERENCES

1. Kreyszig, E. Advanced Engineering Mathematics, 8th ed. Wiley, New York,

1998.

2. Amundson, N. R. Mathematical Method in Chemical Engineering. Prentice-

Hall, Englewood Cliffs, NJ. 1966.

3. Wylie, C. R. Advanced Engineering Mathematics. Wiley, New York, 1966.

4. Duncan, W. J., Collar, A. R., and Frazer, R. Q. Elementary Matrices,

Cambridge University Press, Cambridge, 1963.

Appendix B

Numerical Methods

Analytical solution methods are restricted to linear, low-dimensionalproblems. For nonlinear problems and/or high-dimensional systems, anumerical solution must be used.

B.1 SOLUTION OF ALGEBRAIC ANDTRANSCENDENTAL EQUATIONS

To find the roots of an equation f ðxÞ ¼ 0, we start with a known approx-imate solution and apply any of the following methods.

B.1.1 Bisection Method

This method consists of locating the root of the equation f ðxÞ ¼ 0 between aand b. If f ðxÞ is continuous between a and b and f ðaÞ and f ðbÞ are ofopposite signs, then there is a root between a and b. For example, let f ðaÞbe negative and f ðbÞ be positive. Then, the first approximation to the root isx1 ¼ 1

2ðaþ bÞ.If f ðx1Þ ¼ 0, then x1 is a root of f ðxÞ ¼ 0. Otherwise, the root lies

between a and x1, or x1 and b according to whether f ðx1Þ is positive ornegative. Then, we bisect the interval as previously and continue the processuntil the root is found to the desired accuracy.

In Figure B.1, f ðx1Þ is positive and f ðaÞ is negative; therefore, the rootlies between a and x1. Then, the second approximation to the root is

x2 ¼ 12ðaþ x1Þ. If f ðx2Þ is negative, the root lies between x1 and x2. Then, the

third approximation to the root is x3 ¼ 12ðx1 þ x2Þ, and so on.

B.1.2 Newton–Raphson Method

Let x0 be an approximation root of the equation f ðxÞ ¼ 0. If x1 ¼ x0 þ h isthe exact root, then f ðx1Þ ¼ 0. Therefore, expanding f ðx0 þ hÞ by Taylor’sseries gives

f ðx0Þ þ hf 0ðx0Þ þh2

2!f 00ðx0Þ þ � � � ¼ 0

Because h is very small, neglecting h2 and higher powers of h, we get

f ðx0Þ þ hf 0ðx0Þ ¼ 0 or h ¼ � f ðx0Þf 0ðx0Þ

ðB:1Þ

Therefore, a closer approximation to the root is given by

x1 ¼ x0 �f ðx0Þf 0ðx0Þ

Similarly, starting with x1, a still better approximation x2 is given by

x2 ¼ x1 �f ðx1Þf 0ðx1Þ

In general,

xnþ1 ¼ xn �f ðxnÞf 0ðxnÞ

Figure B.1 Bisection method.

This technique is known as the Newton–Raphson formula or Newton’siteration formula.

Geometrical Interpretation of Newton–Raphson Technique (seeFig. B.2)

Let x0 be a point near the root � of the equation f ðxÞ ¼ 0. Then, theequation of the tangent point A0½x0; f ðx0Þ� is y� f ðx0Þ ¼ ½ f 0ðx0Þðx� x0Þ�.This tangent cuts the x axis at x1 ¼ x0 � f ðx0Þ=f 0ðx0Þ. This point is a firstapproximation to the root �. If A1 is the point corresponding to x1 on thecurve, then the tangent at A1 will cut the x axis at x2, which is closer to � andis, therefore, a second approximation to the root. Repeating this process, weapproach the root � quite rapidly.

B.2 SOLUTION OF LINEAR SIMULTANEOUSEQUATIONS

Simultaneous linear equations occur in various engineering problems. Thereader knows that a given system of linear equations can be solved byCramer’s rule or by the matrix method. However, these methods becometedious for large systems. However, there exist other numerical methods ofsolution which are well suited for computing machines. The following is anexample.

Figure B.2 Geometrical interpretation of the Newton–Raphson method.

Gauss Elimination Method

In this method, the unknowns are eliminated successively and the system isreduced to an upper-triangular system from which the unknowns are foundby back substitution. The method is quite general and is well adapted forcomputer calculations. Here, we shall explain it by considering a system ofthree equations for sake of simplicity.

Consider the following equations,

a1xþ b1yþ c1z ¼ d1



ðB:2Þ

Step I. To eliminate x from second and third equations. Assuminga1 6¼ 0, we eliminate x from the second equation by subtractinga2=a1 times the first equation from the second equation. Similarly, weeliminate x from the third equation by eliminating ða3=a1Þ timesthe first equation form the third equation. We thus obtain the newsystem


b 02yþ c 02z ¼ d 0

2

b 03yþ c 03z ¼ d 0

3

ðB:3Þ

Here, the first equation is called the pivotal equation and a1 is calledthe first pivot.

Step II. To eliminate y from the third equation in Eq. (B.3). Assumingb 02 6¼ 0, we eliminate y from the third equation of Eq. (B.3), by

subtracting ðb 03=b

02Þ times the second equation from the third equa-

tion. Thus, we get the new system


b 02yþ c 02z ¼ d 0

2

c 003 z ¼ d 003

ðB:4Þ

Here, the second equation is the pivotal equation and b 02 is the new

pivot.

Step III. To evaluate the unknowns. The values of x, y, and z arefound from the reduced system (B.4) by back-substitution.

Note: Clearly, the method will fail if any one of the pivots a1, b02, or c 003

becomes zero. In such cases, we rewrite the equations in a different order sothat the pivots are nonzero.

B.3 SOLUTION OF NONLINEAR SIMULTANEOUSEQUATIONS (MULTIDIMENSIONAL NEWTON–RAPHSON METHOD)

Consider the equations

f ðx; yÞ ¼ 0 and gðx; yÞ ¼ 0 ðB:5ÞIf an initial approximation ðx0; y0Þ to a solution has been found by thegraphical method or otherwise, then a better approximation (x1; y1) canbe obtained as follows:

Let

x1 ¼ x0 þ h and y1 ¼ y0 þ k

so that

f ðx0 þ h; y0 þ kÞ ¼ 0 and gðx0 þ h; y0 þ kÞ ¼ 0 ðB:6ÞExpanding each of the functions in Eq. (B.6) by the Taylor’s series to first-degree terms, we get approximately

f0 þ h@f

@x0þ k

@f

@y0¼ 0

g0 þ h@g

@x0þ k

@g

@y0¼ 0

ðB:7Þ

where

f0 ¼ f ðx0; y0Þ;@f

@x0¼ @f

@x

� �x0;y0

and so forth

Solving Eqs. (B.7) for h and k, we get a new approximation to the root:

x1 ¼ x0 þ h and y1 ¼ y0 þ k

This process is repeated until we obtain the values of x and y with thedesired accuracy.

