....
B. Wayne Bequette
Process Dynamics Modeling, Analysis, and Simulation
Prentice Hall International Series in the Physical and Chemical Engineering Sciences
PRENTICE IIALL INTERNATIONAL SERIES IN THE PHYSICAL AND CHEMICAL ENGINEERING SCIENCES NEAL
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Chemical !'.lIgilleaing ThennodYl/tllllics Process Oynamics: A1odcling, Analysis, and Simulatio/l BWUU.:R, CiROSS\-\i\NN, AND WESTERBERG Sy.l"tcmafh' iHethods rd' Chemica! Process J)esij{11 CROWL and 1,()(JVA1{ Chemica! Process Sq(ety DFNN Process' Fl/lid lvlec!wlI/c"S [;'OGLER {:lemcfI!s q(Cllcmical Rcaction Engineering, 2nd I~'dition HANNA AND SANDA[J COll/putatiolla! Methods ill ChclIIical Engineering I-IJ~IMI~LBLAU Basic Principles {{lid C(llcl/larions ill Chemical Engincering, 6th editioll HINES i\!>il) MADDOX Mass Tnl/lsjt'r KYLE Chemical and Proces..,. '1henllodyl1amics, 2nd edition NEWMAN Efectmchc/Ilical Systcms, 2nd edition PR;\tJSNll/., LICIITFNTI-lMJ',R, and DL A/.l'VEDO Molecular Thermodynamics (~rFlllid-PIUlse Eqlli/ihria, 2nd cdition PHrNTlC:E Elec!rochcmical Ellgincering Principles STEPHANOI'OULOS Chemical Process COlltrol TI~STER AND MOllF.!J. Thermodynamics (lnd Its /\pplicatio/ls, 3rd edition TURTON Analysis, SrI/thesis, and Design (~rChell/iCflI Processes Bi\LZlllSER, S;\:-'·IUEJ.S, Mm 1-::l.uASSEN
Br',QUFTrl::
PROCESS DYNAMICS Modeling, Analysis, and Simulation B. Wayne Bequette Rensselaer Polytechnic Institute
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Catalogil1g~il1-PllblicatitmData
Bequette, B, Wayne. Process dynamics: modeling, analysis, and simulation / B, Wayne Bequette. p.
ern.
Includes bibliographical references and index, ISBN O,-11-206889~3 1. Chemical procc%es. I. Tille. TPI55,7.B45 1998 (j6()',2X4'OI185---dc21
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CONTENTS
Preface
XVII
SECTION I PROCESS MODELING 1
Introduction
1.1 1.2 1.3
1.4 1.5
1.6
1
3
Motivation 3 Models 4 1.2.1 How Models Are Used 5 Systems 10 1.3.1 Simulation 10 1.3.2 Linear Systems Analysis 11 1.3.3 A Broader View of Analysis 11 Background of the Reader 12 How To Use This Textbook 12 1.5.1 Sections 12 1.5.2 Numerical Solutions 13 1.5.3 Motivating Examples and Modules 13 Courses Where This Textbook Can Be Used 13 Summary 14 Further Reading 14 Student Exercises 15
vii
Contents
viii
2
Process Modeling
2.1 2.2
2.3 2.4
2.5
2.6 2.7 2.8 2.9
16
Background 17 Balance Equations 17 2.2.1 Integral Balances 18 2.2.2 Instantaneous Balances 20 Material Balances 20 2.3.1 Simplifying Assumptions 25 Constitutive Relationships 27 2.4.1 Gas Law 27 2.4.2 Chemical Reactions 27 2.4.3 Equilibrium Relationships 28 2.4.4 Flow-through Valves 29 Material and Energy Balances 30 2.5.1 Review of Thermodynamics 31 Distributes Parameter Systems 34 Dimensionless Models 35 Explicit Solutions to Dynamic Models 36 General Form of Dynamic Models 37 2.9.1 State Variables 37 2.9.2 Input Variables 2.9.3 Parameters 38 2.9.4 Vector Notation 38 Summary 40 Further Reading 40 Student Exercises 41
SECTION II
3
Algebraic Equations
3.1 3.2 3.3
3.4
3.5
NUMERICAL TECHNIQUES
49
51
Notations 51 General Form for a linear System of Equations 51 Nonlinear Functions of a Single Variable 54 3.3.1 Convergence Tolerance 55 3.3.2 Direct Substitution 55 3.3.3 Interval Halving (Bisection) 58 3.3.4 False Position (Reguli Falsi) 60 3.3.5 Newton's Method (or Newton-Raphson) 60 MATLAB Routines for Solving Functions of a Single Variable 3.4.1 fzero 63 3.4.2 roots 64 Multivariable Systems 64 3.5.1 Newton's Method for Multivariable Problems 66 3.5.2 Quasi-Newton Methods 69
63
ix
Contents
3.6
4
MATlAB Routines for Systems of Nonlinear Algebraic Equations Summary 71 Further Reading 72 Student Exercises 73 Appendix 78
Numerical Integration 4.1 4.2
4.3
4.4
80
Background 80 Euler Integration 81 4.2.1 Explicit Euler 81 4.2.2 Implicit Euler 82 4.2.3 Numerical Stability of Explicit and Implicit Methods Runge-Kutta Integration 88 4.3.1 Second Order Runge-Kutta 89 4.3.2 Fourth-Order Runge-Kutta 93 MATlAB Integration Routines 94 4.4.1 ode23 and ode45 95 Summary 97 Further Reading 97 Student Exercises 98
SECTION III
5
LINEAR SYSTEMS ANALYSIS
103
linearization of Nonlinear Models: The State-Space Formulation 5.1 5.2
5.3 5.4
5.5 5.6
State Space Models 106 5.5.1 General Form of State Space Models 108 Linearization of Nonlinear Models 109 5.2.1 Single Variable Example 109 5.5.2 One State Variable and One Input Variable 5.2.3 Linearization of Multistate Models 112 5.2.4 Generalization 114 Interpretation of Linearization 117 Solution of the Zero-Input Form 119 5.4.1 Effect of Initial Condition Direction (Use of Similarity Transform) 120 Solution of the General State-Space Form 127 MATlAB Routines step and initial 127 5.6.1 The MATLAB step Function 129 5.6.2 The MATLAB initial Function 129 Summary 130 Further Reading 131 Student Exercises 131
83
110
105
70
Contents
x
6
Solving Linear nth Order ODE Models
6.1 6.2
6.3 6.4 6.5
7
7.4 7.5 7.6
8
Background 143 Solving Homogeneous. Linear ODEs with Constant Coefficients 145 6.2.1 Distinct Eigenvalues 146 6.6.2 Repeated Eigenvalues 148 6.2.3 General Result for Complex Roots 152 Solving Nonhomogeneous. Linear ODEs with Constant Coefficients 152 Equations with Time-Varying Parameters 154 Routh Stability Criterion-Determining Stability Without Calculating Eigenvalues 157 6.5.1 Routh Array 157 Summary 160 Further Reading 161 Student Exercises 161
An Introduction to laplace Transforms
7.1 7.2 7.3
142
168
Motivation 168 Definition of the Laplace Transform 169 Examples of Laplace Transforms 170 7.3.1 Exponential Function 170 7.3.3 Time-Delay (Dead Time) 172 7.3.4 Derivations 173 7.3.5 Integrals 173 7.3.6 Ramp Function 174 7.3.7 Pulse 174 7.3.8 Unit Impulse 176 7.3.9 Review 177 Final and Initial Value Theorems 177 Application Examples 178 7.5.1 Partial Fraction Expansion 179 Table of Laplace Transforms 185 Summary 187 Further Reading 187 Student Exercises 187
Transfer Function Analysis of First-Order Systems
8.1 8.2
8.3 8.4
Perspective 191 Responses of First-Order Systems 191 8.2.1 Step Inputs 193 8.2.2 Impulse Inputs 198 Examples of Self-Regulating Processes 200 Integrating Processes 206
190
xi
Contents
8.5
9
Transfer Function Analysis of Higher-Order Systems 9.1
9.2 9.3 9.4 9.5 9.6
10
11.5 11.6 11.7 11.8
215
247
A Second-Order Example 248 The General Method 251 MATLAB Routine ss2tf 252 10.3.1 Discussion of the Results from Example 10.2 Summary 257 Student Exercises 257
Black Diagrams 11.1 11.2 11.3 11.4
209
Responses of Second-Order Systems 216 9.1.1 Step Responses 217 9.1.2 Underdamped Step Response Characteristics 221 9.1.3 Impulse Responses 222 9.1.4 Response to Sine Inputs 224 Second-Order Systems with Numerator Dynamics 226 The Effect of Pole-Zero Locations on System Step Responses 228 Pade Approximation for Deadtime 230 Converting the Transfer E;unction Model to State-Space Form MATLAB Routines for Step and Impulse Response 234 9.6.1 step 234 9.6.2 impulse 236 Summary 236 Student Exercises 237
Matrix Transfer Functions 10.1 10.2 10.3
11
Lead-Lag Models 208 8.5.1 Simulating Lead-Lag Transfer Functions Summary 211 Student Exercises 211
255
261
Introduction to Block Diagrams 262 Block Diagrams of Systems in Series 262 Pole-Zero Cancellation 263 Systems in Series 267 11.4.1 Simulating Systems in Series 267 Blocks in Parallel 269 11.5.1 Conditions for Inverse Response 271 Feedback and Recycle Systems 273 Routh Stability Criterion Applied to Transfer Functions 11.7.1 Routh Array 277 SIMULINK 278
276
232
xii
Contents
Summary 279 Student Exercises
12
280
Linear Systems Summary 12.1 12.2 12.3
12.4 12.5
282
Background 283 Linear Boundary Value Problems 283 Review of Methods for Linear Initial Value Problems 287 12.3.1 Linearization 187 12.3.2 Direction Solution Techniques 288 12.3.3 Rewrite the State-Space Model as a Single nth Order Ordinary Differential Equation 289 12.3.4 Use Laplace Transforms Directly on the State-Space Model 290 Introduction to Discrete-Time Models 290 12.4.1 Discrete Transfer Function Models 291 Parameter Estimation of Discrete Linear Systems 295 Summary 297 References 297 Student Exercises 298 SECTION IV NONLINEAR SYSTEMS ANALYSIS
13
Phase-Plane Analysis 13.1 13.2 13.3
13.4
14
301
303
Background 304 Linear System Examples 304 Generalization of Phase-Plane Behavior 311 13.3.1 Slope Marks for Vector Fields 313 13.3.2 Additional Discussion 315 Nonlinear Systems 316 13.4.1 Limit Cycle Behavior 323 Summary 324 Further Reading 324 Student Exercises 324
Introduction Nonlinear Dynamics: A Case Study of the Quadratic Map 14.1 14.2 14.3
14.4 14.5
Background 332 A Simple Population Growth Model 332 A More Realistic Population Model 334 14.3.1 Transient Response Results for the Quadratic (Logistic) Map 335 Cobweb Diagrams 339 Bifurcation and Orbit Diagrams 342 14.5.1 Observations from the Orbit Diagram (Figure 14.14)
342
331
Contents
14.6
14.7
14.8
15
360
Motivation 360 Illustration of Bifurcation Behavior 361 Types of Bifurcations 362 15.3.1 Dynamic Responses 365 Summary 376 References and Further Reading 376 Student Exercises 377 Appendix 379
Bifurcation Behovior of Two-State Systems 16.1 16.2 16.3 16.4
17
Stability of Fixed-Point Solutions 344 14.6.1 Application of the Stability Theorem to the Quadratic Map 344 14.6.2 Generalization of the Stability Results for the Quadratic Map 346 14.6.3 The Stability Theorem and Qualitative Behavior Cascade of Period-Doublings 348 14.7.1 Period-2 348 14.7.2 Period-4 351 14.7.3 Period-n 14.7.4 Feigenbaum's Number 353 Further Comments on Chaotic Behavior 354 Summary 354 References and Further Reading 355 Student Exercises 356 Appendix: Matlab M-Files Used in This Module 358
Bifurcation Behavior of Single ODE Systems 15.1 15.2 15.3
16
xiii
Background 381 Single-Dimensional Bifurcations in the Phase-Plane limit Cycle Behavior 384 The Hopf Bifurcation 388 16.4.1 Higher Order Systems (n > 2) 391 Summary 392 Further Reading 392 Student Exercises 392
Introduction to Chaos: The Lorenz Equations 17.1 17.2 17.3
381
395
Introduction 396 Background 396 The Lorenz Equations 397 17.3.1 Steady-State (Equilibrium) Solutions
398
382
346
xiv
17.4
17.5
17.6 17.7
Contents
Stability Analysis of the Lorenz Equations 399 17.4.1 Stability of the Trivial Solution 399 17.4.2 Stability of the Nontrivial Solutions 400 17.4.3 Summary of Stability Results 301 Numerical Study of the Lorenz Equations 402 17.5.1 Conditions for a Stable Trivial (No Flow) Solution 17.5.2 Stable Nontrivial Solutions 403 17.5.3 Chaotic Conditions 405 Chaos in Chemical Systems 407 Other Issues in Chaos 409 Summary 410 References and Further Reading 410 Student Exercises 411 Appendix 412
SECTION IV Module 1
Introduction to MATlAB M1.1 M1.2 M1.3 M1.4 M1.5 M1.6 M1.7 M1.8 M1.9 M1.10 M1.11 M1.12 M1.13 M1.14
Module
2
REVIEW AND LEARNING MODULES
Background 416 Using This Tutorial 416 Entering Matrices 417 The MATLAB Workspace 419 Complex Variables 421 Some MATLAB Operations 421 Plotting 422 More Matrix Stuff 426 FOR Loops and IF-THEN Statements 427 m-files 428 Diary 430 Toolboxes 430 Limitations to Student MATLAB 431 Contacting Mathworks 431 Student Exercises 432
Review of Matrix Algebra M2.1 M2.2 M2.3 M2.4
415
435
Motivation and Notation 436 Common Matrix Operations 436 Square Matrices 440 Other Matrix Operations 448 Summary 450 Student Exercises 450
413
402
xv
Contents
Module 3
Linear Regression M3.1 M3.2 M3.3 M3.4 M3.5
Module
4
6
M6.1 M6.2 M6.3 M6.4 M6.5 Module 7
471
Background 490 The Dynamic Model 491 Steady-State Analysis 495 Step Responses 496 Unforced Behavior 501
Introduction 506 First-Order Irreversible Reaction Van de Vusse Reaction 516
Biochemical Reactors M8.1 M8.2 M8.3 M8.4
478
490
Isothermal Continuous Stirred Tank Chemical Reactors M7.1 M7.2 M7.3
Module 8
466
Introduction 471 Developing the Dynamic Model 472 Steady-State Conditions 475 State-Space Model 475 Laplace Domain Model 477 Step Responses: Linear versus Nonlinear Models Unforced System Responses: Perturbations in Initial Conditions 483
Absorption
456
464
Introduction 464 Transfer Function-Based Simulation Printing SIMULINK Windows 469
Stirred Tank Heaters M5.1 M5.2 M5.3 M5.4 M5.5 M5.6 M5.7
Module
Motivation 452 Least Squares Solution for a Line 454 Solution for the Equation of a Line Using Matrix-Vector Notation 455 Generalization of the Linear Regression Technique MATLAB Routines polyfit and polyval 458
Introduction to SIMUlINK M4.1 M4.2 M4.3
Module 5
452
529
Background 529 Modeling Equations 530 Steady-State Solution 534 Dynamic Behavior 535
507
506
Contents
xvi
M8.5 M8.6 M8.7 M8.8 Module
9
Diabatic Continuous Stirred Tank Reactors M9.1 M9.2 M9.3 M9.4 M9.5 M9.6 M9.7 M9.8 M9.9
Module
10
617
540
559
Background 560 The Modeling Equations 561 Steady-State Solution 563 Dynamic Behavior 565 linearization of Dynamic Equations 567 Phase-Plane Analysis 571 Understanding Multiple Steady-State Behavior Further Complexities 580 Dimensionless Model 584
Ideal Binary Distillation M10.1 M10.2 M10.3 M10.4 M10.5 M10.6 M10.7
Index
linearization 536 Phase-Plane Analysis 539 Understanding Multiple Steady-States Bifurcation Behavior 549
572
597
Background 597 Conceptual Description of Distillation 599 Dynamic Material Balances 601 Solving the Steady-State Equations 603 Solving the Nonlinear Dynamic Equations 605 State-Space Linear Distillation Models 606 Multiplicity Behavior 608
PREFACE
An understanding of the dynamic behavior of chemical processes is ilnportant from bolh process design and process control perspectives. It is easy to design a chemical process, based on stcady~state considerations, which is practically uncontrollable when the process dynamics arc considered. The current status of computational hardware and software has made it easy to interactively simulate the dyo<:unic behavior of chemical processes.
It is comJllon for process dynamics to he included as the introductory portion of a process control textbook, however, there afC a number of limitations to this approach. Since the emphasis of most or the textbooks is on process control, there is too little space to give adequate depth to modeling, analysis, and sinwlatioll of dynamic systems. The focus tends to he on transfer function~based models that arc used for control systenl design. The prime motivation for my textbook is then to provide a more comprehensive treatment of process dynamics, including modeling, analysis, and simulation. This textbook evolved from notes dcvelopcd for a course on dynamic systems that I have becn teaching at Rcnsselaer since 199 I. Vie have been fortunate to have a two~semestcr se~ qucnee in dynamics and control, allowing more depth to the coverage of each topic. Topics covered here that arc not covered in a traditional text include nonlinear dynamics and the usc of MA TLAB for numerical analysis and simulation. Also, a signifi~ cant portion of the text consists of review and learning modules. I~ach learning module provides model development, stcady~statc solutions, nonlinear dynamic results, linear~ ization, state space and transfer function analysis and simulation. The motivation for this approach is to allow the student to "tic~togelher" all of the concepts, rather than {rcat~ ing the III independently (and not understanding the connections between the different methods).
xvii
Preface
xviii
An important feature of this text is the usc of MATLAB software. A set of In-files used in many of the examples and in the learning nlodules is available via the world wide wcb at the following locations: http:/wwwhpi.cdu/-hel}ueb/Process_Dynamics hHp:/www.mathworks.com/education/thirdparty.htm I Additional learning modules will also be available at the RPI location. A few acknowledgments arc in order. A special thanks to Professor Jim Turpin at the University of Arkansas, who taught me the introductory course in process dynamics and control. I-lis love of teaching should be an inspiration to us all. Many thanks to one of my graduate students, Lou Russo, who not only made a numher of suggestions to improve the text, but also sparked an interest in many of the undergraduates that h;:lVe taken the course. The {(Isk of developing a solutions manual has heen carried out by Venkatcsh Natarajan, Brian Aufderheide, l
B, Wayne RequeUe
SECTION
I PROCESS MODELING
INTRODUCTION
1
This chapter provides a motivation for process modeling and the study of dynamic chemical processes. It also provides an overview of the structure of the textbook. After studying this chapter, the student should he able to answer the following questions: What is a process model? • Why develop a process model?
• What is the difference between lumped parameter and distributed parameter systoms? • What numerical package forms the basis for the examples in this text? • What arc the major objectives of this textbook? The major sections arc: 1.1
Moti valion
1.2
Models
1.3
Systems
1.4 1.5
Background or the Reader How to Use This Texthook
1.6 Courses Where This Textbook Can Be Used
1.1
MOTIVATION
nze
Robert Reich in Work {~lNations has classified three broad categories of employment in the United States. In order of increasing educational requirement these categories are: 3
Introdu ction
4
Chap. 1
tic services. To diroutine production services, in-person services, and symboli c-analy rectly quote from Reich: l1ng symbols. They sim"Symbo lic analysts solve, identify, and broker problem s by manipula ented with, COlllexperim juggled, ed, rearrang be call that inmges abstract plify reality into reality. The mainto hack municat cd to other spc<:iaJists, and then, eventual ly, transfor med tools Illay be The ce. experien by d nipulatio ns arc done with analytic tools, sharpene s, psycho-· princiule .rcief/rifle s, gimmick financial lts, lIlathematical algorithms, legal argmllcl n, or any deductio or n inductio of systems amuse, to or persuade to how logical insights ahoUl s) emphasi for other sct of techniqu es for doing conceptu al puulcs. " (italics added
. Process enginee rs Engineers (and particularly process enginee rs) are s)'mholic analysts that charact erize usc fundamental scientif ic principles as a hasis for mathem atical models physical variables, such the behavio r of a chemic al process. Symbol s are used to represe nt d and numerical alas pressure, tempera ture or concentration. Input informa tion is specifie . Process: enginee rs gorithm s arc llsed to solve the rnodels (simula ting a physical system) endatio ns regarding analyze the results of these simulat ions to make decisions or recomm the design or operation of a process. es in one way Most chemic al enginee rs work with chemic al manufactUling pmcc:':is l troubles hooting in or another. Often they arc process ergince rs responsible for technica ~Ire respons ible 1'01' dethe day-tn-deiY operatio ns nf a particld ar chemic al process. Some as tempera ture or pre.ssigning feedback control systems so that process variables (such ihle for redesigning a .sure) can he maintai ned at desired values. Others may be respons ibilities require an unchemica l process to provide more profitahility. All of these respons processes. derstan ding of the time-de penden t(dynam ic) behavio r of chemic al
1.2
MODELS an underst anding of 'fhe primary objectiv e of this textboo k is to assist you in develop ing g the dynami c bethe dynami c behavio r of c1\
Didiona ry Definitions:
Mode!
as a noun, it means "a small Model is derived fro1l11he Latin modus, which means a measure . Used y). Dictiof/w World New r's represen tation of a pl
Sec. 1.2
Models
5
Notice that both definitions stress that a model is a representation of a system or ohject. In this textbook, when we usc the term model, we will be referring to a mathematical model. We prefer to usc the following definition f~)r model (more specifically, a process model).
Working Definition:
Pmccss Model
A process model is a set of equations (including the necessary input data to solve the equations) that allows us to predict the behavior of a chemical process system.
The emphasis in this text is on the development and lise offundamental or first-principles models. By fundamental, we mean models that are based on known physical-chemical rc~ lationships. This includes the conservation of mass and conservation of energy, I as well as reaction kinetics, transport phenomena, and thermodynamic (phllse equilibrium, etc.) relatiomdlips. Another common model is the empirical model. An empirical model might be used if the process is too complex for a fundamental model (either in the fonnulation of the model, or the numerical solution of the model), or if the empirical model has satisfactory predictive capability. An example of an empirical model is a simple least squares fit of an equation lo experimental data. Generally, we would prefer to use models based on fundamental knowledge of chemical-physical relationships. Fundamental models will generally he accurate over .a mueh larger range of conditions than empirical models. Empirical models may be useful for "interpolation" but are generally not useful lilr "extrapolation"; that is, an empirical model will only be useful over the range of conditions used for the "fit" of the data. It should be nOled that it is rare for a single process model to exist. A model is only an approximate representation of an actual process. The complexity of a process model will depend on the final usc of the model. If only an approximate answer is needed, then a simplified model can often be lIsed.
1.2.1
How Models Are Used
As we have noted, given a set of input data, a model is used to predict the output "response." A model can be used to solve the following types of problems: Marketing: If the price of a product is increased, how much will the demand decrease? • Allocation: If we have several sources for raw materials, and several manufacturing
101' course, the real conservalioll law is that of mass-energy, but we will neglect the interchange of mass and energy due to nuclear reactions.
Introduction
6
Chap. 1
plants, how do we distrihute the raw materials among the plants, and decide what products each plant produces?
Synthesis: What process (sequence of reactors, separation devices, etc.) can he used to manufacture a product? • Design: What type and size of equipment is necessary to produce a product? Operation: What operating conditions willmaxirnizc the yield of a product? Control: How can a process input be manipulated to maintain a measured process output at a desired value? Safety: If an equipment failure occurs, what will he the impact on the operating personnel and other process equipment? Environmental: IIow long will it take to "biodegrade" soil contaminated with hazardolls waste? Many or the rnodels cited above arc based on a steady-state analysis. This book will cxtend the steady-state material and energy balance concepts, generally prescnted in an introductory textbook on chemical engineering principles, to dynamic systems (systems where the variables change with time). As an example of the increasing importance of knowledge of dynamic behavior, consider process design. In the past, chemical process design was based solely on steady-statc analysis. A problem with performing only a steady-state design is that it is possible to design a process with desirable steady-state characteristics (minimal energy consumption, etc.) but which is dynamically inoperable. Hence, it is important to consider the dynamic operability characteristics of a process during the design phase. Also, batch processes that arc commonly used in the pharmaceutical or specialty chemicals industries arc inherently dynamic and cannot bc simulated with steady-state models. In the previous discussion we have characterized models as stcady-state or dynamic. Another characterization is in ter111s 01" lumped porarneter system.)· or distributed parameter systems. A lumped parameter systcm assumes that a variable or interest (temperature, for example) changes only with one independent variable (time, For cxamjJlc, but not space). A typical example of a lumped parameter system is a perfectly mixed (stirred) tank, where the temperature is uniform throughollt the tank. A distrihuted parameter system has more than one independent variable; for example, temperature may vary with both spatial position and time.
EXAMPLE 1.1
A Lumped Parameter System Consider a perfectly insulated, well-stirred tank where a hot liquid stream at 60 uC is mixed with a cold liquid stream at lone (Figure 1.1). The wc11~mixed assumption means that the fluid temperature in the tank is uniform and equal to the temperature at the exit from the t<mk. This is all example of a lumped parameter system, since the temperature docs not vary with spatial pOSition.
Sec. 1.2
7
Models Hot
Cold
iii
Outlet
FIGURE 1.1
Stirred lank.
Consider now the ~leady-stale behavior of this process. If the only stream was the hot fluid, thell the outlet temperature wOl/ld be eqllal to the hot fluid temperature if the tank were perfectly insulated. Similarly, if the only stream was the cold stream, then the outlet temperature would be equal to the cold lluid temperature. A combination of the two streams yields an outlet tClJlpcrutufc that is intermediate between the cold and hot temperatures, as shown in Figure ] .2.
60
o
f?
50
'0
@ .2 40 ~
'"
Q.
~ 30
l-
ID
8 20 10
_~~------.l.-
o
0.2
------l~~~~~~~~
0.4
0.6
0.8
Hot Flow, fraction
FIGURE 1.2
Relationship between hot flow and outlet temperature.
We sec that there is a linear steady-state relationship between the hot flow (fraction) and the outlet temperature. Now we consider the dynamic response to a change in the fraction of hot flow. Figure 1.3 compares outlet temperature responses for various step changes in hot flow fraction at t :::: 2.5
Introduction
8
Chap. 1
minutes. We sec that the changes afC symmetric, with the same speed of response. We will find later that these responses afC indicative of a lincar system.
+20%
45
o
40
'" 35 f-~"-~~~~----Q~-o~---~-~~ ~ . .- ---ci:
E
230
-20%
25
20
, ..
o
5
10
15
,.--'-----
20
25
30
time, min
FIGlJRE 1.3
EXAMPLE 1.2
Response of temperature to various changes in the hoI llow fraction.
A Histributed Parameter System
A simplified representation of a counterflow heat exchanger is shown in Figure lA. A cold water stream flows through one side of the excbanger and is heated by energy transfcrcd fro!ll a condensing steam ,,,trcam. This is a distributed parameter systcnl because the tClnperature of the water stream can change with lime and position.
Water in
~I-------F=: FIG-URE 1.4
Steam in
Counterflow heat exchanger.
The steady~slate tell1perature profile (water temperature as a function of position) is shown in Figure 1.5. Notice that rate of change of the water temperature \vith respect to distance decreases as the water tcmpenlturc approaches the steam teinpcrature (I OOOe). This is because the tcmper-
Sec, 1,2
9
Models
ature gradient rOf heat transfer decreases as the water temperature increases. The outlet water temperature as a function of inlet water temperature is shown in Figure 1.6.
100
80 0
'" ""ci.
60
E
JE
40
20
a
0.2
0.6
0.4 Z,
Flf;LJlU: 1.5
0.8
distance
W::tter temperature as
it
function of position.
100 98 96 0
'" "'" J-' E
94 92
~
0
90 88 86
a
20
40
60
80
100
inlet temperature, deg C
FIGURE 1.6
Outlet water temperature;Is a function of inlet water temperature.
Introduction
10
Chap. 1
Mathematical models consist of the following types of equations (including combinations) Algebraic equations Ordinary differential equations Partial differential equations The emphasis in this textbook is on developing models that consist of ordinary differential equations. These equations generally result from macroscopic balances around processes, with an assumption of a perfectly mixed system. To find the steady-state solution of a set of ordinary differential equations, we must solve a set of algebraic equations. Partial ditTerential equation models result from microscopic b
1.3
SYSTEMS We have been using the term system yery loosely. Consider the following definition.
Dctinition:
Svstem
A combination of several pieces of equipment integrated to perform a specific function; thus a fire (arlillery) control system may include a tracking radar, computer, and gun (McGraw-Hill Dictiolimy (?f5;ciellt!fic and Technical Tenns).
The example in the definition presellted is of interest to electrical, aeronautical, and rnilitary engineers. I'or our purposes, a system witl be composed of chemical unit operations, such as chemical reactors, heat exchangers, and separation devices, which arc used to producc a chemical product. Indeed, we will oftcn consider a single unit operation to be a system composed of inputs, states (to be defined later) and outputs. A series of modules in Section V of this textbook cover the behavior of a number of specific unit operations.
1.3.1
Simulation
One of the goals of this textbook is to develop numerical analysis techniques that allow liS to "simulate" the behavior of a chemical process, Typically, steady-state simulation of a lumped parameter system involves the solution of algebraic equations, while dynamic simulation involves the solution of ordinary differential equations. We must be careful when using computer simulation. First of all, we must he able to say, Do the results of this
Sec. 1.3
11
Systems
simulation make sense? Common sense and "back of the envelope" calculations will tell us if the numerical results arc in the ballpark.
1.3.2
linear Systems Analysis
The mathematical tools that arc lIsed to study linear dynamic systems problems me known as linear systems analysis techniques. Traditionally, systems analysis techniques have been based on linear systems theory_ Two basic approaches arc typically used: (i) Laplace transforms afC lIsed to analyze the behavior of a single, linear, nth order o("ciinary differential equation, and (ii) state space techniques (based on the linear algebra techniques of eigenValue and eigenvector analysis) arc used to analyze the behavior of multiple first-order linear ordinary differential equations. If a system of ordinary difreren~ tial equations is nonlinear, they can be linearized at a desired steady-state operating point.
1.3.3
A Broader View of Analysis
In this textbook we usc analysis in a broader context than linear systems analysis that may be applied to a model with a specific vallie for the pararnetcrs. Analysis means seeking a deeper understanding of a process than simply performing a simulation or solving a set of equations for a particular set of parameters and input values. Often we want to understand how the response of system variablc (tempera.ture, for example) changes whcn a parametcr (c.g., heat transfer coefficient) or input (fJowrate or inlet temperature) changes. Rather than trying to obtain the understanding of the possible types of behavior hy merely running many simulations (varying parameters, etc.), we must decide which parameters (or inputs or initial conditions) are likely to vary, and use analysis techniques to determine if a qualitative change ofhehavior (number of solutions or stability of a solution) can occur. This qualitative change is illustrated in l';'igure 1.7 below, which shows possihle steady-state behavior for ajackcled chemical reactor. In l<'igure 1.7a there is a monotonic
Jacket Temperature
a. Monotonic
Jacket Temperature
b. MUltiplicity
FIGlJRE l.7 Two qualitatively differclll types of input/output behavior. Steady-slale reactor temperature as a function of steady-state jackct lelnperalurc.
Introduction
12
Chap. 1
relationship between jacket temperature and reactor temperature, that is, as the s(cady~ state j;:lckct temperature increases, the steady-state reactor temperature increases. Figure 1.7b illustrates behavior known as output multiplicity, that is, there is a region of stcadystate jacket temperatures where a single j'H.:kct temIJcraturc can yield three possible reactor temperatures. In Chapter 15 (and module 9) we show how to vary a reactor design pa~ wllleter to change from one type of behavior to another. Engineering problem solving can be a comhination of art and science. 'rile complexity and accuracy of a solution will depend on the information available or what is dc~ sired in the final solution. If you arc simply performing a rough (back of the envelope) cost estirnatc for a process design, perhaps a simple stcady~state material and energy bal~ ance will suffice. On the other hand, if an optimum design integrating several unit operations is required, then a more complex solution will be involved.
1.4
BACKGROUND OF THE READER It is assumed that the reader of this textbook htls a sophomore- or jUllior~levcl chemical engineering background. In addition to the standard introductory chemistry, physics, and mathematics (including dilTerential equations) eourse,S, the 'student has taken an introduction to chemical engineering (material and energy haJ;:mces, reaction stoichiometry) cour,Se. This textbook can also be used by an engineer in industry who nceds to develop dy~ namic models to perform studies to improve a process or design controllers. Although a few years may have lapsed since the engineer took a differential equations course, the review provided in this text should be sufficient for the development and solution of models based on differential equations. Also, this textbook can serve as review material for firstyear graduate student who is interested in process modeling, systems analysis, or numerical methods.
1.5
HOW TO USE THIS TEXTBOOK The ultimate objective of this textbook is to be able to model, simulate, and (more importantly) understand the dynamic behavior of chemical processes.
1.5.1
Sections
In Section I (Chapters ] and 2) we show how to develop dynamic models for simple chemical processes. Numerical techniques for solving algebl'
Sec. 1.6
Courses Where This Textbook Can Be Used
uks 1 through lO) consists or a number of learning modules cussed in Sections I through IV.
1.5.2
13 10
reinforce the concepts dis-
Numerical Solutions
It is much casier to learn a new topic by "doing" rather than simply reading about it. To understand the dynamic behavior of chemical processes, one needs to he able 10 solve differential equations and plot response curves. We have used the MATLAB numerical analysis package to solve equations in this text. AMATL.AB learning module is in the set
of modules in Section V (Module I) for readers who arc not
l~ulliliar
with or need to be
reintroduced to MATLAB. MATLAB routines arc detailed within the chapter that they arc used. Many of the examples haveMA"rLAB m-files associated with them. It is recom~ mended that the reader modify these Ill-files to understand the effect of parameter changes
on the numerical solution.
1.5.3
Motivating Examples and Modules
This textbook contains III any process examples. Often, new techniques arc introduced in the examples. You are c.ncouraged to work through each example to understand how a palticular technique can be applied. There is a limit to the complexity of an example that can he used when introducing a new technique; the examples presented in the chapters tend to be short and illustrate one or two numcricaltcchniqucs. 'fhere is a set of modules of t11odeb, in Section Y, that treats process examples in much more dclail. The objective of these modules is to provide a more complete treatment of modeling and simulation of a specific process. In each process modehng module, a numher of the techniques introduced in various chapters of the text arc applied to the problem at hancL
1.6
COURSES WHERE THIS TEXTBOOK CAN BE USED This textbook is based on a required course that 1 have taught to chemical and environmcntal engineering juniors at Renssclaer since 1991. This dynamic systems course (originally titled lumped parameler syslems) is a prerequisite to a required course on chemical process control, normally taken in the second semester of the junior year. This textbook can be used for the first term of a two~tcrm sequence in dynamics and control. I have not treated process control in this text because 1 feci that there is a need for more in-depth coverage of process dynamics than is covered in most process control textbooks. This textbook can also be lIsed in courses such as process modeling or numerical methods for chemic111 engineers. Although directed towards undergraduates, this text can also be used in a first-year graduate course on process modeling or process dynamics; in this case, much of the focus would be on Section IV (nonlinear analysis) and in-depth studies of the modules.
Introduction
14
SUMMARY At this point, the reader should be able to define or characterize the following • Process model Lumped or distributed parameter system
Analysis Simulation
FURTHER READING The textbooks listed below are nice introductions to material and energy balances. The Russell and Denn book also provides an excellent introduction to models dynamic systems.
or
Felder, R.M., & R. ROllsseau. (1986). Processes, 2nd cd. New York: 'Wiley.
ElementalY Principles
ql Chemica!
Hinmlclblau, D.M. (1996). Basic Principles and Colclllation.'; ill Chemical E'/lJ;i/leering, 6th cd. Upper Saddle River, NJ: Prentice-Hall. Russell, T.R.F., &M.M. Denn. (1971). Introduction to Chemico! !'.'nginecrin/!.
Ihudysis, New York: Wiley. A comprehensivc tutorial and reference for MA'fLAB is provided by Hanselman and Littlefield. Thc books by l~ttcr provide many excellent examples Llsing MATLAB to solve engineering problems. lIanselman, D., & B. Littlefield. (1996). Maslerill,\!. MATLAB. Upper Saddle River, NJ: Prentice-Hall. [~lter,
n.M. (199]). Engineering ProlJlem Soh'ing Ivilh MATLAB. Upper Saddle River, NJ: Prenticc-Ilall.
E~tter,
D.M. (t096). Introduction to MA TLAU ,Ii)r engineers (/nd c)'CiCllfists. Uppcr Saddle Rivcl', NJ: Prentice-rlalt.
The following book by Dcnn is a graduate-level text that discusses the more philosophical issues involved in process modeling.
Denn, M.M. (1986). Process iV/odcling. New York: Longman. An undergraduate control textbook with significant modeling and sitnulation is by Luyben. Numerous FORTRAN examples are presented. Luyben, \V.L. (1990). Process Modeling, 5'imulatioll alld Control for Chemical EI1Rineers, 2nd cd. New York: McGraw-HilL
Student Exercises
15
A nmnber or control textbooks contain a limited amount of Illodeling.
[~xal11p1cs
include:
Seborg, D.E., TJ'·. Edgar, & D.A. Mellichamp. (1989). Process Dynamics and COIltrol. New York: Vhley. Stephanopmdos, G. (l9X4). Chemical Process Control: An Introduction to Theory and Pract{(·e. r':nglewood Clift's, NJ: Prentice-Hall. The following book by Rameriz is more of all advanced undergraduHtelfirst-ycar graduate student text on numerical methods to solve chemical engineering problems. The emphasis is on FORTRAN subroutines to be used with the IMSL nmlleriea! package. Rameriz, W.F·, (1989). Computational 111ethods for Process SiJllulalioJl. Boston: Butterworths. Issues in process modeling are discllssed by IIimlllelblau in Chapter :1 or the following book: Bisio, A., & R.I.. Kabel. (1985). S'c'oleujJ
(~l ('hemicol
Processes. Nc\v York: V.,Iilcy.
The following book by Reich provides an excellent perspective on the glohal ecollomy
and role played by U.S. workers Reich, R.B. (1991). The Work qFNatiolls. New YOJ'k: Vintage Books,
STUDENT EXERCISES 1. Review the matrix operation:-; module (Section V). 2. \Vork through the MATLAB module (Section V),
3. Consider example 2. If there is a sudden increase in stearn prc.l-;surc (and therefore, temperature) sketch the expected eoJd~sjde temperature profilel-; at 0, 25, 50, 75, and loor;;;) or the distance through the heat exchanger.
PROCESS MODELING
2
In this chapter, a methodology for developing dynamic models of chcmicL11 processes is presented. After studying this chapter, the student should be able to: \Vritc balance equations using the integral or instantaneous methods. Incorporate appropriate constitutive relationships into the equations. Determine the state, input and output variables, and parameters for a particular model (set of equations). Determine the necessary in!~)tmatjon to solve it system of dynanlic equations. Define dinlcnsionlcss variables and parameters to "scale" equations. 'file major sections arc: 2.1
16
Background
2.2
Balance Equations
2.3
Material Balances
2.4
Constitutive Relationships
2.5
Material and Energy Balances
2.6 2.7 2.8 2.9
DistribulcdParamclcr Systems DimcllsionlcssModels Explicit Solutions loDynamic Models General ItOI'm
or Dynamic Models
Sec. 2.2
2.1
17
Balance Equations
BACKGROUND Many reasons for developing process modeLs were given in Chapter 1. Improving or unm(~jor overall objective for developing a dynamic process model. These models arc often used for (i) operator training, (ii) process design, (iii) safely system analysis or design, or (iv) control system design.
derstanding chemical process operation is a
Operator Training. The people responsible for the operation of a chemical manufacturing process arc known as process operators. A dynanlic process model catl be lIsed to perform simulations to train process operators, in the same fashion that flight sjm~ ulators arc used to train airplane pilots. Process operators call learn the proper response to upset conditions, before having to experience theill on the actual process. Process Design. A dynamic process model call be llsed to properly design chcmical process equipment for a desired production rate. F'or example, a model of a batch chemical reactor can be used to determine the appropriate size of the reactor to pro~ duce a certain product at a desired rate. Safety. Dynamic process models can aLso be used to design safety systems. For example, they can be llsed to determine how long it will takc after a valve Ltils for a system to reach a certain pressure. Control System Design. Feedback control systenls arc used to maintain process variables at desirable valucs. For cxample, a control system may measure a product temperature (an output) and adjust the steam Ilowratc (an input) to maintain that desired temperature. r'or complex systems, particularly those with many inputs and outputs, it is necessary to base the control system design on a process model. Also, before a COI11plex control system is implemented on a process, it is normally tested by simulating the expected performance u~ing computer simulation.
2.2
BALANCE EQUATIONS The emphasis in an introductory matcrial and energy balances tcxtbook is on steady-state balance cquations that have the rollowing form:
mass
I F~qlU1tion
01·
--mass or e'l1crgy :
energy]
cntcnng a systcm
-
leaving (l
=
0
(2. I )
systcm
(2,1 )is deccptively simple becausc there may be many ins and outs, particulmly for component balances. The in and out terms would then include thc gcneration and con-
Process Modeling
18
Chap. 2
version of species by chemical reaction, respectively. In this text, we arc interested in dynamic balances that have the form:
1-,
rate of mass (~r CT~crgyl rate of mass or accumulation 111 = .energyentering [ [ a system a system
-
ale of mass or)
enelgy leavmg a system
(2.2)
The rate of mass accumulation in a system has the form dM/dl where M is the total mass in the system. Similarly, the rate of energy accumulation has the form dEJdt where E is the total energy in a system. H Hi is llsed to represent the moles of component i in a system, then dN/dt represents the molar rate of accumulation of component i in the systclll. \Vhcn solving a problem, it is important to specify what is meant by system. In some cases the systcl1lmay be microscopic in nature (a e1i/Terential clement, for example), while in othcr cases it may be macroscopic in naturc (the liquid content of a mixing tank, for example). Also, when developing a dynamic model, we can take one of two general viewpoints. One viewpoint is based on an ifltegral balance, while the other is based on an instml{(lIIcoll.\· balance. Integral balances arc particularly useCul when developing models for distributed parameter systems, which result in partial differential equations; the focus in this text is on ordinary differential equation-based models. Another viewpoint is the instantaneous balance where the time rate of change is written directly.
2.2.1
Integral Balances
An integral balance is developed by viewing a system at two different Sll
mass or energy inside the system
[
atl+tJ.1
I
[mass or energy -
I
inside the system all
. mass or energy entering the system from t to t --I- 6.t
r
-
1
mass or energy ] leavmg the system
It om I to I
+
(2.3)
tJ.1
The rnean-value theorems of integral and differential calculus are thcn used to reduce the equations to differential equations. F·'or example, consider the system shown ill F'igure 2.1 below, where one boundary represents thc mass in the systcm at time t, while the other boundary represents the mass in the system at t + i1t. An integral balance 011 thc total mass in the system is writtcn in the form:
Sec. 2.2
Balance Equations
19
mass in system at time
)If
,r
=:
t
mass in system at time:= t+!\ t
mOUf (f) FIGURE 2.1 Conceptual material balance prohlem.
n~ass contained III
the system
I rn~ass -
In
all+Llt
I
contained the system
mass culcI ing the system
=
r
from t to (
all
j-
D.,
Ir -
mass ICdving the system
I
hom I to t -I nl
Mathematically, this is written: I I ~I
fiLii
MI, '"' - MI, J ';'i" iiI - J 11/"", lif or 1-1-,1/
MI'i"' - MI,
=
J(,izi" - ,iz,,,,,l dl ,
(2.4)
where M represents the total mass in the system, while filill andli1()111 represent the mass rates elltering and leaving the system, respectively. We can write the righthand side of (2.4), using the mean value theorem of integral calculus, as:
f,
1+ !:J./
(';/ill - filma> dl
where 0 < (X < 1. Equation (2.4) can now be written:
dividing by dt,
MI"",__ MI, = ( . . II D..f 111'-1/ -
lJlo"t
l!nM
and using the mean value theorem of dUferentia! calculus (0 < ~ < I) for the Icfthand side,
MI"",~ />1
MI,
,1M,
;it
[+IUI
20
Process Modeling
Chap. 2
which yields dM ~i;- I I ~
. .
I
= (1l1 iJi - In""J II
lIj.f
Taking the limit as At goes to zero, we find dM dl
.
= mill -
.
(2.5)
111,,,,(
and representing the total mass as M = VP"h ifl as FillPill andlilo!ll as FOII/p, where p is the mass density (mass/volume) and F is a volumetric nowrate (volume/time) we obtain the equation: (2.6)
Note that we have assumed that the system is perfectly mixed, so that the density of material leaving the system is equal to the density or material in the system.
2.2.2
Instantaneous Balances
Here we write the dynamic balance equations directly, based on an instantaneous change:
I I
l
the laic 01 accumulatIOn of
mass
III
the system
·
I
rate of mass entering
=.::
I
the system
]
1.. -
rate of
.1
mass leaving
the system
ratc~of~
(2. 7)
which can be written directly as,
dM
df ---
J11 in .- l1I o l1!
(2.8)
or dVp ~ d/
(29)
which is the same result obtained using an intcgral balancc. Although the integral balancc takes longer to arrivc at the same resull as the instantaneous balancc method, the integral balance method is probably clearer whcn dcvcloping distributed parameter (partial differential equation-hased) models. An cxnmplc is shown in Section 2.6. Section 2.3 covers material balances and Section 2.5 covers material and energy balances. Section 2.4 discusses constitutive relationships.
2.3
MATERIAL BALANCES 'l'he simplcst modeling problems consist of material halances. [n this section we usc several process exarnples to illustrate the modeling techniques used,
Sec. 2.3
21
Material Balances
-~----~----
EXAIVIPLE 2.1
----
----~-------------
Liquid Suq.:e Tank
Surge tanks arc often Llsed to "smooth" f10wratc fluctuations in liquid streams f!mvillg between chemical processes. Consider a liquid surge tank with one inlet (flowing from process I) and one outlet stream (flowing to process Ii) (Figure 2.2). Assume that the dellsity is constant. Find how the volume of the tank varies as a function of time, if the inlet and outlet flowratcs vary. List the s((ltc variables, [Jarometers, as weI! as the il/fJut and oll/Im/voriables. Give the necessary information to complete the quantitative solution to this prohlem.
v •• F
_ _ _ _ _- L
FIGUnE 2.2
L,iquid surge tank.
The system is the liquid in the lank, the liquid surface is the top boundary of the system. 'rile following notation is used in the modeling equations: F,
~
inlet volumetric r10wrate (volume/time)
F
~
outlet volumctric f10wratc
V
volumc of liquid in thc tank
P
liquid dcnsity (mass/volullle)
Integral i\'lethod Consider a fillite time intcrval, !J.l. Performing a material balance over that time interval,
J11dSS 01 \Vdlel] [tlldSS 01 WdtCl ] lllslde the lank - lllSldc the td.nk
!J.t
elt I
mass of water] [lllass of walcr entering lank leaving tank
froJ11 ( to I -f- !:it
from I to t 1- ill which we can write InalhcmaticaHy as: (', .3.1
Jr:p dt
,
ESCOL/\ Dc Ei CIULlO
i
1\':'1
J
,
Fpdl
(2.10)
Procoss Modeling
22
Chap.2
Bringing the righthand side terms under the same integral
J(r,:p
1;"\/
vpl, ..\< .- vpl, -
I-i,) rlr
(2.11 )
We can lise the mean value theorem of integral calculus to write the righthand side of (2.11) (where 0 So: 0' :::; 1) as: I !j{
J(1';1' --- I'p)
dl
(2.12)
Vpll'''' vpll
(2.13)
,
Substituting (2.12) into (2.11)
Dividing by f),.t, we obtain
vpl, ,', __ JlPI, ~ (r- r 11 b.t
IP
PI","
(214)
and using the lucan value theorem of differential calculus. as !\.r' ;. 0
dVp dt
(2.15)
InstantaneOllS Method Here we write the balance equations based on an inslautancOU:i rate of-change:
-the rate of ch,~ngc of -I = I.mass n~l\vratc (}C..I Imass flowratc of-I 1mass of water tank _water lllln tank water out of tank_ III
The total mass of water tn fhe tank is Vp, the rate of change is dVp/dt, <:Inc! the density of the outlet stream is equal to the tank contents:
dyp dt
(2.16)
which is exactly what we derived using the intcf{ral method. Given the same set of assumptions the two methods should yield the sam<.~ rnodel. You should usc the approach (integral or instantaneous) that makes the most sense to YOL!o In this. text we generally usc the illstantaneous approach since it requires the fewest number of steps. Notice the implicit assumption that the density of water in the tank docs not depend upon position (the perfect mixing assumption). This ai>sumption allows an ordinary differential cqlJa~ tion (ODE) formulation. \Ve refer to any system that can be modeled by ODEs as lumped parameIer systems. Also notice that the outlet stream density must be equal to the density of water in the tank. This knowledge also allows us to say that the density terms in (2.16) arc equal. This equation is then reduced 101 lit might he tempting to the reader to begin to direuly write "volume balance" exprt'ssions that look similar to (2.17). We wish to make it clear that therc is llO such thing as a volume balance and (2017) is only correct hecause of the constant density assumption. It is a good idea to always write a mass balance expn:ssion, such ::1S (2.16), before making as~;umpti()ns about the fluid density, which Jllay lead to (2.17).
Sec. 2.3
Mntcrial Balances
23 dV
1-; -- F
dt
(2.17)
Equation (2.17) is it lineDr ordinary dilTcwntial equation (ODE), which is trivial to solve if we know the inlet and outlet flowmtes as a fllnction of time, and if we know all initial condition for the volume in the wnk. In equation (2.17) we refer to Vas a slale variable, and F; nnc! F as input variah!cs (even though Fis an outlet stream flowratc). If dellsity remained in the equatioll, \VC would refer to it as a /NlrWJlctcF. tn order to solve this problem \ve nlUst specify the inputs F/t) andF(t) and the initial con·
dition \1(0).
Exanlple 2.1 provides all introduction to the notion of states, inputs, and parameters. This cxarnple illustrates how an overall material balance is llsed to find how the volume of a liquid phase system changes with time. It may he desirable to have tank height, II, rather than tank volume as the slate variable. If we aSSUlllC a constant tank cross-sectional area, ll, we call express the tank volumc as V = Aft and the modeling eqU<:ltion as dli dl
F
A
(2.18)
it
If we also know that the flowrate out of the tank is proportional to the square ront of the height of liquid in the tank, wc can usc the relationship (sec student exercisc 21) F = [3VIi
(2.19)
where 13 is a J'lmv coefficient, to find dli
[3VIi
dt
A
,( -I
A
(2.20)
For tbis model \ve refcr to h as the stole variablc, inlet nowrate (F) as the input variable and f3 and A as parameters. Notice that a single system (ill this case, the liquid surge tank) can have slightly different modeling equations and variables, depending on assumptions and thc objectives used whcn dcve-Jopiug the model.
EXAi\iIPI.,E 2.2
An Isothernwl Chemi<:al Rcactor
Assume that two chemical species,;\ and 8, arc in a solvent feedstream entering a Iiquid~phase chemical reactor that is maintained at a constant tcmperature (l"igurc 2.3). The two spec'lc::, react irreversibly to form a third species, F. Find the reactor concentration of each species as a ftlllClion of time.
Process Modeling
24
Chap. 2
F. I
C
I
F C
FIGlJRE 2.3
Isothermal chemical reactor.
Overall Material Balance The overall mass balance is (since the tank is perfectly mixed)
(2.21) The liquid phase density, p, is not a function of concentration, The tank (and outlet) density is then cqllallo the inlet density, so:
ASSlIIllption:
Pi
P
(2.22)
and we can write (2.21) as:
dV tit
(2.23)
Component IVIaterial Balances It is convenient to work in molar units when writing component balances, particularly if chemical reactions arc involved. Let (,'1\' ('/I. and C" represent the molar concentrations of A, H, and P (moles/volume). Assume that the stoichiometric equation for this reaction is
The component material balance equations are (assuming no component P is in the feed to the reactor):
(Ive A dt dVC II dl dVC,. dl
(2.24)
(2.25)
(2.26)
Sec. 2.3
25
Material Balances
Where 1~4' rr;, and I'p represent the rate of generatioll of species;\, II, and P per unit volume, and (~li and C'/li represent the inlet coneentallons of species A and N. Assume that the rate of reaction of A per unit volume is second-order and 11 function of the concentration of both A and B. The reaction rate can be written (2.27)
where k is the reaction rate constant and the minus sign indicates that A is consumed ill the reaction. Each mole of A reacts with two moles of If (frotH the stoichiometric equation) and produces olle mote of I), so the rates of generation of Band P (per unit volume) arc: (2.2X)
(2.29) Expanding the [cftlland side of (2.24),
dVC A
f'
C dV
df
'/I
dt
(2.30)
combining (2.23), (2.24), (2.27), and (2.30) we find:
(2.31) Similarly, the concentrations of n aud P call be written
dell
(2.32)
df
(2.33) This model consists of four differential equations (2.23, 2.31, 2.32, 2.33) and, therefore, four state variables (V, c~\, C n. and C,,). To solve these equations, we must specify the initial conditions (V(O), CA(O), C/:I(t», and C,,(O)), the inputs (Fi . C Ai, and e/li) as a function of time, and the parameter (k).
2.3.1
Simplifying Assumptions
The reactor model presented in Exalnple 2.2 has four differential equations. Often other simplifying assumptions arc made to reduce the number of differential equations, to make theln easier to analyze and faster to solve. For example, assmning a constant volume (dVldt:::: 0) reduces the number of equations by one. Also, it is common to feed an excess of one reactant to obtain nearly completc conversion of another reactant. If species B is maintained in a large excess, then CHis nearly constant. The reactioll rate cqual'101l can then be expressed: I'.1
(2.34)
26
Process Modeling
Chap. 2
where kl
The resulting differential
equation~
=
(2.35)
kel/
are (since we assumed dVldt and dClldt::: 0)
de"
(2.36)
cit
(2.37) Notice that if we only desire to know the concentration of species A we only need to solve one differential equation, since the concentration A is not dependent on the concentration of P.
or
EXAMPLE 2.3
Gas Surge Drum
Smgc drums are oncn used as intermediate storage capacity for gas streams that arc lransfclTcd hctween chemical process units. COl1::-idcr a drum depicted in Figure 2.4, where (fi is the inlet Inola!" Ilowratc and q is the outlet molar tlowrate. Here we develop a model that describes how the pressure in the tank varies with time.
q. I
{
)--
p
FIGliRE 2.4
q
Gafi surge drum,
Let V = voJurne of the drum and V::: molar volume of the gas (volome/mole). 'rhe total amount of gas (moles) in the tank is then VI\/.
Assumption:
The pressure-volume relationship is characterized hy the ideal gas law, so PV= NT
(2.3H)
where P IS pressure, T is temperature (absolute scale), and R is the ideal gas constant. Equation (2.38) can be written
1
P
if
RT
(2.39)
and, therefore, the total amount of gas in the tank is
v V
PV
NT = total amount (moles) of gas in the tank
(240)
the rate of accumlation of gas is then d(PV/NT)/dt. Assume that T is constant; since Vand Rare also constant; then the molar rate of accumlatioll of gas in the tank is:
Sec. 2.4
27
Constitutive Relationships V tiP RT dt ~ 'Ii - 'I
(2.41)
where q,: is the molar rate of gas entering the drum and q is the molar rate of gas leaving the drum. Equation (2.41) can be written
dP dt
RT V (qi-q)
(2.42)
To solve this equation for the state variable P, we must know the illfJll1s lfi and q, the parameters R, 1; and V, and the initial condition P(O). Once again, although (j is the molar rate Ollt of the drulll, we consider it an input in terms of solving the model.
2.4
CONSTITUTIVE RELATIONSHIPS Examples 2.2 and 2.3 required Illore than simple material balances to define the modeling equation::;. These required relationships are known as constiflltive equations; several examples of constitutive equations are shown in this section.
2.4.1
Gas law
Process systems containing a gas will normally need a gas-law expression in the model. The ideal gas law is commonly used to relate molar volume, pressure, and temperature:
PV= RT
(2.43)
The van del' Waars FVT relationship contains two parameters (a and b) that are systcm~ specific:
(P + \12a) (V, - b) =
RT
(2.44)
For other gas laws, sec a thermodynamics text such as Smith, Van Ness, and Abholl (1996).
Chemical Reactions
2.4.2
The rate of reaction pCI' unit volumc (mollvolume*timc) is usually a function of the con~ centration of the reacting species.I;or example, consider the reaction A + 2B --> C + 3D. If the rate of the reaction of A is first-order in both A and B, we use the following expression: (2.'15)
where rA
k CA CjJ
is the rate of reaction of A (mol A/volume*time) is the reaction rate constant (constant for a given temperature) is the concentration of A (mol A/volume) is the concentration of B (mol B/volume)
Process Modeling
28
Chap.2
Reaction rates arc normally expressed in terms of generation of a species. The minus sign indicates that A is consumed in the reaction <:lbovc. Il is good practice to associate the units with all parameters in a model. For consistency in the units for 1"1\> we find that k has units of (vol/mol B* time). Notice that 2 mols of B react for each mol of A. Then we can write 1'8 ,--
2rA
=
r/!
- 2k C~,Cll k C't\C: n
rc
~
rf)
-- JrA c. 3
=
k C"C"
Usually, the reaction rate coefficient is a function of temperature. The most commonly used representation is the Arrhenius rate law k(T) = A expC -10/ /IT)
(2.46)
where k(T) :::: reaction rate constant, as a function of temperature A ;::; frequency nlctor or prccxponcntial factor (sanle units as k) 10 = activatioll ellergy (cai/glllol) R :;:;: ideal gas constant (1.987 caJlgmol K. or another set of consistent units) T :::; absolute temperature (dcg K or dcg R)
The frequency factor and activation energy can be estimated data of the reaction constant as a function of reaction temperature. Taking the natural log of the Arrhenius rate 1mv, we find:
in k
InA - F'
=
/I
(I)T
(2.47)
and we sec that A and E can be found from the slope and intercept of a plot of (In k) versus (117).
2.4.3
Equilibrium Relationships
The relationship between the liquid and vapor phase compositions of component i, when the phases arc in equilibrium, can be represented by:
v,
.
(2.48)
where Xi
::;:
liquid phase mole fraction of component i
Yi :;;: vapor phase mole fraction of component i
Ki
:;:;:
vapor/liquid equilibrium constant for component i
The equilibrium constant is a fUllction of composition and temperature, Often, wc \vill sec a ('otis/an! relative volatility aSSUlllption made to simplify vaporlliquid equilibrium models.
Sec. 2.4
Constitutive Relationships
29
In a binary system, the relationship between the vapor and liquid phases for the light component often used is: (xx l' =
.
I + (
(2.49)
x :::: liquid phase mole fraction of light component y ::;: vapor phase mole fraction of light component (X :::: relative volatility Co. >. I)
Heat Transfer
2.4.4
The rate or heal transfer through a vessel wall separating two fluids (a jacketed reador, for example) can be described by (2.50)
where
Q U ;\ fiT
:::: rate of heat trallsfcrcd from the hot tluid to the cold fluid ;:;;;; overall heat transfer coefficient ::::: area for heat transfer = difference between hot and cold fluid temperatures
The heat transfer coefficient is onon estimated from experimental data. At the design stage it can be estimated from correlations; it is a function of fluid properties and velocities.
Flow-through Valves
2.4.5
The flow-through valves are often described by the following relationship: F
( ',.t.(.\') \1JAP,. , s.g.
(2.51 )
where
F Cl' x
:::: volumetric f10wrate valve coefficient fraction of valve opening 0./\ :::: pressure drop across the valve s.g. :::: specific gravity of the fluid J(x) :::: the flow characteristic (varies from () to 1, as a function of .r) Three common valve characteristics arc (i) linear, (ii) equal-percentage, and (iii) quickopening.
Process Modeling
30
Chap. 2
0.8 ~ 0
~
quick-opening 0.6
15 c 0
t .:g
0.4
linear
02
equal·percentage
FIGURE 2.5 r~'low characteristics of control valves. (I::::::::' 50 for cqualpercentage valve.
0 0
0.2
0.4
0.6
0.8
fraction open
For a linear valve
f(x)
0"
x
For an equal-percentage valve
f(x)
,,'
1
For a quick-opening valve
f(x)
~
',Ix
The three characteristics arc compared in Figure 2.5. Notice that for the quick~opcning valve. the sensitivity of flow to valve position (fraction open) is high at low openings and low at high openings; the opposite is true for an cqual~pcrcclltagc valve. 'fhe sensitivity of a linear valve docs not change as a function or valve position. The equal-percentage valve is commonly used in chemical processes. because of desirable characteristics when installed in piping systems where ;:1 significant piping pressure drop occurs at high Ilowrates. Knmvledgc of these characteristics \vill be important when developing !cedback control systems.
2.5
MATERIAL AND ENERGY BALANCES Section 2.3 covered models that consist of material balances only. These arc useful if thermal effects arc not important, \vhere systcm properties, reaction rates, and so 011 do not depend on temperature, or if the system is truly isothennal (constant temperature). Many chemical processes have important thermal effecls, so it is necessary to develop material and energy balance Illodels. One key is that a basis must always be selected when evaluating an intensive property such as enthalpy.
Sec. 2.5
2.5.1
31
Material and Energy Balances
Review of Thermodynamics
Developing correct energy balall(:c equations is not trivial and the chemical engineering literature contains many incorrect derivations. Chapter 5 of the book by DCl1Il (1986) points out numerous examples where incorrect energy balances "vert used to develop pnlCcss models. The total energy erE) or a system consists of internal (U), kinetic (KE) and potential energy (Pf{):
TE
U -I KE
+ PI'
where the kinetic and potential energy terms arc:
1 , 2 111\'-
I'E
mgh
Often we will lise cncrgyltnolc or energy/mass and write the following A
U
no""~
TE
+ KE
A
-I 1'10'
U -I K I,' -I I' E
where II and represent per mole and per mass, respectively. The kinetic and potelltial ellergy terms, on a mass basis, arc 1
KE PE
,
2u ~
gil
[,'or most chemical processes where there arc thermal effects, we will neglect the kinetic and potential energy terms because their contribution is generally at least two orders of inagnitude less than thai of the internal energy tcnn. When dealing with flowing systems, we willuslIally work with enthalpy. Total enth;:Jlpy is defined as:
II
U -I pV
11
U I pV
while the enthalpy/mole is
and the enthalpy/mass is (since p::::: 1/( \I)
II
If -I pV
U I·
p
i'
we will make usc
or these relationships in the following example.
Process Modeling
32
EXAMPLE 2.4
Chap. 2
Stirred Tank Heater
Consider a perfectly mixed stirred-tank heater, with a single feed stream and a single product stream, as shown in Figure 2.6. Assuming that the f10wratc and temperature of the inlet stream can vary, that the tank is perfectly insulated, and that the rate of heat added per unit time (Q) can vary, develop a modcllo find the tank temperature as a function of time. Slate your assumptions.
F. I
~
T. I
V
T F T Q
FIGlJRE 2.6
Stirred tank
hl~ateL
Material Balance accumulation
=--=
in - out
dVp
(252)
dt
Energy Balance accumulation = in by flow - out by flow -+- in by heat transfer -+- work done on system dTE •• r;p,T Ei -FpTE + Q + W, df Here we neglect the kinetic and potential energy:
dU dt
~
FIIlI., I · I FI'U ,
f).' ~
WI
(2.53)
We write the total work done on the system as a combination or the shaFt work and the energy added to the system to get the-fluid into the tank and the energy that the system performs on the surroundings to force the fluid out.
(254) This allows us to write (2.53) as:
j.
Pi) _l'jJ(Uj p) + Q + W, Pi
P
(2.55)
Sec. 2.5 and since II
33
Material and Energy Balances "=
U + P\/, we can rewrite (2.55) as
tit
dpV tit
"'Pi Hi -- Fp 11 -I- Q -1-- W,
(2.56)
Since, rip Vld, = V dpldl + P dVldt, if the 1'011111/(-' is cOl/slall! and tile mean pressure change can be neglected (a good assumption for liquids), we can write
(2.57)
We must remember the assumptions that went into the development of (2.57): 'fhe kinetic and polential energy effects were neglected.
The change in p\l term was neglected. This is a good assumption for a liquid system, provided I\p is not too large and constant volume is assumed. The total enthalpy term is:
II
Vp 11
and assullling no phase change, we select an arbitrary reference temperature U:_,j) for enthalpy: r
ii(T)
~. Jc,,
'.
Often we aSSU!llc that the heat capacity is constant, or calcuated at an average temperature, so
ll;
1;,,)
c,,(7'
II =
(2.58)
c>( T; - 7~,,-,)
(2.59)
We now write the energy balance (2.57) In !he following fashion:
d(Vpc"CI' dt
T",))
(2.(>0)
lIsing the asslIIJlplio/ls of constant density and volume (so Fi
V'"
='
0::
F; from (2.52». we filld
F pc'I(T -- Trei.) - (T-- T rei )1 )· l
j
.
+
/il
(2.61 )
~
or
F V (/;- T) I 0::
Q Vpcf
-I !
w
(2.62)
Veep
dTidt. Also, neglecting WI' we can write
F
dt
V
/i (1-1')1" '
.
(2.63)
\fpc"
In ordcT to solve this problem, we must specify the parameters \I, p, (as a function of time), and the initial condition '1(0),
c[J'
the inputs F. Q, and Ti
Process Modeling
34
2.6
Chap. 2
DISTRIBUTED PARAMETER SYSTEMS In this section we show how the balance equations can be used to develop a model for a distributed parameter system, that is, a system where the state variables change with re~ spect to position and time. Consider a tubular reactor where a chemical reaction changes the concentration of the fluid as it moves down the tube. Here we usc a volullle clement 6. V and a time clement Lll. The total moles or species A contained in the clement 6. V is written (6. V)Ck The amount of species A entering the volume is FcAI v and the amount of species leaving the volume is FC) F-,-:.W' The rate of A leaving by reaction (assuming a first-order reaction) is ( k C~,)n
v.
The balance equation is then: I + ill
(n V) C:, I, ,'" - (n V)CAI,
f,
=
[FC"lv - Fc"lv"v - kC"n V] £It
Using the mean value theorem of integral calculus and dividing by D.f, wc find: (2.64)
Dividing by D.. V and letting ti.t and L.i. V go to zero, we find:
aFC --" av - kCA
iiCA
at
(265)
Normally, a tube with constant cross-sectional area is used, so elV;:::: Adz and F :::: Av z' where V z is the velocity in the z-dircction. Then the equation can be written: g(~A iJt
av_c, --'--'- - kC iJz
A
(2.66)
Similarly, the overall material halance can be found as: rJp
at
~Vi:P iJz
(2.67)
If the fluid is at a constant density (good assumption for a liquid), then we can write the species balance as
'1'0 solve this problem, we must know the initial condition (concentration as a function of distance at the initial time) and one boundary condition. For example, the following boundary and initial conditions C~,(z,
t = 0) = CAO(Z) CA(O, t) = CA;,,{t)
(2.69)
Sec. 2.7
Dimensionless Models
35
indicate that the concentration of A initially is known as a function of distance down the reactor, and that the inlet concentration as a function of time must be specified. In deriving the lubular reactor equations we assumed that species A left a volume element only by convection (bulk now). In addition, the molecules can leave hy virtue or a concentration gradient. For example, the amount elltering at V is (2.70)
where is the diffusion coefficient. The reader should be able to derive the following
reaction-diffusion equation (sec exercise 19). (JC~\
(27])
at
Since this is a second-order PDE, the initial condition (C~\ as a function of z) and two boundary conditions must be specified. Partial differential equation (PDEs) models are much more difficult to solve than ordinary differential equations. Generally, PDEs arc converted to ODEs by discretizing in the spatial dimension, then tcchniques for the solution of ODEs can be used. 'rile rocus of this text is on O[)Es; with a grasp or the solution of ODr~s, one can then begin to develop solutions to PDEs.
2.7
DIMENSIONLESS MODELS Models typically contain a large number of parameters and variables that may differ in value by several orders of magnitude. It is orten desirable, at least for analysis purposes, to develop models composed of dirnensionless parameters and variables. 'To illustrate the approach, consider a constant volume, isothcnnal CSTR moue led by a simple first-order reaction:
It seems natural to work with a scaled concentration. Defining x ~ C/C,,/0 where C AID is the nominal (steady-state) feed concentration of A, we rind
where Xj= CN/CAIO ' It is also natural to choose a scaled time, T = //1< where I'" is a scaling parameter to determined. We can usc the relationship df = f"'dT to write:
be
36
Process Modeling
Chap. 2
dx {:;:dr
A natural choice for 1*' appears to be VIF (known as the residence time), so
dx =x·(Vk) 1+ .···x I F
dT
The term VklF is dimensionless and known as a Damkholcr Ilmnber in the reaction engineering literature. Assuming that the feed concentration is constant, .'j::;: I, and letting Ci;::: Vk/F, we can write: dOl I ~X -+- (Xx (IT which indicates that a single parameter, (X, can be used to characterize the behavior of all first-order, isothermal chemical reactions. Similar results arc obtained if the dimensionless state is chosen to be conversion
2.8
EXPLICIT SOLUTIONS TO DYNAMIC MODELS Explicit solutions to nonlinear differential equations can rarely he obtained. The most common case where all analytical solution can be obtained is when a single differential equation has variables that are separable. This is a very limited class of problems. A main objective of this textbook is to present a number of techniques (analytical and numerical) to solve more general problems, particularly involving many simultaneous equations. In this section we provide an example of problems where the variahles are separable.
EXAlVIPLE 2.5
Noulincln Tank Height
Consider it Lank height problem where the outlet flow is a nonlinear function of lank height:
tilt d/
r~
Il
1\
i\
v1/
Here tbere is not au analytical solution because of the nonlinear height relationship and the forcing function. 1'0 illustrate a problem with an analytical solution, we will assume that there is no inler flow to the tank:
dh df \Ve can see Ihat the variables are separable, so
tilt
v'h
Il d/ i\
37
General Form of Dynamic Models
Sec. 2.9
II
dll
(
Vii
h,.
!
-j
IJ A ell
which has the solution ~
2 v'Ji ,- 2\/h"
A (1-
U
or letting
t,,;:::
0, and squaring both sides, \ve obtain tile solution
I fl' vii --~ "
iI(I)
"
0
2/\ .
This analytical solution call be used, for example, to determine the lime that it will take for the tank height to reach a certain level.
2.9
GENERAL FORM OF DYNAMIC MODELS The dynamic models derived ill this chapter consist of a set of first-order (tncaning only first derivatives with respect [0 time), nonlinear, explicit, initial value ordinary differential equations. A representation of" a sel first-order differential equations is
or
XI =
II(.tl" .. ,_rll,uj,... ,ltm,Pt,···,fJr)
);"} --: l:(x1"",x1I'ut,···,ll/ll' PI'''''!)/") (2.72)
where Xi is a state variable, IIi is an input variable and Pi i:-; a parameter. The notation.< is used to represent drldt. Notice that there arc II equations, II state variables, III inputs, and r parameters.
2.9.1
State Variables
A stale variable is a variable that arises naturally in the accumulation tefm of a dynamic material or energy balance. A state variahle is a Tncasurable (at least conceptu;dly) quantity that indicates the state of a systelll. For example, tern perature is the COll'lIllOn state variable that arises from a dynamic energy balance. Concentration is a state variahle that arises when dynamic component balances are written.
38
2.9.2
Process Modeling
Chap. 2
Input Variables
An input variable is
2.9.3
Parameters
A parameter is typically a physical or chemical property value that must be specified or known tomathcmatically solve a problem. Parameters arc often fixed by nature, that is, the reaction chemistry, molecular structure, existing vessel configuration, or operation. Examples include density. viscosity, thermal conductivity, heat transfer coefficient, and mass-transfer coefficient. When designing a process, a parameter might be "adjusted" to achieve some desired performance. For example, reactor volume lllay be an important design parameter.
2.9.4
Vector Notation
The set of differential equations shown as (2.72) above can be written more compactly ill vector form.
x =, f(x, II, II)
(2.73)
\vhere x::;;: vector of n state variables u = vector of In input variables p :::: vector of r parameters Notice that the dynamic models (2.73) can also be used to solve steady-state probJenls, since
x=
()
(2.74)
that is,
f(x,u,p)
=. ()
(2.75)
for steady~state processes. Numerical techniques (such as Newton's method) to solve algebraic equations (2.75) will be presented in Chapter 3. The steady~state state variables from the solution of (2.75) arc often used as the initial conditions for (2.73). F'requently, an input will be changed froni its steady-state value, and (2.73) will be solvcd to understand the transient behavior of the system. 'rhe numerical solution of ordinary differential equations will bc prescllted in Chapter 4. In thc example helow we show Example 2.2 (chcmicalrcactor) in state variable form.
Sec. 2.9 ------~
EXAMPLE 2.6
39
General Form of Dynamic Models
-~~------------- ---~-----
State Variable Form for Example 2.2
COlls'lder the modeling equations for Example 2.2 (chemical reuctor)
dV dl
~F-F
(2~23)
j
(2.31 )
(2.32)
(2~33)
There are four states (V, C~\, Cu' and C p ), four inputs (Fi , F, CAi, ('/Ii), and a single parameter (k). Notice that although F is the outlet fJowrate, it is considered all input to the model, because it must be specified in order to solve the equatiolls.
v CA F
Ii (CHi -- Cli) -
2 kC,C"
F
VC/, + kCAC/I
or
X2
x,
II,
Xl
i:
(lI J -
(11 4 --
11,
x 2)
-" P.X2 X;l
x.,) -- 2
i, x4 +
P,X 2X3
.
l)l X1X;;
I'I(X,II,P)]
___ ti x,U,IJ)
- [ f,(X,II,p) J 4(x,u,p)
Process Modeling
40
Chap. 2
SUMMARY
or
A number material and energy balance examples have been presented in this chapter. The classic assumption of a perfectly stirred tank was generally used so that all models (except Section 2.6) were IUlTlped-parameter systems, Future chapters develop the analytical and Ilmllcrical techniques to analyze and sinudatc these models. The student should now understand: that dynamic models or lumped parameter systems yield ordinary differential equations.
that steady-state models of lumped parameter systcrns yield algebraic equations. The notion of a state, inpul. output, parameter. A plethora of models arc presellted in modules in the final section of the textbook. More specifically,. the following modules arc of interest: Module 5. IJcatedMixing'rank Module 6. LillcarEquilibrium Stage Models (AbsorpLioll) Module 7. Isothermal Continuous Stirred 'Tank Reactors Module 8.
Biochemicall
Module 9. Diabatic Reactor Models Module 10. Nonlinear Equilihrium Stage Models (Distillation) Each of these modules covers Illodel development and presents examples for analytical and nUlllerical calculations.
FURTHER READING A nice introduction to dlClnical engincering calculations is provided by: Felder, R.M., & R. Rousse
L'lcmcnfmy Principles r!/ C'hemicrtl
Excellent discussions of the issues involved in modeling a mixing lank, incorporating density crfects, and energy balances is provided in the follO\:ving two hooks: Denn, M.M. (1986)./Jrocess Modeling. New York: Longman. RusselL T.R.F., & M.M.Denn. (1971). Introduction Allrdysis. New York: Wiley. An introduction to chemical reaction engineering js:
lo
Chemical Engineerillg
Student Exercises
41
Fogler, H.S. (1992). Elements (d' Chcmh'ol ReaUiol/ Engineering, -'.l1d eel. Englewood Cliffs, NJ: Prentice-Ilall. An exccllenttextbook for an introduction to chemica! engineering thermodynmnics is: Smith, 1.M., ftC. Vall Ness, & M.M. Abbott. (1996). Chcmhxd Engineering Ther· fIlorfvllomics, 5th cd. New York: McGraw-HilI. The following paper provides an advanced treatment of dimensionless variables and parameters: Ads, R. (1993). Ends and beginnings in the mathematical modelling or chemical engineering systems. Chemica! l-:.'ngineering 5'cience, 48( 14). 2507--2517. 'I'he relationships for mass and heat transport arc shown in textbooks on transport phenomena. The chernicaI engineer's bible is Bird, R.B., WJ~. Stewart, & E. Lightfoot. (1960). Transpurt Phcl/oIJ/('!la. New York: Wiley. The predator-prey model in student exercise 16 is also known as the Lotka- Volterra e(luations, after the researchers that developed them in the late 1920s. A presentation of the equations is in the following text: Bailey, J .E., & DJ<'. Ollis. (1986). Biochemical Engilleerinf!, Fundamentals, 2nd cd. New York: McGraw-HilI.
STUDENT EXERCISES r~xaJ1lple 2.1 it was assumed that the input and output f10wratcs could he independently varied. Consider a situation in which the outlet flow rate is a function of the height of liquid in the tank. Write the modeling equation for tank h~ight assuming two different constitutive relationships: (i) F:::: f3h, or Oi) F:::: f3 \/h, where [3 is known as a flow c;oefficicnL You will orten sec these relationships expressed as F = h/R or F =V/ZIR, where R is a flow resistance. List the state variables, parameters, as well as the input and output variables. Give the necessary information to complete the quantitative solution to this problem. If the f10wrate has units of liters/min and the tank height has units of meters, find the units of the flow coefficients and flow resistances for (i) and (Ii).
I. In
2. Consider a conical water lank shown below. Write the dynamic material balance equation if the flow rate out gr the tank is a function of' the square root of height of water in the tank (Fo :::: f3Yh). List state variables, input variables and parameters. (Hint: Usc height as a state variable.)
ESCOLA Dc Et'-JGEi'-IHARIA BIBLI01ECA
42
Process Modeling
,
•
R
Chap, 2
•
F
i
H
~
•
Fa
3. Extend the model developed ill Example 2.2 (isothermal reaction) to handle the following stoichiometric cqu;:llion: A + B --> 21'. Assume that the volume is constant, but the change in concentration of component B cannot be neglected. 4. 13xtcnd the model developed in Example 2.2 (isothermal with first-order kinetics) to handle multiple reactions (assume a constant volume reactor).
A
2A
+ B - -> 21'
(react;on I)
+ I' - -> Q
(reaction 2)
Assume that no P is fed to the reactor. Assume that the reaction rate (generation) of A per unit volume for reaction I is characterized by expression -k1 C'A
fA
en
where the minus sign indicates that A is consumed in reaction 1. Assume that the reaction rate (generation) of A per unit volume for reaction 2 is characterized by the expreSSIon I'll
=
- k2
c/\
C/,
[I' the concentrations are ex.pressed in gmolllitcr and the volume in liters, what arc the units of the reaction rate constants? If it is desirable to know the concentration or component Q, how many equations must be solved? If our COncern is only with P, how many equations must be solved? Explain.
5. Model a mixing tank with two reedstreams, as shown below. Assume that there arc two components, A and B. C represents the concentration of A. (C, is the mass concentration 01';\ in stream I and C2 is the mass concentration of A in stream 2). Model the following cases:
F C
43
Student Exercises
a. Constant volullle, constant density. h. Constant volume, density varies linearly with concentration. c. Variable volume, density varies linearly with concentration. 6. Consider two tanks in series where the flow out of the first tank enters the second tank. Our objective is to develop a model to describe how the height of liquid in tank 2 changes \vith time, given the input flow rate fi~I(t). Assume that the flow out of each tank is a linear function of the height of liquid in the tank (F J :;:;: I3lhl and F2 :;:;: l3i12) and each tank has a constant cross~sectional area.
•
h,
to. h2
7.
8.
9. 10.
11.
F,
:b=L
F 2
A material balance around the first tank yields (assuming constant density and P, ~ (3,11,) Two liquid smge tanks (with constant cross~seclional area) arc placed in series. Write the modeling equations for the height of liquid in the tanks assmning that the f!owrate from the first tank is a function of the difference in levels of the tanks and the f10wrate from the second tmik is a funclion of the level in the second tank. Consider two cases: (i) the function is linear and (ii) the function is a square root relationship. State all other assumptions. A gas surge drurn has two components (hydrogen and methane) ill the rcedstreHlll. Let Yi and y represent the mole fraction of lnethanein the feedstream and drum, respectively. Find dP/df and dy/dt if the inlet and outlet f]owrates can vary. Also as~ sume that the inlet concentration can vary. Assume the ideal gas law for the effect of pressure and composition on density. Consider a liquid surge drum that is a sphere. Develop the modeling equation using liquid height as a state variable, assuming variable inlet and outlet /lows. A car tire has a slow leak. The fJowratc or air out or the tire is proportional to the pressure of air in the tire (we ;Ire using gauge pressure). The initial pressure is 30 psig, and after five days tile pressure is down to 20 psig. flow long will it take to reach 10 psig? A car lire has a slow leak. The flowratc of air out of the tire is proportional to the square root of the pressure of air in the tire (we arc using gauge pressure). 'rhe initial pressure is 30 psig, and after 5 days the pressure is down to 20 psig. How long will it lake to reach 10 psig? Compare your results with problem 10.