Solved Example B.1

Solve the system of nonlinear equations

x2 þ y ¼ 11 and y2 þ x ¼ 7

Solution

An initial approximation to the solution is obtained from a rough graph ofthe given equation as x0 ¼ 3:5 and y0 ¼ �1:8. We have f ¼ x2 þ y� 11 andg ¼ y2 þ x� 7, so that

@f

@x¼ 2x;

@f

@y¼ 1;

@g

@x¼ 1;

@g

@y¼ 2y

Then, Newton–Raphson’s equation (B.7) will be

7hþ k ¼ 0:55 and h� 3:6k ¼ 0:26

Solving the above set of algebraic equations, we get

h ¼ 0:0855 and k ¼ �0:0485

Therefore, the better approximation to the root is

x1 ¼ x0 þ h ¼ 3:5855 and y1 ¼ y0 þ k ¼ �1:8485

Repeating the above process, replacing ðx0; y0Þ by ðx1; y1Þ, we obtain

x2 ¼ 3:5844 and y ¼ �1:8482

This procedure is to be repeated till x and y are located to the desiredaccuracy.

B.4 NUMERICAL SOLUTION OF NONLINEARORDINARY DIFFERENTIAL EQUATIONS

A number of numerical methods are available for the solution of nonlinearordinary differential equations of the form

dY

dx¼ f ðx;YÞ with given Yðx0Þ ¼ Y0 ðB:8Þ

B.4.1 Euler’s Method

Consider the equation

dy

dx¼ f ðx; yÞ with yðx0Þ ¼ y0 ðB:9Þ

Its curve of solution through Pðx0; y0Þ is shown in Figure B.3. Now, we haveto find the ordinate of any other point Q on this curve.

Let us divide LM into n subintervals, each of width h at L1;L2;L3; . . .so that h is quite small. In the interval LL1, we approximate the curve by the

tangent at P. If the ordinate through L1 meets this tangent in P1ðx0 þ h; y1Þ,then

y1 ¼ L1P1 ¼ LPþ R1P1 ¼ y0 þ PR1 tan ¼ y0 þ hdy

dx

� �P

¼ y0 þ hf ðx0; y0Þ

Let P1Q1 be the curve of the solution of Eq. (B.9) through P1 and let itstangent at P1 meet the ordinate through L2 in P2ðx0 þ 2h; y2Þ. Then

y2 ¼ y1 þ h f ðx0 þ h; y1Þ ðB:10Þ

Repeating this process n times, we finally reach an approximation MPn ofMQ given by

yn ¼ yn�1 þ hf x0 þ ðn� 1Þh; yn�1ð Þ

This is the Euler’s method of finding an approximate solution of (B.9).

Note: In Euler’s method, we approximate the curve to solution by thetangent in each interval (i.e., by a sequence of short lines). Unless h isvery small, the error is bound to be quite significant. This sequence oflines may also deviate considerably from the curve of solution. Hence,there is a modification of this method, which is given in the next section.

Figure B.3 Geometrical representation of Euler’s method.

B.4.2 Modified Euler’s Method

In Euler’s method, the curve of solution in the interval LL1 is approximatedby the tangent at P (see Fig. B.3), such that at P1 we have

y1 ¼ y0 þ h f ðx0; y0Þ ðB:11ÞThen, the slope of the curve of solution through P1 [i.e.,ðdy=dxÞP1

¼ f ðx0 þ h; y1Þ] is computed and the tangent at P1 to P1Q1 isdrawn, meeting the ordinate through L2 in P2ðx0 þ 2h; y2Þ.

Now, we find a better approximation yð1Þ1 of yðx0 þ hÞ by taking the

slope of the curve as the mean of the slopes of the tangents at P and P1; thatis,

yð1Þ1 ¼ y0 þ

h

2f ðx0; y0Þ þ f ðx0 þ h; y1Þ½ � ðB:12Þ

As the slope of the tangent at P1 is not known, we use y1 as computed fromEq. (B.11) by Euler’s method and insert it on the right-hand side of Eq.(B.12) to obtain the first modified value of y

ð1Þ1 . Equation (B.11) is, therefore,

called the predictor, whereas Eq. (B.12) serves as the corrector of y1.Again, a corrector is applied and we find a still better value of y

ð1Þ1

corresponding to L1:

yð2Þ1 ¼ y0 þ

h

2f ðx0; y0Þ þ f ðx0 þ h; yð1Þ1 Þh i

We repeat this step until two consecutive values of y agree. This is taken asthe starting point for the next interval L1L2.

Once y1 is obtained to the desired degree of accuracy, y correspondingto L2 is found from the predictor,

y2 ¼ y1 þ hf ðx0 þ h; y1Þand a better approximation, y

ð1Þ2 , is obtained from the corrector,

yð1Þ2 ¼ y1 þ

h

2f ðx0 þ h; y1Þ þ f ðx0 þ 2h; y2Þ½ �

We repeat this step until y2 becomes stationary. Then, we proceed to calcu-late y3 as above.

This is the modified Euler’s method, which is a relatively simplepredictor–corrector method.

Solved Example B.2

Using the modified Euler method, find an approximate value of y whenx ¼ 0:3 for the following differential equation:

dy

dx¼ xþ y and y ¼ 1 when x ¼ 0

Solution

Taking h ¼ 0:1, the various calculations are arranged as shown in Table B.1,Hence,

yð0:3Þ ¼ 1:4004 approximately.

B.4.3 Runge–Kutta Method

The class of methods known as Runge–Kutta methods do not require thecalculations of higher-order derivatives. These methods agree with Taylor’sseries solution up to the terms in hr, where r differs from method to methodand is called the order of that method. The fourth-order Runge–Kuttamethod is most commonly used and is often referred to as the ‘‘Runge–Kutta method’’ only.

Table B.1 Calculations of Solved Example B.2

x xþ y ¼ y 0 Mean slope Old yþ 0:1 (mean slopeÞ ¼ new y

0.0 0þ 1 — 1:00þ 0:1ð1:00Þ ¼ 1:10

0.1 0:1þ 1:1 12ð1þ 1:2Þ 1:00þ 0:1ð1:10Þ ¼ 1:11

0.1 0:1þ 1:11 12ð1þ 1:21Þ 1:00þ 0:1ð1:105Þ ¼ 1:1105

0.1 0:1þ 1:1105 12ð1þ 1:2105Þ 1:00þ 0:1ð1:1052Þ ¼ 1:1105

0.1 1.2105 — 1:1105þ 0:1ð1:2105Þ ¼ 1:2316

0.2 0:2þ 1:2316 12ð1:2105þ 1:4316Þ 1:1105þ 0:1ð1:3211Þ ¼ 1:2426

0.2 0:2þ 1:426 12ð1:2105þ 1:4426Þ 1:1105þ 0:1ð1:3266Þ ¼ 1:2432

0.2 0:2þ 1:2432 12ð1:2105þ 1:4432Þ 1:1105þ 0:1ð1:3268Þ ¼ 1:2432

0.2 1.4432 — 1:2432þ 0:1ð1:4432Þ ¼ 1:3875

0.3 0:3þ 1:3875 12ð1:4432þ 1:6875Þ 1:2432þ 0:1ð1:5654Þ ¼ 1:3997

0.3 0:3þ 1:3997 12ð1:4432þ 1:6997Þ 1:2432þ 0:1ð1:5715Þ ¼ 1:4003

0.3 0:3þ 1:4003 12ð1:4432þ 1:7003Þ 1:2432þ 0:1ð1:5718Þ ¼ 1:4004

0.3 0:3þ 1:4004 12ð1:4432þ 1:7004Þ 1:2432þ 0:1ð1:5718Þ ¼ 1:4004

Working Rule

Finding the increment k of y corresponding to an increment h of x by theRunge–Kutta method from