Process Modeling
44
Chap. 2
12. A small room (IOn X Ion X lOft) is perfectly scaled and contains air at I atm pressure (absolute). There is a large gas cylinder (100 nJ ) inside tbe room tha( contains helium viith an initial pressure of 5 atm (absolute), Assume that the cylinder
valve is opened (at t = 0) and the molar f10wratc of gas leaving the cylinder is pro~ portional to the difference in pressure helween the cylinder and the room. Assume HUll room air docs not diffuse into the cylinder. Write the differential equations that (if solved) would allow you to find how the cylinder pressure, the room pressure and the roorn mole fraction of helium change with time. Slate all assumptions and show all of your work.
13. A balloon expands Or contracts in volume so that the pressure inside the balloon is approximately the atlllOspheric pressure. a. Develop the mathematical model (write the differential equation) for the V01UIllC of a balloon that has a slow leak. Lct V rcpresenllhe volume of the balloon and q represent the molar flowratc of air leaking from the balloon. State all assumptions. List state variables, inputs, and parameters. h. The following experimental data have been obtained for a leaking halloon. t (minutes)
o
r (em) I (J
5
7.5
Predict when thc radius of the balloon will reach 5 cm using two different assumptions for thc molar rate of air leaving the halloon: (i) ·rhe molar rate is constant. (ii) 'fhe molar rate is proporlionallo the surface area or the balloon. Reminder: The volumc of H sphere is 4IJ
'ITr 3
and the ,1rea of a sphere is 47Tr2.
14. Often liquid surge tanks (particularly those containing hydrocarbons) will have a gas "blanket" of nitrogen or carbon dioxide to prevent the acculllulation of cxplosive vapors above the liquid, as depicted below.
liqUid
• F
Develop the modeling equations with gas pressure and liquid volume as the state vari'lble;.;. Lel Cfr(lnd Cf repre;.;cnt the inlet and outlct gas molar flowrates, F"rand F the liquid volumetric f1owratcs, V the constant (total) volume, Vjthe liquid volume, and P the gas prcssure. Assume thc ideal gas law. Show that lhe modeling equations arc:
riP dl
P
V~V(Ff-F) I
and statc any other assumptions.
Student Exercises
45
15. Most chemical process plants have a natural gas header that circulates through the process plant. A siInplified version of such a header is shown below.
P
I
V,
From source
valve i Plant piping, represented as a perfectly mixed drum
Gas drum for a boilerhouse unit
To furnaces
Here, the natural gas enters the process plant from a source (the natural gas pipeline) through a control valve. It fltnvs through the plant piping, which we have represented as a perfectly mixed dnnH for simplicity. Another valve connects the plant piping to the gas drum for a boi!erhouse unit. Gas passes through another valve to the boilerhouse furnaces. Write modeling equations asslllning that the pressures in drums I and 2 arc the state variables. Let the input variables be h J (valve position 1),11 2 (valve position 2), and Pi (source pressure). 16. 'file Lotka- Volterra equations \Vere developed to mode! the behavior of prcdatorprey systems, making certain assumptions about the birth and death rates of each species. Consider a system composed of sheep (prey) and coyotes (predator). In the following Lotka- Volterr~1 equations Xl represents the number of sheep and X2 the number of coyotes in the system.
Discuss the meaning of the parameters model.
0', ~,
-y,
E
and the- assumptions made in the
17. Consider a perfectly mixed stirred-tank heater, with a single liquid feed stream and a single liquid product stream, as shown below.
F
i
T;
_
~
L j - - - - t - ' - - - - -.. F T
Process Modeling
46
Chap. 2
Develop the material and energy balance equations that describe this process. Fi is the volumetric nowrate into the tank, F is the volumetric rJOWfi:ltC oul of the lank, '1;is the temperature or the fluid entering the lank, r is the lcrnperalurc or the fluid in the tank, II is the height of liquid in the tank, and Q is the rate of energy added to the tank. State assumptions (such as constant density, etc.). _ Assume that the volume Call vary with time and that F is proportional to \1/1. How lllany differential equations docs it take to model this system? What arc the state variables'! What afC the parameters? What <:11"( the inputs? List the information necessary to solve this problem.
18. Consider a gas surge drum witb variable inlet and outlet molar flowratcs, {jr and q, res[Jectively. Asslilne that heat is being added to the tank at a rate, Q. \Vrite the modeling equations lhat describe how the temperature, and pressure, fl, vary with time. r)o not neglect the p V tenn in the energy balance. 19. Derive the reaction~dilJusionequation
1:
dC.\
at
= - v
neil ilz.
using the same method to derive lhe tubular reactor model in Section 2.6. As:mme that a chelnical species enters a v()lllIne clement via convection (bulk flow) and a concentration gradient (diffusion):
leaves by convection and a concentration gradiellt:
and also leaves by a first-order reaclion.
20. Consider the nonlinear tank heipht model
"II
F,
"I
A
13 VII
A
tmd define the dimensionless variables
II :::: F/F:I . and x:::: hlhs.. 'Where Fs and h. 1 are the stcady-stateflowrale and height. respeclively (VI' :::: ~\Ih). Definc the dimeIl+ sionless limc, T, that will yield the follO\ving dimensionless equation:
dx elT
=
--
\/x
+
U
21. I)erive the constitutive relationship F:::: fjVh by considering a steady-state energy balance around a tank with a constant flowrate. Use P :::: 1\, + pgh for the pressure al the bottom of the lank, where Po is the atmospheric pressure (pressure at the top surface), 11 is the height of liquid in the tank, p is the density of fluid. Assume lhat
47
Student Exercises
the cross-sectional area at the surface is much larger than the cross-sectional area of the exit pipe.
22. Consider the isothermal CSTR Inodel with first-order kinetics: dC,
dt
Use T =- kl as the dimensionless time. [)evclop the dimensionless cquation for t\",O cases: (i) _y ;;;:;: C,/(\l' and (ii) x =- I --- C,/(~\r Compare and contrast the resulling equations with the example in Section 2.7.
23. Semibatch reactors are operated as a cross between batch and continuous reactors. A semibatch reactor is initially charged with a volume of material, and a continous feed of reactant is started. There is, however, no outlet stream. !)eve(op the modeling equations for a single first-order reaction. The state variables should be volume and concentration of reactant A. 24. Pharmacokinetics is the study of how drugs infused to the body are distributed to other parts of the body. 'rhe concept of a cOlnpartmental model is orten used, where it is assumed that the drug is injected into compartment I. Some of the drug is eliminated (reacted) in compartmcnt I, and somc of it diffuses into cmnpartrnent 2 (the rest accumulates in compartment I). Similarly, some of the drug that diffuses into compartment 2 diffuses back into compartment I, while some is eliminated by reac~ tion and the rest accumulates in compartment 2. Assume that the rates of diffusion and reaction arc directly proportional to the concentration of drug in the compartmenl or interest Show thal the following balancc equations arise, and discuss the meaning of each parameter (k{i' units of min-I) dX I
dt
df
where Xl and x2 =- drug concentrations in compartments I and 2 (/-Lg/ml), and u =: ratc of drug input to compartment I (scaled by the volumc of compartment I, /-Lg/ml min).
SECTION
II NUMERICAL TECHNIQUES
ALGEBRAIC EQUATIONS
3
The purpose of this chapter is to introduce rllcthods to solve systems of algebraic equations. Arter studying this module, the student should be able to: Solve systems of linear algebraic cqua.lions. Solve nonlinear functions of one variable graphically and fllll11crically. Usc thcMAI'LAB function fzero to solve a single algebraic equation. Discuss the stability of iterative techniques. Usc thcMATLAB function f solve to solve sets of nonlinear algebraic equations.
The major sections in this chapter are:
3.1
3. J
Introduction
3.2
General r,'onn for a Lincar System of Equations
3.3
Nonlinear Functions of a Single Variable
3.4
MATLAB Routines for Solving Functions of a Single Variable
3.5
MuJtivariablc Systems
3.6
MATLAB Routines for Systeills of Nonlinear Algehraic Equations
INTRODUCTION In Chapter 2 we discussed how to develop a model that consists of a set of ordinary differential equations. To solve these problems we need to know the initial conditions and how the inputs and parameters change with time. Often the initial conditions will be the steady-state values of the process variables. To obtain a steady-state solution of a system of differential equations requires the solution of a set of algebraic equations. The purpose of this chapter is to review techniques to solve algebraic equations. 51
Algebraic Equations
52
Consider a set of 11 equations in
11
Chap. 3
unknowns. The representation is
fl(XI,xz,
x,,) ~ ()
flex I ,x2'
).:)
:=
0 (31 )
/;/(.\;\,X2' ... x,)::: 0
The objective is to solve f\1r the set of variables, Xi' that force all of the functions, .fi, to zero. A solution is called aJixed point or all. equilibrium poinf, Vector notation is used for a compact representation I'(x)
~
(}
(3.2)
where rex) is a vector valued function. Notice that these can be the same functional relationships that were developed as a set of differential equations, with x:= rex) := O. The SO~ lution x is then a steady-state solution to the system of differential equations. Before we cover techniques for systems of nonlinear equations, it is instructive to review systems of linear equations.
3.2
GENERAL FORM FOR A LINEAR SYSTEM OF EQUATIONS Consider a linear system with n equations and allxl
unknowns. The first equation is
11
+ "12x'2 + 0nrl + ... + all/XII = b l
where the a's and b's are known constant parameters, and the x's are the unknowns. The second equation is: anx I +
([22
x2 + ([2J~rJ +. . + ([2Ir\:1/ =
/;2
while the nth equation is:
The coefficient, aU' relates the jilt dependent variahle lo the itlt equation.
a 12
au
a21
an
aLl
alii
aliI
(lId
I"
"1 llZIl
XJ
b,
Xz
b2
x:'\
bJ
(3.3)
a/l/l
XII
Or, using compact matrix notation: Ax~b
(3.3a)
The goal is to solve for the unknowns, x. Notice that (3.3a) is the same form as (3.2), with f(x)
~
Ax - b
Sec. 3.2
General Form for a Linear System of Equat'lons
53
Prclllultiplying each side of (3a) by A--l we find: A-IAx=bu and since,
(3.4)
provided thallhe inverse of A exists. [I' the rank or A is less than 11, then A is singular and the matrix inverse docs not exist. If the condition number of A is vcry high, then the solution may he sensitive to model error. The concepts of rank and condition number are reviewed in Module 2 in Section V. Equation (3.4) is used for conceptual purposes to represent the solution of the sel of linear equations, In practice the solution is not obtained by finding the matrix inverse. Rather, equation (3.3a) is directly solved Llsing a numerical technique such as Gaussian elimination or LU decomposition. Since the codes to implement these techniques are readily availahle in any Ilumericallibrary, wc do not rcview thcm hcrc. The next example illustrates how MATLAli can he used to solve a system of lincar equations.
EXAI\ilPLE 3.t
Linear Ahsor(}tion Model, Solved Using l\:lATLAB
Consider as-stage ahsorption column (presented in Module 6 in Section V) that has a model of the following form (x is a vector of stage liquid-phase compositions and u is a vector of column feed compositions):
O=1\x+llu or Axc;::-~HII
the solution for x is x c;:: - A-I n 11 Thc valucs of A, n, and 11 are:
a -0.3250 0.2000 0 0
0
O.12'SO
o
a
-0.3250 0.2000 0 0
0.1250
o
-0.3250 0.2000
--0.3250
0.1250
o
0.2000
-0.3250
b
0 .2000 0 0 0 0 u
-
0 0 .1000
0 0
a 0 0.2500
0.1250
o o o
Algebraic Equations
54
Chap.3
The following MATLAB command can be used to solve for x
»x
~inv(a)*b*u
x
0.0076 0.0]98 0.0392 0.0704 0.1202
Usc of the MArLAB left-division operator (\) yields the same resull more efficiently (faster computation time), using the LU decomposition technique:
3.3
NONLINEAR FUNCTIONS OF A SINGLE VARIABLE Functions of a single variable Call be solved graphically by ploUing./(x) for many values of.t and finding the values of x whcrcj(x) :::;; O. This approach is shown in Figure 3. J. An interesting and challenging characteristic of nonlinear algebraic equations is the potential for mnltiple solutions, as shown in Figure 3.1 b. In fact, for a single nonlinear algebraic equation it is often not possible to even know (without a detailed analysis) the number of solutions that exist. The situation is easier for polynomials because we know that an nth order polynomial has 11 solutions. Fortunately, many chemical process problems have a single solution that makes physical sense. Numerical techniques for solving nonlinear algebraic equations arc covered in the next section. A graphical representation will be used to provide physical insight for the numerical techniques. Numerical methods to solve nonlinear algebraic equations are also known as iterative techniques. A sequence of guesses to the solution are made until we arc "close enough" to the actual solution. To understand the concept of "closeness" we must use the notion of convergence tolerance. f(Xi
f(Xi
x
°
x
The solution
a. Single Solution
FIGURE 3.t
b. Multiple Solutions
Graphical solution tof{x) "'" O.
Sec. 3.3
3.3.1
55
Nonlinear Functions of a Single Variable
Convergence Tolerance
A solution to a problcmj(x) :::; 0 is considered converged at iteration kif:
where 1::: is a tolerance that has been specified, !/(xk)! is lIsed to denote the absolute value of fUllctionJ(x) evaluated at iteration k, and x k is the variable value at iteration k. At times it is useful to base convergence on the variable rather than the function; for example, a solution can be considered converged at iteration k if the change in the variable is less than a certain absolute tolerance, E(I:
'I'he absolute tolerance is dependent on the scaling of the variable. so a relative tolerance specification, l,;r' is orten lIsed:
The relative tolerance specification is not useful iLr is converging to O. A combination the relative and absolute tolerance specifications is often used, and can be expressed as
or
The iterative methods that we present for solving single algebraic equations are: (i) direct substitution, (Ii) interval halving, (iii) false posilion, and (iv) Newton's Jnethod.
3.3.2
Direct Substitution
Perhaps the simplest algorithm for a single variable nonlinear algebraic equation is known as direct substitution. We h.ave been writing the relationship for a single equation in a single unknown as f(x) = 0
(3.5)
Using the direct substitution technique, we rewrite (3.5) in the Conn x = g(x)
(3.6)
this means that our "guess" ft.)!' x at iteration k+ 1 is based on the evaluation of M(X) atiteration k (subscripts arc used to denote the iteration) Xk+ l :::: M(Xk )
(3.7)
If formulated properly, (3.7) converges to a solution (within a desired tolerance) x* = g(x*)
If not formulated properly, (3.7) may diverge or converge to physically unrealistic tions, as shown by the following example.
(3.8) solll~
56
Algebraic Equations
EXAMPLE 3.2
A Reactor with
Chap. 3
Second~Ordcr Kinetics
The dynamic model for an isothermal, constant volume, chemical reactor with a single sccondorder reaction is: ~{~;, df
Find the
slcady~statc
concentration for the following inputs and parameters:
F/\I:;;;; I min-I, CAf '=. I gmolliiter, k:::: I litcr/(gmol min)
At steady-state, riCA/til
:=
n, and substituting the parameter and input vatues, we find I
~
Clis ~ C'As ~ 0
where the subscript s is used to denote the steady-state solution. For notational convenience, let .r:;;: CA" and write the algebraic equation as
We can directly solve this equation using the quadratic formula to find x:::: -·1.6IX and 0.61X to be the solutions. Obviously a concentration cannot be negative, so the only physically meaningful solution is x::;;: 0.618. Although wc know the answer llsing the quadratic formula, our objective is to illustrate the behavior of thc direct substitution method. To use the direct substitution method, we can rewrite the function in two different ways: 0) x 2 :;;;; ~J_- + I and Oi) x:;;;; --x 2 + J. We will analyze (I) and leave (Ii) as an exercise for the reader (see student exercise 4). (i) Here we rewrite JCi,.:) to fino the following direct substitution arrangement
x ~
v=:;:-+) ~ g(x)
Or, using subscript k to indicate the kth iteration
For a first guess of xo :;;;; 0.5, we find the following sequence
x,
~
x,
~
V - 0.5 +
1
0.7071 + I
~
0.7071
~
0.5412
+ 1 ~ 0.6774
XJ ~
0.5412
X4 ~
0.6774 + I
~
0.5680
'[his sequence slowly converges to 0.618, as shown in Figure 3.2.
57
Nonlinear Functions of a Single Variable
Sec. 3.3
0.75 0.7 0.65
'"
.".
0.6 0.55 0.5 0
20
15
10
5
k
FIGOH.E 3.2 The iteration XI; sequence converges to 0.61 XO.
I I with Xo
11
0:;;:
0.5. This
Notice that an initial guess of xo:::O () or ! oscillates between 0 and 1, never converging or diverging, as showil in Figure 3.3.
0.8 0.6
'".".
0.4
I I I 0.2 I I 0 0
\ \ \ \ \
2
4
6
10
FIGURE 3.3 The iteration xI; i [ = ~--::t~- 1 Wilh Xo::= 0 (dashed) or (solid). Thi::; sequence oscillates between 0 and 1.
As noted earlier, this problem has two solutions (x* :;:::; --1.618 and x~: :;;:: 0.(18), since il is a second-order polynomial. This tall be verified by plotting x versus J{x) as shown in
Algebraic Equations
58
Chap.3
Figure 3.4. From phy:;ical reasoning, we accept only the positive solution, since a concentration cannot be negative.
~
a/V 3:
/
-1 f··················· , .
.
,............
\ .
\[\
-2
-3 L...".
-2
FIGURE 3.4
.i.._ _
.._--"~._
_-"- .
-1
2
Plot ofj(x) versus x to find whcrcJ(x)::::; O.
Example 3.2 illustrates that certain initial guesses may oscillate and never yield a solution, while other guesses may converge to a solution. It turns out that the way that a direct substitution problem is formulated may eliminate valid solutions from being reached (sec student exercise 4). These problems exist, to a certain extent, with any numerical solution technique. The potential prohlems appear to be worse with direct substitution; direct substitution is not generally recommended unless experience with a particular problem indicates that results arc satisfactory, Direct substitution is often the easiest numerical technique toformulate for the ,,,>olution of a single nonlinear algebraic equation. If a numerical technique does not converge to a solution when the initial guess is close to the solution, we refer to the solution as unstable. The stability of iterative methods is discussed in the appendix.
3.3.3
Interval Halving (Bisection)
The interval halving technique only requires that the sign of the function value is knowll. The following steps arc used in the interval halving technique:
1. Bracket the solution by finding two values of x, one wherej{x) is less than zero and another whereJ{x) is greater than zero. 2. Evaluate the function,./C-r), at the midpoint of the bracket. 3. Replace the bracket limit that has the same sign as the function value at the midpoint, with the midpoint value. Check for convergence. If llot converged, go back to step 2.
Sec. 3.3
Nonlinear Functions of a Single Variable
59
f(x)
initial lower limit
x 2 initial upper limit
FIGURE 3.5
Illustration of the bisection technique.
An example of the interval halving approach is shown in Figure 3.5. Notice that the solution was bracketed by
1. Findingx\, whcrcj(xl) is negative andx2 wheref(x2) is positive. 2. The midpoint between xl and x2 was selected (x]). The function value at x3,}(x3)' was negative, so 3. xl was thrown out and x3 became the lower bracket point. 4. The midpoint between x 3 and .:rz was selected (x4)' The function value at --'"4' .!{.r4), was positive, so 5. '\:2 was thrown out and x 4 became the upper bracket point. 6. The midpoint between x 3 and x4 was selected (x s)' The function value al x 5, j(x s )' was negative, so 7. x3 was thrown out and x5 became the lower bracket point You can see that the midpoint between x5 and x4 will yield a positive value for}(x(), so that the x4 point will be thrown out. You can also see that this process could go on for a very long time, depending on how close to zero you desire the sol11tion. Engineering judgement must be used when making a convergence tolerance specification. The advantage to interval bisection is that it is easy to understand. A disadvantage is that it is not easily extended to multi variable systems. Also, it can take a long time to reach the solution since it only uses information about the sign of the function values. The next technique is similar to interval bisection but uses the function values to determine the variable value for the next iteration.
60
Algebraic Equations
3.3.4
Chap.3
False Position (Reguli Falsi)
The false position or rcguli falsi technique uses the function values at two previous iterations to determine the value for the next iteration. The technique of false position consists of the foJlowing steps:
1. Select variahle values xk and x k+ to bracket the solution. 2. Draw line hctwccnj{xk) andf(xk + 1) and find xk+2' 3. Evaluate f(xk+2)' Replace the bracket limit that has the S
the sign o(l{xk+2)' An example of the false position approach is shown in J1'igurc 3.6. The next step would be to draw aline from.f(x'j) to.!(x2) and find -'"4- Continue until a certain tolerance is met. The false position method generally converges much more rapidly since it uses known function values to determine the next "guess" for the variable. We have sho\vn graphically how each technique is used. You willlHlVe an opportunity in the student exer~ ciscs to write all algorithrn to implement the two techniques.
3.3.5
Newton's Method (or Newton-Raphsonl
The most commonrnethod for solving nonlinear algebraic equations is known as Newton's method (or Newton-I
I(x
+ Llx)
= f(x)
+ I'(x)Llx + rex) (Llx)2 + f"'G') (Llx)' + 2
6
=0
(3.9)
f(x)
initial lower limit
X2
initial upper
limit
FIG-URE 3.6
Illustration of the false position technique.
------------------~~~-
Sec. 3.3
Nonlinear Functions of a Single Variable
61
where
and so
011.
Neglecting the second-order and higher derivative terms and solving for 1{,I:+~t) = 0, we ohtain
-/(x)
L'.x
=
(3.10)
F(x)
Since this is an iterative procedure, calculate the guess YOI x at iteration k+l as a function of the value at iteration k:
defining from (3.1 0)
(3.11 )
=At,)
dX k + I
f'(x,)
from (3.11)
F(x,)
}\x,)
= x, -/"( _ x.k )
(3.12)
(3.13)
(3.14)
Equation (3.14) is known as Newton's method for a single-variahle problem. Notice that we can obtain the following graphical representation {{Jr Nc\vton's method (Figure 3.7). Statting from the initial guess of Xl' we find that -':2 is the intersection off'Cr 1) with the x-axis. [~valuatcJ(.1:2) and draw a line with slopcf'(x2) to the x-axis to find x 3 . This procedure is continued until convergence.
I(x) I(x,)
I(x,) I(x,)
I(x)
~
'SlOpe" f'(x,)
-1-------",'"""'7."---,----
x,
x,
ESCOLA Dc ENC:::NHARIA BIBLIOTLCA
Flf;{JRE 3.7 mclhod.
Illustration of NeWlon's
62
Algebraic Equations
Chap. 3
Advantages to Newlon's method include quadratic convergence (when close to the solution) and that the method is easily extended to l1lultivariablc problenls. Disadvantages include the fact that a derivative of the function is required, and that the method may not converge to a solution or may not converge to the nearest solution. Foran example of nonconvcrgencc, consider the fUllction shown in l"igure 3.8. Here the initial guess is at a point where the derivative of function is eqllal to zero (f'(xo ) -:;: 0), therefore there is no intersection with the x-axis to determine the next guess. \Vc also sec from (3.14) that there is no finite value for the next guess for x. Another probIcm is that the solution could continuously oscillate between two values. Consider/Cx) =::.x3 -- x. A plot of Newton's method for this function, with an initial guess o1'xo::= ~1/\/5,is shown inl<'igure 3.9. Notwithstanding the problems (possible diviSion by zero or continuous oscillation) that we have shown with Newton's method, it (or some variant of Newton's) is still the most commonly llsed solution technique for nonlinear algebraic equations. Notice that Newton's method essentially linearizes the nonlinear model at each iteration and therefore results in successive solutions of linear models. Notice that increasing amounts of information were needed to use the previous techniques. Interval halving required the sign of the function, reguli falsi required the value of the function, and Nevvton's method required the value and the derivative of the function. We also found that not all solutions to a nonlinear equation arc stable when direct substitution is used. Next, we show that all solutions are stable using NeWlon's method.
1
0.5 f(x)
a ______
_
.>-..-L~_"'
-0.5
horizon1al 1arlgel
never meet! x-axis -1
-1.5
-1
-0.5
o
0.5
1
x FIGURE 3.8
Problem with NeWlon's method whcllf'(r)::;;: O.
1.5
Sec. 3.4
63
MATLAB Routines for Solving Functions of a Single Variable
1
0.5 f(x)
l'
a
,
-....... ~_
l
'~_
,
,
J,I
-0.5 -1
-1.5
a
-0.5
-1
0.5
1
1.5
x FIGURE 3.9
3.4
OscillatIon of solution between two values.
MATLAB ROUTINES FOR SOLVING FUNCTIONS OF A SINGLE VARIABLE MA'fLAB has two routines that can solve for the zeros of a function
or a single v.:triable.
FZERO is lIsed for a general nonlinear equation, while ROOTS can be used if the nonlinear equation is a polynomial.
3.4.1
FZERO
The first routine that we usc for illustration purposes is [zero. [ZI<;1:'O uses a combination of interval halving and false position. In order to usc f ZE,;ro, you must first write a MATL,AB m-filc to generate the function that is being evaluated. Consider the function.!(x):::: x 2 - 2x ~ 3 :::: O. The following MATLAB m-filc evaluates this function (the Ill-file is named fcnl . m):
function y y
~
:=
x A 2 - 2*x
fcnl(x) 3;
After generating the m-filc tcnl. m, the user must provide a guess for lhc solulion to the fzero routine. The following command gives an initial guess of x:::: O.
y
fzero{'fcnl' ,0)
64
Algebraic Equations
Chap.3
MATL.AB returns the answer: -1
y
For an initial guess of x = 2, the user enters
z
~
fzero( 'fenl' ,2)
and MATLAB returns the answer
z
3
~
These results arc consistent with those of Example 3.2, where we found that there were two solutions to a similar problem (we could usc the quadratic rOl11mla to find theln). Again, the solution obtained depends on the initial guess. A third argument allows the user to select a relative tolerance (the default is the machine precision, eps). A fourth argument triggers a printing of the iterations.
3.4.2
ROOTS
Since the equation that we were solving was a polynomial equation, we could also lise the MATLAB routine roots to find the zeros of the polynomiaL Consider the polynomial function:
The user must create a vector of the coefficients of the polynomial, in descending order. e~[1-2-3]'
Then the user can type the following command roots(e) and MATLAB returns
ans :::: 3 ~l
Again, these are the two solutions that we expect.
3.5
MULTIVARIABLE SYSTEMS Tn the previous sections we discussed the solution of a single algebraic equation with a single unknown variable. We covered direct substitution, bisection, reguli falsi, and Newton's method. In this section, we will discuss the reduction of a multivariab1e problem to a single-variable problem, as well as the multivariab1c NeWlon's method. Consider a system of 1l nonlinear equations in Jl unknowns f(x) = 0
Sec. 3.5
65
Multivariable Systems
'There arc some special cases where 11 - 1 variables can he solved in terms of one vari~ ablc----lhen a single variable solution technique can be llsed. This approach is shown in the following example.
EXAMPLE 3.3
Reducing a
two~variahlc
problem to a
single~variablc
problem
Solve the following system of nonlinear cqllati(Hls. fl(x 1,X 2):::::
4x~-
XI
(3.15)
x Jx 2 :::O
(3.16) From (3.15) we can sol ve for x 2 in terms of x I to find: -'2:;:;;1-4x 1
(3.17)
I +3x,-ZBxf::::O
(3.18)
Substituting (3.17) into (3.16), we find:
which has the two solutions 1'1)]
XI
(from the quadratic formula)
The corresponding values of x 2 (1i'on} (3.17)) arc X 2 ;:::
OJ) and
x2:::;:
1.5714
Or, writing these solutions in vector form:
.
solutIOn 1 is
Question:
X.co"
[0.25]. . 0 . ' while solution 2IS x .
=
-0.1429] [ 1.5714
Arc we certain that there are only two solutions?
Observation: We can also see by inspection that the origin (sometimes cJlled the trivial solut'lOn), x =: tgl, is also a solution. Another solution that is slightly Jess obvious is x;::;; [~]. which we can see by inspection satisfies (3.15) and (3.16). Question:
How did we miss the other two solutions?
Observation: Perhaps we will find the olher solutions if we solve (3.16) for Xl in tCrIns of -\2' then substitute this result into (3.15) to solve for x 2 . When this is done, we obtain the result that
x-i -- 4 x
2
+
4 = 0
4 /116-16 dnd llSlllg the l(UMh,ttlC 100nJuid \} ~ 2 :1: \
--2
2
which gives us the solution x:;::: l~l. Notice that we are still missing the trivial solution, x ;::;;
rgl.
Algebraic Equations
66
Chap. 3
The mistake that we made was back at the fifst step, when we solved (3.15) for J.-2 in terms of xl to find (3.17). We must recognize that (3.15) is quadratic in Xl' therefore there arc two solutions for Xl in terms OLt2' This is more dear if we write (3.15) .'lS ~4
x7 + xI (1 - x2) + 0 ::: 0
and solve for xl to find xl -:;:;:. 0 or Xl ::: 1/4 (I - x2). The reader should show that substituting these values into (3. 16) will1cad to the four solutions:
x·~ 1.°' °11°.11°.25.]. 1- 01.5714' .1429.1 2 ' . ° ' dnd The previous example illustrates the care that must be taken when using reduction techniques to solve several nonlinear algebraic equations. If a problem cannot be reduced to a single variable, then a general multivariablc strategy (such as that discussed in the next section) must be used,
3.5.1
Newton's Method for Multivariable Problems
Recall that we arc solving the general set of equalions
rex) = 0
(3.19)
That is, a set of n equations in n unknowns
fl(x}l"t zl ' . , x,J
0
f,(x\,x" ... x,,)
0 (3.20)
t"(x,,x,, ... x,,)
0
The oQjectivc is to solve for the sct of variables, Xi' that forces all of the functions,};, to zero. We can use a Taylor series expansion for eachf;:
t;(x + Ax)
=
,f,(x) +
:s ar !H + higher order terms II
-c"
j
(3.21)
j ' l dX j
Neglecting the higher order terms and writing in matrix form
f(x + Ax)
=
f(x) + J Ax
(322)
where.J is known as the Jacobian
ilf,
iii,
iii,
ax!
dxz
dX"
.J =
(3.23)
aj;/ ax}
iJj;/
iI/;/
OX2
aXil
Sec. 3.5
67
Multivariable Systems
Now, since we wish 10 solve for Ax such that r(x + Ax):::: 0, then (from 3.22)
f(x) + J Ax
~
0
(3.24)
Solving for Ax at iteration k
(3.25) but Ax is simply the change in the x vector from the previolls iteration
=
AX k
xk
xkl- 1 -
(3.26)
Substituting (3.26) into (3.25) (3.27) Remember that xI, is a vector of values at iteration k. Notice that for a single equation (3.27) is:
.
x, , I
_.
xk
-
f(x k )
x, )
-"(
j
which is the result that we obtained in Section 3.3.4. Comment: In practice the inverse of the Jacobi;:l11 is not actually used in the sollilion of (3.25). Actually, (3.25) is solved as a set of linear algebraic equations, using Gaussian elimination or LU decomposition.
where
Jk and f(x k) are known at iteration k.
EXA1VIPLE 3.3
Revisited. Newton's method 1;(-'"x,)
f;(X"xJ
or
Xl -
=
4
2 x2-
xj
XjX? =
0
x~ -I 3 X1X2 = 0
f(x)·· [- x l -4xf
-X,X'j
1 - 8x 1 ~
J?·""'·······=-x L iJx .l
[OJ
-- 2 x 2 ~ xi + 3 x j x 2 _ =0
The Jacobian elements are
JJI
=
ill; .. ()J:
= r
so the Jacobian is written
X2
fJ/;
2
Algebraic Equations
68 Consider guess
all
initial guess of ,:rj :;:; ; --1 and
---J. Let x(O) represent the vector for this initial
x2 ;;;;;
The value of the Jacobian at this initial guess is
.I(x(O))
10
~_ [
3
T'hc inverse of the Jacollian is
~I]
1 .I I(X(O»
~13
[ .3 13
10 13 .
13
The value of the function vector for the initial guess is -
-- 1
4 I
f(x(O» ~ [ 2 (- I)
I[
+3
[-~]
and the guess for x( I), \-vherc x( I) represents the vector at iteration I, is
x(1) ~ x(O) ~.1 '(x(O)) f(x(O)
r~610 ~ r~ I [ ~ [.. 3 :-.d] 10 1 U
~I
x( I)
13
x(1 )
[
13
~053851 0.3846
Continuing with iterations 2 through 7 we find the foJlowing results
Iteration 0 1 2 3 4 5 6 7
Chap. 3
x,
XI ~I
~0.5385 ~0.3104
0.2016 ~0.156]
-0.1439 -0.1429 --0.1429
-
]
0.3846 1.0688 1.3952 1.5317 1.5683 1.5714 1.5714
Sec. 3.5
Multivariable Systems
69
The sequence of iterations for an initial guess of xl
::::
.I and x2::::::1 is
Iteration
x,
X2
0 1 2 3 4 5 6
1 0.6190 0.3893 0.2870 0.2542 0.2501 0.2500
1 0.0476 0.0081 0.0008 0.0000 0.0000 0.0000
Noticethataguc:'isofx:::o [-1 -1]' convergcdtox [-0.1429 1.5714.1' aftcrsix iterations, and a guess of x ::: [ l 1J' converged to x '" [ 0.2500 0.0000 I ' aftcr six iterations. Other initial guesses may lead to the other two known solutions that were determined analytically.
The previous example illustrates that, for systems that have ITIuhip1c solutions, the solution obtained depends on the initial guess.
3.5.2
Quasi-Newton Methods
Most computer codes actually implement some variant of Newton's method; these arc referred to as quasi-Newton methods. Remember that Newton's method is guaranteed to convcrge only if thc system is nonsingular and we arc "close" to the solution.
DAMPING FACTOR Often it is desirable to "dampen" the change in the guess for xk+l' to make Newton's method more stahlc. Applying a damping factor, a, to (3.27), we write xu, = x k - a J k ' f(xkl
(3.28)
where a is chosen so that II f(x k+1) II < II f(Xk) II and () < 'x S; I (we lise the II f(x k) I1l10tation to represent the norm of the vector f(x k Often IX is selected to minimize II f(x k + l ) II using a search technique l that is, IX is adjusted until II f(x k + 1) II is minimized. It should be noted that the single-variable equivalent to (3.28) is
».
(3.29)
HANDLING SINGULAR (OR ILL-CONDITIONED) JACOBIAN MATRICES Notice that the Newton method with ot without the damping factor requires the inverse of the Jacobian matrix (or the solution of a set of linear algebraic equations) to determine the
Algebraic Equations
70
Chap. 3
value for the next iteration. If the Jacobian is singular, it cannot he inverted. One method that avoids this problem is known as the L,evcnhcrg~Marquardl method:
xk + t
_
~
xk
~
T
(J k J k
+ [31) .... ] Jk T f(x k )
(3.30)
where T represents the matrix transpose and [3 is an adjustable parameter llsed to avoid a singularity. The single variable equivalent to (3.30) is (3.31 )
Notice that jf j3 = 0, the standard Newton algorithm results.
WHEN ANALYTICAL JACOBIAN MATRICES ARE NOT AVAILABLE If an analytical Jacobian is not available, a numerical approximation to the Jacobian mllst be used by the quasi-Newton techniquc.A backward differences approximation for the Jacobian is (3.32)
where oxJ-(k) is a small perturbation in variable x·.I at iteration k. A problcrn with this approaeh is that an n-variabfc problem requires 11 + 1 evaluations of the function vector at each iteration. There are other techniques that rely on infrequent function evaluations to update the Jacobian matrix.
3.6
MATLAB ROUTINES FOR SYSTEMS OF NONLINEAR ALGEBRAIC EQUATIONS 'fhe MATI,AB routine fsolve is used to solve sets of nonlinear algebraic equations, using a quasi-Newton method. The user must supply a routine to evaluate the function vector. It is optional to write a routine to evaluate the gradient of the function vector. As another option, the user can select tbe Levenburg-Marquardt method.
EXAMPLE 3.3
Reconsidered. Using MATLAB
The tn-file used to implement Example 3.3 using fsolve is:
function f = nle(x) [(1)= x(1)-4*xll)*xl1)-xll)*xI2); f{2)= 2*x(2)-x(2)*x(2)+3*x(1)*x(2); which is placed in an tn-file called nle. m The initial guess is entered xO
[1
1]';
Summary
71
and we obtain the solution by entering X
fsol ve ( 'nle' ,xO)
::0
which gives us the expected results x
=
[0.2500
0.0000] ,
Computationally faster results will he obtained if the analytical Jacobian is used.
2 'fhe following function file gcnerates the analytic,tI Jacobian for this problem.
[unction gf '=' gradnle(x) gf(1,1)=1~8'x(1)-x(2);
gf(l,2)=-x(1) ; gf(2,l)=3'x(2);
gf(2,21=2-2*x(2)+3*x(1); which we place in all Ill-file called gradnle. m. We can then solve this problem by entering xO =
[l
1]';
options (5) c-,-O; x fsolve( 'nle' ,xO,options, 'graclnle') 'fhe options vector can be used 10 seleel the Levcnberg-Marquanlt method by setting options(5)'='1;
SUMMARY In this cbapter we have presented a number of techniques to solve nonlinear algebraic equations. Each technique bad a number of advantages and disadvantages-the approach that you use may depend on tbe problem at hane!. If you have the option, it is a good idea to plot the function,f(x), to see if the results agree with your numerical solution. This option may not be available with l11ultivariablc problems. Notice that if a particular problem has multiple solutions, the actual solution obtained will depend on the initial guess. There are many actual chemical. processes that have "multiple solutions," that is, there are several possible "steady~states" that the process may operate at. In practice, the steady-state obtained wiU depend on dynamic considerations, such as the way that the process is started up. These issues will be ad~ dressed when we discuss differential equation~hased models. Much research is being done in applied mathematics to develop numerical tecbniques that yield everyone of the multiple solutions, without having to make many initial guesses. Among these is "homotopy continuation." These techniques arc not welldeveloped and are heyond the scope of this text. For now, the student must he willing to
Algebraic Equations
72
Chap. 3
try a numher of techniques, with a number of initial guesses, to find all of the solutions to a problem. It is recommended that you generally lise commcrically availahle routines for solving these problems. The numerical techniques covered were Direct substitution Interval bisection Rcguli falsi Newton's method Newton's method (and some variants) is the most commonly llsed multi variable technique. The reader should he able to understand the notions of • Jacobian Fixed or equilibrium points • COllvergence and stability
Tolerance Ilcration The MATLAB routines that were introduced in this chapter are f801 ve: Solves a system of nonlinear algebraic equations, using quasi-Newton and Levenberg-Marquardt based algorithms fzero:
Solves for a single equation
roots:
Solves for the roots of a polynomial equation
A I1lnnber of other numerical rOlltines are availiable through thcMATLAB NAtj Toolbox.
FURTHER READING More detailed treatments of numerical methods given in the textbooks by Davis, Finlayson, and Riggs: Davis, M.E. (1984). Numerical Methods ol1dModeling for Chell/ieal Engineers, Ncw York: Wiley. Finlayson, B.A. (1980), Nonlinear Analysis in Chonit'al Fngineering. New York: McGraw-Hili. Riggs, J.B. (1994). Numerh..al Techniques for Chemical Engineers, 2nd cd. Lubbock: Texas Tech University Press.
Student Exercises
73
Example numerical techniques arc also presented by Felder and Rousseau: Felder, R.M., & R.W. Rousseau. (1986). Elementary Principles Processes. 2nd cd. New Yark: Wiley.
(~r
Chemical
The following book is more of ail advanced undergraduate/first~year graduate student text on I1lllncrical methods to solve chcmical engineering problems. The emphasis is on FORTRAN subroutines to be used with tbe IMSL (FORTRAN-based) package. Rameriz, w.r-<. BUtlerworths.
«( 989).
Computational Methods for Process 5'imulalion. Boston:
STUDENT EXERCISES Single Variable Methods 1. Real gases do not normally behave as ideal gases except at low pressure or high tcmpcrature. A number of equations of state have been developed to account for the non-idealities (van del" Waal's, RedliclyKwong, Peng-Robinson, etc.). Consider the van cler Waal's (YO T relationship
where Pis pressure (absolute units), R is the ideal gas constant, T is temperature (absolute units), {, is the molar volume and a and b arc van der Waa!'s constanls. The van der Waal's constants are often calculated from the critical conditions for a particular gas. Assume that a, h, and R are given. We find that if P and T arc given, there is an iterative solution required for V. a. How many solutions for V arc there? Why? (Hint: Expand the PVT rclationshlp to form a polynomial.) b. Recall that direct substitution has the form Vk+J :::: gV k)' Write three different direct substitution formulations for this problem (call thcse 1, H, and Ill). c. What would be your first guess for V in this prohlem?Why? d. Write the MATLAB m-files to solve for V using direct substitution, for each or the three formulations developed in b. Consider the following system: air at 50 atm and -1 ()(PC. The van der Waal's constants arc (Felder and ROllsseau, p. 201) a :::: 1.33 altn Iitcr2/gmo!2, b = 0.0366 Iitc/Jglllol, and R = OJl8206 liter alm/glllol K. Solve the I{Jl1owing problems numerically. e. Plot V as a function of iteration number for each of the direct substitution methods in b. Usc 10 to 20 iterations. Also usc the first guess that you calculated in c. Discuss the stability of each solution (think about the stability theorem).
Algebraic Equations
74
Chap. 3
f. H.carrangc (I) to the form of/CO") :::: O-and plot./n/ ) as a function of V. How many sol uti OilS arc there? Docs this agree with your solution for a? Why or why
not? g. Usc the MA'rLAB function roots to solve for the roots of the polynomial de-
veloped in a. h. Write and lise an m-fi1c to solve for the equation in a using Newton's method. The nUlllerical Solulion is eonsidercd converged when ,,~
IVk~ Vk11
< ". Lel
k-- [
0.0001.
2. Consider the van der Waals relationship for a gas without the volume correction lerm (Ii ~ 0)
(P + V2a) V· A
A
a. IIow many solutions for
RT
if arc there? Show the solutions analytically.
b. Which solution is correct? 3. A process furnace is heating 150 Ibrllol/hr of
vapor~phase
mllmonia. The rate of
heat addition to the furnace is 1.0 x J 0 6 Btu/hI'. The ammonia fcedstrcam tempcra~ ture is 550 0 R. Use Newton's method to find the tcmperature of the aJl1monialeav~ ing the furnace. Assumc ideal gas and use the following equation for heat capacity at constant pressurc:
Btu
Ibmol "R Ii
ell
=
a + bT + cT- 2
~ 3.33
X
to
3
Btll lOS Btu "R Ibmol "R' c = - 1.20 X Ibmol
T
and remember that Q c.
,i r,,'·'" C" dT
where Ii is the molar rlowmtc of gas and Q is the ratc of heat addition to the gas pCI' unit timc. l-Iow did you detcrrnine a good first gucss to lise? 4. Consider Example 3.2 , .ltx) = _x2 ~ X + 1 = 0, with the direct substitution method jCll'lllulated as.x = -x 2 + 1 :;;:: g(x), so that the iteration sequence is Xk
I
c=
g(x k ) = - x~ + I
Try several different initial conditions and show whether these convcrge, diverge, or oscillate bctwcen values. Discuss the stability of the two sollitions .x*:;:: 0.618 and x*:;;:: -1.618, based on an analysis ofg'(.I;*). 5. Show why the graphical Newton's method is eqlJivalent to xk+l = x k ~-J(xJ.Yf(xk)'
6. Develop an algoritlnl1 (sequence of steps) to solve an algebraic equation using intcrval halving (bisection).
75
Student Exercises
7. Develop an algorithm (sequence of steps) to solve an algebraic equation using reglilt falsi (false position). Compare and contrast this algorithm with Newton's mcthod. S. A componcnt material balance around a chemical reactor yields the following steady-slale equation
0= F
where V
.=
F V
C- kc' ftC>
Ibmol
and k = 0 . 0 5 , . IbmoV ITIm a. How many steady-state solutions arc there? b. Write two different direct substitution methods and assesS the convergence of each. c. Perform two iterations of Newton's method using an initial guess of C:::: 1.0. 0.1 min " Cj"
1.0
9. Consider the following direct suhstitution problem, which results from all energy balance problem
T
-I'
k+ I -
15.04 0.716 _ 4.257 X 10
7~
'I'
Willlhis method converge to the solution or T = 443.571? Why or why not?
IO. Consider the following steady-stale model for an exothermic zero-order reaction in a continuous stirred-tank reactor. The variable x is the dimensionless reactor temperature.
lex)
0.42204 cxp
x
+
x
-1.3x=O
20
I'here arc two solutions to this equation, as illustrated in the figure below. The solu!ions mc x = 0.55946 and 2.00000. a. Formulatc a direct substitution solution to this problcm. Will your direct substitution tcchnique converge 10 0.55946'1 Will your direct substitution technique converge 10 x::: 2'1 Use a rigorous mathematical argument in each case. h. For an initial guess of x = 1.35, would you expect Newton's method to have any trouble with this problem? Explain. c. What is the next guess from the reguli falsi lechnique if your first guess was x;:;: I and your second guess was x:::::; 2.5 (show this mathematically)? d. Write a function routine to solve this problem using theMATLAB function f zero. Also show the commands that you would give in the MATLAB command window to run fzera.
76
Algebraic Equations zero-order rxn - dimensionless
0.5
~
Chap. 3
0
-0.5
a
2.5
2
1.5
0.5 x
Plot of x versus f(x) for problem IO.
11. A component material balance around a chemical reactor yields the following steady-state equation
o VF C';,,- FV C c
F where V
=
IU min'!, C;"
=
1.0
Ibmol 1 - and k fl'
=
k(·25
If'5
0.05 . ..... I ' lbmol "- min
Write two different direct substitution methods and assess the convergence of each to the steady-state concentration of 0.75354. b. Perform two iterations of Newton's method using an initial guess of C:::: 1.0. i:1.
12. Consider the dimensionless equations for an exothermic CSTR (continuous stirred tank reactor) shown in a module in the final section of the textbook.
= -'!>XIK(X,) + (I - XI) = 0 f,(x"x,) = j3X IK(X,) + (I + o)x, = 0 fl(xI,x,)
where K(X ) 2
= exp
2 .) (_X 1 + x,I'Y
Uscft (x['X2) to solve for xl in terms of x 2 and substitute intoh(xl,x2) to obtain the single equation
I)
I«X,) - (1 + o)x, ~ .
=
0
Consider the parameter values ~ :::: 8, (V : :;: 0.072, 'Y :::: 20, 0 = 0.3. Determine the number of solutions to this problem (graphically), Find the xI and .x2 values for each solution, using the single variable Newton's method.
Appendix
77
Multivariable Methods 13. Use the multivariabJe Newton's method to solve the following problem. Solve this as a two-variable problem. Do not reduce it to a single-variable problem via substitution.
f,(x)
=
2x,-x2 -5
=
0
J;(x) =-x, -x2 + 4 .•~ 0 How many iterations docs it take to converge to the solution? Explain this result conceptually.
14. A simple bioreactor model (assuming
where JJ..mi]X ::::: km = ::::: k, ::::: c :::::
stcady~statc operation)
is
0.53 0.12 0.4545 0.4 4.0
is the biomass concentration (mass of cells) and xl is the substrate concentration (food source for the cells). Find the steady-state values for xI and x2 if f)::::: 0.3 (There are three solutions).
MATLAB.
f,(x"x,)
CC
-
q>X,,,(xz) + (1 - x,)
= 0
j;(x"x,) = l3q,x,,,(x,) - (1 + I\)x z .c. 0 where ,,(x,) = exp (. . . ) I j- x,h and the following parameters arc used
13 =
8
'Y =
20
(I' = 0.072 1\ = 0.3
Algebraic Equations
78
Chap.3
APPENDIX Stability of Numerial Solutions-Single Equations If the iterates (xk) from a numerical algorithm converge to a solution, we refer to thal solution ns being stable. ..
-----._----------~----_
_--~-~--~---~-_._--------~
Definition 3.1 Let x''' represent the solution (fixed point) of x*
=:
S(x'''), or get''')
yO" =:
O. --------~-
Theorem 3.1 x~'
.IS a stahle soI uUon ' 01 x'::' . K(.r""), 0::
g ,- ~ g(,"'). if lu l (JX If
1~lgl
'1'li1gl - I
1
--~------
. .IS an 11llsta) II c SOIutHlll ' eva I llatcd at -r"'r';' () I'
<
llx
t evaluated at x*
!, then no conclusions can be drawn. For simplicity in notation, we generally usc
rJx
(jv
f
g to represent 1.':>1_ dx
\Vc continue with Example 3.2 to illustrate the numerical stability of direcl substitution.
- - - - - - - - - - - - - - - - . _ - -. ._ - - EXAMJ'LE 3.2 Contimlcd.
Stability of din~ct substitution
Consider the Case I formulation,
x
~
g(x)
x+1
which has the derivative
ag
ax II
I I
- I
. g' -
x
I 1
I, then x*' is a stable solutioH.
1. For xo;' := 0.618, we find that
lex ")
:::;0
O.61X +- I
O.X090
I.Sox''':=O.6ISisa
stable solution, that is, all initial "guess" for X o close to 0.618 will convergc
-I 2. [-'or x'"
0::::
-1.618, wc find I g'(x"') I 0::::
.618
f
J
10
xO;o 0::::
0.618.
0.30(]O,' 1. So _y*' :::; . . ~ 1.618 should
be a stable solution, that is, an initial "guess" frw Xu close to - 1.61 g should converge to 0:::: -1.618.
x'"
c
Appendix Question:
79 Why did we find previollsly that an Initial guess close to -1.61 Xdid not converge?
Observation: We must realize that the square root of a positive number has two values. For example, the square rooL of I call either be + I or -I (after all (-1)2 ::0:: In. --------You may be questioning the utility of a stability test for the direct substitution method, which requires that the solution be known to apply the test. Our purpose is mainly to show that the direct substitution method can he unstable. Newlon's rncthod guarantees stable solutions, if the "guess" is close to the solution.
Stability of Newton's Method for Single Equations Here we use 'rhcorcm 3.1 to show that Newton's method is stable. 'to/e see that NewLon's lllcthod can be written in the form of
where
fix,) f'(x,J Then we can find that tbe derivative, //Ctk) is
or g'(x,) At the solution,
~
lex,) n\,)
II' (\,J I'
x"" we sec that g' (x*)
/(x*) r(x"')
II'(x*)]'
And sincc.f(x*') ::::: 0, we find that the stability constraint is satisfied
g'(x*) = () as long as/ex"~') *- O. This shO\vs that Newlon's method will converge to the solution, provided an initial guess close to the solution.
NUMERICAL INTEGRATION
4
Most chemical process models arc nonlinear and rarely have analytical solutions. This chapter introduces numerical solution techniques for the integration of initial value ordi~ nary differential equations. After studying this material, the student should be able to: Understand the difference between explicit and implicit Euler integration. Write MATLAB code to implement fixed step size Euler and Runge-Kutta techniques. Usc thcMATLAB ode45 integration routine. The major sections
4.1
or this chapter arc:
4.1 4.2
Background Euler Integration
4.3
Runge-Kulla Integration
4.4
MATLAB Integration Routines
BACKGROUND Thus far we have developed modeling equations (Chapter 2) and solved for the steady states (Chapter 3). One purpose of developing dynamic models is to be able to perform "what if' types of studies. For example, you may wish to determine how long a gas storage tank will take to reach a certain pressure if the outlet valve is closed. This requires in80
Sec. 4.2
81
Euter Integration
tegrating the differential equations from given initial conditions. If the dynamic equations are linear, then we can generally obtain analytical solutions; these techniques will be presented in Chapters 6 and 8 through 10. Even when systems are linear, we may wish to use numerical methods rather than analytical solutions. At this point itis worth reviewing the difference between linear and nonlinear differential equations. An example of a linear ordinary differential equation is:
dx = -r dt • since the rate of change of the dependent variable is a linear function of the dependent variable. An example of a nonlinear differential equation is:
dx dt
=
~-x
2
since the rate of change of the dependent variable is a nonlinear function of the dependent variable. Although this particulary nonlinear equation has an analytical solution, this will not normally be the case, particularly for sets of nonlinear equations. Notice that the following equation is linear:
since the only nonlinearity is in the independent variable (t). The purpose of this chapter is to introduce you to numerical techniques for integrating initial value ordinary differential equations. The first numerical integration technique that we will present is the Euler integration. In the next section we discuss two algorithms, the explicit and the implicit Euler methods.
4.2
EULER INTEGRATION Consider a single variable ODE with the form
dx. dl = x
=f
( ) x
(4.1 )
We consider two different approximations to the derivative. In Section 4.2, I we consider a forward difference approximation, which leads to the explict Euler method, In Section 4.2.2 we consider a backwards differences approximation, which leads to the implicit Euler method.
4.2.1
Explicit Euler
If we use a forward difference approximation for the time derivative of (4, I), we find
dx dl
=
t(k
+ I _. + I) ~ t(k)
(4.2)
Numerical Integration
82
Chap. 4
where k represents the kth discrete time step of the integration. Now, assume that)(.\:) is evaluated at x(k). We will refer to this fuuclion ast(x(k)), and can write (4.1) and (4.2) as x(k+ 1)...- x(k) = j(x(k)) I(k + 1) - 'rk) Normally we will use a fixed increment of time, that is, t(k + 1) - t(k) ~
(4.3)
at,
where!:;;'/ is
the integration step size. Then we write (43) as x(k
+
2t- x(k)
Cc
J(x(k))
(43)
Solving t{)r x(k + I)
x(k
+ I)
= x(k)
+ D.tJ(x(k)) K,plicit Euler
(44)
We can view (4.4) as a prediction of x at k + 1 based on the value of x al k and the slope at k, as shown in Figure 4.1. Equation (4.4) is the expression for the explicit Euler method for a single variable. The general statement for a lllultivariable prohlcm is
x(k + 1)
=
x(k) + 6.lf(x(k))
(45)
Where x(k) is a vector of state variable values at time step k and f(x(k» is a vector of functions evaluated at step k. Equations (4.4) and (4.5) are explicit because the statc variable value at time step k + 1 is only a function of the variahle values at step k. 'I'his method is straightforward and easy to program on either a handheld calculator or a cornputer. A major disadvantage is that a small step size must he llscd for accuracy. llowevcr. if too small of a step size is used, then numerical truncation problems may result. Explict Euler is not often used in practice, but is covered here for illustrative purposes.
4.2.2
Implicit Euler
This method uses a backwards difference approximation for the derivative in (4.1). The function (or vector of functions) is evaluated at time step k + 1 rather than time step k:
+I)-x(k)
6.1'
=
j(x(k
+ 1))
(46)
x(k+ 1) based on x(k) and f(x(k))
x(k)
----~ k
k+1
slope, f(k), evaluated at step k
FIGURE 4.1 Pictorial representation of the explicit Euler method.
Sec. 4.2
Euter Integration
83
which can be written
x(k + J) = x(k) + Iltf(x(k + 1)) Implicit Euler
(4.7)
Equation (4.7) is implicit because the value x(k + I) must be known in order to solve for + 1). What this generally requires is a nonlinear algebraic solution technique, slIch as Newton's method. For a linear system, equation (4.7) can be explicitly solved to obtain the form
x(k
x(k + 1) = g(x(k)) The following section compares the explicit and implicit methods for a single linear ordinary differential equation. In particular, we compare how the integration step size (At) affects the stability of each method.
4.2.3
Numerical Stability of Explicit and Implict Euler Methods
Consider the tank height problem covered in Example 2.1. There was no inflow, and the outlet flowratc was assumed to be linearly related to height, which gave us the following equation: dx
1
dt
T
x
Ax
co
= f(x)
(4.8)
where 'T = A/13, A = - lIT and the state variable x is the tank height. Since the variables are separable, the reader should show that the analytical solution is:
= x(O)e,j· = x(O)e"
x(t)
(4.9)
Next, we compare this analytical solution with the explicit and irnplicit Euler solutions.
EXPLICIT EULER The function value at step k is:
f(x(k»)
J Oc
x(k)
-
T
and the state variable value at the next time step is:
x(k + J) = x(k) + - Ilt T x(k) =
(t1t) J - .~. x(k)
IMPLICIT EULER The implicit Euler method evaluates the function at k + 1 rather than k:
I(x(k
+ 1))
=
-
I T
and the state variable at slep k + lis:
x(k
+ 1)
(4.10)
Numerical Integration
84
x(k + I)
x(k) + -
~
ill
Chap. 4
x(k + I)
(4.1 J)
T
Notice that, since this is a linear problem, a nonlinear algebraic equation solver is not needed for (4.11). We can rewrite (4.11) as
x(k + I)
1
~
ill x(k)
+
(412)
T
In the next section we compare the numerical stability of the explicit and implicit Euler methods.
NUMERiCAL 5TABILITY The explicit Euler solution, written in terms of the initial condition, is (from (4.10»):
x(k + I) which will be stable if
~
(1 - flt)"
I
T
x(O)
11- AlIT 1< 1. This is the same re~Htlt if we usc the representation: x(k + i)
and the stability requirement that
=
g(x(k))
~
(I - ~1)X(k)
Ig' I< I. The explicit Euler method is then stable if: -1<1-
ill
T
and will oscillate for:
flt
-1< 1-- < 0 T
These criteria lead to the explicit Euler stability condition of:
0
T
o<
ill < T
The implicit Euler solution, written in terms of the initial condition, is (from (4.12)):
x(k +
'"
( )
I)~ I;~I
x(O)
th
Euter Integration
Sec. 4.2
85
which will be stable if II/l+~t,'T1 < 1. This is the same result if we usc the representation
x(k + I) = g(x(k)) = (
1
I
+ !it
) x(k)
7
and the stability requirement that 11/ ::to < 1. Notice that the implicit Euler method is stable for any value of D.f (as long as the sign of D.l is corrcct) and will not oscillate.
EXAMPLE 4.1
Numerical Compm'json of Explicit and Implicit Euler
Let x(O) ::::: 4, T :::: 5, and tlt ;:: I. Table 4.1 compares the exact solution (4.9) with the explicit Euler (4.10) and implicit Euler (4.12) methods.
TABLE 4.1
x,
Linear First Order Example (. = 5, dt;:: 1)
exact
()
4.()()()()
1 2 3
3.2749 2.6813 2.1952 1.7979 1.4715
4
5
x, Explicit Euler 4.0000 3.2000 2.5600 2.0480 1.6384 1.3107
error
-2.YkJ ·-4'yf{-)
-6.7(70
---8.99() -10.9%
x, Implicit Euler
errol"
4.0000 3.3333 2.7778 2.3148 1.9290 1.6075
J.(-H(' 5.49;
The results shown in Tahle 4.1 are illustrated graphically by the curves in Figure 4.2.
implicit explicit
FIGlJRE 4.2 D.t = I.
Comparison of the exact solution with the explicit and implicit Euler for
1.81;1,
7.Yfri
9.2(;{,
86
Chap,4
Numerical Integration
at : : :
Larger Integration Step Size. We have seen the well-behaved response for I (which is 7/5). Consider now a larger D..t. We cun sec from (4.10) that the explicit l-:u1cr method predicts.r::::: 0 for all time after time 0, if at;;:::; T. Indeed, the explicit Euler solution is oscillatory for fj,t > 7, for this process. For example, let at = 6 for this problem. The results arc shown in 'fable 4.2. These results arc illustrated more graphically by the curves plotted in Figure 4.3. The implicit Euler techniquc has monotonic behavior and more closely approximates the exact solution. We sec that the implicit Euler method call tolerate it larger integration step size than the explicit Euler technique. Linc)]r First Order Example (r;:: 5, ilt;:: 6)
TABLE 4.2
,~~---~
x, exact
x, Explicit Euler
4JJOOO U04S 03629 0-1093 (W329
4JJOOO -II-SOOO 0,1600 ,(HmO OJI064
°
6 12 IS 24
-Y,
Implicit Euler 4,0000 LSIS2 11-8264 IU757 IU70S
4
3
2
implicit
~
exact
0 explicit
-1 0
FIGURE 4.3
5
10
15
20
25
Comparison of exact solution with implicit and explicit Euler for D.l;;o.
30
o.
We have seen that thcre is a limit to how large a step size can bc tolerated by the explicit Euler method before it goes unstable, while the Implicit Euler Inethod rcmains stable for any step size. This is true for a simple linear ordinary differential equation. The following example illustrates an important issue when solving nonlinear equations. That is. the implicit Eulcr method requircs an iterativc solution at each time step.
Sec. 4.2
87
Euter Integration ---------------~-- -
EXAjVII.JIJ~ 4.2
....
_----------~----
Explicit and Implicit Eulcl"-Nonlinear System
Consider a surge tank with an outlet f!owratc that depends on the square rool of the height of liquid in the tank. Whclllhcrc is 110 inlet flow, the model has the following form fix
(4.13a)
dl
WA (:::: flow coefficicntJcross-scctional
where x is the tank height and (/:::;;
Analytical solution (exact).
'l'he analytical solution is
[\1,(1»
x(l)
lit
/21'
Next, we compare tbis solution with the explicit and implicit
Explicit Euler.
area).
(4.13bl r~Lllcr
solutions.
The explicit Euler method yields the following equation
,(k
+
I)
(4.14)
x(k) - tH lI\le(k»
l<'or the numerical example of ([;;;; 0.8, curves for 3 different D.ts are shown in Figure 4.4.
2
3
4
5
F1G{][{E 4.4 Explicit Euler solution. !:1t:::: O.()I, 0.1, and LO. The 0.01 and 0.1 step sizes yield virtually identical results. while there is a significant error ill a step size of 1.0.
Implicit Euler.
The implicit Euler method yields the following equation
,(k + I)
~
x(k) -- uta\lx(k + 1)
(4.15)
R.ecall that when we were dealing with a linear equation we were able to rewrite the implicit Euler equation to yield an explicit calculation of the state variable at the next time step. Here we sec that this is impossible for a nonlinear equation. Rewriting (4.15),
,(k + I) + ula\lx(k + I)-x(k)" 0
(4.16)
we sec that (4.16) requires an iterative solution. That is, at time step k + I we must usc a numerical method that \vill solve a nonlinear algebraic equation. We know from Chapter 3 that a num-
Numerical Integration
88
Chap.4
bel' of techniques, including Newton's method, can be used. To illustrate clearly one approach that we can take, let y represent tbe value of x at step k + I for which we desire to find the solulion. We can rewrite (4.16) as
y + b\/v - c = 0
(417)
where y -= x(k + I), b "" 1:11 (/, and c :;;: x(k). If Newton's method is used, we can write:
MCr(i»
c.
y(i) + 1,v)'O) - c
(4.18)
where (i) is an index to iudicatc the ith iteration of Newton's method.
yO +
I) ~ rei) _ M(YO» . g' (yO»
(4.191
where
g'(y(i»
~
1
iJ
+ --7'
(420)
2VY(I)
We call write (4.19) as (from (4.18), (4.19), and (4.20»
yO+
I),~y(i)-
Iln,'(i) iJ
(\1)'0)]
(4.21 )
\lv(i) + 2 Equation (4.21) is then iterated to convergence. Summarizing, at step k+l of an integration, we must iteratively solve for the value of x at k+ I. That is, (4.21) is iteratively solved to convergence, in order to find x(k + 1) in (4.16).
COMMENT ON IMPLICIT INTEGRATION TECHNIQUES We have seen that the implicit Euler method is more accurate and stable than the explicit Euler method. We also noted that, for nonlinear systems, a nonlinear equation must be iteratively solved at each time step. The implicit Euler method allows a much larger time step, hut some or the computational savings Inuch be sacrificed in the inlcralive solution of the nonlinear algebraic equation at each lime step_ There arc a Ilumber of more advanced implicit methods that arc used in a number of commercially availahle integration codes. In this text we emphasize e.rplicit techniques, which are used by the MATLAB routines ode23 and ode4 5. SIMULINK has choices of some implicit integration routines. An explicit technique that is more accurate than the explicit Euler technique is known as the Runge-Kutta method and is shown in the next section.
4.3
RUNGE-KUTTA INTEGRATION This technique is an extension of the Euler method. In the Euler method, the derivative at time step k was used to predict the solution at step k+ 1. Runge~Kutta methods use the Euler technique to predict the x value at intermediate steps, then use averages of the slopes at intcmediatc steps for the full prediction from the beginning of the time step.
Sec. 4.3
4.3.1
89
Runge-Kutta Integration
Second-Order Runge-Kutta
The first Rungc-Kutta approach that we discuss is the second-order Rungc-Kutta method, which is also known as the midpoint Euler method, for reasons that will become clear. The Euler technique is first used to predict the state value at I:3.t12. The derivative is evaluated at this midpoint, and used to predict the value of x at the end of the step, l:i.t, as shown in r'igufc 4.5. Let Tnj represent the slope at the initial point and lJl 2 represent the slope (dddt) at the midpoint:
n1,
~
n1,
= [
x(k + 1)
I(t (k), x())
~
(4.22)
At At) ( t(k) + 2' x(k) + 2 n1,
x(k) + n1,At
(4.23) (4.24)
For autonomOllS systems, the derivative functions arc not explicitly functions of time, so (4.22) and (4.23) can be written:
n1,
[(x(k))
(4.22a) (4.23a)
or
n1 2
•••
[(x(k) +
~ !(X(k))
(4.23b)
x(k+l) based on x(k) and slope at t k + D. t 2 x(k) ----------~ t k+ 1
FIGURE 4.5
Pictorial representation of sccolld~ordcr Runge-Kutta (midpoint Euler) technique.
Numerical Integration
90
Chap" 4
Equation (4.24) can now be written:
x(k + I)
~ x(k)
+ L1t r (r(X(k) +
~tr(X(k))))
(4"24a)
which is of the form:
x(k + I)
~
g(x(k))
It should be noted that the second-order RUllgc- Kutta is accurate to the order of :i1 2 ,
while the explicit Euler is accurate to the order of dt.
Second-order Rnngc~Kutta (Midpoint Euler) Tedmique
EXAMPLE 4.3
Consider again the first-order process: dx
dt
~ {(x) ~
1
-
x
(4"8)
T
Ill, ~
r(x(k))
~
( -"
I
T
x(k) )
~
-I T
x(k)
(4"25)
(4"26)
For t1t::;;;; 1,
T
From (4"25)
=
5, and x(O) = 4.0: 111[=
~
-4 --5
-1( I) x(O) - -15 (I - OJ)(4) I - """" 5 10
From (4"26)
[:rom (4.24)
-1 x(O) 5
x(l)
~
x(O) +
x(1)
~
4"0 - OJ2 ~ 32800
111 ,
~
H 5
bot
Compare this with the analytical result of 4.0 e -1/5:::: 3.2749 Notice that the error is O.16 1YtJ. Contrast this with an error of-2.3% from Table 4.! for the Euler method. For the same step size, Rungc-Kutta techniques arc more accurate than the standard Explicit Euler technique.
Thus far, we have used single variable examples. The next example is for a able system.
two~statc vari~
Sec. 4.3
EXAMPLE 4.4
91
Runge-Kutta Integration
Two~statc
Variable System, Second~ordcr Rungc~Kutta Method
Consider two interacting tanks in series, showll in Figure 4.6, with outlet tlowrates that are a fun~~,igl_~_ ~!f the square root of tank height. Notice that the He)w from tank 1 is a function of while the flowratc out of lank 2 is a function of Vl1
VII I - il 2,
2·
F
FIGURE 4.6
Interacting tanks.
The following modeling equations describe this system
(4.27)
For the following parameter values: ft2.5
13,
= 2.5
min
13 2
5
=·~C
fl2S
A, = 5 ft' A, = to ft'
\/6 min
and the input: F = 5 ft 3 /min the steady-state height values are:
Numerically, we can write (4.27) as:
dh c] i t ' . ) 1,(h"h 2 dh, 11;(h"h 2) 1 [ dt
~
[
, ! ·.---1-0.5vh,-h, 2< O. -,
=
_I '2
_1. ;,-
]
(4.28)
2\16 \ h,
Since this system is autOnomous (no explicit dependence on time), we can leave (oul of the arguments:
m,
=
111
~ 2
.
I
];(h,(k), h2(1e))] j(h(le)) = j;(h,(k), h,(k))
.ti(h,(Ie) + at2 [1;(h,(k) + at2
""I'
Ill",
at]
h,(k) +2 h2(k)
fI1 21 )
M
+ 2
fI1 21 )
Numerical Integration
92
+ IlJ + 1l ~ [h,h '((k l ~ [hh '((klJ l + k + 1 2 k
h(k
Chap. 4
1I1,C,t
Let the initial conditions be 11 1(0) = 12 ft and hiD):= 7 ft. Also. let!J.! 0::: 0.2 minutes. Por k = 0, we find
. 05
YI2-~7
I ..
-7-,IY7-
2v6
-0.118034.] 1. 0.018955
so
I
h,(O)] hiO)
j
h,(O) +
!!.t 2
. 12 +2 0.2 ( 0.118(34) • ~ 111.9881971
----m
~t 111" ~
h,(O) + !!.t 111" 2'
"'" •. t;(h,(O) ~
"'" .0
.=
0.2 7 + -;C (0.018955)
'
+
~
12 +
7
f·
7.001896
0~2 (_ 0.118(34) ~
0 .2 (0.018955) 7_
~t "'". h (O) + ~t "''') 2
~
11.988197f1
7.001896 fl
1;(11.988197.7.001896)
1-0.5vI1.988197-7.001896 ~ -0.116501
t;(h,(O)
+ ~. "'", h,(O) +
~t "''') ~ /,( 11.988197, 7.001 R96)
0.25 vi (.988197 -7.(J()lWJ6-
",(I)
~
",(0) +
",( I) - ",(0) +
m"c,t
~
v7.001896
~
0.OIH116
12 + - 0.116501 (0.2) - 11.976700 It
m"c,t - 7 + 0.01 R116 (0.2) .= 7.1l03623 It
and we call continue for the next time step, k = t. A plot of the response of this system is shown in Figure 4.7. The response of Ii] actually increases slightly hefore dccrcasing------this is missed became of the scaling. Notice that when hi is greater than h 2 , the flow is from tank I to tank 2; while when hi is less than 1l 2, the flow is from tank 2 to tankl (although this cannot occur at steady-state). Since we have assigned a positive value to F[ when the flow is from tank I to tank 2, then a negative value of F I indicates the opposite flow. Care must he taken when solving this problem numerically, so that the square root of a negative number is not taken. For this pU'lJOse, the sign function is used
Sec. 4.3
93
Runge-Kutta Integration
::~ 10 1
- - - -
gL
81
7r-------61o
20
--==========~! 40
60
80
100
time (minutes)
FIGlJUE 4.7
Transient response of the interacting lank example, using second-order
Runge- Kutta.
The idea behind the second-order Runge-Kutta can be extended to higher-orders. The most commonly used method is the .f(Jllrth-order Runge-KaNa method as outlined in the next section.
4.3.2
Fourth-Order Runge-Kutta
Using this method, the approximations arc more accurate than explicit Euler or sccondorder Runge-Kutta. The idea is to use the initial slope (m J ) to generate a first guess for the state variable at the midpoint of the integration interval. This first guess is then used to find the slope at the midpoint (m 2 ). A "corrected" midpoint slope (m 3 ) is then found by using mz. A final slope (m 4 ) is found at the end of a step using 111 3- A weighted average of these slopes is then used for the integration. The algorithm is fill ~
[(I(k), x(k))
fIl 2 =
[~(k)) + ~ D.I, x(k)
(4.29)
~ fIllD.I)
(4.30)
~ [(I(k) + J D.I, x(k) + 1 11I D.I) · 2 2 2
(4.3 J)
+
fll,
fIl 4 = [(I(k)
t
+ D.I, x(k) + 11I,D.t)
(4.32)
94
Numerical Integration
x(k
+ i)
~
x(k)
j
I
~!lL
-I-
IJ}z
11
3
+ tn, + m•4 3
11
!,
uf
Chap.4
(433)
To become rnore familiar with integration techniques, yOll should solve sOllle of your initial problems using the explicit Euler method. Make certain that your step size is small enough so that the errors do not build up 100 rapidly. As you find a need for more accuracy, you should then lise the fourtlHlrdcr Runge-Kulla method. In Section 4.4 we introduce the MAI'LAB routines that arc available for numerical integration.
SELECTiON OF INTEGRATION STEP SiZE Generally, integration step size must be "small" for Euler, call be larger for second-order RUllgc~KLltta (as far as accuracy is concerned), and can be still larger for fourth-order Rungc-Kutta. Particularly for Euler, step sizes that are (00 large can be unstable or inaccu~ rate. Step sizes that arc too small may waste computer time or have numerical truncation errors sillce the state variables Inay not change much from step to step. If the student L1ses a fixed step si/.e, then it is generally a good idea to try larger and smaller step sizcs to sec if the results change significantly. Generally, you will want to use as large a step size as possible, A particular challenge is from "stiff' systelns (time constants that span a wide range), where a commerical code specifically for stirI' systems should be Llsed. One welJknown implicit mcthod for stitT systems is (Jear's method, which is available in SIMULINK. Implicit methods will only work wel! for systems that arc continoLls. IT discontinuous systems are simulated (for example, step changes at certain times), then Runge-Kutta methods should be used. Most eommerical integration packages use a variable step size. The integration step size is automatically chosen and varied from step-to-skp to assurc accuracy while minimizing computation time. The integration routines in MA'rLAB use a variable integration stcp size.
4.4
MATLAB INTEGRATION ROUTINES Thc primary purposc of thc previous sections in this chapter was to review simple numcrical techniqucs for integrating initial valuc ordinary differential equations. We havc illustrated the techniques with some simple numcrical examples, implemcnted as m-files in MATLAB. In practicc, we do not recOlnmend that you write your own integration routincs. You will spend much tinle debugging these routines and they will generally not be as po\verful as existing academic or commcricalintegration routines. Your goal should bc to providc the correct formulation of the model, spccifying the corrcct initial conditions and parameters. You should generally usc a well-documented, commerical or public domain integration code to implement your simulation. MATLAB has several routincs for numerical integration; two arc ode23 and ode45. ode2:3 uses second-order and odC'4:) uses founh-order RLlngc~Kutta integration. Both routines usc a variable integratioll step size (Lit is not constant). The integration
Sec. 4.4
MATLAB Integration Routines
95
step size is adjusted by the routinc to provide the necessary accuracy, witholll taking too !llllch computation limc-,
4.4.1
ode23 and ode45
To usc od(-~23 or ode4~-), the reader must first generate an m-filc 10 evaluate the state
variable derivatives. Theil the student gives the cOlllmand:
It, xl
ode45 ( , xpr ime '
I
rto, t
t ] xO )
where xprimc
is a string variable containing the Halne of the Ill-file for the derivatives
to
is the initial time is the fjnal time is the initial condition veclor for the state variables (usually a column vector)
t t
xO
The arrays that arc returned arc
x
a (column) vcctor or lime an array of state variables as a function of time (column I is Slate I, etc.)
For example, jf the lillle vector has 50 elements, and there arc three state variables, then the stale variable vee lor has the 50 rows and three columns. After the integration is performed, if the student wishes to plot all three variables as a fUllction of time, she/he simply types
plot(t,x) If you only want to plot the second state variable. then the command plot (t, x ( : ,2) ) is given.
EXAMPLE 4.4
Revisited Solution Using MATLAH Routine ode45
rirst. the following file titled twotnk. m was generated:
function hdot
Lwotnk{t,h) ;
consL~(1/{2*sqrtI6)));
hdotll) = 1-0.5*sqrtlh{l) h(2)); hdot(2)
"" O.2')"c3qrtlhll)--h{2))-cow,t*:;qrt(h(2));
'fhcn, the following command is entered ill the !VIATLAB command window: [t, h 1"'ode4S { T twotnk' , [0100] , [12 7
J ' };
Numerical Integration
96
Chap. 4
Notice lhat we arc generating two arrays, ( and h, and using ode45. The function file is named twotnk. m. The initial time is to :; :;: 0 and the final time is ~r:;;:: 100. The initial condition IS 110:;;:;: fl2 Tj'. At the MATI,AB prompt (») the following commands were given: plot(t,hl"l)) plot (t, hi, ,2) )
The transient responses are shown in Figure 4.8.
12
11.5
£
11
10.5
10
o
20
40
60
80
100
80
100
time (minutes) a. Height of Tank 1
7.2 7 6.8 N
-c
6.6 6.4 6.2 6 0
20
40
60
time (minutes)
a. Height of Tank 2 FIGURE 4.8
Transient response curves for interacting tank example.
Notice the tremendous reduction in effort when compared with generating your own RungeKuua code.
Further Reading
97
Often it is desirable to know the state variable valucs at a particular timc or at fixed time steps. A variable stcp size algorithm yields variable valucs that are not at a fixed step sizes. One has two options. If the variable step size is smaller than that of the variable step, then we could reduce the step slze. The major disadvantage is that computation time will increase. The best option (and that recommended by MATLAB) is to use a spline fit to interpolate or extrapolate the values to desired points. The routine used is interpl.
SUMMARY It is important for the student to understand the Euler, as well as the second and fourth order Runge-Kutta integration techniques. When using your own fixed step size integration code, be careful with the selection of !1t. In practice, it is preferable to use a commercial integration code, which automaticalJy selects the integration step size. The MATLAB routines llsed were ODE23:
Variable step size, second-order Runge-Kutta
ODE45:
Variable step size, fourth-order Runge-Kutta
FURTHER READING A nice treatment of numerical integration is provided by: Parker, T.S., & L.O. Chua. (J 989). Practical Numerical Algorithms .ftJr Chaotic S'.vstems. New York: Springer- Verlag.
A treatment of integration techniques with chemical engineering applications is presented by Davis. Davis, M. E. (1984). Numerical Methods and Modeling for Chemical Engineers. New York: Wiley. The following book is more of an advanced undergraduate/first-year graduate student text on numerical methods to solve chemical engineering problems. The emphasis is on FORTRAN subroutines to he used with the IMSL (FORTRAN·hased) package. Rameriz, W.P. (1989). Computational Methods ./iJr Process Simulatioll. Boston: Butterworths.
Numerical Integration
98
Chap.4
STUDENT EXERCISES 1. Consider the scaled predator-prey equations.
dy,
··-=a(l-y)y dt .2 I
dy,
;il
The parameters are
<X
=
r3
=
-13(1 - y,)y,
= 1.0 and the initial conditions arc ."1(0) :;;;; 1.5 and
y,(O) = 0.75. The time unit is days.
a. Solve these equtions using explicit Euler integration. Compare various integration step sizes. What b.t do yOll recommend? In addition to transient responses (t versus y, and Y2), also plot "phase-plane" plots (y, versus yz). b. Solve these equations using the MATLAB integration routine ode45. Compare the transient response curves with the Euler results. c. How do the initial conditions effect the response of YI and Y2? Please elahorate.
2. Consider a CSTR with a second-order reaction. Assurne that the fate of reaction (per unit volume) is propottional to the square of the concentration of the reacting component. Assuming constant volumc and constant density, show that the modcl~ ing equation is:
dC dt Use the following parameters: V 5 . p=.mll1
k 2 = 0.32 flO Ibmol' min-'
and a steady-state inlet concentration of C i , = 1.25 Ib11101 fl ' Calculate the steady-state concentration of C, = 0,625 Ibmol ft-3 Assume that a step change in thc inlet conccntration occurs at t:::: o. That is, C i changes from 1.25 Ibmol fl-} to 1.75 Ihmol ft-} atl = 0 minutes. Use ode45 10 simulate how the outlct concentration changes as a function of time.
3. Analyze the stability of the fOliJ1h ordcr Runge-KuHa method for the classical order process.
dx
tit
first~
-x
What is the largest integration step size before the numerical solution becomes unstable?
t
Student Exercises
99
4. A gas surge drum has two componcnts (hydrogen and methane) in the feedstream. Let Yi and y represent the mole fraction of methane in the fcedstream "mel drum, respectively. Find dP/dt and {I,v/dt if the inlet and outlet now rates can vary. Also assume that the inlet concentration can vary. Assume the ideal gas law for the effect of pressure and composition on density. Assume that the gas drum volume is 100 liters. The temperature of the drum is 3 J.5 deg C (304.65 K). At steady-state the drum pressurc is 5 atm, the molar flow rate in and out is 2 gmollmin and the concentration is 25Cff! mcthane, 75% hydrogen. Usc Eulcr integration and ode45 to solve the following prohlems. Discuss the effect of integration step size when using Euler integration. In all cases, you arc initially at steady-state. a. Assume that the molar llowrates remain constant, but the inlet methane concentration is changed to 509h. F'ind how pressure and composition change with time. b. Assume that the molar f10wrate out of the drum is proportional to the difference in pressure between the drum and the outlet header, which is at 2 atm pressure. Perform a step change in inlet concentration to 5(Y./o methane, simultaneously with a step change in inlet tlowrate to 3 gmollmin. c. Assume that the MASS f10wrate out of the drum is proportional to the square root of the difference in pressure between the drum and the outlet header (which is at 2 atm pressure). Again, perform a step change in inlet concentration to 501ft" methane, simultaneously with a step ch;:mgc in inlet flowratc to 3 gmol/min, d. Assume that the MASS rlowrates in and out are proportional to the square root of the pressure drops. Assume that the steady-state inlet gas header is at 5 atm. Perform a step change in inlet concentration to 50W" methane, simultaneously with a step change in inlet pressure to 6 atm. 5. Pharmacokinetics is the study of how drugs infused to the hody arc distributed to other parts of the body. The concept of a compartmental model is often used, where it is assumed that the drug is injected into compartment I. Some of the drug is eliminated (reacted) in compartment 1, and some of it diffuses into compartlnent 2 (the rest accumulates in compartment 1). Similarly, some of the drug that diffuses into compartment 2 diffuses back into compartment 1, while some is eliminated by reaction and the rest accumulates in compartment 2. The rates of diffusion and reaction arc directly proportional to the concentration of drug in the compartment of interest. The following balance equations describe the rate of change of drug concentration in each compartment.
Numerical Integration
100
Chap. 4
drug concentrations in compartments J and 2 (fLg/kg patient where x J and x2 weight), and u = rate of drug input to compartment 1 (scaled by the patient weight. [Lg/kg min). Experimental studies (of the response of the compartment I concentration to various drug infusions) have led to the following parameter values: ;:;0
(k lO + k 12 ) = 0.26 mini
(k 20 + k21 ) = 0.094 min-I k l2 k 21 = O.(1I5 mio- I for the drug atracurium, which is a I1lll.<';clc relaxant. Notice that the parameters have not hccnindcpcndcntly determined. Show (through numerical simulation) that all of the following values lead to the same results for the behavior of xl' while the results for x2 afC different. Let the initial concentration he 0 for each compartment and as~ slime a constant drug infusion rate of 5.2 fLg/kg min.
a. kJ2~k2l b. k 12 =2k 21 C. k l2 = 0.5 k21 Discuss how the concentration of compartment 2 (if measurable) could be Llsed to dctcrmine the actual values of k l2 and k 21 . Usc the MATLAB [unction ode45 for your simulations. 6. A stream contains a waste chemical, W, with a concentration of I moilliter. To Illcet EPA and statc standards, at least 90% of the chemical must be removed by rcaction. The chemical decomposes by a second-order reaction with a rate constant of 1.5 liter/(mol hr). The stream f10wratc is 100 liter/hr and two availahle reactors (400 and 2000 liters) have been placed in series (the smaller reactor is placed hcforc the larger onc). a. Write the modeling equations for the concentration of the waste chemical. Assume constant volume and constant density. Let ell'l ~ concentration in reactor 1, mol/liter C w2 ~ concentration in reactor 2. mol/liter F' ::::: volumetric tlowrate, liter/hr
Vj V2
liquid volume in reactot 1, liters liquid volume in reactor 2, liters k : : : second-order rate constant. liter/(mol hI') :::::
b. Show that the steady-state concentrations are 0,33333 mol/liter (reactor I) and 0,0')005 mol/liter (reactor 2), so the specification is met. (I-lint: You need to solve quadratic equations to ohtain the concentrations,) c. The system is not initially at steady-state. Write a function file and use ode45 for the following:
Student Exercises
101
(i). If C",I(O) = 0.3833 and C",2(0) = 0'()9005, fine! how the concentrations change with time. (ii), [I' C",I(O) = 0.3333 and C",2(0) = 0.14005, find how the concentrations change with time.
7. Consider a batch reactor with a series reaction where component A reacts to form the desired component B reversibly. Component B can also react to form the undesired component C. The process objective is to maximize the yield of component B. A mathematical model is used to predict the time required to achieve the maximum yiele! of B. The reaction scheme can be characterized by
Here k]fand k tr represent the kinetic rate constants for the forward and reverse reactions for the conversion of A to B, while k2 represents the rate constant for the con~ version of B to C. Assuming that each of the reactions is first-order, the reactor operates at constant volume, and there arc no feed or product streams, the modeling equations arc:
where CA, Cu' and C c represent the concentrations (mol/volume) of components A, B, and C, respectively. a. For k,{= 2, k l ,. = I, ane! k 2 = 1.25 h,l, use ode45 to solve for the concentrations ~IS a function of time. Assume an initial concentration of A of CAO := t mollliter. Then plot the concentrations as a function of time. For what time is the concentration of B maximized? b. Usually there is some uncertainty in the rate constants. If the real value of k2 is 1.5 hr- l find how the concentrations vary with time and compare with part a.
SECTION
III LINEAR SYSTEMS ANALYSIS
LINEARIZATION OF NONLINEAR MODELS: THE STATE-SPACE FORMULATION
5
Many dynamic chemical processes arc modeled by a set of nonlinear, first-oruer differential equations that generally arise from material and energy balances around the system. Common analysis techniques arc based on linear systems theory and require a ,\·tate~space model. Also, most control system design techniques are based on linear models. The purpose of this chapter is to provide an introduction to state-space Inodcls and linearization of nonlinear systems. After studying this chapter the reader should be able to: Write a linear model in stale-space form. • Linearize a nonlinear model and place in state-space form. Use thcMATLAB eig function to analyze the stability of a state-space model. • Develop the analytical solution of state-space models. Understand stability and transient response charactcrstics as a function of the eigenvalues. • Understand the importance of initial condition "direction". Be able to use the MATLAB routines step and initial for simulation of statcspace models. The major sections in this chapter are: 5.1 5.2 5.3 5,4 5.5 5.6
State-Space Models L,inearization of Nonlinear Models Geometrical Intell1retation of Linearization Solution of the Zero-Tnput Form Solution of the General State-Space Form MATLAB Routines step and initial 105
106
5.1
Linearization of Nonlinear Models: The State-Space Formulation
Chap. 5
STATE·SPACE MODELS Thus far in this text we have discussed dynamic models of the general form:
x=
f(x,u)
('ill
where f(x,u) is, in general, a nonlinear function vector. A linear model is a subset of the more general modeling equation (5.1). 'I'he fonn of linear model that we discuss in this chapter is known as a statc-space model. First. \vc show how to write statc~spacc models for systems that arc inherently linear. Tlll'.IL \VC show how to approximate nonlinear systems with linear models. [~xamplc 5.1 illusln:ttcs the form of a state-space model.
EXAi\IPLE 5.1
NOllintcracting Tanks
Consider two tanks in series where the flow nut of the first tank enters the second lank (Figure 5.1). Our objective is to develop a model to describe how the height of liquid in tank 2 changes with lillIe, given the input fJowratc F,,(t). We aSsumc that the flow out of each lank is u linear function of the height of liquid in the tank. Fo
t h1
tbi. tt=L. F1
h2
FIGURE 5.1
F
2
Noninteracting tanks.
A material balance around the first tank yields (assuming constant density and F] :;;; [1 IlIl)
(5.21 where A J is the constant cross-sectional area (parameter), I) I is the flow coefficient (paranlt'((TL Po is the fJowrale into the tank (input), and hi is the tank hcight (state). Writing a materi
F. J- _ ../3)' II A, 2
A,
Sec. 5.1
107
State-Space Models
where A 2 is the constant cross-sectional area for tank 2 (parameter), ~2 is the now coefficient (parameter), F J is the flowratc into the tank, and /1 2 is the tank height (state). In this case, F I is not an independent input variable that can be manipulated, since F 1 =. {3Jh J " We can write the previous equation as
dI',. /3, df - A
~2 h
h
2
A
1 -
2
2
(5.3)
Notice that we can write (5.2) and (5.3) in the following matrix form:
(5.4)
which has the general form:
x=Ax+Bu
(5.5)
where:
/3, A,
/3, A2 The state and input vectors arc (notice that the input is a sC:;llar):
x
= r~:J
anel
II
= Fo
The additional equation that is nonnally associated with a state space model is (5.6)
y"c:Cx+[)u
where y is a vector of output variables. Generally, output variables arc variables that can he measured (at least conceptually) or arc of particular interest in a siumlation study. Here, we will consider the case where both tank heights arc outputs. Let output I be the first tank height and output 2 be the second tank height
The matrix-vector form is: y
[I 0] [h'l
~ o
I
h,
cc
where:
c
oj
I [o I
ex
108
Linearization of Nonlinear Models: The State-Space Formulation
Chap, 5
/fwe also consider the input, F o' to be the third output variable, we have the following relationship:
y
~
which is the form of (5.6), with
5.5.1
'l I [] 1
0
0
1 [;;'] +
0
0
1
o
'
2
()
[F;,]
General Form of State Space Models
Example 5.1 illustrated a specific case of a state-space model. In general, a state-space model has the f{)lIowing form: dX I
-I-
dt
(JIll X;,
-I- b ll
H)
-I-
-I- h Jill
-I-
lim
dim U IlI
which has n state variables (x), m input variables (ll) and r output variables (y). This relationship is normally written in the matrix form:
["] x"
['
(lui
(112
(1,,2
'or] [" """
x"
+
b",
[" I"[" '1'],I' ell
Yr
(r1
(;,<2
em
x"
tid
btl
bill
"r] b~!Il1 I:",
d l2
da
"]1 "] d~lIl ltl/l
Linearization of Nonlinear Models
Sec. 5.2
which has the general
109
(state~space) form:
x~Ax-l-Bu y~Cx
-I-Du
(5.7)
where the dot over a stale variable indicates the derivative with respect to time. A~ shown in Section 5.4, the eigenvalues of the Jacobian matrix (A) determine the slabihty of the system of equations and the "speed" of response. The aij coefficient relates state variable j to the rate of change of state variable i. Sillli~ larly, the bij coefficient relates inputj to the rate of change of state variable i. Also, cij relates state j to output i, while dij relates input) to output i. We can also say that the klh row of C relates all states to the ktll output, while the kth column of C relates state k to all outputs. In this section we have shown how to write modeling equations that arc naturally linear in the state-space form. In the next section we show how to linearize nonlinear models and write them in the state-space form. Linear models are easier to analyze for stability and expected dynamic behavior.
5.2
LINEARIZATION OF NONLINEAR MODELS Most chemical process models are nonlinear, but they arc orten linearized to perform a stability analysis. Linear models are easier to understand (than nonlinear models) and are necessary for most control system design methods. Before we generalize our results, we will illustrate linearization for a single variable problem.
5.2.1
Single Variable Example
A general single variable nonlinear model is: d-,,; dl ~f(x)
(5.8)
The function of a single variable, j(x), can be approximated by a truncated Taylor senes approximation around the steady-state operating point (.-r):
. ~ f(xJ -I- .. iif j(x)
I (x - xJ -I- I ..,a rI (x - x,)' -I- hIgher . order terms 2
dx x,
2 dx
(5.9)
x,
Neglecting the quadratic and higher order terms, we obtain:
. = j(xJ . j(x) -I- af . I (x - xJ dx
(5.10)
x,
Note that:
dx y dl
() ~jx, ~O
(5.11 )
110
Linearization of Nonlinear Models: The State-Space Formulation
Chap. 5
by definition of a steady-state, so:
I
dx = f(x) == 'iii . dt rh
(x- x,)
.1,
where the notation 0f1(}ti.t is used to indicate the parlial derivative ofJ(x) \vith respect to .t, s evaluated at the steady-state. Since the derivative of a constant (1;) is zero. we call write: dx
'J
d(x dl
dl
I) 13)
which leads to:
I (x - x,)
d(x - xJ ,.= ill .' dl dx
() 1-II
-I,
ThcTcaSOll for using the expression above is that we arc orten interested in deviations in '-1 stale from a steady-stale operating point. Sometimes the symbol is llsed to represellt dc viat;rm variu!JIes, x' ::::: X -'_:y We can sec that a deviation variable rcprcscllls the CIWll).!C or perturhation (dcviatirm) 1'1'0111 a stcady-state value. I
() 1)1
'Ihis can be written in
state~space
Corm: dr' (f
dt
(516)
x'
where {/ :::: iH!dxx s . We have shown how to linearize a single variable equation. Next, system with one state and one input.
5.2.2
\VC
consider a
One State Variable and One Input Variable
Similarly, consider a function with one state variable and one input variable dx
x
dl
~
f(X,II)
() 171
Using a Taylor Series Expansion Corf(x,u): x
. /(X"II,) + ;'ill I d.x
.t,.Il,
.iY'/1 .:
dxtJU
(x - xJ
+ {iu .'ill I. (II - IIJ .1,.1/,
(x - X,)(II - II,) + l ;i'il 2 1 (II ~.1 II,)' .1,.11,
2
rJlr
.\\.11\
+ higher order terms
b
Sec. 5.2
Linearization of Nonlinear Models
111
and truncating aftcr the linear terms, we have:
x.
~~
. It ) fix
,
s, s.
I
iI f +. (Jx
(x - x )
.\",.(1,
S
-
t-
ill ()ll
I
IJ',
(It -
ItJ
(5.18)
and realizing thatj(x,I"u) :::: 0 and dx/dt = d(x ~ .r)/dt:
d(x dl
xJ .~. ilf I ax
(x
~. x) +
',.11,
,
Using deviation variables,./ ::::.r - xI" and u' =
It -
ilf
all
I
(li
',,II,
.~
uJ
11,1':
which can be written:
dx' dt
= (/ x'
j--
b u'
(5.19)
where ({:::: cYIc-1rLr,\' and b::::; ~fl()u[x,\"u.\, If there IS a single output that is a function of the states and inputs, then:
g(x.lI)
Y
(5.20)
Again. performing a Taylor series expansion and truncating the quadratic and higher terms:
(5.21 ) Since g(x,I"u\) is simply the stcady-state value of the output Cv), we can write:
(5.22) or
y - y, ~ where c:::: aJ;J{)xi.\. \" /I s and d
co (x - xJ + d (li - lIJ
= dg/au:,_'-
.\',II.\'
Using deviation notation:
y' ,,::. l~xalnplc
system.
C
x'
+ d II'
5.2 illustrates the application of linearization to a
(5.23) one~input,
one-slate nonlinear
Linearization of Nonlinear Models: The State-Space Formulation
112
EXA \IPLE 5.2
Chap. 5
Consider H Nonlinear 'rank lleight Probll'l11
\vhcrc Ii i\ tlie
sjalC
dh
I·
13 \ II
.1
.'1
\ariilhlc. F is the input \'ariahlc. 13 Hild A arc parameters. The riglJlhand side is:
r
I(ld)
I ,
1(l1F) I
hi'
Ij \
,I
\
A
[3
I-
2/\ \ h,
III
(:,),251
rile ftl'\l terll] ell] the righthand side is lern. because the lincari/
I
I
dl!'
()
.1
J
W(, call jll)\'\ \\Tite:
hJ
d(h
II, 11.1
I'
1,,\ \ II
and lI\ing dC\ialiul1 \ariahk nutatioll (1/::0: Ii - h and j
dh'
I'
2/\\ 11,
I +
II'::::
A
r
II and dmpping the
II" ,I
I.'
!--'\)r \:UIl\CllicllCC (simplicl1;' in llotatj()!11 we often urlJJ! the (') notation and assume that x ~l!1d II
arc dc\j,Hi(1I1 \(lriahles Cr:::: Ii - h,.
/I::::
r
(,i
dIld
write:
til:
IJ
I
dl
2/\ \' h,
I
"
which i" in the :sta!('-sp;\l'l' form
dr
dr
5.2.3
ux-l-bu
(527)
Linearization of Multistate Models
The prC\ious cx,-ullples sho\\ed how to linearize single-stale vari~lhlc systems. In this seclion \VC geJlcralizc the technilJue for any' I1umber slates. lkfnre vvc generalize the techniquc. it is worthv,hile to consider an example system with Ivve) states, olle input and OIlC output.
ur
-
••
~ . ._ - - - - - - - - - - - - _
Sec. 5.2
EXAMl'LE 5.3
113
Linearization of Nonlinear Models
Two~statc
S.ystCJll dX j ~ 1;(.r, ,X"ll) lit
(5.28)
(5.29)
(5.30) Performing a Taylor series expansion of the nonlinear functions, and neglecting the quadratic and higher terms:
From the linearization about the steady-state:
j;(Xj"Xh,U) = J;(X il ,X2",Il,)
=
0
and:
g(X I."X2"U) ,=-, y, Since the derivative of a constant is zero:
"\x,
-=-or,,} and
dl
we can write the state-space model:
d(x] - x 2 )
lit
Linearization of Nonlinear Models: The State-Space Formulation
114
Chap. 5
,II,
,ll""1
(Ju
f-
_.
[ df e (JII
,
I
\\hid)
i~
,II~'
dg
d.t]
t .. " . "
ilx,
I·
(5.311 XI,
t"
lhe form or a st:l!t'-SP:lCl' IlHldel:
x'
A x' ex'
y'
+ H u' + D u'
\\here (') indicates dC\iation variahles.
5.2.4
J
Generalization
NO\\ consider the tll'ncral nonlinear Illodel \\'hcrc x is
x,
~
f,(.r,. ... \'",u, .. ·1/111)
xn
..
1;,(x 1
... .x/I·u I'
lJ
is
(j
, .. lI m )
-V ,
gJ (.r j ....-\"".11 J' .If",)
Yr
g,(.\l"
,X",II], .. .11 1)
In \-ector notation:
f( X,II)
15 ..\.11
g(x,u)
(';,\41
F killen!" of the linearization matrices arc defined III the follll\ving fashion: /\ 1.1
"1.1
(lX,' __ Ii
<11/,
I .. "
b
Sec. 5.2
Linearization of Nonlinear Models
115
I
(5.37)
c,'.I
= iJu '''i ~ dAj
i)u D = -"-" II
After linearization, we have the
.1,_1<,
I
aU
j
statc~spacc
"/'>
(538)
form:
x'
A x' + H u'
y'
Cx'+Du'
Generally, the () notation is dropped and it is understood that the model is in deviation variable form:
x=Ax+BlI y=Cx+[)lI Usually, the measured (output) variable is not a direct function of the input variable, so it is Illore common to sec the following state~spacc model: x~Ax+BlI
This procedure is applied in
EXAMPLE 5.4
[~xample
5.4.
Interacting Tanks
Consider the interacting tank height problem shown in Figure 5.2:
FIGURE 5.2
Interacting tanks.
AsslIlne that the flowratc out of tanks is a nonlinear function of tank height. The f]owratc out of tank one is a function of the difference in levels between tank I and lank 2.
116
Linearization of Nonlinear Models: The State-Space Formulation
Chap. 5
Also, assume that only the second tank height is measured. The output, in deviation variable form is
Notice that there arc two state variables, one input variable, and one outptH variable. Let
= F- F,
u
The clements of the A (Jacobian) and R matrices «5.35) and (5.36» arc:
"fl
I
ilf ah'}
I
lJil 1 Au
=
2
IJ I 2A I Vhl,\~--h~;
hJ;
~I
2A1VhL\-=7i~\
h F ".<
~2
~J
2A2v'hl;~-li;_;
li
= II
ill I ilF hJ',
2A,vh"
1
i
Al
=0 Since only the height of the second tank is measured, y
ell
and the state-space model is:
=
0:::
(jj;; k.", ~~ 0
g(h 1,h 2,F)::o::: h 2 --
ilL,
(from (5.37)):
Sec. 5.3
117
Interpretation of Linearization
Il,
(!li'] [
--- 2/tl\,/h l:;- :~)17s cc
Il,
{fJ:}
2A2V/hl,--~-h2'
dt
t
y
Ul}UI
[O!]
1~:1
where:
In this section we have shown how to linearize a nonlinear process model and put it in slate-space form. The states in this model arc in deviation (perturbation) variahle form; that is, the states arc perturbations from a nominal stcady~stat:c. A stale-space moclel provides a good approximation to the physical system when the operating point is "close" to the linearization point (nominal steady-state).
5.3
INTERPRETATION OF LINEARIZATION In Section 5.2 we illustrated the method of linearization of models into slalc~space form. The objective of this section is to illustrate what is meant hy linearization of a function. Consider the single tank height problem, which has the following model: dh dl
~
F
J(h,l')·c
A
-
f3 II
...
vh
(5.39)
I/V5
for a system with A -= 1 n2 , !Is :::: 5 ft, and fj ;;:; ft L5 /min the steady-state f10wrale is 3 F s ;;;;: I ft /min. To focus our analysis 011 the meaning of the linearization with rcspect to the statc variable, consider the ease where the input is constant. Then, from (5.39) and the given parameter valucs:
f(hYJ
c.c
I ~
1
vh
performing the lincarization:
J(h.FJ = f(h,,!',)
+:"If I . (h ~ h,) dh
h,)"s
(5.40)
118
Linearization of Nonlinear Models: The State-Space Formulation
Chap, 5
0,5 ~
"'"
linear
o
-a,s
o
4
2
6
10
8
x
FIGlJRE 5.3 Basic idea of linearization. 'rhe linear approximation is exact for the steady-state value of x;::o: 5.
for our parameter values (h,FJ '" 0
+
2\15
(h -
hJ
or I
('(hF)=O--(h-h) 's . 10'
(54 I)
.
10
8
'" 'q; '" .c
6
1"
.' ..
4
'
2
a
o
10
20
30
40
lime, min
FIGURE 5.4
Comparison of linear and nonlinear responses for two different
initial conditions.
50
Sec. 5.4
Solution of the
Zero~lnput Form
119
We can see how good tbe linear approximation is by ploning both the nonlinear function (5.40) and the linear function (5.4l), as shown inlijgurc 5.3 011 p. liS. Here we have used x to represent tank height and .f(x) to represent the nonlinear and linear functions. Notice that the linear approxiruation works well between roughly 3.5 to 7 feet. or course, the two functions arc exactly equal at the steady-state value of 5 [eet, which was the point at which the Taylor series expansion was performed. H.calizc that .f(x) is dx/dt, which is the rate of change of tank height. It makes sense that the rate of change is positive at a tank height less limn 5 feet, because the system "seeks" to achieve a steady-state level of 5 feel. Similarly, for a tank height greater that 5 feet, the rate of change of tank height is negative, because the level "desires" to decrease to 5 feet. We can also see that the linear system wiJl be slower than the nonlinear syslem, if the tank height is less that 5 feet, but will be faster if the height is greater than 5 feet, as shown in Figure 5.4.
5.4
SOLUTION OF THE ZERO·INPUT FORM We have previously written the general state space Inodel in the following form: x~Ax+llll
where x and u are deviation variable vectors for the stales and inpulS, respectively. In this section we assume Ihat the inputs are held constant at their steady-state values, but thaI the states may be initially perturbed from steady~state. The "zero-input" form of the state space model is then:
.j [ [ \1
all
t'/I
{lnl
{/lll
(/1I2
llll/l
·] 1r XI
.fn
or
x
Ax
(542)
This form is used to analyze the stability of a system and to understand thc dynamic behavior of a system that has had its statcs perturbed from the steady-state values. Recall that the single variable equation:
X
ax
has the solution:
X(I)
,,'" x(O)
which is stable if a < O. In a similar fashion, the solution to (5.42) is
X(I) ~ e"'x(O)
(5.43)
and the solution to (5.43) is stable if all of the eigenvalues of A are less than zero. The response of (5.43) is oscillatory if the eigenvalues arc complex.
Linearization of Nonlinear Models: The State-Space Formulation
120
Chap. 5
There arc many differenl ways to calculate the exponential of a matrix; in this chapter we discuss only the similarity tr{[n.~1"orm method.
Recall that the eigenvector/eigenvalue problem is written (sec Module 2 for a review):
AV=V;\
(5.44)
l
V = where
S,
=[~~:]=
S2 =
[~~~]
[s,
S2J
I
v" =. v"
"12) "22
(5.45)
first eigenvector (associated with AI)
= second eigenvector (associated with 1.. 2)
and the following eigenvalue matrix: ;\ =
I I I.., 0 1..0
(5.46)
2
Multiplying (5.44) nn the right side hy V·] wc find:
A=V;\V '
(5.47)
multiplying by the scalar t and taking the matrix exponential, we find: eM
= V
ei\1
V-I
(5.48)
(5.49)
where and we sec imlllediately why Ai
< () is required for a stable solution. The solution for x(t) is x(t) = V e'" V-I x(O)
(5.50)
An interesting result is that an initial condition vector in the same direction as ~i has a response in the direction of ~i with a "speed or response" of Ai" This is shown by the foll{)w~ ing analysis.
5.4.1
Effect of Initial Condition Direction (Use of Similarity Transform)
Recall that we arc solving the following model
x=
Ax
(5.42)
Define a new vector z, such that
x = Vz
(5.51 )
Sec. 5.4
121
Solution of the Zero-Input Form
or
z = V--- J X
(5.52)
and notice that (from (5.51 )):
x=
V Z
(5.53)
Substituting (5.53) and (5.51) into (5.42):
Vz=AVz
(5.54)
z = Vi A Vz
(5.55)
V'AV=!\
(5.56)
or, left multiplying by V-I
But, from (5.44) A V = V A so we can write:
which yields (from (5.55) and (5.56)):
z=
(5.57)
!\ z
But A is a diagonal matrix (sec (5.46», so we have:
[il] 'A0I "-20]1"'_ ] Z2
=
_
Z2
(5.58)
Notice that (5.58) represents two independent equations:
z]
= "-IZ\
12 =
"-2 Z2
(5.59) (5.60)
which have the solutions:
z,(t)
=
Z,(O) e'"
(5.61)
z,(I)
= Z2(0) e'i
(5.62)
"e'0 " e',1O][Z](O)] .Z2(O)
(5.63)
z(O)
(5.64)
and we can write:
[ ZI(t)1 z,(t)
=
or
z(t)
=
eAl
Notice that, if the z(O) vector has the form: (5.65)
ESCOLA Dc EHGENHARIA BIBLIOTECA
122
Linearization of Nonlinear Models: The State-Space Formulation
Chap. 5
then:
z(t)
(5.66)
and, if tile z(()) vector has the form: (').67)
then,
(568)
that is. initial condition,,; of z{O) :::
AI' v\ihile initial conditions ofz(O):::
will yield a ';spced of response" associated with
"
will yield a "speed of response" associated with
Al • 'I'llis means that state variable initial conditions in the "direction" of the first eigenvector will have a speed or response associated with the first eigenvalue:
(569)
and state variable initial conditions in the "direction" of the second eigenvector will have a speed or response associated with the second eigenvalue
x(t)
v z(t) (5.71l)
Knowing the effect of the initial condition "direction" is impOitant. If a randOlll case study approach was taken, then we might arbitrarily select initial conditions that \verc '"fast.'" \vhilc other (missed) initial conditions could cause a much slo\\'cr response. We \vill show two examples of tbe c-free! of initial condition: Example 5.5, where the syS!elll is stable, and r':xample 5.6, \\'here the system is unstable.
t
Sec,5A
I<;XAMI'LE 5.5
123
Solution of the Zero-Input Form
A Stable System
Consider the following system of equations
+ x2
XI = - 0.5 XI
(5,71 ) (5,72)
Using standard state-space notation
x = Ax
(5A2)
The Jacobian matrix is A
= [. -1l.5
1 '.1
°
(5,73)
-2
the eigenvalues are the solution to det(H - A) :;;:; 0, which yields A
det ([.
+ 05
°
J)
- 1
+ 05)(A + 2)
= (A
A+ 2
~
°
so
A,
- 0,5
= -
A,
2
and the eigenvectors are
<, = [-11.5547 '1 0,8321 Note that Sl is the "slow" subspace, since it corresponds to Al = -0.5 and space, since it concsponds to'\2 :::;::-2, The numerical values of (5.50) for this problem are
X(I)
= V
~2
is the "fast" sub-
e'" V 'x(O) O
-0,5547J[.e 0,8321
°
"
011' 0,6667] x(O) 1.2019
e"
°
(5,74)
If the initial condition is in the direction of ~l' that is
x(O)
~ [~J
(5,75)
we find the following slate solution (from (5.74) and (5.75»:
X(l)
=
[I e-~5'1
(5,76)
If the initial condition is in the direction of S2' that is,
X(O)
= I. - 0.5547.1 0,8321
(5,77)
124
Linearization of Nonlinear Models: The State-Space Formulation
Chap. 5
we find the following state solution (from (5.74) ami (5.77»:
- 0.5547
X(f)
(5.781
0.8321
Note that x(O) :::; ~l"'" f.lll is the slow initial condition and x(O) :;; ~2 :::;; I---~;_:~-_:~
I is
the fast
initial condition, as shown in Figures 5.4 and 5.5. The initial conditions in the fast subspace IWH' reached the steady-state in roughly 2.5 minutes (Figure 5.6), while the initial conditions in the slow subspace afC roughly 7SCYo complete in 2.5 minutes (J-'igurc 5.6').
11'IGlJRE S5
Transient response for initial condition in the slow
subspace.
The expm (matrix exponential) function from MATLAB can be llsed to verify these simulations. Using t :;; 0.5 and the fast initial condition, we find
»d
;:::
[~O. ~j
I
1;
0, '-2] ;
»x -;;; expm(a*O.5)*[--O.5547 x
0.8321]
;:::
-0.2040
0.3061
t
Solution of the Zero-Input Form
Sec. 5.4
which agrees with the plot
~hown
125
in I,\gurc 5.6.
fa.:H suospace
x(O) = [- 0.5547 ] 0.8321 -1
o'-------------------=' 2 345
FIGURE 5.6
Transient response for initial condition in the fast
subspace.
The previous example was a stable system. The next example is an unstable system.
EXAMPLE 5.6
An Unstable S}'stem (Saddle)
Consider the following system of equations; .\:) =
~\'2
The Jacobian matrix is
A
1.22
=
2x 1
+ Xl
= 2x, -. Xl
II
- I
the eigenvalues are A,
= -1.5616
A2
I
<2 -
=.
2.5616
and the eigenvectors are
<1 =
...
0.270:1 .1 -O.962g .
o.R719]
l0.4896
s[
since Al < 0, is a stable subspace; since A2 >. 0, S2 IS an unstable suhspace. The solution for this system is: Xl
__._ ["
() -
o.R71911e
0.2703 -0.9628 0.4896
15(,lr"
0
°11 0.5038
e 1.56 \(il.O.9907
If the initial condition is in the direction of ~l' that is:
8972 -0. 1 x(O) 0.2782
126
Linearization of Nonlinear Models: The State-Space Formulation
Chap. 5
O,'7(H!
x(O)
--
O.~h2-f-; I
we find the fulJrming :--1,l1C SolUli(l1l O.270J ('
xl t) ~I\
which is a stablt' solution.
shown in
~
o
r"i)!UH:
.".7
stable sUbspace
~~-~--===-------I
~~
x -0 5
- O,9h28 ('
f /,.
1/ "
x(O)
~[
-
0.2703] 0 9628
-I'L----:----------::---;;-----o 2 3
FlCa"IU: 5.7
Inlli,t1l'(11Hli\i(l1l in till' \[:lhk SUh"p:ICl'.
If lhe initi;d condilinll is in the dirl'ctioll pI x(lI)
that is,
X7191
0 [ (l'-iS96
we find the following state \o]lIliun: X(I)
rUnl9 (-"';('11,' [O.-iX(l() eo' ,",h'
which is all lHl.s[,lhJe SO!llliulL as ShU\\'ll lJl FlgUIT 5.S.
o
o
FIGUI-U: 5.8
02
O~
06
08
Initial umditioll in tilt' unstahle \uhspau_'_
Il should be noted lhal if the initial conditioll is Ill)! ('.Y(lU/\' in the stahle slIh"pacc. the solUliull \\'illl.w~jll to di\l'I'!,!c and hCCOllll' unslahk. Thut is. if thl' initlal cunditi()11 i" olT by. "a;, 10 !p. thl' rl'Sp()Il"l~ \\,ill l'\,cntually lxconw Llllbuumkd.
t
Sec. 5.6
5.5
127
MATLAB Routines step and initial
SOLUTION OF THE GENERAL STATE-SPACE FORM
b 12
;.]
",-
b n2
b nm .
_U~iI
Now, consider the general form:
[] ["
__u:
"T] l"
at?
11
(lll/!~
ali?
b"l
\ /I
or x~Ax
Illu
(5.79)
Recall that the single variable equation:
x=a.\l-bu
(5.80)
has the solution:
X(I)
~c
b en, x(O) I (e'" - I) 1/(0) a
(5.8 I)
when u(f) = constant = {f(O).
In a similar fashion, the solution to (5.79), for a constant input (u(t) = t::::: {)
nCO»~
from
to tis
X(I)
~
P xeD)
+
Q u(O)
(5.82)
where
(583)
Q
and F:quation (5.82) can be used to time step by using:
ttl
~
(I' - I) A 'II
(5.84)
solve f{}I' a system where the inputs change from time step
X(I + AI)
+ Q U(I)
(585)
Px(k) + Qu(k)
(5.86)
P X(I)
More often this is written as
x(k + I)
~
where k represents the kth time step. Often a general purpose nuttlcrical integration tcch~ niquc (such as one presented in Chapter 4) will be used to solve (5.79).
5.6
MATLAB ROUTINES step AND initial We show the usc of step and 1. nit i a 1 by way of the following example.
128
EXA!\'1PLE 5.7
Linearization of Nonlinear Models: The State-Space Formulation
Chap. 5
A Linearized BiOl"Cactol" l\lodel
Consider the follO\ving: lincarilcd form of a biorcactor model with suhstrate inhibition kinetics (see tvloduJe 8 for details):
x=Ax-fBu
)'=Cx+Du where:
o
A=
- 0.7500
009056-] 2.5640
B _[- 1.5302 [ _ 1.8255
[oI 0]I J) -
[~;I
Enter the state space model: »
a =
[0,
0.9056;--0.7500, -).')640]
<::1
o
0.9056
-0.7500 -2, 'j640 »
b
b
c
co::
[--1.5302;3.8255)
-1.5302 3.8255 [l,O;l,OJ
>-,.c
1
0
o
1
[0;0)
»c1
d
o o
t
Sec. 5.6
MATLAB Routines
~;tep
and initiiJl
129
Check the stability »
eig(a)
ans : :; : ~().3000
-2.2640
The system is stable. Assume the process is initially at steady-state. Since this model is in deviation variable form, the initial condition is the zero vector.
5.6.1
The MATLAB step Function
The MATLAB step [unction assumes a deviation variable form (the initial conditions are zero). The cOlllmands arc.: »
[y,x, t]
stcp{a,b,c,d,l);
" plol(t.,y) which yields the plot shown in Figure 5.9. Notice that the ::3 te-p function automatically determined the length of the Ii me vector. You may also provide an equal~spaccd time vector and usc the following command [y,x] = step(a,b,c,d,l,t)
5.6.2
The MATLAB initial Function
The MATLAB ini t ial function assumes a deviation variable form, with the initial conditions perturbed from zero. The commands arc:
FIGURE 5.9
Plot of outputs, for a step input change.
130
Linearization of Nonlinear Models: The State-Space Formulation » »
[Y,x,tJ
=
Chap. 5
initial(a,b,c,d/l);
ploL(t,y)
Notice that the band d matrices are not really used by' the initial function. since it is assUllled that there is no input change.
SUMMARY In this chapter we have developed a statc~spacc model of a chemical process thm is in.. hcrcntly linear (e.g., the tank height example). \\ic have also \ho\vn I1mv to linearize models that are nonlinear. The models obtained in this fashion arc based on r/eriulioll l'ariables, that is. the states and inputs arc perturhotiolls from the steady-state operating point where the linearization is pcrfonncd. The stahility of a nonlinear syqcrn is determined from the eigellvalues of the Jacobian matrix in the linearized model (statc-sracc form ). Several important concepts were presented in this chapter.
For unforced systems (zero input). the initial conditiclll vector \vil! determine the "speed" (If response. For stable systems (all A 0), the eigenvector associated \vith the largest rnagnitude A is the fast direction, while the eigenvector associated with the smallest A is the s\()\v direction. Although it is possible for a system with both Ilegatin; (stable) and positi\e (unstable) eigenvalues to have stable behavior if the initial condition is In the stable subspace, this is impossible in practice. Any perturhation from the stable trajectory \-vill cause lhe solution to become unbounded (unstable). The I,,:1ATLAB routines that \-vere used include: exprn:
step:
Matrix exponential Step response of a state-space (or transfer function) model
State-space rnodels can be transformed to Laplace transfer function fOnl1, which is particularly useful for control sy'stcm design. Applications of Laplace transforms \vill be presented in Chapters 7 through 10. Eigcnvector/cigenvalue analysis \vill be useful in pe-rfonning phase~p!anc analysis. which is covered in Chapter 13. The reader should understand the fol\O\ving terms: state-space Jacobian deviation or perturbation variable eigenvalue
Student Exercises
131
eigenvector linearization stability Taylor series
FURTHER READING L,inearization is discussed briefly in most books on process control, including: Luyben, W.L. (1990). Process Modeling, Simulation and Control for Chelnical Engineers, 2nd Ed., New York: McGraw-Hill. Marlin, T.E. (1995). Process Control. Designing Processes and Control Systems for Dynamic Performance, New York: McGraw-HilI. Ogullnaike, B.A. and W. H. Ray (1994). Process Dynamics, Modeling and Control, New York: Oxford University Press. Seborg, D.E., T.F. Edgar, and D.A. Mcllichamp (1989). Process Dynamics and Control, New York: Wiley. Stephanopoulos, G. (1984). Chemical Process (:ol1trol: An Introduction to Theory and Practice, Englewood Cliffs, NJ: Prentice-Hall.
STUDENT EXERCISES 1. As a process development engineer you arc working on a process with three continuous-stirred-tank reactors (CSTRs) in series. A constant volumetric H(l\vrate (flowratc docs not vary with time) is maintained throughout the system, howcver the volume in each reactor is different (but constant). Since thc temperature varies from reactor to reactor (but is constant in an individual reactor) the reaction rate parameter is different for each reactor. The molar concentration of the inlet stream varies.
2
Assume that the density of the streams remains constant (independent of concentration). The reaction is a first-order (irreversible) decomposition (11 --> B). Molar rate of decomposition of A (per unit volume) = k CA
Linearization of Nonlinear Models: The State-Space Formulation
132
Chap. 5
tI. \\irite the::; dynamic model equations. h. \Vritc the statc-space model oCt") ::: A x + B It c. The values or the parameters and variables arc
F::::: 1 nJ/min C(I::: I IhmoJlft.'l
V, = I() ft1 V, = ft' V, = 5 fl ' :;:; O.O:n3 mini k::.::::: 0.2 min- 1
kl
k:;:::
0.55 mirr· 1
i. Find the steady-state concentrations in each reactor it Evaluate the A matrix (Jacobian) and find the eigenvalues
2. Consider a chemical reactor with bypass. as shO\vn below. Assume that the rcnction rate (per unit volullle) is first-order (r::: kC j ) and ('I is the concentration in the reactor (the reactor is perfectly' mixed). Assume that the volume in the reactor (V) and the feed flowrate (F) remain constant. crhc fraction of feed bypassing the reactor is (l -a)F and that entering the reactor is aF. Assume that the fraction bypassing the re~ actor does not change. The inlet concentration (C i) is the input vmiable and the mixed outlet strcarn composition (e 2 ) is the output variable. \\lrite this model in state-space form (this model is inherently linear, so deviation variahles are not neededj.
A :( + n If \"=C'x+-Dl/
o(.{-)
3. C\msidcr the follc)\\.;ing set of series and parallel reactiolls A
k. I
.~)
A + A
k.-, [i ~,
k,
C
.)
D
Material balances on cOlnponents A and B yield the following t\-\/O equations F , V ((",1/ - CAl - k. (", - k, (":,
d('., dl dC"
F
V ( .. ("/1)1 (k,
dl
where
k,
5 6
min
(Ii
I
Ie 10 mol
liter
5 ]
(A -
k: elll
.
I liters 6 mol min
nlln- I k, =----:-
,
(" ii'
~
mol
.,-~
.
I iter
a. Find the steady-state dilution raIl' UIVj and concentration of B (show all units). h. Linearizc and put in state-space hlrm (find the numerical values of the A. B, and (' matrices), assuming that the manipulated vari,l!,!e is dilution rate (F/I/) , and the output variable is CI). c. r;'ind the eigenvalues (shmv units). d. Find perturbations in initial conditions that arc in the fastest and slo\-vcsl directions.
h
133
Student Exercises
4. A chemical reactor that has a singlc second-ordcr reaction and an outlct r10wrate that is a linear function of hcight has the following model: dVC -= Fe dt In
- FC - WC
2
III
dV = F -F dt III
(5.87)
(5.88)
where the outlct f10wratc is linearly related to the volume of liquid in the reactor (F == 13V). The parameters, variables and thcir steady-state values are shown below.
== inlet llowrate (I liter/min) C in = inlet concentration (l gmolliiter) C == tank concentration (0.5 gmollliter) V = tank volume (I liter) k == reaction rale constant (2 liter/(gmol min)) f3 == I min-- l Fill
Equations (5.87) and (5.88) can be written in physical state variahle form as dC = Fi" (C _ C) _ kC 2 dt V -
(5.89)
(1l
dV
dt = F - "V t..J
(5.90)
11/
l---ist the states, outputs, inputs and parameters for the nonlinear equations (5.89) and (5.90). b. Linearizc (5.89) and (5.90) and write the state space modcl (find the numeric_al values for the A, B, and C matrices), assuming that thc inlet tlowratc is the input variable and that both states arc output variables. Define thc deviation variahles for states, inputs, and outputs. 5. Find the "fast" and "slow" initial conditions for the foJlowing model H.
6. Find the stable and unstable subspaccs for the following system of equations
Plot the transient responses for initial conditions in both the stable and unstable subspaces. Show that a small perturbation from the stable initial condition will lead to an unstable solution.
134
Linearization of Nonlinear Models: The State-Space Formulation
Chap, 5
7. The l1ol1intcracting tank model I:) (sec Example 5. () 1
13
"-
0
AI,
_
A,
A,
13 1
I
1
13,,
I" I ] ["1\ + AI, ",'. 0
.
I';,
.
Consider a system where the stcady~statc f10wratcs arc 5 ftJ/min, and the following cross-sectional areas and steady-state heights: Al ~ ~
hi We find (1"')Ill F I ~
2n'
A, ~
2.5 It
h,
~
lon' 5 It
I3 lh l and F, ~ 13 2h 2), then, that: 2 1'1 nz 13, ~ 2 , I', ~ 1 rnm
nlln
and the state-space model (in physical variables) becmHes:
1,
0 '1Ihh, ,I+[05I F l
--1
0,2 - OJ
:J [o ()
0
"
(:1[" 1,1 + [~;- [F I I
",
()
0
"
1
a. Work in deviation variable form and find the fast and slow subspaces. Use ini-· tial to simulate the unforced deviation variable systcrn (input deviation remains constant at 0), from initial conditions in both the fast and slow suhspaces. h. Usc the results from part <1, and convert to the actual physical variables. c. Work in physical variable form. Usc ini tial to simulate the unhJrced deviation variable system (input remains constant), from initial conditions in both the fast and slow subspaccs. Show that the results obtained afe the same as those in part b. 8. As a chemical engineer in the pharmaceutical industry you are responsible for a process that uses a bacteria to produce an antibiotic. The reactor has been contaminated with a protozoan that consumes the bacteria. The predator-prey equations arc Llsed to model the system (b ::::: bacteria (prey), jJ ::::: protozoa (predator)). The time unit is days.
dh dl dl' dl
u
b
'Y bp
" 'Y hp-
13 I'
Student Exercises
135
a. Show thatthc steady-state values are
b. Usc the scaled variables, tv and z, to find thc following scaled modeling equations Ii rill'
d/
dz d/
,,(I =~
p
z=
ll'
~z)w
fl (1
~
w)
Z
c. [<'ind the eigenvalues of the Jacobian matrix for the scaled equations, evaluated at W s and zr Realize that ll'.I ' and z,\. are 1.0 by definition, Find the eigenvalues in terms of ct and f3. d. "rhe parameters arc (~ :::0 f3 ;:; : 1.0 and the initial conditions arc w(O) ;:;;: 1.5 and z(O) = 0.75. i. Linearize and write the state-spacel~ml1 (let the state variables be Xl :::0 HI ,- It\- and );:, ;;;;; Z - z).Find the initial condition vector xo ;;;;; [,,".,«0)) [, to use . ~ . 'x,D with in.it_ial. ii. Solve the state space model from (I) lIsing lsim and plot the transient response of Xl and x2 as a function of time (plot these curves on the same graph), simulating to at least t ;:;;: 20. iii. Show a phase-plane plot, placing Xt on the x-axis and x2 on the y-axis. iv. WhItt is the "peak-to-peak" time for the bacteria? By how much time docs the protozoan "lag" the bacteria? 9. Consider the state-space model
Ii'l I. ~ 1.0 "2
4.0
~
0.011 X'i .\,
5.0
a. Find the "fast" and "slow" initial condition directions. 10. Consider the foJiowing system of two reactors.
~
Linearization of Nonlinear Models: The State-Space Formulation
136
Chap. 5
Assullle a firsHmJcr decornpositi()tl of ;\-->8. Assume that all flo\\TatC\ arc C(lll· stant (volumes arc consUlllt). a. \\;'ritc the moueling equations for concentration or A, using either the instantaneous or integral method. h. \Vritc these in state-space form: X~AX4
Bu
c. Given the following constants. calculate the steady-slale concentrations:
I·;,
cu
=
_ kgnlOl .j
IJ.3.1.13.1 hr
L
I
111
V~ = l)m
VI'" 15nr'
d. Find the eigenvalues of the A matrix. Discuss the swbility of this Sy'stClll. e. The inlet concentration.
('Ii'
is changed from 1.5 to 1.75 at {:::: O. Usc ,:;l:cp tel
simulatc the behaVior of this system. 11. ;\ stirrc,d lank heater is used to supply a chemical proct:~ss \vith a fluid at a constant temperature. The heater receives fluid from an upstrcam pnll'l'SS unit \vhich may cause the f10wrate or tcmperature to change. Consider the diagram of the stirred tailk heater sho\vn belt)\:!". \vhere the tank inlet stream is received from another process unit. /\ heat transfer fluid is circulated through ajacket to heat the fluid in the tank, Assume that no change of phase nccurs in either the tank liyuid or the jacket liquid. Tank inlet
F,
T,
~
~
V
Tank Jacket outlet
Jacket
r-
T
---
Jacket inlet
F
T Tank outlet
Part 1 a. Write the dynamic modeling equatiolls to find the tank ami jacket tcmperaturcs. Do not usc any numerical values----Jeavc these equations in terms of the process parameters and variahles. State any additional assumptions needed [0 sol ve the problem.
Student Exercises Assume:
137
Constant level. Perfect mixing in both the tank and jacket.
The tank inlet flowratc, jacket flowrate, tank inlet temperature, and jacket inlet temperature may change. The rate of heat transfer from the jacket to the tank is governed by the equation Q ;:;; UAClj ~ T), where U is the overall heat transfer coefficient and A is the area for heat exchange. b. State the nl(~jor objective of this process. c. What do you consider the most important measured variable? d. What is a likely input variable variahle that you would use to maintain a desired tank temperature?
Part 2 Assume that both the tank fluid and the jacket fluid are water. The steady-state ues of this system variables and some parameters arc:
I'l' min
1; = Tjill
=
p(~, = 61.3
Btu
OF
p/'~)j = 61.3
50"F
Toe 125°F
v c=
20(Y'F
7; = 150"F
VJ = 1 ft'
10
val~
Btu OF
n'
e. Find Fj and UA (show units) at steady-state. f. Linearize the set of two nonlinear ODEs obtained in problem a, to obtain the state space form:
x=Ax+Uu
where
x=
=
u=
state variables
input variables
1'- l' ]
Y = 1T- 7: J J.I
= output variables
Determine the A, H, and C matrices (symbolically and numerically) g. Find the eigenvalues of A.
Linearization of Nonlinear Models: The State-Space Formulation
138
Chap. 5
Part 3 h. Simulate the system of state-space equations for a step change in the jacket flowratc from F:i =: 1.5 ft 3/min to l'j =1.75 ft 3/rnin F at time::: 5 minutes (work in deviation variables, but remember to convert back to physical variahles before plotting). What is the final value of the states, in the physical variables (T and 7/? Plot the response. i. Perform some simulations with step changes on some of the other input variables. Comment on any different behavior that yOll Illay observe. 12. Consider the following model of 2-stagc ahsorption column:
dzdl ~, (L) w _ (L + Vii) "+ V ". MM· M'I where IV and z arc the liquid concentrations on stage I and stage 2, respectively. L and V arc the liquid and vapor molar flow rates. 'Zr is the conccntration of the vapor strcam cntering the column. Thc steady-state input values arc L := 80 gmol incrt liquid/min and V:::: 100 gmol inert vapor/min. 'fhe paramcter values arc M:::: 20 glnol inert liquid, a = 0.5, and 7f= 0.1 gmo] solutc/gmol inert vapor. H. Find the steady-state values of wand z. h. Lincahzc and find the stale space model, assuming that L and V arc the inputs. c. Find the cigenvalucs and eigenvectors of A (Jacobian). d. l,'ind thc expected "slowest" and "fastcst" initial conditions (perturbations from steady-statc).
13. Most chemical proccss plants have a natural gas header that circulates through the process plant. A simplified version of such a header is shown below.
Pi From source
valve i Plant piping, represented as a perfectly mixed drum
Gas drum for a boilerhotlse unit
To furnaces
Here, the natural gas enters the process plant from a source (the natural gas pipeline) through a control valve. It flows through the plant piping, which we have represcnted as a perfectly mixed drum for simplicity. Another valve connects thc
Student Exercises
139
plant piping to the gas drum for a hoilerhouse unit. Gas passes through another valve to the hoilerhouse furnaces. The objective of this problem is to develop a linear model that relates changes in valve position to changes in drum pressures. a. Write modeling equations assuming that the pressures in drums I and 2 arc the state variables. Let thcinput variables- be (I) valve position I, (2) valve position 2, and (3) sourcc prcssure. b. Solve for the steady-state conditions and write the modeling equations in lincar, deviation variable form.
x=Ax+Uu y = ex
x=
,
rP2p , -- 1"'1 Ph
=
:;::
\I
state variables
[
change in valve position 1 change ~n v<.llve position 2
J = input
variables
change III sourCe pressure
output variables
y
c. Study the effect of step changes in each input on each tank pressure.
HINTS: For simplicity, assume that the following equations can be used for the flow through the valves: fJj
== (Xi hi (pi - PI)
=::
flow through valve i
C/I :::: ('(, hi (PI - P2) == flow through valve I
q2 ::;: 0'2 11 2 (P2 ~ p}) == flow through valve 2
where the f10wrate is in Ibmol/min, h is the fraction that a valve is open (varies between 0 and I), and Ct is a valve coefficient.
STEADY STATE DATA: gas flowratc Pis:;::;;
=:
lOOO std ff~/min
250 psig, PIS::::: 50 psig, P 2s ::;: 30 psig, P 3s == 5 psig
assume that each valve is 1/2 open under these conditions (his:::: h lA,:::: h 2s == 0.5)
140
Linearization of Nonlinear Models: The
State~Space Formulation
Chap. 5
CONSTANTS: V, ~ I 135 n3, V, ~ 329 ft3, Temperature ~ 32 "F
R (gas constant)
~
10.73
psia ft J IbmoloR
MAGNITUDE OF STEP CHANGES: Make separate step changes of 0.1 (10(*)) in the valve openings, and! 0 psia in the inlet pressure. Simulate for t:::; 0 to I = 15 minutes. 14. A strealll contains a waste chemical, W, with a concentration of 1 mol/liter. To Illeet EPA and slale standards, at least 9()fJ{) of the chemical must be removed by reaction. The chemical decomposes by a second-order reaction with a rale constant of 1.5 liter/(mol hr). The stream nowratc is 100 Iiter/hr and two available reactors (400 and 2000 liters) have been placed in series (the smaller reactor is placed before the larger onc). a. Write the modeling equations for the concentration of the waste chemical. A')sumc constant volume and constant density. Let ('wi
C w2 F
V, V2 k
~
~
~
~
~
~
concentration in reactor 1, mol/liter concentration in reactor 2, mol/liter volumetric flowrate, liter/hr liquid volume in reactor I, liters liquid volume in reactor 2, liters second-order rate constant, liter/Cmol hI')
b. Show that the steady-stale concentrations are 0.33333 mol/liter (reactor I) and 0.09005 mol/liter (reactor 2), so the specification is met. (Hint: You need to solve quadratic equations to obtain the concentrations.) c. L,inearize at steady-state and develop the state space model (analytical), of the form:
x
Ax-t-Bu
where: u~
d. Show that the A and B matrices are: - - 1.25 A = [ 0.05
o
- 0.320 IS
I
Il
.=
[0.0016667 O.25J 0.0001216 o
(also, show the units associated with each coefficient) c. i. Find the eigenvalues and eigenvectors using the MATLAB eig function. ii. Find the eigenvalues by hand, by solving det(Al- A) ~ o.
141
Student Exercises
f. The system is not initially at steady-state. Solvc the following for the linearized model, using the MATLAI3 function initial (first, convert the physical variables to deviation variablcs) i. If Cw,(O) = 0.3833 and C w2 (0) = O.(J9005, find how the concentrations change with timc. ii. If Cw,(O) = 0.3333 and CwiO) = 0.14005, find how the concentrations change with time. Relate these responses to the eigenvalues/eigenvector analysis of c. Discuss the differences in speeds of response (you should find that a perturbation in the first reactor concentration responds more rapidly and a perturbation in the second reactor concentration). The MATL.AB ini tial function needs you to create the following matrices before using it:
c
=
[1o 011
0=[0 01 °°
g. Solve f for the nonlinear equations, using ode4 5. Compare the linear and nonlinear variables on the same plots (make certain you convert from deviation to physical variables for the linear results).
h. Now, consider a step change in the flowrate from 100 liters/hour to 110 liters/hour. Assume the initial concentrations arc the steady-state values (0.3333 and ().(l9005). Compare the IincHr and nonlinear responses of the reactor concentrations. Is the removal specification still obtained? i. Would better steady-state removal of W be obtained if the order of the reaction vessels was reversed? Why or why not? (Show your calculations.)
ESCOLA Dc ENGENHARIA BIBLIOTLCA
6
SOLVING LINEAR nTH ORDER ODE MODELS
The purpose of this chapter is to review methods to solve solve linear 11th order ()[)F~s, After studying this material, the student will be able 10: Transform a linear state-space model with differential equation.
1/
states to a single nth order ordinary
Solve an 11th order constant coefficient coefficient homogeneous ODE. Solve an 11th order constant coefficient coefficient heterogeneous ()DE. Solve a first-order ()DE with a time-varying coefficient. Usc the H.outh stability criterion for stability analysis.
The nwjor sections in this chapter arc:
142
6.1 6.2
Background Solving Homogeneous, Linear ODEs with Constant Coefficients
6.3
Solving Nonhomogeneous, Linear OIJEs with Constant CocfTicicnts
6.4 6.5
Equations with Time- Varying Parameters Routh Stability Criterion ,.. Determining Stability without Calculating l:':igenvalues
Sec. 6.1
6.1
Background
143
BACKGROUND A model composed of a single, nth order linear ordinary differential equation has the lowing form:
dx + ",,(I) x til du + b,(I) til + b,,(I)
fol~
... + {/,(t)
(6.1 ) It
where the state variable is x and the input variable is u. 'rhis general model is linear because the state (x) and input (ll) and all of their derivatives with respect to time appear linearly. Notice thallhc coefficients do not have to be linear functions of time, however.
Models of the form of (6.1) do not arise naturally when chemical processes arc modeled. As shown in previous chapters, dynamic chemical process models arc generally sets of first-order (either linear or nonlinear ordinary differential equations. The advantage of the form of (6.1) is that there exist a number or·tcchniques to obtain analytical solutions. In this chaptcr we show how t(? transform scts of lincar, first-order differclltial equations to a single nth order differential equation. We then review several techniques for solving this type of equation. For motivation, we use Hbatch reactor example to illustrate each of the tcdllliques. It should be noted that there arc many good mathematics tcxts that covcr each of these techniques in more depth (see Boyce andlJiPrima, 1992, for cxampic). Our goal here is to provide a concisc overview of some morc lIscful tcchniques to solve dynamic chemical process problems.
EXAMPLE 6.1
Hatch Chemical
R(~ador
Cunsidcr a hatch chemical reactor, where there is no flow in or out of the vessel. The reactor is initially charged \vith a liquid of volume \I and an initial concentration (mol/liter) of reactant A of" C~,IO'
We consider a series reaction where component A reacts to lImn the desired component B. Component B call rurther relet to form the undesired component C. Each of the reactions is irreversihle, so A can react 10 rorm R, hUI B does not react to form A. A
k
--,>1
k 11 -}- C
Here k l represents kinetic rate constant (lime l ) for the conversion of /\ to Ii, while k] represents the rale constant II)r the conversion of U to C. Since component 8 is the desired product, we would like to know how long to run the feaction in order to maximi/.c the amount of 8 produced. If the reaction time is too long, all or B will eventually be converted to C,
Develop the Modeling Equations. Assume that each of the reactions is first-order. Since the volume is constant (dVldt ;:;:; 0), and there is no flow in or out, the modeling equations arc (the reader should be able to derive these, based on material balances on each component);
Solving Linear nth Order ODE Models
144
dC""j dt
Chap. 6
(6.2)
k 1 Cit
~{_Cll dt
(6.3)
dec _ . '
-'--- Ie,
f
(6.4)
(II
{/
where C.',\' C'fj, Hnd C( represent the concentrations (mol/volume) of components A. fl, and C, respectively. The units for the rate constants (k 1 and "-2) arc LIllie' I. Notice that the time faLl' of change of component A is only a fUllction of the COllccntnllion of /\. 'fhcH equation (6.2) call be solved, since C~4 and t arc separable, (0 find (6.5)
where C;\O is the initial condition for the concentration of 1\. If we define the conversion of A as \: :;;:; (CAO - C;\)/('"o and the dimensionless time T :::: kJt. (6.5) can be represented by the single
curve shown in r;igurc 6.1 (x:::: [
~ ('7).
0.8
c
0.6
.Q
"'>
ill c
g0.4
0.2
o o
2
3
4
5
dimensionless time
FIGlJRE 6.1
Conversion of;\ as a function of the dimensionless timc,
Now wc wish to find a single diJlerelltial equation to solve for Cflo
Reduce to a Single Equation for CB. Here we have (wo different ways to solve for Method I, Substitute (6.5) into (6.3) to obtain the expression
ell' (6.6)
Sec. 6.2
Solving Homogeneous, Linear Odes with Constant Coefficients
145
Equation (6.6) is a linear, constant coefficient, heterogeneous differential cquatiOiL ]t is hetenJgcneolls because of the "forcing function" on the righthand side. Heterogeneous equations are solved in Section 6.3. Method 2. Here we can rewrite (6.3) to solve for ell in terms of en: (6.7)
Taking the first derivative of (6.7) with respect to time, we find: (6.8)
Substituting (6.7) and (6.8) into (6.2), we find the second-order equation:
(!2~~'-'i ' {It
+ (k, + k,)
deft
til
+ k,k,
('/I
C
(6.9)
0
Notice that (6.9) has the form (6.10)
Equation (6. J 0) is known as a linear, constant coefficient, homogeneolls differential equation. The term homogeneous means that there is no "forcing function" on the rigbthand side. In Section 6.2 we cover the solution of these equations.
6.2
SOLVING HOMOGENEOUS, LINEAR ODES WITH CONSTANT COEFFICIENTS Homogeneous nth order linear differential equations have the form
d/lx lin
tit'f
+ an _ I
{pr--1x dt'l-l
dx
I + aox + ... + ", {/
=
0
(6.11)
To solve equation (6.11) we replace all (JiX/df i terms by AJ (6.12)
Equation (6.12) is called the characteristic equation. The n roots of the characteristic equation are called eigenvalues (in control textbooks the rools are often called poles). The eigenvalues arc used to solve (6.1 I). Two related methods are used, depending on whether the eigenvalues are distinct (all are different) or repeated (some are the same),
Solving Linear nth Order ODE Models
146
6.2.1
Chap. 6
Distinct Eigenvalues
We sce that (6.12) is ,jJl nth order polynomial that will have arc distinct (not repeated), the solution to (6.11) is
JI
roots, Ai' If all of the roots (6.11)
or
where each the constants ('I through en is found from the initial conditions, \iO). !lx ll " I (0 l!dIIl··1 . In (lrder [0 find the coefficients, c i' \VC rl)ust know the illitia/ cOlldifiollS for r and its derivatives.
6.1 Continued. ~c('
Solution for C:nmponent A
l'nllll equation (6.2) that the concentration of /\ does not depend on the vahle:; or U
e (h,14)
til lIle
l'haracteri'ilic
cqlW(IOII is
A+ k]
II
and [he eigenvalue is A:o: -- k l The solution is then C\(I)
e ,(I)
=c
J
C
e",e
(6.16)
k!
(h.17)
I.,
\\hich is the ",[llle result nhlaillcd in (h.)) usil1g Sl'lMratioll of variablcs and integnilioJ1. Now. let's cuntilluc and use lhe gel]cnd procedure 10 so!vc a sccond-order differential eqllalion
EXX\IPLE 6.1 ContimH'(I,
Solution for Component B
r\.t'ctl! lhatthc cyuatin!l fur the concentration of B i.s: (I
(h.! 01
t
Sec. 6.2
Solving Homogeneous, Linear Odes with Constant Coefficients
147
which can be written: (6.19)
So the eigenvalues arc:
and the solution call be written:
CJI)
~
c, exp(- ki) + (', cxp( . ki)
(6.20)
We need two initial conditions, ell(O) and dCjj(O)Jdt, to evaluate the constants, (:1 and ("2' We assumed that there is no component B in the reactor initially, so ell(O) :::: O. From (6.20) we then find:
From (3) we sec that: (6.22)
Taking the derivative of (6.20) and using (6.21) and (6.22), we find: (623)
C,,(t)
This expression can be used, for eX
Dimensionless Equation, It should also be noted that (6.23) call he made dimensionless by defining the following variables: x
Cfic'!W = conversion of A to B =
dimensionless time
= rate constant ratio to find:
I n
~-
1
[expCT) _. exp(-uT)1
which is shown in Figure 6.2, for a "'" 0.5 and 2. Notice that when the first rcw:tion is faster than thc sccond (ex "'" 0.5), there is a highcr concentration of B than when the first reaction is slower
Solving Linear nth Order ODE Models
148
Chap. 6
than the second (0: = 2). When the second reaction is faster than the first, component B reacts further to form C before a substantial amount of H is formed. 0.5 ............
_-~
-alpha - - alpha
co 0.4
~ 0
0.5 2.0
'0 C
"'"c0 ""' "'
'"
0.3 :'~~~~,
.:
C
.Q
"'c
0.2
,
,
'"
:'
E '6
: 0.1
0
"'
.. '. ", ",
"
0
.... 345
2
6
7
dimensionless time
FIGLJRE 6.2 Concentration of B as a function of time. When the rate for the second reaction is faster than the first «(~ = 2), the peak concentration of B is lower.
We notice in the previous example that (6.23) cannot be used if k 2 :::: k l . This is a case where the eigenvalues are repeated. The procedure for repeated eigenvalues is shown next.
6,2,2
Repeated Eigenvalues
If a particular root in the solution of (6.12), Ai' occurs r times, then the corresponding terms in the solution to (6.11) are: (6.24)
EXAMPLE 6.1 Continued.
Repeated Roots
The equation for the concentration of B is when k2
~t':~_:.11_ rIl
2
:::::
kl
;;;;;:
k is: (6.25)
Sec, 6,2
Solving Homogeneous, Linear Odes with Constant Coefficients
149
and the characteristic equation is: A'
+
2 k A + k2 ~ 0
(A
+ k) (A + k)c., 0
(6,26)
which can he factored as:
so the eigenvalues (roots) are: A,
A2
~
~
-k
The solution can be written:
C,,(t)
~
(e, + e, I) cxp( - k,)
(6.27)
Notice that we can find cl from the initial condition for C n- From (6.27) at t = 0,
But Cn(O);;;;; 0, since there is no II initially, so:
C,,(I)
~
e, t cxp( -
kl)
(6,2H)
The derivative of (6.28) with respect to time is:
at t = 0,
We also know from (6.3) that
"e,,(O) , dt - k CAO so:
C,,(I)
k C AO ' exp( '" I,,)
If we define lhe conversion or A to H as x = (6.29) can be written:
x(i)
c,/eAO ' ilmllhe dirncnsionlcss TexpC T)
(6.29)
tiulC as T;::: kt. then
«(dO)
Solving Linear nth Order ODE Models
150 which is shown version of A to
0.4
ill
Chap. 6
Figure 6.3. The reader should be able to find the 1l.1<1xiunJJn value for the
COI\~
n and the reaction time required for this conversion. ----,.-
0.35 0.3 0.25 0.2 0.15 0.1 005
o o
________-'
-'--
L.
2 3 dimensionless time
4
5
FIGURE 6.3 Conversion of A to B as a fUllction of dimensionless time (T "" kt), for the case of equal rate constants.
The previolls example illustrated the solution for systems with real roots. The next pie illustrates a system with complex roots.
- - - - - _.. EXA MIJl..E 6.2
cxam~
_ ~ -...
Complex Roots
Consider the second-order equation:
dx dt
+x
()
(6.31 )
The characteristic equation is:
A'
o
+ A+ I
Solving for the roots lIsing the quadratic formula, we find that the rools mc complex:
±
-4
2
A~_l±V.3. 2
2
J
(6.32)
Solving Homogeneous, Linear Odes with Constant Coefficients
Sec. 6.2
151
I. The solution is
where j
,(t)
c .1
e(-
2I
+ V3) 2 j t + c 1 e (J ,- 2 -
Vi) --2' -- j t
(6.33)
We can use the following Euler identities:
ejll elo and the property that
eXtI' "" eXe Y
10 ce'-cos ' 1
x(t)
1
x(t) = e
_
l
•
=
=0
cos f:l
jSln f:l
(6.34)
cosO -jsinH
«(dS)
j-
to write (6.33) as:
V3 (--I 2
J'sill .
Vii [Vi Vi I t -1- ' c ('. '/'"cos 2 . 2 1-- J" .sin 2 I
., [c 1 COS V3 2 t
(6.36)
(6.37)
/-
which can be written:
x(t)
e ,/2
[e c ,
x(t)
e
vi
Ie,) ]cos . 2 t e (e, -- ( 2 ) j sin
[V cos 2i t + c
2 lie;>
4
sin Vi 2 t
~3 tj
«(d8)
I
(6.39)
Again, initial conditions for x(O) and ~\:(O) can be used to determine {'3 and ('4- The student should verify that jf x(O}:::: 1 and ~~(O) = I, then c J :::: \.0 and (.'4 ::::1.5. A plot is shown in Figure 6.4.
1.5
::.:;
0.5
a 0.5
o
2
FIGURE 6.4
4
6
8
Plot of (6.39) with x(O) : :;:; I and ~r((»:::: I.
10
Solving Linear nth Order ODE Models
152
Chap. 6
Chemical process system models with complex roots include some exothermic chemical reactors. Also, models including feedback control will often have complex roots (leading to oscillatory behavior).
6.2.3
General Result for Complex Roots
We can now generalize the results of Example 6.2 for any equation that has pairs of complex roots. For each pair of complex roots, A ;: : ; At ±j Ai' where Ar and Ai arc the real and imaginary portions, the solution is:
x(/)
co
(N
lei cos A.,f + c, sin Ai II
(6.40)
In the previous example the real part of the complex roots was negative (stable). Notice that the state variables decayed to zerO with time. Notice from (6.40) that there will be no decay (simply a continuous oscillation) if the real portion is O. We can also sec from (6.40) that a positive real portion of the complex root will lead to an ever growing (unstable) solution. This behavior is shown by studellt exercise 19. Most chemical processes are stable; however, some exothermic chemical reactors have unstable operating points. Also, improperly tuned feedback control systems call be unstable. Thus far in this chapter we have solved the homogeneous problems. Homogeneolls problems result from models that are "unforced," that is, there is no input. This usually occurs when the process model is in deviation variable form, and there is no change in the input variable. They are based on a perturbation from steady~state in the state variable values. In Section 6.3 we will solve nonhomogeneolls problems llsing the rnethod of undctcnnined coefficients. These types of problems arise when there arc input changes to a process.
6.3
SOLVING NONHOMOGENEOUS, LINEAR ODES WITH CONSTANT COEFFICIENTS In Section 6.2 we solved homogeneous problems with constant coefficients: (6.41)
In this section we will solve nonhomogeneous problems with the following form:
+ aox = '1(/) lIsing the method (4 undetermined coefficients, which is outlined below.
(6.42)
Sec. 6.3
Solving Nonhomogeneous, Linear Odes with Constant Coefficients 153
Method of Undetermined Coefficients The method of undetermined coefficients consists of the following steps:
1. Solve the homogeneous problem to find
2. Solve for the particular solution by determining the coefficients of a trial function (sec Table 6.1) that satisfy the non!1o!11ogcncOlls equation
X,,(I)
:t
Combine the two solutions for
X(I)
~
Xu(l)
+ X,.(I)
TABLE 6.1 Trial Functions for Method of Undetermined Coefficients (noyee and DiPrim~l, 1992) Forcing FUllction
Trial Function
A (a constant)
B (a constant) BeHI 8] cos ctl + B} sin of 11,/1/ + Bn_1lll--J + ... +
Ac(~(
A cos ca or A sin at A ttl
flo
We illustrate the method hy usc of an illustrative example.
EXAMPLE 6.1 Continued.
First Order Heterogeneous System
Notice that we can take the solution for (A as a function of time
cAt)
(6.14)
= CAUe-k,1
and substitute it into (6.3) to obtain (6.43)
Step 1.
The homogeneous solution to (6.43) is (6.44)
Step 2. Since the forcing function is k] CAU e tion for the particular solution: Xp(t)
k,I,
we use c2 e
k,f
c). e kl !
(Table (),1) as our trial func(6.45)
substituting this solution into the original equation (6.47),
---k l
C e- k ,!
z
+ k 2 Cz e--- k,f
=
kl
CAO
e- k,f
(6.46)
Solving Linear nth Order ODE Models
154 which we can solve for
('2:
c,
Step 3.
;\
call
k,
c--
evaluate
('I
=
c e 1
(6.47)
k2 ~ k,
Now find the complete solution as xU)::::: (' (f)
We
Chap. 6
I..,,'
XII(t)
+ _'(/,(1)
+ k, __ k e l
(6.48)
kit
from the initial conditions, em!:::: 0: (.'\ =
and the lotal solution is:
which, of course, is the same result obtaineu previously (0.23) by solving the second-order homogeneous equation in CII _
We have lIsed a single first-order equation to illustrate the procedure for heterogeneous equations. The same procedure is used for higher-order equations.
6.4
EQUATIONS WITH TIME·VARYING PARAMETERS Consider a first-order equation with the following form: dx
1
({
(649)
-+ pet) x ." q(t)
Notice that the coefficient is timc~varying and the equation is heterogeneous. One approach to solve thiS type of problem is to use an illtcf.iratingfactor. Let the integrating factor be represented by 1J.(r)
,,(t)
~ exp [fp(t) dl]
(6.50)
Equation (6.49) IS solved by multiplying each term by the integrating factor:
,,(t)
I
dx -+ ,,(I) p(t) x dt
exp lJ l'(t) dt «11; -+ exp
If
,,(I) q(t)
p(t) dtl p(t) x
~ exp
(65 I)
If
pet) dtl q(t)
(6.52)
Notice that the lefthand side of (6.52) is simply the expansion of:
(~t
[X(t) exp {Jp(t) dt}]
~ exp lJp(t) dtl
(:;; -+ exp lJp(l) dt] p(I)X (6.53)
b
Sec. 6.4
155
Equations with Time-Varying Parameters
so we can write:
::t [X(I) exp
U
P(t)dt}]
[f pet) iltl q(t)
= exp
(6.54)
which is a separable equation. Separating and integrating, we find:
x(t) exp
[f pet) dl]
=
fq(t) exp
IJP
(I) dtl ill
+
C
(6.55)
and, evaluating c using the initial conditions,
X(t)
EXAMPLE 6.3
f pet) dl] {X(O) + fq(t) exp
= exp [-
II
p(t) dt] ill}
(6.56)
Semi-batch Reactor
Consider the case where the batch reactor is being filled. Assume a single, first-order reaction (A -)B) and a constant volumetric flow rate into the reactor (F), with no llow out of the reactor.
The modeling equations arc: (6.57) (6.5X) Expanding the LHS of (6.58) as:
dve 1 dt
---~.
dV + v-_·_ dell = c., _. _ ··· A
dt
(6.59)
dt
we find: (6.60)
If the f10wralc is constant and the initial volume is 0, then:
v = Ft
(6.61)
and: (6.62)
Let:
fp(t) dt
exp
IJ
pet) ilt]
~
fG + k l ) dt
~
=
exp [In t + kit +
=
c, I exp [kll]
In t + kit + c i
c,l ~
exp [In t] exp [kltl exp
(6.63)
[cd (6.64)
Solving Linear nth Order ODE Models
156
Chap. 6
Multiplying through on each side of (6.64) by ('2! ('xp Ik1tl and dividing by c_'
exp [k,l]
(6.6.'1)
and noting that the lefthand side is simply
exp [k,11 C ,] dt
"
(6.66)
we multiply by dt and integrate to find
k multiplying by exp I -kill and dividing hy
C
.,
t.
k,1
,
[exp [k,ll- I I
(6.67)
we find the solution
II
exp [
k,111
The division by t is bothersome at f;;;;; O~ the reader should llSC L Hospital's rule to sl1(1\\ that the correct initial condition is obtained with this expression. NOlice that we can define a dimensionless coucentration and time as
to find V(T)
, 11 - exp [
,II
(6.69\
which
IS shown in Figure 6.:). Notice that this solution hold.s while the rcacf(l1" IS being "fed" After the feed is stopped the model is simply dei/ill:;:: ~kICi\ with appropriate intinl conditi(lI1s (sec student exercise ~f))
1 - _ . . ~---
«
o
,----
0.8
o
c
o ~ 0.6 ~
~
co
~
~ 0.4 E
is
0.2
o
o FIGURE 6.5
2
4 6 dimensionless time
8
Solution, dilncnsionless sernihatch reactor prohlem.
t
Sec. 6.5
6.5
157
Routh Stability Criterion
ROUTH STABILITY CRITERION-DETERMINING STABILITY WITHOUT CALCULATING EIGENVALUES The stability of the characteristic equation is determined from the values of its roots (eigenvalues). This is easy for first and second order equations (and not too hard for third) since there is an analytical solution for the roots of polynomials through third order. If the polynomial is fourth order or higher, the roots must be determined numerically. There is a method for determining if any of the roots arc positive (unstable) without actually calculating the roots (RoLlth, 1905} This method involves all analysis of the coefficients of the characteristic polynomial by setting up the Rowh Array. The test of the coefficients in the ROllth Array is called the Routh ,S'tability C'riterioll. The Routh Stability Criterion is based on the characteristic equation that has the following polynomial form
(670) We can arbitrarily assume that all > O. If an < 0 then multiply (6.73) by
~1.
A neces-
salT condition for stability is that all of the coefficients in (6.70) Inust be positive. If any
of the coefficients are negative or zero then at least one eigenvalue (root of the characteristic equation) is positive or 7.ero, indicating that the equation is unstable. I~ven if all of the coefficients are positive, we cannot slate that the system is stable. What is needed is a sl4licicllt condition for stahility. To determine that the system is stable, we must construct the ROLlth array and use the ROllth stability criterion, which provides necessary and sufficient conditions for stability. Sometimes we simply wish to determine if a particular system is stable or nOl, without actually evaltltlting the eigenvalues. This is partieuJtlrly true if we wish to determine values of system parameters that will cause a systelll to lose stability. This approach will be useful in performing a bifurcation analysis in later chapters,
6.5.1
Routh Array
If "lll of the coeffients of the characteristic eq uation (6.70) arc PO"ilive, the llccessaJ)' condition for stability is satisfied. The following Routh array (Seborg, Edgar, & Mcllichamp, 1989) is developed to test for the s/!llfcicnt conditions for stability:
Row an
(/11 ___ '2
{[Il-!
{[n-3
3
h,
4
ci
h2 c2
2
n+l
a n--4 an 5
h3
Solving Linear nth Order ODE Models
158
Chap.6
where II is the order of the characteristic polynomial. Notice that the first two rows consist of the coefficients of the characteristic polynomial. 'fhe clemenls of the third row arc calculated in the following fashion:
hI =
({1I-I(ln--2 -" ti,/In
h2 = {III
and so
011.
I
Elcments of the fourth and larger rows arc calculated in a similar fashion: bj({/I
(:1
hl{[/I S -
({nlhZ
) -
lltJJ}J
('2
h,
hi
and so Oil,
Routh Stability Criterion A necessary and sufficient condition for all roots of the characteristic polynomial to have negative real parts is that all of the coefficients of the polynomial are positive and all of the elements ill the len column of the Routh array are positive.
EXAi\lPLE (,.4
Sccond~ordcl'
Charadcdstic Equations
Consider the second-order ODE: (Px (/2 tlt2
+ ttl
dx
((J.7 I)
til
The characteristic polynomial is:
al
;>..?
+
al A
+
an
0
(6.72)
If all of the coefficients (12' (II' and (10 arc positive, then the necessary condition is satisfied. We call form the ROllth array to test for the sufficient condition: ROll'
a)
2 3
(10
(ll
an
Since the left column consists of the pc}lynomial coclTicienls, if all of the coefficients in the second order system are positive, the. system is stahle.
Notice that, for second-order sysLems, a lesL for positive coetJicienLs is necessary and sufficient for stability.
Sec. 6.5
EXAMPLE 6.5
159
Routh Stability Criterion
ThirdMonlcl' System
The system:
rex
riP
+2
(Px
dP
+3
dx dt
+x
= 0
has the characteristic polytl(l!llia]:
A'
+ 2 A' , 3 A + I
~ 0
and the following Routh array: ROH'
j
2 3 4
2 5/2 I
3 I
All of the coefficients of the characteristic polynomial arc positive and all of the elements in the left column of the Routh ,-trray afC positive, so the system is stable. - - - _ . ~ ~ ~ ~ - - ~ -_._---~-~
The Routh array is particularly useful for determining how much a parameter call vary before a system loses stahility. The following cxan1pJe illustrates such a system.
EXAIVIPLE 6.()
ThinJ·order, System 'Vith a Variable Parameter
The system:
d\d 2x dx. , f-Cx _. 0 dt' -I 2 dr; +3 dt has the characteristic polynomial: A'
2 A'
+
3 A , f-C ~ 0
where 1.1 is a parameter that may vary. The Routh array is:
2 3 4
I 2
h, ('I
3 f-C
160
Solving Linear nth Order ODE Models
Chap. 6
where b l :::: J -- f.L/2 and c j ::::: IJ... From the characteristic polynomial, we sec that /L > 0 is required. The same result holds true for the requirement of c) > O. We notice that h] will be posj~ live only if f.L < 6. F'rom these conditions, we find that the stability requirement is 0 < /L < 6. for complex, high order (3 or greater), it is not uncommon for a system to have parameters that stabilize the system only over a certain range of parameter values. 'rhis is particularly true of feedback control systems.
SUMMARY We have reviewed techniques to solve homogeneolls and nonhomogeneous (heterogeneous) 11th order ODEs.