dy

dx¼ f ðx; yÞ; yðx0Þ ¼ y0

is as follows:Calculate successively

k1 ¼ hf ðx0; y0Þ

k2 ¼ hf x0 þh

2; y0 þ

k12

� �

k3 ¼ hf x0 þh

2; y0 þ

k22

� �

and

k4 ¼ hf ðx0 þ h; y0 þ k3Þ

Finally, calculate

k ¼ 16k1 þ 2k2 þ 2k3 þ k4½ �

which gives the required approximate value y1 ¼ y0 þ k.

Notes:

1. k is the weighted mean of k1, k2, k3, and k4.2. One of the advantages of these methods is that the operation is

identical whether the differential equation is linear or nonlinear.

Solved Example B.3

Apply the Runge–Kutta method to find an approximate value of y forx ¼ 0:2 in steps of 0.1 for the following differential equation:

dy

dx¼ xþ y2 and y ¼ 1 when x ¼ 0

Solution

Here, we have taken h ¼ 0:1 and carry out the calculations in two steps.

Step 1

x0 ¼ 0; y0 ¼ 1; h ¼ 0:1

k1 ¼ hf ðx0; y0Þ ¼ 0:1f ð0; 1Þ ¼ 0:100

k2 ¼ hf x0 þh

2; y0 þ

k12

� �¼ 0:1f ð0:05; 1:1Þ ¼ 0:1152

k3 ¼ hf x0 þh

2; y0 þ

k22

� �¼ 0:1f ð0:05; 1:1152Þ ¼ 0:1168

and

k4 ¼ hf ðx0 þ h; y0 þ k3Þ ¼ 0:1f ð0:1; 1:1168Þ ¼ 0:1347

Finally, we have

k ¼ 16k1 þ 2k2 þ 2k3 þ k4½ �


k ¼ 16 0:100þ 0:2304þ 0:2336þ 0:1347½ � ¼ 0:1165

giving

yð0:1Þ ¼ y0 þ k ¼ 1:1165

Step 2

x1 ¼ x0 þ h ¼ 0:1; y1 ¼ 1:165; h ¼ 0:1

k1 ¼ hf ðx1; y1Þ ¼ 0:1f ð0:1; 1:1165Þ ¼ 0:1347

k2 ¼ hf x1 þh

2; y1 þ

k12

� �¼ 0:1f ð0:15; 1:1838Þ ¼ 0:1551

k3 ¼ hf x1 þh

2; y1 þ

k22

� �¼ 0:1f ð0:15; 1:194Þ ¼ 0:1576

and

k4 ¼ hf ðx1 þ h; y1 þ k3Þ ¼ 0:1f ð0:2; 1:1576Þ ¼ 0:1823

Finally, we have

k ¼ 16k1 þ 2k2 þ 2k3 þ k4½ �


k ¼ 160:134þ 0:1551þ 0:1576þ 0:1823½ � ¼ 0:1571

giving

yð0:2Þ ¼ y1 þ k ¼ 1:2736

Hence, the required approximate value of y is equal to 1.2736.

Predictor–Corrector Methods

In the methods explained so far, to solve a differential equation over aninterval ðx1; xiþ1Þ only the value of y at the beginning of the interval wasrequired. In the predictor–corrector methods, however, four prior values arerequired for finding the value of y at xiþ1. A predictor formula is used topredict the value of y at xiþ1 and then a corrector formula is applied toimprove this value. We now explain one such method.

Milne’s Method

Given

dy

dx¼ f ðx; yÞ; yðx0Þ ¼ y0

To find an approximate value of y for x ¼ x0 þ nh by Milne’s method forthe above-mentioned differential equation, we proceed as follows:

The value y0 ¼ yðx0Þ being given, we compute

y1 ¼ yðx0 þ hÞ; y2 ¼ yðx0 þ 2hÞ; y3 ¼ yðx0 þ 3hÞby Picard’s or Taylor’s series method. Next, we calculate

f0 ¼ f ðx0; y0Þ; f1 ¼ f ðx0 þ h; y1Þ; f2 ¼ f ðx0 þ 2h; y2Þ;f3 ¼ f ðx0 þ 3h; y3Þ

Then, to find y4 ¼ yðx0 þ 4hÞ, we substitute Newton’s forward interpolationformula

f ðx; yÞ ¼ f0 þ n�f0 þnðn� 1Þ

2�2f0 þ

nðn� 1Þðn� 2Þ6

�3f0 þ � � �

in the relation

y4 ¼ y0 þðx0þ4h

x0

f ðx; yÞ dx

We get

y4 ¼ y0 þðx0þ4h

x0

f0 þ n�f0 þnðn� 1Þ

2�2f0 þ

nðn� 1Þðn2Þ6

�3f0 þ � � ��

dx

Setting x ¼ x0 þ nh and dx ¼ h dn, we get

y4 ¼ y0 þ h

ð40

f0 þ n�f0 þnðn� 1Þ

2�2f0 þ

nðn� 1Þðn� 2Þ6

�3f0 þ � � ��

dn

Thus, we get

y4 ¼ y0 þ h 4f0 þ 8�f0 þ20

3�2f0 þ

8

3�3f0 þ � � �

� �

Neglecting fourth- and higher-order differences and expressing �f0, �2f0,

and �3f0 in terms of the function values, we get

y4 ¼ y0 þ4h

32f1 � f2 þ 2f3ð Þ

which is called a predictor. Having found y4, we obtain a first approximationto f4 ¼ f ðx0 þ 2h; y4Þ.

Then, a better value of y4 is found by Simpson’s rule as

y4 ¼ y2 þh

3f2 þ 4f3 þ f4ð Þ

which is called a corrector. Then, an improved value of f4 is computed and,again, the corrector is applied to find a still better value of y4. We repeat thisstep until y4 remains unchanged. Once y4 and f4 are obtained to the desireddegree of accuracy, y5 ¼ yðx0 þ 5hÞ is found from the predictor as

y5 ¼ y1 þ4h

32f2 � f3 þ 2f4ð Þ

and f5 ¼ f ðx0 þ 5h; y5Þ is calculated. Then, a better approximation to thevalue of y5 is obtained form the corrector as

y5 ¼ y3 þh


We repeat this step until y5 becomes stationary and then we proceed tocalculate y6, and so on.

This is Milne’s predictor–corrector method. To ensure greater accu-racy, we must first improve the accuracy of the starting values and thensubdivide the intervals.