Homogeneous problems arc solved using the roots of the characteristic equatioll, forming the solution IIS a sum of exponential terms. Homogeneous equations generally occur if the system is unforced, but there is an initial deviation from steadystate in the state variables. The method (~l undetennilled c()(dlicients is lIseful for solving nonhomogeneous (heterogeneolls) problems. These generally occur if the system is forced by a chang~ ing input. The integrating factor method was useful for solving a first-order heterogeneous equation with a time-varying coefficient. The Routh array was llsed to test for the stahility of a differential equation. This is lIseful for finding vailles of a paranletcr that cause a system to lose stability, such as in feedback control system design or bifurcation analysis. The type of dynamic behavior of an nth order differential equation is a function of the eigenvalues (roots of the characteristic equation). Eigenvalues that arc further in the left half plane arc "fast." The larger the ratio of the imaginary portion to real portion of a complex eigenvalue, the more oscillatory the response. Stability is detennined by the real portion of the complex eigenvalue. If all eigenvalues have a real portion that is negative, then the system is stable. If any single eigenvalue has a real portion that is positive, then the systern is unstable. Often engineers study the dynamic hehavior of processes by starting out at stcady~state, then applying a changing input to the process. Although the method of undetermined coefficients can be used to solve these problems, the Laplace transform technique is used more often. The Laplace transform method is introduced in Chapter 7.
Student Exercises
161
FURTHER READING Boyce, W., & R. DiPrima. (1992). Ordinary Differential Equations and BoundmJ Value Problems, 5th cd. New York: Wiley. Routh, EJ. (1905). Dynamics ofa 5'ystern (~lRigid Bodies, Part Il. London: Macmillan. Seborg, D.E., TF. Edgar, & D.A. Mellichamp. (1989). Process Dynamics and Control. New York: Wiley.
STUDENT EXERCISES I. Consider the state-space model:
Ix, Ix,
I
1- 1I
.~ 211 XI] x2
0
a. Find the second order ODE in terms of XI' b. Find the second order ODE in tenns,-of x'2' c. For xI(O)::::: ~l and x2(O)::::: 1, ohtain the analytical solution for xl(t) and x 2(t)· d. Usc ode45 or initial to solve the set of two differential equations, given the initial conditions in part c. 2. Consider the following linearized form of a hioreactor model with substrate inhibilion kinetics (sec Module 8 for delails): x=Ax+8u y=Cx+DtI where: ~.
I
=
1 0 1
A~
C
0
-0.7500
0.9056.·.1 -2.5640
0I 1
I D= I;:]
·.~ 1.53021 3.8255
a. Find the second-order ODE 1n terms of XI' assuming u::::: O. b.Find the second-order ODE in terms of x 2 , assuming u::::: O. c. For x,(O) =- I and x 2(O) = I, oblain the analytical solution for x, (I) and xP)· d. Use ode45 or initial to solve the set of two differential equations, given the initial conditions in part c. 3. Consider the slate space model for a two-state system:
x= Ax [Xl] = .1 all '-\:2
ESCOLA Dc ENGEN HARIA BIBLIOTECA
(121
Solving Linear nth Order ODE Models
162
a. Lcty
Chap. 6
= xl' and derive the following relationship:
y -[all
+
a 22 ]
Y+
[a ll a 22 - a2\ad
Y = 0
which has the characteristic equation:
1\2 - [all
+ a 22 ]
1\
+
[a ll a 22 - a2\a 12 ] =
0
Recall that the eigenvalues of the A matrix arc calculated by:
det(H - A) = 0 Show that det(l\I- A) = 0 applied to this general two-state example, yields: 1\2 - [all
+ an]
1\
+
[allan - a 2I a 12 ] =
0
4. Solve the following differential equation with the given initial conditions:
d 2y dy ,+5········+6y=O dx dx dy(O) yeO) = 0 and dx
I
5. Find the particular solution of the differential equation: d'y ~i.~'
dy
- 3 ;I:;' - 4 y =
. 2 sm x
6. Consider the following second·order homogeneons ODE:
,l'x dx ····--3-+3x=O 2 dt dt a. Write the characteristic equation for this ODE. b. Find the solution (solve for any constants), x(t), if the initial conditions arc x(O) 2.0 and x(O) 3.0. c. Discuss the stability of this system.
=
=
7. Consider the following first-order heterogeneous ODE:
3
dx + x = 2(1 - e''') dt
a. Write the characteristic equation for the homogeneous part of this ODE. b. Find the solution to the heterogenous problem. Show all steps. The initial condition is x(O) = 2.0. 8. Consider the following state-space model that results from a linearization of the
predator-prey equations:
Student Exercises
163
-1.0] [XII 0.0 X2
with initial conditions x,(O) = 0.5 and x 2 (0) = -0.25. a. What are the eigenvalues of the A matrix? Use both MATLAB and your own analytical solution. b. Write the second-order ODE that corresponds to xI. Use the method of Section 6.2 to solve for x,(t). Plot x,(t). c. Write the second-order ODE that corresponds to x2. Use the method of Section 6.2 to solve for xlI). Plot x2(t). d. Compare the results from band c with those obtained by integrating the statcspace equations using cither ode45 or ini tial. e. Show a phase-plane plot (Xl versus Xl)' placing XI on the .:r-axis and Xl on the y-axis. 9. Consider a system described hy the following third-order ODE:
d'y dy -(Py. . + 1.5, + .... + Zy dt-
dt'
dt
=
0
Is the system described by this equation stable? Why or why not?
10. For a general third-order polynomial: aJ
Po.
3
-I- 112 Po.
2
-I- ill
°
A -I- a o = 0
show that {lj > 0 and ([la2 ~ {loa 3 > are necessary and sufficient for stability. 11. For a general fourth-order polynomial: ([4 Po.
4
+ llJ Po. 3
-I- a2 Po. 2 -I- a l
show that aj > 0, a2a3 ~ {l\a 4 > 0, and {l1{l2{l3 sufficient for stability. 12. Consider the following third-order ODE:
d'y
;/ti
d 2y
Po.
- (l4ar
-I-
ao = 0
~ aoaj > 0 are necessary and
dy
+ Ziti + (0: -I)d{ + o:y
= 0
where ('{ is a parameter. Find the range of ('{ that will cause this equation to he stable. 13. Consider the «lllowing second-order ODE:
d'y dt2 which has eigenvalnes of -I y(t).
±
dy dt
+ Z-- + Z Y
= 0
Ij and initial conditions y(O) = 2 and y(O) = -Z. Find
Solving Linear nth Order ODE Models
164
Chap, 6
14. Consider the series of two tanks, where the levels interact. O
F
- - .
tHJ21~h2
. F2
F,
a. Assuming that the flow from the first tank is linearly proportional to the difference in the tank heights (P'l :::.: ~l (h t - h 2 )), the flowrate from tank 2 is proportional to the height in tank 2 (F2 ;:;;; rJ 1 liz), and the tanks arc or COllstant crosssectional area (A I and A 2 ) show that the modeling equations arc
dh 13J (h _ h ) -13, .... 2 '0 ' h dl A2 I 2 A 2 . 2 b. Reduce these two equations to a single second-order equation in 11 2 . c. Assume that the steady-stale f10wratc is 3 [(3/min, and the steady-state tank heights for tanks 1 and 2 arc 7 and 3 feet, respectively. The constant eros,\;sectional area is 5 ftl for each tank. The initial conditions arc h\(O) :::: 6 fect and h 2(O) = 5 feet. Solve for the heights of tanks 1 and 2 as a function of time, Plot the tank heights as a function or time. Discuss your resulls. d. Write a MATLAB Ill-file and lise ode45 to integrate the two equations shown above. Show that the numerical integration agrees with your solution in part c. 15. For the batch series reaction (Example 6.1): k
k
A-}IJ-fC H.
Find the reaction time that maximizes the production of B. Recall that the tion for the concentration of B is:
solu~
k I C,lO[ cxp( - kit) - cxp( eli'() I = 'k--=-;- k] 21) 2
/(j
and that the maximum occurs when the condition den/dt:::: 0 is satisfied. b. For k j ::::: I and k 2 :::o 5 min-I, find the maximum conversion of A to B (express as Cn/CAo) and the tinle required for this conversion. c. In practice there is uncertainty in the rate constants. If the actual vallie of k, is 7.5 min-I, and the reaction time from b is used, find the actual c()nvcrsion~or A to Ii, d. Use the MATLAB routine ode4 5 to integrate the three state variable equations and solve for C;\> C n' and C c as a function of time, for the parameter values in
Student Exercises
165
h, with C~\O := 1.5 and em> Ceo::::: 0 mol/liter. Make a comparative plot for the parameter values in c. What do you observe about the concentrations of A, B, anel C? 16. For the batch series reaction with ilTeversible reactions (A --.--.7 B --.--.7 C): a. Find the reaction time (tllla,,) that maximizes the conversion of A to B for the case INhere k 2 ::::: k j ::::: k. Also find the value for the maximum conversion of A to B. Recall that the solution for the conversion is :0;:
X(I)
= CH(I)/ C AO = k I cxp( - kl)
b. Assume that the reactor is run for the period {max found in a. Now consider the effect of an error in the reaction rate constant of +50%. What is the actual conversion orA to B obtained at t max ? 17. Consider a batch reactor with a series reaction where component A reacts to form the desired component B reversib(v. Component B can also react to form the undesired component C. The reaction scheme can be characterized by:
k ll --> B 1<, C A f---> k lr
Here kjr"nd k 1r represent: the kinetic rate constants for the forward and reverse reactions for the conversion of A to H, while k 2 represents the rate constant for the conversion of B to C Assuming that each of thc reactions is first-order and constant volumc, tbe modeling equations are
where CA , C u' and Cc represent the concentrations (mol/volume) of cOlnponents A, B, and C, respectively. Using the following definitions: Dimensionless time, Conversion orA, Dimensionless concentration of B, Ratio or rate constants, Ratio of forward and reverse rate constants,
'T
= kit
= (C';jO ~ CJ\)/CAO x 2 = C u/ CAO a = k z/ kIf Xl
p=
k ir / kIf
Solving Linear nth Order ODE Models
166 ll.
Show that the equation
l~)r
(Px) (/72
Chap. 6
the dimensionless concentration of B is
+ (1:\ +
<X
+
I) dX 2
d'T
and that the roots of the characteristic equation can never be complex or ullstable (assuming that the rate constants arc positive). h. Solve the previous equation to find x2 as a function of T and ex and r3. c. For kif:::: 2, k 1r = ], and k2 :::: 1.25 hr"], find the maximum conversion or A to lJ and tlie reaction time required for this conversion. d. Usually there is some uncertainty in the rate constants. If the real value of k 2 is 15 11I~1 and the reaction is fun for the time found in c, what will be the actual conversion of A to B? 18. Consider the series reaction: le l k? C'~--->'f) k~ A-lIJ-j
The modeling equations for a constant volume batch reactor arc
a. Show that the third-order ODE descrihing the concentration of Cis:
[Hint: Solve for en fWln the third equation and take the derivative to find
deBldt.1
h. Assuming that all of the kinetic parameters are positive, show that this system is stable.
19. Consider the second-order equation:
Por initial conditions x(()) = I and x(O) =1, find the analytical solution and shm\' that the following plot describes how x changes with time.
Student Exercises
167
150 100 50 K
0 50 -100 150 2
4
6
8
10
20. Consider a semibatch reactor (Example 6.3) with a first-order kinetic parameter of
k:::: I hr- I . I\x a nowrate of 10 liters/hour, a feed concentration of 5 mol/liter, and arced time of 2 hours, find (and plot) how the concentration changes fronl 0 to 10 hours.
7
AN INTRODUCTION TO LAPLACE TRANSFORMS
Arter sludY'ing this chapter, the reader should hc able to:
Define the Laplace transform alld apply it to several example
fUl1c1iollS.
Usc Laplace transforms to convert an nth order ()DL·: to the Laplace donwin.
Manipulate the algebraic equations by! performing a partial fraction expansion. ·'Invert" the Laplace domain functions to obtain the time domain solution. Use the final v,due theorem to compute the lOllg+:r1n hehavior of a system. The important sections of this chapter arc:
7.1
7,1
Motivation
7.2
I.kfinition of the Laplace Transform
7.3
Ex,unples of Laplace 'rransforllls
7.4
Final and Initial Value Theorems
7.5
Application Examples
7.6
T;:lblc of Laplace Transforms
MOTIVATION In this chapter \vc introduce {\ mathematical tnoL the Lllp/ace fnlllsji irlii , which i:-; \cr.\ useful ill the analysis of linear dyn,unic -.;y-.;terns. crill' purpose of tile Laplace transform. as used in this textbook, is to converl linear differential equations inLo algehraic equa,
168
------
, ~ d'i i t
Sec. 7.2
Definition of the Laplace Transform
169
lions. Algebraic equations arc much easier to manipulate than differential equations. An analogy is the usc of logarithms to change the operation of multiplication into that of addition. Laplace transforms arc lIseful for solving jinear dynamic systems problems, particularly nonhomogeneous (heterogeneolls) problems (Le., where the input to the process system is changed), and are commonly used in process cOlltrol system design and analysis.
7.2
DEFINITION OF THE LAPLACE TRANSFORM
Definition:
Laplace transform
Consider the time domain function/(t). The Laplace fransform ofj{f) is represented by qj{t)J and is defined as 1.[j(l)1
= F(s)c
J/(t) e "
(71 )
n
This operation transforms a variable from the time dornain to the s (or Laplace) domain. Note that some texts use an overbar or capital leucrs for the transformed variahle. In this initial development wc will let/C!) represent the time dOlnain function and F(s) represent the Laplace domain function. Later we may be more relaxed in our notation and let .f(s) represent the Laplace domain function. The Laplace transform is a linear operation, as shown below.
L[a,Jj(t)
+ a,Nt)1
J [a,Nt)
=
+ ",Nt)] c "tit
<)
Oc
]a,I;(t) (J
= aJ'f(t) ()
,
e~"
Ja,/;(t) e" tit
()
e" tit + a,!/;(t) e" tit
(72)
0
Equation (7.2) satisfies the definition of a linear operation. In (7.1) we used Ll/{t)1 :so; F(s) to define the transform of a time domain function. [I' we wish to transform a Laplace dornain (sometimes called the s-domain) function to the time dornain, we lise the notion of an inverse transform
L'IF(s)1 =/(1) Although not emphasized in this text, Laplace transforms can also be used to solve linear partial diffcrenlial equations (PDEs).
170
7.3
An Introduction to Laplace Transforms
Chap.7
EXAMPLES OF LAPLACE TRANSFORMS In this section we develop transforms of some functions that commonly occur in the solution of linear dynamic problems. These functions afC: 0) exponential function, (ii) step function, (iii) time-delay, (iv) derivatives, (v) integrals, and (vii) impulse.
7.3.1
Exponential Function
Exponential functions commonly arise in the solutions of linear, constant coefficient. ordinary difTcrcntial equations: J(t) = e,n'
A plot of this function is shown in Figure 7.1. Recall that the transfoml is defined for t > 0 (we also use the identity that e X + 1'
L[e-,n] =
JHt) e- n dt
=
o
I
= _ S
LIe
,a] .-
+ (l
J e-,n e- n dt
=
()
J e-(da l ' dt 0
[(((.\'+O)lj'''' 0
=
1
I
s+a
s+a
'-[0-1]=-
I
s+a E\pOllential
L
1[' .s
1 ....
+
a
J = eO'"~
0.8
1ji 0: x w
0.6 OA
I i
0.2 0 0
2
3
at
FIGURE 7.1
Exponential function.
4
5
;:::
e·1c\)
Sec. 7.3
171
Examples of Laplace Transforms
Notice that the way we have solved for the limits of integration is only rigorously true for -a; we will assume that this condition is always satisfied.
{/ > O. For a < 0 the solution still holds for s > 7.3.2
Step Function
The step function is used to solve dynamic problems where a sudden change in an input variable occurs (a flowratc could he rapidly changed from one value to another, for example). The step function is defined as 0 before t = 0 and A after [ = 0, as shown in Figure 7.2, Ofort 0 We must use the "more precise" definition of the Laplace transform, because of the discontinuity at t = 0:
L[t(t) I = lim + 1'---+0
T----'>7;
Since the transform is defined for t
L[A] =
T
I }(t) e-" dt "
> 0,
J Ae-·" dt =
-
A
0'
L[A]
s
[0 - i]
A s
= ;\ s
5;tep
Notice that the same expression is used for the Laplace transform of a constant.
A Step Function
o
t = 0 FIGURE 7.2
Step function.
172
7.3.3
An Introduction to Laplace Transforms
Chap. 7
Time-Delay (Dead Time)
This is important for systems with transport delays (flow through pipes, etc.), or delays due to measurements. l"ct ttl represent the time delay. If the undclayed time domain function isj(t), then the delayed function is/(t - l
Llf(t - t,,)1
=
Jo ](1 - I,Je-
=
e-·w,Jo
W
dt
](t') e"" dt' =
e""
1'(.1')
Notice that the lower limit of integration did not change with the change of variable, because the function/ttl is dehned as jet) = 0 for 1 < 0,
Llf(t - td ) 1 ~
1"'"
P(I')
Time~De!ay
The transform of a delayed function is simply function.
c---stt!
times the transform of the undclaycd
I(t)
function
delayed
"FIGURE 7.3
function
Delay funclion.
~-----------_.
Sec. 7.3
7.3.4
Examples of Laplace Transforms
173
Derivatives
This will he important in transforming the derivative (accumulation) term in a dynamic equation to the Laplace domain.
LI.tiM]." 1 dl(dtt) e "dt . dt_ 0
Using integration by parts II ~,
Lei
U adv:::: uv - f
vdu)
e wand v'·, fell
[..[{.IJ.(t).!_- J" dill) or/t
dt
e." dl ~ [e'" j(l)]'
• + Jf(t) se" dt
DO
., + s JJ j ' (t) . e'" dt ,. shes) - flO)
dl(t)] . L [ ' . = [0 - j (0)] dt .
0
Derivative
t, 'lsF(s) -f(O) 1
dIet) fit
Since we often work with deviation varinhlcs,j(O) := 0 in many cases. In general, you should he able to show the following (sec student exercise I):
nth Order Deriv(ltive 11
7.3.5
initial conditions arc neededf(O), ... ,./(n-I)(O)
Integrals
This is often used in process control, since many controllers use information about the integral of the error between the desired value (sctpoint) and the measured value:
{I
J(t) dll
=
I~
J(l) d l " dl I
Again, intcgratc by parts, lIsing u = e--,\-I dt and v::::: I./U)dt, to find
o
An Introduction to Laplace Transforms
174
Chap. 7
b
FIGUH.E 7.4
I F(s) s
7.3.6
Ramp functioll_
Integral
Ramp Function
Consider the following ramp function:
f(l)
bI
as depicted in I·'igurc 7.4. You should be able to shmv (sec exercise 2) that
R(/Illp
L
7.3.7
'IINI
Pulse
Consider the pulse function in Figure 7.5, which consists of a step from () to A at [::::: 0, and a slep back to 0 at t;;::;; tp.fiind lhe Laplace transfer function for this pulse. There arc two ways to solve this problem.
FIGURE 7.5
Pulse fUllction.
Sec. 7.3
Examples of Laplace Transforms
175
ONE METHOD The pulse function is defined over the following two time intervals: f(t) = A for 0 < I < I" .I(t) = 0 for I > I"
and we can write the Laplace transform as:
F(s)
=
f/(l) e" dl •.c f/(I) e" dl + ff(l) e'" dl ( l O l l '
,
f A e" dl + fOe" dl- ~ ~
o
F(s) = ~
A. A [e""~ I] = [I s s
~e
A s
[e"I'"
0
"'I
'file usc of A::::: I is the unit pulse.
Llunit pulse of duration
'pi ,--
e- I"1 I
I [1 s
Unit Pulse
A SECOND METHOD Consider that the pulse is simply the sum of two step changes, as shown in Figure 7.6. 1'hat is, it is the sum of a positive step change at f = 0 and a negative step change at t ::::: tv Lctfl(t) fcprCScllt the step change at t :::: 0, and lit) represent the negative step change at I :::: If!>
f(l) ,•• .1;(1) + NI) F,(s) + F,(s)
F(s)
.c
L[I(I)] ,,- Llt;(I) + f,(I))
but notice thatl2(t) = ~f,(t ~ II')
A
TL
/, (I)
o
FIGURE 7.6
+
Pulse function.
An Introduction to Laplace Transforms
176
and that (from the stcp fUBction):
Chap. 7
s
and (from the delay function): -('
1.-"
A .I'
1'(.1')
SO \VC call \vrite
/1 s
[] .. e'·']
\vhich is consistent \vith the previous derivation.
7.3.8
Unit Impulse
In Figure 7.7, consider the pulse function as the pulse time is decreased, but the pulse area remains the same, as ShOVv'1l by the dashed lines below. l'he unit impulse fUllction is a special case of the pulse function, with zero width (t/J -t 0) and unit pulse area (so A ::::. llt/,'!. Taking the limit and applying L' llnpiLa]'s
rule:
,
,
~
:- ..
~
~
.
~,-'
-'--'------'---- - - - - - - -
FIGURE 7.7
Impuhc function.
Final and Initial Value Theorems
Sec. 7.4
L[o]
lim
177
- c".'] .... lim
(!'--.,o
11')0
-I S
I -sc "1"1
Unit Impulse
7.3.9
Review
'rhus far we have derived the Laplace transform of a number of functions. f:;'or example, we found:
LIe "']
~
s+a
If we have a LapJace domain function, such asl/(s + a), we can "invert" it to the time domain.For example,
LII,
I
al.~e'"
Althollgh the student should he able to derive Laplace transforms of any lillle domain fUllction, that is not our major objective. Our major objective is to usc Laplace transforrns as a tool to solve clynarnic problems. The l-,aplacc transforms of many time-domain functions have been derived and compiled in various tables and handbooks. Already, we can construct a table of eight (exponential, step, time-delay, derivative, integral, ramp, pulse, and impulse) time-domain functions along with their Laplace domain functions. Addi-
tional Laplace transforms aJ'e provided in Table 7.1 in Section 7.6.
7.4
FINAL AND INITIAL VALUE THEOREMS The following theorems arc useful for determining limiting values in dynamics studies. They will be used frequently to find the shortAenn and long-term behavior. The 1011g term (final valuc) of a timc domain function can be found hy analyzing the Laplace domain behavior in the limit as the s variable approaches zero. Thcinilial value of a time domain function can be found by analyzing the Laplace domain behavior in the limit as s approaches infinity.
Final Value Theoretl1
lim
Is F(s)]
(7.3)
lim
is 1'(.1')]
(7.4)
,1',0
Initial Volue Theorem
lim f(t) {-)O
~
)' __Pc
An Introduction to Laplace Transforms
178
Chap. 7
If we have lransforlnc<J a time domain function to the s domain, \ve Call still find out the
value of the time domain function as it goes to or the Laplace dornain fUllction as
orems is shmvn in
EXA1VIPLE 7.1
L~xaTl1ple
S _Nt
steady~statc (l -~) (0)
by finding the value
O. All application of the final and initial value the-
7.1.
Applkatioll of Final and Initial Valu(' Thl'on'UlS to the Exponential Function
(\lllsider the exponential function: j(1) which had the Laplace transform:
\" + (/ Finu/ Vulue Theorem. \Vc first find:
lim s F(s) ;(1
o
lim >it,-
()
which checks 'Ivith
lim !(T) as IOllg as
1/
Jim
c!l"
()
is positive.
Inilia! Va/Ill.' Theorem. We first find ~I
\ t-
II
\vbich checks with
Illn
."
!t t)
lim e'"
'"
which is safisfied for any fillite (/. . ... --~~----------~--------~ ~ ~ ~ ~ ~
One pOJl1lnot often made in texthouks is that the final value theorem only hold", for stahle sy,'",tems (0 > 0).
7.5
APPLICATION EXAMPLES crhe following is a checklist for solving dynamics probkms using Laplace transt'onns.
Stcp I. Stcp 2.
Start \vith a linear ordinary differential equatioll and initial conditiolls. Transform each of the tiJ\le domain fUllctions to the Laplace dumain. gen~ erally by using a tabk of Laplace transforms.
Sec. 7.5
Application Examples
Step 3. Step 4.
7.5.1
179
Usc algebraic manipulations to solve for the transformed variable. The partial fraction expansion approach is particularly useful. "Invert" to the time domain, by using a table of Laplace transforms.
Partial Fraction Expansion
The partial fraction expansion approach is based on representing a ratio of two polynomials as a sum of simpler terms. L,ct N(s) and D(s) represent numerator and denominator polynomials, respectively.
N(s) D(s)
C
D,(s)
+
+
D,(s)
.. UJs)
Ci arc constants and D i are lower order (typically J) polynomials. The [our-step procedure is llsed in each of the following examples. The partial fraction expansion is first used in [~xample 7.3.
gXAMPI.J~
7.2
Step 1.
Homogeneous First-order Problem Consider the simple IwmogeneotlJ' (unforced) first-order problem: dx dt
+ 2x
~
0
(7.5)
subject to the initial condition:
x(O) Step 2.
~
4
(7.6)
Rccall the following transforms:
II:';;] ~ "iilxj
~
s X(s)·· x(O) al jxl c.
1/
Xes)
Then wc (an take the Laplace transform of (7.3) aud (7.4) as:
II ~; I
~0
+ 2Lix]
s Xes) - x(O) + 2 Xes) -, 0 s X(s) - 4 I 2 X(s)
Step 3.
~
0
(7.7)
Solving (7.7) for Xes):
4
X(.,) .I'
j--
2
(7.8)
An Introduction to Laplace Transforms
180 Step 4.
Chap.7
Inverting each clement back to the time domain:
L '[Xes))
CII
crt)
(791
41 ~4e"
17.10)
4 e"
(7. II)
.s+ 2.
and the solution is
x(t)
Indeed, using the lllcl!lod in Example 7.2, equation:
dr
tit
\VC
+ax
can show that the general first-order
cc;:()
with initial condition x(O) x(t)
has the solution
cc
x(O) e '"
which, of course, is the same solution ohtained by separating the variables and integrating. The real power of Laplace transfonns is in· solving heterogeneous problems, as illustrated in Example 7.3.
EXAMPLE 7.3
Illustration of the Partial Fradion Expansion
Tc(~hnitluc
Step 1. Consider the simple heterogeneous first-order problem: dx dt
-+- 2x
4.5
=
(7.12)
with the initial condition
x(O) Step 2.
~
4
17.1\1
Taking the Laplace transform of each c1cmcnt:
s X(s)- x(O) + 2 Xes)
~
4.5 s
which can be written (since -,_-(0) "'" 4):
(sl 2) Xes)
Step 3.
-c
4'
4.5
s
Solving for the transformed varlahle
Xes)
~
4
s
f
2
+
4.5 s(s + 2)
(7.14)
Sec. 7.5
181
Application Examples
We would like to invert (7.14) to the time domain, however we do not know how to invert the last term 4.5/s(s + 2). We will use the approach known as apartialFaetion expansion. That is, write:
4.5 s(s \ 2)
A .\'
+
Ii S
(7.15)
+2
to find A, finit multiply (7.15) by s:
4.5 s-+-2
"C
A
+ lis
s+ I
then set s:::o;: 0 and solve for A:
A
~
2.25
To find fl, first lllUltiply (7.15) by s + 2:
A(s + 2) + IJ s
4.5 s and set s = -2 to solve for B:
~.
IJ
2.25
which yields:
4.5 .I(S + 2) and
We
2.25
- 2.25
+
s
s -+- 2
(7.16)
can write (7.15) as:
Xes)
Step 4.
~
4
s
+-
2
\
2.25
s
- 2.25
+
s +2
(7.17)
Inverting element by clement ill (7. J7) we find
x(t) = 4
C- 21
+- 2.25 +- - 2.25 ell
or
X(I)
~ 1.75
e" + 2.25
(7.1 ~)
the reader should verify that this solution satisfies the initial conditions and the differential equation.
Examples 7.4 and 7.5 provide additional illustration of the partial fraction expansion technique.
EXAJ\tlPLE 7.4 Write
Find the Inverse Laplace Transform of lI(s + ales + b) 1 (s \' ales + b)
Multiply (7.19) by s + a, set s =::
-0
to find:
A
s+a
+
B
s+b
(7.19)
Chap.7
An Introduction to Laplace Transforms
182
.\ \lultil'l;. 7.19) hy.\ -+- f),
~l'l
s
h
--u
--/1 tu find:
(/ - h
I.
L
1~ "II, "I 1 II, I I
h
h)
(1)(0,
u I h
17 ..2UI
10 ~ hi,
- 101]
(1)(.\
I h
(7..~
II
The Illl'thod in the prc\iou" c\i1lllple failed ir the rouLS n( the LttpLICC domain rlll1l~!i()l1 were equal. "fhc ldhming c\arnplc "IH'\'" hO\\ 10 pl~rrorll1 a partial fraction C\\"KlIbioll fur I"Cj)('(lfcd I"oofs.
EX';\\IPLE 7.5
Consider the Following Transfer Function with Repealed Roofs
(s
([) :'{.\
hi
\
101
(s
Here we
C,lll!lU{
llluitiply b;. \ -+-
multiply 0,20)1 by
(I
,p1(1 set
;:c:
(7,231
(s
S =-i1,
because we \'\luld find
11l1buUlllkd (nULS.
Fir"L
(s -+- (1)2.
C(s
01
i\( ,\
and C;d ,\
(
\-+-i!
I' --'--
h
-u lu find Ii
\lultiply 17.23\ hy ,\ -+- hand -;ct s
0:::
11
(/
b 10 filld
--------------------- ....
Application Examples
Sec. 7.5
183
Notice that we have solved for two of the coefficients of (7.23). Now, we can solve for olle equation in one unknown, by setting s:;:;: any valuc-. For simplicity, choose s = O. Froln (7.23):
A
I
1
(b -
a
we can reduce the solution for A to
(a
bJ'h
-I
A =
(a-b)'
We have solved for A, H. and C' in (7.23), So we can perform an clcmcilt-by-clemcnt inversion of
(7.23) to find the time domain function:
L
,[.I(a=;))2] + l(b~a),1 + l(a~b)'I]· ~ + + + (s
a)
(s
a)'
(s
- I
e"l
h)
and
\VC call
t em -+-
J
_j
(a,·
h-a
I
... e hi
(a_b)'
write:
L
'I (, +
+ b)
I
e"l -+
(il '
b~a
t c-- il ' -+-
( , ,,11/ (a-b)'
(7.24)
As an alternativc-, we tan JJnd a COllUllon denominator for the right hand side of (7.23) and write:
+ b)
A(s+ a)(s + b) + B(s (s + a)'(s
1
b)
1
b)
I·
e(s + (I)'
(725)
then expand the numerator and solve for the coeJTicienls A, lJ, and C such that the righthand side is equal to the Ie-niland side. See student exercise 13. L.-
.
~
,
~
The previous examples were for ODEs with real roots. This next example is a problem with complex roots.
EXA1\lPLE 7.6
Step 1.
A
Second~ordcr
System \vith Complex Roots
Consider the homogeneous problem:
d~r
i+
dt'
dx dt
+x~O
(7.26)
with the initial conditions:
\(0)
Step 2.
,(0)
(7.27)
From the table of Laplace transforms:
LI d"xi ~ s" Xes) _ S"IX(O) _ . dt il
.
S"
2 .\'(0) - ... -
S Xl"
2'(0) - xl" "(0)
An Introduction to Laplace Transforms
184
Chap,7
So, for a second derivative:
L', d"J ~ S2 XeS) - ,\' x(O) - x(O) dt
I
and, for a first derivative: dx
L [.. ··
l = sX(s)-x(O)
df .
We can now write the Laplace transform of (7.26) as:
s' Xes) - s x(O) -teO) + s Xes) - x(O) + Xes)
Step 3.
=
0
Attempting to isolate Xes) on the LHS:
(S' + s + I) Xes) ~ s x(O) + x(O) + x(O) dividing by (,52 +
S
+ 1):
X(s)
S
x(O)
=-----
.\'2+ s +1
x(O) + x(O) + -------s2+.\"+1
and from the initial conditions: 2
,\'
= +s+ + +s+ J /2 ± V3]2 j (from the quadratic fOl1TIula): Xes)
the roots of (.\"2 + s + I) are -
(7,28)
Notice another way that we can write (s2 + S + 1) is:
-Z'( '2I)' + (Vi)2
(S2 + s + 1) ~ s + which means that we can write (7.28) as:
2
s
Xes)
= (
1); (\(3)-; + -(-'1-)2(\/3); s+'+--2
Step 4.
2
s+-+················ 2 2
Notice from a table of Laplace transforms that: L[e-/JI sill wt] = --------~-----;
(729)
--------~--~~---- (s + Ii)' + ",2
(7,111)
(s + Ii)' + ",'
Lle-'"
cos wI} =
and we should maneuver (7.28) into the form of (7.29) and (730). Notice that we can write (7.30) as:
Sec. 7.6
185
Table of Laplace Transforms
s+ X(S)
2
I.S
(SI i)'1
(7.31)
(\~7
and we invert each clement of the RHS of (7.31), using (7.29) and (7.30): '
l x(t)=el"cos
\/3 J.5 Ii' \/3 t+ e-""sin t 2 2
2
ttl)
)
V"3
• e 'I':. sin V3 ( v3 2
e 'I ~cos. 2 l
which has the lime domain response sho\vll in Figure 7.8. As we Iloticcdin Chapter 6, complex roots give oscillatory responses. We see in Chapter 9 that this type of response is called llndcrdamped. 1.5
:..::
0.5
o -0.5
o
2
4
6
8
10
time
FIGURE 7.8
7.6
()sciJlatory response due to complex roots.
TABLE OF LAPLACE TRANSFORMS For your convenience, selected Laplace transforms arc presented in Table 7.1. If yOll desire to transform a function from the time domain to the Laplace domain, then look for the time dOlnain ('unction in the first column (j(t» and write down the corresponding Laplace domain function in the second column (F(s)). Similarly, if you arc trying to "invert" a Laplace domain function to the time domain, then look for the Laplace domain function in the second column and write down the corresponding time domain function from the first column.
186
An Introduction to Laplace Transforms
TABLE 7.1
Chap, 7
Laplace Transforms for Selected Time-Domain Functions 1'(.1')
fit) (j(t) (unit impulse)
,
S(t) (Uillt step)
{Ofoft 0
,I'
Ah cos F(s)
A (constant)
f(t--O) (time delay)
1
t (ramp)
-I
.v" df dt
(derivative)
.1'1'(.1') -.1(0)
d'~f
sllFCs)
dtil
~
SIl-l.!W) - s"-2j'(I)(())
- .1./,"-21(0) -.I''' e-- il !
s+a I
(e--Oll -- e- Ojl )
aj
-
"(0)
(.I'
{/2
~~L_=~lJ
s -JCs -+- llj)(S -+- (/2)
c,,,,r +~lJ - (12 e"/
(J2~al
+ "')(.1' + "2)
al~a2
I
s(n + I) W
sin wt
s2
+ S
cos wt
+ (,)
(.I' e-Il!
cos
s+a
(UI
1+
(.I' + (1'1
1'2 -
+
C [J)'
e-
t/
T
,
~ 1'2
f
1 .1'(7,.1' + 1)(7,.1' + I)
e---Il,,)
1'1
(1 _ 1
I 2;; 7S
+ where w
,(V
:c:.:c
+
I)
= tan---- 1
7
I
+
_'l"~-_:'l"1 e-I/T) Tl~T2
+'l"J .---'1"2
1- (1 - ::J e--t!'d
1'2-1'1
e
+1
liT, .1'(7,.1'
+
Til.\' --j--
.1'(7".1'
j-
1)(7",
+ I)
1
I)
___________________ •• TtII'"
Student Exercises
187
SUMMARY We have defined the Laplace transform and applied it to several functions that commonly appear in the solution of chemical process dynamics problems. Although the Laplace transform concept seems quite abstract at this point, in the chapters that follow you will find it extremely llseful in solving differential equation models. The final (7.3) and initial value (7.4) theorems will be useful for checking the long-term (steady-state) behavior and the initial conditions for a particular problem. A number of examples were provided to illustrate the power of the Laplace transform technique for solving ordinary differential equations. We noted that the tcchnique allows us to convert the ODE problem to an algebraic problem, which is casier for liS to solve. After performing algebraic manipulations in the Laplace domain, often with the usc of a partial fraction cxpansion, we then look up inverse transforms to obtain thc time domain solution. In the chapters that follow, we use Laplace transforms to analyze the dynamic behavior of different types of linear process models.
FURTHER READING Many differential equations and process control textbooks provide details on Laplace transforms. Some examples arc:
Boyce, W.• & R. DiPrima. (1992). Ordinary Differential Eqaatiolls alld BOlll/dary Value Problems, 5th cd. New York: Wiley. l~uyben, W.L. (1990). Process Modeling, Simulation and Controlf(Jr Chernical En" R;,/{:ers, 2nd cd. New York: McGraw-HilI. Seborg, I).E., T.P. Edgar, & D.A. Mellichamp. (1989). Process D.vnmnics {{nd Confrol. New York: Wiley. Stephanopoulos, G. (J 984). Chemical Process Control: An Introduction 10 ThcOf)l and Practice. Englewood Cliffs, NJ: Prentice Hall.
STUDENT EXERCISES 1-5. The student should derive the Laplace InmsfoJ'm for the following functions; L
d'~l
dt"
2, '/(1) 3, '/(1)
~
bl
~
t2
188
An Introduction to Laplace Transforms
5. ttt):::; cos
Chap.7
(I)!
Although you can solve question:) using integration by pans. you may \vish usc the Euler identity cos (I){:::; 1/2 (ei'dl + e-F'II).)
(flilll: 10
6. Find the Laplace transform, u(sL o!' the following input function: 2
u(11
o
o
10
20
35
7. Find the Laplace transform of the function ,md initial conditions:
d\
4
dt'
,(OJ
.)
dl
y(l)
d\"
dl
-'-
lhal satisfies the differential equation
2y
d\(O)
d\(O)
dt
dt,1
2
0
8. Solve the dlllcrcntial equation:
\ (0)
2.0
9. A process input has the following Laplace transform: u(s)
2 S
\Vhat is the lillle domain input.
6
~
It(t)?
\
C
S
Find this analytically".
Sketch the time domain Input.
10. Find the tillle dornain solution yO) for the Laplace domain transfer function (\\'ilh ~ <: J):
Student Exercises
189
Y(s) =
I
, S(T"S'
+
2~TS
+ I)
11. Derive the time domain solution yet) for the Laplace domain transfer function: +1 1)(T,s+l) 12. Derive the time domain solution y(l), for the Laplace domain transfer function:
Y(s) =
+1
.'(TI'+I) "
13. Consider Example 75, involving the following transfer function with repeated roots:
(.I' +
+ h)
A s + a
=--+
Ii
(.I' + a)2
C
+-.I' + h
Find a common denominator for the righthand side:
(.1'+
+b)
A(s + a)(s + h) + B(s + h) + C(s+ (.I' + a)2(s + h)
af
then expand the numerator and solve for the coefficients A, B, and C such that the righthand side is equal to the lefthand side.
TRANSFER FUNCTION ANALYSIS OF FIRST-ORDER SYSTEMS
8
Arter studying this chapter. the reader should 11l1ck:rsland: 'fhe responses oj" first-order systems to step and illJpubc inputs.
flow chemical reactions change the tillle constant of a stirred tank. The behavior or all integrating procesc,. How to compare the long-term beh;lvior of a nonlinear process \viLll tll<'ll of' a linear process \vitholll integratlng the nonlinear modeling cquaticllls. The responses of first-order + ti!ll\>clcla)! models. HO\\ to estimate the par~HT1Clcrs o( first-order and first-order + tillle-debl) lr;:1I1sfcr fUllctions by apply'ing step input changes. "rile response of a lead/Jag IlH)c!C] to a step input.
The important scctiolls in this chapter arc: 8. [
190
Perspective
82
Responses nf First-Order Systems
B.3 SA
l':xarnplcs of Sdf-J{cgulmillg Processes
B.5
Lcwl-[,ag rVlodcls
Integrating Processes
Responses of
Sec. 8.2
8.1
First~Order Systems
191
PERSPECTIVE One of the powers of the-Laplace transform technique is the ease with which it handles heterogeneous (forced input) problems. It is most uscftl1 when the models arc separate fronl the type of input imposed (step, ramp, etc.). The models that arc developed arc called tl'mqlerjl/I/('lioll models and will be llsed frequently in control system design. Process engineers often learn ITlLlch about the behavior of a process by changing the inputs and seeing how the outputs respond. The goal of this chapter is to illustrate the typical responses of first-order models to step changes in inputs. Knowledge or these types of responses will allow an engineer to determine a good approximate model for the process, including the best parameter values, based on measured data fronl the process.
8.2
RESPONSES OF FIRST-ORDER SYSTEMS The equation for a linear
first~order process
dy
T.....
is generally written in the folIo\\ling form
-+- Y ~. ~ k u
dt
(8.1 )
where the parameters (T and k) and variables (y ;:lnd u) have the following nailles: T
k
::::
time constant (units of time) process gain (units of output/input)
y :::: output variable
u
input variable
The rllodel (8.1) is sometimes derived by linearizing a nonlinear model about a given steady~state and then placing the resulting linear Illodel in deviation variable fortlL For this reason, we assume that the initial conditions arc y(O) ::;: 0 <mel u(O) ::;: O. The input, U and the outputy afe functions of time; uU) must be specified to solve for y(t). In the next example, we show how a standard first~ofdcr process model arises,
EXA.MPLE 8.1
A mixing tank
Assume that a chemical compound, A. is in a ICedstream cntering a mixing lank. Assume tbat there is no reaction, and that the concentration of A has no effect on the density of the fluid (this is true for trace components ill water, for eXlIlnpk). Also assume that the f!owrate is constant
Transfer Function Analysis of
192
First~Order Systems
Chap.8
F C
;
F C
FIGURE 8.1
Mixing tank.
Overall Material Balance dV = F - F = 0 dt
(from problem statement)
Component Material Balance dVC
----~
tit
FC-FC
'
since V is constant
de
F
F
v C' vC
tit
(X.2)
First of all, we can solve for the initial steady-state concentration. At steady-state, dC/dt = 0, so from (8.2) we find:
where Csis the steady-state tank outlet concentration and Cis is the steady-state tank inlet concentration. Now, since -FiV Cis + FIV C s ::::: 0, we can add this to (8.2). Also, since C~ is a constant. dC/elf = d(C - C)/dt, and we can write:
d( C - C,)
----,It'
F
_.
F.
.
~v (C, - C,J - V (C - C,)
(X.3;
or
vF d(C= S,) + (C- C).\ ~ (C- C) dt I
J.\
(X.4)
Equation (8.4) is identical to the first-order equation:
dy T--+y=ku dt
(X I)
with-r= VIF, k= 1,y:::: C- Cs' It =: Ci - Cis Notice that the time constant in this case is simply the residence time of the tank, that the average amount of time that a molecule stays in the lank.
j~.
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ " ~~fPI
Sec. 8.2
Responses of First-Order Systems
193
Notice that for linear systems, we can directly write the deviation variable Illodel directly from the physical model, skipping several intermediate steps. Also. since deviation variables arc defined on the basis of a steady~st:ate operating condition, if the process is initially at steady-state. then yeO) = 0 and lI(O) = O. Taking the Laplace transform of (8.]) we find:
Tis yes) - yeO)] + yes) Ie lI(s) TS yes) + Y(s)c Ie lI(s) (n + J) Yes) .c k U(s)
(8.5 )
which is most commonly written: Ie
Y(ll
TS + I lI(s)
(8.6)
yes)
g(s) lI(s)
(8.7)
or,
Ie
g(s)
where:
(88)
'fhc reader should become familiar with this type or representation. In general terms, /:(s) is known as a !ral1.~Ierfimc[i()lI. In this specific case. g(s) is a first-order transfer fUllction. You \vill often see a block diagram representation of Ut7) as shown in Figure 8.2 One nice thing about (8.6) is that it holds for any rirsHmJer process (with zero ini~ tial conclitions)- -we have not had to usc any knowledge (yet) about the input u as a fUllction of time. Once we know u(t), we can usc Laplace transforms to find U(s) to solve the problem. We will sec later that block diagrams and transfer functions arc easy [0 work with, when we have a complex system that is composed of a number of subsystems. Before we deal with such systems. we will first understand the behavior of first-order systems to different types of inputs.
8.2.1
Step Inputs
Themosi common input forcing function is the step input. For this problem, asSUllle a step input of magnitude 6.U altimc I ~ O. We know that the Laplace transform of a step input is (from Chapter 7):
M!
LIMII
V(s)
--1 FIGlJRE 8.2
(8.9)
s
g(s)
Block diagram.
Yes)
Transfer Function Analysis of First-Order Systems
194
5
S-
0.8
C 0
·w
I I I
I I I
I I I
I
I I I
I I I
I I I
I I I
I I 1
I I I
I~~-----I------T------
I I
------1-------1------1------
0.6
---r------l-------r
0.4
c
ill
E '5
I I I
_
I
"0 00 00 ill
I
I _~_~_~l
Chap. 8
-T-----I I I
-----r------l-------r------1-
0.2
I
I
I
0 3
2
0
4
5
Vlau
FIGURE 8.3
Dimensionless output step response of a first-order process.
and we can then write (8.6) as:
k
Y(S) = 'IS
Y(s) =
ill)
+1 s
k!W S(TS + I)
(8.10) (8.11 )
From the table of Laplace transforms in Chapter 7 (the reader should be able to derive this result, using a partial fraction expansion):
L
I]
I [. S(TS
+ 1)
=
1_ 1'-';'
(812)
and the solution to (8.11) is then:
y(t)
= k",U[I-e
'I']
(8.13)
Notice that we can represent the solution of (8.13) with a single plot, by dividing (8.13) by k!1U to obtain the dimensionless output: (8.14)
A plot of (8.14) is shown in Figure 8.3, where we have used fIT as a dimensionless time.
I
'X,,,,,.'" ~,,'~.
As a numerical example, consider the case where V:;;;:: 5 ft\ F:;;;:: 1 f(\/min, and the steady~state concentration (inlet and olltlet) is 1.25 Ihmol/ftJ . Consider a step change in inlet concentration from 1.25 UHno]/ft J to] .75Ibmol/ft J . Theil:
Sec. 8.2
Responses of First-Order Systems
U(s)
~
!>.if
~
s
Y(s) .~
1
5s
+
0.5 s
195
(illl(t) ~ J .75 ~ 1.25 ~ 0.5 Ibll1ol/ll')
0.5 I s
(8.15)
which has the time domain solution:
y(l) ~ 0.5 [1 ~ e '/5/
(8.16)
Since we desire to find the actual concentration, we can convert back to the physical variables, froHl the relationship:
y
~
C
~
C, => C(I)
~
C, • y(l)
(R.17)
and (8.17) can be written:
C(I) ~ 1.25
+ 0.5 [I - e '/51
Ibll1ol/ft ,
(8.18)
Notice that C(t -----0> DO) """ J .75, as expected. This can also be obtained by applying the Final Value Theorem to (8.16) (mel using (8.18). A plot of (8.18) is shown in Figure 8.4.
1.7 1.6
f'
~0
1.5
E
.0
G
1.4 1.3 0
5
15
10
20
25
I(min)
}i'IGUU.E 8.4
Transient response of mixing tank.
PARAMETER ESTIMATION FOR FIRST-ORDER PROCESSES H.cturning to the general model for a first-order process, we sec that tllcrc arc two parameters of interest the process gain and the process time constanl.
yet) == kMJ [I -- c ';'1 tests
(813)
Process engineers often find process gains and time conslanls by performing step processes.
Oil
Transfer Function Analysis of First-Order Systems
196
Chap. 8
GAIN ESTIMATION \VL' '-;l'l' 1'rOn] (0.1-+) thaI :lftCI termined:
I»
k..
=
T. the eli, ternl approachc" O. 'fhe \aluc or t:. can he de\(1) as I -) (large)
(~I'J)
::,.C
ie.,. the pn)("cs" gain ie., thL' clldllgC in output {as it appnwcl1('" \id('d hy the c1wngl' ill illpLll.
lililt
it
nC\\ stead\ -stall') di-
TIME CONSTANT ESTIMATION \\\' CIJl find lhe time CUllst,-lllt for '-1 firs!-(mkr prucc\s in the folhl\\'ing LlShioll. Apply a "tep input to the process aL [;:::: O. From O';.I..J.). we set' that y(l) goes to a \(due of k.::"Ll as r"7 \\'lll'Il the timc is l'qual to lhe time COlhlanl (f;;;; ,j. from (S 13): ,YO.
I
(I)
UU [I
(' 'I = O.67,2k::"U
hy finding thl' [inll' where thl' output. 1'(Tl. IS at \It'ad> -\«ltC) 'fhi" rule is al\o O)WiOllS b> 1()O~in~;l1
l!Jdl is. the lime constant cail he dClCl"Illilll'd h.1.~(,1
of the ultimate
l:i~lll'C
S.3: \\hCll/I.:::: I,
1'i.-'''pOIlSC (I1CW
\(lllk~L!:::: O.6.r2.
You "Iwuld 1)(' ('~Hd\tI. Ik'l';HIS(' t11l'; is onl) trllc fur firsl un!.:'r jll\)(\:>SI.:''' \Iith \I<:'p Illput at I ::;: U,
If
till' PI'UCl'S\ j" \ccolld·unlct PI" the
input is
11111 a
Sll'll
Ill)
tillll'-lkl;l) ~Illd :1
Ch;lll,~L, elL:..
thi" 6.\.2 1,;
\alul' \\ill nul hc currcct
rOll s!wuld get ill Ilu' !whil of i1ssocioring ullits \I·/tli ({II (~r f/ie \'(/rioblcs, ()[l\iou"ly. the pmccss time cOlhlan!. •. 1T1ll\1 h,\\c units or limc hecausc c 1/7 must he diull'llsionk"s. /\ho, thc prOl'css gain, k. must havc units of Oulput/input to he dimcIlsioll;dl)- C01FistCili.
SLOPE METHOD
or
c"tinl(llil1~ the tilllc COllstant is to rcnli/c tl1
All allnnali\T Illethod
till'
dl(1)
k::"( .
ill
7
and l'\alllatillg at (:::: O. we find
ilql
ill
II)
[,'
Sec. 8.2
Responses of First-Order Systems
197
'5 Q. '5 0
"'"' C
0.6
ill
0
'ii;
0.4
c
ill
E '6
0.2 0
2
0
3
4
5
tltau
FIGURE 8.5 Slope method for time constant estimation (dimensionless output;;;; ylk!1u).
If we extrapolate this slope to the final value of the output that is achieved, we find the time constant 1", as shown in Figure 8.5. This is a dimensionless plot, so the intersection at tIT::;: I indicates an intersection at t::;: 'T in physical time. Parameter estimation for first~ordcr processes llsing a step response is illustrated by the next example.
EXAMPLE 8.2
Parameter Estimation of a First-Order Procc"is
A process operator makes a step change in an input from 20 to 17.5 gal/min (gpIll) and finds that the output eventually changes from an initial value of SO psig to 55 psig, as shown in Figure 8.6 below. Find the process gain and time constant for this system.
56 55 0> 'ii;
54
Q.
'5
53
Q.
'5 0
52 51 50 0
5
10
15
20
time, min
FIGURE 8.6
Slope method for time constant estimation.
25
Transfer Function Analysis of First-Order Systems
198
Chap. 8
\Ve call inlll1cdialely calculate the process gain from k =' .lr/:J.u "'" 55 SO psig/17.5 20 gplll ::;: --2 psig/gpl1l. \VC can cdculatt' the time constant in a 1ll11l1hcJ' uj' different \"a)'s. (lnt'
\\,<1Y is to find thl' time where the output change is
6_~2(;'i
(If tht' rinal ch;lI1gc. I'hi:,> occurs v.. hen
the output is SO + 0.632())::;: :'13.2 psig. From the plo\. this occurs at I :::::.5 minutes. Another \\it) 10 find the time constant is to e\lWPOJatc the initial slope of the response III the final \,due. This occurs at r:::=:'1 minutes, as Sh(1\\'I1. Th.' identified pnKcss transff'l" function is thell:
-- 2
Ss + I NOlin: thallhc gain (-2 pSig/gpllll and timc cOIIQanl C'1 min) haH' units associated with tilt'llL
8.2.2
Impulse Inputs
C:nnsidcr a fir:-;t-order process \vith an impulse input of magnitude A. The tran:-;form of a unit impulse (0) is 1. so L[A0]:::= A. The first order Laplace domain respomc is: Y(s)
k
.~ TS
. .
U(s)
+ I
kA cc
is
(820)
+
the linK domain response is: kA c
\,(1)
(821 )
i-
Dividing by kit \ve find the dimc.nsionlcs:-; OUq1ul respollse shO\\'n ill the 1-"igUl\~ 8,7 belo\\. The prime characteristic of a first-order :-;)-,':-;tem is that there is an immediate rcspome to an impulse input. In practice it is difficult to actually implement an impulse function. /\ close approximatioll can be made by implementing a pul:-;e input over a shon period of time. as SI1O\\11 in the next cxamp1t:'"
"i
~ ~
0.8 0.6
~
c0
~
0.4
c
w E D
0.2
0 0
2
3
4
5
t/tau FIGURE 8.7 [lnpulse Response for output is \'(f)/k.·\.
it
first-order process. The dimensionless
Sec. 8.2
19U
Responses of First-Order Systems
EXA-~-il-I-'I->-E-8-.-3--(-:0-,-n-')-"-"-is-o-n-I-"-'-h-n-p-n-'-sc-' -,,-n-d-I-',-,-ts-c~np-ll-ts-'-----~------~-~-~---
--j
In the previolls example an impulse of magnillldc A was applied to the process. Consider il pulse input, where an input value of D..1l is applied for tfl units of time, as shown in Figure ~L8. The tolal applied input is then A :;;:; till 1{"
I
o FIGURE 8.8
Pulse jnpuL
Frolll Chapter 7 we find that:
!:111 [I - e s
U(s) ~
"'I
So, the output for a firsHmJer process with unit gain, is:
llu
yes) yes)
=
I
e li,"'j
11
S
+1 !lll e (,.' TS
D.ll
I)
s(Tsf
s(Tsf I)
which has the time-domain solution (Chapter 7):
1'(1)
co
£'." [I - e 'I'] - £'." [I -- e
where fl(t)::: 0 for l < '" nnd I for t 2 The i rnpuJ sc response i,~:
11'
(H)h]
(8.22)
11(IXpulse)
and the total input applied over the If! lillle units is !:if( fV
(impulse)
yet) The pulse and impulse responses are compared in Figure 8.9 for
tf'::;:;
O. IT and A
(8.231 .:0:
1.
1!
0.8 :;
'i
0.6
0.4
'!: 02 ;
,i o
o
". 2
4
3
5
Vlau
FIGURE 8.9 responses,
1 1.
Comparison of pulse (dashed, tp
.:0:
O.1T) and impulse (solid)
200
8,3
Transfer Function Analysis of First-Order Systems
Chap, 8
EXAMPLES OF SELF·REGULATING PROCESSES The standard first-order rnodel presented in the previous section is a typical scir-rcglllat~ ing process. If the input is changed to another value. the output eventually comes to a Ile\\' steady-state. Contrast this "vilh \lon-self-regulating systems \vhcrc the output continues
10
change forC\'cr after a step input change. Self-regulating hehavior is shown by the systems presented in the following example. One key idea to nntc is that a chemical reaction
changes the time constant of a standard mixing 1,mk model.
EXA\IPLE S.4
A C:STR with a
First~()rder
Readion
Nu\\, ('x [end the F':Xamplc S.l to include () single dccnmpositilln reaction. The component material balance i~:
d\ (
FC,
cit \\'hcrc f:. j ie; the rcadlr1!1 r;llc constant.
SlllCC
V
F, \'
F(
Je; c()n.~1;1ll1
r(
dC
\'
rtt and wc
(',Ill
calculate the "ready "ratc concentration" from dC/ilr,,-, 0
I' V I I' Thl' dnialinn "ariahle form uf our
d)-narnil~
nwdc! ic;
,I
d((
df OJ
l
1
r
V
d( (
~
(
IC
(It
k.;
~
(,)
I 1
~
(C~c,,)
I I'
'\!2
I
F
V C
~
C
1
\'
I
F I'
I'
;]llt!
I,
r
"' r k,
C 1\1
\ and
11
C
~
I
(
r k)
(S.27)
Sec. 8.3
201
Examples of Self-Regulating Processes
I
,mel thCldOlC, we know the soluuon tOJ d step change IllIllJet cOllccnlldllOn dt I:::: 0
Notlcc IhM the gdlllS and tlme consldnls 10]
stilled ldnk wIth ICdcllOll ,Ile less thdll those to! d stilled !dllk WIthout lCdC!1{)/l TillS means tlMI dB 1Iliel composltlon chdllgC has d L1S(CI dy nalHlC eltcelIll a system \Vah ChCI1llCdl lCdClltHl 'Ihlll III d system With Just mlXll1g <1
~~--~~---
Note that the previous examples \vcrc linear because the flow rate was constant. If the tlowratc weJ'e changing (i.e., was considered an input), the models \vollid be nonlinear (actually bilinear), because of the terms where an input llluitiplics a slate variable. The linearization techniques developed ill Chapter 5 must then be llsed before a j,aplacc transform analysis can he performed. fn the following example, linearization must hCLIsed be~ cause of the second-order reaction term.
EXAiVlPLE 8.5
A CSTR "dtll a Second-Order Reaction
Hcre we cxtend the previous examplc to include a second-order reaction problem. We will as~ Slime that the rate of reaction (per Ulli! volume) is proportional to thc square of the concentration of the rcacting componcnt. An example would he A +;\ ---} H. As beforc, \ve afC making the simplifying assumption that the fluid dcnsity is not a function of the concentration. Again, assume that Ci Is thc Input. The component material balance is:
dve
FC,- FC-k, VC'
d/
(B.2B)
where k2 IS the reaction rate constant. Since V is constant, dC
(B.2~)
d/ ,md we can calculate the steady-state concentrations from dC/rtf;:;;o 0
k ~ C' -,
,~
F V Cs
(B.JO)
Notice that (X.30) is LJuadratic in C\_,
d(C - CJ . dl
"II.
de
(C - C,)
.<\
+iill. (C,~ C,J dC'.H
to find that:
/' I
F+ ,C) ( V 2k ,-,
d(C-CJ+(C d/
CJ
V
2 k,
C,)
(C,
C,J
(B.J I)
Transfer Function Analysis of First-Order Systems
202 .'\gam.
Chap,8
kl\c a first-onJer. linear relationship. \\lhen::,:
\-Vt'
prucess grlln
,
/ I'
-
I'
2" C,
/,J
\,
I
I
time COnsli\nt
t lk, C ') (I I' ",
(I
- - - - - - -,~~--~
SUllllllarizin§!. the pmCll1lcters for each of the previolls cxalnplcs arc shown in Table R.],
"fA BI.E S.I
Summary of Pannll('ters from Examples ~~-
t,x.0.1
Lx.. K.4
\'Jixing Tank
CSTR First-Order Rxn
No Rxn PnlCCY';
I
Crain. k
\'
F
Ex. S.5 CSTR Secolld-()rder Rxn
, I
\'
Prnccss Time
(~onslal1L
\'
..
I \'
I
F 1;1
EX.-\:\IPLE 8.6 Here
A Numerical Stud.v of Examples Sol, 8.4. 8.5
will perform a numerical study. using the following values'
\\'l'
\'
,
F
." min
k,
o.~ min I O.J2 1'1' lhlllOl I min
C"
1.25 Ih11101
r1WIl. \\C l'
AJJ case"
,
foll()\\'ill~
(S1'R with first-mdt'l RXII 1
n steady
1.25 Ihrnol fr,' 0.625 IhllHlI rt 0.625 lbll)o! 1'r"'
sldll!
CSTR with sccund-ordcr Rxn All cases concentratiollS: Ivlixinp tank with no Rxtl CSTR with first-OHler Rxn (STl~ \Vilh sCC\llId-onkr Rxn
Sec. 8.3
Examp les of Self-Re gulating Process es
Mixing Tank
No Rxn Process Gain. k Process Time Constan t,
203
CSTR First-Or der Rxn
Second-Order RXll
0.5 2.5
0.5 2.5
I T
(min)
5
CSTR
For all of the example s, assume that a step change in thcinJet concentr ation occurs at t:::: O. Tbat is, Cj changes from 1.25 lbmol flu] to 1.75 IbmoJ fr J at t::;: 0 minutes. In terms of deviatio n variables, this means that u increase s from 0 to 0.5 Ihmol ft-3 at t::;: O. Recall that the solution for a fifst~order system with it step input change of magnitu de A is:
kA [J - e'hl
Y(I) und since
y(t) -, C(t) - C,
our solution is
C(t) ~ C, +kA
For the mixing tank
C(t) ~ 1.25
r"or the CSTR with first-ord er Rxn
C(I)
For tile CS'fR with second-o rder Rxn
e(r)
0.625
+
II
_e'h]
0.5 [I -
(8.32)
e,j']
(8}3)
+ 0.25 [I - e -,j' 5]
(8.34)
= 0.625 + 0.25 [I
-- c 1/).51 (8.35) Notice that solution s for the 11lixing tank (8.33) and the CSTR with first-ord er Rxn (8.34) are exact because these systems arc inherent ly linear. The solution to the CSTR with second-o rder Rxn (835) is only approxim ate, because it is based on a Iincarize d approxim ation to a nonlinea r model. The actual rcsponse of the nonlinea r model (using ode45) is compare d with the linear solution (8.35) in Figure 8.10. Notice that the initial response is similar, but the long-ter m rc-
t)
0.9
linear
0.8
nonlinear
0.7
0.6
o
5
10
15
20
t(min)
FIGUR E S.lO Reactor concentr ation responsc to conccntr ation, for a second-o rder reaction.
7
it
step iucrca,se in inlet
204
Transfer Function Analysis of First-Order Systems
Chap. 8
spO!1se of the lincar model deviates significantly from the nonlinear model. Indeed, we can cal" culatc the long-term response without doillg any lllllllcricaJ integration, as showll below, ~
0.625 + 0.25
Lim'ar Model (8.35) as {---->=
C("j
Nonlinear rVlodel (8.30) as
C~ + \lk, e',
F
!--'V'C)
F
~
Vk~ (
O.S750
o
C; t- 0.625 C, - 0.625( 1.75)
0
C(,,) ~ 0.7790
'filc solution that makes physical sense is:
InF;x
EXA:VlPLE 8.7
First Order + Dcadtimc
l'hc most common rnodel for process control studies is known as a first-order + dcadtilllc process 11l0dcL and is written in the following form T
dt -I- ."
=
k u(t ,-- 0)
(8.3(»)
where 0 is known ,'IS the time delay. A.ssul1lc that )'(0);;;;; 0 and 1/(0);;;;; O. The input. put yare functions of timc; 1/(1) must be specified to solve for I'{t). To understand how this equation might arise, sec r;igure 8.11.
F
II
lind the
OLlt~
c',
c,
F C
FIGlJRE S.li
rVlixing tank.
Notice that if the inlet pipe Iws a sigllit'icant volume, there will be a delay between a change in the concentration at the inlet pipe and the concentration at the outlet of the pipe. The delay can be calculated as:
Sec. 8.3
Examples of
205
Self~Regulatjng Processes
H ,- V" F
where Vp is the volume of the pipe. The relationship between the concentration at the exit of the pipe and the inlet of the pipe can be found by: ~
C; (t)
Cj(t - 0)
That is, the concentration at the exit of the pipe is equal to what the conccntnllion at the outlet of the pipe was () time units in the past. The modeling equation is:
F
dC
F
V C , V q(t)
dl
which can be written: 1("
F
cit
V
( ..~ -
+
C
F
V
Cit - 0) I
which is equivalent to (8.36) when written in dcviat'lon variable form, where:
V
c,
'j=
F
Taking the Laplace transform of (8.36) we find: Tis
Vis) - y(O) I + Vis) ~ k TS V(s)
(TS
+ Y(s)
CO"
0_ k e
0,
U(s) U(s)
(X.37)
+ I) Vis) - k eO' U(s)
which is most commonly written: k e°l'
Vis) -
'is -l I
(X.3X)
Vis)
or,
Vis) ~ Mil) Vis)
K(S) -
where:
k
C---- fl ,
TS
+
(X.39)
(X.40)
I
Assume a step input of magnitude flu at time t::;;:; O. We know that:
L IClu] -
Clu
(XAI)
s
and we can then write (8.38) as: Y(s) ~
Vis)
~
Vis)
~
keos 0/1 , I s
(XA2)
TS
kou
elf,
(XA3)
S(TS + I) Mu c'"
[Is ~
T 'TS
1,,]
+- .
(XA4)
Transfer Function Analysis of First-Order Systems
206
I'{t)
()
fOJ
U
H
1.II"j
Id" [I
,It I
Chap. 8
l'or!
\!uticc lh:ll (x ....J-."i) 1\ merely a traml~lri(ln urlhe first-(lnkr n>,p(ln~l' h~ 0 time Ul1ll\. Cumidcr till' (olhl\\in,'; pin! (FIgure S.121 of lile reSpmlS(' nt" ~\ -.y\tcrn [I) a step input Ch:Ulf'-l~ o( llwgnitude U,S at lilll<.' I ~: O. VV'L' '.t'e illll1lClIiaiely that the llllw ,klu) i\ 0 =::; Illinlllt..'\ Sinn' tile d1aIl~~' ill \)utpUI ,rfrer (l long penoel of time i\ ..1" = I k ..111, \\C \C'C thaI I. :: (ulli(\ Dj il1put/(-'UlpU[), The' prUCl'\" lime COIl\!
matl'l: 5 minutc",
0.8 06 ~
0.4 02 0 0
FHiCRE K12 Sl<:.'P input
8.4
~.lll::;:
5
10
r,~C\Plill\C of a fit'\t-ordn
15
+ dC:llllill1l' 15
20
25
tilllL' lJlllhlll1nl!c'l[u a
0,
INTEGRATING PROCESSES 'rhe P1T\ioLiS e\~l!llplcs were for sclr-re,~ula[il1g proCCSSL:", If ,Ill input changed. [!lC)] Illl' pnJCc"" output camc [0 a Ill'\\ Slc
EXA \IPI.E X,X
An Integrating Systelll
Cunsidcr a \\;ller SIOr(igc lank \vith itllet and OUI!ct SII"(',\I11\ lh,il C(in bl' incll'pcndcntly
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ z ....
Sec. 8.4
207
Integrating Processes dV
~
cit
~
P,
E':,
(8.46)
where F j and F o arc the inlet and outlet flowratcs, and V is the tank volume. Assuming a constant cross-sectional area: d1l
(8.47)
dt
To satisfy steady-slate relationships Fis
:;;:;
Fps ' so we can
L1SC
the following deviation variable
form:
hJ
d(h '.
~ (P,
dl
For simplicity, lei's assume that d(h
l;~)
1
P,,)
A
•
( E'
"
•
~f) m
(8.48)
is constant, then:
~
~ J(F~F)' A
dl
I
(8.49)
(\
which has the form: (~Y
dt
where y:;;:; Ii
~ hI'
k:;;; 1//\ and
1/:;::
(8.50)
ku
Pi ~ Pi,I" Taking Laplace transforms, we find:
sY(s)
~
yeO)
k U(s)
where yeO) ~ h(O) - hI" Assuming that we are, starting from a stcady-slate, y(O):;:; !I(O) So we can write (8.51) as:
s yes)
~
(851 ) hI';::: O.
k U(s)
Of,
k
yes)
s
U(s)
(852)
Using the notation D.U for tllC magnitude of the step increase:
and
U(s)
~
yes)
~
Lll/
(8.53)
s
k
Liu
(8.54)
Taking the inverse Laplace transform:
V(I) ~ L'[Y(s)] ~ L y(i)
kilnl
,I.k Lll/I
(8.55)
(8.56)
Substituting hack for the physical variables, (8.57)
208
Transfer Function Analysis of First-Order Systems
nr.
h. I
I II
1t=.:J.ft
Chap. 8
j,F I ' (6 ~ 5) ai/min
100 ft'
[
Solving for Ii ;::: I () I"t ( lOft
100 ft' 4 ft) . . \
(6~5)n/mjIl
60n ruinutcs
10 hnms Since the step change was madc at I :00 pm. the tank \vill overflo\\' at II :O(j pm. i\ plot of tank height versus time is slwwn in r;igurc 8.IT
10 9 8
S
"
7
6
5 4
0
100
200
300
400
500
600
time (min)
FIGL'RE 8.13
Intcgmtillg .system.
j
Notice in equation (8.52) t11at the process transfer fUllction has a pole at s ;::: O. This is a characteristic oj" an integrating system.
8.5
LEAD-LAG MODELS Some dyuarnic systems. particularly involved \vith process controL have the follO\ving fonn for a lrallSfcr fUllction modcl:
Y(s) Consider a slt'p input change
T S
-I- I
i,r"
+ 1
k"
U(s)
or magnitude t111
'---------
." _.,,,,h I
Sec. 8.5
209
Lead·Lag Models
Y(s)
~
k
TIlS
7,/'.
+ I All + I s
(8.59)
The reader should find that time domain response is (see student exercise II)
)'(1)
~ k c,u [ 1- (I - ~~) e '/'J
(8.60)
A plot of (8.60) is shown in Figure 8.14, for kAu :::: I. Notice if 'Til> T d the immediate in~ crease in the OlItput is greater than the ultimate steady-state increase, while if 'III < 'T d > the immediate increase in the Olltput is less than the ullimatc steady-state increase. !
8.5.1
Simulating Lead/Lag Transfer Functions
We have derived the step response for a IcadlIag transfer function. This transfer function does not usually arise in the modeling of a physical system, but it often arises in control system design. Our desire in this section is to show how to convert a lead/Jag transfer function to state-space form, so that a general simulation package can be used to integrate the corresponding ordinary differential equation. Multiplying through by the denominator term in (8.58), we find:
(T,t'
+ 1) Y(s)
~
(T"S
+ 1) 1I(s)
(8.61 )
2 10 1.5
Lead-Lag for 'd = 5
7.5
Various values of 'n
5
I
2.5 Y
kt>.u
0.5 1 0 -5
-0.5 -1
0
5
10
FIGURE 8.14
15
Lead/lag response.
20
25
210
Transfer Function Analysis of First-Order Systems
'~illg
Chap. 8
the I,apL\cc trallsl'orm relationships.
_ (.<1\ 'j
"dr
_ (.
.\(II))! \ .
!!(II))
'/I "dt
!!
and we KllO\\ that to obtain the tr,\llsfer function fu rill , the illiti,tI conditions of aJI \'ari, abies \\('fC, ,!'isul1ll'd to he I(TO. so: ,/\
!!
f Y
I,
" til
We canllol sohc (XJd) 11) using a gc'ncra] purP{hC integrator. because it j" not in the "tall LIard forlll uf dr/dl ;::: j(x). Our gual Jl{)\\ is tn ddinc a !lC\V variahle that ,,,,'ill ,illow LIS to usc' ,I standard intci2r;llur Rl'arrangc (:-\.63) [0 rind:
ill'
(111 'II
dr
T,i\
and "incc ',t (lud to find:
'11
!
(x
6-1!
(X.h:"1
Lake the c!cl"j\aLi\t' oj 18,(5) with rcspccIlO lIl1h.'
dr T
!!
- \'
i,,/I
;11'(' CUI1\t
=
rill ., dl
dt
(8.661
Suhstitutill,l! the righthand side of (8,66) for the Icftiwild sidc of (8.6..J-). wc find: \ t-
(S.671
II
t\\l\\. \\c must soh'C (8.65) tu find r as a function 01 .r, tu ohtain
(8.681
r =
\\'hich \\(': substitute intI) (X.67) to find:
dr dr and
\\l'
-
~
.\ +
I,i
rI
.. : )
".
II
(S J}<)I
1<1
see lhat wc h,1\'e the standard slale-space !"orm:
x=cAx ex
1I11 Du
except that (8.68) and (8.691 consist of scalars:
----------------- "-
Student Exercises
211
dx
cc=aX+bll
(X.70)
y-=cx+du
(8.71 )
dt
where
h
a=
(I
T"
d
C
Td
Td ) -
~d
Tn T"
We wilJ sec in Chapter II how (8.70) and (8.71) can be used within the context o1'a block diagram.
SUMMARY 'We have studied the response of a number of processes that have denominators of transfer function models that arc first-order in the Laplace variable, s. The systenls were: firstorder, first-order + deadtime, integrating, andleadllag. Most chelnical processes can be represented by a cascade of these types or modes. We found that stilTed lank chemical reactors arc linear first-order processes, as long as they have first-order kinetics (or no reaction) and the input 1'lowrate is not changing. For first-order and first-order + time-delay transfer functions, we disclissed how to e!ilimatc the parameters (which always have units associated with theJ,n) by applying a known step input to the process and observing the response. First-order + tirne-dclay models arc commonly used in control system design. In lhe next chapter we study the transient response behavior of seconcl- and higherorder systems.
STUDENT EXERCISES 1. As a process engineer, you are attempting to estimate the model parameters for a process that you believe is first-order (with no deacltime). At 3:00 pm, you lnake a step input change to the process. At 4:00 pm, the process output has reached 8(v;,{; of its final change. What is the time COllstant of the process? 2. Consider a water storage tank with inlet and outlet streams that can be illdepen~ dently adjusted. The storage tank has a cross-sectional area of 100 1'f2. Initially, the flow in is equal to the flow out, which is 5 ft:'ljlnin. The initial height of water in the lank is 4 I't and the height of the tank is J 0 1'1. At 1 ::::- 0 a ramp increase in the inlet flowrate is made, so that Fi(f) ::::- 5 + 0.251 where the nowrate units are nJ/rnin.
Transfer Function Analysis of First-Order Systems
212
Chap 8
Hm\' long docs it take the tank to overlhm"? Solve using Laplace transforms. ()bti.lin a general CXjJl'cssioll for SystClllS llJO(ll'lcd in Lin iatic)11 \ ,niahlt' (orlll hy:
dy
III I)
\\'here:
(l [
3. \\,'rile a eli/Terential equation \vllich corresponds to the following input-olltput transfer function rclatioJlshirJ:
\(s)
k
Tn,\"
I
I
',jS I I
II(S)
4. Consider a chemical reactor that has fern-order kinetics. that is. the rate of reactioll
per unit volume is a constant (a fero-order Kinetic pararnell'r) that does not depend on concentration. Compare this model with that of a stirred tank mixer. and a stirred
tanK reactor with first-order kinetics. Perform a numerical study'. related to ExamplL' X.5. by finding the 7ero-order paramcter thal yields the sanK' slcad;..'-statc conc(':l1tration as the first-order kinctic modeL 5. A process operator rnakes a step change on an input variable at 2:00 pIll and discu\,ers no output response is observcd until aflcr 2,10 p111. She finds that the output i" tHY>;: the way to its final steady-state aL 2:4::1 pm. YULl beliCH' Lbat this is a fir"L order + dcadtirnc process.
or
JI1put
output
2001h/hr
IIIO"! IOO"! IIIO''!' I OO"!' IIIII"F 9j f T:
tlllli..'
1:00pm I: .10 pnl I ::')9 pill 2:00 pm 2: 10 pill
2:..:1-5 pm
atter 5:00 pm (i)
cOO lh/hr 200lb/hr 225 Ib/1lr 225lh/hr 2251h/hr 225 Ib/1l1
90"F
What is the deadt]mc for this process (show ullits)"
(ii) WhaL is the tinlc constant for this pnKcss (sho\\' unit.s)"?
(iii) What is the process gain (sho\\' units)"?
6. As the process engineer for an operating ullil ill a process plant. you are trying to g('t a "feel" for the dynamic characteristics of a particular process. You have a discussion \vitb the operator about a process (w'hieh .ynt! feel is J"irsl--orderJ that uses stealll f]owratc as all input variable. and process ternpcralurc ,IS a measured variable. After the steam f]O\-'vTatc is increased from 1000 lb/hr to lion Ib/hr (quickJy), lhe process fluid temperature changes from I O(tr~' (the initiaJ steady-state) to 11 O"F in 30 [ninutes. The temperature e\Tlltual1y renches a ne\\' steady-state \',duc of J 2()"F.'. 0) Find the process gain (show ulliIS). (ii) Find the process time constant (show units).
213
Student Exercises
7. A process input is: u(t)=O
!')rt
u(t) = ! - e' I')r t > 0 The process transfer function is: 2.5
g(s)
12s
+
Find the time domain output, y(f). Plot both the input and the outpul. 8. Consider the mixing process shown below, where a portion of the feed stream bypasses the mixing tank. 3. Show that the process has a lead/lag transfer function, if the input is Cj- and the output is C\. (Hint: Write a dynamic balance around the tank and a static balance around the mixing point (after the tank outlet). Use deviation variable form.) b. Let F::::: 2 m3/min, F I ::::: 1 m3/min, Cr: : : I kgmollm3, and V::::: 10 !TI 3 . Find the state-space model and the transfer function representing this system. c. Consider a step increase of C to 1.5 kgmollm 3 . Find the response in C 3 to this change.
I
F,
F
C,
F,
C,
c,
C3
9. Compari.wm qllmpulse and Pulse Re.s])(JIlses. Consider a tank with constant crosssectional area, At::::: 1 m 2 , and assume that the flow out of the tank is a linear function of the height of liquid in the tank. 'The steady-state values of tank height and flowrate are I meter and 1 m3/hr, respectively. Find the impulse response of tank height if 1 m 3 (in addition to the constant steady-state flow) is instantanollsly dumped in the tank. Compare this with several pulse responses, where the additional I m3 is added over 0.05,0.1, and 0.15 hour periods. IO. Consider a chemical reactor where a step change in coolant flow rate frolll 10 gal/Inin to 12 gal/min (at t::::; 0) causes the change in reactor temperature shown in the figure below.
Transfer Function Analysis of First-Order Systems
214 155 150
u..
'" ill
145
"0
ci
E 2'
140 135 130 0
20
40
80
60
100
time, minutes
I"inel the gain, lime constant, and time-delay for this system.
U. For step response of the lead/lag transfer function:
Y(s) = k
+ T,t', +
s
Show that time domain response is v(
I)
= k ""
[I _ ...
(I _T,,) T,j.
120
Chap.S
TRANSFER FUNCTION ANALYSIS OF HIGHER-ORDER SYSTEMS
9
After studying this chapter, the reader should be able to: Understand the dynamic behavior of sccond~ordcr systems. Understand the effect of poles and zeros on the response for higher-order systems. • Usc the Padc approximation for time-delays. Understand the concept of inverse response. • Understand how to simulate transfer function models using ODE solvers that require sets of fiL':'I-ordcr ODEs. • Usc theMATLAB routine tf2ss to convert from transfer funclion to state-space
fOrln. Usc the MATLAB routines step and impulse. The 1l1,\jor sections are:
9.1
Responses of Second-Order Systems
9.2 9.3
Second-Order Systems with Numerator Dynamics The Effect of Pole-Zero IJlCations on System Step Respollses
9.4
Pade Approximation for Deadtimc
9.5
Converting the Transfer l'\mclionModel to State-Space Form
9.6
MATLABRoutines for Step and Impulse Response
215
216
Chap. 9
Transfer Function Analysis of Higher-Order Systems
The dYJ1atnic behavior of./lrSHJI"dc!" systems was studied in Chapter R. In this chapter, \ve
present results for higher-order systems and shmv I1mv to usc standard nutlll'l"ical integration routines for tinlc domain sirllulation of these ITHldcls. V,ic first study second-order s)'ste111s, thcll generalize our results to higher-onler systems.
9.1
RESPONSES OF SECOND-ORDER SYSTEMS Consider a linear second-order ()I)E, with constant parameters: (ty 1./,
h II( I)
"',
- dc
This is
UftC.Il
(9.1 )
\vrittcn in the form: (h
nT- dr \vhcrc (obviollsly
o()
!r~kll(l)
(9.21
* 0): (/,
(/,
2'7
T'
an
(ill
k
h ali
where the parameters arc:
k (
T
:::0;
;;;:0
:::0;
gain (units of output/input) damping factor (dimensionless) natural period (units of time)
We discussed in Chapter () that single nth order ODEs do not naturally arise in chemical processes, The second-order model shmvn in (l).l) or (9.2.) gCllerally m'ist's by changing a set of t\)".'o first~()rder equations (state-space model) to a single second-order equ;ltioll.Fol a given second-order ODE, there are all infinite numher of scts of two firSl--nrdcr (statc~ space) models that arc equivalellt. Taking the Laplace transform of (9.21: T'
lv'
Y(v) ~ sv(O)i,(O)]
+ 2CT I.IY(.I)
~ v(O)]
+ Y(s)
\vhere Y(s) indicates the Laplace transformed variablc. Assuming initial conditions arc zero. that is .\'(0)::: ."(0) :::: O.
Vie
kU(v)
find:
Y(v)
(9.3)
\vhich can he represented as:
Y(I)
cc
g(s) U(s)
T'he choruC!crisfi( cquariol/ of the second-order transfer function is T~S~ + 2~T can find the roots (also known as the poles) by Llsing the quadratic formula:
S
+ 1. \Ve
Sec. 9.1
Responses of Second-Order Systems
217
Characteristic Behavior of Sc('()nd~Ontel'Transfer Functions
TAHLE 9.1
Damping ['actor
Case
Pole Location
I II III
Characteristic Behavior
2 real. distinct poles 2 real, equal poles 2 cOInp]cx conjugate poles
ovcrdumped critically damped underdamped
- 2\: TJ \/4\:'T2 - 4T2 fJi=
(9.4)
2T 2
which yields the following values for the roots:
1',
~
fJ 2
...
\: T
+
\: 27
2
V\:2_ I
·1 -_.. _ - - - - -
T
(9.5)
T
\:
v\:' .. I
T
T
(9.6)
The following analysis assumes that ~ > 0 and T > O. This implies that the real portions of PI ancljJ z arc negative and, therefore, the system is stable. The three possible cases arc shown in Table 9.1.
9.1.1
Step Responses
Now, we consider the dynamic response of second-order systems to step inputs (U(s) :::; !>u/s):
Yes)
~, , TT
k
+
.
2\:TS
!>u
+
I s
(9.7)
where Llu represents the magnitude of the step change.
(:ASE 1
Ovcrdamped (t
Since
'>
~"
1)
1, we can see that the two roots will be rcal and distinct. Also, since we assumed that
:> 0, the system is stable (the roots arc less than zero, since we are assured that \/~2~--J< '). We btctor the polynomial T 2S 2 + 2'TS + 1 into the following form;
T
(98)
We sec immediately from (9.8) that the poles (values of s where the polynomial,;: 0) arc; (9.9)
1',
from (9.5), (9.6) and (9.g) we find: 1',
-liT, ~ ..
(iT + V(('
which gives the following value for the first time constant:
1)/T
(
/\hu.
\\l~
(9.10)
-I
-
find lhe :--ccond pole
_ \
~c
which gin:::-- the following \allle fur the second time cunstantT
(911 )
r\panding the rlghthand side of (9,S). 2~ TS
T,\-
\\'e
(9.12 )
·t-
call \,Tile:
\\lllCh
,,'
lead to the relatiollships T
T1
2\
"
(i). 1.:1-)
Oi"'
\Ve call derivc (sec student excrcisc la) the fuIIO\\'ing solution 1'01 step ["eSpll!1SeS ()fllVenlalllpcd SystClllS
fhcnlmnpl'cl.
If f) v,here
"
unll
S>
] " c
1'1('
and (
-
ft).
I5)
T T
T,--
\
Nlltc thaL as in thl' case llf first-order systems. we can divide hy k":'l1 to develop a dimensionless UUlpUL Also, the dimctl\ioliJcss timc is th and we can plu( Cllr\'es f(n dimellsionless outPUl as a fUllction of i;. Thi:-- IS donc III Figure 9. I. \v-hich includes the critically damped l:asc. as discllssed ncxt. \los( chcmical processes t'xhihit ovcnJalllped behav-ior. The critIcally dampl'd .step l'eSpUIlSC is also .ShUV..-rl ill ['igurc 9.1 (curn: with ~:::: I).
0.8
1.5
0.6
0.2 Various values 01 damping factor
o o
10
5
i5
t/lau
FIGLRE 9.1
Step ''''I,onse of a second-order rller"damped system.
Sec. 9.1
CASI-: 2
Responses of Second-Order Systems
Critically Damped
(~
219
= I)
The transition between ovcrdampcd and underdmnpcd is known as critically damped. We can derive the following for the step response of a critically damped system (see student exercise I b)
Critically (hlmpcd,
~ :=
I [Repeated polesl (9.16)
Notice that the main difference between overdampcd (or critically damped) step responses and first-order step responses is that the second-order step responses have an "S" shape with a rnaximum slope at an inflection point, whereas the first-order responses have their maximum slope initially.
The initial behavior for a step change is rcally dictated hy the relative order of the system. The relative order is the difference betwccn the orders of the numcralor and denominator polynomials. If the relative order is I, then output response has a non-zero slope at the linle of the stcpinput; the step response of a system wilh a relative order greater than I has a zero slope at the time of the step input.
CASE 3 For
Undcrdamped
S <:
(~
< 1)
l, from (9.5) and (9.6), we find that the poles are complex:
fJ= ..
~I T
T
which is wriUcn in terms of the real and imaginary contributions:
P=(X±jf3 where:
ex
~
=
f)
T
~
T
We can derive the following step response for all undcrdamped system (sec student exercise I):
lJndcnJampcd, (t < 1) [Complex poles 1 V(I)
where
f)
=
:-=
k!:l.u (I
V'(
T
('
= tan
(1(.
I
~
,,,in (I3t +
(9.17)
Transfer Function Analysis of Higher-Order Systems
220
Chap. 9
Again, dividing by kD.lI, we can produce the plot shown in Figure 9.2. 2nd order underdamped
1.6
1.4
/
/:~
,-
/ I?': ,\\
1.2
//og 75'\>" '/.
i
I factor
varia JS values 0
!
08
.0
.
/.
\
\ /
\
1/
/
•
t,\
0.4
...
./
/
0.6
"
)/
...
·
•
i
t
0.2
•
•
o
o
2
6
4
8
10
12
14
Utau
FIGURE 9.2 Step response of a second-order undcrdampccl system as function of the damping factor (0.
it
A number of insights call be obtained from r;igure 9.2 and an analysis of the step response equations. Notice that the poles for the second-order SystCll1 can be wrillen:
P - [-(
.
.. _ I
+ jV(I-(')! T
Observe that, for smaller S, the respollse is more oscil1atory. For S < I, the ratio of the imaginary portion to the real portion of the pole is: imaginary
real
S
As the imaginary/real ratio gets larger the response hecomes more oscillatory. We also notice that a decreasing T corresponds to a larger negative value for the real portion. As the real portion becomes larger in magnitude (more negative) the response becomes faster. We use these insights to interpret po1cl7.ero plots in Section 9.3.
Responses of
Sec. 9.1
Second~Order Systems
221
time to first peak
l c
rise time
I.-oscillation period of j
. b decay mIlO = -
•
overshoot ratio =
%
FIGURE 9.3
Time
9.1.2
Step response character-
istics of undcrdamped second-order processes.
Underdamped Step Response Characteristics
The following C0111mon measures of underdampcd second-order step responses arc shown on r7 igurc 9.3 and defined below: (I) risc timc, (2) timc to first peak, (3) overshoot, (4) decay ratio, (5) period of oscillation.
Rise lime. The amount of timc it takes to first reach the new steady-state value. Time to jirst peak. The timc required to reach the first peak. Notice that there arc an infinite number of peaks. Overshoot. The distance between the first peak and the new steady-state. Usually expressed as the overshoot ratio, as shown in the figure. Decay ratio. A measure of how rapidly the oscillations are decreasing. A h/a ratio of 1/4 is commonly called "quarter wavc damping". Period (d'osc'illation. The tilnc betwecn successive peaks. The following example shows how to usc Figure 9.2 to estimate these values.
EXI\l\;lPLE 9.1
Undcrdampcd Second-Order System
Consider the following transfer function, subject to a ullit step (L\.u ::::: 1) input change (assume time units arc minutes):
g(s)
~
5
, 4s"1 0.8s
+
I
Find the (1) rise time, (2) time to first peak, (3) overshoot, (4) decay ratio, (5) period of oscilJa· tion, (6) value of yet) at the peak time,
ESCOLA Dc H,GENHARIA BIBLIOTECA
222
Transfer Function Analysis of Higher-Order Systems Our first step is tu calculalt' the system parameters. \Ve
Cilll
Chap,9
sec that;
k~5 7 2 ::;:
2i::'T -
:=
4 so
T
0.8 so
:::
2 II.K
D,S
t::: '" ::: ----- : : : 0.2 -.!T.:.1-
\VC usc Figure 9.2 as lhe lJasis for the follmvil1g calculatio!ls,
1. rhe risc lime fur
~ =
I
0.2 is' ~
::: 0.2
15J I
=--::
2. The time to first peak for
3. The
o\Trsh()()(
ratio is
1)1, so
I 15
4. The decay ratio is lSI
I
rile \'aluc pf YUf,l is
1"
is'~
l,g
T
1.2. so
= J.6 minutes
'II:::
3.2 ..
;.c;
6.4 Illinu(t's
(l.5J.
I I ~ IJ,\
5. The ptTiod or oscillation is';' (..
II"::::;;
,
k~1I = J .53.
9.6 - 3.3. so
f(,,, ::::
6.3 ... ;;;; ] 2.h rninutt's.
so \' = 1 5.1 kJII::: 1.53(5) ::: 7,(\5.
Although equation (1).17) C
9.1.3
Impulse Responses
NO\v, \ve cO[l;-.;idcr the dynamic response of second-order systems to impulse inputs.
Y(s)
k
-+-
2~7S
+
J
where 1\ represents the lnagllitude of the jmpulse.
CASE I
O\'crdamped
(~
1)
The lime domain solution for the ovcrdamped
T
C<.l.SC
is (see sludent exercise 2a):
(,~I ~ sinh
\I,'hne dt)/k/\ is the dimensionless output and lIT is
thl~
(: \1 - ('
~)
dimensionless rimt'.
Sec. 9.1
CASE 2
223
Responses of Second-Order Systems
(:riticallJ' Damped
(~:::::
I)
The impulse rcspOilsC for the critically damped case is (see student exercise 2b): y(t)/kA = - 7 e
liT
T
CASE 3
Vndcrdampcd
(t / 1)
The time domain solution for the nnderdamped case is (see student exercise e-U/' sin
Y(I)/kA T
2(~):
(VII
The impulse responses as a function of {are shown in F'igure 9.4. 0.8
0.6
'5
%
'···f-'r·.,··.+······················,·················
;
,.....................
o
6
8
!
0.4
0
w w
'"
0.2
'"E
0
C 0
'wc
i5
·0.2
·0.4
2
4
tltau
FIGURE 9.4
hnpulse response as a function
of~.
10
224
Transfer Function Analysis of Higher-Order Systems
Chap. 9
CI'Iw underdampcd responses show characteristic oscillator)' behavior.
9.1.4
Response to Sine Inputs
Consider the case where the input is a sine fUllction. with amplitude A and frequency (,):
lI(r)
A sin
c:::C
(ll!
The Laplace transform is:
when applied to the second-order transfer function and inverted response after a long period of lime is the periodic function: y(l)
[0
kA sin (rof + (l))
+
\
the
lillle
domain. the
(9181
\vhcrc:
(9191 (sec student exercise 3). The amplitude of the output is:
kA
and the phase angle is (b. Often, system behavior is discussed in tenns of an amplitude
ratio, which is the amplitude of the output eli vided hy the amplitude of the input. The amplitude ratio is: k
These relationships are used in ExarnpJe 9.2.
"'--~-I
EXi\IVIPLE 9.2
Sine Forcing of Se(~ond~OrderSystems
C:unsider the fol1owing system:
g(s)
I
\.: -+- n,:?:s +- 1
A Imv frequency sine forcing ((,) ::::; 0.1 min-- I) yields the input/output response shown 111 Figure 9.5. Notice that the output lags slightly behind the input. and the amplitude of the output is slightly smaller than the input amplitUde. Contrast this result with the folhnving case 01" a higll frequency input.
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ ."':,£'"ifi
225
Responses of Second-Order Systems
Sec. 9.1
1.5
-0.5 -j
-1.5
a
20
40
60
80
JOO
time, min
FIGURE 9.5 A high frequency sine forcing (l') Ufe 9.6.
"
5 min I) yields the input/output response shown ill Fig-
u
0.5
"
0:::
Low frequency sine input response.
a
y
~
-0.5
-j
a
3
2
4
5
time, min
FIGURE 9.6
High frequency sine. input response.
Notice that the output lap significantly behind the input, and the amplitude of the output is much smaller than the input amplitude. A particularly interesting type of behavior that can occur with second-order undcrdamped systems is known as reSOll(///(.'C pe(lking, which occurs in intermediate frequency ranges as shown in Figure 9.7, where a frequency of 1 fad/min is llsed. Here the output amplitude is significantly higher than the input amplitude although the input/output gain is I. At lower ([;igme 9.5) and higher (Figure 9.6) frequencies the output had a lower amplitude than the input, while al an intermediate frequency (Figure 9.7) the output had a higher amplitude than the input. This phenomcna can only happen in systcms with complex roots.
226
Transfer Function Analysis of Higher-Order Systems
5
10
15
Chap.9
20
time, min
FIGlJRE 9.7
R.t'sonam;c peaking phenomenon.
The concept of phase angle is illustrated by Figures 9.5 through 9.7. At low frequencies (Figure 9.5) the output barely lags the input, and therefore has a phase lag of almost {) deg. At inlt'Tmcdiate frequencies (Figure 9.7) the output Jags the input by (jO", and ill high frt'qucncics IFigure 9.(l) the output lags the input hy almost ISO". Also noll' that the notion of "high,"' "intcnncdialc.'· and "lil"/' frequencies is relative (dependent Oil T). Lm',!, medium. and high frt'ljUCJKit'S cnllcspol1d roughly to (,jT 0::; 0.1, J, and 10. re::;pcctivcly.
T'he method of sinc-forcing a system is llsed in the analysis of feedback control systel11s and is knO\vn as frequency response analysis. Bode diagrarns arc llsed 10 plot the amplitude and phase angle as a function of rrcqlll~ncy. Vv'e do not provide further analysis here. but refer the reader to any textbook on process control for 1110re detai 1.
9.2
SECOND-ORDER SYSTEMS WITH NUMERATOR DYNAMICS "rhe previous discussion involved pure second-order systems. COil sider
\10\\'.
a sccond-
order system with numerator dynamics with the gain/time constant form: s(s)
k(T"S + J) .//(1) (",.1+ 1) (T!S+ I)
The pole-zero form is: s(S)
(s -1',)(.1 - 1',)
//(s)
W201
Sec. 9.2
227
Second-Order Systems with Numerator Dynamics
where: PI
=
The gainltimc constant form has the following time domain response to a step input (sec student exercise 4): (9.21 )
The reader should show that, if Til:::: T2' the response is the same as a first~ordcr process.
EXAMPLE 9.3
Consider the Following Transfer Function
yeS)
+J lI(S) (3s + 1)(15, + I)
(9.22)
The step responses are shown in Figure 9.X. Notice that negative numerator time constants yield a step response that initially decreases hefore increasing 10 the final steady-stale. This type of response is known as inverse n\"jJollse and causes tough challenges for process control systems.
"1 = 3 "2 15 varying
"n
60 FIGURE 9.8
Step responses of a second-order system with numerator dy-
namics.
Notice also that a numerator time constant that is greater than the denominator timc constant causes ovcrshoot before settling to the hnal steady-state. Also notice that the inverse response becomes "decr>er" as the process zero (-117 11 ) approaches a value of zero from the right.
Transfer Function Analysis of Higher-Order Systems
228
9.3
Chap. 9
THE EFFECT OF POLE-ZERO LOCATIONS ON SYSTEM STEP RESPONSES There (lfC it numher 01 difkrcnt \va)'s to represent process tra/lsfer [unctions. The "t!,lintime constant" form is: k(TII1S
g(,,)
(T,I'" WI1Cl\' TII/'
-+- I
-+- J). ,('''''/ -+- ]
+
+
I )(T,I'"
is (\ nUlllerator time conslanl and
I) .. (Td,,"
+
(9.23)
I)
Tdi is a nUlllCl"(l[or time constant.
The "poIY'l1omiaJ" form is g(.l)
hm
(b,!,SIll --_.
(1,/'!'
hiS (f
([II
IS
I hll ) I do
(9.24 )
The \'alues of \' that cause the numerator oj' (9.23) or (9.24) [0 equal zero arc known as the "zen},,·' of the transfer function. 'fhe val LIes of s tll,l! cause the denominator of (CJ.23) or
(lJ.24)
to equal /cro arc knowll as the "poles" of the transfer function. The "pole-/el'o" for111 is:
cJ ("-
C, )(.1 g/,(I)
II, )(" - pJ ("
\\Ilcrc:
cm ) p.,l
. II
k
i
,
I', ) J
i'i , (-:)
The notation il ( fl,) is shorthand for ( -fll)( -P2) ··(Plll. Notice ~ll~() Ihm the poles ;m: Ill:::
(l).271
I,ll
and the Icrn is (9.281
and that complex poles (or Zt~ros) must occlIr in cornplcx conjugate pairs.
EXA\II'I,E 9,4
Comparison of Various Transfer Function Forms
('nnsidcr a trallsl"cr fUllction \\ith lhe follt\\\ini! E!
+ 1)( 15.1
~~~~~
t J)
~ "'_-Itj-:.l;:~
Sec. 9.3
The Effect of Pole-Zero Locations on System Step Responses
229
The polynomial form is:
g(.,) -
"- 20s + 2 -1- ISs
+
The gain-polynomial form is:
g(s)
lOs 1- I)
~
+
18s 1 I)
and the pole-zero f(mn is:
4) (, - io) ( N,Js) ~- 9 (s +1)(s + I) 3 The zero is O. J, and the poles are
15
1/3 and -1 lIS.
Notice that the zero for Example 9.4 is positive. A positive zero is called a right-halfplane (RHP) zero, because it appears in the right half of the complex plane. Right-halfplane zeros have a characteristic inverse response, as shown in r'igurc 9.9.
Also notice that the poles are negative (left-half-plane), indicating a stable process. Right-half-plane poles (positive poles) arc unstable. Recall that complex poles will yield an oscillatory response. A pole-zero plot of the transfer function in Example 9.4 is shown
in Figure 9.10 (the pole locations arc (-1/3.0), (-1115,0) and the zero location is (0.1,0); the coordinatc~ arc (reaJ,imaginary)). For this system, there is no imaginary component and the poles and zeros lie on the real axis (Figure 9.10).
2
1.5
'"
0.5 0 0.5 -1
0
20
40
FIGUU.E 9.9
60
Inverse response.
80
100
Transfel' Function Analysis of Higher-Order Systems
230
Chap. 9
Imaginary axis
Real axis
F[G(:RE II.! 0
P"k-/cr" locatio" plot
((I)' Example 9.-4 (x-j1olc\. {l-!l'ro)
As poles move further 10 the left they yield a ra...;ter rcspnnsc. while illcrcclsing the n1
Figure 0.11. RcC<.·q also that a process with a pole at the origin (and nOlle
in the right-halt'-planc) is kllO\vn ,-IS an intcgmting system. that is the sy-stcrn nevCl setllts [() a stcady-state when a step input change is made. Mulli[J1c right-hair-pjtmc zeros «lUSt' multiple "changes in direction" for e\ample. \vilh two RIIP leros. the step respollse is initially ill nile dirceti(HL switches dirL'ctioll. then ::-witches back to the initial direClion.
f),tI
P,I.WE APPROXIMATION FOR DEADTIME Recall eJat the Laplace transfer function rllr a pure linl(>dela) is (,-HI where 0 is the tinH> delay'. This i::- an irrational transfer function: an approximation that is rational and often pro\'idcs an ad'LJuak representation of the deadtilllt' is known as the Pack approxilnaliol1.
more oscillatory
r
Imaginary aXIs
x Inverse response zeros Real axis
faster response
unstable poles
x
1 FIGUHE 9.11
[(Teet of polc-/l'J'o location (lJ) dynamic hd1avlor (\-P(llt"'. As poles hCc(l1l\t' more negative. !Ill' rt'SPC1I1SC is la\lt'l'. As the 111Ia~il1dry/n::nl ratio increa'ics. lhe response becomes mme uscill~llor:
n-/cm).
Sec. 9.4
231
Pads Approximation for Deadtime
1.5 1
.J
Second-Order Pade' Approximatlon
0.5 y
Pure Time-Delay
0 -0.5 -1
Fir.ll-0rder Pade' Approximation
0
r-
10
5
!,::>
time FIGURE 9.12 Comparison of step responses for pure time-delay with order and second-order Pade approximations. Dcadtimc::: 5,
rirst~
The first-order Pad6 arr)foxil1lation is
B e
2
0-,=
+
0 2
s (9.29)
s
The second-order Padc approxilnation is
c
0,,=
o \' + 2·
02
12
, y'
2 o 0 " 1 + ,. + 12·'·· 2.
(9.30)
A comparison of the step responses of first and second-order Padc approximations with pure time delay arc shown in r"igurc 9.12,
EXAiVIPLE 9.5
Comparison of the Pade Appmximations for Dcadtimc
Consider the foUowing
first~order
+ deadtimc transfer function
xes)
~
e
Ss
5s -+- I
Transfer Function Analysis of Higher-Order Systems
232
Chap.9
The first.-order P,ldc appruximatioll yields tht' fullowill(! lnulsfcr functiol\
-- 2.Ss 7.Ss and the second-order Pade appro\i11latio/l .yields
2.0SJJ\: -- 2S1
1041
j
1-1-.58.\3,' -+ 7.51'
~(s). gl(s) and g~l(S) is "llUwn in [:igurc 9.] J. Notice tllat the first-urdcr aPllW\llllati(Hl has an inverse response. while the sl'l'())Jd~ordcr appmximatill)l has a "double inverse 1'l:.'SP(1!1"t'." The reader should I'ind that th,'I\' 1\ it sin,!,'Jc Ilo\itin,' len) fur NI{S) and there arc t\\'() po.siti \'c. complex-conjugate lcros of the J111I11Cra\(lr tr,lIlSfcr functi(lI1 \)f g ~(.\).
a cOlllparison {If the step response.s (If
0.5 y
o
L
FIGURE 9.13
5
10
15
20
25
Cumparisun uf first-urdcr + tlcadtill1c rcspulhC with fll'st- and
~ '_c_c_"_n_r1_-,_"_.d_c_r_l'_a_r1_c approximations 101- dcadtimc Most ordinary dirferenti,ll equation llurllcrical illtcgrat()rs (including c)(le4~-)) fCCjU1IT pure differcntialcquations (with no time-delays). II' yCIl! h,H-e a system of differential equations \vhich has time-delays, the Pack approximation call bc used to convert them tll delay-frec dillerentiaI equations, \vhieh call then be numerically inll'grmed. See studeIlt exercise 2S as an example. One urthe mallY advant
9.5
CONVERTING THE TRANSFER FUNCTION MODEL TO STATE-SPACE FORM Til this section \ve sho\-v One way to convert the input-output transfer fuul'liOll model to st;ltespace form. Although the Laplace domain is llsed for mwlysis, the state-space form \-\'ill nor mally be used for tillle domain simulaliolls. Consider the transfer function relationship:
Sec. 9.5
Converting the Transfer Function Model to
y(S)
State~Space Form
233
k
=
_I 2{7S 1_ I lI(s)
which arises from the following equation: 7
, d'" C dl i
dy
+
2{7· dl
+ Y c.c k U(I)
(931 ) (932)
y
LeI:
.r] ::::
and:
X2 ::::x 1
(9.33)
y
(9.34)
so: Divide (9.31) by
72
::::;~-rl::: ·\:2
to obtain: d'y
dt 2
+.2{ dy.. + tit
1"
which we can write as: 2{ dy I k ._- ;Y+,U(I) 'T dt 'T~ . 'T or,
and since:
and we can write in the slate-space fonn
(935)
y =
[1 0]
1::1
(936)
The student should show that defining y::;;: x 2 leads to the following state-space model: .
I.
XII X,
=
I
-
2{
1
7
-7'
I
()
.
l.xll +
k
I I] .',
7' [u]
(937)
()
(9.38)
Transfer Function Analysis of Higher-Order Systems
234
Chap. 9
J'vlATLAH has routines for converting fron] transfer function fonn to stalc-spi.]CC form (tf2ss) and vice versa (ss2tf). tL2ss is llsed in LxarnpJe 9.5.
EXAiVIPLE 9.5
i\lATLAB Routine tf2ss
C'onsidcr the following second-order system: j
1 I/(s)
U.7071 \
First define the Ilumerator and denominator arret)'S by:
and
ClllCl"
num -
J
den
).
0.7071
the command: la, b, c,dJ
Lt2soc; (num, den)
I'vlATL \13 returns the SUite-space matrices:
a b --:
0.3535 -0.5000 i.(lOUO 0 ] (1
c-
(J
c.1
()
1.S000
Notice that the slale space models ill Exmnplc 9.5 are dif{l~rent than the malices that arc obtained from (9.3."1) and (9.36) or (9.:n) and (9.38), hut the differellt forms vvould all yield the same results for the output variable via simulation. Rcmcmber that a transfer function relates inputs to outputs but docs not represent the actual stales of the s)'stern. There arc an infinite number of statc-space models that \vill yield the sank' input/output lllodel. After finding the state-spacc form for a transfer function, we can use any available numerical integrator to solve problems. MATLAB rOlltines of interest include od(:, 4 (), in.itiaL and step.
9.6
MATlAS ROUTINES FOR STEP AND IMPULSE RESPONSE MAITAB has routines for step and ilnpulse response of either transfer function models or state-space models. In the following, we sllmV' hmv these routines arc used for transfer function models.
9.6.1
step
A quick \vay to generate stCJ1 responses is 10 use the MATLAB function step. This call be used \vith either a state-space or a Laplacc domain model.
235
MATLAB Routines for Step and Impulse Response
Sec. 9.6
Consider the following Laplace dOlnain model:
g(s)
~
+ I) + 15 s +
2(IOs
--50 s'
3
\vhich can be written:
g(s) ••
50
20s
+2
+
15 s
+
3
The following MATLAB commands arc llsed to generate the response shown in I-;igure9.J4. Dum den
= ,0:
[20 2] i [50 15 3J;
[Y,x,L] = step{num,den) plot(t,y) Notice that a tinlC vector is automatically generated, wilh a length close to the settling lime of the process. The same plot could be generated from the state-space form by using:
[y,x,tJ = step(A,B,C,D,l) pIot(t,y)
n,
where A, C, and f) arc the Slate-space matrices and' I' indicates the first input. Although state variables arc calculated, only the output variables arc of interest. We could supply an equally spaced time vector and use:
[y,x] = step(num,deD,t)
12
0.8
"
0.6
OA 02 0 0
10
20
30
time
FIGURE 9.14
Step response for the example system.
40
Transfer Function Analysis of Higher-Order Systems
236
Chap,9
0.4 0.3
0.2
"
0.1
0 01
0
20
10
30
40
time FI(~LRE 9.15
Impulse rcsp(lnsc for lhe example sy:'lClll.
for tilt.' Sll'p response of a transfer fllllction IllUdel. The nUlllhet' or urguJl1CnL" determine,,> \\'hether a tntl1.';fcr function or statc space rnodel is used by lllC step function, and whether the lime \ ('c\(X has hel-'ll specified or not.
9.6.2
impulse
The output and time veclors arc generated lIsing: [y,x,tJ
=
impul~;e(num,clcn);
the plot is obtained from
plot (t,Y) 'rhe plot is
SIl0\Vll
in Figure 0, 15 above. Notice that an impulse has an immediate
(disl'(Hl-
titlllOlIS) dfcct Oil the output. because this is a relative order olle sy'slcm.
\Vc could also supply an equally spaced time [y,x]
Vl'([Or
and usc:
= impulse(num,den,t); SUMMARY
The step responses of the classical seconu order system (o\'erdampcd, critically' damped, and undcrdampedl were presented. In addition. \ve sho\ved the effect of ll11rncrator dynamics (and particularly right-half-plane zeros) 011 the rc-"ponse or a sccond-order sy'stenL The Pad0 approxilnations for dL'
Student Exercises
237
The MATLAB routines used were transfer function to state space step response impulse: impulse response tf2ss:
step:
Critical concepts from this chapter include: damping factor natural period numerator dynamics Padc approximation for time~dclay relative order
STUDENT EXERCISES 1. Derive the step responses for the following second-order systems. a. Overdamped b. Critically damped c. Underdamped 2. Derive the impulse responses for the following second-order systems. a. Overdamped b. Critically damped c. Underdamped 3. Consider a sine input with magnitude A and frequency w. Solve for the time domain value of the output for the following second-order systems. a. Overdampcd h. Critically damped c. Underdamped 4. For a second-order system with numerator dynamics, find the step response for the following. H. Ovcrdampcd. b. Underdamped. c. Critically damped. 5. A second~order system has the following Laplace transfer function form:
Y(s)
=
2.5
+ 5s +
I
U(.\)
where the time unit is hours. The initial steady-state value for the output is 20 psig and the input is 4 gpnl. At t::;;: 0, a step input decrease is made, from 4 gpm to 3 grm.
238
Transfer Function Analysis of Higher-Order Systems
Chap, 9
a. \Vhat is the final value of the outpUt'? h. \\,'hcn docs the Olltput first reach this final \,-tllIe',J ('. \\:hal is the minimum value of the output',)
d. \VhCll docs the output hit this minimulll \aluc'? e. PInt the response. Ol)F~:
6. ('ullsickr the follO\ving second-order
d\ 'II'
dt'
(T
l
with the initial conditioll"
d\' - ill
-I- T,)
f
,,'(OJ::: dO) ::: 1f'(O)
=::
k
/I
u(()) ::: ()
a. Find the Laplace lransfmm of the elitTerential equatioll. \Vrilc this cxprC'\SI(ll1 in the form of \'(s)
::::: ,l,'(s) II(S}
b. Nnw. assullle that a step clwngc of magnitude .-\ !l1 tilt' variahk /I Dccur" at tinlc::: 0, Find the lime domain resulL y(l), c. Nm\'. assume that a step change or lllaf,:JliWc!c ,\ in the \
'II
7. C:ollsider the fol]()\\'ing two first-ordcl' ODE'.;: '[
d.r I I til
dr,
"
til
+
,[
"
,[
II
ell
and the static rclation"hip y:::::: XI + x:~ where XI and x::, arc two state \'ariahles. r is the output \,lriable, and II is ;m input \ariahlc. a. Show that the twn equations call he combined to yield a single ODL in the Corm of problem 6. Find k and I I I a" a rUlIction of k I' k::.. 'I' '::,. b. No\\'. assume that a step change of magnitude ~u in the \ariahlc II OCCULS :H time:::: O. Find the time domain result. y(fl. hy ming a partial fraction l'.\jlaIlSloll and solving for the iJl\erse Laplace transform by hand. c. Plot_l.'t(l). x:,(t) and YU.l if j,l/:::: I. k 1 :::: ~I. k 2 :::: 2. 11:::0; J. and '2:::; 10. R. As a proct'ss engineer with the Complex Pole Corporation. you arc as"igned a unit \vitll an cxotlll'nnic chemical reactor. In order to learn Innre ahout the dynilll1ic" of the process, ),'OLJ decide to make a step change in the input variable. which is ('()ulant lenlperaturc. from 10"e to [SOC, Assume that the reactor \vas inItially at a srl'~ldy slate. You obtain the follmvingplot for the outpUl variable. \vhich is reactor tCJllpcr~ aturc (notice that the reactor temperature is in OF).
----
~~rg.iF,r
Student Exercises
239
290
I I I
285 u.
'"
-...J--____
--- ---
280
ID "0
~ 0-
~
''""
I
I
270
I
I I
I I
I I
1--
I-
I I __ I I I I
255 0
I
I _____1
I I
I
I
I I --!---I
I I I
I I ---- --t-I I
___ I
r I I
-- - ---t -
250
I I
~I
260
I I 1
---~----
I I
265
--~t-----
I I I
l-----~---r--------I---------f----
----- -----
E
2!
----~-
I I 1
275
~
iiiID
I I I
I I
I I I
I I
1
-1--
I I I
I I I
I I
20
40
60
- -t---I I
I
80
100
time, minutes
a. What is the value of the process gain? (show units) h. What is the value of 1''1 (show units) c. What is the value of~? (show units) d. What is the decay ratio? c. What Is thc period of oscillation? (show units) f. Write the second-order transfer function. (). A process is described by tile f,Jifowing finear ordinary differential equation; 4
d~'
d1
~
2
+
1~2
dv ~
tit
+y
2.5
d 2l{
where y is the Olltput and u is the input. Assume that: tlr( 0)
c
»(0)
til
also, assume that at time tionship
1 :=
du(O) dl
~~ ~
0
0, the input begins to increase with the following rela-
U(I)
1
2
2I
The units for time are minutes. a. What arc the values of the poles of this process (give units)? b. When docs the output of the process reach a maximurn value? c. What is the maximum value or the process output? 10. A process has two poles and one zero. The poles are located at -I ± 0.5) and the zero is located at 0.5. Sketch the type of responsc that you expect to a step change in
Transfer Function Analysis of Higher-Order Systems
240
Chap.9
input. Explain. F,'ind the transfer function and verify these results assuming a gam of one.
(1,0.5)
x
(-1,-0.5)
(0.5,0)
x
11. Consider the following state-space model (from Module 7):
[ ~: I [
2AOS 0.833
Y -
o
-2.238
II x, [ + [ x,
7 [ -1.117 /I
10 11 [x,'[ Xl.
a. b. c. d.
Find the transfer function g(s) where yes) Find the poles and zeros. Plot the response to a unit step input. Plot the response to a unit impulse input.
~ g(s) lI(s).
12. A process engineer responsible for the operation of a complex chemical reactor has the process operator make a step change in the coolant f10wrate from 10 gpm to 15 grm to the reactor at 2:00 pm. The reactor temperature is initially 150"F at 2:00 pm and drops to a low of 115°P at 2: I0 pill. Eventually the reactor temperature cornes to a final stcady-state temperature of 125"F. Assuming that the response is secondorder (k!T 2 \.2 + 21;TS + I), find k, 1;, T (show units). 13. The output of a sccond~order, 1I1lderdamped system has a rise time 01" I hour. and a maximum value of 15°[" (in deviation variables), aner a stcp change at time { =: o. After a long period of lirne. the output is 12"F' (again ill deviation variables). a. What is the value of 7'1 b. What is the value of C c. What arc the poles? (also, show their location in lhe complex plane) 14. A step change of magnitude 2lb/min is applied to the input of a process. Thc resulting oulput response, in dcviation variables, is shown in the figure below.
_____________________________
~_,=_jf-_-_kLl'*;1t
Student Exercises
241
40
30
20
10
/
17
o o
/
\ >,,_/ ,
----.....
/
.............
100
50
/-
--
150
200
250
300
time (min) Step response of a physical system, in deviation variables.
a. Pind the period of oscillation, rise time, and time to first peak, for this system. Show your work. h. Find parameters (show the units) in the transfer function, g(s) :::::: k/(T 2S2 + 2~TS + I), by using the dimensionless plot, Figure 9.2. Show your work. IS. Consider the following third-order transfer function, where ~ is a parameter. Find the conditions on the parameter [3 that will give an inverse response.
g(s)c
(5.1'
I-
(2.1'2+.1'+/3) 1)(3.1' + 1)(2.1' + I)
Show your work and explain your answcr. 16. Consider the following transfer function:
11(.1')
S2
+s- 2
s2+4s+3
a. Find the poles and zeros for this transfer function. b. A unit step change is made at f = O. Find thc value of the outpul, using the final and initial value theorems: i. After a long timc. ii. Immediately after the step change. c. Verify your results in b by finding (analytically) the time domain solution. d. Verify the results in h using the MATLAB function step. 17. Consider the following state-space model:
I~; I I
x, + 10.00155111 0.00248
-6.5 2.511 X'I 4 - 6.5
ESCOLA Dc ENGE~~HARIA B I BL I 01 cCA
242
Transfer Function Analysis of Higher-Order SystElms
Chap. 9
Find the lrallsfn l"unctiulls n.·lating the input tu c
IS. A unit step change in input is lTl
ill
011 a 11lJlnhcr of processes (f--lV), rile '\~'illjtillg the plot below. Associate each process with a rCSpn!1Sl' ClIl'\t'
('line (kucl'l fr(jIll Pint
21'
L
<';(,\)
II
C:(.\ )
III
.l::(.\)
=
IV.
(;'(.1)
=
2\
.:s
I
I
2\
-1-(.\
-
!)
Is
2
"
d
a
15
0.5 0
05 0
19. ('ollsidcr (\
SC(nIH!
2
4
8
6
10
order transfer flllll'lion \vilh IlUlllcralm dYIl<.Ullil"';:
I)
11(' )
let 11 represent the qmtlkr denominator time constant. ,,\S';';\lllle a step change ill inpuL. Shc)\\' thal a max.imulll ill \'(l)/kj./{ occlirs if 'II T, and tkJl a llliIllJ1lLl111 (illdicating ill\ erst' rcspllllsC) occur.., if Til O. i\l..,o 'ibmv th,lt there is IlO L'.\lrellW ill Lhe '-tep respoll"e if 0 '11 To. (llim: Reali/e LhaL a lTw.\imUIll or rninilllllill oel'I.I!"" at \/1/1 ~ 0.) 20. ('ol1sidcr the tran"fer fU1Jction
g,,(s)
12.\1 2 -L I
Student Exercises
243
gJs)
a. Write the gain-polynomial form
b. Write the gain-time constant fonn g,,(s) r; (s') =
c. Write the gain-pole-zero form
'I"
k(T"S + I) r\,l + 2STS + k(TIJ ,"
h.l' +
+
I
1)(7 2.1' + I)
(.1'-1',)(.1'-1'2)
21. The reader should show how the firsl- and second-order Pade approximations relate to a Taylor series expansion. The 'I'aylo1" series approximalion to a time-delay in the Laplace domain is ,'--
1 - IJ.I' I-
(J2.S·2
O\J
2'
:V
+
O·ts 4
05S 5
4'
5'
+
06",6 61
-
Use long division of the first- and second-order Pade approximations and cOinmenl on the number of terms thai arc consistent with the Taylor series expression. 22. Consider the following interacting lank problem. Assume that the flow between tanks 2 and I is linearly proportional (rJ 1) to the difference in lank heights and that the outlet flow from tank 2 is proportional (f3 2 ) to tank height 2. Develop the transfer function models relating the inlet Ilowrate to both tank heights.
~
23. Consider an exothermic chemical reactor that has the f{Jllowing transfer function relationship hetween the inlet flowrate (input) and the reactor tcmpcrature (output). ~
2( - 2.5.1' I- I) v = '.' •. ...... () 9s 2 + 3s ,I, I
<'
The units of the input arc liter/min and the output is in deg C. a. Find the values of the zeros and poles. Is tbis systcm lIndcrdamped or overdamped? b. For a step input change of +3 liter/min, find how the Olltput changes with time. How much docs the temperature decrease before increasing? Compare plots of your analytical solution with those obtained using the MATLAB function step.
c. What is the ultimate change in temperaturc aftcr a long period or time'! d. If the steady-state input and output values (in physical terms) arc 10 liter/min and 75°C respectively, what arc the physical values of thc results in band c'!
244
Transfer Function Analysis of Higher-Order Systems
Chap. 9
c. If a step decrease in the input of ~3 liter/min is made, what would he the results in b, c, and d? 24. Consider a CSTR with a first-order irreversible reaction A --> B. The modeling equations are:
dCA~_ (F) ·+kCA + FCAr dl
V
V
dC" F dl ~ k C" ~ V CII
The following parameters and steady-state input values characterize this system: F
V k
= 0.2 min- l
cc-c:
0.2 min-- ' 1.0
gmol
liter
The input is CAl and the output is Cn. You should be able to show that the stcadystate values of (~ and CjJ arc 0.5 gmolliitcr. a. Show that the transfer function relating the feed concentration of A to the COIlcentration of B is:
y(.I") =
0.5
(5.1" + 1)(2.5.1" + I)
. . gmol B/lilcr .... 15---------------, and the tllne lIl11! IS Il1lIlutcs. gmol A/liler
where the gam
b. At time t 0, the input begins to vary in a sinusoidal fashion with amplitude 0.25 and frequency 0.5 min -J; that is, :;:0
1/(1)
~,
0.25 sin(O.s I)
Using Laplace transforms, find how the output varies with time. c. Compare your results in b with the integration of the modeling equations using the MATLAB integration routine ode45. Remember to use the correct initial conditions. Also, remember that the transfer function results arc in deviation variable form and must be converted back to physical variable values. d. Discuss how the amplitude of the output changes if thc input frequcncy is changed to 5 mitr-- t . 25. Oftcn higher-order process transfer functions arc approximated by lower-order transfer fUllctions. Considcr the following sccond-order transfer function:
245
Student Exercises 1
1)(11 1) Find the value of 'T in a first-order transfer function, Ij('TS + 1), which best approximates the step response of this second-order transfer function, in a least-squares sense. (Hint: Define an error as a function of time as e(t):::: Y2(t) - YI(I), where }'2 and YI are the step responses of the second- and first-order responses respectively. Find I \vhich minimizes e2(t) when t --> inf'.)
26. Consider a criticaIJy damped second-order system: ~(\) "
"
1 (71
+
+
1)( TI
1)
a. POI' a unit step input change (Au = 1), find the time at which the rate of change of the output is greatest (i.e., find the inflection point). b. Compare this rate of change with a unit step response of a first-order systelll with the following transfer function:
1 g(l) ~ (27,V + I) c. Plot the step responses for a and h, for sponses.
'T
= I. COinpare and contrast the n>
27. Pharmacokinetics is the study of how drugs infused to the body are distributed to other parts of the body. The concept of a cOinpartmental model is often used, where it is assumed that the drug is injected into compartment 1. Some of the drug is eliminated (reacted) in compartment I, and some of it diffuses into compaltment 2 (the rest accumulates in compartment I). Similarly, some of the drug that diffuses into compartment 2 diffuses back into compartment I, while some is eliminated by reaclion and the rest acculllulates in compartment 2. Assuming that the rates of diflu~ sion and reaction are directly proportional to the concentration of drug in the COI11partment of interest, the following halance c<.J.uations arise: ~ -(k,o
+ k 12 ) x, +
k 21 x,
+
II
where XI and x2 ;::: drug concentrations in compartments I and 2 (f.1g/kg patient \vcight), and u rate of drug input to compartment I (scaled by the patient weight. /-Lg/kg min). Experimental studies (of the response of the compartment I concentration to various drug infusions) have led to the following parameter values for the drug atracurium, which is a muscle relaxant: ::;0;
Transfer Function Analysis of Higher-Order Systems
246
(k lO
+
k,,) ~ 0.26 min
(k 2ll
+
k 2l ) ~
Chap.9
I
0.094 min
I
0.015 min
I
a. [,'inc! the poles and zeros of the transfer function that relate the input,
II,
to the
olltput, xl'
b. Find the response of the concentration in compartment I, Xl' to a step input (If I /-Lg/kg mill. What is the value at 10 minutes? What is the value after a long pc~ rind of time? c. Find the response of the concentration in compartment I, Xl' to an iII/pulse input of 10 f.Lg/kg. What is the value at [::: 0'1 'What is the value at 10 minutes'! 28. Consider the following delay-differential equations: dX 1
dl
~
-x,(1 - 8) I 1/(1) dx)
...
dt
~
-2x J
using the first-order Pacte approximation for dcadtimc, write the corresponding (approximate) pure differential equations. (llin!: define a new variable x3;:;;: xz(t - H).) Solve the equations using ode45, for an initial condition in all states, and a value of 1 for the input
or ()
MATRIX TRANSFER FUNCTIONS
10
Chapter 6 presented simple examples for transforming a state-space model to a single 11th order ditlcrcntial equation. Once the single differential equation was obtained, the methods of characteristics and undetermined coefficients (Chapter 6) or Laplace transforms (Chapters 7-9) could be used to obtain a solution. A general method for converting a slate-space model directly to the Laplace domain is presented in this chapter. With the transfer function representation, one can easily obtain the corresponding single nth order differential equation. After studying this chapter, the reader should be able to: Convert a state-spate model to a transfer function model analytically. Convert a slate-space model to a transfer function model using the MATLAB routine Sf32tf. Discuss interesting effects from pole-zero cancellation, The
rm~ior sections
arc:
10. J
A Second-Order Example
]0.2
The General Method
10.3
MATLAB Function ss2tf
247
Matrix Transfer Functions
248 The goal of this chaptn is 10 take
il
Chap, 10
gCllt'ral stdte-space model: x
Ax+Bu
y~CxjUll
and convert it to the matrix transfer function form:
yes)
<;(s) lies)
and usc this mode] to solve for the response;.; of each output to each input. \Vc will also usc this technique to easily find the 11th order differential equation corresponding to l'
10.1
A SECOND-ORDER EXAMPLE Consider the follo\ving tv\'o-statc, single-input. single-output model: dX I
=
dl
(111'\"1
-+
(1 1.',1.":,
-+
( I0 I ,
hi] II
dr")
{ 10.21
dl
( IO.JI
'faking Laplace transforms of (IO.l) through (10.3), \ve find:
s X,es)
',(0)
"" X,(') + II,. )(,(s)l iJ" (J(I)
s X.(s)
x,(o)
!I"
X,(I) + II" X.(s)
iJ", (J(I)
{ 10.) I
c,. X,es)
d" (J(s)
( 10(; I
Yes) _. "" X,es) + Assuming .\)0)
~
( I ()·l,
.1.':;(0) ;;:;: O. and rearranging:
(s -- II,,) X,es) - II" X.(s) (s ..
lI,el X,es)
-!I"
X,es)
iJ" U(s)
( 10.71
iJ" U(s)
( I O,~ I
III order [0 generalize this procedure later. we \vritc (10.7) and (10.8) in matrix form: (,,)]_.] 0" ..
(/: 1
(1)1 ~ (/., I I.\:,',,(.,)
11,",1 \
hill U(s)
1 h"
or, (/,Ii
1
(/) I
0',',1,\f [X,(S)] (/.. X.(I)
h" lues)
Ih
( I () _C) I
21
~~~~-~~----- ~,
Sec. 10.1
A Second-Order Example
249
and (10.6) is written in matrix form:
yes)
[e"
=
I
X,es) e 12 ] [X,(s) + <1" U(\")
(10.10)
We scc (10.9) is of the form: ~
(sl
A) Xes)
~
II U(s)
(10.1 I)
with the solution for Xes):
Xes) • (sl ~ A) 'II U(s)
(10.12)
C X(I) + D U(s)
(10.13)
and writing (10.10) as:
yes) combining (10.12) and (10.13):
yes)
c"
[C (sl ~ A) 'II
+ D] U(s)
(10.14)
recall that often D::::: 0, in which case (10.15) is written:
yes) •..• [C (sl
~
A) 'II] U(s)
(IO.IS)
or,
yes)
~
G(s) U(s)
(10.16)
In this example, since there is a single input and a single output. G(s) is a single transfer function, which we call g(s). The transfer function is the ratio of a numerator and a denominator polynomial:
g(s) •..•
H(s) D(s)
(IO.l7)
The reader should show that the polynomials in (10.17), based on (10.15) exercise 4):
H(s) ~
"l S
+ "0
afC
(sec student
(1O.18a)
D(s) = s' + <1, s + do
(HU8b)
where the polynomial coefficients, in terms of the matrix coefficients, arc: 11 1
=
flo =
ell btl
elJ [0 12
+ en b I, bll
-
{In
(IO.19a)
blJl +
C l2
ra
21
bl ] ~
alJ
h21 1
( 10.1%)
(lO.19c) (lO.19d) Since the inpuHmtput relationship is wrillen:
Matrix Transfer Functions
250
Chap. 10
Nix) . lJ(s) lis)
Y(s) We call further write:
Dis) Y(s) = N(s) Uis)
or.
The corresponding differential equation is:
till II,
dl
V·'/e no\\ have all automated procedure to find the transfer function for a single-input. single-output. two-slate S}'stel\1. An example is shmvl1 below.
Example 10.1
Linear Biol"l'ador \todd
('oll"idcl
rhe statc-space matrices arc
A=
l
II -II 7';1111
B
c
~
[)
[II I] II
Using the follt1\\ing steps [0 rind Gel) A)
(sl
c:;;:
C (sf - A)
!O.7~O(J
J
B: O.9().')()
I
s + 2.."640_
Recalling the simple method for ill\'t:'lling a 2 x 2 matrix. \\'e find: (s)
( (d
A)'
A)
J
s
I
j
2.:'i640 IJ.7SIII1
2..'1640 s
o 7';00\ I, \
0.7';011 II [
"l;l~21 S
0.67920
U.6792(J
2.:'1640 s
. .).XL).J
f
'
t
2.5640 s I O.h7920
Sec. 10.2
The General Method
251 3.RZ55.\ 1 1.14765
C(sl-A) 'I!
" 2.5640 s
+ 0.67920
so,
yes)
~
s/
3.RZ55 sl 1.14765 U(s) t- 0.67920
-t-- 2.5640 s
and we easily find that
d\' , de
+ Z.5640
dy ' dt
,j
0.67920 \' --
3.RZ55
~
dll dt
+ 1.14765
II
We generalize this procedure in SectioIl 10.2.
10.2
THE GENERAL METHOD Consider a general slate-space model with 11 stales, tit"1
=
dt
(lIIX I
+ (/I2X2 -+- ... +-
JJ1
Ol/J·T"
inputs, and r outpuls (see Chapter 5):
-+- h ll
u 1 -I ...
+
bllJlu m
(10.20)
which can he written in matrix fonn as:
i,
I.~" .,
[all
I =
_a",
a""
I [""
x"
+ -"", (10.21 )
Y',1 ~ Ie,,' [ \' c .-,
a'''.lx,
II
,'' 1[''] (,/1
_
\11
Id" d'l
d'2
252
Matrix Transfer Functions
Chap, 10
which has the form: x~Ax+Bu
(10.22) y~Cx+Du
where the dot over a slale variable indicates the derivative with respect to time. Recall from Chapter 5 that the eigenvalues of the Jacobian matrix (A) determine the stability of the system of equations and the "speed" of response. Now, taking the IJaplacc transform of (10.22):
Xes)
(.II
~
A) 'lIli(s)
Y(I) ~. lC (.II ~ A) 'II
+
DIV(s)
If D ; : ; 0 we can write:
Yes) "' G(s) lies) where:
(I'
G(s) ~ X Ill) (I'
C(sl~A)'ll X
n)(n
X
n)(n
X
Ill)
The trailsfcr function matrix, G(s), is:
gll(S)
G(s)
r
g" (s)
g pes)
g,,(s)
g'm(s)
I
R,m(s)
Notice that G(s) is square if r:::: m (number of outputs::: number of inputs),
10.3
MATLAB ROUTINE ss2tf The routine ss2 t £ can be used to convert a state-space model to a transfer function
model. After entering the A, 13, C, and D matrices, the command: [num,denJ~ss2tf(A,B,C,D,m)
will generate the numerator and denominator Laplace domain polynomials for the transfer function between input number m and the outputs, in descending order of s.
EXAMI'LE 10.2
Example 10.1 Using MATLAII ss2tf
Here we consider the linearized biorcactor model, with two inputs. The first input is dilution rate, the same input llsed above. The second input is the substrate feed concentration. We will also consider both state 1 and state 2 to be outputs, and modify the C and 0 matrices so that ss2 tf provides the transfer functions between the input and both outputs.
Sec. 10.3
MATLAB Routine [3,,2,,[ »A
253
[0,0.9056;-0.'15,
=:;
2.5640]
A -
a 0.9056 -0.7500 -2.5640 »
B=-
[-'1.5302,0;3.8255,0.3]
13
" c c
··1.5302
o
3.825'3
0.3000
[1,0; 0, ] ]
-
c:o
0
1 0
"
D
D
""
1 [0,0; 0,0]
0
0
0
0
Input I The numerator and denominator polynomials relating the first input to the two outputs arc found using the following command: »
[nuffi,den] "",ss2tf (A,B,C,D, 1)
num = -1 .5302 J .8255
[]
0
-0.4591 1.1476
den
1.0000
2.5640
0.6792
where the first row of the num matrix is the coefficients of s in the gil (s) polynomial. in dccreas~ ing order from left to right. Similarly, the second row of the num matrix is the coefficients of s in the 1;'21(8) polynomial, in decreasing order from left to right:
-1.5302 .I' - 0.4591 + 2.5640.1' + 0.67920 )',(.1') ~ g2l(s)
"l') ~
.1"
11 (,) l'
3.8255.1' + 1.14765 + 2.5640.1'1- 0.67920 ",(.I')
We realize that the eigenvalues of the A matrix and the poles of the transfer fUllctions will be the same. 'fhis is verified by the rooLs and eig cOll1mands
Matr"ix Transfer Functions
254
Chap. 10
(den)
((lot-,; dll;:;
O.30()O::1 »
eiq (al
an,; . 3 () CJ CJ
.) . /. 6 ,} ~J We
call
ahcl
\\I'ilt'
the Ir;lTlsfcr
fUIlCliulls
ill
POk-/t~l\)
fOHn"
I ,,"iJ02 (s +- O.J)
(s 1\\herc
\~C
+- 2.2(40)(0"
+-
3)Q55 (s
0._')
is +
',-.11 1(,\) 11.,1
!J'''.
~.jA(I)(.,
ha\'c the interesting result (11:11 tile lern
III(S)
0.3)
canccl~
(jill' uf the pll1cs ttl yield firsHll'clc1
sy"kIl1S
(s
j.
).5J02 2.2(40)
3./{255 (s + 2.1(40)
/I
(s)
1_
II,(S)
0.6759 0.4417.1'1 1 11 1(.1') 1.()S()7 1/,
(1·)·)17 \
(s)
1
\Vi.-' \I.ould notice the Il'ro-pok cancellation if we ;\Is() lIsed the TOC)t~3 cOll1llwnd \() find the of the IHlIJ1Crator polynomial
fOOls
q'Tot:-? (nurc(l,:))
- () . 3 C1 (I 0 '>rU(I\ '; ([;1 :Ti{~;, :))
-().~jOCC
and
\\"l'
sec that tbe rool of the Jlllll1craloJ pulynomi;l! is the same a" one of the
nOlllinalor p()Jynomial.
nl(l!s
of the de-
Sec. 10.3
255
MATLAS Routine ",,2tf
Input 2 The numerator and denominator polynomials relating the second input to the two outputs arc found using the following command:
»[num,den]=ss2tf(a,b,c,d,2) num : : :
o
o
0.2717
o
0.3000
o
den
2.5640
1.0000
0.6792
,-tnd we have the result that;
v,(s)
= g,,(s) lI,(s)
y,(s)
= gn(s) lI,(s)
+
0.2717 2.5640.':1 -t- 0.67920
( ) 1/
2
0.3 s
+ 2.5640 s
s
( ) I ().t,7 ')7l) _ 1/, s
The relationship between the second input and the second output is particularly interesting. The secoodinput has no steady-stale effect on the second output, as can he seen from the final value theorem. Assume a step change of magnitude !YJ.u z in input 2.
y(t-,
10.3.1
~) =
s yes -) 0)
0.3 s au} -- 0 ( 2.5640 s I 0.67Y20 s
Discussion of the Results from Example 10.2
THE FIRST INPUT We noticed that the transfer functions with respect to the first input had pole-zero cancel~ Jation. This created an input-output relationship where the step response is faster than would be expected, because the slow pole was canceled by the process zero
y,(s) = (.I'
- 1.5302 (.I'---+ 0.3)... !I, (.I' ) + 2.2640)(.1' + 0.3)
3.8255 (.I' + 0.3) ( ) Yo (.I' ) = (sf 2.264(1)(;+ ()3) !I, s This can also be seen using the
gain~time
constant form:
····0.6759 \.,._,_,_,_, .I' + yJC,\') = (().44·17 ",. + 1)(3.3333 s + '1) !I,(s) 1.6897 " ..._., ....
y,(s)
=
.I'
+
1
(0.4417.\. + 1)(3.3333 s + I) u,(s)
Matrix Transfer Functions
256
Chap. 10
2 ,- --- --- - -- ---1.5
:;,..,
y,
0.5
o -0.5 -1
o
1.5
0.5
2
time
FIGURE 10.1
Unit step change in input I.
or,
- 0.6759 y,(s) = 0.4417 s + I LI,(S) 1.6897 y,(s) ~ 0.4417 s + I fl,(S) The step responses for a unit step input change are shown in Figure 10.1.
0.4
y, 0.3
2:
"
0.2
0.1
a
0
5
10
15
20
FIGURE 10.2 Unit step change in input 2. Notice that the- steady-state value of Y2 docs not change.
Student Exercises
257
THE SECOND INPUT Notice that input 2 docs not have a steady-state effect on output 2, only a dynamic effect. This can be seen hy using the MATLAB step function, then plotting the results (sec Figme 10.2).
Dum 0 0
0 0.3000
0.2717
1.0000
2.5640
0.6792
o
d.en »[y,x,t]=step(nuffi,den)
plot(t,Y)
SUMMARY We have shown how to converl a state-space model to a transfer function model, for IllUItiple inputs and outputs. We have also seen some interesting results regarding pole-zero cancellation. One has to be particularly careful with pole-zero cancellation if a pole is UIlstable (positive), as will beshowll in Section 11.3. The following MATLAB routines were used:
ss2tf:
converts state space to transfer function form
eig:
matrix eigenvalues
roots:
roots of a polynomial
STUDENT EXERCISES 1. Compare the step responses of the following three transfer [unctions: 1 a. gt(s) ~ (0.4417 s + 1)(3.3333 s + I) 1 b. g2(s) ~ 0.4417 .\' + I c. g,(s) .
~
1 3.3333 s
+
1
Which has a faster step response? Why? 2. Consider the following state-space model (a 5-stage absorption column)
Matrix Transfer Functions
258
A=
B=
- 0.325 0.2 0 0 0 0.2 0 0 0 0
OJ25 -0.325 0.2 0 0
0 0.125 - 0.325 0.2 0
0 0 OJ25 - 0.325 0.2
Chap. 10
0 0 0 O. I25 0.325
0 0 0 0 0.25
a. Convert this model to transfer function form, assuming that all of the states arc outputs, using ss2tf.
b. Find the response of all of the states to a unit step in input I. Usc the fUllction step. c. Find the response of all of the states to a unit step in input 2. Usc the function step. d. Compare and contrast the curves from band c. 3. Consider the following model f()f an isothermal CSTR with a single ltTcvcrsi ble reaction (sec Module 7). Find the transfer function matrix relating bOlh inputs to both
states.
dX,] ai dX2 dt
I
x,
~ O.2IU + 1- 0.5
-OA = [ 0.2
O'
0.5
0.211' o //'.1 II)
2~state, single-input, single-output process, derive the relationships shO\\1l in (10, I 8) and (10.19).
4. I"or a
5. For the following state-space model, find the transfer function matrix rcl'llill,!.! all four inputs to both outpulS,
A=
-OA
I3 B_[0 - SD
C= [Io
D.3.1
-4.5
-7.5
D.l
D
D
D
1.5
I
259
Student Exercises
6. Consider the following state-space Illodel:
at-
III
5
_ 1
=
10
OJ 1 2
\,
[\,1 + [
5
7 ] 2 25
II
}'::::: x 2
Show that the eigenvalues of the A matrix are -liS and 112, so the system is unsta~ hie. Also, plot the step response. Derive the transfer function relating u(s) to yes) and show that the unstable pole is cancelled hy the positive zero. This problem will be analyzed in more detail in Chapter 11. 7. Consider a chemical reactor with bypass, as shown helow. Assume that the reaction rate (per unit volume) is first-order (r = kC 1) and C 1 is the concentration in the reactor (the reactor is perfectly mixed). Assume that the volume in the reactor (V) and the feed fJowrate (F) remain constant. The fraction of feed bypassing the reactor is (l ~ a)F and that entering the reactor is of. Assume that the fraction bypassing the reaclor docs not change. 'l'he inlet concentralion (C in ) is the input variable and the mixed outlet stream composition (C2 ) is the output variable. Write this lllodel in state-space form, using deviation variahles.
x'-:::Ax--I-Bu
y=Cx+Du
(1_(<)
F (X
F
F
C,
Find the transfer function relating
If
C2
to y.
For the following parameters, simulate a unit step response.
F = J() lim in, V = 1001,
("n
= 1 gmolil, ,,= 0.5, k = 0.1
min-I.
260
Matrix Transfer Functions
Chap. 10
8. Consider the follmving set of series and parallel reactions (from Module 7)
kj
A+A~f)
Material balances
OIl
components A and B yield the following two equations:
dell F .. ~ ('C 1,) + (k l C..I 'k,'cI') dt V ' \vhcrc the rate constants are: 'i
k = ' 1nio' I 6
j
.1
Itllll
liters
k,
6 mol min
and the steady Slate feed and reactor concentration of component A arc: III
IlHl]
liter
CA\
1
mol
. liter
n
a. Find the steady-state dilution rate U/V) and concentration of (shO\v all unih). h. Linearize and put in state-space form (find the numerical values of the A, B, and C matrices), assuming that the manipulated variables afC dilution rate (FIV) and
feed concentration and that hoth states are outputs. c.
l~'ind
the eigenvalues (show units).
d. Find the transfer functions relating each output to each input. Find the poles and zeros for each transfer function and make plOlS of the responses to unit step changes in each input. Comment on your results.
11
BLOCK DIAGRAMS
The objective of this chapter is to introduce block diagram analysis. After studying this chapter, you should he able to: Analyze the stability of a block diagranl system.
Understand how inverse response processes can arise. Understand potential problems with
polc~zcro
cancellation.
Write a set of differentia] equations to simulate systems modeled by transfer fUllctions in series.
Usc the MATLAB routines roots.
sE~ries,
paral1(~1 J
fC'cdback, conv, and
Use Sll'vIUIJNK for block diagram simulation.
Major sections of this chapter arc: 11.1
Introduction to Block Diagrams
11.2
Block Diagrams of Systems in Series
11.3
Pole-Zero Cancellation
! 1.4
Systems in Series
11.5
Blocks in Parallcl
11.6
Feedback and Recycle Systems
11.7
Routh Stability Criterion Applied to Transfer Functions
t 1.8
SIMULINK
ESCOLA DE ENGEN HARIA 13IBLIOTECA
261
262
Block Diagrams
(
)
FIGl'llE 11.1
y(s)
Chap. 11
Hinck diagram n:,prl'
~cntatlun.
\\'\, han: shown 110\\- Laplace transforms (irc Llsed to reduce differcntialcljuatiollS to algehraic l"clatiullships. Algehraic equations arc mllch casier to Irlanipul ethan difrl'rt'l1liall~qll{\tiol1s. Similarly, hlock diagrams allO\v us to easily Illanipu . complex modds that ,Ire cumpo...;cd of subset'; of simple illudels.
iNTRODUCTION TO BLOCK DIAGRAMS ('oll\idcr a standard first-ordn process
1ll()(k~l:
!lr(i) T
!II
+
r(l)
k
( I I. I I
11(1)
\\hich h,j':; the transfl'r fUllction Conn:
r(s)
( lUI
g(s) 11(.\)
( 11.31
g(s) 'TS+
engineers usually try' ,1Ilel solve problems b) 'ike-telling diagrams to undC)";-,lal1d rL'1iltionships. Process control engineers usually lise block diagrams to urllkr.:;lllIJ the illpu["output relation"hip" in a dynamic system, A block cli,lgrarn represcntation \1!'(11.2)isshowninFigure 11.1. \\\~ can sec that II(S) is tIll' input to the trallsfer functiun block
lilpLiL-OUlpUI
!iLOCK DIAGRAMS OF SYSTEMS IN Sl:RIES ('ullsidcr 1l0\\ the block diagram rcprc"entation of t\\'o prucesses in "eric" i ;gme 11.2.
u(s)
--.~I FIGliRE 11.2
9, (5)
I
2(5)
~I
-9, (5)
If------.
Block dirlg'ill11 olIW" I"nccsses ,n
"""'5.
Y(5)
,IS
shown ill
Sec. 11.3
Pole-Zero Cancellation
1'hc input/output transfer function of Figure 11.2 is:
y(s)
0
g,(s) z(s)
cO
g,(s) g,(s) 1/(.\)
or,
y(s)
~
g(s) I/(s)
( 115)
g,(s)g,(s)
(Ii,.)
where:
g(s)
c=
If the two transfer functions arc first-order: (Iii)
and ( 118)
then the overall process is second order:
g(s)
I< (T,sf I )(T2S + 1) I<
where:
~
( 1191
1<,1<2
'rhe same idea can be continued for any Dumber of transfer functions in series. The ~~;tl!~ dent should notice that the poles of 1.1 system composed of many lran:':lfcr functions in series arc simply the poles of each transfer function. This leads to the following conclusion about tbe stability ofsysfems lvi/II tran.~ler.fllnctirms in series:
If([ system is composed (?(traJl.~r('rli/ll('ti()lIsin series, (llId ifal! r?!'t!wse tralls(cr/il1l(,t;o/iS (/ie s/(/ble, then the overall system ;,<, stable,
Also, the 7.eros of a system or transfer functions in series are simply the zeros of the inchvidual transfer functions.
11.3
POLE-ZERO CANCELLATION Again, in this section we consider two blocks in series, as shown in Figure 1 J .2:
y(s)
Oc
g,(s) z(s)
~
g,(s) g,(s) 1/(.1')
(1l.IO)
Block Diagrams
264
Chap. 11
If WC ;lre not careful. we Gill O\t'rIUUK possihle prublems \\ith sy'>lCllh in scries, if \\c look onl> at the u\'crilil input/output rclati()lhiJip. 111 the Ile\! cX<JI11pic \\c slw\\' problems \\-jtll f)ofL <em ('(f!lcellu{ioJ).
--------Lead/Lag in St'ril's with Lnstahle First-Order Systl'lll
EX"'IPLE 11.1
Cu[]sidt'f the' follO\\illg lead/la~ in s(Tic" \\ itll <111 L11l.\['lhk' first-order' ": :'lCllI
I
2.\ g:(s)
(11.11 )
I
).,
( Il.l .2)
2\ -
'I 1/(\)
I·
"1'(,\)
g{i) 11(\)
I
I
II(S)
\\'(' Intl"t ['caJIIl' Ill:]! tllese tr,lIlsl"cT fUJlctiolh UllimiltL'l: ['cprt'SC111
(l
I I I _~ )
phY\lcd )'r'uccss III pracrin'.
ph) "iedl j!iH,II11('tCI'S <.-'annul hl' kll(1\\'ll pnl't~(11) Whal tillS I11C,II1S h Ill,]l gt'll(~r(lll) the 1l1l111Cf<.llu]
llf g (\\ \,ill I1Pl ('\;It:ll) cmCc'l the dCllOminatur of g~(sl. in pr:lctin'.
COllsjtkr II n:ali:-li . . . (,;dSI:\ \\ h.. . re .t.':(si ha) ,[ sl,iglH t:rrOl in the \'alul' (If till' poll' t;,(s)
thell \\ C
find
I -- :U)()O!l -'--- I
ill 14)
thdl
2\ I'{s)
.:.()OUll
'\Olll..:(' th,ll \\ IlL'll \\l' do
nnt
11d\<"
I J
--.:\
I{S) "---' g(s) 11(\)
I I ~I
I (),()Ons\~
pnfcd p\lkllt:n\
Cdlll..:l'llilIHlIl.
illput!PlItPlit relationship. \(sl= ,L:II) 11(1 I. Our goal t1ll\\ i\
\(J
thnc
i~ :111 ull\labk pok in tlh'
C()lllp;tl'l: tll(' I"l'SPOIlSC\ 01
thc 1\\0
11HldL'I\ I 11 1J) dllt! ( I 1.1,"'\), LCl II (\ ) I"cpl"CS,'lll lhl: (lutpul ill ( 1 I.I,~I ;\Ild \'JI) rl']1rL'~cnt tile \lUlpUl in I
11
I:' I
\\~lllllillg:l unll \kP illpUl. 11(1);;::
\(5s
\'(1)
Ill.
I)
I I I 1(1)
111 17)
Pole-Zero Cancellation
Sec. 11.3
265
Also,
y,(s)
~
-2\'
s(
+]
~I (J.60lJSs' -:;:
2.9999s + I)
(lUg)
___ 0.0001 7.0001
(11.l9)
which has tbe time domain solution (Chapter 9):
y( t ) " I 2
7 e'/5 7.(JOOl
er/2.(l{)(Il
and we call sec that, at low t, (ll.19) is almost identical to (11.17). As time increases, however, the unstable exponential tcrm in (11.19) begins to dominate. This is shown dearly in Figure 11.3.
Y1' stable
0.5
o
-0.5
o
5
15
10
20
25
time
FIGUREl1.3
Comparisollof(11.17)and(1119).
Note that if we had used the state-space form for the model represented by equations (ll.JO) through (Il.J 2), we would have discovered the instahility, cven for the perfect parameter case. The following example analyzes the state-space form of l~xamplc 11.1.
EXAMPLE 11.1
Continued State-Space Aualysis
Refer to z(s) as the output of the lead/lag block. FrotH Chapter 8 we find the following statespace realization of the lead/lag:
dx dt
z;::~
x+'T!'u 'Trt
'T(/
Block Diagrams
266
T"S
+
T,t\
+ [
Chap.11
, so:
5s ~ . I dl." dl
+
\
)
7
(11.20j
II
)
2 ,\
(11.2 J)
II
5
)
(Illd lhe stall' space rcali/ation of the ullstable lag is:
,I
dv dl
\'
Suhstiturill12 (11.21) into (11.22). we find
I
dy
\'
dl
III
2
x
+
( 11.23J
II
)
rf we usc notation
r 1 :=:.\ Xl
c::;;:
Y
\\c can \\rile ( II 20) and (11 2J) in the following form:
7 5
dt j 5
dt dx-.
I 10
2 x,
dl
Xi
(1 J .24)
II
)
( 11.2'1)
II
Using the usual :,talc-space notation:
x
Ax
Y
(.'
Bu
xl l) u
\Ve \tTlle
(hi
dl
:"
dx,
I I()
1_ dl
~]i:' I
7
I
1J
( Il.2hl
II
\\it' c
Sec. 11.4
267
Systems in Series
The previous example illustrates the importance of not cancelling an unstable pole with a right-half-plane zero. It also shows how slalc~space analysis can always be used to address the stability of a system.
11.4
SYSTEMS IN SERIES The dynamic behavior of chemical processes can often be represented as a scries of simple models, such as first-order transfer functions. As an cxmnplc, consider the following process, which is characterized as II first-order processes with a gain of 1 and a time conslant of 5:
I g(s)c (5.1 + I)"
(I 1.27)
The step responses for 11 I to 5 arc shown in Figure 11.4. Notice the characteristic S-shape for all orders greater than 1 and the additional lag associated with each higher order. ::::;0
11.4.1
Simulating Systems in Series
Although we analyze processes lIsing transfer functions, to obtain time domain responses we must usc a IllHllcrical integration package. Consider a system of II first-order processes in series, as shown in Figure! 1.5. IJere we write the sct of ordinary differential equations that describe this process.
'fhe ODE describing the first process is: dX 1
dl
T,
Xl
+
k,
T,
u
( I 1.28)
1 (53 + 1)'
o~L.L~~~~--c:::---:'::------:c:::------:l o 5 10 15 20 25 30 time
FIGURE 11.4.
Step responses of first-order systems in series.
Block Diagrams
268
XIL-lS)r----'XIL(S) )~S)
gutS)
FIGURE 11.5
n processes in series.
Notice that we can think of the output of the first process as the input to the second
process: k
72
x)
+ 7 z X'
( 11.29)
j
2
and so on through the nth process: ( 1130)
To solve (Il.28) through 01.30) we can lise the numerical integration techniques developed in Chapter 4 or the analytical expressions developed in Chapter 6. We can also write (11.28) through (! 1.30) in the following state-space form:
dX j dt dx z dt
dX n dt
dX n
,Ii
-
1 T,
,
0
k2
1
T2
T2
0
x,
k
0
0
x2
0
1
0
0
0
0
,
0
Tn
T,
+
,
kl/
X"
Tn
u
( 11.26)
0 Tn
-
X"
,
0 0
In Example 11.2 we show how to usc the MATI,AB routines series and conv to find a transfer function that represents two blocks in series.
EXAMPLE 11.2
Two Transfer Functions in Series
Consider two processes, g](.I') and g2(s), in
g,(s) ~
,2
SCrlCS,
~
where:
1.5
2, + I
3 (s) - 4s +
g(s)
~
g/,) g,(s)
Sec. 11.5
Blocks in Parallel
269
We use the following MATLAB commands als for each transfer function:
(0
enter the numerator and denominator polynomi-
»numl
[1.5]
j
>:> denl
[2 1]
j
»num2
[3];
» den2
-
[4 l];
the series command generates the llurucrator and denominator polynomials for the transfer
function g(s):
»[num,den]
:'3cries (numi, denl, num2, den2)
Dum :::
o
0 4.5000
den : : : 8 6 1
which iildicatcs that
M(s)
~
4.5 0
gs~
1- 6s 1- 1
cony cony is llsed to multiply two polynomials. Using the previous example, we multiply the numerator polynomials to find: »
num
cony (nwnl num2) J
nwn 4.5000 and the denominator polynomials to find: »
den
den
=
COIlV (deni , den2 )
~
861
11.5
BLOCKS IN PARALLEL Sometimes the behavior or a chemical process can be modeled by transfer functions in parallel as shown in Figure 11.6.
270
Block Diagrams
Chap. 11
y, (5)
9, (5) +
u(s)
Y(5) +
9, (5) Y2 (5)
For this sy.'stem
\\'C
can
\\Tite
FIGLREI1.6
System"
the total output. r(sL as the sum of two outputs,
In
parallcl.
\'1(.1)
+ Y:::(.I), (II ,I) {1112}
or, 1"(S)
{ I I .11}
g(s) u(s)
\\'here:
( 11 ,3~1
Consider the case \vhere
gl(s)
and
g2(s)
arc fir'>t-ordcr lramJcr functions: (1115)
'loS
-+
( II,ft) 1
so
fl..
g(s)
(1117)
TjS
Developing a common denominator.
\\'C
find:
(] J
.38)
Notice that (11.38) has the form (sec Chapter 9): g(s) =
_ (T,\
kiT,s
i
I)
',I
+
1) (T,S
+
( 111'))
I)
5ec.11.5
271
Blocks in Parallel
where:
" ", ~
T"
",
",
(I lAO)
I 1<2 j
k2T l
I
"2
(I IAI)
We will assume thallhc transfer functions gl(s) and g2(s) arc stable, so Tl and T2 >, O. The goal of this section is to show a system where inverse response (discussed in Chapter 9
and Example 9.3) behavior can occur. 11.5.1
Conditions for Inverse Response
Recall that a transfer function will have inverse response only if there is a right-half-plane (positive) zero. Since the zero is "-117/1' this system will have inverse response only if Til < O. We find that Til <:; 0 only if k j and k2 are of opposite sign. We can arbitrarily assume that k 1 :> 0, which means that k2 <: 0 is necessary for inverse response. For inverse response, the condition:
means that
kl
+ k 2T l <0 + k]
or,
"
<:
", + ",
", + ",
which yields the following conditions for inverse response.
1. If k J + k2 ::> 0, thcn k[I2 <
kzlk[ for
m~
which implies that 7/11 must be -:> -kik] for
in~
k211' which implies thal 12 fT [ musl be./
verse response. 2. If k J + k 2 -< 0, then kJI2 ::> verse response.
~~k211'
Physical examples of systems with inverse response include: steam drum level, rcboilers in distillation columns, chemical and biochemical reactors. A reason that inverse response behavior is important is that it creates tremendous challenges for tight process control. We can usc theMA'fLAB routine parallel to simulate lwo systems in parallel, as shown by the next cxample.
EXAMPI,E 11.3
'1\\'0 syskms in parallel
Considcr Ihc following systcm of (wo first-order processes in parallel (Figure 11.7):
g,(s)
2 5s +1
Block Diagrams
272 8,(S) -
g(s)
~
Chap. 11
-I lsi 1
2
5s +
-1
f
Is
+1
»num!
[2] ;
» denl =
[5
1];
»
num2 -
[~lJ;
»
den2
[1
1];
The following command is used to fiud the new transfer function: »
[num, den]
;0:
parallel (numl,denl,Ilum2,den2)
nulU = 0 -3 1
den 5 6 1 »
stE~p(num,c1en);
[y,x, t]
»[yl/xl]
sLep(numl,denl,t);
»[y2,x2]
step(num2,den2,L);
»
p1ot(t,y,t,yl,t,y2) 2
y,
y
y,
2
o
FIGURE U.7
5
10
15
20
25
30
Two systems in parallel that have an inverse response when
added together. ••..•__•••
1
Sec. 11.6
Feedback and Recycle Systems
273
This previous example has shown that inverse response occurs in systems where the gain of the "s(ow process" (larger time constant) is larger in magnitude (but opposite in sign) than the "fast process" (smaller time constant).
11.6
FEEDBACK AND RECYCLE SYSTEMS r,'ccdback systems arc common in engineering. Examples include chemical and biochcllli~ cal reactors, where a certain portion of the product stream may be recycled to the fccdstream. Feedback naturally occurs in most "self-regulating" models where, for example, the rate of change of a state variable (say, concentration of A) is a function of the same or another state variable (say, concentration of B). The entire field of process control is based on the concept and theory of feedback systems. Our goal with this section is to introduce feedback analysis and, in particular, stability analysis of feedback systems. A block diagram of a feedback system is shown in I·'igure 11.8. In this figure, the input to the feedback system is res) and the output is yes). Here we develop the relationship between res) and yes).
yes)
~
g lv) /1.(.1')
(I 1.44)
but
1/(.1')
~
r(s) + z(s)
(l1.4S)
and
z(s)
~
g,(s) yes)
(I 1.46)
So we can write (11.44) as:
yes)
~
g,(s) (r(s) + g2(s)y(S»
( 11.47)
yes)
~
g,(s) .. res) I - g,(s)g,(s)
(11.48)
y(s)
~
g,., (.I') res)
(I 1.49)
Solving for .y(s) we find:
Notice that we can view this as:
g (\) " '
where
u(s) r(s)
+
+
~
'--
z(s)
~
g (s)
g,(s)
---------I - g,(s)g,(s)
(11.50)
I--
1
g2(s)
f4FIGURE t 1.8
Feedback diagram.
Block Diagrams
274
u(s)
r(,) +
L....-
z(,)
r(s)
g (,)
y(')
1
+
gz(,)
Chap. 11
f-
----..I_g_cl(_S)---.JI-----.
y(S)
FIC;.(JU.E 11.9 diagraills.
Equivalent block
and we know that if the poles of gcl(s) arc stable, then the feedback system is stahle. \Vc realize that the two block diagrams shown in l·'igurc 11.9 arc equivalent.
EXAiVlI'LE 11.4
Feedhack syst('Jll
Consider two first-order process transfer functions:
(11.51 ) k2 '2S
( 11.52)
_of
glv) 1i,(s)Ii,(·I)
I I) - k,k)
kl(.;s (T,S ,- I )(T)S k,
('1T,S2 -1 ('I
(T).I -j
,
I)
'2)sl
k1k!)
(11.5.1 )
and Xd(s) is stable if the roots of '1'2S2 + ('I] + T2 )S + J - k 1k 2 arc stahle. We recall frum the I{outh stahility criterion that aJi of the roots of a quadratic polynomial afC negative if the l'oelli cients of the polyllOlniaJ are positive. If we assume that 'I and 1"2 arc positive, then (11.53) will be stable if I k1kz is positive. For stability, then, k 1k 2 must be less than I. [,d's consider the folJowing IlUinerical example:
Sec. 11.6
275
Feedback and Recycle Systems
g,(s)
k] lOs 1
Since k l ;;;;; 2, then k 2 must be less than 0.5 for stability. As a Ilumerical chc(k, lei k 2 :;;:; ~ I. Solving for the roots of T1lzS
l
+
(T I
+ 7))S
1
-j
k/<2
0
we find
50."J + 15s-1 3
0
which has the roots (using the quadratic formula)
-- 0.151 0.1936j (we can verify this result using the MATLAB routine roots) Since the real part of the rools is negative. the system is stable. This is verified in the MA'rLAB simulation presented in Figure I J.I 0, where the response of the output to a unit step change in r is presented.
1.2
0.8
'"
0.6 0.4 0.2 0 0
20
10
30
40
time
FIGURE 11.10
Step response for the example feedhack system.
We can also Lise the MATLAI:3 feedback function to obtain the closcd"Joop transfer function, as shown below.
2 5s
-I lOs +
Block Diagrams
276 numl
-
»den]
=
"
den?
Chap, 11
] ;
[5
'-'
1];
1I ;
and. using lhe rtlutillC fcC'db,)ck " nUllt
[nurn,c]cn] -
den
S (j '\)v'l'
T ')
3
use the routine :;tep to find the step respollse: [y, .":, L I
~,;t ':~'I)(nutn,dcn)
which gives the plut shuwl1 in Figure 11.1 n.