Solved Example B.4

Apply Milne’s method to find a solution of the differential equation dy=dx ¼x� y2 in the range 0 � x � 1 for the initial condition y ¼ 0 at x ¼ 0.

Solution

Using Picard’s method, we have

y ¼ yð0Þ þðx0

f ðx; yÞ dx

where f ðx; yÞ ¼ x� y2. To get the first approximation, we put y ¼ 0 inf ðx; yÞ, thus giving

y1 ¼ 0þðx0

x dx ¼ x2

2

To find the second approximation, we put y ¼ x2=2 in f ðx; yÞ, thus giving

y2 ¼ðx0

x� x4

4

!dx ¼ x2

2� x5

20

Similarly, the third approximation is

y3 ¼ðx0

x� x2

2� x5

20

!224

35 dx ¼ x2

2� x5

20þ x8

160� x11

4400ðiÞ

Now, let us determine the starting values of the Milne’s method from (i) bychoosing h ¼ 0:2; we get

x ¼ 0:0; y0 ¼ 0:0000; f0 ¼ 0:0000

x ¼ 0:2; y1 ¼ 0:0200; f1 ¼ 0:1996

x ¼ 0:4; y2 ¼ 0:0795; f2 ¼ 0:3937

x ¼ 0:6; y3 ¼ 0:1762; f3 ¼ 0:5689

Using the predictor,

y4 ¼ y0 þ4h

32f1 � f2 þ 2f3ð Þ

we get

x ¼ 0:8; y4 ¼ 0:3049; f4 ¼ 0:7070

and the corrector is given by

y4 ¼ y2 þh


which gives

y4 ¼ 0:3046; f4 ¼ 0:7072 ðiiÞAgain using the corrector, y4 ¼ 0:3046, which is the same as in (ii).

Now using the predictor,

y5 ¼ y1 þ4h

32f2 � f3 þ 2f4ð Þ

we get

x ¼ 1:0; y5 ¼ 0:4554; f4 ¼ 0:7926

and the corrector

y5 ¼ y3 þh


which gives

y5 ¼ 0:4555; f5 ¼ 0:7925

Again, using the corrector, y5 ¼ 0:4555, a value which is the same pre-viously. Hence, yð1Þ ¼ 0:4555.

B.5 SOLUTION OF SIMULTANEOUS FIRST-ORDERDIFFERENTIAL EQUATIONS

The simultaneous differential equations of the type

dy

dx¼ f ðx; y; zÞ and

dz

dx¼ �ðx; y; zÞ

with initial condition

yðx0Þ ¼ y0 and zðx0Þ ¼ z0

can be solved by the methods discussed earlier, especially Picard’s orRunge–Kutta methods. The method is best illustrated by the followingexample.

Solved Example B.5

Using Picard’s method, find the approximate value of y and z correspondingto x ¼ 0:1 given that

dy

dx¼ xþ z and

dz

dx¼ x� y2

with yð0Þ ¼ 2 and zð0Þ ¼ 1.

Solution

Here,

x0 ¼ 0; y0 ¼ 2; z0 ¼ 1

Also,

dy

dx¼ f ðx; y; zÞ ¼ xþ z and

dz

dx¼ �ðx; y; zÞ ¼ x� y2

Therefore,

y ¼ y0 þðxx0

f ðx; y; zÞ dx and z ¼ z0 þðxx0

�ðx; y; zÞ dx

Firstly approximations are

y1 ¼ y0 þðxx0

f ðx; y0; z0Þ dx ¼ 2þðx0

ðxþ 1Þ dx ¼ 2þ xþ x2

2

z1 ¼ z0 þðxx0

�ðx; y0; z0Þ dx ¼ 1þðx0

ðx� 4Þ dx ¼ 1� 4xþ x2

2

Second approximations are

y2 ¼ y0 þðxx0

f ðx; y1; z1Þ dx ¼ 2þðx0

xþ 1� 4xþ x2

2

!dx

¼ 2þ x� 3

2x2 þ x3

6

z2 ¼ z0 þðxx0

�ðx; y1; z1Þ dx ¼ 1þðx0

x� 2þ xþ x2

2

!224

35 dx

¼ 1� 4xþ 3

2x2 � x3 � x4

4� x5

20

Third approximations are

y3 ¼ y0 þðxx0

f ðx; y2; z2Þ dx ¼ 2þ x� 3

2x2 � x3

2� x4

4� x5

20� x6

120

z3 ¼ z0 þðxx0

�ðx; y2; z2Þ dx ¼ 1� 4xþ 3

2x2 þ 5

3x3 þ 7

12x4

� 31

60x5 þ x6

12� x7

252

and so on.

When

x ¼ 0:1

y1 ¼ 2:015; y2 ¼ 2:08517; y3 ¼ 2:08447

z1 ¼ 0:605; z2 ¼ 0:58397; z3 ¼ 0:58672

Then

yð0:1Þ ¼ 2:0845 and zð0:1Þ ¼ 0:5867

correct up to four decimal places.

B.6 SECOND-ORDER DIFFERENTIAL EQUATIONS

Consider the second-order differential equation

d2y

dx2¼ f x; y;

dy

dx

� �

By writing dy=dx ¼ z, it can be reduced to two first-order differential equa-tions:

dy

dx¼ z and

dz

dx¼ f ðx; y; zÞ

These equations can be solved as explained in the previous method (forsimultaneous first-order differential equations).

Solved Example B.6

Using the Runge–Kutta method, solve the following second-order differen-tial equation for x ¼ 0:2, correct up to four decimal places:

d2y

dx2¼ x

dy

dx

� �2

�y2

with initial conditions x ¼ 0, y ¼ 1, and dy=dx ¼ 0.

Solution

Let

dy

dx¼ z ¼ f ðx; y; zÞ

Then,

dz

dx¼ xz2 � y2 ¼ �ðx; y; zÞ

We have

x0 ¼ 0; y0 ¼ 1; z0 ¼ 0; h ¼ 0:2

Using k1; k2; k3; . . . for f ðx; y; zÞ and l1; l2; l3; . . . for �ðx; y; zÞ, Runge–Kuttaformulas become (as shown in Table B.2).