11.7
ROUTH STABILITY CRITERION APPLIED TO TRANSFER FUNCTIONS Recall from Cb,OIptcr A that the purpose of the Routh stability criterion polynomial with the following form has any positin:' roots:
IS t()
determine if a
II Since trallsfer functions that have dCIl()lllinator polynomials in the Laplace transform variable (s) are arc the S,lIl1C form (IS (I J .54). \VC can lise Routh i1naly'sis tu determine the S[dbility or transfer rUlltioJ1s. As before, assume that (/11 > O. If (//1 < 0, then rmtltiply' (J 1.54) by ~ I. A nccl?sswy condition for stability is that all of the l'{)efficicnts in (11.54) must Ix positive. If any of the coefficients arc negative or zero, then at !cast onc pole (root of the characteristic equation) is positive or lero, indicating that the equation is unstable, Even if all of the coefficients arc positive. \vc cannot state that the systcnl is stahle. \Vhat i" needed is a sl{/I/cient condition fur stability'. 'ro deterrnine that the system is stahlc. \\(' must construct the Routh alTay and llSC the Routh stability' criterioll. which provides nccessary and sufficient conditions for stability. Sometimes we simply \vish to determinc if a particular systcm is stable or noL \vithout actually evaluating the eigenvalues. This is particularly true if we Vv,'ish to determine valucs of system panlll1ClCTS that \vill cause a SYSll'll1 to lose sUlbility 'Thi::; appnl,lCh will be useful in performing a bifurcation analysis in later chapters (14 and 15), and in LUlling control systems for chemical processes.
Sec. 11.7 11.7.1
Routh Stability Criterion Applied to Transfer Functions
277
RouthArray
rf all or the coetlients or the characteristic equation (11.54) are positive, then develop the following ROllth array:
Ro\v I
an
(In- __ 2
(ln4
2
{fll_!
an
(111_ )
3 4
", '"
172
-3
h3
(;2
J/+I
where 1/ is the order of the polynomial. Notice that the first two rows consist of the coefficients of the polynomial. The clements of the third row arc calculated in the following hlshion: ({n-2 --
a/PII
(lll_'_
!
{In __ ]
and so on.
[~lcmcnts
of the fourth and larger rows arc calculated in a similar fashion:
and so 011.
A sufficient condition for an rools of the characteristic polynomial to have negative real paris is that ali of the elements in the first colunm of the Routh array arc positive.
EXAMPLE 11.5 Uouth Array to Determine Closed.Loop Stability Consider the block diagram of figure 11.9.
yes) or,
yes)
g,,(s) res)
where:
g,(s) I - g,(s)g,(s)
Block Di8grarns
278 And the trails!''')
fUlll'llOJlS
Chap. 11
are:
I )(1,
( )S
~
I)
A lOs
Our g(lal h ((i r-Iml k, til (lSSUrl' __ lahility nl the clnscd-luup We e:lsil) find the traw,fci l'ullclinil. gi.\):
"111h
g(s)
I} ISs --~
9).\"
( I50s'
s~'-.tl'1ll
2/,;,)
\\hkh has Ill(' c!l,u-ilctcristil' pulYJlolnial
150s' --" 95\ \\ llich
j"
?k,
lSs
of the rlmn
150
IN
,
\)5
I
1>1
II
4
cl
2L
C()I1c1lliull is that all (II > n, Ilhidl is sari'-Jit'd if J - 2k, > 0, u!" k, ().:'i. 'I'll(' sufficient conditiull is s
ilCCC,I.\UrI
(u/
h
1/,11"
a
hi 1."1
-
IX
ISO
90
a)) ,
(1,-,
(III
hi
-
II
2k
2k ,)
II
II
The hi l'lillditioll is satisfied if k~ > :=;) while the ('I cUlldlljOI] is the ditioll. \Ve then kl\l? the folluv,'illg rcstrictioll on k, fur st;lhility'
5.2
11.8
\(lillt'
as the necessary l'(lll
0.5
SIMULINK III the prC\ious sections \VC 11;\\c shown hO\\ \1XfLAB routines can be' usnJ for hlod; dl agr,iIll analysis and simulation. The ohjeetJ\"c of this sec-tirHl is to Lise the bind dl
Summary
279
Clock
time
ff------j
;=;G~s---I~
Step input
.J'J output
g2(s)
FIGlJRE 11.11
SIMULINK block diagnllll for two blocks in parallel.
Consider the block diagram system from Example 11.3. A SlMLILJNK block diagram is shown in l~'igure 11.1 I, Notice the use of step, transfer function, sum, workspace and clock blocks to generate the nccessary input and output information. The parameters menu is used to specify the intcgration type (L1NSIM), final time (30), and minimnll (0.01) and maximum (I) step sizes. The results arC the same as shown in r'igure J 1.7. More information on SIMULINK is provided in Module 4 in the final section of the text.
SUMMARY Block diagram analysis is irnportant because it allows us to think about a system of processes in terms of a cOlnbination of the individual processes. We have shown how to analyze the slability of a bloek diagranl systelll, particularly if there are recycle or feedback processes. \Vc have shown how inverse response processes can arise from systems in paralleL We have also shown potential problems with pole-zero cancellation when analyzing transfer functions in series. The following MArLAR routines were used: series:
conv: parallel: feedback: roots:
two models in series (either transfer function or state space) multiplies two polynOinials two models in parallel (either transfer function or state space) two models in feedback form (either transfer function or state space, and either positive or negative feedback) find the roots (zeros) of a polynomial
SlMULINK has also been used for block diagram simulation.
280
Block Diagrams
Chap. 11
STUDENT EXERCISES 1. Consider a first-order process that has a positive pole (Ilcgati\c tilne co]]stant). indicating that the process is open-loop ullstahle.
5s + It is desirable to design a feedback cOTnpcnsatOJ g2(s), so that the feedback systl'rn is stable. Assume that g2(S) is simply a gain: A'~(s)
J,:,
Find the range of gains that \vill make the following feedback system stable.
u(s) r(,)-..-~
+
+
g (s)
y(s)
1
,(s)
2. Consider the recycle system shown bclmv, where: ]
g,ls) ~ Is ~ l)(s + IJ.5)
g,ls)
c.
k
Find the values or II. (if allyl that \vil! ensure stahility of the system. Shmv and explain y'our reasoning. u(s)
r(s)
+
g (s)
)'OUT
y(s)
1
+
-
z(s)
gz (s)
r.-
3. Find the analytical solution for a unit step applied to the following process:
ge,)
(5.1
1)'
4. Consider the recycle system shO\vn belm\', \vhere:
wurk
281
Student Exercises
u(,) r(')
+
g (3) 1
+
~
Z(3)
g2(3)
I--
Discuss how the values of k2 and 1'2 effect the dynamic behavior of y with respect to a unit step input change in r. Use SIMULfNK and show compare plots for various values or k 2 and 1'2 to illustrate your points.
LINEAR SYSTEMS SUMMARY
12
One purpn-;c or thi" chapter i\ to summari/c the technique'> LhaL ha\c been dnelc)pcd ill Chapters:) through I I to :-,ol\c lincar ordinary dilfCfl'lltial equation;.,. Since the focus 11:1" beell on initial \;due problems. \Vc also intrntlucc techniques to soh'C huundary \U]Ul' ()[)I:: problems ..Also. since the emphasis has hCl'll on cOlllinoLls (dilTncJltial equatiun
h
dl~Haclcri"'lic
equation lT1Cl1lOd to \ol\c hound;uy \alut' linear ODE
Select all appropri,l1c technique to sohc a particular linear initial \<.tlucproblcllJ r:onnulalc lint'ar discrete-time lllodcb
I':stinlalc parameters for linl'ar
di~crcte
time mo(\eh
The major sections of this chapter art':
12.1 Background
12,2 Linear Boundar) V
Discrctc~
rime i\lodcls
12.) Parameter J-~sLilllaLioll of Discrete Linear Systems
282
Sec. 12.2
12.1
Linear Boundary Value Problems
283
BACKGROUND Thus far in this text, all of the problems that we have solved have been initial value ordinary differential equations. '1'0 solve these problems we simply need to know the initial values of the state variables, and how the inputs change with lime. The models arc then integrated to find how the states change with time. Ordinary differential equation models may be constrained to satisfy boundary conditions. Boundary value problems often arise when solving for the steady-state behavior of a dynamic system modeled by a partial dif~ ferential equation. 'T'ypically, a boundary value prohlem has distance as the independent variable and the boundary conditions lhat must be satisfied are tbe values of the state vari~ abies at different locations (typically at each "end" of the system). Recall that in Chapter 11 we required ninitial conditions to solve an nth order initial value ODE. Similarly, we require n boundary conditions to solve an nth order boundary value ODE. Most chemical processes that can be modeled as second-order boundary value problems (e.g., the reaction-diffusion equation) arc two-point boundary value problems. A second-order split boundary value prohlem has a boundary condition at one end and another boundary condition at the other end. If both boundary conditions were at the front end, then our problem would be an inital value problem. U' both boundary conditions were at the rear end, then we would have an initial value problem by simply redefining the independent vari'lble and forming an initial value problem in the opposite direction. In this chapter, \ve first cover linear boundary value problems in Section 12.2 and review methods to solve linear initial value problems in Section 12.3. We provide an introduction to discrete-time models in Section 12.4 and show how to estimate parameters for discrete-time models in Section 12.5.
12.2
LINEAR BOUNDARY VALUE PROBLEMS An analytical solution to boundary value ordinary conditions can be obtained lIsing the method of characteristics when the ODE and the boundary conditions arc linear. Consider the following second order ODE
+ aox
=
0
( 12.1)
where £1 0 _ at and 02 arc constant coefficients, _r is the state variable (dependent) and the independent variable (often distance). The solution to (12.1) will have the form
z is
( 12.2)
where Al and 11. 2 arc obtained by rewriting (12.1) as (/211.
2
--1-- ((I
A+
an
0
( 12.3)
llsing the method discussed in Chapter 6. The constant coefficients (e J and ( 2 ) are obtained [rom the boundary conditions.
-----------------------------------------_.--------------- EXAMPLE 12.1
Second-order Soundar)' Value Problem
Consider the following second··order equation: d)x
dx
I
7 x 4
0)
2
( 12,))
x(z=-I)=1
{ 12,hl
4
j
dz
o
( 12.-1.1
subject to the boundary conditions 'II each end; ~
x(z
We solve for the eigenvalues by using the characteristic equation:
+
A'
4A
+ : _.
( 12.7)
0
which yields (from the quadratic formula):
(lilt!
AI
3.5
(J 2.SI
A, -
0.5
( 12.91
the solution is: ( I 2.]() I
Substituting the boundary conditions results in two equations and two unknowns: 2
c1
=
+
( 12111
c;~
C e:1.S j
I- (2 eO' S
(12, J 2)
which yields: C 1 ""
0.36968
", - 1.63032 A plot of the solution, x:;:;; 0.36968
e -35z
+ 1.63032 eO. 5z , is shown ill figure 12.1.
2
1.8 1.6
1.4
1.2 1
a
0.2
0_6
0.4
0_8
z
FIGURE 12.1
Solution to Example 12.1.
Sec. 12.2
285
Linear Boundary Value Problems
More generally, the boundary conditions may consist of some function of the state variable and its derivative. The more general linear boundary condition is the form:
dx ",
EXAMPLE 12.2
I<"ox=d
dz
Second-Order Boundary Value Problem
Consider the second-order problem from the previolls example;
7
+ 4x = 0
(124)
subject to the new houndary conditions, dx
dz
+x=1
dx dz
=0
al z ,. 0
(12.13)
at z
(12.14)
Since the solution is: ( 12.15)
then the first derivative with respect to z: is:
dx
(12.16)
elz and boundary condition (12.13) yields:
( 12.17) while boundary comlilion (12.14) yields:
3.5 C j e --'5 Solving these two equations for c 1 and
x
c2'
0.5
('2 CO,')
0
(12.IX)
we obtain the solution:
0.37394
C
33, I 0.13032 eO"
(12.19)
Linear Systems Summary
286
Chap. 12
which is showll ill Figure 12,2. 0.1
o X;
0.1
-0.2
0.3
o
0.2
0.6
0.4
0.8
z FIGURE 12.2
Solution to Example 12.2.
We have illustrated how the method of characteristics is llsed to solve lincar bOllndar) value problems. 'rhe solution to nonlinear boundary value problems generally in\'llIH':-' it
crativc methods. For example, consider a single second-order nonlinear problem \vitll boundary conditions at each end. 'We know that the second-order equation can be COI1verted to two first-order equations. Typically, one boundary condition will fix an '"inititd condition" for one of the states. A second initial condition can be iteratively guessed (lising a Quasi-Newton method, for example) until the equations, when integrated. yield the correct value for the end boundary condition. This approach is shol,vn in l~xalllpl(' 12.,~ for the linear system considered in l':xamp1e 12.1.
EXAiVlPLE 12.3 Consider the
Formulating a Boulldal'Y Value Problem as an Itcnttive Initial Valuc ProblcHl secolld~ordcr bOllndary
valuc problem:
7 4
x~O
( 12.-1-1
2
( 12.5)
with thc boundary conditions: x(z
~
0)
x(z
cc
1)
It can be shown that (sec student exercise I), by defining equations arc obtained:
(12,61 xl'"" X
and '\'2 = dy/d:. the follu\\'il1!!
Sec. 12.3
Review of Methods for Linear Initial Value Problems
287 ( 1220) (1221 )
and that one of the initial conditions is
X, (2
~
0)
~
2
( 12.22)
We sec that -\"2(2"" 0) must be "guessed," then the two equations can he integrated from z:::;:; 0 to z cz J. The value of x I at z:::;: I is then checked; if x I (2 ;;;:; 1) is not equal to I (within an acceptable tolerance) then values of .:r2(z ;:;;: 0) arc iteratively guessed until the final value is satisfied. This method is known as the "shooting method." 'fhe reader is encouraged to llSC this approach to
solve exercise I.
12.3
REVIEW OF METHODS FOR LINEAR INITIAL VALUE PROBLEMS rn Chilptcrs .5 through II we presented a number of tcchniqtJcs for solving linear initial value ordinary differential equations. In Chapter 5 we noted that dynamic chemical process models arc oftcn formulated as a set or first-order, nonlinear differential eqLla~ tions, where the initial values are known. I'bcsc equations have the general fonn:
x=
f(x,u)
(12.23)
y ~ g(x,u)
(12.24)
where x is a vector of n state variables, u is a vector of m input variablcs, and y is a vcctor of r output variables:
.:\:/I
= j;Jxj, ... ,X/pll1,···,um)
YI c-~ gj(xl"",xn,uj"",utI/)
12.3.1
Linearization
Elcmcnts of the linearization matrices are dcfincd in tbe following fashion:
288
Linear Systems Summary
Chap. 12
_rJg'l
() .. II
allj
X"II,
where X,I" H,I" and Y.I, represent the steady-state values of the states, inputs, and outputs, which solve:
f(x"uJ
( 12.25)
y, = g(x"uJ
( 1226)
() =
After linearization, we have the state space form:
x'
= A x'
y' = ext
+ B u'
( 12.27)
+
(J2.2n
Dn'
where the deviation variable vectors are: X' = II'
=
x - x,
( 12.29)
,
( 12.10!
U ~ II
Generally, the (') notation is dropped and it is understood that the model is in deviation variable form:
x=Ax+BII
(12.31)
y=Cx+DII
(\ 2.32)
Once the model is in this form, a number of techniques can be used.
12.3.2 3.
Direction Solution Techniques
Solve the zero-input (perturbation in initial conditions) form (Chapter 5):
x(t)
= e A'
x(O)
( 1233)
One way the matrix exponential can be solved is eAI
= V
eAt
V--l
(12.14 )
The MATLAB function for matrix exponential is expm. b. For a constant step input at time zero (Chapter 5):
X(I)
=
eA'x(O) + (eA'_I) A' B u(O)
( 1235!
c. For inputs that are constant over each time step (from t to t + Cit) (Chapter 5): x(t + !J.I) which is often written as:
.~ e A'"
X(I) + (CAM - I) Al B lI(t)
(1216 )
Sec. 12.3
x(k
+ I)
o. e AM x(k)
+ (e A "' - I) A
where k represents the ktll time step_ This represents a discussed in more detail in Section 12.4.
12.3.3
289
Review of Methods for Linear Initial Value Problems I
B u(k)
discretc~timc
( 12.37) model, which is
Rewrite the State-Space Model as a Single nth Order Ordinary Differential Equation
a. Solve the homogeneous problem (Chapter 6);
+ ... +
dx (lJ
tit +
u() J:
=--=
0
( 12.38)
by first writing the characteristic equation: all
All
+
{In_1
AIl -
1
+ ... + G 1 A +
al! = 0
( 12.39)
and solving for the roots (eigenvalues) of the nih order polynomial. 1f the roots arc distinct, the solutionis of the form: ( 12.40) where the coefficients arc found llsing the n initial conditions. h. Solve the nonhomogeneous problem using the method of undetermined coefficients (Chapler 6):
dn-Jx J "",,, J
rt
dx
+ ... +
ii,
d
- .f
+ ""x
=
f(1t(t»
(12.41)
using 11 three-step procedure, i. Solve the homogeneous problem to Find: (12.42) ii. Solve for the particular solution by determining the coefficients of a trial function (see Table 6.1, Chapter 6) that satisfy the nonhomogeneous equation:
xl'(t)
(12.43)
iii. Combine the two solutions for:
x(t)
~"
Xf/(l) + xl'(t)
( 12.44)
c. Use Laplace transforms to solve the nth order equation (most useful for nonhomogeneous equations) (Chapters 7-1 I): (1245) which corresponds to the differential equation:
290
Linear Systems Summary
Chap. 12
(12.46)
The more general case is: (12.47)
which corresponds to the differential equation:
dllu
= b---/I dtll
+
b II-I
ri"-Iu du J+ ... +b J dt +bou dt ll --
(12.48)
For physically realizable systems, b ll = 0, Onen many of the leading hi terms arc zero. If the leading r terms in the b polynomial arc zero, then the systcnl is referred to as relative order r.
12.3.4
Use Laplace Transforms Directly on the State-Space Model
Previously we have assumed that: the state-space model has already been converted to a single nth order differential equation. We can also transform the set of n firsl-ordcrlincar state space equations directly using:
Y(.I) = lC (.1'1- A) 1 R + DI U(.I)
( 12.49)
Generally, the Laplace transforms technique is used for nonhomogeneolls problems, that is, systems that have an input forcing function (such as a step).
12.4
INTRODUCTION TO DISCRETE-TIME MODELS Consider the genera] linear state space model: XI
([1,,] . lX1] . + Ib.
l1
. {{/Ill
. _~XJ!
. _nhlll
or, x~~Ax+Ru
Recall that the single variable equation:
(1231 )
Sec. 12.4
291
Introduction to Discrete-Time Models
x-::;:ax+!Ju
( 12.50)
has the solution:
iJ X(I) = eal X(O) + (eM - 1) U(O)
(12.51 )
({
when u(t) :::: constant::::: u(O). t=
In a similar fashion, the solution to (12.31), for a constant input (u(t):::: lI(O» fronl 0 to tis:
X(I)
c. (I' x(O) + r u(O)
( 12.52)
where: (1253)
G = (t/>-I) A 'B
and
( 12.54)
Equation (12.52) can be used to solve for a system where the inputs change from time step to time step (t to t+6.1) hy using:
X(I + AI)
tP x(t) +
r
u(t)
( 12.55)
+ "u(k)
(12.56)
More often this is written as:
x(kc I)
tI' x(k)
where k represents the kth time step. The output at tillie step k is written:
y(k)
C x(k) + D u(k)
=
( 1257)
The stability of the discrclcstatc-space model is determined by finding the eigenvalues of (J>, If the magnitude of all of the eigenvalues is less than I, then the system is stable.
12.4.1
Discrete Transfer Function Models
Continuous time models transfer function models arc characterized by the Laplace tranform variable, s. Similarly, for discrete transfer function models, a discrete transform variable, 7., is uscd:
Y(z)
c.c
G(z) U(z)
( 12.58)
where:
G(z) =
IC (zl- A) 'n
+- DJ
(12.59)
f,'or the case of a single input""single output system, (;(7.) consists of a numerator and denominator polynomial of the form: lJ
.( ) = bnz
g Z
+ +
bl/_Izl(
Ii.
anz
ali jZ
II
I I
... + h1z + h" + .., + alz + a"
( 12.60)
292
Chap. 12
Linear Systems Summary
The transfer function is normally written in terms of the bachvards shUt operator, Multiplying the transfer function by Z'--II/t---II , we find:
+ hoz- n + {loZ-n
;-1.
(1261 )
The backwards shift operator is defined as;
= z' y(z)
y(k ~ 1) so y(k ~- 2)
:=
[l.V(Z),
( 12(2)
etc. The discrete transfer function notation:
__. b ll -t-
y(z)
-I- ... -I- {'IZ.---n+
1
+ ... + {lIZ'-n
1
j
f /, Z····f{ . " __ 11 aoZ
--f-
( 12(3)
t 1(") 4.
then represents: (an
+
{l1l_IZ-1
(hI] + b ll
_12
+ ... +
+ ..- +
j
+ ((OZ'-II) Y(2)
(ljZ-n+l
btZlll1
+ buz I!) u(z)
( 12(4)
which corresponds to the discrete input/output model:
a"y(k) =
+ a"ly(k ~ 1) + ...
""u(k)
-I-
h"o,u(k·· 1)
-I-
(I"y(k ~ 11 ~ 1)
(I,y(k" 11)
-I- ... -I-
hlu(k ··11)
h"u(k
f
11
~
1)
( 12.(5)
Usually we arc solving for y(k+ I), and without loss of generality we call assume all :::; ].
y(k =
-I-
1)
-I-
"ou(k +1)
{["ly(k) -I-
-I- ... -I-
"o,u(k)
a,y(k
-I- ...
~
n
+ ",a(k
-I-
~
1)
-I-
{[oy(k
~
11)
11 ~ 1) + "ou(k ~ 11)
( 12(6)
Also, for rnost systems there is not an immediate effect of the input on the output, so
h n ;;;;: O. The most common
discretc~til11c model
y(k
-I-
1)
-I-
is fin..;t-order:
I/"y(k)
~c
hol/(k)
( 12671
or,
y(k + 1)
I/oy(k) + h"u(k)
=
which has the transfer function relationship:
y(z)
=
1
boz- J
( 126~)
+ lloZ-
Physically realizable systems will always have at least a z -I factor (unit time delay) in the numerator. A first-order discrete system with N additional units of lime-delay is written
y(k or,
+ 1)
-I-
I/"y(k) = hol/(k
~
N)
( 12N))
Sec. 12.4
293
Introduction to Discrete-Time Models
y(k
-I
I)
-aoy(k)
=
-I
boll(k- N)
which has the transfer function relationship: ( 12.70)
EXAMPLE 12.4
Linear Van de Vnssc Reador Model
Consider a statc-space model from the isothermal chemical reactor module (specifically, the Van de Vussc reaction); A = 1- 2.404R
o
- 2.23Rl
0.R3:n II
1
c·~
[0 I[
1
7.00001 - 1.11 70
1~I
J)
For a sample lillle of 0.1 the discrete Slate-space model is (using (12.52)-( 12.54), or the MATLAB c2d function): (I) =-0
1'-
O.7R63
0 1 0.7995
r0.0661
6222 1 0. ··0.OR49
and the discrete inpul-outpulmodcl is (using (l2.59), or the MATLAB ss2tL function) g(z)~
- 0.0751 z 1 - I.oR57z
I I I
-I
0.100 I? ' ' 0.62R6z -
which has poles of 0.7995 and O.7g63 (which have a magnitude less than J, sO the system is stable). The zero of the numerator polyllOll1iai is 1.3339. The step responses of the continuous and discrete systems arc compared inr'igure 12.3. For a sample time orO.?5, the discrete Slate-space model is (using (12.52) -(12.54), or the MAfLAB c2d function): (I)
r
10. [647 OJ 096 2 43 1 . [4J 0.1164
o ]
n.1 k66
Linear Systems Summary
Chap. 12
0.6
05
04 o
continuous de It := 0.1
S 0.3 0.
S o 0.2 0.1
o 0.1
~~~~~~~~L~ _ _
o
FIGURI'; 12.3
,old
thl~
2 time
3
4
Step respOllse of continous and dis<:rete (0.1::::; 0.1) models.
discrete input-output model is (using (12.59), or the MA"llJ\B
g( z)
0.1564z
I
0.3513z
~3E;2tt
funClion):
0.240Xz ' ,. 0.1I307z
whidl has pole; at 0.1866 and 0.1647, indicating stability. The zero of the numerator polynomial IS
15199.
A comparisoll of the step responses of the continuous and discrete models is ShO\Vll in Figure 12.4. Notice that the discrete sample time is too large to capture the "inverse respollse" hehavior of the continuous systelll.
0.6
05
r-.. continuous [OdeJt 0.75
0.4
:=
S 0. S 0
0.3
0.2 0.1
0 -0.1
0
FIGURE 12.4
2 time
3
S1l'p response of continolls :md discrete (.::\.1::::; 0.75) models.
4
Sec, 12,5
12.5
Parameter Estimation of
Discrete~LinearSystems
PARAMETER ESTIMATION OF DISCRETE LINEAR SYSTEMS Often when discrete linear models arc developed, they afC based on experimental ,-:y.';[CllJ responses rather than converting a continuous model to a discrete model. The e~;tillla{ioi1 of parameters for discrete dynamic models is no different than the linear regression ,maly sis presented in Module J. Please review Module 3 to understand the not:ctioll ,mel ideas hehind linear regression. The measured inputs and outputs arc the independent variables, and the dependent variables arc the outputs. For simplicity, consider the folIowing single inpllt~singlc output model:
y(k)
~
(/,y(k -- 1)
~
(/oy(k ·'·2) + h, lI(k
~
1)
1
hO/(k
~
2)
(12.71 )
Now, 1'01' the SystClll of N data points we call write: Y~(I)H
(IU2)
where,
Y~
l
Y(I).J
['1'0)1.'] 0 ellee .
,
,
y(N)
. -", I [ --: U o
( 12.73)
hi bo
[y(k ~ I) y(k ~ 2) lI(k --I) u(k ~ 2)1
( 12.74)
'rhc solution to this problem is: (12. I';)
_
.. ~ ~ - - ~ . _ ~ - - ~ . ._ ~ ~ ~ - -~ . .
EXAMPLE 12.5
Parameter Estimation
A unit step input is made to a system at time t::;: 0 (k:;;:; 0). The sample lime is response data are shown below and plotted in figure (2.5. k
r(k)
o
0
I
0,1564
2
0.45l2
3
11,5513
4
0.5770
5
0.5830
D.I:;;:;
0.75. The step
Chap. 12
Linear Systems Summary
296 0.6
o 0.5
o 0.4
'5 J3.
"0
0.3 0.2 0
0.1
o --------o
J__
~
~
.L_
__'__
2
5
4
3
sample time
FIGUREl2.S
y
y(2) J.'( I) y(3)
1'(4) 1'(5)
I
Step response data (some as l'igurc 12.4).
O 15fJ4 0.4522 0.5513 . 1 0.5770 0.5830
r
y(O)
y(4)
1/(0)
y( ·1)
u(4)
y(3)
1/(-1)]11(:)
o
IJ
O.15fJ4 0.4522 0.5513 IJ.5770
o 0.1564 0.4522 0.5513
The solution is:
:::,]
[- ~:~~~]~]
ho __
0.2409_
hi
0.1564
which afC the same parameters that we found for the discrcle transfer function model converted from the contiullOU\' Illodel with a sample time of 0.75.
thal W:lS
References
297
This sirnplc cxample illustrated the step response of a perfectly modeled system (no measurement noise). The approach can also be applied to a systenl with arbitrary inputs and with noisy measurements. The data was analyzed in a batch fashion, that is, all of the data were collected before the parameter estimation was performed. There are other approaches that are useful for estimating model parameters in real time, often using the model parameters to modify feedback control laws. These approaches arc beyond the scope of this textbook. The MATLAB System Identification Toolbox is useful for these types of problems. A good reference is the text hy Ljung.
SUMMARY There were multiple objectives to this chapter. The first was to introduce analytical solution techniques for boundary value ordinary differential equations. The second was to provide a concise review of techniques to solve linear initial value ordinary differential equations. The final ohjective was to introduce discrete-time modcls and discuss parameter cstimation for these models. For continuous-time models, the eigenvalues of the state-space model must have negative real portions for the system to be stable. Equivalently, the poles of the continu~ ous transfer function models must be negative (the eigenvalues of the state-space m(){icl are equal to the poles of the transfer function model). Analogollsly, the eigcllvalues of the discrete state-space model mLlst have a magnilUde less than one to be stable. Also, the poles of a discrete transfer function model must have magnitudes less than one to be stable. Continuolls-timc input/olltput (transfer function) models with zeros that arc positive exhibit inverse response. Similarly, discrete transfer function models with /,cros that have a magnitudc greater than one (yet have a negative real portion) exhihit inverse response.
REFERENCES The following text provides methods to solve houndary value ordinary differential equation models. Rameriz, W.P. (1989). Computational Methods j(Jr Process SimulatioN. Boston: Butterworths. Many process control textbooks provide coverage of linear discrete-time models. Sec, for example: Ogunnaike, B.A., & W.H. Ray. (1994). Process Dynamics, Modeling and Control. New York: Oxford. System identification (model parameter estimation) is covered in the text by Ljung Ljuog, L. (1987). System NJ: Prentice-Hall.
fdent{flcati()Il-,~The()ry for
the User. Englewood Cliffs,
Linear Systems Summary
298
Chap. 12
STUDENT EXERCISES 1. Consider the following second-order boundary value problem:
J!x
dz 2
+ 4 dx + tlz
7 x 4
.-
0
where:
x(z
0) = 2
x(z
1)
=
Show that, by defining x I ::::; x and -1:2::::: dx/dz., the following equations arc obtained:
dx,/dz
cc
x2
dx,/dz = ..
~
7 4 x,
~
4x 2
and that one of the intial conditions is:
We see that x2(;: = 0) must he "guessed," then the two equations can be integrated (using ode4 ~j) from? :::: 0 to ? :::: I. The value of x I at z ::::: 1 is then checked: if xIV:::: I) is not equal to J (within an acceptable tolerance) then value,,> ofx2L~ :::;: 0) arc iteratively guessed until the final value is satisfied. This method is known as the "shootingmclhod". Usc [zero to solve for the initial condition that satisfies the end boundary value. 2. Consider the reaction/dispersion equation
ac;\ iJl
CA
let:
=
dimensionless concentration
=
dimensionless axial distance
C'AO
and:
y
define:
'T -:::
z I, D.,/{ I}
p = "
and:
D
dimensionless time Peclct Illllnber
/),,/
k /,'
II
--- Damkohler number
J),l7
to show that:
ilC
PI'
ilC
ay
+
(1221
299
Student Exercises
f"ind the dirncnsionlcss form of the Danckwerts boundary conditions at steady-state:
dC" (L)
and:
=0
dz
u. Perform steadY-Slate calculations (analytically) using lhe Danekwerts boundary conditions for: i. P(':::;I, D II ::;:;:; I, 10, 25 (compare on same plot) ii. 1',.= IO,D o = I, 10,25, 100 (compare on samcplol) iii. Pe = 25, D(/ :::: I, 10,25, 100 (compare 011 same plot) iv. 1',. = 100, Do = I, 10,25,100 (comparc on same plot) 3. Consider the following continuous state-space model:
A 0'
Ii=
-3.6237 1. 0.8333
=
)
5.50511
I.
-- 1.2660
C . = [0 D
o
- 2.9588
II
0
a. Find the continuous transfer function model. b. For a sample time of 0.25, find the discrete state-space and transfer function models. c. Compare the step responses of the continuous and discrete 11lOdcls. What do you observe'! 4. Consider a unit step change made at k:::: 0, resulting in the output response shown in
the plot and table below.
1.2
o
o
c.;--0
o
o
o
o
0
0
0
o
0
o
0
o
0.8
'5 -50,6
0
0
0.4
0
02 0
0
_._.L
0
20
_
------.-1-
L
40 60 sample time
80
100
Linear Systems Summary
300 I 0.1044
II 10 9 7 8 6 3 4 5 2 0.3403 0.6105 0.8494 1.0234 1.1244 1.1616 1.1531 1.1184 1.0746 I.OJ3h
k Y
0 0
k
20 18 19 17 15 16 13 14 12 1.0023 0.9828 0.9744 0.9742 0.9790 0.9860 0.9929 0.9985 1.0022 Estimate the parameters for a discrete linear model with the fonn:
y
Chap. 12
-f-
g(z) ~I _
boz--- 2 - ({oZ2
SECTION
IV
NONLINEAR SYSTEMS ANALYSIS
PHASE-PLANE ANALYSIS
13
'rhe objective of this chapter is to introduce the student to phase-plane analysis, which is one of the most important techniques for studying the behavior of nonlinear systems. After studying this chapter, the student should be able to: Usc eigeIlvalues and eigenvectors of the Jacobian matrix to dUlracterizc the phascplane behavior
Predict the phase-plane behavior close to an equilibrium point, hased on the linearit.ed model at that equilibrium point Predict qualitatively the phase-plane behavior of the nonlinear system, when there arc multiple equilibrium points The major sections of this chapter arc: (3.1
Background
13.2
Linear System Examples
13.3
Gcncralit,ation of Phase-Plane Behavior
13.4
Nonlinear Systems
303
Phase-Plane Analysis
304
13.1
Chap. 13
BACKGROUND Techniques to find the transient (time domain) behavior of linear state-space models \verc
discussed in Chapter 5, Recall that the response characteristics (relative speed of response) for unforced systems were dependent on the initial conditions. Eigellvalue/eigenvector analysis allO\vcd us to predict the fast and slow (or stable and unstable) initial COIlditions. If we plotted the transient responses based on a number of initial conditions, there would soon be an ovcnvhclming Ilumber of curves on the transient response plots. Another way of obtaining a fecI for the effect of initial conditions is to usc a pIUlS('~p/(/IiC plot. A phase-plane plot for a two~slatc variable system consists of curvcs of OIlC-SUltC variable versus thc state variable (xl(t) versus xi!)), where each curve is hased on a dif~ fercnt initial condition. A phase-space plot can also be made for three-state variahks. where each curve in 3~spacc is based on a different initial condition. Phase-plane analysis is one or the most important techniques for studying the behavior of nonlinear systems, since there is usually no analytical solution for a nonlinear systcm. It is obviously important to understand phase-plane analysis for lincar sysll'!1E before covering nonlinear systems. Section 13.2 discusses the phase-plane behavior of linear systems and Section 13.3 covers nonlinear systems.
13.2
LINEAR SYSTEM EXAMPLES Nonlinear systems often have multiple steady-state solutions (sec Modules X and l) for examples). Phase-plane analysis or nonlinear systcms provides an understanding oC \vhicll steady-state solution that a particular set of initial conditions will converge to. The local behavior (close to one of the steadY-SUite solutions) c,Im be understood from a lincar phase~plane analysis of the particular steady-state solution (equilihriwIl point). In this section \\le show tile different types of phase-plane behavior that can be CX~ bibilCcl by linear systems. The phase-plane analysis approach will be shown by \vay (ll' a number of examples.
EXAJ\ilPLE 13.1
A Stable Etll1ilihrilllll Point (Node Sink)
Consider the system of equations: (I '.1)
( 11.2)
The reader should find lhat the solution
[0
(13.1) and ( 13.2) is:
xl(r)
= Xl" ('I
(13,4 )
i
5r.
Sec. 13.2
305
Linear System Examples
where xI" and x 20 arc the initial conditions for xI and x2' We could plot Xl and x 2 as a function of time for a large Humber of initial conditions (requiring a large number of time domain plots), hut the same information is contained 011 a pllase-plane plot as shown in Figure 13.1. Each curve corresponds to a different initial condition. Notice that the solutions converge to (0,0) for all initial conditions. The point (0,0) is a stahle equilibrium poiot for the system of equations (13.1) and (13.2)~thc plot shown in Figure 1:1.1 is often called a ,I'table node.
'1
FIGURE 13.1 stable node.
EXAMPLEI3.2
Phase-plane plot for Example 13.1. The point J:T
:;::
(0,0) is a
An Unstable Equilibrium Point (Saddle)
Consider the system of equations:
X, -tl
(13.5) =
4 Xl
(13.6)
'fhe student should find that the solution to (13.S) and (13.6) is:
X1(1) xit)
l
(13.7)
x 20 e"l
(13.8)
= x lo e
The phase~plane plot is shown in Figure 13.2. If the initial condition for the x2 state variable was 0, thell it trajectory that reached the origin could be obtained. Notice if the initial condition x2(1 is just slightly different than zero, then the solution will always leave the origin. The origin is an unstable equilibrium point, and the trajectories shown in Figure 13.2 represent a saddle point.
Phase-Plane Analysis
306
Chap. 13
The x j axis represents a stahle subspace and the Xl axis represents an unstable subspace ["or Ihi\ problcrn. 'l'he term saddle can he understood if you view the x J axis as the line (ridge) hct\\ccn the "horn" and rear of a saddlc. A baJJ starting at the horn could cOllceptually roll d(l\vn the sdddie and remain exactly on the ridge between the horn and the rear of the saddlc. In practice, d small perturbation from the ridge would calise the ball to begin rolling to olle "stirrup" or [Ill' other. Similarly, a small perturbation in the initial condition from the Xl axis in l~xal1lpk L\2 would cause the solution to diverge in the unstable direction.
FIGUnE 13.2 saddle point.
Phase-plane plot for Example 11.2. The point xl' ::::: (OJ)) is a
Figures 13.1 and 13.2 clearly show the idea of separaticcs. A separatrix is a line in the phase-plane that is not crossed by any trajectory. In l-<'igure J 3,1 the sepnF{/liccs are the C\lordinate axes. A trajectory that started in any quadrant stayed in that quadrant. This is he cause the eigenvectors arc the coordinate axes. Similar behavior is observed in Figure 13,2, except that the xJ coordinate axis is unstablc. Solving the cquations for Examples 13.1 and 13.2 and the phase-plane lrajccl(ll'il's were straight-forward and obvious, because the eigenvectors where simply the cnordin,th' axes. [n general, eigenvalue/eigcnvector analysis must be llsed to determine the stahle dnd unstable "suhspaccs." The eigenvectors arc the separaticcs in the general case, Examplc 13.3 shows how eigenvaluc/eigenvector analysis is used to rind the swllie and unstable subspaccs, ,md to define the separatices,
Sec. 13.2
Linear System Examples
EXAMPLE 13.3
AnothcrSHddle Point Problem
307
Consider the following system of equations:
i
j
i1
=
2x 1 + x 2
=
2x l
( 13.9) (13.10)
-X 2
Using standard state-space notation:
x=
Ax
(13.11 )
The Jacobian matrix is:
the eigenvalues are:
-1.5616 and the eigenvectors arc: ', t
,0.27031 - 0.9628
8719 0. 1 0.4896 1
Since A] < 0, ~I is a stable subspace; also, since 1.. 2 > 0, ~2 is an unstable subspace. A plot of the stable and unstable subspaces is shown in Figure 13.3. These eigenvectors also define the sepa·· ratit:cs that determine the characteristic behavior of the state tn~jeetories.
Unstable Subspace
Stable Subspace FIGlJRE 13.3
Stable and unstable subspaces for Example 13.3.
Phase-Plane Analysis
308
Chap. 13
'file lime domain solution to (13.11) is:
X(I) ~ e A ' x(O)
(JJ.12)
which is often solved as (sec Chapter 5):
x(l)
~
V e'\' V
I
x(O)
(13.13)
which yields the following solution for this system:
X(I ~ )
I
0.2703 0.9628
(U~Tl91
Ie
0.4896
LS(,I(,t
0
5038 . 011 00.9907
('2.561(;/
- o. 81J72 1 x(O) 0.2782
Recall that the solution 10 (13.[4), if.\'(O) "" ~l,isx(l):::: ~Ie'\ll, so
if x(O)
0.2703
_ I - 0.9628
Ithen
( I
xt)=
0.2703 ] ._ e _- 0.9628
The phase-plane plot is shown in F'igure 13.4, where the separatrices clearly define the phascplane behavior.
FIGURE 13.4
Phase-plane plot for Example 13.3,
Sec. 13.2
309
Linear System Examples
'fhe previous examples were for systems that exhibited stable node or saddle point behavior. In either case, the eigenvalues and eigenvectors where real. Another type of behavior that can occur is a spiral focus (either stable or unstable), which has complex eigenvalues and eigenvectors. Example 13.4 is an unstable focus.
EXA1VIPLE 13.4
lJn~tHhlc
Focus (Spiral Source)
Consider the follO\ving system of equations: XI
=x l
1
2 x.,
( 13.15) (13.16)
Using standard state-space notation:
x=
A x
The Jacobian matrix is: i\~
I
I
-- 2
with eigellvalues I ± 2j. This systenl is ullsfablc because the real portion of the complex eigenvalues is positive. 'file phase-plane plot is shown in r:igufc 13.5.
FIGURE 13.5
Example l3.4, unstable focus (spiral source).
310
Phase-Plane Analysis
Chap. 13
Another type of linear system behavior occurs when the eigenvalues have a zero real portion. That is, the eigenvalues arc on the real axis. This type of system leads 10 closed Cllnes in the phase-plane, and is known as ccnter behavior. Example 13.5 illustrates center beha\'ior.
EXAMPLE 13.5
Center
Consider the following system of equations: --Xl
X(
(1117 )
iLUS) The Jacobian matrix is: ;1~
I
~
I
_ 4
and the eigenvalues arc () ± 1.732 Ij. Since the real part of the eigenvalues is zero. there is a peri odic solution (sine and cosine), resulting in a phase-plane plot where the equilibriul11 point j" ~\ cenlcr, as shown in Figure 13.6.
FIGURE 13.6
Example 13.5, eigenvalues with l,cro real portion arc centers.
Sec. 13.3
Generalization of
Phase~Plane Behavior
311
Examples 13.1 to 13.5 we have provided an introduction to linear system phase-plane behavior. We noted the important role of eigenvectors and eigenvalues, and how these relate
to the concept of a separatrix. Section 13.3 provides a generalization of these examples.
13.3
GENERALIZATION OF PHASE·PLANE BEHAVIOR We wish now to generalize our results for second-order linear systems of the form: (13.19) where the Jacohian matrix is:
A=
{[II
( 13.20)
Ian
Recall that the eigenvalues arc found by solving det(A./- A) = 0:
del(A! - A)
=
(A - ",,)(A - aD) - {/" a 2l
CC
()
( 13.21)
which can be \vrittcn as:
del(H-A)
==
A2 -1r(A)A
+ del(A)
()
( 13.22)
The quadratic formula call be used to find the eigenvalues:
A = Ir(A)+ V(lr(A»' - 4 dellA) 2
( 13.23)
or, expressing each eigenvalue separately,
A, =
lr(A) - v(lr(A»)2 - 4dcl(A) 2
and.
tr(A) r \/(lr(A»)2 - 4 dellA) 2 V./c notice that at least one eigenvalue will be negative if 'r(A) < O. We also notice that the eigenvalues will be complex if 4 de-teA) > tr(A)2.Remcmbcr that the different behaviors resulting from Al and A2 are: Sinks (stable nodes):
Rc (A,) < 0 and Rc (A 2 ) < ()
Saddles (unstable):
Rc (A,) < 0 and Rc (A2) > ()
Sources (unstable nodes):
Rc (A I) > 0 and Re (A z) > 0 Al and A2 arc complex complex conjugates. If Re(AI) < 0 then stable, i r RC(A,) > () then unstable.
Spirals:
We can then usc Figure 13.7 to find the phase-plane behavior for second-order linear ordinary differential equations as a function of the trace and detenninanl of A. In Figure
312
Phase-Plane Analysis
Chap. 13
de1(A)
(tr At= 4 det A
spiral source
Center node
node
sink
source
saddle
FIGURE 13.7 DYllarnic behavior diagram for second-order linear systems, Thcx
13.7, the x-axis is the trace orA and the .v-axis is the determinant of A. For example, sider Example 13.1, where A
tr(A)
- I 1
()
COlJ-
°1
-4
-'i
dct(A) c= 4 'fhe point (-5,4) lies in the second quadrant in the node sink sector, as expected. since the two real eigellvalues arc negative (indicting stable node behavior).
Sec. 13.3
Generalization of Phase-Plane Behavior
313
l<jgurc 13.8 shows the phasc~planc behavior as a function of the eigenvalue locations in the complex plane. For example, two negative eigenvalues lead to stable node behavior.
13.3.1
Slope Marks for Vector Fields
A qualitative assessment of the phasc~planc behavior can be obtained by plotting the slope marks for the vector field. Consider a general linear 2-statc system
Eigenvalues
Phase-Plane Plot
1m
""i.. . ~ iRe
3table node
1m
~ ~e x i ....
FIGlJRE 13.8
3table fOCIJ3
Phase-plane behavior as a function of Eigenvalue location.
314
Phase-Plane Analysis
Eigenvalues
f~ ~
~e
~
Chap. 13
Phase-Plane Plot
unstab1JJ tocos
Jm
saddle
He
r
m _
"
~e
FIGURE 13.8
Continued
( /3.24) X2 = all x I
+ an x 2
( 13.25)
We call divide (13.25) by (/3.24) to find how x 2 changes with fCS]lcotto x,: dX 2
a 21
dx]
al1xj+1l12X2
Xl
-I-
(/326)
±LA
Sec. 13.3
Generalization of
Phase~Plane Behavior
315
and we can plot "slope marks" for val lies of xl and x2 to determine an idea of how the phase plane wil! look. Let us revisit Example 13.3. The slope marks can be calculated from (13.27): dX 2 dXj
2x, 2 Xl + x 2
( 13.27)
Figure J 3.9 shows the slope marks for Example 13.3. These arc generated by forming a grid of points in the plane, and finding the slope associated with each point; short line seg~ mcnls with the slope calculated arc then plotted for each point. Notice that one can lise the slope marks to help sketch state variable trajectories, as shown 111 Figure 13.10. Saddle point behavior found in Example 13.3 is clearly shown in l<'igurc 13.10.
13.3.2
Additional Discussion
Phase-plane analysis can be used to analyze autonomous systcms with two statc variablcs. Notice that state variable trajectories cannot "cross" in the plane, as illustrated by the following reasoning. Think of any point of a trajectory as being an initial condition. The model, when integrated from that initial condition, must have a single tn~jectory. If two tn~jcctories crossed, that would he the equivalent of saying that a single initial condition could have two different tnyectories. If a system was nOlH1utonomous (for examplc, if there was a forcing function that was a function of time) then state variable trajectories could cross, because a model with the same initial conditions but a different forcing function would have different tn~jectories. An autonomous (unforced) system with n stale variahles cannot have tn~jectories that cross in n-space, hut may have trajectories that cross in less than n-space. For exam-
FIGURE 13.9
Slope marks for the vector field of Example 13.3.
Phase-Plane Analysis
316
FIGIJRE 13.10
Chap. 13
Slope marks with trajectories for Exarnplc 13.3,
pic, a third-order autonomous system cannot have trajectories that cross in 3-spacc, but the trajectories Illay cross when placed in a two-dimensional plane.
13.4
NONLINEAR SYSTEMS In the previous sections we discovered the types of phase-plane behavior that could be observed in linear systems. In this chapter we will find that nonlinear systems v,'ill orten have the same general phase-plane behavior as the rnodellincarizcd about the equilibrium (steady-state) point, when the system is close to that particular equilibrium point. In this section we study two examples. Example 13.6 is based on a simple bilinear model, while Example 13.7 is a classical bioreactor model.
EXAMI'LE·13.6
Nonlinear (Bilinear) System
Consider the following system:
rtf dZ 2
tit
z,(z, + I)
( 13.28)
(13.29)
Sec. 13.4
317
Nonlinear Systems
which has two steady-state (equilibrium) solutions:
F'rJllilihrillm I: trivial
1. 2s
0
Equilibrium 2: nontrivial Linearizing (13.28) and (13.29), we find the following Jacobian matrix:
A
Zl'~ +
]j
<-'1.1'
.
In the following, we i.lnaly/,c the stability of each equilibriulH point.
Equilibrium 1 (Trivial) The Jacobian matrix is:
A
and the eigenvalues arc:
11. 1
[~ (~] = -- v'3
11.) = \/3
We know from linear system analysis that equilibrium point one is a saddle poim, sillce one eigenvalue is stable and the other is unstable.
0.5j
I.0.866 _
The stable eigenvector is:
The unstable eigenvector is:
~,
0.5
I0.866
I
The phase-plane or the JincarJ/cd model around eqnilibriulll point one is a saddle, as shown in I"igurc D.l I. The linearized model is
Phase-plane of the [~xample 13.6 model lillcarized around trivial equibrium point. This point is a saddle point.
FIGlJH.E B.ll
Phase-Plane Analysis
318
x 1
where x = z -
0 3
Chap. 13
11 x
0
z,
Equilibrium 2 (Nontrivial)
'rhe Jacobian matrix is:
;1
the eigenvalues arc:
AI
-3
0 011
·3
A2
So, we know from linear system analysis that equilibrium point eigenvalues arc stable.
[\\10
is a stable /lode, since both
The "fast" stable eigenvector is The "slow" stahle eigenvector is
The phase-plane of the linearized model around cquilhrium point two is a stable node, as shown in Figure 13.12.
FIGURE 13.12
Phase-plane of the EX
trivial cquibriulH poinL This point is a stable node.
The phase--plane diagram of the nonlinear model is shown in r;'igllrc /3.13. Notice h()\\, the linearized models capture the behavior of the nonlinear model \Vhell close to one of the equiIbrium points. Notice, however, that initial conditions inside the "right" saddle "blow up," while initial condititllls inside the left saddle are attracted to the stable point. Slope-field marks are shown in F'igurc 13.14.
Sec. 13.4
319
Nonlinear Systems
FI(;lJRE 13.13 Phase-plane of [~xaJnplc 13.6. Tn~jcclorics (except those of the righl side of the saddle) leave the unstable point and arc "attracted" 10 the stable point.
,~~
FIGURE 13.14
Slope-field marks and some trajectories for Example 13.6.
Phase-Plane Analysis
320 ...
EXAMPLE
n.?
Chap. 13
_---~---~-
Biorcactol' 'with Monod Kinctks
Consider a model for a bioreactor with MOllOd kinetics (sec rVIoduJe 8): dx(
D)
tit
x,
03.3(1)
/-lX 1
(.'i - x,) IJ
Y
JJ-m;nXI
km +
(13.31 )
(13 ..J2)
~y!
where: /-tIll:)X
y
k", == 0.12
== 0.53 ~
11.4
4.11
is the biomass concentration and x2 is the substrate concentration. There arc (wo steady-state (equilibrium) solutions for this set of pararneters.
Xj
J:quililniulII I: trivial
Xl.>
xb
1.4523
L'quilibriu/II 2: nontrivial
= 4.0
0.3692
Lineari/.ing (13.30) and (13.31) we find the following Jacohian matrix:
/-L,\
where we have defined I-l'
X:!.\( kIll ·f
x1.J
E()uilihrium t (Trivial) The Jacobian matrix is:
A ~
I 0.114563 - 1.286408
o
-IIA
I
with eigenvalues of; ~l = 0.114563 and A] == -0.4 indicating that the steady-state is unstable (it is a saddle point). The unstahle eigenvector is:
The stable eigenvector is:
Sl
0.3714]
I. - 0.9285
I~I
This steady-state is known as the "wash-out" steady-state, because no hiomass is produced and the suhstrate concentralion in the reactor is equal to the feed suhstrate concentratioll.
Sec. 13.4
321
Nonlinear Systems
Equilibrium 2 (Nontrivial)
The Jacobian matrix is:
A
with eigenvalues of
~
1
°
3.215929] 8.439832
0.4
and
- 1
indicating 1hal the stcady-state is a stable node.
~, ~
The "slow" stable eigenvector is
1
0.99241 - 0.11234
[.°.0.9285
3714
The "fast" stable eigenvector is
1
The phasr>plunc plot of Figure 13.15 sbows that the trajectories leave unstable point I (0,4) and go to stable point 2 (1.4523,0.3692). More detal! of the phase-plane around the unstable point is sho\Vn in Figure 13. J 6, while Figure 13.17 shows more detail around the stable point.
s 4.0 3.0
2.0
1.0
1.0
2.0
FIGURE 13.15 Phase--planc for bioreactor with Monnd kinetics. x is biomass concentration and s is substrate concelltration.
ESCOLA Dc ENCEi'JHARIA BIB L 10 'I L '"" 1\
x
322
Phase-Plane Analysis
0.0 - - - - - - - - - FIGUIU~
13.16
Phase~planc
Chap. 13
- - - - - - - - - 0.05
behavior ncar the unstable point (0,4) (L:quiJi-
britlln 1).
0.4
0.3 1.4 - - - - - - - - - FIGURE 13.17
(Equilibrium 2).
1.5
Phase-plane behavior ncar the stable point (1.4523, 0.36(2)
Sec. 13.4
Nonlinear Systems
323
0.5
x,
0 -0.5
·1 ··1.5 ··1
0
FIGURE 13.18 behavior.
X2
Example of center
In the previous examples the system trajectories "left" an unstable point and were "at~ tracted" to a stable point. Another type of behavior that call occur is limit cycle or periodic behavior. This is illustrated in the following section.
13.4.1
Limit Cycle Behavior
In Section 13.2 we noticed that linear systems that had eigenvalues with zero real portion formed centers in the phase plane. The plwsc-planc trajectories of the systems with celltel's depended on the initial condition values. An example is shown in Figure 13.18. A somewhat related behavior that can occur in nonlinear sYStCll1S is known as limit cycle behavior, as shown in Figure J3.19. The major difference in center (Figure 13.1 R) and limit cycle (Figure 13.1 <)) hehavior is that limit cycles are isolated closed orbits. By isolated, wc mcan that all inital pcr~
1.5
x,
r-~------~--,
0.5
o 0.5 -1
1.5
L-
--'
1
0
x,
FIGlJRE 13.19 cycle behavior.
Example of limit
Phase~Plane Analysis
324
Chap. 13
wrbatiol1 from the closed cycle eventually returns to the closed cycle. Contrast that ,",vith center behavior, where a perturbation leads to a different closed cycle. Limit cycle behavior will he discussed in more detail in Chapter 16.
SUMMARY phasc~plane
As j10tcd earlier,
analysis is a useful tool for observing the behavior or
llOll~
linear systems. We have spent. time analyzing autonomous linear systems, because the nonlinear systems will behave like a linear system, in the vicinity the equilibriulll point (where the linear approxiIl1ation is most valid). A qualitative feci for phasc~planc bdl
or
FURTHER READING Strogatz, S.ll. (1994). Nonlinear Dynmnics and Chaos. Reading, MA: Addisoll
Wesley.
STUDENT EXERCISES linear Problems r~
1.
x
~I
l
°1
_2 x
325
Student Exercises
2.
x
I·· ~
3.
x
I~ ~IX
4.
x
I:
5.
x=
6.
x
7.
Compare
8.
x
9.
x
10.
x
~] x
~IX
1: -I
,::.~.
1
-I
1 2
~I
31 x
-2
-I
2
-2J -2 x
It.
=:-
1
_~] x 0 2 0
r; -3": r: ] 0
I~
-0. 25 1 2 x
-OSI x
0
x
-.0.251 x with x. ,,--2
x··· [-:
0 2 0
o
x
0
_~
x
12. A process engineer has linearizcd a nonlinear process model to obtain the following state-spacc model and given it to your boss. Your boss has forgotten everything he learned on dynarnic systcms and has asked you to study this modcl using linear systcm analysis techniques.
x
Ax
where:
A
10 1.0
- 1.0 1 0.0 .
326
Phase-Plane Analysis
Chap. 13
with initial conditions x,(O) ~ 0.5 and x2(0) ~ - 0.25. What are the eigenvalues of the A matrix? Use bothMATLAB and your own analytical solution. b. Show a phase-plane plot, placing xl on the x-axis and x2 on the y-axis. 13. Consider a process with a state-space A (Jacobian) matrix that has the following eigenvalues and eigenvectors. Draw the phase-plane plot, clearly showing the direc~ tion of the trajectories. The eigenvalues arc: 3.
and the eigenvectors are:
14. An interesting example of phase-plane behavior is presented in the book hy S(ro~ gatz (1994). He develops a simple model for love affairs, using Romeo and Juliet to illustrate the concepts. Consider the case where Romco is in love with Juliet, but Juliet is fickle. The more that Romeo loves her, the more that Juliet resists his love. When Romeo becomes discouraged and backs off, Juliet becomes more attracted to him. Let: J.:!
= Romeo's love/hate for Juliet
x2 = Juliet's love/hate for Romeo where positive state variable values indicate love and negative values indicate hate. The model for this relationship is:
'itl cit = dx _____ f.
dt
=
axz ~
bx
'
where a and b arc positive parameters. Show that this model has center behavior and discuss the meaning from a romance perspective. 15. Consider a more general formulation of the RomcolJuliel problem in 14 above. In this case, let:
dX J dt
Hf
327
Student Exercises
wherc the paramcters lljj can bc either positive or negative. The choice of signs specifics the romantic "styles." For each of the following cases (parameters a and b arc positive), determine the phase-plane behavior. Interpret the meaning of the results in terms of romantic behavior.
a.
(i~L ::0;
ill
a x]
+ b x2
dx, ~ b x dl
b.
dx]
ill
t ::o;~ax]+hx2
dx')
dt e.
4::o;bx-ax 1
2
dx] ilt
(L'lz rll
Nonlinear Problems 1().
As a chemical engineer in the pharmaceutical industry you are responsible for Ii process that uses a bactcria to produce an antihiotic. The reactor has been contaminated with a protozoan that consumes the bacteria. Assume that predator-prey equations are used to model the system (x1 ~ bacteria (prey), x2 ~ protozoa (predator)). The time unit is days. =axI-'YxIXZ
a. Show that the nontrivial steady-state values are:
h. Use the scaled variahles, YI and Y2' Yt = -~ to find the scaled modeling equations:
Y2
~
X
Z,
Phase-Plane Analysis
328 dYI
Chap. 13
,,(I~y,)y,
"I
dh
,11=
~13(I-y,)y?
c. l·'ind the eigenvalues of the Jacobian matrix for scale equations, evaluated at YJs and Yzs. Rcali:r,c that Yt.v and Y2s arc 1.0 by definition. [,'inc! the eigenvalues in teflllS of (X and ~3. d. The parameters are (X :;::: ~1 ::::: 1.0 and the initial conditions arc )'1 (0) : : : ; 1.5 and Y2(0) = 0.75. i. Plot the transient response of)'J and )'2 as a function of time (plot these curves on the same graph using MA'l'LAB). Using your choice of integration methods, simulate to at least ( ::::: 20. ii. Show a phase-plane plol, placing)'1 on the x-axis and)'2 011 the y-axis. iii. \Vhal is the "peak-tn-peak" lime for the bacteria? By how much time docs the protozoa "lag" the bacteria? c. Now consider the trivial sleady-state (xis::: -r2.1':;:::: 0). Is it stable? Pcrfonn simulations when xj(O) 0 and x2CO)"* 0, What do yOll find?
*
f. What if" (0) " 0 and x2(O) = 0'1 g. What if,,(O) = 0 and '2(0)" 0'1 17. Consider the bioreactor Illodelused in Example 13.7 with substrate inhihition rather than MOllod kinetics (sec Module 8 I'or more detail)
"X.'•
"I
=
(iL~ I ) )
X
'
where: J-Lma\:
y k,
'-'-
0.53
km
0.12
0.4
Sf
4.0
0.4545
and Xl is the biomass concentration and Xl is the substrate concentration. Assumc- that the steady-slate dilution rate is D.I ,::: 0.3,
a. J"ind the steady-state (equilibrium) solutiOllS (Hint: I'here arc three).
h. Analyze the stahility of each steady-state, l·'ind the Jacohian, the eigenvaluc-s, and the eigenvectors at each steady-state.
Student Exercises
329
c. Construct a phase plane plot. What do you observe about the unstable steadystale?
d. What would you do if it was desirable to operate the reactor at an unstable steady-state? 18. Perform some time domain plots related to the phase-plane plots for Example 13.7. Discuss how these plots relate to the phase-plane results.
19. A chemical reactor that bas a single second-order reaction and has an outlet f10wrate that is a linear function of height has the following model where the outlet f10wrate is linearly related to the volume of liquid in the reactor (F:;:; f3 \I). F ~'(Ci" ~ C) ~ kC'
dC dl
dV
Lit
=
F 1/1
~
'"V t-'
The parameters, variables and their steady-state values arc shown below: inlet flowrate (I liter/min) gmollliter) C::::: reactor concentration (O.S gllloi/liter) V = reacto volume (I liter) k = reaction rate constant (2 Iiter/(gmol min)) f3:;:: I min-] Fill :;;: ('ill
= inlet concentration (I
Perform a phase-plane analysis and discuss your results. 20. Consider two interacting tanks in series, with outlet flowrates that arc a fUllction of
the square root of tank height. The flow from tank I is a function of \/h l -ill' while the tlowrate out of tank 2 is a function of \Ih}
F
The following modeling equations describe this system:
dh' dl . f,(h"h,,!'} dh, = IMh' ,h,,Ii)
I
dl
I
Phase-Plane Analysis
330
I-'or the following parameter values
13,
ft25 c_
and the input
2-5
5
13 2 = V6
_ mill
F
-~~
=
5 [(2
[('
5 111m
The steady-slale height values arc
hJs
10
Perform a phase-plane analysis and discuss your results.
A 2 cclOlt'
Chap_ 13
INTRODUCTION TO NONLINEAR DYNAMICS: A CASE STUDY OF THE QUADRATIC MAP
14
''['his chapter provides an introduction to bifurcation theory and chaos. After studying this chapter, the reader should be able to:
Sec the similarity between discrete time dynamic models and lluillcric;:1! methods Dctcnninc the asymptotic stability of a solution to the quadratic map Unclcrstand the concept of a hi fllrcation
Understand how to find pcriod-2, pcriodA, .. _, period-a solutions or the universal number 4.6691 96223
• Understand the significance
When a parameter of a discrclc~timc model is varied, the llUlnbcr and character of solu~ lions may changc---lhc parameter that is varied is known as a bU;tfcutiol1 parameter. For some values of the bifurcation parameter, the dynamic model may convcrgc to a singlc valuc aftcr a long valuc of timc, while a slnall change in the bifurcation parameter may yield periodic (continuous oscillations) solutions. For some discrete equations, valucs of the parameter may yield solutions that appear random --these are typically "chaotic" solutions. Chaos can occur in a single nonlincar discrete equation, whilc threc autonomous (no explicit depcndencc Oil time) ODEs arc required for chaos in continuous models. The major sections in this chapter arc:
14.1
Background
14.2 14.3 14.4
A Simple Population Growth Modcl A More Rcalistie Population Model Cobweb Diagrams 331
Introduction to Nonlinear Dynamics: A Case Study
332
14.1
14.5
Bifurcation and Orbit Diagrams
14.6
Stability of l'"ixed-Point Solutions
14.7
Cascade ofPeriod-Doublings
14.8
Purthcr Comments on Chaotic Behavior
Chap. 14
BACKGROUND Many engineers and scientists have assumed (at least. until roughly twenty years ago) [hill
simple models have simple solutions and simple behavior, and lhat Ihis behavior is predictable.lndeed, the main objective for developing a rnodel is Llsually to be able to predict behavior or to match observed behavior (measured data). During the past thirty years, a nurnhcr of scientists and engineers have discovered simple 11lOLlcis \v!lerc the short-term behavior is predictable, hut sensitivity to initial conditions make the long~tertlJ predictillil impossible. By initial conditions. \\,ie mean the value of the variables at the beginnillg of the integration in time. All example is the simple weather prediction model of Loren/. (1963). which is a system of three nonlinear ordinary differential equations; the Lorell/. 1l1Odcl is covered in more detail in ChaptGr 17. Another example is the population grO\vth model used by May (1976), which is a single nonlinear discrete time equation. ·fhis population lllodel is the topic of this chapter. The cOllllllonly accepted term for the dynamic hehavior a system that exhibits sensitivity to initial conditions is clwos.Tcrms for the branch of mathematics related to chaos include nonlinear dynamics, dynamical systems theory. or nonlinear science. \in\' chaos books, written for a general audience, appear frequently: some of the more interesting olles arc referenced at the end of this chapter. This chapter will not make you an expert on nonlinc.'lr dyn<1111ics, but it will help you understand what is meant by sCl1sitil'itr to initiol ("ollditiollS and practical limits to long-term [Jredictability.
or
14.2
A SIMPLE POPULATION GROWTH MODEL Assume that the population of a species during one time period is a functioll of the prcvi~ nus time period. Perhaps we are interested ill the number of bacteria cells that arc gnnving in a petri dish. or mayhe we arc concerned ahout the population of the United States. In either case. the Inathematical model is: (14.1 )
wherc
Ilk
~
hk die.
~
population at the beginning of tilne period k number of births during tinlc period k number of deaths during time period k
Now assume that the number of births and deaths during tillle period k is proportional to the population at the beginning of time period k.
Sec. 14.2
333
A Simple Population Growth Model
( 14.2) (14.3)
where (Xli and
(Xd
arc birth and death constants. Then: (14.4)
llk+1
which \ve call write as: (14.5)
(l+r)II,
or,
where
r;::: (Yb ~ eXd'
(14.6)
Eqn. (14.6) call be simply written as: Ilk I I
where ex ::::; I + r;::: I + ex/} - c"-r/ (obviously, The analytical solution to (14.7) is:
0'
=
( 14.7)
0' fl k
> 0 for a physical system) ( 148)
where /lois the initial condition. 1"1'0111 inspection of (14.8) we observe lhal
if (X < 1 The population decreases during each lime period (converging to 0). if (X > I The population increases during each time period (-----+ 00). if ex;::: I
The population remains constant during each time period.
These results arc also shown in Figure 14.1 3 2.5
D
+ alpha
2
0
~
c 0
1.5
~
+
"3
+
+
+
1.1 +
+
+
+
alpha
10
0-
0
0-
0.5
alpha
0 0
2
4
6 time step
FIGURE 14.1
Simple population growth model.
8
10
334
Introduction to Nonlinear Dynamics: A Case Study
Chap. 14
These results arc easily rationalized, since birthsdcalhs, and births == deaths for the three cases. The result for 0: > I is consistent with Malthus. who in the nineteenth century predicted an exponential population growth. The result that the population increases to ex) for (X > J is a bit unrealistic. In practice, the amount of natural resources available will limit the total population (for the bacteria case, the mnounl of nutrients or the size of the Petri dish will limit the maximum number of bacteria that can be grown). In the next example, we show a simple model that "constrains" the maximum population.
14.3
A MORE REALISTIC POPULATION MODEL A common model that has been used to predict population growth is known as the logistic equation or the quadratic map (May, 1976). ( 14.9)
Here, .r" represcnts a scaled population variablc (sec student excrcise 3). Note the similarity of (14.9) with the numerical methods presented in Chapter 3: Xk , I ~
g(x k )
( 14.10)
Recall that direct substitution is sometimes used to solve a nonlinear algebraic equation. The next guess (iteration k+ 1) for the variable that is being solved for (x) is a funetioll of the current guess (iteration k). Equation (14.9) shows how the population changes from time period to time period--·--that is, it is a discrete dynamic equation. Since (J4.9) is the same f011n as (14.10), we will learn a lot about the quadratic map frOTH analysis of the direct substitution technique and vice versa. You will also note that many numerical integration techniques (Euler, Rungc-Kutla) have the fOHn of (14. [0). Since (14.9) is a discrete dynamic equation, we can determine the steady-state behavior by finding the solution as k ------;. 00. Writing (14.9) ill a more explicit form, (X
(14.11 )
xlc.
as we approach a steady-state (/Lred-poillt) solution, xk+l ::::: xk- so we can write: Xs
0'
.:r, -
(X
.Y./"
(14.12)
which can be written:
"X,' - (" -
I) x, ~
(J
(14.13)
We can use the quadratic formula to find the steady-state (fixed-point) solutions:
x,
~
0 and
1
u
(14.14)
(y
It is easy to see from (14.9) that if the initial population is zero, it will renwin at zero. For a n0I1-2erO initial condition, one would expect convergence (steady-state) of the population to (0.' - l)/cl'_. We will use a case study approach to show that the actual long-term
Sec. 14.3
A More Realistic Population Model
335
TABLE 14.1 Parameters and Non· zero Solutions for Four Cases Case
"
X ,I
I
2.95 3.20 3.50 3.75
0.6610 0.6875 0.7143 O.n33
2 3 4
(stcady~statc)
behavior can be quite complex. Table 14.1 shows the is expected from (14.14).
(X
parameter and the
non-zero
stcady~stalc that
14.3.1
Transient Response Results for the Quadratic (Logistic) Map
Each case presented in Table 14.1 has distinctly different dynamic behavior. As shown in the following sections, case 1 illustrates asymptotically stable hehavior, cases 2 and 3 illustrate periodic behavior, and case 4 illustrates chaotic behavior.
ASYMPTOTICALLY STABLE BEHAVIOR Let Xo represent the initial condition (the value of the population at the initial time) and .tk represent the population value at time step k. For case I we find the following values, using the relationship xk+l := 2.95 -'-"k (1 - ·l"k): Stepk
xk
0 1 2 3 4 5
0.1 0.2655 0.5753 0.7208 0.5937 0.7116
0.2655 0.5753 0.7208 0.5937 0.7116 0.6054
w
0.6610
0.6610
Xk+!
The transient response for case 1 is plotted in Figure 14.2 for an initial condition of 0.1. Notice that the responsc converges to the predicted steady-state of 0.6610. This type of rcsponse for continuous models is usually called asymptotically stable hehavior since the output converges to the steady-state (fixed-point) solution.
PERIODIC BEHAVIOR The transient response curve forcase 2 is shown in Figure 14.3. The curve oscillates between 0.513 and 0,800, while the predicted resull (equation 14.14 and Table 14.1) is
Introduction to Nonlinear Dynamics: A Case Study
336
population model, alpha
0.8
Chap. 14
2.95
0.7 c 0
.~
0.6
-S
0-
0
0-
0.5
m m
"
C
0.4
0
.~
c
"E '5
0.3 0.2 01
0
10
20
30
40
50
time step
FIGlJllE 14.2
crransicnt population response, easel; cOllverges to single
steady-state.
population model, alpha::: 3.2
0.8 0.7 c 0
~
0.6
-S
0-
0
0-
0.5
m m
"
C
0.4
0
.~
c
"E '5
0.3 0.2 0.1
0
10
20
30
40
50
time step
FIGURE 14.3 Transienlpopulation response, case 2; oscillates between values (pcriod-2 hehavior).
(\\,0
A More Realistic Population Model
Sec. 14.3
337
0.6875. 'This type of response is known as pcriod-2 behavioL In case 3 the transient response oscillates between 0.383, 0.827, 0.501, and runs as shown in r'igurc 14.4. This is known as pcriod-4 behavior, since the system returns to the same slale value every fourth time step.
CHAOTIC BEHAVIOR For Case 4, the transient response never settles to a consistent set of values, as shown in Figure 14.5; rather the values appear to be somewhat "random" although a dctcnninislic equation has been used to solve the problem. Figure 14.6 shows that a slight change in initial condition from 0.100 to O. J 0 J leads to a significantly different poinHo-poinl rcsponsc--this is known as sensitivity to initial conditions and is characteristic of chaotic systems.
WHERE WE ARE HEADING At this point, you arc probably wondering how to predict the type of behavior that thc quadratic map is going to have. Changes in the (X parameter have led to many different types of behavior. The purpose of Section J4.4 is to show how to predict the type of bc~ haviorthat will he observed using cobweb diagrams. Section 14.5 will thcn introduce bifurcation plots, which reduce the long-term result.s from many transient plots to a single plot. Section 14.6 introduces linear stability theory for discrete systems. Section 14.7 shows how to find the period~l1 points.
population model, alpha == 3.5
0.9 0.8 co
0.
0.7
.~
"5
0. 0. 0.
w w
'"0.
0.6 0.5
C
.~
0.4
co
'"
E '6
0.3 0.2 0.1
0
10
20
30
40
time step
FIGURE 14.4 Transient population responsc,case 3; oscillates between four values (periodA belmvior).
50
Chap. 14
Introduction to Nonlinear Dynamics: A Case Study
338
population model, alpha:o= 3.75
0.8 c 0
~ "3
0. 0 0.
0.6
"'"' ill
C 0 iii c ill
0.4
E '5
0.2
o
o
40
20
60
80
100
time step
FIf;LJRE 14.5
Transient population response, case 4; chaotic behavior,
population model, alpha:;;;: 3.75
1\ c
0.8
0
:g "3
0. 0 0.
"'"'ill C 0
!
!
I
I
\1
I
\
, I
c
\
I
I
\
!
\ I
1/
\i
ill
I I
I
\ I
I
0.4
iii
E '5
I I
\ !
0.6
" "
\'
0.2
0 75
80
85
90
100
95
time step
FIGURE 14.(j Transient population response, case 4; chaotic behavior (Solid Line-- -initial condition of Xo ;:;:: O. L Dashed Iinc----initial condition of X o 0.10 I). 'This illustrates the sensitivity to initial conditions.
::;0;
Sec, 14.4
Cobweb Diagrams
339
Xn + 1 1
0,75
°5 0,25
°"'-
°
14.4
-L
0,25
--'-
°5
-'
--'
0,75
1
x n
FIGlJRE 14.7 Cobweb diagram for the quadratic map problem. The initial point is -'0::;:;; 0.1.
COBWEB DIAGRAMS Insight to the behavior of discrete singlc~variablc systems can be obtained by constructing cobweb diagrams. Cobweb diagrams are generated by plotting two curves: (i) g(x) versus .r and (ii) x versus x; the solution (fixed-point) is at the intersection of the two curves. For exampic, consider the case 1 parameter value of (y ::::: 2.95 and an initial guess, ):0 ::: 0, I, The xlI+1 = g(x) = 2.95 .tll (I -.1:) curve is shown as the inverted parabola in I<'igurc 14.7. Since the X o value is 0.1, the x J value is obtained by first drawing a vertical line to the g(x)
Xn +l
1
0,75
0,5
0,25
o i<::
o
--'0.25
-' 0.5
"0.75
--l
1
x
n
FIGlIlU: 14.8 C"se I (n ~ 2,95) rnap, convergence to a single solution (x = 0.6(1); corresponds to the transient response in Figure 14.2.
340
Introduction to Nonlinear Dynamics: A Case Study
Chap. 14
Xn +l
I
0.75
0.5
025
FIGORE 14.9 Case 2 (LV:::: 3.20) map, oscillates between x:::: 0.5130 'Illd "-
-'-
0.25
-'
0.5
'-
0.75
-' x n 1
0.7995 after initial transient; initial condition ol'xo:::: 0.1- -corresponds to
the transient response in figure 14.3.
i
i curve to rind g(.ro) :::: 0.265, then drawing a horizontal line to the x :;:;; .·r curve (since xI :::; g(xo); therefore, ·-tt :::: 0.2(5). A vertical line is drawn to the g(x) curve lto obtain g(x t ):::: 0.575), then a horizontal line is drawn to the x:::: X curve (so, xl:::: 0.575), "fhcsc
initial steps arc shown in Figure 14.7. l'igurc 14.8 shows that this process converges to the fixed-point of x,~, :::: 0.661, for the case 1 parameter value of 0' :::: 2.95. Figure 14.9 shows that the iterative process eventually "bounccs" between two solutions for the case 2 paramctcr value of a;:;;:; 3.2. This is shown more dearly in Figure 14.10 where an initial guess of Xo = 0.5130 leads to solu· tions of 0.5130 and 0.7995 (pcrioet-2 behavior). Case 3 has pcriod-4 behavior. as shown in Figurcs 14.11 and 14.12. Case 4 (Figure J 4. J 3) is an example of chaotic behavior. where the sequence of iterates never repeats.
Xn +l
1
0.75
0.5
0.25 FIGlJRE 14.10
o "-----'-----'-----"'-----' xn
o
0.25
0.5
0.75
1
Case 2 (0 ~ 3.201 map, oscillates het\Vccll x = 0..') l_lO and 0.7995; initial condition or Xo = n.5130 (period-2 behavior).
)i
Xn+l
I
0.75
0.5
025
o "'-
-"-
o
0.25
-'. 0.5
'-0.75
-' x 1
n
FIGURE 14.11 Case 3 (n ~ 3.50) map, oscillates between x:;:;o 0.3828, 0,8269,0.5009, and 0.8750, The initial condition is xo:;: 0.1. This COITcsponds to the transient response in Figure 14.4 (pcriod-4 behavior),
Xn +l I
0.75
05
0.25
o "'-
-'-
o
0 _25
-' 0.5
'--__--' x 0.75
1
n
FIGURE 14.12 Case 3(n ~ 3.50) map, oscillates between 0.3828, 0.8269,0.5009, and 0.8750, initial condition of Xo :;: 0.3828 (period-4 behavior).
0.75
0.5
0.25
o
~
o
...L
0.25
-'
0.5
.L..
0.75
.J x n 1
FIGURE 14.13 Case 4 (" ~ 3.75) l.nap, chaotic behavior; corresponds to the transient response in Figure 14.5. 341
ESCOLA D2 tNGENHARIA IJIt3L10Tt:CA
342
14.5
Introduction to Nonlinear Dynamics: A Case Study
Chap. 14
BIFURCATION AND ORBIT DIAGRAMS vYhcn a parameter of a discrete-time model is varied, the number
or
14.5.1
Observations from the Orbit Diagram (Figure 14.14)
There is a single steady-state solution untillY::;:: 3, where a bifurcation to two solutions occurs. 'fhe next bifurcation point is {Y ;:: 3.44949, where four solutions emerge. A period~X bifurcation occurs at 0' ::: 3.544090, period-J6 at (X ::: 3.564407, j"Jeriod-32 (:It Ct ::: 3.568759, and period-64 at (X ;:: 3.569692. Chaos occurs at 0'. :::;; 3.56995. Notice that there are some interesting "windows" of periodic behavior, after the onset of chaos. For example, at (X ;:: 3.83 we find a window of period-3 behavior. The period-3 behavior occurs after approximately 60 time steps, with an initial condition of 0.1, as shown in Figure 14.15. This behavior is shown more clearly in Figure 14.16 which is simply the data from
0 9
o. 8 0 .7
/-~/'---~~ 1----.
o. 6 x
'
.....--.......--.......
~---------./-----~
0 .5
'~-
0 .4
o. 3 0.2
o. 0
2 9
3
3.2
3 4
3.6
3.8
U.
FIGURE 14.14
Orbit diagram for the quadratic map.
0'
is the bifurcation parametcl".
4
Bifurca tion and Orbit Diagram s
Sec. 14.5
343
populatfon model, alpha
;:=
3.83
C
0.
'fg S
0.
0.
0. if> if>
ill
C g if>
C ill
E i3
0.1
o
40
20
80
60
100
time step
3.83, initial FIGUR E 14.15 Period-3 behavior (after inllial transient) for Ci. 7. 0.95741 and , ,0.50466 0.15615 x;:;;: are values Periodic 0.1. ;;:;; condition =0
c
0.
~ S
0. 0. 0.
I
I I
0.6
+
C
i
I I
0.2 0 75
+
+ I
Y 80
I
'"
\
+ / I I I
I
/
/
"
\f
\/
I
~
+
+
Ii
'\
I
I
I
\
I
I
\
I' I
I
I
I
I \
I /
/
I
/
/
\
I
\
/ 'I
I
I
,"
/ I
I
/
I
0.4
C ill
E i3
\
f
ill
I I I
I I I
0.8
I, /
+
tI
t
t
i' 'I
1\
I
if> if>
0. ijj
+
\'
!I
85
I
I
,
+
\
\ I
\
"
+
I
+ /
/
\
I I
\
" I "/
90
, I
\
I
/
I'
4-
I I
I
I
\
/
t
95
time step
Pcriod-3 behavior for (~ ::;: 3.lB. Values arc x ;:;: 0.15615, FIGlJR E 14.Hj . 0.957417 and 0.50466,
100
344
Introduction to Nonlinear Dynamics: A Case Study
Chap. 14
Xn +l
1
0.75
05
0.25
o &L_ _--' o 0.25
-'-
FIGURE 14.17 Pcriod-J bclwvior at 3.83. Values arc x:;:::; 0.156\ S. 0.50466. and 0.957417.
-'-_ _---' X
05
0.75
(X
n
1
0;::
Figure 14.15 plotted between 75 and 100 time ;.;tcps. The cobweb diagranl of Figure 14.17 also shows the pcriod-3 behavior. Research has shown that pcriod-:, behavior implies chaotic hehavior.
14.6
STABILITY OF FIXED·POINT SOLUTIONS When we performed our case study, we found that case I converged to the predicted fixed point, while the other cases had periodic (or chaotic) solutions that were not attracted to the fixed points. We wish now to usc an analytical method to determine when the solutions will converge to the fixed-point solution ~ -that is, when is a fixed-point stable? The follo\ving stability theorem is identical to the stability theorem used for the numerical analysis in Chapter 3. '~~"'-'--"
Definition
Theorem
14.6.1
Let x'~ represent theJixed-fJoillf solution of x'o: :;::: g(x*'), or g(x''') -- x~':;::: O.
.:r* is a stable solution of x* :;::: g(x*), if ilg _
Idx
I
<~
1 whcn evaluatcd at ).:* .
Application of the Stability Theorem to the Quadratic Map
Here we will make lISe of this theorem to determine the stability or a solution to the quadratic map:
g(x)
~
(XX( I - x)
= (l'X - (U'
(14.15)
Sec. 14.6
345
Stability of Fixed,Point Solutions
So, iJg
.=<¥-2,xx=<¥(1
2x)
dx
For simplicity in notation, we will usc g' to represent i)g/(Jx. r~valLJatcd at x*, we have:
=
g'(x*)
2exx* = ev (1
2x*)
(14.16)
Therefore, from (14.16) and the stability theorem, if the following condition is satisfied:
I,x (1
2x*)
I<
I
( 14.17)
Theil'. x* is a stable solution. Remember from (J 4.14) that there arc two solutions to x* :::: t¥x*(l-.-t*):
x*
c. 0
x*
or
=
<¥
Momentarily we will generalize the stability results !()r any value of cc First, we will study the four specific cases.
CASE 1
(Y
= 2.95
At one fixed point solution, x*
:=
0, we find:
Ig'(x··) 1·1 <¥ (I
- 2x*)
I····· 2.95 >
I
which indicates that the fixed point is unstable. At the other fixed point solution, x*::::
Ig'(x*)1
=
I
0' -
I/O':::: 2.95 - 112.95 = 0.6610, we find:
- 2x*)I··12.95 (I - 2(0.6610» 1= 1-0.9499 I
= 0.9499 < 1 \vhich assures that the second fixed-point is stable. 1;'01' 0' := 2.95, we expect the numerical solution to converge to the stable fixed point, x*:= 0.6610, since the olher fixed point (x*;:;;;; 0) is unstable. The stability results for cases I through 4 are compared in Table 142. Notice thaI case I is the only one of the four cases where lhere exists a stable solution. The reader . should verify that an initial guess ofx arbitrarily close to zero (but not exactly 0), will not converge to zero for any of the four cases.
TAIlLE 14.2 Case
Stability Results for Cases 1-4 (Y
r"'
Ig'(x*) I
condition
2.95 3.20 3.50 3.75
unstable unstable unstable unstable
x'~
Ig'(x*) I
conclition
0.6610 0.6875 0.7143 0.73"
0.9499 1.2000 1.5000 1.7500
stable unstable unstable unstable
~-~.
2 3 4
2.95 l.20 3.50 3.75
0
0 0 0
Introduction to Nonlinear Dynamics: A Case Study
346 14.6.2
Chap. 14
Generalization of the Stability Results for the Quadratic Map
Notice thai we have been quite limited in our study, since we have only considered four with 2.95 S ex :S. 3.75. Now we will consider the general results for any ('( > O. Again, recall that there arc two fixed-point solutions for a given value of (X
ca~:;cs
x*
~
()
or
x*
At the risk of complicating our notation, let and
xI =
Our goal is to determine how the stability of x{~ 01
x'o AS A
STABILITY OF
Since x(~:::: 0 and g'(xfj)
then
Ig' (x~)1
=
(X -
.'.:;"
(X-
lX
changes as a function of 0',
FUNCTION OF lY 2nO ::::: (X
I ex I
Also, since III < 1 is required for stability, then x{~ is a stable solution only as long as -1 < 0' < J. Otherwise, x('; is unstable (rccalIthal Ci < 0 docs not make physical sense).
STABILITY OF x~ AS A FUNCTION OF lX SincexT~(lX-I)/lxandg'(xT~lX-ZlY(lX-I)lcx~lY-Z(lY-I)~-lX+Z
Ig'
I , I-c, zi
(x'D + which indicates stability for I < (X < 3. Otherwise, is unstable. These results arc shown on the bifurcation diagram of Figure 14.18 for 0 < (X < 4. Generally, solid lines will be Llsed to represent stable solutions and dolled lines will be used to represent unstable solutions. As discussed above, a change of stability for x;j occurs at 0' ::::;: I. Also, changes of stability for x't occur at 0' ;:::: I and (Y :::: 3. The values of 0' where the stability characteristics change arc known as bifurcation points. The bifurcation that occurs at 0' = 1 is commonly known as a ';lranscritical" bifurcation (sec Chapter 15)-an exchange of slability between the two solutions has occurcd. Notice that Figure 14.18 is a h(fl-lrcatioll diaRrarn based on a linear stability analysis. It differs frol11 an orhit diagram (sllch as Figure 14.14), because it docs not show the periodic behavior obtained from solving the nonlinear algebraic equation for the popula~ tion growth model. An orbit diagram cannot disp(ayullstable solutions, however.
then
14.6.3
x1'
The Stability Theorem and Qualitative Behavior
The theorem states that if lag/ax I at the fixed point, the fixed point is stahle. Further, if ag/ox is negative, then the fixed point solution is oscillatory. If dg/ax is positive, the be~
Stability of Fixed-Point Solutions
Sec. 14.6
347
bifurcation diagram
lr---r---,----r---.-----,----.---~-...._,
"I
0.5
0
x
...-------
.. stable
.. Xo
---
UIlBtable
..
X
stable
1
o 1--------------------------------UIlBtable
I f
I I
-0.5
I
I
UIlBtable
I .. IXI
I I -11..---'---~--~-~~-~--~--~---'
o
0.5
1
FIGlJRE 14.18
1.5
2
2.5
3
3.5
4
Bifurcation diagram based on linear stability analysis.
havior is monotonic. We can then develop the fo]Jowing table of results from the stability theorem
i)g
stability
response
unstable
oscillatory
<.: 0
stable
oscillatory
I
stable
monotonic
unstable
monotonic
itr <~l
- 1< 0< > I
ilf;
Dg
ax
a.r <
Although the linear stahility analysis is useful for determining if a fixed-point is stable, it cannot he used directly to understand possible periodic behavior. This is the topic of the next section.
Introduction to Nonlinear Dynamics: A Case Study
348
14.7
Chap. 14
CASCADE OF PERIOD-DOUBLINGS We have noted that there appears to be a series of period doublings in route to chaos. The limitation to the method presented in Section 14.6 showed that it could predict that a particular fixed point was unstable, but could not identify the type of periodic behavior that might occur. In this section we will show how to find these period-doubling bifurcation points and the respective branches shown in Figure 14.13.
14.7.1
Period-2
When period doubling occurs, the [lopulation value at time step k is equal to the value at time step k -- 2. This can be represented by (141 X)
or
using the notation ( 14.19)
then (14.20) (14.2 I) (14.22)
Warning:
Do not confuse the g2(rk ) notation with that
or the square of the operator Ig(xk ) ]2.
r:or the quadratic map, we call develop the relationship shown in (14.22): Xk+2
and substituting xk-tl
:=
a ·X"k
(I
~ J.~k)
=
aXkl1
(1 ~xk\l)
(14.23)
into (14.23), we find: (14.24)
Since (from 14.19):
we can write (14.24) as ( 1425)
Sec. 14.7
349
Cascade of Period-Doublings
Expanding (14.25), ( 14.26) or
g'(x,) ~ a' [---a x/ + 2ux", - (1 + u)x,' + x,l
(1427)
Notice that there arc four solutions to the fourth-order polynomial. We can find the solutions graphically by plotting g2(x) versus x, as shown in Figure 14.19 (for (X 3.2). If you closely observe the plot, you will find the following four solutions for pcriod2 behavior: ;:0;;
x"
~
0,0.5130,0.6875, and 0.7995
We can sec graphically that the solutions x* ::::: 0 and 0.6875 arc unstable, since the slope of g2(x*) is greater than L (A period-2 solution is stable if 1()(;;2(x»/d.\:1 < 1.) Also, notice that the solution for x ::::: g(x) will always appear as one of the solutions for x ::: g2(x). If a solution for x ::::: g(x) is unstable, it will also be unstable for g2(x). We can sec that x ;::; 0.6875 is the solution for both .Y ::::;; g(x) and x : : ; g2(x), by observing [!igure 14.20. At this point it is worth showing the results of x versus g(g(.:r) for case I, which we know has a single, asymptotically stable solution. Figure 14.21 shows that there is a sin~ gle stable solution of X ::::;; 0.6610. This makes sense, because as k -) we know that x k :::: 0.6610; this Illeans that xk+2 -= x k+ 1 -= :r k ::::: 0.6610. We can sec fromF'igure 14.22 that n ;;;;; 3.0 is a bifurcation point, since absolute values of the slope of g(x) and g2(J::) ::::: I at x::::: 0.66667. Figure 14.22 is clearly a transition point between l<'igurc 14.21 and Figure 14.20. 00,
1,
/.
08!
('
,""
~
I
I
0.4'
unstable
I
//~
.I 0.2;J
il
/f\
j1". . , " \ / /Vi
~
A
./
stable
\
i' Ii
X /
//~
\!Ii
unstable
i
o
02
!
I I
/
/v oV
\
OA
OB
x
DB
VlGlJRE 14.19
Plo! of g(K(X))
versus.r 10 find the period-2 values for 0';::;:
3.2.
Introduction to Nonlinear Dynamics: A Case Study
350
Chap, 14
0,8
"'"
0,6
0;
0.4
CT
"" >{
02
°°
02
0,6
0.4
0,8
FIGURE 14.20
x
gl(x) versus.x for
Plots of g(x) and (t ""
J.2.
0,8
"'" ""
0,6
CT
0;
0.4
>{
02
0.4
0,6 x
OB
FIGtJRE 14.21
Plots of g(x) and 0:;;;:; 2.95. No
g(g(x» versus x for
pcriod-2 behavIor.
Sec. 14.7
Cascade of Period-Doublings
351
0.8
s:
0.6
"
0.4
'& 0 0;
" 0.2
0 0.2
0
0.4
0.6
0.8
FIGURE 14.22 Plot of g(x) aod g'(x) ven,lIs x for n "'" 3.0.
x
14.7.2
Period-4
When pcriodA behavior occurs, the population value at time step k is equal to the value at lime step k - 4. This can be represented by: X k +4
= xk
(14.28)
Using the same arguments that we used for pcriod-2 behavior, we can find that since xk+ I :;;;; g(..r k),
(14.29) ~ g(g(x kU »
~ g(g(g(xk+'») = g(g(g(g(x k))))
.
_g'(x )
Xk-I-4 --
k
( 14.30)
We can obtain the solutions to (14.30) by plotting g4(x) verSllS x as shown in Figure 14.23 for C/. = 3.5. Again, do not confuse g4(x) with Ig(x)]4 g4(x) is an eighth-order polynomial with eight solutions as shown in Figure 14.23. Figure 14.24 shows that the solutions for x:::: g(x) and x :::;: g2(x) arc also solutions (although unstable) for x = g4(x).
Chap. 14
Introduction to Nonlinear Dynamics: A Case Study
352
alpha:= 3.5
u 0.8
s
s u
-;;-
0.6
;,-
'"~ 0,4
u
s
0.2
u 0 0
0.2
0.6
0.4
0.8
x FIGURE 14.23
Plots of g4(x) versus x to find the pcriod-4 values for Ci "" 3.5.
alpha = 3.5
, "
::
0.8
'\ ..,, \ .,, ,
-;;-
t;
'~;\ ::'I
C5 0.6
...."
:s '0,
\.
;,
•
~OA
'"
V\
~-x
-g(x) ·••••..·9'(x) ._-- 9"(x)
0.2
______ ___ L._
0.2
0.4
0.6
0.8
x
FIGURE 14.24
,
\ \
Plot.s of R(X), !,:lex) and ,t,JA-(x) versus x for (.~
0:::
3.5.
Sec. 14.7 14.7.3
353
Cascade of Period-Doublings
Period-n
By analogy to tnc pcriod-2 and pcriod-4 hehavior, we call sec that for any period have the following relationship
1/,
we
(14.31 ) v Ak -lll
.~
( 14.32)
""(.",.) () ./i
Note that gll(Xk) will he a polynomial that is order 2n, and there will be 2n solutions, n of which arc stable.
14.7.4
Feigenbaum's Number
The quadratic map exhihits a period doubling route to chaos. As the hifurcation paralnetcr 0'. is increased, model goes through a series of period doublings (IJcriod-2, perindA, period-S, period-16, etc.). Feigenbaum noticed that the quadratic map had a consistent change ill the bifurcation parameter between each period doubling. Indeed, he found that any "one-hump" (sec any plot of g(x) for the quadratic map) model will have a cascade of bifurcations which will yield the /"'·eif;cn!Jaum nurnbcr. The r"eigenbaum Humbcr is calculatcd by comparing (X valucs at each sllccessivc hifurcation point in the following fashion -" Ct
lim
'1 ~ 4.669196223
( 14.33)
C'ii+ [ - O:i
where C'ii represents thc parameter value at thc i th period doubling point, where the period is 11 ::::: 2 i . To obtain a rough cstimate of the Feigcnbaum number, usc thc valucs of (X for period-16 (2 4 ), pcriod 32 (2 5 ) and pcriod-M (2 6 ) e
3.56R759
~
3.5044IJ7
3.569692
~
3.56R759
~
4.6045
which is close to 4.6692 A summary of thc bifurcation points is providcd in 'fablc 14.3. Chaos occurs when the pcriodis 00 (state sequcnce never repcated) at
TABLE 14.3
I 2 3 4 5 6 ,j'j
Values of (\~ at Bifurcation Points period
"
2
3.0 3.44949 3.544090 3.564407 3.568759 3.569692 356995
4 8 16 32 64 oc
0' :::::
3.56995.
354
Introduction to Nonlinear Dynamics: A Case Study
Chap. 14
FURTHER COMMENTS ON CHAOTiC BEHAVIOR We have llsed the quadratic map (a model of population growth) to introduce you to nonlinear dynamic behavior. This model consisted of a single discrete nonlinear equation. Dynamic behavior similar to period-2 call result from a sct of two nonlinear ordinary dir~ fcrential equations. Exanlplcs of period behavior in continuous systems include the Lotka-Voltcrra model used to predict the populations of predator and prey species. The change in a bifurcation parameter that causes a limit-cycle to form in a 2 OIJE system is known as a !ioplb{fllrcatiOIl, and will be covered in Chapter 16. Chaos is possible in a sd of three autonomous nonlinear ordinary difJcrcntial equations. This behavior was discovered by Lorenz in a simple (reduced~order) model of a \veather system (really a model of natural convection heat transfer) and will he detailed in Chapter 17. Lorenz coined the phrase "butterfly effect" to describe a system of equations that is sensitive to initial condilions (hence chaotic). He stated conceptually that a butterfly flapping its wings in ·froy. New York could cause a monsoon in China several months later (<Jr something similar~l. Some of the earliest results of what is now known as chaos were really discoverclJ by Poincare in the late nineteenth century, involving the three-body problem. He found that it was easy to determine the planetary motions due to gravity in a system \\lith t\VO bodies, but when three bodies were considered, the system of equations became nonintcgrable-leading to the possibility of chaos. We sec turbulence throughout our daily lives, from the waleI' flowing from our hlUcels, to the effect of wind blowing through our 11;:\ir as we ride our bicycles. to the boil, ing \vater on our stoves. Many researchers have tried to model turbulence by adding stochastic (random) terms to our models of physical behavioL It has only been realized in the past three decades that a good physical (nonlinear) model can simulate the effects of tur hulence through chilDS. Nwnerical mcthods arc used to solve thc vast majority of chemical process models. Angclo Lucia (sec references) has found that chaos can occur in the solution of some tiler Inodynamic equations of state if the numerical lllethods arc not formulated carefully. It is likely that many people have obtained similarly bad solutions before reformulating thcll1 correctly.
SUMMARY A lot of material has been presented in this chapter. You may be wondering how discretc maps and bifurcatiolltheory tics in with applications in chemical engineering. There arc at leastlwo important reaSOns for studying this material: We have shown that the quadratic map problem is conceptually identical [0 numer! cal methods that can be used to solve a nonlinear algcbraic equation. Since lhe qua dratic map problem cxhibits exotic behavior undcr certain values of the paramcter n. this tells liS that a poorly posed I1lHllerieal method lllay have silnilar problcms. f3c can~/ltl when llsing J/umerical methods.'
References and Further Reading
355
We noticed that a discrete population growth model is represented by the quadratic map problem. This population growth lnodel is a simple example of a discrete dyn,
REFERENCES AND FURTHER READING The prinlary reference for the behavior of the quadratic map is (this has been reprinted in a number of sources) is by -May. May, R.M. (1976). Simple mathematical models with vcry complicated dynamics, Nalure 261: 459-467. 'rhe general field of chaos theory was introduced to much of the public in thc popular hook hy Gleick: Glcick, J. (1987). Chaos: MakinR a lYeH' .')'ciencc. New York: Viking. Lorenz is given credit for the discovery of "sensitivity to initial conditions": Lorenz, E,N. (1963). Deterministic non periodic flows. }ourJ/o! (!{ Atmospheric Sciellce. 20: 130-141. Software for the MacintosllO that was packaged with the following hook was used to generate some of the quadratic map diagrams. ·fhis book also docs an excellent job of discussing tbe quadratic map and dynamical systcms theory. TufiUaro, N.B., T. Abbott, & J. ReiJly. (1992). An I,,);pcrimefltal Approach to linear DYllarnics and Chaos. Redwood City, CA: Addison-Wesley.
NO/l-
Chaos can appear in numerical solutions to chemical engineering problems, such as phasc equilibrium calculations, as shown in the following paper: I.ucia, A., X. Guo, PJ. Richey, & R. Dercbail. (1990). Simple process equatiot/s. .fixed [Join! methods, and chaos. American Institute of Chemical Engincers Journal (lIICI1£.I.), 36(5): 641~654.
356
Introduction to Nonlinear Dynamics: A Case Study
Chap. 14
The book by Strogatz is an excellent introduction to nonlinear dyiHunics: Strogatz, S.H. (1994). Nonlinear Dynamics and Chaos. Reading, MA: Addison Wesley.
STUDENT EXERCISES I. \Vhy arc the results for the simple quadratic map problem important to understand, when chemical and environmental process models arc obviollsly much more complex (hascd 011 ODEs)') 2. Usc MATLAB to generate transient responses for the quadratic map, for various values of n. Explore regions of single st:cady~statc solutions, as well as regions of periodic and chaotic behavior. Use Figure 14.14 to try and find regions of periodic behavior in the midst of chaotic behavior. 3. Derive the scaled logistic equation (14.9) from the following unscaled model for population growth.
where r and I. arc constants (Hint: [)efinc the scaled variablc, x::::: nr/( I + 1')1.). What is the physical significancc of L'!
4. Consider the "constant harvesting" model for population growth, where 'Y is a term that accounts for a constant removal rate per unit timc period (e.g., hunting deer or removing cells from a petri dish),
How docs 'Y effect the cquilibrium population values? (Show calculation, and COllsider stability of the equilibrium.) tAct tV ::::: 3.2. What 'Y values
5. Consider the "proportional harvesting" model for population growth, whcre the removal rate per unit timc pcriod is proportional to the amount of population
How docs 'Y cffect the equilibrium population values? (Show calculation, and consider stahility of the cquihbriuln.) Let (\' ::::: 3.2. What 'Y valucs arc required for 0 (the trivial solution) to bc a stable equibriul11 solution? What 'Y valucs are required for a stable nontrivial solution'? 6. Consider the pcriod-2 behavior that occurs at a value of 0' ;:::: 3.2. Show that the values of x ~ 0 and x ~ 0.6875 are unstahle. (flint: Let h(x) ~ X(X(x» and show thai I/{(x) I ;:> I at those values.)
Student Exercises
357
7. Using MATLAB construct the orbit diagram (Figure 14.14) for the quadratic map. S. F'ind the (real) fixed points 01'.\""+1::: \/t/,' and analyze their stability. Also, develop a cobweb diagram for this problem. 9. Consider the nonlinear algebraic equation, .f(x) ::: _.1;2 - X + I ::: O. Using the direct substitution method, formulated as x::: _.r 2 + I ::: g(x), the iteration sequence is
x kl
J
::0:::
g(x k ) = - xl
--j-
1
Try several different initial conditions and show whether these converge, diverge or oscillate between values. Discuss the stability of the two solutions x';: ::: 0.618 and x* :::;; 1.618, based on an analysis Of{I,/(x~:).Deveiop a cobwcb di<-:tgram for this SYStCllL
10. Consider the nonlinear algebraic equation,f(x)::: -x 2 - x + I ::: O. Usillg Newton's method,
write the iteration sequence in the form of:
Try several different initial conditions and show whether these converge, diverge, or oscillate between values. Discuss the stability or the two solutions .r:t: ::: 0.618 and x":::: 1.618, based on an analysis ofg'(x*'). Develop a cobweb diagranl for this system. 11. Consider the scaled Lotka-Volterra (predator (Y2)-prey (VI» equations, \vhere dv
•• , = a (I ~ v») V,
til
.- .
The parameters are O~ ::: ~ ::: 1.0 and the initial conditions arc Yj(O) ::: 1.5 and YiO) ::: 0.75. The time ullit is days. Integrate these equatioJls lll11ncrically (using odc4 5, for example) to show the periodic behavior. 12. The I--Icnon Inap is a discrete model that can exhibit chaos: x, (k
+ I)
x)(k)
x,(k
+ I)
h x,(k)
-j
I ~ax,(k)'
For a value of!J::: 0.3, perform numerical simulations for various values of u. Try to rind values of a (try a > 0.3(75) that yield stahle period-2 behavior. Show that chaos occurs at approximately a:= 1.06. 13. Read the paper by Lucia ct al. (1990) and use cobweb diagrams to show different types of periodic behavior that can occur when direct substitution is llsed 10 find the volume roots of the SRK equation-of-stale for the multicomponent mixture (CH"I' C ZH4, and C,H(,(J).
Introduction to Nonlinear Dynamics: A Case Study
358
Chap. 14
APPENDIX: MATLAB M-FILES USED IN THIS MODULE
% % % %
function rtime,x] ~ pmod(alpha,xinit,n) population model (quadratic map), prnod.m 29 August 1993 (e) B.W Bequette revised 20 Dec 96 input data:
%
alpha: growth parameter
% %
n: nUlnber of time steps xinit: initial population (between 0 and 1)
(between 0 and 4)
%
%
% % % %
clear x; clear k; clear time; x(l) ~ xinit: time (1) ~ 0: for k ~ 2:n+1; time(k) ~ k-1: x(k) ~ a1pha*x(k-1)*11-x(k-1)): end run this file by entering the following in the command window [time/x] ~ pmod(alpha,xinit,D); with proper values for alpha, xinit and n then enter the following plotltime,x) function [x,g,g2,g3,g4j
~
gn_qmapla1pha):
%
% finds g(x), g"2(x), gA3(x) and g"4Ix) % the quadratic map problem %
% Ie) B.W. Bequette
% 23 july 93 % modified 12 Aug 93 % revised 20 Dec 96 %
x g
zeros (201, 1): x: g2 ~ x: g3
x: g4
%
for i~1:201: xli) ~ (i-1) *0.005: gli) ~ a1pha*x(i)*(1-x(i)): g2li) ~ a1pha*gli)*11-g(i)):
x:
functions for
359
Student Exercises
g3(i) g4(i)
~
-
alpha*g2(i)*(1-g2(i)); alpha*g3(i)*(1-g3(i));
end
% can plot, for example % plot(x,x,x,g,x,g2,
1~I,x,g4,
'_, ')
BIFURCATION BEHAVIOR OF SINGLE ODE SYSTEMS
15
The goal of this chapter is to introduce the student to the concept of bifurcation behavior. applied to systems modeled by a single ordinary differential equation, Chapters 16 and 17 will involve systems with more than one state variable. After studying this chapter, the student should be able to DClcnnine the bifurcation }Joint for a single ODE Determine the stability of each branch of a bifurcation diagram Determine the numher of stcady~statc solutions near a bifurcation point 'fhcmajor sections in this chapter afC: 15.1
15.1
Motivation
15.2
[IIustratioll of Bifurcation Behavior
15.3
Types of Bifurcations
MOTIVATION Nonlinear systems can have "exotic" 11chaviof such as multiple stcady~statcs'and transitions from stable conditions to unstahle conditions. In Chapter 14 we presented the quadratic map (logistic map or population model), which showed how a discrete-time systeln could move from a single stable steady-state to periodic hehavior as a single parameter was varied. This would he considered a dynamic bifurcation of a discretc-tinle system, where the behavior changed from ()symptotically stable to periodic.
360
Sec. 15.2
Illustration of Bifurcation Behavior
In this chapter we introduce bifurcation behavior of
361 conLinllous~timc systems.
A
steady-stale bifurcation occurs if the number of steady-state solutions changes as a system parameter is changed. If the qualitative (stable ys. unstable) behavior of a system changes as a function of a parameter, we also refer to this as bifurcation behavior. This chapter deals with systems modeled by a single ordinary differential equation. Bifurcation analysis is particularly important for complex systems such as chemical and biochemical reactors. Allhough only single variable examples arc used in this l1lod~ ule, the same types of bifurcation behavior afC also observed in chemical and biochemical reactors.
15.2
ILLUSTRATION OF BIFURCATION BEHAVIOR Here a simple polynomial equation will be used to illustrate what is meant by h(jllrcafio!l behavior. Assume that the following cubic polynomial equation describes the steady-state behavior of a system.
f(x.[L) = [Lx - x' = 0 The solution can be obtained by plotting the function and finding the valucs of or where fCql,) :;;:: O. A plot of this function for fL :;;:; ~ I, 0 and J is shown in Figure 15.1 below. We sec that the number of real solutions (f(.q..L) :;;:; 0) for fL = -I is one, while the number of real solutions for f.L = 1 is three. The curve for fL :;;:: 0 is a transition between the two cases. We say that: f.L = 0 is a bifurcation point for this system, because the number of real solutions changes from one to three at this point. We will see in the next section that this behavior is characteristic of a pitchfork bi~ furcation. We will also fino that the number of solutions is always three for this problcln; somctimes two of the solutions are complex, and other timcs thc solutions arc all thc samc value.
x
FIGURE 15.1
Polynomial behavior as a function of /L.
ESCOLA Dc ENGEN HARIA GIBLIOrLCA
Bifurcation Behavior of Single ODE Systems
362
15.3
Chap. 15
TYPES OF BIFURCATIONS 1'he types of bifurcations that will be presented hy way of examples include: 0) pitchfork, Oi) saddle-node, and (iii) transcritical. We will also cover a form of h.vsteresis behavior and show that it involves two saddle-node hifurcations. Before we cover these specific hifurcations, we will present the general analysis approach. Consider the general dynamic equation: (15.1 ) where x is the slale variable and !.L is the bifurcation parameter. The steady-slale solution (also known as an equilibriulll point) of (15.1) is:
()
f(.qL)
( 15.2)
A bifurcation point is where the both the function and its first derivative are zero:
iJf .I(X,fL) = iJx = ()
(15.3 )
Notice that thcrirst-dcrivalivc is also the Jacobian for the single-equation model. Also, the eigenvalue is simply the Jacobian for a single equation syslem, so the eigenvalue is () at a bifurcation point. The number of solutions of (15.2) can be determined from cafastmehe t!lcory. [~quatioll (15.2) has k solutions, if the following criteria arc satisfied:
f(X,fL) = ()
af ... a'f ()x
- ax?
(15.4)
and ( 15.5) In Example 15.1 this method is applied to a system thaI exhibits a pitchfork bifurcation.
EXAl\lPLE 15.1
Pitchfork BiflHTation
Consider the single variable system shown previously ill Section 15.2. x
f(X,fL) ~ fLx- x-'
The equilibrium point is:
the solutions
to.f(x,p~);:;;: 0
are
-r,,()
=
x,.j =
x,
=
0
V;~
V~
(15.6)
Sec. 15.3
Types of Bifurcations
363
Notice that if 11, < 0, then xl' ;;::: 0 is the only physically meaningful (real) solution, since \/~ is complex if fJ. < O. The Jacobian is
Since the Jacobian i.o, a scahH, then the eigenvalue is equal to the Jacobian:
If Po. < 0, then the systcm is stahle. If Po. > 0, then the systcm is unstahle. NO\v, we can find the sta bility of the system, as a function of the hifureation panunetcr, /-L.
I. ~ < O.
The only real equilibrium solution is '\,0;;::: 0, so the value of the cigenvalue is:
A - fL" which is stahle, since fJ. < O.
II.
~
> O.
POI'
lise the notation
this case, there are threc real solutions; we will analyze each one separately. We X/'ll' X;'I' ami x e2 to indicate the three different solutions.
3 x?: -1 /J."
lIllstahlc
I-l c
\1 fJ'
-:1- .r e2
2
I-l" -
stable
- VfL
-+- IL,.
-3 .J
I-l"
-+-
~,.
2 J.-l"
- stable
It is eomOlon to plot the equilihriulll solutions on a bifurcation diagram, as shown in Figure 15.2.
For I-l <0, therc is a single real solution, and it is stable. r"m I-l > 0 there arc three real solutions; two arc stablc and one is unstable. A solid line is used to represent the stable solutions, while a dashed line indicates the unstable solution. Notice that a change ill the number of equilibrium solutions and the type of dynamic behavior oecured at 1.L;::;: O--the hUim'atioll poinl. The bifurcation point satisfies the conditions in (15.3):
!(X"fL,.)
x',:
1.L;x,.
- 0
{lnd
iJ/1
ax
j \,.l'
-I
1.L"
0
364
Chap. 15
Bifurcation Behavior of Single ODE Systems
The slalc and parameter values that satisfy these conditions simultaneollsly are: fl.,.
()
X"
=0
and The higher-order derivatives at the bifurcation point are:
a'I? I
=6x""'0
ax""!,,
and
alII
ax}~",l'~
(, I ()
This analysis indicates that the mnnber of solutions is three in the vicinity of the bifurcation point (sec (15.4) and (15.5)),
x x~ stable
/.5
A~ -21'
x~
0
A=
I'
-----+---- -------------- ---- -- -- ---- --- --- ------ -- A~
I'
unstable
I'
stable
FIGlJRE 15.2
Pitchfork Bifurcation Diagram---Example 15.1.
It should be noted that there arc actually three solutions to the steady-state equation throughout the entire range of l..L values. For fJ... < 0, two of the solutions for Xc arc C'Hllplcx and uncis real. r:or fJ.. = 0, all three solutions for J:e arc zero. For f.L > 0, all three solutions for Xc are real.
Sec. 15.3
Types of Bifurcations
365
2 r -_ _~_-=E:::xam=p<.:le:..l:..:.~)l'-...-=_-..:l _ _~_ _...,
-2 L-_-~----....-.,---:-----,J o 2 3 4 5 FIGURE 15.3 Transient response for Example 15.1, /.L::::-1. time
15.3.1
Dynamic Responses
Figure 15.3 shows the transient response for fJ,::;;: -I for two different initial conditions; both initial conditions converge to the equilibrium solution of x = I. Figure 15.4 shows the transient response for JJ, = 1 for two different initial conditions; the final steady-state obtained depends on the inlial condition. Notice that an initial condition of X o :::: 0 would theoretically stay at x = 0 for all time, however, a small perturbation (say J 0--9 ) would eventually cause the solution to go to onc of the two stable steady-states. Example 15.1 illustrates pitchfork bifurcation behavior, where a single real (and stable) solution changes to threc real solutions. Two of the solutions are stable, whilc one is lln~table, It is easy to find cases where a (subcritical) pitchfork occurs, that is, where a single unstable solution branches lo two unstable and OIlC stable solution. For example, consider the system j:
Example 1
"0 =
= fL<
+ x3
J' = I
0.Q1
1
0.5
x
j(x,!-'l
0
i
-0.5
-I
"0 = -
-1.5 0
2
0.01
6
~
time
8
10
FIGlJRJ1: 15.4 Transient response for Example 15.1, IJ. =: l. The final steadystate reached depends ()n the initial condition.
366
Bifurcation Behavior of Single ODE Systems
Chap. 15
The reader is encouraged to find the bifurcation behavior of this system shown below (see student exercise 5). 2 r-----.,-
~
0 --1
2
-21.5 --1
1.5
·0,5 0 0.5 rou
2
Also, a perturbation of the pitchfork diagram can occur with the following system:
i "0 f(x,/L,u) ,~ u + /LX -- x 3 which can have a diagram of the form shown below (see student exercise 7). u;;:: 1
4
2 ~
0
2 4
6 rou
EXAMPLE 15.2
o
Saddle-Node Bifurcation (Turning Point)
Consider the single variable system: ( 15,7)
The equilibrium point is:
The two solutions are: XI'! =
x,,):
=
\/;J: --
YJJ.
The Jacobian (and eigenvalue) is:
The bifurcation conditions, (15.4) and (15.5), are satisfied for: IJ.,. = x" = 0
Sec. 15.3
367
Types of Bifurcations
The second derivative is:
which indicates that there are two solutions in the vicinity of the bifurcation point. Now, we can find the stability of the system, as a function of the bifurcation parameter, j..t,
I.
/.l <
O.
II. !J- > O.
From
Xci
=
± V~, we see that there is no real solution for !J- < O.
There are now two real solutions; we will analyze the stability of each one.
a. For solution J;
the eigenvalue is
which is stable. b. For solution 2:
the eigenvalue is
which is unstable. The bifurcation diagram (saddle-node) is shown in Figure 15.5.
x
L, Jl = a
.1l>.ble
X= JlO.~ >2 Jl 0.5 ,,--
, uns1l>.ble
FIGURE 15.5
Saddle-node bifurcation diagram, Example 15.2.
Bifurcation Behavior of Single ODE Systems
368
Chap. 15
Notice that there are actually two steady-state solutions for .:re throughout the entire range of f.-l' For f.-l < 0 both solutions for xI' arc complex; for f.-l;;;; 0 both solutions for xI' flrc 0; for f.-l > () both solutions for xl' arc real.
Dynamic Responses. Transient response curves for /L:= 1 are shown in Figure 15.6, for two different initial conditions. Initial conditiolls .ro > -1 converge to a steady-slate of x;;;; I, \vhilc .:ro < ~-1 approach x = ~oo. It should he noted that a consistent physical (or chemit:al) ~bascd model will not exhibit this sort of unbounded hehavior, since the variables will have some physical meaning and will therefore be bounded.
Example 2.
"0= -
=1
0.99
1
0
•
Jl
_I
i
-2
-3
"0=-
1.01
3
2
0
4
5
time FIGURE 15.6 Transient response for Example 15.2, !L;;;: 1. Initial conditions or Xo > -I converge to a steady-state of x;;;: 1, while Xo < -I "blows up",
EXAIVIl']"E 15.3
Transcritical Bifurcation
Consider the single variable system:
.r The equilibrium point is:
The solutions arc
The Jacobian is
~ !(x,p.) ~ p.x - x'
( loY)
Sec. 15.3
369
Types of Bifurcations
The eigenvalue is also
A
~
jJ.-2x,.
The bifurcation point is !(x, .....) = df/dx = 0, which occurs at
-2
fJ,
= x e = O. The second derivative is:
*0
which indicates that there are two equilibrium 1'>0Iu60I1s. Now, we can find the stabil1ty of the system, as a function of the bifurcation parameter, IJ-.
a. One solution is:
with an eigenvalue:
A = fl. - 2 x,.
fl.,
which is stable (since IJ-" is negative). b. This equilibrium solution is:
which has the eigenvalue: A = IJ- - 2x" = J...le - 2IJ-e = -2!J..e
which is unstable (since I-t e is negative).
II.
>0 a. One solution is
jJ.
which has the eigenvalue:
which is unstable. b. Another solution is:
which has the eigenvalue
A = JJ.. - 2 x" = fJ.,. - 2 IJ-"
-2 }L"
which is stable. These results are shown in the bifurcation diagram of Figure J5.7, which illustrates that the number of real solutions has not changed; however, there is an exchange of stability at the bifurcation point.
Bifurcation Behavior of Single ODE Systems
370
Chap. 15
x
L,
x= j!, ,\ =-j!, stable
x=o
x=o
----------of-----------------,\=j!, ,\=j!,
stable
UIl.'ltable
,\ =-j!,
UIl.'ltable FIGIJRE 15.7
Transcritical bifurcation, Example 15.3.
Dynamic Responses. Transient response curves for the transcritical bifurcation arc shown in Figures 15.g and 15.9. Notice that the transient behavior is a strong function of the initial condition for the slate variable. For some initial conditions the state variable eventually settles at a stable steady-state, while for other initial conditions the state variable blows up.
Example 3.
j!,
= -I
"0=- 0.99 -0.5
1
-1 X
i
-1.5
"0 = -
-2
1.01
-2.5 -3
0
2
6
8
10
time
Flf;(JRE 15.8
Transient response for Example 15.3, importance of initial conditions.
J.-L
---I. Notice the
Sec. 15.3
Types of Bifurcations
371
-1
time
FIGUH.E 15.9 Transient response for Example 15.3, fJ. ;:;;; 1. Notice the importance of initial conditions.
The next example is significantly dille-rent from the previolls examples. Here we allow Iwo parameters to vary and determine their effects on the system behavior.
EXAMPLE 15.4
Hysteresis Behavior
Consider the system: ( 15.9) which has Iwo parameters (u and /1-) that can be varied. We think of If as an adjustable input (manipulated variable) and f.L as a design-related parameter. We will construct steady-state input-output curves by varying II and maintaining /1. constant. We will then change /L and see if the character of the input-output curves (x versus II) changes. We first work with the case fL ~-1.
I. fJ-;:;;; -1.
The equilibrium point (steady-state solution) is:
!(X,•. fL) ~ 0 ~
II -
x,. - x,:
(15.10)
The steady-state input-output diagram, obtained by solving (15.10) is shown in Figure 15.10. This curve is generated easily by first generating an x" vector, then solving If = xl' ,I x~. The stability of each point is found from:
ill I
ax
XJl c
which is always negative, indicaling that there arc no bifurcation points and that all equilibrium points are stable for this system. Contrast this result with that for /L 0;;; I, shown next.
Chap. 15
Bifurcation Behavior of Single ODE Systems
372
,pie 3.
)l
=1
01
6
8
10
time
InpllHJUtput diagram 1\,[ Example 15.41\"·,, =-1.
FIGURE 15.10
II. IJ-;:;;; 1.
The equilibrium point (steady-state solution) is:
f(X,,,ll) ~ 0 ~ II + x, - x; Notice that this is a cubic equation that has three solutions for xI' for each value of ple, consider u ;:;;; O.
(15.11) ll.
For exam-
At u;:;;; 0:
so,
x"
=
1,0, or I.
The stability of each solution can be determined from the Jacobian:
The eigenvalue is thcn A = 1 - 3 "-I, = 0,
XI';:;;;
XC
x,,= 1,
x;. For the three solutions, we find:
A= 1-3=-2 A=I A= 1-3=-2
which is stable. which is unstable. which is stable.
Now we can vary the input, If, ovcr a range of values and construct a steady-state input-output curve. These results are shown on the diagram of Figure 15.10 (the easiest way to generate this figure is to create an Xc vector, and then solve Itt' = ~xc + x;. See student exercise 2).
Sec. 15.3
373
Types of Bifurcations
mu
~
1
-~-- ---- ---,. r----------T---- - --------.-------
2
\
'" O·
/
-1
.. - ----,-----
.. - ~------- ....-.-----.--------
\
~~ -2 -4
.~
-2
-3
FIGUrU~
15. t I
-1
0
u
2
4
3
Input-output diagram for Example 15.4 with l-t
0=
r.
Notice that r<'igurc 15.11 cOiltains two saddle-node (or turning poioI) bifurcatio!l points (sec Example 15.2), The bifurcation (singular) points can be determined frolll the solution of: ( 15.12)
'fhe bifuf(.:atioll points arc theu:
1 3 or,
x" = :!.::
(15.13)
whicll call he seen to be the x values at the upper and lower fuming points. Substituting (1:=;,13) into (15.11), we find thal the bifurcation points oc;_<;ur at the inp~!! values of lie ::::; ± 2/3\1\ as shown in Figure 15.10. N~)lice that for II. < -2/3V3 or II > 213\/3 there is only a single,. stable solution, while for -2/3"'/3 < II < 213Y'3 there arc three solutions; two arc stable and one is Ullstable. We have referred to the behavior of this cxarnplc as hysteresis behavior now let us show why.
Starting at Low Values ofu. Notice that if we begin with a low value of If (say, -3) a single, stable, steady-state value is achieved. If we increase II a slight amount (to say, -2.9), we will achieve a slightly higher steady-state value for x. As we keep increasing It, we will cO!l.!illue to achieve a new stahle steady-state valuc for x for each u. This continues untilll :::0 213V3, where we find that the stable solution "jumps" to thc top curve. Again, as we slowly increase II, the stable steady-state solution remains on the top eurvc.
374
Bifurcation Behavior of Single ODE Systems
Chap.15
Starting at High Values of u. Notice that if we begin with a high value of u (say, 3) a single stable steady-state value is achieved. If we decrease Ii a slight amount (to say, 2.9), we will achieve a slightly lower steady-state value for x. As we keep decreasing Il, we will con~inuc to achieve a new stable steady-state value for x for each u. This continues until /l = -2/3Y...1 where we find that the stable solution "jumps" to the bottom curve. As we slowly decrease Ii further, the stable steady~state solution remains on the bottom curve. This is termed hysteresis behavior, because the trajectory (path) taken by the state variable (x) depends on how the system is statted-up. Ajump discontinuity occurs at each "limit" or "turning" point (the saddle-node bifurcation points). Discussion. Notice that there is a signific.1I)t difference between the inpUH)utput behavior exhibited in Figures 15.10 and 15.11. For fJ.'::::: ~1 (Figure 15.9), there is monotonic relationship between the input (Ii) and the output (x). For fJ.:::: 1 (Figure 15.10), there is a region of multiplicity behavior, where there are three values of the output (x) for a single value of the input (u). There has been a qualitative change in the behavior of this system as fJ. varies from ~ I to L The value of fJ. where this occurs is a hystersis bifurcation point. At this point the following conditions are satisfied (since there are three solutions in the vicinity of the bifurcation point):
f(x,l")
~
0
~
.•af ) ..
~
<X
and
a'f
iJ.tj
*0
The equations are:
fJ. ~ 3x~
~O
~6xe
~
-6 It is easy to show that, for a value of Ii
::::
0
*0
0, the bifurcation conditions are satisfied at:
x"
= /L" =
0
The stcitdy-statc input-output curve for thi.<; situation is found by solving:
which yields the plot in Figure 15 12, which is clearly a transition between Figures 15.10 and 15.1 L A three~dimensional plot of x versus u as a function of /L is shown in Figure 15.13. The behavior represented by this diagram is commonly known as a cusp catastrophe. At low values
Sec. 15.3
375
Types of Bifurcations
mu= 0 T~~--'·l-----~U--.-------
" 0 -1
FIGURE 15.12
Input-output diagram for Example 15.4.
of /L we observe monotonic input-output behavior, with a transition to multiplicity (hysteresis) behavior at JL :::: O. The turning points in Figure 15.13 can be projected to the /J--Il plane to find the bifurcation diagram shown in Figure 15.14. A saddle-node (turning point) bifurcation occurs all along the boundary of the regions, except at the "cusp point" (J.L:::: 0, u:::: 0), where a codimcnsion-2 bifurcation occurs. The term "codimensioll-2" meal1~ that two parameter.~ (/J-,u) are varied to
2 1 " 0
-1 -2 -2
o mu u :FIGURE 15.13
"Cusp catastrophe" diagram for Example 15.4.
376
Chap. 15
Bifurcation Behavior of Single ODE Systems
achieve this hifurcation (Stwgatz, 19(4). The reader is encouraged to construct this diagrmn (sec student exercise 6).
1.5
0.5 ~
one fixed-point three fixed-points
a 0.5 -1 -1.5
-1
a
-0.5
0.5
1.5
2
mu FIGURE 15.14
Two~pannnctcr (f.-l,1I)
I
bifurcation diagram for Example 15.4.
SUMMARY We have studied the bifurcation behavior of some example single nonlinear ordinary
ferential equations of the form.i:
c=
f(x,f.l), where x is the state variable ancl
dir~
is the bil'llrfound by solving ./(X,.,/-L,.) = O. f.l
cation parameter. The equilibrium (steady-slate) points afC The stability is determined by finding the eigenvalue, 'A, which is simply the Jacobian, ~flax lx,,,tLe' for a single equation system, JJ A is negative, the equilibrium point is stable. If "A.. is positivc, thc equilibrium point is unstable. A bifurcation diagram is drawn by plolting the equilibrium value of the state variahle I.r,_·I.\·( as a function of the 11ifurcation parameter. [rthc c(]uilibrium ])oint is stable ("A..;:o;: D//fh . < 0), a solid line is drawn. IT the equilibrium point is unstable, a dashed line is dr;'l\vn. The bifurcation points can be found by solving for i)/IJx.I.Y,.,jL" = 0 where l(x",J.-l,,) O. Thcse same techniques can also be applied to systems of several equations, particularl) if the cquations can be reduced to a single steady~slate algebraic equation (in a single stale variable). This can he done for many simple chemical and hiochemical reactor problems.
REFERENCES AND FURTHER READING The following texts provide nice introductions to bifurcation behavior: Hale, J.K., & H. Kocak. (1991). Dynamics and BUitrca{;o/ls. New York: Verlag.
Springcr~
377
Student Exercises
Jackson, E.A. (1991). Perspectives bridge University Press.
(~f Nonlinear
Dynamics. Cambridge, UK: Cam-
Strogatz, S.H. (1994). Nonlinear Dynmnics and Chaos. Reading, MA: AddisoIlWesley.
STUDENT EXERCISES 1. r'or the system in Exmnple 15.4:
x ~ f(X,fL)
.~
II +
x'
[LX -
with II ::::: 0 and fL = I, perform transient response simulations (using MATLAR) to show that the final steady-state obtained depends on the initial condition. 2. For the system in Example 15.4, we found that there are ranges of u where there are three equilibrium solutions for x (when fL =1). When solving for the roots of a cubic polynomial, either a complex analytical solution (see any math handbook) or a root solving routine (such as the MATLAR routine roots) must be used, Show how .t' can be considered the independent variable and u the dependent variable to obtain an easier analysis of this problem. Then, simply plot x versus u. 3. For the system in Example 15.4:
x ~ f(X,fL) with fL ::::: 1, show that the "turning points."
saddle~node
~
II +
[LX -
x'
bifurcation conditions are satisfied at the
4. Consider the constant harvesting model of population growth (Hale & Kocak, 1991):
.r
~ f(x,k,c,h) ~ k x -
C
x2
-
h
where all of the parameters are positive. h is the rate of harvesting, while k and care intrinsic growth rate parameters, The problcm is, for fixed k and c, to determine the effect of the harvesting on the population. Since the population density cannot be negative, we are interested in solutions where x 2:: O. For a positive initial population density (xo) the population is exterminated if therc is a finite value of t such that x::::: O. Without finding explicit solutions of the differential equation, show the following: a. If h satisfies 0 < h <; k2/4c, then there is a threshold value of the initial size of the population such that if the initial size is below the threshold value, then the population is exterminated, If the initial size is above the threshold value, then the population approaches an equilibrium (steady-state) point. b. If h satisfies h > k2!4c, then the population is exterminated regardless of its initial size, c. Comment on the physical ramifications of palis a and b. Should models be used by State Fish and Game authorities to determine proper hunting and fishing Iim~ its?
378
Bifurcation Behavior of Single ODE Systems
Chap. 15
S. Show that the following system exhibits a pitchfork bifucation, with three real solutions (one stable, two unstable) for iJ- < 0 and a single unstable real solution for fL > O.
.< =
f(X,fL) = fLx +
x'
2
, o
,
-------~-------------,---------------------
-1
-2
1.5
2
-1
-0.5
o
0.5
1.5
2
mu 6. Consider the system shown in Example J 5.4:
x = f(X,fL) = II +
[LX -
xJ
Develop the cusp bifurcation diagram shown below. Find the values of if and /L on the boundaries between the one and three fixed-point solution behavior. .5
one fixed-point
0.5 "
three fixed-points
0 -0.5
-1.5
-0.5
1
o
0.5
1.5
2
mu
.
~~~_ • . • •_-.
lIIIblllll.
>r,_ _
Student Exercises
379
7. C\)llsiderthe systcm shown in Example 15.4: "" .~ f(X,fL,U) ~ U + fU -
x'
For a value of u :::: 1, develop the steady-state bifurcation diagram shown below. Find the values of x and jJ. where the saddle-node (turning point) bifurcation occurs. Notice that this is a perturbation of a pitchfork bifurcation. This type of behavior can occur, for example, in exothermic chemical reactors when the feed llowrate is varied whilc maintaining a constant jacket temperature (a so-ca.lled isola forms).
u ::;: 1
4
2
0
~
-2
-4 ·4
0
-2
4
2
mu
APPENDIX h
cusp diagram
% b.w. bcqucLtc 'I.
1~
90
~;olvcs
dec 96
the problem
f(x,u,mu)
'f;
::;: u + mu"'x
x"3
0
" with x varying between -2 and 2 '1 mu varying from -2 to 2 and '!, whatever u' s r,<~ul t clear x; clc,ar u; clear rnu; x -2;{)'05:2; u
x. "3 +2*x;
plot3 (u, - 2 "'ones (size (u)) ,x, 'w' )
hold on
6
8
10
380
Bifurcation Behavior of Single ODE Systems
mu ~ ~1.875:0.125:2; for i "" 1:32; u = x. A 3 - mu(i) .*x; plot3(u,mu(i)*ones(size(u») ,x, 'w'} end
hold off »view(15,-30)
Chap. 15
BIFURCATION BEHAVIOR OF TWO-STATE SYSTEMS
16
The goal of this chapter is to introduce the reader to limil cycle behavior and the HopI" bifurcation. After studying this chapter, the reader should be able to Find that many of the same types of bifurcations that occur in single-state systems also occur in two-state systems (pitchfork, saddle-node, transcritical) Understand the difference between limit cycles (nonlinear behavior) and centers (linear behavior) Distinguish between stable and unstable limit cycles Determine the conditions for a Hopfbifurcation (formation of a limit cycle) Discuss the differences between suhcritical and supcrcritica.1 HopI' bifurcations '1'he major sections in this chapter are:
16.1
16.1
Background
16.2
Single-Dimensional Bifurcations in the Phase-Plane
16.3
Limit Cycle Behavior
16.4
The Hopf Bifurcation
BACKGROUND In Chapter 15 we presented the bifurcation behavior of single-state systems. 'A'e found that a number of interesting bifurcation phenomena could occur in these systems, including transcritical, pitchfork, and saddle-node bifurcations. We find in this chapter that these 381
382
Bifurcation Behavior of Two-State Systems
Chap. 16
types of bifurcations can also occur in highcr~order systems. This is the subject of Section 16.2. In Section 16.3 we review limit cycle behavior, which was initially presented in Chapter 13 (phase-plane analysis). In Section 16.4 we present a type of bifurcation that can only occur in second- and higher-order systems. In a Hopi' bifurcation, we find that a stable node can h(furcate to a stahle limit cycle if a parameter is varied; this is an example of a supercritical Hopf bifurcation. This phenomena has been shown to occur in a number of chemica! and biochemical reactors. Before turning to the interesting Hopf bifurcation phenomena, we will discuss single dimensional bifurcations in the phase plane.
16.2
SINGLE·DIMENSIONAL BIFURCATIONS IN THE PHASE·PLANE Consider the two-variable system (notice that the two equations are decoupled):
XI = f,(X,fL) = fLXI - xi x, =
f2(X,fL) = -x,
(16.1 ) (16.2)
The equilibrium (steady-state or fixed~point) solution is:
IfLX~:,:t:,]
f(X,fL) =
=
[~]
There are three solutions to f(x,f,t) ::::: O. The trivial solution is:
and the two nontrivial solutions arc: XI'
=
and Xc
=
[~;:] = [ - :~I
Notice that only the trivial solution exists for j.J.. < 0, since we will assume that equilibrium values must be real (not complex). We can detenninc the stability of each equilibrium point from the Jacobian, which is:
°1
-I which has the following eigenvalues:
= fL A2 = -I Al
3xj"
Sec. 16.2
Single~Dimensional
Bifurcations in the
Phase~Plane
383
Since the second eigenvalue is always stable, the stability of each equilibrium point is de~ termincd by the first eigenvalue. Here we consider three cases, fJ, < 0, f.L:::::: 0, and fJ, > o.
J.L
II
J.L
=0
The equilibrium solution is x k
:::::
0, so: ~I =
J.L - 3xt
=
0
which is stable; the system can be shown to exhibit a slow approach to equilibrium by observing the analytical solution to the differential equations.
III
J.L > 0
The eigenvalue for the trivial solution (xl e ::::: 0):
is unstable since p.. > O. The eigenvalues for the nontrivial solutions (±\Iii) are:
and
So the nontrivial solutions arc stable for f.L> O. This means that a saddle point (trivial solution) is bounded by two stable nodes for this case, since the three solutions are:
x, = [:;;] = [ -
x ,. =
x" Ix"1
IxX".j
=
., = 2,<
:~] with ~ = [~;]
0 ~ =~2 IOJ·1 I~'J WIt 1
=
_I, [J.L]
IV~]. WIth A = [~11, = 0
= [
~2tj =
_
stable node
saddle pomt .
I' -2J.L1 I
~2,
=
=
stable node
We notice the following phase-plane diagrams (Figure 16.1) as f.L goes from negative to positive.
384
Bifurcation Behavior of Two-State Systems mu= --1
mu= 1
0.5 ~~
Chap. 16
0.5
a
~~
a
-0.5
-0.5
-1
-1
x, a.
IJ
x, <
a
b.
u >
a
FIGURE 16.1 Pitchfork bifurcation behavior in the plane. There is a single stahle node for /L < 0, and two stable (0) nodes and a saddle point (+, unstable) for 1-1 > 0,
16.3
LIMIT CYCLE BEHAVIOR In Chapter 13 we noticed that linear systems that had eigenvalues with zero real portion formed centers in the phase-plane. The phase-plane trajectories of the systems with cellters depended on the initial condition values, as shown in Figure 162 below. Different initial conditions lead to different closed-cycles. In this section, and the rest of this chapter, we are interested in limit cycle behavior, as shown in Figure 16.3. The major difference in conter (Figure 16.2) and limit cycle
0.5
x, -0.5 -1 -1.5 -1
a x,
FIGURE 16.2 havior.
Example of center be-
Sec. 16.3
Limit Cycle Behavior
385
0.5
x, 0 -0.5 -1 -1.5 1
FIGURE 16.3 cycle behavior.
0
x,
Example of limit
(Figure 16.3) behavior is that limit cycles are isolated closed orbits. By isolated, we mean that a perturhation in initial conditions from the closed cycle eventually returns to tbe closed cycle (if it is stable). Contrast that with center behavior, where a perturhation in initial condition leads to a different closed cycle.
EXAMPLE 16.1
A Stable Limit Cycle
Consider the following system of equations, based on polar coordinates: j. ~ ,.
II
(I - ,.')
~
(16.3) (16.4)
Notice that these equations are decoupled, that is, the value of ret) is not required to find OCt) and vlee versa. The second equation indicates that the angle is constantly decreasing. The stability of this system is then determined frotH an analysis of the first equation. The steady-state solution of the first equation yields two possible values for r. The trivial solution is r;o;; () and the nontrivial solution is r"'" 1. The Jacobian of the first equation is:
of ar
I - 3 r'
We see then that tbe trivial solution (r:::;: 0) is unstable, because the eigenvalue is positive (+1). The nontrivial solution is stable, because the eigenvalue is -~2. Any trajeeto'·y that starts out close to r"'" 0 will move away, while any solution that starts out close to r;o;; I wiH move towards r;o;; I. The time domain behavior for x [ is shown in Figure 16.4, Notice that we have convelicd r cos 0, x2 "'" r sin 0). The phase-plane behavior is the states to rectangular coordinates (XI shown in Figure 16.5. Initial conditions that are either "inside" or "outside" the limit cycle converge to the limit cycle. ;0;;
Bifurcation Behavior of Two-State Systems
386
Chap. 16
1.5
··0.5 -1
o
20
10
30
40
time
FIGVRl\ 16.4
Stable limit cycle behavior (Example 16.1).
1.5
0.5 .;;:
0
-0.5
·1
F1GUH.E 1().5 Stahle limit cycle behavior (Exi1mpleI6.1).
X,
The previous example was for a stable limit cycle. It is also possible for a limit cycle to be unstable, as shown in Example 16.2.
EXAMPLE 16.2
An Unstable Limit C)'cle
Consider the following system of equations, based on cylindrical coordinates
r (j
···r(l-r') I
(16.5) (16.6)
Again, notice that these equations are decoupled. The second equation indicates that the angle is constantly decreasing. The stability of this system is then determined 0"0111 an analysis of the first equation.
Sec. 16.3
387
Limit Cycle Behavior 1.5
/
...-
................
l ~!
0.5
-0.5
---..
-1 -1.5
~.~
....
o
-1
x, FIGURE 16.6
Phase~plancbchaviorfor all
unstable limit cycle.
The steady-state solution of the first equation yields two possible values for r. The trivial solution is r ::;: 0 and the nontrivial solution is r::;:: I. The Jacobian of the first equation is: iif ()r
~1+3r2
We sec then that the trivial solution is stable, because the eigenvalue is negative (--1). The nontrivial solution is unstable, because the eigenvalue is positive (+2), Any trajectory that stalts oul less than r = I will converge 10 the origin, while any solution that starts out greater than r ::::1 will increase at an exponential rate. This leads to the phase~planc behavior shown in Figure 16.6. The time domain behavior is shown in Figure 16.7. 1.5 --
0.5
o o
_-------1...--._
.. L
2
L..
3
4
5
6
time FIGURE 16.7 Time domain behavior for an ullstable limit cycle. An initial condition of 1'(0) = 0.9 converges 10 0, while an initial condition of r(O) = 1.05 blows up.
Bifurcation Behavior of Two-State Systems
388
Chap. 16
Examples 16.1 and 16.2 have shown the existence of two different types of limit cycles. Tn the first case (16.1) the limit cycle was stahle, meaning that all trajectories were "attracted" to the limit cycle. In the second case (16.2) the limit cycle was stahle, and all trajectories were "repelled" from the limit cycle. Although hoth of these examples yielded limit cycles that were circles in the plane, this will not normally be the casco Usually the limit cycle forms morc of an ellipse. Now that we have covered limit cycle behavior, we arc rcady to determine what types of system parameter changes will cause limit cycle behavior to occur. That is the subject of the next section.
16.4
THE HOPF BIFURCATION In Chapter 15 we studied systems where the number of steady-state solutions changed as a parameter was varied. The point where the number of solutions changed was called the bifurcation point. We also found that an exchange of stability generally occurred at the bi~ furcation point. A Hop/hifurcation occurs when a limit cycle forms as a parameter is varied. In the next example we show a supercritical Horf bifurcation, where the system moves from a stahle steady~slatc at the origin to a stable limit cycle (with an unstable origin) as a paramcter is varied.
EXAMPLE Hi.3
SupercriticallIopf Bifurcation
Consider the systern:
i] =
.<,=
Xl
+ Xl (p" -
xf - xi)
-x, +x2 (/L-X;-xi)
( 16.7) ( 16.8)
This can be written (see student exercise 4) in polar coordinates as:
i·
~ r
o~
(fJ. - r 2 )
-I
Since these equations are decoup1cd, thc stability is dctcrmined from the stability of: ;. =
fer) = r (fJ. - ,2)
the Jacobian is:
af ar
/.L - 3,.2
The equilibrium (stcady-statc) point is/{r):;;:; 0, which yields, r(fJ.-r 2 ) = ()
(16.9)
(16.10)
389
The Hopf Bifurcation
Sec. 16.4
which has three solutions: r ,= 0 (trivial solution)
r
\ifL
(not physically realizable)
For /-L < 0, only the trivial solution (r:;::: 0) exists. For I-l < 0,
which is stable, since /-L < O. 1<'01' /-L::::O 0, all of the steady-slale solutions
,lrC r:;:::
0. aod tbe Jacobian is:
·3 r' which is stable, but has slow convergence to r";=: O. For 11, > 0, the trivial solution (r::::o 0) is unstable, because:
The nontrivial solution (r:;::: \/f-L) is stable because:
iJ{ " . ' = p, -- 3 r'
-2f-L
(),.
and we find the following phase-plane plots shown in Figure 16.8. The bifurcation-diagram for this system is shown in Figure 16.9.
1.5
0.5
~\.
-~)/
-0.5
1.5
1.5~~······
·1
0
x,
a.
f"~-1
-1.5
1
0
x,
b'I.c~ 1
F1G-lJRE 16.8 Phase-plane plots. As J.-l goes from -1 to I, the behavior changes from stable node to a stahle limit cycle.
390
Bifurcation Behavior of Two-State Systems
Chap. 16
1.5
l....
0.5
-----------------------
o··_~-~-
-0.5 ~-1
~~~~-
-2
--.l---~--~---~---~--
-1
1
0
4
3
2
mu FIGURE 16.9 Bifurcation diagram. Indicates that the origin (r 0) is stahle when I-l < O. When I-l > 0 the origin becomes unstable, but a stable limit cycle (with radius r::::: \/iL) emerges. :;:;;0
Here we analyze this system ill rectangular (xI point or equilibrium) solution to (16.7) and (16.8)
coordinates. The only' 15:
Linearizing (16.7) and (16.8):
atl _
2
__~~1-l-3Xl-X2
2
~
ax,
dX j
2 Xl
=-1-2x x; J
dX)
We find the Jacobian matrix:
[ ~-'
-xL
-1-2\:I,X'1
J
-2x l "X 2,
l-l-x~,.--3
which is, for the equilibrium solution of the origin:
The characteristic polynomiaL from det(,\l- A) ::: 0, is:
,'- 2~' + ~' + 1 ~ 0 \\lhich has the eigenvalues (roots):
2
{,
steady~stalc
(fixed-
Sec. 16.4
The Hopf Bifurcation
391
We see that when ~ < 0, the complex eigenvalues are stable (negative real portion); when f.L::::: 0, the eigenvalues lie on the imaginary axis; and when f.L> 0, the complex eigenvalues are unstable (positive real portion). The transition of eigenvalues from the left~half plane to the right-half plane is shown in Figure 16.10.
+++ ~ <
~ =
0
0
~ >
0
FIGURE 16.10 Location of eigenvalues in complex plane as a function of fL. A Hopf bifurcation occurs as the eigenvalues pass from the lefthand side to the righthand side of the complex plane.
Example 16.3 was for a supercritical Hopf bifurcation, where a stable limit cycle was formed. We leave it as an exercise for the reader (student exercise 6) to show the formation of a subcritical Hopt" bifurcation, where an unstable limit cycle is formed. We have found that the Hopf bifnrcation occurs when the real portion of the complex eigenvalues became zero. In Example 16.3 the eigenvalues crossed the imaginary axis with zero slope, that is, parallel to the real axis. In the general case, the eigenvalues will cross the imaginary axis with non-zero slope. We should also make it clearer how an analysis of the characteristic polynomial of the Jacobian (A) matrix can be used to identify when a Hopf bifnrcation can occur. For a two-state system, the characteristic polynomial has the form: a 2(fl.) ),,2
+ a,(fl.) )" + ao(fl.) = 0
where the polynomial parameters, ai' are shown to be a function of the bifurcation parameter, fl. (It should also be noted that it is common for a2 = I). Assume that the a/fl.) parameters do not become 0 for the same value of fl.. It is easy to show that a Hopf bifurcation occurs when a,(fl.) = 0 (see student exercise 7).
16.4.1
Higher Order Systems (n > 2)
Thus far we have discuss Hopf bifurcation behavior of two-state systems. Hopf bifurcations can occur in any order system (n ;:::: 2); the key is that two complex eigenvalues cross the imaginary axis. while all other eigenvalues remain negative (stable). This is shown in Figure 16.11 for the three state case.
Bifurcation Behavior of Two-State Systems
392
+
o o
~t
p < 0
:=
Chap. 16
~f--
a
p > 0
FIGURE 16.U Locatioll of eigenvalues in complex plane as a function of /-L. A Ilop!" bifurcation occurs ns the eigenvalues pass from the lefthand side 10 the righthand side of the complex plane.
SUMMARY In this chapter we have shown that the same bifurcations that oeemed in single-state sys~ tems (saddle-node, transcritical, and pitchfork) also occur in systems with two or more states. We have also introduced the Hopf bifurcation, which occurs when cmnplcx cigcn~ values pass from the left-half plane to the right-half plane, as the bifurcation parameter is varied. A HopI' bifurcation can also occur ill systems with more than two stales. F,'or a SllpcrcriticalHopf bifurcation, two complex conjugate eigenvalues cross from the left-hall' to the right-hall' plane, while all of the other eigenvalues remain stable (in the left-half plane).
FURTHER READING The following sources provide general introductions to bifurcation theory: Hale, 1., & H. Kocak (1991). DYl/mnics and H{flfrcatiol1s. Nc\v York: SpringerVerlag. Strogatz, S.H. (1994). Nonlinear Dynamics and Chaos. Reading, MA: AddisonWesley. 'I'he following textbook shows a complete exmnplc of the occurance of I[opf bifurcations in a 2-state exothermic chemical reactor model: Varma, A., & M, Morhidclli. (1997). Mathematical Methods in Chemical Engineering. New York: Oxford University Press.
STUDENT EXERCISES 1. Show that the two-variable system ·\1 ~ J;(x,f.l-)
.r,
~t;(x,f.l-)
IL - xi
- -x,
exhibits saddle-node behavior in the phase plane.
393
Student Exercises
2. Show that the two-variable system
x, = f,(x,fL) = x,
=
/LX, - xl
f,(x,fL) = -x,
exhibits tnmscritical behavior in the phase plane. 3. Show that the two-variable system:
x, = f,(x,fL) = x, = J;(X,fL) =
Ii
+ x, -
xi
-x,
exhibits hysteresis behavior in the xl state variable. This means that, as u is varied, folJows an S-shaped curve, which exhibits the ignition/extinction behavior shown in Chapter 15. 4. Show that: x\e
can be written: ;. = ,.
(fL - r')
0=-1 if Xl = r cos () and x 2 = r sin O. 5. Consider a generalization of Example 16.3, which was a supercritical HopI' bifurcation (a stahle limit cycle);
;. = r (fL - r')
e=
(f)
+ b r2
Discuss how w affects the direction of rotation. Also, discuss how b relates the frequency and amplitude of the oscillations. 6. Consider the following system, which undergoes a subcritical Hopi' bifurcation:
r= Ii =
/-L
r +
w
+
,3 -
r5
b,.'
Show that, for IJ- < 0 an unstable limit cycle lies in between a stable limit cycle and a stahle attractor at the origin. What happens when fL = 0 and fL > O'! 7. Show that the condition for a HopI' bifurcation for the f()Uowing characteristic equation
is a l (!J-) = O. This is easy to do if you realize that a HopI' bifurcation occurs when the roots have zero real portion and write the polynomial in factored t~)rll1.
394
Bifurcation Behavior of TWQ+State Systems
Chap. 16
Relate this condition to the Jacobian matrix, A(/-t), realizing that: ),2 _
s.
tr(A(fL») ),
+ dct(A(fL» = 0
Consider a Hopf bifurcation of a three-state system. Realizing that one pole is negative (and rcal) and that the other two poles arc on the imaginary axis, relate the Hopf bifurcation to the coefficients of the characteristic polynomial of the Jacobian matrix arc:
You can assume, without loss of generality, that a 3 (/-L) = 1. How do the conditions on the polynomial coefficients relate to the conditions on the Jacobian matrix (trace, determinant, etc.)?
INTRODUCTION TO CHAOS: THE LORENZ EQUATIONS
17
The objective of this chapter is to present the Lorenz equations as an example of a system that has chaotic behavior with certain parameter values. After studying this chapter, the reader should be able to: Understand what is meant by chaos (extreme sensitivity to initial conditions) • Understand conceptually the physical system that the Lorenz equations attempt to model. Understand how the system behavior changes as the parameter r is varied. 'fhe major sections in thi:-; chapter arc', 17.1
Introduction
17.2
Background
J 7.3
The Lorenz Equations
J 7.4
Stability Analysis of the Lorenz Equations
J 7.5
Numerical Study of the Lorenz Equations
(7.6
Chaos in Chemical Systems
17.7
Other [ssucs in Chaos
395
396
17.1
Introduction to Chaos: The Lorenz Equations
Chap. 17
INTRODUCTION In Chapter 14 we presented the quadratic ll1i:lp (logistic equations) and found that the tran~ sient behavior of the population varied depending on the growth parameter. Recall that when the qualitative behavior of a system changes as a function of a certain parameter, we refer to the parameter as a b(lurcalion parameter. As the growth parameter was varied, the population model went through a series of period-doubling behavior, finally becoming chaotic at a certain value of the growth parameter. At that time we noted that chaos is pos~ sible with onc discrete nonlinear equation, but that chaos could only occur in continuous (ordinary differential equation) models with three or more equations (assuming the model is autonomous). In this chapter we study a continuous model that has probably received the most attention in the study of chaos-the Lorenz equations. Before we write the equations, it is appropriate to give a brief historical perspective on the Lorenz model. For a more complete history, sec the book Chaos by James Cjlcick (1987).
17.2
BACKGROUND In 1961, Edward Lorenz, a professor of Meteorology at MIT', Was simulating a reducedorder model of the atmosphere, which consisted of twelve equations. Included were functional relationships between temperature, pressure, and wind speed (and direction) among others. He performed numerical simulations and found recognizable patterns to the behavior of the variables, but the pattel11s would ncver quite repeat. One day he de~ cided to examine a particular set of conditions (parametcr values and initial conditions) for a longer period of time than he had previously simulated. Instead of starting the entire simulation over, he typed in a set of initial conditions based on results from midway through thc previous run, started the simulation and walked down the hall for a cup of coffee. When he returned, he was shocked to find that his simulation results tracked the previous run for a period of time, but slowly began to diverge, so that after a long period of time there appcnred to be no con·clation between the runs. His first instinct was to check for a computer error; when he found none, he realized that he had discovered a very important aspect of certain types of nonlinear systems~-that of extreme sensitivity to initial conditions. When he had entered the new initial conditions, he had done so only to a few decimal places, whereas several more decimal places were carried inter~ nally in the calculations. This small difference in the initial conditions built up over a period of time, to the point where the two fUllS did not look similar. This discovery led to the realization that long-term prediction of certain systems (such as the weather) will never bc possible, no matter how many equations are used and how many variables are measured. In order to learn more about the behavior of these types of systems, he reduced his model of the atmosphere to the fewest equations that could describe the bare essel1tials~ this required three equations. Here we discuss the "physics" of the three equations, while Section 17.3 presents the equations and discusses the equilibrium solutions and stabilty of the equations.
Sec. 17.3
The Lorenz Equations
397
T,
FIGURE 17.1 Convection rolls due to a temperature gradient in a fluid where density decreases as a function of temperature (1'1 > 72),
T,
Consider a Huid maintained between two parallel plates, as shown in Figure 17.1. When the top plate temperature (T2) is equal to the hottom plate temperature (T,), there is no flow and the system is in equilibrium. Now, slowly increase the bottom temperature. At low temperature di ffcrcnces, there is still no llow because the viscous forces are greater than the buoyancy forces (the tendency for the less dense Huiet at the bottom to move toward the top and the more dense lluid to move toward the bottom). Finally, at some criti(al temperature difference, the buoyancy forces overcome the viscous forces and the nuid begins to move and form convection rolls. As the temperature difference is increased, the fluid movement becomes more and more vigorous. Although the following point may be less clear to the reader, for some systems there 1sa value of temperature difference that will cause the smooth convection rolls to break up and become turbulent or chaotic-. One can think of the speed of these convection rolls as wind speed in a miniature "weather model" and the direction of the convection rolls as wind direction. In the next section, we analyze the Lorenz equations, which attempt to model the ilow pattern of Figure 17.1.
17.3
THE LORENZ EQUATIONS The Lorenz equations are:
x, =
(17.1)
r Xl - X 2 - X I X 3 X3 = hx] + X l X 2
(17.2) (17.3)
X2
=
Notice that the only nonlinear terms are the bilinear terms x Jx 3 in (17.2) and xlx2 in ( 17.3). The state variables have the following physical significance: proportional to the intensity (speed) of the convective rolls x2 = proportional to the temperature difference between the ascending and descending currents proportional to distortion of the vertical temperature profile from linearity = x3 XI
398
Introduction to Chaos: The Lorenz Equations
Three parameters,
(T,
Chap. 17
r, and b have the following physical significance:
Prandtl number (ratio of kinematic viscosity to thermal conductivity)
(J
::::::
r
::::: ratio of the Rayleigh number, Ra, to the critical Rayleigh number, Rae
b
a geomctic factor related to the aspect ratio (height/width) of the convection roll
The Rayleigh number is:
Ra =
galP!::.T vI<
where:
a = coefficient of expansion 11 = distance between plates
gravitational acceleration g !::.T = temperature difference between the plates (T 1 ~ T 2) v = kinematic viscosity I< = thermal conductivity For a fixed geometry and tluid, Ra is a dimensionless measure of the teo1peraturc difference between the plates. For 0 ::; r < I (Ra < RaJ the temperature difference is not large enough for the buoyancy forces to overcome the viscous forces and cause motion. ['or r> 1 (Ra> Rae) the temperature difference is large enough to cause motion.
17.3.1
Steady-State (Equilibrium) Solutions
The Lorenz equations have three steady-state (equilibrium) solutions under certain condi~ tions. First, we present the trivial solution, then the nontrivial solutions. In Section 17.4 we will determine the stability of each equilibrium solution.
TRIVIAL SOLUTION
By inspection we find that the trivial solution to (17.1 )-(17.3) is XL, = x2, = Xl.> = 0 (17.4) The trivial condition corresponds to no convective flow of the fluid.
NONTRIVIAL SOLUTIONS From (17.1) we find that: (175) Substituting (17.5) into (17.3), we find thatx" =
~ xi" or: (17.6)
Sec. 17.4
Stability Analysis of the Lorenz Equations
TABLE 17.1
State Variable
399
Summary of the Equilibrium Solutions Nontrivial I
Nontrivial 2
Trivial Solution
(r> 0 required)
(I' > 0 required)
o o
Vb (r ~ I)
Vh(r~ 1)
Vj)~ r -- I
-Vb(~-~ I)
o
Substituting (17.6) into (17.2) at
steady~statc, we
r
1
find',
(17.7)
and substituting (17.7) into (17.6) and using tbe results of (17.5):
x,," x2, =1 V';(r -
I)
(17.8)
For rcal solutions to (17.8), condition r ~ I must be satisfied. This Incans that for r < 1, there is only one real solution (the trivial solution), while for r > I there arc three rcal solutions. This is an example or a pitchfork bifurcation. For Ra < Rae' there is no convective flow. The equilibrium behavior is slIlllllwrizcd in Table 17.1.
17.4
STABILITY ANALYSIS OF THE LORENZ EQUATIONS Linearizing (17.1 )-(17.3) around the stcady-statc we find the following Jacobian matrix: j
( 17.9)
which we will analyze to determine the stability of the equilibrium solution.
17.4.1
Stability of the Trivial Solution
For the trivial solution, XIs
= x2.\,:::: x:h:::: 0 , the Jacobian matrix is:
(17.10)
and the stability is determined by finding the roo Is of dCl(A1 ~ A):::: O.
Introduction to Chaos: The Lorenz Equations
400
I':,"
"-A del(A/- A)
cc
Chap. 17
o o
---
A+
o
A+h
(A + h) IA.+
I
-
A+ 1
--I'
(17.11 )
~
(A + h) [(A +
A2
~
+ 1 - V(
j
1)2 - 4
2
( 17.13)
(17.14) Clearly, the first eigenvalue is always stable, since b > O. It is also easy to show that the second and third eigenvalues can never he complex. The second eigenvalue is always negative and the third eigenvalue is only negative for r < I. We then sec the following eigenvalue structure for the trivial (no flow) solution
r < 0: all eigenvalues are negative, trivial solution is stahle r> 0: saddle point (one unstahle eigenvalue), trivial solution is unstable For,. > I, then f{o > Rae' which means that flow will occur. Notice that Ra is proportional to b.T. This mcans that once fj,T is increased beyond a certain critical fj,1~:, convective flow \vill begin.
17.4.2
Stability of the Nontrivial Solutions
Here we find the roots of det(AJ - A) = 0 for the nontrivial solutions. Starting with:
AI- A
I:,' '; -- ):2s
-
XIs
~XIs
A+ h
o ]
( 17.15)
and solving det(AJ - A):;:: 0, you should find:
A' + "21\.2 + hi A + IJ o
(17.16)
Sec. 17.4
401
Stability Analysis of the Lorenz Equations
where:
( 17.17) (17.18) ( 1719)
b2 -lr(A) IT+b+1 b,'··(r+lr)b bo = ~- det(A) = 2lrb(r - I)
Recall that real nontrivial solutions only exist for r > I.Thc coefficients b 2 , hi' and ho arc then all positive, satisfying the necessm)' condition for stability. The Routh array must be lIsed to check the sl{1I1Cienf condition for stability. As derived in the appendix, the critical r for stability is: r·11
=-
+
(IT
~
h + b - I)
(17.20)
If r > /)/, then the stability condition is not satisfied. This is an interesting result, because
it Incal1s that nOlle of the cquibrium solutions (trivial, nontriviall, nontrivial 2) is stable for r> rl!- The subscript His llsed in (17.20), because a Hopf bifurcation forms at that
value (sec student exercise 1). If a supercritical Hapf bifurcation occurred, a stable limit cycle would form, yielding periodic behavior for the nontrivial solutions. It turns out that .<1 suhcrificalllopf birfurcation is formed, that is, the limit cycle is unstable (sec Strogatz for a nice discussion). Since there arc no stable equilibrium points for r > 1'/1, and no stahle limit cycles, the solution "wanders" in phase space, never repeating the same trajectory. This behavior is known as chaos and the solution is said to be a strange atfracfor.
17.4.3
Summary of Stability Results
We have seen that for r < 1 there is only one real solution, the trivial solution, and it is stable. When r =1 there is a pitchfork bifurcation, yielding three real solutions for r> 1. The trivial solution is unstable for r > 1, while the nontrivial solutions arc stable for 1 < r < rHo This behavior is shown clearly by the bifurcation diagram shown in 1,'igurc 17.2. The formation of the unstable limit cycle at I' = 1'/1 is discussed by Strogatz.
15 unstable limit cycle
10 5
"
c:
. .•. •.•. .•. . . .•.•.•~..• -
0 5
-----
-10
----- ~7---unstable limit cycle
15
25
30
r FIGURE 17.2 Bifurcation diagram for the Lorenz equations. Based on parameters in Section 17.5.
17.5
Chap. 17
Introduction to Chaos: The Lorenz Equations
402
NUMERICAL STUDY OF THE LORENZ EQUATIONS Lorenz used the following values to illustrate the chaotic nature of the equations IT ~ ]() b~
8
3 r c. 28
from (17.22) we calculate that I'll ::;: 470/19 ; : : ; 24.74, indicating that all of the equilibrium points arc unstable, since the value of r::::: 2S is greater than rJl" Before we continue with the set of parameters that Lorenz used to illustrate chaotic behavior, we will first pcrfonn sinlulations for two other cases. ]n the first, we show a set of conditions where the trivial solution is stahle. In the second, we show a set of conditions where the nontrivial solutions are stable.
17.5.1
Conditions for a Stable Trivial (No Flow) Solution
We have found that the trivial steady-state is stable for 0 s: r < I. l-lerc we use the u and b parameters used by Lorenz, but set r:::: 0.5 ror a stable trivial steady~state. (J
~
/,
10
~
X 3
r
~
0.5
As in future simulations, we assume an initial condition oLto ;;;: 10 1 Ol"r, A time dornain plot for all three-state variables is shown in Figure 17.3, and a phase-plane plot (x3 vs. tl) is shown in r<'igure J7.4. The convergence to equilibrium occurs rapidly.