Hence at x ¼ 0:2,

y ¼ y0 þ k ¼ 1� 0:0199 ¼ 0:9801

y 0 ¼ z ¼ z0 þ l ¼ 1� 0:1970 ¼ 0:803

Table B.2 Calculations of Solved Example B.6

k1 ¼ hf ðx0; y0; z0Þ¼ 0:2ð0Þ ¼ 0

l1 ¼ h�ðx0; y0; z0Þ¼ 0:2ð�1Þ ¼ �0:2

k2 ¼ hf x0 þh

2; y0 þ

k12; z0 þ

l12

� �

¼ 0:2ð�0:1Þ ¼ �0:02

l2 ¼ h� x0 þh

2; y0 þ

k12; z0 þ

l12

� �

¼ 0:2ð�0:999Þ ¼ �0:1998

k3 ¼ hf x0 þh

2; y0 þ

k22; z0 þ

l22

� �

¼ 0:2ð�0:0999Þ ¼ �0:02

l3 ¼ h� x0 þh

2; y0 þ

k22; z0 þ

l22

� �

¼ 0:2ð�0:9791Þ ¼ �0:1958

k4 ¼ hf ðx0 þ h; y0 þ k3; z0 þ l3Þ¼ 0:2ð�0:1958Þ ¼ �0:0392

l4 ¼ h�ðx0 þ h; y0 þ k3; z0 þ l3Þ¼ 0:2ð�0:9527Þ ¼ �0:1905

k ¼ 16ðk1 þ 2k2 þ 2k3 þ k4Þ

¼ �0:0199

l ¼ 16ðl1 þ 2l2 þ 2l3 þ l4Þ

¼ �0:1970

Appendix C

Analytical Solution of DifferentialEquations

C.1 ANALYTICAL SOLUTION OF LINEAR ORDINARYDIFFERENTIAL EQUATIONS

C.1.1 Homogeneous

d2y

dt2þ a

dy

dtþ by ¼ 0 ðC:1Þ

The solution will be a linear combination (addition) of terms in the follow-ing form:

y ¼ e�t e.g., y ¼ C1e�1t þ C2e

�2t� � ðC:2Þ

From Eq. (C.2), we get

dy

dt¼ �e�t and

d2y

dt2¼ �2e�t ðC:3Þ

Substituting Eqs. (C.2) and (C.3) in Eq. (C.1), we get

�2e�t þ a�e�t þ be�t ¼ 0

On simplifying, we get

e�t �2 þ a�þ bÞ ¼ 0�

which gives

�2 þ a�þ b ¼ 0

The solutions of the above quadratic equation is given by

�1;2 ¼�a�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 � 4b

p

2ðC:4Þ

So the solution is given by

y ¼ C1e�1t þ C2e

�2t

where C1 and C2 are obtained from the initial or boundary conditions.

C.1.2 Types of Roots (k1; k2)

1. �1 and �2 are real and distinct.2. When a2 � 4b ¼ 0, we get repeated roots

�1;2 ¼ � a

2¼ �

Then, the solution is given by

y ¼ ðC1 þ C2tÞe�2t

3. When 4b > a2, we get complex roots

�1 ¼ � a

2þ i! and �2 ¼ � a

2� i!

where

i ¼ffiffiffiffiffiffiffi�1

p

! ¼ffiffiffiffiffiffiffiffiffiffiffiffiffib� a2

4

r

The solution is given by

y ¼ C1eð�ða=2Þþi!Þt þ C2e

ð�ða=2Þ�i!Þt

On rearranging, we get

y ¼ C1 e�ða=2Þtei!t� �

þ C2 e�ða=2Þte�i!t� �

Recalling the complex function analysis, for a complex numberz ¼ xþ iy, we can write

ez ¼ exðcos yþ i sin yÞ

So for our case, we can write

z1 ¼ � a

2tþ i!t and z2 ¼ � a

2t� i!t

We can also write

ez1 ¼ e�ða=2Þt cos!tþ i sin!tð Þ � e�1t

and

ez2 ¼ e�ða=2Þt cos!t� i sin!tð Þ � e�2t

Thus, the solution y ¼ C 01e

�1t þ C 02e

�2t becomes

y ¼ C 01 e�ða=2Þt cos!tþ i sin!tð Þh i

þ C 02 e�ða=2Þt cos!t� i sin!tð Þh i

On simplification, we get

y ¼ e�ða=2Þt C 01 þ C 0

2

� �cos!tþ C 0

1 � C 02

� �sin!t

�which can be rewritten as

y ¼ e�ða=2Þt C1 cos!tþ C2 sin!t½ � ðC:5ÞThus, Eq. (C.5) is the solution of the differential equation.

Special Cases

When a > 0, we have decaying oscillations, as shown in Figure C.1 (clearlywhen a < 0, we have oscillations with increasing amplitudes).

When a ¼ 0, we have sustained oscillations, as shown in Figure C.2. Thesolution is of the form, y ¼ C1 cos!tþ C2 sin!t.

Figure C.1 Decaying oscillations.

C.2 THE TRANSFORMATION OF HIGHER-ORDERDIFFERENTIAL EQUATIONS INTO A SET OFFIRST-ORDER DIFFERENTIAL EQUATIONS

For example, let us consider the following second-order differential equa-tion:

d2y

dt2þ a

dy

dtþ by ¼ 0

Let us define

x1 ¼ y

from which we can write

dx1dt

¼ dy

dt

and

x2 ¼dy

dt

from which we can write

dx2dt

¼ d2y

dt2

Thus, our main second-order differential equation can be rewritten as

dx2dt

þ ax2 þ bx1 ¼ 0 anddx1dt

¼ x2 ðC:6Þ

Figure C.2 Sustained oscillations.

The set of two first-order differential equations (C.6) can replace the originalsecond-order differential equation. Thus, we have

dx1dt

¼ 0x1 þ x2

dx2dt

¼ �bx1 � ax2

Let us define the following vectors and matrix:

X ¼ x1x2

� �

(this is the state vector, i.e., vector of state variables).We can also write

dX

dt¼

dx1dtdx2dt

0BB@

1CCA

and

A ¼ 0 1�b �a

� �

In matrix form, we can write

dX

dt¼ AX

with initial conditions

Xð0Þ ¼ Xo ¼ x1ox2o

� �

C.3 NONHOMOGENEOUS DIFFERENTIAL EQUATIONS

d2y

dt2þ a

dy

dtþ by ¼ f ðtÞ

The solution of the above nonhomogeneous differential equation is of theform

y ¼ yh þ yp

where yh is the solution of the homogeneous equation (complementaryfunction) and yp is the particular integral depending on f ðtÞ. Some of thevalues of the particular integral depending on the nature of f ðtÞ is tabulatedin Table C.1.

C.4 EXAMPLES OF SOME SPECIAL CASES

Example C.1

d2y

dt2þ 4y ¼ 8t2

First, we calculate for the homogeneous solution yh:

d2y

dt2þ 4y ¼ 0

We get

�2 þ 4 ¼ 0

The solutions are

�1 ¼ 2i and �2 ¼ �2i

Thus, the solution is given by

yh ¼ C1 cos 2tþ C2 sin 2t

Now, to calculate the particular integral yp. From Table C.1,

yp ¼ k2t2 þ k1tþ k0

and we get

d2yp

dt2¼ 2k2

As yp must satisfy the differential equation, we get

2k2 þ 4 k2t2 þ k1tþ k0

� � ¼ 8t2

Table C.1 Choice of yp

f ðtÞ Choice of yp

ke�t Ce�t

ktn (where n ¼ 0; 1; 2; 3; . . .Þ kntn þ kn�1t

n�1 þ � � � þ k1tt þ k0

k cos!t or k sin!t K1 cos!tþ K2 sin!t

ke�t cos!t or ke�t sin!t e�tðK1 cos!tþ K2 sin!tÞ

Reorganizing the equation with all the terms with same tn together gives

ð4k2 � 8Þt2 þ 4k1tþ ð2k2 þ 4k0Þ ¼ 0

For the above equation to be correct for all values of t, all of the coefficientsof t must be equal to zero, so we get

4k2 � 8 ¼ 0 ) k2 ¼ 2

4k10 ¼ 0 ) k1 ¼ 0

2k2 þ 4k0 ¼ 0 ) k0 ¼ �1

Thus

yp ¼ 2t2 � 1

So the total solution is given by

y ¼ yh þ yp ¼ C1 cos 2tþ C2 sin 2tþ 2t2 � 1

Example C.2

d2y

dt2� 3

dy

dtþ 2y ¼ et

First, we calculate for the homogeneous solution yh,

d2y

dt2� 3

dy

dtþ 2y ¼ 0

We get

�2 � 3�þþ2 ¼ 0

The solutions are

�1 ¼ 2 and �2 ¼ 1

Thus, the solution is given by

yh ¼ C1et þ C2e

2t

Now, to calculate the particular integral yp. From Table C.1, we usuallychoose yp ¼ Cet, but now et is a part of yh. So, in this case, we must choose

yp ¼ Ctet

To obtain the value of C, we perform the following calculations:

dyp

dt¼ Cðtet þ etÞ

d2yp

dt2¼ Cð2et þ tetÞ

Substituting the above values into the differential equation gives

Cð2et þ tetÞ � 3Cðtet þ etÞ þ 2Ctet ¼ et

On reorganizing, we get

etð2C � 3C � 1Þ þ tetðC � 3C þ 2CÞ ¼ 0

We finally get

C ¼ �1

So the particular integral is equal to

yp ¼ �tet

The total solution is given by

y ¼ yh þ yp ¼ C1et þ C2e

2t � tet

Note: Notice not to have in yp a term which is same as a term in yh (not thesame in functional form). The reader is advised to consult his mathematicsbook for details.

Appendix D

Table of Laplace Transforms ofSome Common Functions

Solution of differential equations using Laplace transformation is given inSection 5.6.6.

Table D.1 Table of Laplace Transforms

f ðtÞ ff ðsÞ

1 1 1

s

2 t 1

s2

3 tn � 1

ðn� 1Þ!1

sn

42

ffiffiffit

rs�3=2

5 tk�1 �ðkÞsk

ðk 0Þ

6 tk�1e�at �ðkÞðsþ aÞk ðk 0Þ

7a e�at 1

sþ a

f ðtÞ ff ðsÞ

7b 1

�e�t=� 1

�sþ 1

8 1� e�t=� 1

sð�sþ 1Þ

9a te�at 1

ðsþ aÞ2

9b t

�2e�t=� 1

ð�sþ 1Þ2

10 1� 1þ t

�

� �e�t=� 1

sð�sþ 1Þ2

11a 1

ðn� 1Þ! tn�1e�at 1

ðsþ aÞn ðn ¼ 1; 2; . . .Þ

11b 1

�nðn� 1Þ! tn�1e�t=� 1

ð�sþ 1Þn ðn ¼ 1; 2; . . .Þ

12 1

ðb� aÞ e�at � e�bt� � 1

ðsþ aÞðsþ bÞ

13 1

ðb� aÞ be�bt � ae�at� � s

ðsþ aÞðsþ bÞ

14 1

ab1þ 1

ða� bÞ be�at � ae�bt� ��

1

sðsþ aÞðsþ bÞ

15 sin bt b

s2 þ b2

16 sinh bt b

s2 � b2

17 cos bt s

s2 þ b2

18 cosh bt s

s2 � b2

19 1� cos bt b2

sðs2 þ b2Þ

Table D.1 Continued

f ðtÞ ff ðsÞ

20 e�at sin bt b

ðsþ aÞ2 þ b2

21 e�at cos bt sþ a

ðsþ aÞ2 þ b2

22 e�atf ðtÞ ff ðsþ aÞ

23 f ðt� bÞ;with f ðtÞ ¼ 0 for t < 0 e�bsff ðsÞ

Figure D.1 Relation between s and t domains.

Appendix E

Orthogonal Collocation Technique

The orthogonal collocation technique is a simple numerical method which iseasy to program and which converges rapidly. Therefore, it is useful for thesolution of many types of second-order boundary-value problems (it trans-forms them to a set of algebraic equations satisfying the boundary condi-tions), as well as partial differential equations (it transforms them to a set ofordinary differential equations satisfying the boundary conditions). Thismethod in its simplest form as presented in the appendix was developedby Villadsen and Stewart (1). The orthogonal collocation method has theadvantage of ease of computation. This method is based on the choice of asuitable trial series to represent the solution. The coefficients of the trialseries are determined by making the residual equation vanish at a set ofpoints called ‘‘collocation points,’’ in the solution domain (2).

E.1 APPLICATION OF THE SIMPLE ORTHOGONALCOLLOCATION TECHNIQUE

When the Fickian diffusion model is used, many reaction–diffusion pro-blems in a porous catalyst pellet can be reduced to a two-point boundary-value differential equation of the form of Eq. (E.1). This is not necessarycondition for the application of this simple orthogonal collocation techni-que. The technique, in principle, can be applied to any number of simulta-neous two-point boundary-value differential equations

r2Y þ FðYÞ ¼ 0 ðE:1Þwhere

r2 ¼ d2

d!2þ a

!

d

d!ðE:2Þ

where a ¼ 0 for a slab, a ¼ 1 for a cylinder, and a ¼ 2 for a sphere, with thefollowing boundary conditions

At ! ¼ 1,

dY

d!¼ � YB � Yð Þ ðE:3aÞ

At ! ¼ 0;

dY

d!¼ 0 ðE:3bÞ

where YB is the value of Y at bulk conditions and � can be Nusselt orSherwood numbers depending on whether we are using the heat or massbalance equations, respectively.

For interior collocation, the approximation function is chosen so thatthe boundary conditions are satisfied. A suitable function is the following:

Y ðnÞ ¼ Yð1Þ þ ð1� !2ÞXn�1

i¼1

aðnÞ1 Pið!2Þ ðE:4Þ

in which Pið!2Þ are polynomials of degree i in !2, that are to be specified,and also a

ðnÞi are undetermined constants. The even polynomials, Pið!2Þ,

automatically satisfy the symmetry condition at the center of the pellet[Eq. (E3.b)].

A set of ðnÞ equations is needed to determine the collocation constantsai. We note that once Y ðnÞ has been adjusted to satisfy Eq. (E.1) at n collo-cation points !1; !2; . . . ; !n, the residual either vanishes everywhere or con-tains a polynomial factor of degree (n), whose zeros are the collocationpoints.

Villadsen and Stewart (1) pointed out that the equations can be solvedin terms of the solution at the collocation points, instead of the coefficientsaðnÞi . This is more convenient and reduces the Laplacian and the first-order

derivative to

r2Y��!¼!1

¼Xnþ1

j¼1

BijYð!jÞ ¼ BY ðE:5Þ

dY

d!

��!¼!1

¼Xnþ1

j¼1

AijYð!jÞ ¼ AY ðE:6Þ

respectively, where i ¼ 1; 2; 3; . . . ; n is the number of interior collocationpoints.

Integrals of the solution (which are useful for the calculation of theeffectiveness factor) can also be calculated with high accuracy via theformula

ð1:00

FðYÞ!2 d! ¼Xnþ1

i¼1

!ðnÞi FðYiÞ ¼ W FðYÞ ðE:7Þ

A, B, and W can be calculated as follows:First, the values of Y can be expressed at the collocation points by

Y1

Y2

..

.

Yn

0BBBBBB@

1CCCCCCA

|fflfflffl{zfflfflffl}¼

!01 !2

1 � � � !2n1

!02 !2

2 � � � !2n2

..

. ... . .

. ...

!0n !2

n � � � !2nn

0BBBBBB@

1CCCCCCA

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

d1

d2

..

.

dn

0BBBBBB@

1CCCCCCA

|fflfflffl{zfflfflffl}Y C d

ðE:8Þ

which can be simply written as

Y ¼ C d ðE:9Þ

Then, the collocation constants are

d ¼ C�1Y ðE:10Þ

On differentiating the trial function Y , we get

dY

d!

� �!1

dY

d!

� �!2

..

.

dY

d!

� �!n

0BBBBBBBBBBBBB@

1CCCCCCCCCCCCCA

|fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl}

¼

d!0

d!

!!1

d!2

d!

!!1

� � � d!2n

d!

!!1

d!0

d!

!!2

d!2

d!

!!2

� � � d!2n

d!

!!2

..

. ... . .

. ...

d!0

d!

!!n

d!2

d!

!!n

� � � d!2n

d!

!!n

0BBBBBBBBBBBBBBBB@

1CCCCCCCCCCCCCCCCA

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

d1

d2

..

.

dn

0BBBBBB@

1CCCCCCA

|fflfflffl{zfflfflffl}

DY R d

ðE:11Þor simple, we can write

DY ¼ Rd ðE:12ÞFrom Eqs. (E.10) and (E.12), by substituting for d, we get

DY ¼ RC�1Y ðE:13ÞLet us express

DY ¼ AY ðE:14Þwhere,

A ¼ RC�1

Also, let us express

r2Y ¼ T d ¼ BY ðE:15Þwhere

T ¼

r2!0� �

!1r2!2� �

!1� � � r2!2n

� �!1

r2!0� �

!2r2!2� �

!2� � � r2!2n

� �!2

..

. ... . .

. ...

r2!0� �

!nr2!2� �

!n� � � r2!2n

� �!n

0BBBBBB@

1CCCCCCA

From Eqs. (E.10) and (E.15), by substituting for d, we get

r2Y ¼ T C�1Y ¼ BY ðE:16ÞThus,

B ¼ T C�1 ðE:17ÞAlso, let

FðYÞ ¼Xnþ1

i¼1

Ci!2ði�1Þ ðE:18Þ

Then,

!2FðYÞ ¼Xnþ1

i¼1

Ci!2ðiÞ ðE:19Þ

Therefore,ð1:00

!2FðYÞ d! ¼ð1:00

Xnþ1

j¼1

Ci!2ðiÞ d! ¼ I:F: ðE:20Þ

The integrated function I.F. will be

I:F: ¼ W F C ðE:21Þwhere

W F ¼ 13

15� � � 1

2n þ 1

� �ðE:22Þ

and

C ¼C1

C2

..

.

Cn

0BBB@

1CCCA ðE:23Þ

Also,

FðYÞ ¼ QC ðE:24Þwhere

C ¼

!01 !2

1 � � � !2n1

!02 !2

2 � � � !2n2

..

. ... . .

. ...

!0n !2

n � � � !2nn

0BBBBBB@

1CCCCCCA

ðE:25Þ

From Eq. (E.24), we get

C ¼ Q�1FðYÞ ðE:26Þ

From Eqs. (E.21) and (E.26), the integral can be expressed as

I:F: ¼ W F Q�1FðYÞ ¼ W FðYÞ ðE:27Þ

We can calculate W as

W ¼ W F Q�1 ðE:28Þ

Applying the simple case of one internal collocation point ðn ¼ 1, and ! atthis collocation point is !1), the Laplacian at the interior collocation pointcan be written as

r2Y��!¼!2

¼ B11Yð!1Þ þ B12Yð!2Þ ðE:29Þ

and the derivative at the surface boundary condition (! ¼ !2) takes theform

dY

d!

��!¼!2

¼ A21Yð!1Þ þ A22Yð!2Þ ðE:30Þ

and, therefore, the boundary condition can be written as

A21Yð!1Þ þ A22Yð!2Þ ¼ �ðYB � Yð!2ÞÞ ðE:31Þ

Thus, we get

Yð!2Þ ¼�YB � A21Yð!1Þ

A22 þ �ðE:32Þ

Substitution of Eqs. (E.29) and (E.32) into Eq. (E.1) gives

A11Yð!1Þ þ B12

�YB � A21Yð!1ÞA22 þ �

� �¼ FðYð!1ÞÞ ðE:33Þ

which is a single equation in the single variable Yð!1Þ at the collocationpoint !1. Solution of this equation gives Yð!1Þ, which can be substituted inEq. (E.32) to give Yð!2Þ, the surface of Y . The values of Yð!1Þ and Yð!2Þcan be used to evaluate the integral as follows:

ð1:00

!2FðYÞ d! ¼ W1Fð!1Þ þW2Fð!2Þ ðE:34Þ

The values of B, A, and W for a single internal collocation point using theLegendre polynomial, for a spherical catalyst pellet, are

B ¼�10:5 10:5

�10:5 10:5

!

A ¼�2:291 2:291

�3:5 3:5

!

W ¼0:2333

0:1

!

The values of A, B, and W for large number of collocation points are givenin a number of references (1,2).

E.2 ORTHOGONAL COLLOCATION TECHNIQUEAPPLICABLE TO PARABOLIC PARTIALDIFFERENTIAL EQUATIONS

The concept of orthogonal collocation for ordinary differential equationscan be easily extended to solve ‘‘parabolic partial differential equations.’’The difference is that the application of orthogonal collocation method onthe two-point boundary-value differential equation discussed earlier resultsin a set of algebraic equations, whereas application of orthogonal colloca-tion method on parabolic partial differential equations results in a set ofordinary differential equations.

Let us assume the parabolic partial differential equation to be of thefollowing form:

@Y

@tþ r2Y ¼ FðYÞ ðE:35Þ

The second term of Eq. (E.35) can be replaced as shown in Eq. (E.5),whereas the partial derivative term of Eq. (E.35) can be converted to anordinary derivative term at each collocation point to give the following set:

dYð!iÞdt

þXnþ1

j¼1

BijYð!jÞ ¼ FðYÞ!iðE:36Þ

where i ¼ 1; 2; 3; . . . ; n and n is the number of interior collocation points.Thus, Eq. (E.36) will give a set of n ordinary differential equations to besolved in order to solve a parabolic partial differential equation.

For elliptical and hyperbolic partial differential equations, the reader isadvised to consult engineering mathematics texts (3–6).

Partial differential equations can also be solved numerically using themost commonly used method, the ‘‘method of finite differences’’ or the

Crank–Nicholson method. The reader is advised to consult engineeringmathematics texts (3–6).

For higher-order systems, finite element techniques can be used, butthis is beyond the scope of this undergraduate book. Any interested readercan consult Refs. (3–6) to master those techniques.

REFERENCES

1. Villadsen, J. and Stewart, W. E. Solution of boundary-value problems by

orthogonal collocation. Chem. Eng. Sci. 22, 1483–1501, 1967.

2. Finlayson, B. A. The Method of Weighted Residuals and Variational Principles.

Academic Press, New York.

3. Kreysig, E. Advanced Engineering Mathematics, 8th ed. Wiley, New York, 1998.

4. Masatake, M. The Finite Element Method and Its Applications. Macmillan, New

York, 1986.

5. Strikwerda, J. C. Finite Difference Schemes and Partial Differential Equations.

CRC Press, London, 1989.

6. Strauss, W. A. Partial Differential Equations: An Introduction. Wiley, New

York, 1992.

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Appendix F

Some Software and ProgrammingEnvironments

We provide brief details of some commonly used software and program-ming environments used for solving the mathematical models and visualiz-ing them. We emphasize that the choice of any particular software orprogramming environment depends on the user.

F.1 POLYMATH 5.X

POLYMATH (http://www.polymath-software.com) is a proven computa-tional system, which has been specifically created for educational or profes-sional use. The various POLYMATH programs allow the user to applyeffective numerical analysis techniques during interactive problem-solvingon personal computers. Results are presented graphically for easy under-standing and for incorporation into papers and reports. Students, engineers,mathematicians, scientists, or anyone with a need to solve problems willappreciate the efficiency and speed of problem solution.

The main options available in POLYMATH are the following:

. LIN: Linear Equations Solver: Enter (in matrix form) and solve asystem of up to 200 simultaneous linear equations (64 with con-venient input interface).

www.polymath-software.com

. NLE: Nonlinear Equations Solver: Enter and solve a system of upto 200 nonlinear algebraic equations.

. DEQ: Differential Equations Solver. Enter and solve a system ofup to 200 ordinary differential equations.

. REG: Data analysis and Regression. Enter, analyze, regress, andplot a set of up to 600 data points.

F.2 MATLAB

F.2.1 Unified, Interactive Language, and ProgrammingEnvironment

The MATLAB (http://www.mathworks.com) language is designed for inter-active or automated computation. Matrix-optimized functions let you per-form interactive analyses, whereas the structured language features let youdevelop your own algorithms and applications. The versatile language letsyou tackle a range of tasks, including data acquisition, analysis, algorithmdevelopment, system simulation, and application development. Languagefeatures include data structures, object-oriented programming, graphicaluser interface (GUI) development tools, debugging features, and the abilityto link in C, Cþþ, Fortran, and Java routines.

F.2.2 Numeric Computing for Quick and Accurate Results

With more than 600 mathematical, statistical, and engineering functions,MATLAB gives immediate access to high-performance numeric computing.The numerical routines are fast, accurate, and reliable. These algorithms,developed by experts in mathematics, are the foundation of the MATLABlanguage. The math in MATLAB is optimized for matrix operations. Thismeans that you can use it in place of low-level languages like C and C þþ,with equal performance but less programming.

MATLAB mainly includes the following:

. Linear algebra and matrix computation

. Fourier and statistical analysis functions

. Differential equation solvers

. Sparse matrix support

. Trigonometric and other fundamental math operations

. Multidimensional data support

F.2.3 Graphics to Visualize and Analyze Data

MATLAB includes power, interactive capabilities for creating two-dimen-sional plots, images, and three-dimensional surfaces and for visualizing

www.mathworks.com

volumetric data. Advanced visualization tools include surface and volumerendering, lighting, camera control, and application-specific plot types.

F.3 MATHCAD

Mathcad (http://www.mathcad.com) provides hundreds of operators andbuilt-in functions for solving technical problems. Mathcad can be used toperform numeric calculations or to find symbolic solutions. It automaticallytracks and converts units and operates on scalars, vectors, and matrices.

The following is an overview of Mathcad’s computational capabilities:

. Numeric operators perform summations, products, derivatives,integrals, and Boolean operations. Numeric functions apply trigo-nometric, exponential, hyperbolic, and other functions and trans-forms.

. Symbolics simplify, differentiate, integrate, and transform expres-sions algebraically, Mathcad’s patented live symbolics technologyautomatically recalculate algebraic solutions.

. Vectors and Matrices manipulate arrays and perform variouslinear algebra operations, such as finding eigenvalues and eigen-vectors and looking up values in arrays.

. Statistics and Data Analysis generate random numbers or histo-grams, fit data to built-in and general functions, interpolate data,and build probability distribution models.

. Differential Equation Solvers support ordinary differential equa-tions, systems of differential equations, and boundary-valueproblems.

. Variables and Units handle real, imaginary, and complex numberswith or without associated units. A high-performance calculationengine provides speed and sophisticated memory management tohelp find solutions faster.

F.4 DYNAMICS SOLVER

Dynamics Solver (http://tp.lc.ehu.es/jma/ds/ds.html) solves numericallyboth initial-value problems and boundary-value problems for continuousand discrete dynamical systems:

. A single ordinary differential equation of arbitrary order

. Systems of first-order ordinary differential equations

. A rather large class of functional differential equations and sys-tems

www.mathcad.com

. Discrete dynamical systems in the form iterated maps and recur-rences in arbitrary dimensions

It is also able to draw complex mathematical figures, including manyfractals. Dynamics Solver is a powerful tool for studying differential equa-tions, (continuous and discrete) nonlinear dynamical systems, deterministicchaos, mechanics, and so forth. For instance, you can draw phase spaceportraits (including an optional direction field). Poincare maps, Liapunovexponents, histograms, bifurcation diagrams, attraction basis, and so forth.The results can be watched (in perspective or not) from any direction andparticular subspaces can be analyzed.

Dynamics Solver is extensible: Users can add new mathematical func-tions and integration codes. It also has all the advantages of Windowsprograms, including the ease of use, the ability to open several outputwindows simultaneously and a larger amount of memory, which allowsanalyzing more complex problems. Moreover, there are many ways toprint and export the results.

F.5 CONTROL STATION

Control Station (http://www.ControlStation.com) is both a controllerdesign and tuning tool, and a process control training simulator used byindustry and academic institutions worldwide for

. control loop analysis and tuning

. dynamic process modeling and simulation

. performance and capability studies

. hands-on process control training

The Case Studies feature provides hands-on training by challengingthe user with industrially relevant process simulations. The software letsusers manipulate process and controller parameters so that they can‘‘learn by doing’’ as users explore the world of process control.

The Controller Library feature lets users use different controller set-tings (e.g., P-only, PI, PD, PID, cascade, ratio control, and so forth). DesignTools is a powerful controller design and analysis tool that can also incor-porate dead times and the Custom Process feature can let users implement aprocess and control architecture to their own specifications.

www.ControlStation.com

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