Lorenz Equations, r "" 0.5
0.8
0.6 ~
0.4
0.2
0 0
FIGURE 17.3
2
4
6
8
10
Lorenz equations under conditions for a stable trivial solution.
Sec. 17.5
403
Numerical Study of the Lorenz Equations Lorenz Equations, r := 0.5
0.2
0.15
'P 0.1 0.05
o
o
0.1
0.2
0.3
0.4
Q5
0.6
0.8
0.9
Xi
FIGURE 17.4
Phase-plane under stable conditions for the trivial solution
(x] - xJ plane).
17.5.2
Stable Nontrivial Solutions
We have found that the nontrivial steady-states arc stable for 1 < r < r l1 " I-Icrc we use the and b pannnctcrs used by I"orenz, but set r::: 10 (recall that r ll ::: 24.74 for these values of (J and b) to show that the nontri vial steady-states arc stable. IT
iT
= to
h=
8 3
r = 21
For the trivial steady-state, x::;:; 1000F', the eigenvalues arc:
2.67 A} =- 20.67 A, = 9.67 A,·e -
as expected, the trivial steady-state is unstal?h.:~(~saddlc_l?9Jrlt). The.nontrivial steady-slates, x = lV160/3 V160/3 201 1 and x -- vu,0/3 20r', have eigenvalues of:
A,
=
-
V160/3
13.4266
1.. 2 = --0.1200
A;
= [-
+ 8.9123j
-O.l200-8.9123j
verifying that the nontrivial steady-states are stable. Time domain plots for x J are shown in Figure 17.5, for r:::; 21 Hnd two difTerent initial conditions. Notice that convergence to a particular equilibrium point depends on the initial condition, that is, plot a converges to one equilibrium point. while plot h converges to a different equilibrium point. Also notice that plot b exhibits what is known as transient
404
Introduction to Chaos: The Lorenz Equations
Chap. 17
r" 21, Xo " [0;--1 ;0)
20 15 10 5
"
0 ~5
10 ~
15 20
0
40
5 time 8.
20
r 21, Xo =- [0
OjT.
0:;:.
r"21,xo "
[-~20;10;10)
15 10 5
-5 10 -15 200 -----5---io-i·5-~---20time
25
30
35
40
FIGURE 17.5 Lorcnz cquations under conditions for stable nontrivial solutions.
chaos. Thc initial trajcctory appears chaotic, but eventually the tr(~jectory converges to an equilibrium point. In othcr systems thc systcm can exhibit transient chaos and settle into periodic behavior. The phase plane diagram of Figure 17.6 also clearly shows the effect of two different initial conditions. In curve a thc trajectory almost immediately goes to the equilibrium point on the right (positive value of xl)' In curve b the trajectory first winds around the left equilibrium point, switches to the right equilibrium point, and (after going back and folth a fcw times) evcntually winds around the left equibriulll point, slowly converging.
Sec. 17.5
Numerical Study of the Lorenz Equations
405
r= 21. Xo = [0;-1;01
35
30
,
25 20 15 10
5
o
-----~~~~-~
-20-15
10
o
-5
5
15
10
Xi
35
r = 21. ~
Xo
= [-20;10;1 OJ
.~~~.
30
25 20
!i' 15 10
5
o
-20
--·-:-15-----:--.,lnO~~--5"---O ·--5~·-~-TD---t5----
-20
Xi
b.r=21.xo =[20
17.5.3
10
10l T
FIGURE 17,6 Phase plane under sl
Chaotic Conditions
The parameters used for this case are (f
~
lO
b~
8 3
r
~
28
Recall that all of the equilibrium points are unstable, since the value
than
FJ!>
or r::::: 2X
is greater
406
Introduction to Chaos: The Lorenz Equations
Chap. 17
For the trivial stcady~state, x -;;:; 10 0 OfT, the eigenvalues arc: /.., ==
2.67
/.., =-22.83 /..3 co 11.83 as expected, the trivial stcady~statc is unstable (a Xor the nontrivial steady-states, x =27]1', the eigenvalnes are: =
sa441c poiJlt).
[Vn vn
vn
271' and x
[- V72
/.., - - 13.8546 /..2 = IW940 =- 10. I 945j /..3 (W940 + H1l945j
indicating that the nontrivial stcady~slatcs arc unstable. Notice that all of the steady-state operating points are unstable. '{'his means that the curves in the three-dimensional "phasecube" plots will not asymptotically approach any single equilibriulll point. The curves may exhibit periodic-type behavior, where the three-dimensional equivalent of a limit cycle is reached. The curves could even have "quasi-periodic" behavior, where the oscillations appear to have two frequency componcnts. It turns out for this set or parameters that. the curves ncver repeat. The curves have a strange affractor because they stay in a certain region of three-space, but never intersect or repeat. This is known as chaotic I)chavior. Figure I?? shows the Lorenz behavior for the "Yr variable under unstable (chaotic) conditions. The initial condition is xo :::: [0 1 Orr. Plots of the other states (x2 and ''\:3) arc similiar. Figure 17.7 was 11 time domain plot for Xl under chaotic conditions. More interesting results arc also shown in the following phase-plane diagram (Figure 17.8). Notice that the trajectory will spend some time "winding around" one equilibrium point, before jump~ ing to the other side and winding around the other equilibrium point for a while. This process goes on forever, with the trajectory never crossing itself (in 3-space). Lorenz E
20
= a 10 I = 50
15
10 5
"
a 5
10 -15
20
a
40
50
time
FJGllRE 17.7
Transient response ofx j under chaotic conditions.
Sec. 17.6
407
Chaos in Chemical Systems Lorenz Equations, t", 0 to t:= 50
50
40
30
20
10
o
20
--15
--10
-5
a
5
10
15
20
x, FIGlJRE 17.8
Phas<>plane or Lorenz equations under chaotic conditions.
'rhe development of the curve in FigUfC17.8 is shown more clearly in the phaseplane plots in Figure 17.9, which show varies "pieces" of time. A three-dimensional plot (phase cube) of this trajectory is shown in Figure 17.10. The reader is encouraged to perform simulations of the Lorenz equations, to be understand concepts such as sensitivity to initial conditions. MATI,AB has a demo titled lorenz. m (simply enter lorenz in the command window) that traces a three-dimensional plot of solutions to the Lorenz equations. Each new run uses a new set of random initial conditions. If you write an m-file to simulate the [,orenz equations using ode45, remember to use a name lhat is different than lorenz .m, to avoid conniels with the MATLAB clemo.
17.6
CHAOS IN CHEMICAL SYSTEMS The Lorenz equations provide a nice example of chaos, because the equations arc reasonably simple to analyze. An even simpler set of equations was developed by Rossler to demonstrate chaotic behavior (sec student exercise 3). Chaos has also been shown to appear in models or chemical process systems, particularly exothermic chemical reactors. Reactors that arc forced periodically Uacket temperature is a sine wave, for example) have been shown to exhibit chaos. Also, a series or reactors with heat integration can exhibit chaotic behavior. [t appears that chaotic reactors may have had low amplitude "oscillations" (say in temperature) that may have been interpreted as measurement and process noise in the past. A comprehensive review of nonlinear dynamic behavior in chemical reactors is provided in the article by Razoll ami Schmitz (1987).
Introduction to Chaos: The Lorenz Equations
408
Chap. 17
-, '" '" ()~---,,-
-7.11
(~O (0
-1.\
-111
5
(~5
"'1-
III
l~
211
'"
IS
JII
"
(010
T
41)
'" :10
'm '~
-111
'" 0-_}II
-15
-10
" (~1O(015
(~15
(020
II ---~~--, -20 -IS -10
--~--~-
HI
"
(~20(025
(~25(o30
.10
In (~
30 (035
FIGURE 17.9
"
(~
35 (040
Phase-plane of Lorcnz equations under chaotic conditions.
\<;
20
Sec. 17.7
Other Issues in Chaos
409
5H
'" )11
'" '"
11)
'"
t=
"
"
40 to 45
t
FIGURE 17.9
17.7
= 45 to 50
C'ol/iit/lled
OTHER ISSUES IN CHAOS Chaos is a complex field with ITiany books and conferences devoted to this simple topic. Clearly, it is impossible to give this topic adequate coverage in a single chapter. Our goal is to provide an introduction {tl, and motivation for, the topic. 'rhe reader is encouraged to consult the many books and articles available 011 the topic.
r
=:
28, Xo
[0;1;OJ
60
40
20
o -50
o ~
50
-20
~--~20 10
~
0
-10
x, FIGURE 17.10 Three-dimensional phase space plot of Lorenz equations under chaotic conditions.
410
Introduction to Chaos; The Lorenz Equations
Chap. 17
[ssucs that may be of particular interest include: How docs onc detect chaos experimentally? One method is to lise experimental data to calculate Lyapwwv exponents. See Strogatz for example. Chaos can be used to encode secret messages. See Cuomo and Oppenheim (199.)). who used ideas presented by Pecora and Carroll (1990).
SUMMARY We have presented an introduction to chaotic behavior by studying the Lorenz convective flow equations. A number of chemical processes have been shown to exhibit similar be~ havior. It is necessary to have three nonlinear autonomous differential equations before chaos can occur. Although not shown here, chaos call occur in a system of two nonlinear IlOIl(lutonO!1lOUS equations (that is, if some type of periodic input forcing is used), Also. we saw in Chapter 14 that chaos can occur in a single discrete nonlinear equation (the quadratic map, or logistic equation),
REFERENCES AND FURTHER READING A nice book on the history at' chaos, written for the general public, is by Gleick,
Gleick, J, (1987). Chaos: Making
II
New Science. New York: Viking.
The first paper to develop the notion of sensitivity to initial conditions is by Lorenz. Lorenz, E.N. (1963). Deterministic nonperiodic flow. Journal ences, 20: 130-141.
(~lAtnlOsphcric Sci-
A number of introductory-level textbooks provide nice introductions to chaos. The"e elude:
in~
Strogatz, S.H. (1994). NOlllinear Dynainics and Chaos. Reading, MA: AddisonWesley. A review of nonlinear dynamic behavior (including Chaos) of chemical reactors is provided by: Razon, L.l'., & R.A, Schmitz. (1987). Multiplicities and instabilities in chemically reacting systems--A Review. Chel1l. Eng. Sci., 42(5): 1005-1047. A large number of papers on chaos have been collect in the following book: Hao, Bai-Lin. (Ed,). (1990), Chaos fl. Singapore: World Scientific Press.
411
Student Exercises
Papers that develop a way of encoding secret messages using chaos arc: Pecora, L.M., & T.L. Carroll. (1990). Synchronization in chaotic systems. Physical Neview Leflers, 64: 821. Cuomo, K.M., & A,v. Oppenheim. (1993). Circuit implementation of synchronized chaos, with applications to communications. Physical Reviell Leflers, 71: 65. J
STUDENT EXERCISES 1. Consider the following parameter values for the Lorenz equations: " ~ 10
r
~ f, ~
470/19
~
24.74
8 b ~-3 For the nontrivial solution, show that a supercritical Hopf bifurcation occurs at this value of r. That is, for r < r e, all eigenvalues arc stahle, for r::::: r c' two eigenvalues arc on the imaginary axis, and for r > rc' two eigenvalues have crossed into the right half plane. 2. Show the sensitivity to initial conditions of the Lorenz equations. Run two simulations with the parameter values shown in the numerical study
" ~ .10 r ~ 28 b ~ 8/3 For the first simulation use the initial condition xo ;;::; [0 I OIT, For the second simulation lise the initial condition Xo ::::: [0 1.0 I Orr. When do the simulations begin to diverge? Run some more simulations with smaller perturbations in the initial conclitions, Also, make perturbatiollsin the initial conditions for the other state variables. What do you find? 3. Consider the Rossler equations (see Strogatz, for example): Xl =
x2 = x, =
-x2 - x J x l +ax2 b + x,(x\
~
c)
which have a single nonlinear terlll. Let the parameters a and h be constant with a value of 0.2. Usc simulations to show that this system has period-I (limit cycle), pcriod-2, and periodA behavior for c::::: 2.5, 3.5, and 4, respectively. Show that chaotic behavior occurs for c :::: 5,
412
Introduction to Chaos: The Lorenz Equations
Chap_ 17
4. The Henon map is a discrete model that can exhibit chaos:
xi(k +- 1) ~ x 2(k) +- 1 - a xi(k)"' x 2 (k -I- I) ~ bxi(k) For a value of b ~ 0.3, perform nUlllerical simulations for various values of a. Try to find values of a (try a > 0.3(75) thai yield stahle pcriocl-2 behavior. Show that chaos occurs at approximately a ;::;: I '()6.
APPENDIX Stability analysis of the nontrivial steady-state using the Routh array:
Row I 2
3 4 Where
I
"2
hi)] - b o h,l.
"0 "0 =2 ,,"(r -
()
hi
I)
~(r+,,)"
b2 =
Since the nontrivial steady-state only exists for 1';:: I, then bo is ahvays positive. It also follows that hi and b 2 will always he positive. The only entry from the Routh array that we must check is the first column in row 3. This entry will he positive if:
h2h J
-
bo > 0
or
h 2b 1 > b o
Making the substitutions for parameter values in the coefficients:
( 2
(A-2)
After some algebra, this can be written: 'T
+- h+-I) r--- -- ,,( 'TI h
-I-
3)
(Al)
or,
('T - b - I) r < "(
(J
(A 4)
> h + I, the condition on r for stability is:
r<
,,(,,+-h+(A-5)
('T - h- I)
Notice from (A-3) that if (J < b + I, then any,. satisfies the requirement will often define the critical value, rc:
fl.)!"
stability. We
" (" +- h +-
(A-6)
(" - {) - J) If r > rc' then the syslem is unstable.
________________
__
~~~
~_~_~
..~A • _ _.......
SECTION
IV
REVIEW AND LEARNING MODULES
INTRODUCTION TO MATLAB
MODULE
1
The purpose of this module l is to review MATLAB for those who have used it before and to provide a brief introduction to MA'fLAB for those who have not used it hefore. This is a hands-on tutorial introduction. Arter using this tutorial, you should be able to:
r';nter I1latriccs Make plots Write script files Perform Inatrix operations Usc MATLAB functions Write function files The m'ljor sections of this module arc:
MI.l MI.2 M1.3 MIA MI.5 MI.6
Background Using This Tutorial Entering Mali"ices The MATLAB Workspace Complex Variahles Some MA'rLAB Operations
IGood references include Introductioll to 11.1;\ 7V\U for E'flf;ineers (ind Scicntisfs by D.M. EtLer (Prentice-Hall, 19(6), and Mas/eriflg MArLA/; byD. Hanselman and 13. Littlefield (PrcnticcHall, 1996).
415
416
Introduction to MATLAB
M1.7 MI.R
MJ.9 M I.J()
MI.II MI.12 MI.13 MI.J4
M1.1
Module 1
Plotting More Matrix Stuff FOR Loops and IF-THEN Statements m-filcs Diary
Toolboxes Limitations to Student MATI,AB Contacting MATHWORKS
BACKGROUND MATL,AB is an interactive program for numerical computation and data visualization. It was originally developed in FORTRAN as a MATrix LABoratory for solving numerical linear algebra problems. The original application may seem horing (except to linear algc~
bra enthusiasts), hut MATLAB has advanced to solve nonlinear problems and provide
de~
tailed graphics. It is casy to usc, yet vcry powerful. A few short commands can accoll1~ plish the same results that required a major programming effort only a few years ago.
M1.2
USING THIS TUTORIAL This tutorial provides a hrief overview of essentialMATLAB commands. You will leal'll this material morc quickly ({YOll use MATLAS interactively as you are reviewing this {lf~ (orial. The MATLAB commands will be shown in the following font style:
Monaco font the prompt for a user input is shown by the double arrow
MATLAB has an extensive on-line help facility. For example, type help pi at the prompt
)) help pi PI
PI c, 4*atan(I)
~
3.1415926535897 ....
so we see thatMATLAB has the number »
'IT
"built-in". As another exalllple
help exp
EXP EXP{X) is the exponential of the elements of X, e to the X.
Sec. Ml.3
Entering Matrices
417
sometimes you do not know the exact command to perform a particular ope-rallon. In this case, one can simply type »
help
and MATI-,AB will provide a list of commands (and m~files, to he discussed later) that are available. If you do not know the exact command for the function that you arc after, another useful cOllltnand is lookfor. This command works somewhat like an index. If you did not know the cOlnmand for the exponential futlction was exp, yOli could type »
lookfor exponential
EXP EXPM EXPMl EXPM2 EXPM3 EXPME
M1.3
Exponential Matrix exponential. Matrix exponential via Pade' approximation. Matrix exponential via Taylor series approximation. Matrix exponential via eigenvalues and eigenvectors. Used by LINSIM to calculate matrix exponentials.
ENTERING MATRICES The basic entity inMATLAB is a rectangular matrix; the entries can be real or complex. Commas or spaces are used to delineate the separate values in a lnatrix. Consider the following vector, x (recall that a vector is simply a matrix with only one row or column): [1,3,5,7,9,11]
»x =
x
1
7
5
3
9
11
Notice that a row vector is the default inMATLAB (in contrast to the default column veclor lL"lcd by this and most other texts). We could have used spaces as the delimiter between columns: [1
»x
3
5
7
9
11]
x 1
3
5
7
9
11
There is a faster way to enter matrices or vectors that have a consistent pattern. ['or example, the following command creates the previous vector: » X
1:2:11
=
x
1
3
5
7
9
11
Introduction to MATLAB
418
Module 1
Transposing a row vector yields a column vector (, is the transpose command in MATLAB):
" y
x'
y ~
1 3 5 7 9 11
If we want to make x a column-vector, we usc a semicolon as the deli meter between rows: "
x
~
[1;3;5;7;9;11]
x
1 3
5 7 9
11 To make x a row vector again, we usc the transpose: » X
x'
0=
Say that we want to creale a vector z, which has clements from 5 to 30, by 5s: »
Z
5:5:30
::::
z 5
10
15
20
25
30
If we wish to suppress printing, we can add a semicolon (i) aftcr any MA TLAB command: »
Z ::::;
5:5:30;
The z vector is generated, but not printed in the command window. We can find the value of the third clement in the z vector, Z (3) , by typing
Sec. M1.4
the MATLAB Workspace
419
" 2(3)
ans 15 Notice that a new variable, ans, was defined automatically.
M1.4 THE MATLAB WORKSPACE We can view the variables currently in the workspace by typing: » who
Your variables are: ans
x
z
y
leaving 621420 bytes of memory free. More detail abollt the size of the matrices can be obtained by typing: »
whos Name
Size
ans
1 1 6 1
x
Y 2
Grand total is (19 * 8)
~
by by by by
Total 1 6 1 6
1 6 6 6
Complex No No No No
152 bytes,
leaving 622256 bytes of memory free. We can also find the size of a matrix or vector by typing: »
[m,n};;;:size(x) 1
n
~
6
where ill represents the number of rows and n represents the number of columns. If we do not put place arguments for the rows and columns, we find:
420 »
Introduction to MATLAB
Module 1
size (x)
ans 1
6
Since x is a vector, we can also usc the length command " length(x) ans 6
It should be noted that MATLAB is case sensitive with respect to variable names. An X matrix can coexist with an x matrix. You may wish to quit MATLAB but save your variables so you don't have to retype or recalculate them during your next MATLAB session, To save all of your variables, usc: »
save
file~Dame
(Saving your variables does not remove them from your workspace; only clear call do thall You can also save just a few of your variables: »
save file__name x y z
To load a set of previously saved variables: » load file name
Sometimes it is desirable to clear all of the variables in a workspace. This is done by simply typing: »
clear
More frequently, you may wish to clear a particular variable, such as x »
clear x
This is particularly true if you arc performing a new calculation of x and the new vector is shorter than the old vector. If the new vector has length fl, then all of thc clemcnts of thc new x greater than x(n) will contain values of the previous x vector.
421
Some Matrix Operations
Sec. M1.6
M1.5 COMPLEX VARIABLES Both i and j represent the imaginary number,
\1=1, by default:
» i
ans »
+ 1.0000i
o
+ 1. OOOOi
j
ans »
o
sqrt (-3)
ans
;=
o
+ 1.'73211
Note that these variables (i and j) can he redefined (as the index in a for loop, for CX~ ample), as will be shown later. Matrices can be created where some of the clements arc complex and the others arc real: »a
[sqrt(4) {
lisqrt(~4),
-5]
a 2.0000
o
+ 2.00001
1.0000 -5.0000
Rendl that the semicolon designates the end of a row.
M1.6 SOME MATRIX OPERATIONS Matrix multiplication is straightforward: »
b
[1 2 3;4 5 6J
~
b
1 4
2 5
3 6
Using the a matrix that was generated in Section 1.5:
ESCOLA DE ENGEN HARIA BIBLIOTECA
Introduction to MATLAB
422 »
C
;:;,
Module 1
a *b
c 6.0000 9.0000 12.0000 -20.0000 + 2.0000i -25.0000 + 4.0000i -30.0000 + 6.0000i Notice again that MATLAB automatically deals with complex numbers. Sometimes it is desirable to perform an element-by-elenwnt multiplication rather than matrix multiplication. I,'or example, d(iJ) =- b(ilj)*CUlj) is performed by llsing the -k command d:::o c. *b
»
1.0e+02 * 0.0600 -0.8000 + 0.0800i
0.1800 -1.2500 + 0.2000i
0.3600 -1.8000 + 0.3600i
(Notice the scaling that is performed when the numbers arc displayed.) Similarly, clement by clement division, b(i\i)Ic(iJ), can be performed using. /: )-)- e :::; b.le
e 0.1667 -0.1980 - 0.0198i
0.2222 -0.1950 - 0.0312i
0.2500 -0.1923 - 0.0385i
Other matrix operations include: (i) taking matrix to a [lower and (ii) the matrix exponential. These arc operations on a square matrix: »
f
=
a/,2
f
4.0000 + 2.0000i o - 6.0000i »g
9
expm{a)
~
7.2232 + 1.8019i -0.4302 + 2.0760i
1\11.7
-3.0000 25.0000 + 2.0000i
1.0380 + 0.2151i -0.0429 • 0.2962i
PLOTTING For a standard solid line plot, simply type: » plot (x, z)
Sec. M1.7
423
Plotting
where x and z have been generated using »x
»z
1:2:11; 5:5:30;
30 25 20 ~
15 10 5 0
2
4
6
10
8
12
x
Axis labels arc added by using the following commands: »
xlabel ('x')
» ylabel ( , z ' )
For more plotting options, type: »
help plot
PLOT
Plot vectors or matrices. PLOT(X,Y) plots vector X versus vector Y. If X or Y is a matrix, then the vector is plotted versus the rows or columns of the matrix, whichever lines up. PLOT(Xl,Yl,X2,Y2) is another way of producing multiple lines on the plot. PLOT(Xl,Yl, ':' ,X2,Y2, '+') uses a dotted line for the first curve and the point symbol + for the second curve. Other line and point types are: solid dashed dotted dashdot
point plus star -
,
circle 0 x-mark x
+
red green blue white invisible arbitrary
r g
b w
i cl
t
c15,
etc.
PLOT(Y) plots the columns of Y versus their index. PLOT(Y) is equivalent to PLOT(real(Y),imag(Y)) if Y is complex. In all other uses of PLOT, the imaginary part is ignored. See SEMI, LOGLOG, POLAR, GRID, SHG, CLC, CLG, TITLE, XLABEL YLABEL, AXIS, HOLD, MESH, CONTOUR, SUBPLOT.
Introduction to MATLAB
424
Module 1
If we wish to plot discrete points, using + as a syrnbol, we can usc the following: »
plot(x,z,'+') 30 - _..
+
25
20
+
15
+ +
10 )
i
5'
+
0
2
4
6
8
10
12
x
'rext call be added directly to a figure using the gtext command. gtext ( 'string ') displays the graph window, puts up a cross-hair, and waits for a mouse button or keyboard key to be pressed. The cross-hair can be positioned with the rnouse. Pressing a mouse button or any key writes the text string onto the graph at the selected location. Consider now the following equation:
Y(I)
4e
01,
We can solve this for a vector of t values hy two simple- commands: » t » y
::::
0:1:50; 4 * exp ( ~ 0 . 1 * t)
;
and we can obtain a plot by typing:
" p1ot(t,y) Notice that we could shorten the sequence or commands by typing:
"
p1ot(t,4*exp(~O.1*t))
4
3
~
2
o
o
10
20
30
40
50
Sec. M1.7
Plotting
425
We call plot the function yet) = »
Y
t
e - lUI by using:
t.*exp(-O.l*t);
=
" plot(t,Y) »
gLexL('hey, this is the peak! ')
" xlabel ( , t ' ) " ylabel ('y') 4
islhe
3
o o
20
10
40
30
50
axis ( square ') will place the plot in a square box, while axis ( normal') will change back to a normal aspect ratio. You can also explicitly set the upper and lower bounds on the plot with: I
I
axis([xlow
xhigh
ylow
yhigh])
For this example we would usc: " v »
~
[0
50
0
4];
axis(v);
Multiple curves can be pJaccd on the same plot in the following fashion. plot(t,4*exp(-0.1*t) ,t,t.*exp(-O.l*t), '--')
3
,, , 2 . , , ,,, , ,, I
oI
o
_.__
~._L_. 10
~~~---=:::::====~ 20
30
40
50
Introduction to MATLAB
426
Module 1
The subplot command can be lIsed to make multiple plots.
»subp1ot(2,1,1), p1ot(t,4*exp(-O.1*t)) »subp1ot(2,1,2), p1ot(t,t.*exp(-O.1*t))
4 ~-----~- - - -
:l_~__~ __~ ~ o
10
20
,-
",=_
30
I
,
-
__
40
:Ln.~. .::.. o
10
20
30
40
50
50
lIcre, subplot(iJ,k) means that there are i "rows" of figures,.i "columns" of figures. and the current plot is the kth figure (counting right to left and top to bottom). To rctum lO single plots, simply enter subplot (1,1,1).
M1.8
MORE MATRIX STUFF A matrix can be constructed from two or more vectors. If we wish to create a lnatrix \'. which consists of two columns, the first column containing the vector x (in colunlll fonn) and the second column containing the vector z (in column form), we can usc the J'olhmlI1g: »vo::o[x',z']
v 1
5
3 5 7
10 15 20
9 11
25 30
If we wished to look at the first column of v, we could use:
.....
~----------------~-_
Sec. M1.9 »
V ( :
J
For Loops and If-then Statements
427
1)
ans 1 3
5 7 9 11
If we wished to look at the second column of v, we could
" v (:
lISC:
,2)
ans 5
10 15 20 2o"j 30
And we can construct the same plot as before, by using (' --
"
I
gives a dotted line):
p1ot(v(:, 1) ,v(: ,2), '--' )
30 25 >
0 0J
20
C
E => <3
"
15 10 5 0
2
4
6
8
10
12
column 1 of v
M1.9
FOR LOOPS AND IF-THEN STATEMENTS A for loop in MA'rLAB is similar to a DO (oop in PORTRAN. The main difference is that the 1"ORTl{AN DO loop must have an integer index variable; for does not have this restriction. An example of a for loop that is virtually identical to a DO loop is:
428
Introduction to MATLAB
[or k tlk) ylk) end
Module 1
l:~)OOl;
»
(k1)*0.01; sinltlk));
Another way of implementing the saille loop is increment ( from 0 to 50 in increments of 0.01: »k »
0
for t
~
0:0.01:50;
k = k + 1; y(k) = sin(t);
end
The developers ofMATLAlj highly recommend that you usc the vectorizcd version of the above for loops: t 0:0.01:50; y - sin I I) ;
since the computation time for this method is over 200 tilnes faster than the nonvcc{(lrized methods.
Ml.l0 M-FILES 'I'hus far we have shown the interactive features of MATL.AB by entering one command at a time. One reason that MA'rLAB is powerful is that it is a language, and programs of MATLAB code call he saved for later lise. There arc two ways of generating your own MATLAB code: (1) script files and (2) function rOlltines.
1.10.1
Script Files
A script file is simply a sequence of commands that could have been entered interactively in the MATLAB command window, When the sequence is long, or must be performed a nurnbcr of times it is much casier to generate a script file. The following example is for the quadratic map population growth model:
where xk represents the value of the population (dimensionless) at the kth time step. We have titled the file popmod. ill and stored it in our directory.
Sec. MUO
%
M-Files
429
poprnod.m
% population model, script file example %
clear x,k n input ( 'input final time step
alpha xinit
')
i
input('input alpha '); input('input initial population ')
i
x(l) - xiniti time ( 1 ) ,. O' for k;::: 2:n+l; time(k) - k-l; x(k) alpha*x(k-l)*(l-x(k-l)); end
plot(time,x) % end of script file example Notice that we have used the MATLAB input function to prompt the user for data. Also note that a percent sign (%) may be used to put comments in a script or function file. Any text after a % is ignored. T'he file is run by simply entering: » popmod
in theMATLAB command window.
M1.10.2
Function Routines
A more powerful way of wiving problems is to write MATLAB function routines. Flllle· tion routines are similar to subroutines in FORTRAN. Consider the previous example.
function [time,x] = pmod(alpha,xinit,n) % population model example, pmod.m clear time; clear x; clpar k; x(l) xinit; time (1) = 0; for k = 2:n+l; time(k) k-l; x(k) = alpha*x(k-l)*(l-x(k-l)); end % end of function file example
We can now "run" this functioll routine (using alpha:::;: 2.8, xini t :::;: 0.1, n :::;: 30) by typing » [tstep,xpop]=pmod(2.8,O.1,30); » plot(tstep,xpop)
Module 1
Introdu ction to MATLA B
430
0.7 0.6 0.5 00 0-
0.4
x
0.3 02 0.1 0
5
10
15
20
25
30
tstep
. This fC,'llurc leads to This function routine can also be called by other function routines and debug. "stl1lcturcd program ming"; structur ed program s are casy to follow through out this MATLA B has many built-in function routines that you will usc and ode45 , to solve a course. -The most tommo nly used rOLltines arc fzero, [solve: ., equatio ns, and a set of single nonline ar algehra ic equatio n, multipl e nonline ar algebra ic nonline ar differen tial equatio ns, respectively.
M1.11 DIARY the sequenc e or COl11When preparin g homew ork solution s, it is oftell necessa ry to save ork. The diary commallels and output results in a file to be turned in with the homew mand allows lhis. and most of the diary f i le._nam e causes a copy of all subsequ ent terminal input off sllspends it. resulting output to be written on the file namcd file__ name. diary statc. Diary riles may be diary on turns it back OIl. diary, by itself, toggles the diary n entries. edited latcr \\'ith a text editor to add comme nts or renlove mistake to assist them in trousession your of file diary a See to wish ants Often the consult bleshoo ting your MATL.AS problems.
M1.12 TOOLBOXES to solve speciali zed MATLA B toolbox es arc a collection of function routines writtcn the Student Edition of problem s. The Signals and Sy.<;lems Toolbo x is distribu ted with s the Symbol ic Algcbra MATL.AB. The newcst edition of Student MATLA B also contain analytic al solution s to toolbox , which is a collecti on of routines that allows onc to obtain in chemica l process algebnd c and differcntial calculu s problem s. One toolbox orten used control is the Control System s Toolbo x.
-------------
Z'"'''''jK> ' Ii • •
]I
kk
~i
Sec. Ml.14
Contacting Mathworks
431
M1.13 LIMITATIONS TO STUDENT MATLAB The Student Edition of MATLAB has a few limitations compared with the Professional version. For example, Student MATLAB (4.0) is limited to 8 I92-element alTays.
M1.14 CONTACTING MATHWORKS MATHWORKS has aholncpagc on the \Vorld Wide Web. The URL, for this home page IS:
http://www.mathworks.com/ You can find answers to frequently asked questions (FAQ) on this homcpagc. Also, there arc a number of technical notes that give more inf<Jnnation on using MA'fL,AB function routines. Suggestions arc made for modifying these routines for specific problcrns. There is also a MATLAB ncwsgroup (bulletin board) that has up-tn-date questions (usually supplied by MA'fLAB users) and answers (supplied by other users as wel! as MATHWORKS personnel). The newgnmp is:
camp.
soft~sY~3. matlab
Berore posting questions on this ncwsgroup it is a good idea to read the FAQ from the MATlIWORKS.
Summary of Commonly Used Commands axis clear clc diary end exp for format function gtext help hold if length
axis limis for plots removes all variables from \vorkspace clears command window save the text of a MATLAB session end of loop exponential function generates loop structure output display format user generated function place text on a plot help function holds current plot and allows new plot to be placed on current plot conditional test length of a vector
Introduction to MATLAB
432
Module 1
lookfor plot
keyword search on help variables
size subplot
size of the alTay (rows, columns)
who whos
view variables in workspace view variables in workspace, with more detail (size, etc.)
plots veclors multiple plots in a figure window
matrix multiplication
transpose suppress printing (also - end of row, when used in matrices)
* ./ %
clement by clement 111l11tplication clement by clement division denotes a column or row in a matrix. also creates a vector placed before comment statements
Frequently Used MATLAH Functious Function
Use
Chapter
eig
eigenvalues, eigenvectors
5
[solve
solve algebraic equations solve a single algebraic equation
3
[zero
impulse
3
9
ode45
impulse response integratc set or ordinary differential equations
polyfit
least squares fit of a polynomial
Module 3 (linear rcgression)
ss2tf
convert state space to transfer function modcl
10
step tf2ss
stcp response
5, 9
convert transfer function to state spacc model
9
4
STUDENT EXERCISES 1. Plot the following thrcc curves On (i) a single plot and (ii) multiple plots (using the subplot command): 2 cos(t), sin(t) and cos(t)+sin(t). Usc a timc period such that two or three peaks occur for cach curvc. Use solid, dashed, and '+' symbols for the
dilTcrent curves. Usc roughly 25-50 points for each curvc. 2. a. Calculatc the rank, detenninant and matrix inverse oUhe following matriccs (usc help rank, help det, and help inv):
Student Exercises
433
A
= [
I - I
2 2
2
4
-~J
[- ~ -n 2
B
4 4
[-~
e 3. Find
2
iJ
4 4
eel, where:
4. 3. Calculate xTx, and h. Calculate xx T, where:
5. Find the eigenvalucs of thc matrix:
-I
D~:
Io
o
()
2 3
()
o
()
I
6. Consider the expression:
with A~
[~
3 ] -1
[ - 11.896 Q _. -17.192 Solve for R.
...- 2(~.3281 -18.856
B ==
[:1
K~
G ~]
434
Introduction to MATLAB
Module 1
7. Find the solutions to the equation j(x) ::::: Jx 3 + x 2 + 5J: - 6 ::::: O. Usc roots alld fzera. 8. Integrate the following equations from t::::: 0 to I ::::: 5:
dX 1
-~
dl
xJ + Xz
dX 2 dl with the initial condition .1:1(0):= x 2(O)
=:;
I. Usc ode45 and plot your results.
9. Write your own function file for the following equation:
k(T)
~
a cxp b (
~
c
'I'
~ In
dT ~ e 'I' +
IT 2 )
for(/~3.33,h~ 13,148,c~5639.5,d~ 1.077,e~5.44x lO-4,I~ 1.125x 10-7 T is in units of Kelvin. Plot k as a function of l' for temperatures from 373 to 600 K. (we suggest 50 points)
10. Find
V( cm~. . ) for the following equation of state: gmol
RT V
~
b
(/
'["'V (V
+ b)
~ 13.76 har, b ~ 44.891 T ~ 333 K, (/ ~ 1.56414 x 10' cm(i bar/gmol 2 KO.5, R::::: ideal gas constant in appropriate units.
for P
cm 3/gmol,
...
L
REVIEW OF MATRIX ALGEBRA
MODULE
2
The purpose of this module is to provide a concise review of matrix operations and linear algebra. After studying this module the reader should be able to: Multiply lnatrices
Find the transpose of a matrix Find the eigenvalues and eigenvectors of a matrix Understand concepts of rank and singularity The major sections of this module arc: M2.1
Motivation and Notation
M2.2
Common Matrix Operations
M2.3
Square Matrices
M2.4
Other Matrix Operations
In this module we review only the essentials of matrix operation necessary to understand material presented in the chapters of this textbook. For more detailed explanation:'! and more advanced concepts, please consult any textbook on matrices or linear algebra. MATLAB is very useful for performing matrix operations. Module 1 provides a review of MATLAB. 435
Review of Matrix Algebra
436
M2.1
Module 2
MOTIVATION AND NOTATION In the study of dynamic systems it is comillon to usc l1l<:ltrix BoUllion. The usc of matrix notation allows the compact representation of a system composed of many variables and equations. Matrices are two-dimensional arrays that contain scalar elements that can be either real or complex. In this module most or our examples involve real matrices. Consider the following matrix, which consists of n rows and In columns:
A
a 2l
(/n
",.
{ll/!
{[n2
{fl/Ill
{{II
r'"
([211I
]
where (lij represents the scalar clement in the ilh row andjth column of matrix A. It is a good habit to denote the numbers of rows and columns beneath the matrix as (n,rn) or (n x tn). For example, the following is a 2 x 3 matrix:
.~ I~ ~ ~I
A
(2
X
3)
A vec!oris a special case of a two~dimcnsionall1latrix. Normally, the use of the tcnn vec~ tor implies a column vector. The following is an example of a column vector of length :; (a 3 x I matrix), where Vi is an element of the vector:
A row vector of length three is W
o:c;
[-WI
'tV!
'l.03J
The convention of this textbook is to usc lower-case bold leUers to represent vectors and upper-case bold letters to represent matrices. Unless stated otherwise, it is assumed that a vector is a column vector.
M2.2
COMMON MATRIX OPERATIONS M2.2.1
Matrix Addition
Two matrices can be added by simply adding the individual elements of the matrix. The matrices must have the same number of rows and columns. The operation:
C=A+B
Sec. M2.2 is simply
437
Common Matrix Operations
clcmcnt~by-e1cmenl addition
to form a new matrix:
(ij
:,,"_c
+ b ij
aij
For example,
c=
I:
2
31 [7
6 + 10
5
X II
91 12
=
IX 14
10
16
Clearly the order of matrices docs not matter for the addition operation (A + n : :; B -I- A).
M2.2.2
Matrix Multiplication
Consider an 11 x III matrix A and an r x nmatrix product of H times A:
C
:::
(r x 111)
n.
An ,. x m matrix C is defined as the
II
A
(rxn)(nxm)
Notice th:'ll the number of columns of B must be equal to the number of rows of A. The clclnenl in the ith row andjth colulllll of Cis: C ij = Hi] A Ii
+ Hi2 A 2j
-1- •••
+
BinA llj
n
Cij:O:;:
2: BikA
k
(M2.1)
kj
.1
Notice that we can view this as an operation on row and column vectors within the matri-
ces. That is, the ith clement in the jth column of C is equal to the scalar value that is obtained from "multiplying" the ith row of B times the jth column of A. C (",111)
Cd
=
H (r.lI)
A (1I,1J1 )
Crm
ijth clemenl
ilh row
i
jlh column
438
Review of Matrix Algebra
Module 2
[,'or example,
c ,= [
~ 1~)j /:
11 (rows,cols)
2
5
12
(3,2)
(2,3)
7(1) + X(4) ==
~
7(2) + X(5)
9(1) + 10(4) [ 11(1) + 12(4) 39 49 [ 59
54 6X
69
X2
lOS
X7
7(3) + X(6) 9(3) + 10(6) 11(3) + 12(6)
9(2) + 10(5) 11(2) + 12(5)
j
(3,l)
j
Clearly the order of the matrices arc important in multiplication. In the example above, BA is a consistent multiplication, while An is not Even for square matrices (where the number of rows is equal to the number of columlls), AU -::F- HA in general.
M22.3 Let D be
Transpose 11
transpose of the matrix A, then:
D
~Af
(M22)
where (1) represents the transpose operation. The ij clement of Dis theji clement of A til}
= {iji
Columns become rows and rows become columns. For example, 2
5
63..1
A
f
~ [~ ~j 3
6
It caIl he shown that the following property holds:
(ABC)f
=
CTBTA T
As noted earlier, we normally think of vectors as being column vectors. The transpose of a column vector is a row vector. Consider an n-dimensional vector x. The transpose of x is xT .
and X tl
Sec. M2.2
439
Common Matrix Operations
The vector transpose is often used to calculate a scalar function of two vectors. [)or CXHrllpic, if wand z are column vectors, of length 11, the operation wTz results in a scalar value:
n
2: ;,,--1
'/l};Zj
Notice that multiplication in the opposite order, zw1~ results in ann x n matrix: 21
22
23
['w,
'W 2
'W 3
IlJ tl
l
Z['lD J
Zj'lV 2
Z l'l()\
Z2~Vl
Zz'!D2.
2 2103
2,,'10 1
ZII'W Z
Zn?D j
r
""'·1 z
Z 1DI/
Zll1IJ il
Zn
(II.! )
M2.2.4
(11,11)
( 1.11)
Diagonal Matrices
Diagonal matrices arc commonly used in weighted least squares (regression) problems. Matrix Q is diagonal if
Qij
for i ~ j 0 for; i j
'Ii ~
For example, Q = diag( I,3, 12) represents the following matrix:
o
oOJ
3
o
12
IDENTITY MATRIX An identity matrix, I, is a diagonal matrix with ones on the diagonal and zeros [ii ~
I if ; o it' ;
'fhe 5 x 5 identity matrix is:
~
*
1 [0=
diag(l,I,I,J,I)
o o o o
j j
0 1
000 000
0 0
I
010
0
001
0
0
off'~diagonal.
440
M2.3
Review of Matrix Algebra
Module 2
SQUARE MATRICES A number of the important matrix operations lIsed in this textbook involve square (number of rows equal number of columns) matrices.
M2.3.1
Trace
The trace of an n
X
n (square) matrix is simply the sum of its diagonal clements, tr A
:;0:
2:" ;
M2.3.2
(M13)
{lii
1.
Determinant
One of the most important properties of a square matrix is the determinant. Consider 2 X 2 matrix:
A
::::0
la" {/ 1 a an. '.2..
21
The determinant of A is
det A
-_0
allan -
{f21{f12
Sometimes the det A is written as IAI. An algorthim for finding the determinant of a larger matrix involves matrix cofactors and is shown below.
M2.3.3
Minors and Cofactors of a Matrix
The matrix formed by deleting the ith row and the jth column of the matrix A is the m;,/or of the clement aU' c1cnolcdM ,j" Consider the 3 x 3 matrix:
A
o
[::: •.1'
au an
{I'll
(l31
(f-12
aD
an
The minor of (lj2 of the matrix is obtained by deleting the first row and the second column. In this case:
ani
o:u__
The detcrminant of the resulting matrix is:
del MI2
0::0
(l21a:n -
(1:-Ha
n
(M2.4)
The cqj'actor is thc "signcd" valuc of the minor. That is, the cofactor is dcfined as:
cy
1)" J del Mq
co ( - -
(M2.S)
Sec. M2.3
441
Square Matrices
The cofactor e l 2 of the 3 X 3 example matrix is: e\2
cc_-., (~lY+jdctM12
=--=
(~1)112(a21a:Bm_{{:l1an)
=
112[ll:n
+
(l:lIa}}
The determinant of a matrix call be found hy expanding ahout any row or column (this approach is sometimes called the Laplace expansion). (i) Expanding around any columu j:
2:"
det A
{lijcij
(M2.6)
{lljCij
(M2.7)
i"\
(ii) Expanding around any row i:
det A :::::
2."
j.-.--.- 1
As an example, consider the 3 x 3 matrix A. The expansion around row 1 is
det A = ([nCII + {f12et2 t- ajJe n del A {[" ( .. I) I , I del M" + II 12( -- I)' ' 2 del M 12 + ([ lJ( - I) I ,; del Mil det A = all (az/l:1,J ~
M2.3.4
(f12alT) --
aI2(a21ay~ ~ ({Jl{l2.1) -l--
{{lJ
(a 21 11:JJ. - ((31022)
Matrix Inversion
The inverse of a square matrix A,is called A
!
and is defined as: (M2.8)
The matrix inverse is conceptually useful when finding x to solve the following problem: Ax~y
Multiplying each side (onlhe left) by A', we find: A-I A x (11 X 11)(11 X 11)(11 X I)
~
A··'
Y
(II X 11)(11 X I)
wbich yields: Tx~A '
y
or,
x
~
An algorithm for finding the inverse of an
A
I
11
A ly x n matrix is known as Cramer's rule:
--, adjA del A
where the
a(~i()int
(M2.9)
matrix of A (adj A) is the transpose of the matrix of cofactofs of A. (M2.IO)
ESCOLA DE E,,~CENHARIA L,,,LI0Tt-:A
442
Review of Matrix Algebra
Module 2
In practice, this procedure is not used because the computational time is quite large t()r large matrices. Generally, a method such as Gaussian Elimination or LU decomposition is used.
EXAMPLE !VI2.1
Inversion of a 2
X
2 Matrix Using Cramer's Rule
all]
= [all a
A
an
21
I
A
adj A detA
-
where the cofactofs arc
Cll
-- (-l)l+ldetM lI
e 12
= (--I)) t2 dctMj2 =
-a:?1
en
-~ (-~"1 f l
cn
= (-ti
(1n
1
det M 21
--
~ (l12
l2
det Mn
=
all
= [~~:
adj A = {c;Jr
=
~~:r' = [~:~
A' ~ adj A = ,dctA
an . ~a21
The formula for the inverse of a 2 x 2 matrix should be committed to memory: A
-
[~ b] d d
- b]
1
~
d...ct A.
a . det A ~ [ E det A
where de! A
00:;
-bA ] det ....... 0 ...
detA
ad - be
That is, ~wap the diagonal terms, take the negative of the off-diagonal terms, and divide by the determinant.
Sec. M2.3
M2.3.5
443
Square Matrices
Other Issues in Matrix Inversion
A singular matrix cannot be inverted. A matrix A is singular if det A :::: O. Also, a singular matrix has a rank that is lower than the dimension of the matrix. The rank of a matrix is the number of independent rows or columns. For example, the matrix:
A
= I~
~]
has rank = I, because the second row is linearly dependent on the first row. The matrix is singular, because
det A = 4 - 4 = 0 We note that the matrix inverse of a singular matrix does not exist. For this example, note the division by 0 in the following calculation
AI =
I-24
1 -2] I [4-2 -21 4 .0 4. detA =
=
.. d undchne
The rank of a square matrix is equal to the number of nonzero eigenvalues. Eigenvalues of a square matrix arc disclissed next
M2.3.6
Eigenvalues and Eigenvectors
Eigenvalue/eigenvector analysis will bc important when studying dynamic systcms. Eigenvalucs dctcrminc how "fast" a dynamic system responds, and thc associated eigenvector indicates the "direction" of that response. This material is used in Chapter 5. Eigenvalue/eigenvector analysis can only be perrormed on square matrices. The eigenvector/eigenvalue problem is: (M211)
where: A = matrix (square, number of rows::::: number of columns) A = cigenvalue (scalar) ~ = eigenvector (vector)
Normally, we will usc the notation ~i to rcpresent the eigenvector that is associated v"ith eigenvaluc Ai' There are n eigenvalues and n eigenvectors associated with an II X 11 matrix. The interesting thing about equation (M2.11) is that a matrix times 11 vector is equal to a scalar times a vector. A scalar multiplying a vcctor docs not change its "direction" in n-dimensional space, only its magnitude. In this case, a matrix A times the cigenvector ~ yields the eigenvector ~ hack, scaled by A. Equation (M2.11) can be written as
(AI -
A)~ = ()
(M2.12)
444
Review of Matrix Algebra
Module 2
where A must be a scalar value for which Al --- A IS singular, otherwise ~ :;;;; 0 (the trivial vector). A requirement for AJ - A to be singular is dctCAJ ~ A) :;;;; 0, where the notation
det(AI - A) is used to represent the dctcnninant of l\!- A. EIGENVALUES The eigenvalues of an n X n matrix A arc the n scalars that solve:
det(l\! - A) •.. 0
(M2.13)
Equation (M2.13) is also known as the characteristic polynomial for the matrix A~ the eigenvalues arc the 1'001,0;; of the c1Ulractcristic polynomial. For an 11 X 11 matrix A, the characteristic polynomial will he nth order, so there will be II roots (n eigenvalues).
EIGENVECTORS The corresponding 11- eigenvectors (~) can be found from
A ~i ~i
l\j ~i
(M2.14)
is the eigenvector associated with the eigenvalue AiFor example, consider the general 2 x 2 matrix: 1I'.l
A
1(In
(Jizi (/22_
l\-II 11
l\I-A 1..
~(/Zl
det(l\I - A) = (l\ - a,,)(l\ - II,,) - all
"21 =. 0
det(l\I- A) = l\' - [1111 + lin] l\ + {alia" - a2l 1l,,{ ~ 0
(M2.15)
Equation CM2J5) can also be written as:
det(l\! - A) ~ l\' - triA) l\
+ det(A)
~ 0
(M2.16)
where:
triA) = all + a 22 dct(A) = allan - anll n Equation (M2.16) is the characteristic polynomial for a 2 x 2 matrix.
It can easily he shown that, if tr(A) < 0 and det(A) > 0, the roots (eigenvalues) of (M2.16) will be negative. Also, if det(A) '= 0, then one root (eigenvalue) will be zero. The reader should derive the following characteristic polynomial for a 3 x 3 matrix. (see student exercise 7).
Sec. M2.3
445
Square Matrices
The characteristic equation for a J x 3 matrix is:
det(Al- A) = A' - tr(A) A' + M A - dCl(A) = ()
(M2.17)
where:
tr(A) = M
(/Il
+ (/" +
{/11
c~
det Mil + del M" + det M 11
:0:::
"ntr:,>:,> -" (lJ211n
det A =0
+ {llla J3 ~
+
{fJl((jJ
(l11(a 22 {f:n - anil lz ) ~ {[12(a n ll J:.
-
a lJ a 22 -
((?,j(ln)
{{12(f21
+ {/u(a2111:12 ~ (l:\1(fn)
The ROllth stability criterion (Chapters 5 and 9) can be Llsed to find the conditions on the polynomial coefficients that will yield negative roots (eigenvalues). The necessary condition is that all of the polynomial coefficcnts arc positive. The necessary condition is then tr(A) < 0, M > 0, and dct(A) < O. The reader can usc the Routh stahility criterion to determine the sufficient conditions for negative roots.
EXAMPI.. EM2.2
Eigenvalue/Eigellvector Calculation
Consider the following matrix:
~ I~
A
dCl(,.I~A) ~
-\ I
A' -[2~ 1[,. 1-[2(-1)-2(1)[ ~ IJ A!
,.~4cO
F'rolll the quadratic equation:
I I
A
- 4( 2
4)
I 2
+
VI7 2
so,
and A,
~
2.5616
The first eigenvector is found from (M2.14):
(M2.1X) where we have lIsed the notation:
446
Review of Matrix Algebra
Module 2
Applying equation (M2.1S),
II
I~ So the following
fWD
1.5616 v 11 'V 21
equations must be satisfied:
2V 11 +V 21 2V 11 -- 021
1.56J6v 11
= - 1.5616 'On
which yield: 3.561 5 ~/)]l
1J 21
If we wish, we
call
arbitrarily select ~,
VII
cc
1.0 and II" I [ 1} } n - - -
'/)21
= 3.56J5
1.0 c I 3..l 6 !5
1M2. J9)
Most complltcl:pal:kag~:"\~il] report cigyllycctors that have a unit norm (length:o:: J). If we divide 1)" (1',,)2 12.6841 3.6992, we rind that:
1M2. J 9) by
V(
V+
VI-:,
o.W)}) 0.9628 An
(~qllivalent solution
is:
- O.2703j. 1 0.9628 The reader should show that: leads to: ~2 -
I. '~I !] tv
0.8719] 1 0.4896
Eigenvector problems arc easily solved used the eig [unction in MA'T'LAB (sec ModuleJ), as shO\vn below. a
v
=
2
1
2
-1
=
0.8719 0.4896 d
2.:')616
o
-0.2703 0.9628
o -1. 5616
Sec. M2.3
447
Square Matrices
Column i in the v matrix is the eigenvector associated with the ilh eigenvalue (which is the ith diagonal clement ill the d matrix). The trace and determinant can also be f(lUnd using MATLAB commands: » an~;
trace(a) 1
»
det (a)
ans ··4
M2.3.7
The Similarity Transform
or
The similarity tran.~I()nn is uscful for understanding the dynamic behavior linear sys[ellls (sec Chapter 5). Recall the eigenvector/eigenvalue problem for a 2 x 2 matrix:
(M2.20) (1112.21 ) Notice that equations (M2.20) and (M2.2l) can be written in the following form:
AV
VA
(1112.22)
where V is the matrix of eigenvectors: (M2.23)
where:
1/)".1 1)21.
~2
and
=
1/)"1 v)~
A is the matrix of eigenvalues: ;\
°1
A} .
(1112.24)
Multiplying (M2.22) on the right side by VI we find
A' V A V
I
(111225)
r~quation (M2.25) will he useful when developing analytical solutions for sets of Iincar differential equations, and can be used for any n x II (square) matrix.
448
M2.4
Review of Matrix Algebra
Module 2
OTHER MATRIX OPERATIONS M2.4.1
Vector Norms
It is quite natural to think of vectors as having a "length" property. This property is called a vector norm. The most common norl11 is the Euclidean norm. In two-dimensional space, the Euclidean norm is:
I xII = V~;
+~l
Sometimes the Euclidian norm is called the 2-nonn, In n-dimensional space,
JJ
x
lb.
I X II, = Vx; + xi + ....•~.\;' = ~± xJ ,coj
notice that the sum of the squares can be written as xTx , lhalis:
and the 2-norm is then:
IlxI12=~ M2.4.2
Orthogonality
A column vector w is orthogonal to a column vector z if wTz = 0.0. For example, consider the 2-D vector
w=
1
l -O.S
"I
and the 2-D vector
Then
xTz = [l
~O.Sl [O;S]
= I(O.S)
+
(-O.S)1
.~
OJ)
Notice that orthogonal vectors arc "perpendicular" to each other, as depicted below.
z (0.5,1)
w (1,-0.5)
M2.4.3
Normalized or Sl'aled Vectors
Many times, we will scale vectors, so that they are !1onna/ized. A vector is normalized if it has a norm (length) of 1.0. Normalization is pcrformed by multiplying each c1ement_ or thc vector by I/vector norm. For a givcn vector, x, the scaled vector, U is u:::: X l/~
EXAIVIPLE M2.3
Vector Normalization
Consider the vector x T
[l
=
Then,
21: [I
XIX
2JI~I~1+4c5
And,
The normalized vector is F
A vector norm
1. 2.
3.
011
RII is a function
0
x:;t:
4
II (XX I I X YiI
I1II : RII 4
I xI > 0 I III xii I xI + I y I (X
.,;
j
I~I ~ ~ [£;]
Common Vector Norms " I. I X Iii ~ 2: I Xi I
R satisfying:
Where
CI..
is a scalar
'rriallglc inequality
l-norm (sum of absolute values)
j······l
2. 3. 4.
I X II, ~ (~ IXi If'
2-norm, Euclidean Norm
I x I ,. (~ Ix, I") 1/,. I x"" ~ max I x, I
p-nonn oo-norm
Notke that the p-norm includes all other vector norms
EXAMPLE M2.4
Comparison of Norms
31 I x Ii, III I 121 + 131 ~ 6 2 !I X II, ~ (iJI' I 12/' + Im '/ ~ 3742 Ilxt ~ max 1111.121, 1311 ~ 3 Ilxll,,, ~ (IJI'" + 1_21 '0 + 131 10)1/10 3.005 Xl
Notice that. as p ..,
~
[I -2
=.11 X II" --> I x IL.
Review of Matrix Algebra
450
Module 2
SUMMARY In this module we have provided a concise review of matrix operations. This material should be sufficient for most of the matrix operations used in the textbook. It is partiCLIlarly important that the reader understand: Matrix notation
How to multiply matrices That an n X 11 matrix has n eigenvalues, which arc the equation: det(~[ - A) ~ 0.
11
roots to the characteristic
That a matrix A is singular if the dct(A):::: O. A singular matrix docs not have an inverse. A singular matrix is not full mnk. That the rank of a square matrix is equal to the number of nonzero eigenvalues
STUDENT EXERCISES
For problems 1-3, usc x"
[~]
I. Find x'. 2. l
3. Find I x Ib 4. Perform the indicated matrix operations for the following matrices:
A ~ a. FindC
b. Find [) c. Find E
~
~
All
B~ [79
3
f1
6 5
IliA A 'll
5. Solve for x as a fUllction of.y where:
v ~ [I2
.
1]
-1
x
6. Find the determinant of:
7. Derive the characteristic polynomial (del (1\.1- A)) for a general 3 x 3 matrix: 1112
A
...
11
{/21
an
11
([-\I
{lJ2
{lJl
[""
((21
]
Student Exercises
451
8. Find the determinant, rank, and eigenvalucs of the following matriccs:
a.A=[~J h. B
= [-11
:31 :21
LINEAR REGRESSION
MODULE
3
The objective of this module is to review linear regression analysis, with a focus on a ma~ tfix algebra formulation and MATLAB routines for linear regression, After reviewing this module, the reader should be able to: Formulate a matrix algebra solution for a least squares problem
• Usc the MATLAB routine polyf i t to find a polynomial to fit data • Usc the MATLAB routine polyval to evaluate a polynomial The major sections of this module are: M3.1 M3.2 M3.3 M3.4 M3.5
M3.1
Motivation Least Squares Solution for a Line Solution for the Equation of a Line Using Matrix~ Vector Notation Generalization of the Linear Regression Technique MATLAB Routines polyf i t and polyval
MOTIVATION The models developed in this text require the values of many parameters such as reaction rate coefficients, etc. It is common to use linear regression (also known as I1near least squares analysis) to estimate thc values of these parameters. To illustrate the basic principles of linear regression analysis, we first consider the equation of a linc. Consider the
452
Sec. M3.1
Motivation
-
0
0 0
y
453 A
Y
a x
+ b
0
0
x FIGlJREM3.1
L~xpcrimcntal data
and linear model prediction.
data shown as open circles in Figure M3.1. We would like to find a line that provides a best fit to the data. Let xi and Yi represent the independellt variable and measured dependent variable for the itil experimental data point. Also, let "Vi represent a lnodel prediction or the dependent variable, given the experimental value of the independent variahle, xi' 'rile sets of dependent and independent variables can be represented as vectors: Xl
Y2
X2
X
=
Y XN
which can be wriuen in the following fashion, to minimize space:
x' y'!
[x, x2 [Y, y,
xNI YN!
where
f represents the transpose operation. Given experimental data, x and y, we wish to find paramelers that allow a model to best fit the data. For a good fit of the data, we would like to minimize the dilTcrence between the data points Hnd the model prediction = ax + b). Our first instinct might be to find the model parameters, ({ and b, so that the sum of the absolute values of the errors is minimized. Herc we have defined the error as the difference between the experimental value and the model prediction. That is, our objective is to find the values of a and b slich that equation (M3.1) is minimized:
(y
(M3.1) The major disadvantage to this approach is that there is not a simple, closed-fonn, analytical solution to this problem. An alternative approach is to minimize the SUlll the squares of the errors. That is, find a and b such that (M3.2) is JniniruizecL
qr
454
Linear Regression
Module 3 (M.'2)
l<'or N data points, we can represent (M3,2) using the more compact notation in (M3.3): N
2.: (y, -
yJ'
(MDl
i= J
or
Besides the analytical solution thal will follow, another advantage to the Slllll squares formulation is that large errors arc penalized 1110re heavily than small errors.
M3.2
the
LEAST SQUARES SOLUTION FOR A LINE Using optimization notation, we refer to (M3.3) as the objective .Iltnc'/ioll. We desire to find the values of a and b (known as decision variahles) that minimize the ohjective fUllction. Let/(a,b) represent the objective function: N
f(a,b) ~
2.: (y, - y,)'
(M.'AI
J
N
f(a,b)
2.: (y, -
ax, - b)2
(M.'.5)
i····· 1
We know from calculus tbe necessary condition for a mininmm of a function I.\lith respect to a variable. The minimum ofj(a,b) is satisfied by (M3,6) and (M3.7),
af(ll,b)
da a/(a,h) ah
o
(M.'6)
o
(M.'.7)
From (M.'.6) N
a/(a,h) ria
22.: (y, - ax, - h) (x,)
0
(MU)
i'cl
and from (M.'.7):
dEa,.')) iJb
N
-22.: (y, - ax, - b) ~. ()
(M.'9)
I
Rcmoving the conslanl value -2, wc find that (M3.8) and (M3.9) call he written as (M.'.I 0) and (M3.11): N
2: (yeti -
ax/ - bxi ) = 0
(M.'.IOl
- 0
(M.'.ll)
i"l
N
2.: (y, - ax, - b) i"l
Sec. M3.3
Solution for the Equation of a Line Using Matrix Vector Notation
455
Rcrlloving the parameters a :md b frorn the summations: :V
N
'\.'
b L.J
Xi
I
i
N
",' +a" L..J xt i=l
L..J
I
+ bN
(M3.12)
1
N II "'" X
'\.~
= L.J Yr\~i
N '\.~ y LJ./
c .•
i.l
(M3.13)
i,1
Notice that we have two equations «M3.12) and (M3.13)) and two unknowns (a and b).
We can easily solve for b in terms of {[ from (MJ.13) to obtain:
1 INN N.2: Yi - a
2:
./-1
Xi
j
(M3.14)
1""1-
Of course, we sec that the terms in (M3.14) arc merely the mean values of y and x.
I
x··
N
2: Xi
N;"J
so that:
b co:c).i -
ax
(M3.1S)
Substituting (M3.1 5) into (M3.12) we find: N
(Y-ll~¥)Nx
+ a LX;
N
=
i=1
2:Y,t:;
(M3.16)
i'l
or, (M3.17)
which leads to:
a=
(i~ y,xi) -Ny i (~X,')-N.t2
(M3.18)
and b can be determined from (M3.15).
M3.3
SOLUTION FOR THE EQUATION OF A LINE USING MATRIX-VECTOR NOTATION The model prediction of each dependent variable can be written as:
Yi =
aX i
+b
(M3.19)
Linear Regression
456
Module 3
\-vhieh can be written in rnatrix-vcctor form, Cor N data poinls, as: t
xj
y}
.x::!
~I'
I;; I Yv_
\'
(Nx 2; (2 x 1)
(Nx I)
and
V·iC
compact
(M320)
can sec that the dimensioning of the rnalrices and vectors is consistent. lIsing t1latrix~vector notation, \ve write (M3.20) as:
Yoe'I_O
(M321)
The objective function is: v
/(0) =
L i
(Y,
-Y,)'
(M322)
t
which can he \vrittcn as:
](0) = x I)
(I
IY - VI'IY - VI (I
x IV)
(N
x
(M3B)
I)
The general optimization statcrnctlt is then: Minimi/.c
((O)
IY YI'IY-Yl
(M3Ll)
Y= (I) 0
subject to
(M321)
whcrcj(O) is the objective function, 0 arc the decision variables. and (1\13.21) is the
cqllal~
ity constraint equation. Since the constraint equations arc linear and arc equality constraints, and the objective fUllction is quadratic, there IS an analytical solution to this problem. 'rhe solution is (l::dgar and Ifilllmciblau, 19RR):
(M324)
M3.4
GENERALIZATION OF THE LINEAR REGRESSION TECHNIQUE It should be noted that the only requirement for parameter estimation using the least squares solution (M3.24) is that the model must be linear with respect to the parameters. "rhere is no limitation to the functionalities with respect to the independent variables. As an example, consider;
Y where the
.y' s
{/ .\1 + b .I:"~:+ c In x,
+ d eX'
arc the independent variables. Represent the kth data point as:
.Y(I<) = IIX,(I<) + by,(k)' I clllxJk) + dc','k! which can be written as:
. _ _ _ _ _ _ _ _ _ _ _ _ _ _
~.
••••
?
•
~
••
b.;.
......
Sec. M3.4
Generalization of the Linear Regression Technique
457
y(k) = 'I'(k)le where:
x 2 (k)'
'I'(k)' = [xl(k) OT
= [a
b
d]
c
Now; for the system of N data points we caIl write: where:
The solution to this problem is (M3.24) and the generalization to any system that is linear in the parameters is clear. A common formulation is the least squares fit of a polynomial. Consider a general nth-order polynomial equation, where Pi is a parameter (coefficient) to be found as a best fit to data:
The parameter vector is:
0=
Pn PIl+l
The matrix of the independent variable functions is:
x(1)" x(2)"
x(l ),,-1 X(2)'"-1
x(l ) x(2)
x(N)"
x(N)"'
x(N)
<1>= 1
The measured variable vector is:
y(l) y(2)
y= yeN)
I
458
Linear Regression
Module 3
where x(i) and ~v(i) represent the independent and dependent variables at the ith data point. The solution to this problem is equation (M3,24).
M3.5
MATLAB ROUTINES polyfit AND polyval 'The MA'rLAB routine polyfit is lIsed to fit data to an 11th order polynomial, and the routine polyval is Llsed to evaluate an nth order polynomial. I,et x ::: independent variable vector, y ::: dependent variable vector, and n := order of polynomial. The best-fit polynomial cocflkientB arc found from:
p = polyfit(x,Y,n) where the clements of the p vector are ordered from the highest power 011 down. Given a polynomial p and an independent vector xl, the resulting dependent vector yl caB be found from:
yl
~
polyval (p. xl)
We show the usc of poly fit and polyval by way of Examplc M3.1.
EXAlVIPLE M3.1
Batch Reactor [I:xample
Consider a batch reactor with a single first-order reaction, A -) H. The model is:
dCA til where CA = concentration of A, k:;:; rate constant, and
l:;:;
time. Separating variables and integrat-
1l1g:
dC;1 C /l In
(.'11 =
k dt In CAO
-
k t
where C;\o is the initial concentration of A. Notice that this is the equation for a line. If we let
y
In C A
1'(1)
~
-k
p(2)
=
In CAn
and we have the form:
y C~ 1'(1)1
+ 1'(2)
Now we lise polyf:i t [0 find the best linear fit of the data. The hatch reactor data are shown in Table M3.1 and Figure M3.2.
Sec. M3.5
MATLAB Routines polyt i
TARLE M3.J
t
459
and polyval
Concentration as a Function of Time
o
time, min (~, kgn101/m J
I 5.00
8.47
2 2.95
3
4
5
I.X2
LOS
11.7 1
10
8
'c" u
6 I
0
u
o 4
o 2
o o
0
---~
2
0
3
4
5
time, min
FIGURE M3.2
Batch reactor data. Concentration of A as a function of time.
»L::: 0:1:5; »
t
c-:;
t';
" CA c [8.47;5.00;2.95;1.82;1.05;0.711; » y ::: log (CA) ; » p ::: polyfit{L,y,l) p
--0.5017
2.1098
'The same values for the parameters call be obtained from equation (M3.24), by performing the following: » phi [t ones ( 6 , 1) ] ; » p ::: inv (phi' *phi) *phi' *'y
P " ~0.5017
2.1098
The parameters from the linear regression arc converted back to the physical parameters:
k CAll
~ 0.502 min I cxp(p(2)) ~ 8.24 kgnlOl/mJ
•• -pCI)
Now, we wish to compare the experimental data with the best fit line (model). The line is generated using the polyval function. The model and expcrirnent are compared in Figure M3.3.
» >.>
ymod polyval(p,t); ploL(L,ymod, t,y, ' 0 ' ) :=0:
2.5 2
..
1.5
0
E
0.5
o
0 ___ L_
-0.5 0
2
3
4
5
time, min
FIGURE M3.3
Semilog plot of concentration data and best fit line.
We also wish to compare the experimental data with the model on a time-concentration plot. » tl ::0: 0:0.25:5; % notice that more point::, are used for »CA_illOd 8.24*exp(~O.502*tl); » plot (tI, CA__ffiOd, t, CA, '0' )
mnoot~hness
The data and model are compared in Figure M3.4. The rate constant has been evaluated at a single temperature. It can be evaluated at a number of temperature and linear regression can he lIsed to evaluate the Arrhenius cons/ants (frequency factor and activation energy) as shown in student exercises 3 and 4.
o c
8
time, min
F[(;{JRE M3.4
Concentration data and model as a function of lime.
Student Exercises
461
REFERENCES AND FURTHER READING Good coverage of least squares analysis, and optimization in general, is provided by the following text: Edgar, T.E, & D.M. Himmelblau. (1988). Optimization New York: McGraw-HilL
(~f
Chemical Processes.
STUDENT EXERCISES 1. Note that a different solution for the best fit of a line is obtained if it is assumed that h is known; this becomes a single parameter estimation problem. Derive the following result for the estimate of a if it is assumed that b is known.
a=
2. Use the matrix algebra approach to solve for the slope of a line, if the intercept b is known. You should obtain the same result as problem ( above. 3. The
Arrheniu~
rate expression is used to find reaction rate constants as a function of
temperature:
k
~
A exp( -EIR1)
Taking the natural log (In) of each side of the Arrhenius rate expression, we find:
In k
~
In A - (EIR) (liT)
where Ris the ideal gas constant (1.987 cal/gmol K). }jnear regression analysis can be used to find A and E. Rate constants as a function of temperature for a first-order decomposition of benzene diazonium chloride are shown below: I
k, min
T,K
I
0.026 313
0.062 319
00108 323
0.213 328
0.43 333
Find A and E using least squares analysis (show units). Show that polyfit and the matrix algebra approach (M3.24) yield the same results. Compare model and experiment by (i) plotting In k versus lIT and (ii) k versus T
IThis data is from Example 3.1 in Fogler, I-LS. (1992). Elements (?l ChemicaL Reaction Engineering, 2nd ed. Englewood Cliffs, NJ: Prentice Hall.
ESCOLA DC: ENGEN HARIA BIBLIOTECA
Linear Regression
462
Module 3
4. 'rhe Arrhenius rate expression is used to find'rale constants as a function of temperature:
Ie
~
A exp( ... E/ R 7)
Taking the natural log (In) of each side, we find: In Ie
In A- (E/R) (1/7)
I.incar regression analysis call be used to find A and E. Rate constants as a function of temperature for a first-order reaction arc shown below.
k, min T,K
1
0.0014 300
0.0026 310
0.0083 330
0.0047 320
0.014 340
0.023 350
0JJ38 360
0.059 370
0.090 380
Find A and E' lIsing least squares analysis (show units). Show that polyfit and the matrix algebra approach (M3.24) yield the same results. Compare model and cxpcriIncllt by (i) plotting In k versus lIT and (if) k versus 1'. 5. The growth rate expression for a biochemical reaction, lIsing a Monod model, is: /--Lm
where f.L is the specific growth rate, /Llllax and kill arc parameters, and x is the substrate concentration. The growth rate relationship can be rearranged to:
I)ata for a particular reactor are shown below. Use linear regression to solve for the pararneters (/-Llllax and kllJ IJ., hr···· 1
x, g/Iiter
0.25 0.1
0.31 0.15
0.36 0.25
0.43 0.50
0.45 0.75
0.47 1.00
0.50 1.50
0.52 3.00
Show that polyfit and the matrix algebra approach (M3.24) yield the saIne re~ suits. 6. The growth rate expression for a biochemical reaction, using a substrate inhibition model, is fL
~
--
km
_ P'!lJ'IX X
+ x + k1
where /-L is the specific growth rate, I-1 milx and kl/l are parameters, and x· is the substrate concentration. The growth rate relationship can he rearranged to
Data for a particular reactor arc shown below. Use linear regression (M3.24) to solve for the parameters (/-LnHlX' kin' and k,). fL, hI'
I
x, g/litcr
0.24 0.1
0.27 0.15
0.34 0.25
0.35 0.50
0.35 0.75
0.34 1.00
0.33 1.50
0.22 3.00
463
Student Exercises
Note that:
x(l)
l/x(l)
x(2)
l/x(2)
x(8)
l/x(8)
y
and the solution is H «1>'1'(1») I(J)Ty Compare the model and experimellt by plotting IL
verSUS.L
INTRODUCTION TO SIMUUNK
MODULE
4
'rhe purpose of this module is to introduce SIMULINK, a block diagranl environment for simulation. After studying this module, the reader should be able to; Construct a hlock diagram for simulation
Set proper simulation parameters
• Print SIMULINK windows The major sections of this module arc:
M4.1
M4.!
Introduction
M4.2
'fransfcr
M43
Printing SIMlJLINK Windows
Function~Bascd SilTiulation
INTRODUCTION '[here arc limitations to using the standard MATLAB environment 1'01' simulation. Engineers (and particularly control engineers) tend to think in terms of block diagrams. Although MA'fLAB functions slIch as parallel, s(.:;ries, and feedback allow the simulation of block diagrams, these functions arc not visually pleasing. SIMULINK proviclcs a much more natural environment (a graphical user interface, (lUI) for simulating systems that arc described by block diagrams. SIMULINK also provides integration
464
Sec. M4.1
Introduction
465
~~~~~~~
Sources
Sinks
Discrete
Linear
Nonlinear
Connections
Extras
SIMUUNK Block library (Version 1.3c)
FIGURE M4.1
SIMULINK block library.
methods that call handle systems with timc~dclays (rather than making a Pad6 approximation for dcadtimc). We will illustrate the use of SIMlJLlNK by way of example. [n the MATLAB command window, enter "simulink" (small letters): »
simulink
The SIMlJLINK block library window appears, as shown in Figure M4.1. Th~re arc many possible SIMULINK functions (icons) available. You will find it very llseful to use the most commonly used blocks, which can be found by doublcclicking on the Frtras icon. 'fhc resulting window is shown in Figure M4.2.
B
Block Library
SIMULINK Demos
Most commonly used blocks
Additional blocks using the mask utility (documented in release notes):
~ ~ ~ ~
Conversion
Flip-Flops
PIO Controllers
Analyzers
Blocks and demos that require toolboxes:
~~~~~~
Controllers
Filters
System 10
FIGURE M4.2
Robust Control Demos
MuTools Demos
SIMULINK Extras block.
Neural Networks
466
Introduction to SIMULINK
Most commonly used blocks:
~
B E B
[i}
00
Signal generator
ill
>/2»
Inport
Gain
I
Transfer Fen
Demux
Scope
~
yout To Workspace
~ Sum
{2] Cutpar!
>~
Auto-Scale Graph
OJ ~
~Sin(U[IJ+ Fcn
Dis. Transfer Fen
Mux
Module 4
XY grapl
Other commonly used blocks:
~ Derivative
)
G
(s-1 ) s(s+ 1)
PID controller
Zero-pole
[]}
(9)
[j}
Step input
Clock
Sine wave
MATLAS Fen
W
0
system
S-fullction
Dis. Zero-Pole
1
~
1;22- 1 Filter
[i} Constant
Rate limiter
Switch
~
Saturation
m
fj ~
~
From Workspace
ffi
Look Up Table
Product
~ Transport Delay
@ Integrator
l/z
p
Unit Delay
Full Block Library
Slmullnk
FIGlJRE M4.3
SIMULlNK most commonly used blocks.
Double-clicking on the Block Lihmlyicon then reveals a window for the most commonly lIsed blocks, shown in l~'igurc M4.3. AltcrJl
M4.2
TRANSFER FUNCTION-BASED SIMULATION Here we provide an exalnple simulation of a first-order process. 'fhe first step is to drag icons frorn the most commonly lIsed block library, as shown in Figure M4.4. 'rhe clock icon is llsed to create a time vector. The To Workspace icons allow the vectors to be writ~ len to the MATLAB workspace for later manipulation or plotting. Notice that each of the variable names, as well as the icon names, can be changed by the lIser, as shown in riigurc M4.5. The variahle namcs are changed by dOllble~clicking on the icon and changing the name in the variable space.
Sec. M4.2
I
C91----~~1 Clock
467
Transfer Function-Based Simulation
yout To Workspace
rnf---..~ITlI------~1 ~
yout To Workspace1
Step input
Transfer Fen
FIGlJH.E lVJ4.4
Blocks for transfer
function simulation.
C91----l~~1 ~t~
Clock
Time
rnf---.~ITlf---~~1 ::::::!Y=:= ~
Step input
Output
Transfer Function
FIGlJRE !\i14.5
Blocks for transfer
function simulation- -variables and icons renamed.
Notice that the default transfer function has a pole at zero, that is, a simple integrator. This is changed by double~clicking on the icon and entering numerator and denominator coefficients. An example or a SystClll with a gain of 2 and a lillie constant of 5 is shown in Figure M4.6. The default step input is to step the input: from 0 to 1 at I :;:;: L This can easily be changed by double-clicking on the 5'tep input icon and changing the specifications. The simulation parameters (start time, stop timc, integration method, etc.) arc changed by using the simulation pulldown menu and selecting Parameters. It is particularly important to change the stop tilllc l which is 999999 by default (this is because SIMlJLINK was initially developed for usc by electrical engineers who often usc a COIl~ tinuous oscilloscope function). Also, you will usually want to change the maximum integration step size; if the step size is too large, the simulation may still be accurate (particularly Lor linear systelns), but the resutting plots may be jagged. 'rhe default integration method is RK-45 (Runge-Kutta), which you may wish to change to Linsim 10 irnprovc the computational speed. You may wish to try severalclifferent integration methods, depending 011 the type or system you arc attempting to simulate. 'fo begin the simulation, simply select Start from the simulation pulldown menu. The t and }' variables are stored in the MATLAB workspace. 'fhe resulting responsc is
C91----------~1 Clock
t
Time
rn---.~[ifuJf----.~1
Step input
5s+1
Transfer Function
y Output
FIGURE IVI4.() Transfer function sinwlation--first-order process.
468
Introduction to SIMULINK -- -r- - -- ---- -
Module 4
-~._----.-----------~-------r
-~---
-
1.5
'5 Q. '5 o
0.5
___-------.l.--__ _
0-
o
5
10
15
----l
20
time
FIGlJRE M4.7 t;e::. I.
SirHulation result for firsHmJer example with a step input at
shown in Figure M4.7. Notice that the input was stepped from 0 to 1 at t = I, that is, thedefault values were used. A major advantage of SIM1JLINK over the standard MATLAB integration routines is the ability to handle systems with liIlle delays. This is dono using the Transport Delay block shown in Figure M4.8. Here we use a timo-delay of 5, obtained by modifying the default value of 1. It should also be noled that the user must supply an initial input value for the transport delay block. In this case the initial value is 0, because we arc dealing with a system in deviation variable form (that is, all or the initial conditions arc 0). The resulting simulation is shown in Figure M4.9. Notlcc that no output change occurs before [ = o. This is duc to the step input changing at t::::: 1, combined with the 5 unit time delay (I + 5 = 6). Although the block diagram feature of SIMULINK is very handy, it would be timeconsuming (and tiring) if every time you wanted to change a set of parameters you had to double-click on the related icons. You can place variahle names, such as k, tau, theta, for a first-order + time delay system directly in the blocks, then simply change the values (of k,
(9----..1 Clock
t Time
rnr--~·f2]--'·1~ I-~·I:::ry~ ~
Step input
FIGLJRE M4.8 cess.
Transfer Function
Transport Delay
Output
Transfer function simulation----fjrst-order + time-delay pro-
Sec. M4.3
469
Printing SIMULINK Windows
2
1.5 '3 Q. '3 0
0.5
a a
10
5
FIGURE iV14.9 with delay of 5.
time
15
25
20
Simulation result. Step input at 1 ~ I for first-order system
tau, theta) in the MATLAB cOllunand window before pcrforrning a new simulation. This block diagram is shmvil in 17igurc M4. J O.
M4.3
PRINTING SIMULINK WINDOWS When using a PC or Mac it is casy to print a \vindow directly to the default printer simply by selecting print from thcJile pulldown mellu. If you arc lIsing a Unix-based version of
MAll"AB/SIMULJNK, or you wish to create a postscript file for a window, you need to enter a command directly in the MATLAB connnand window. Let's say that you titled the example simulation window "tLcxamp" and you want to print the window from a Unix-based workstation. The command that you \vmlld use is: print:
stf_ examp
which will send the figure to the dcfaull printer.
C9----...~1 Clock
rnr--~~I ~
Step input
t
Time
I
' Transfer Function
~I ~ I~~I Transport Delay
y Output
FIGURE [\'14.10 Transfer function simulation. "fhe values of k, fan, and theta can be entered in the MATLAB command window before beginning the SilllU~ Jation.
470
Introduction to SIMULINK
Module 4
SUMMARY
or
The purpose of this module was to provide a brief introduction to the usc SIMULINK for block diagram programming. Other inputs (sources) such as sine waves or ramps can be lIsed. Also, other "sinks" such as an "XY graph" can be Llsed; we generally recommend that variables he written to the MATLAB workspace, allowing the user to usc standard MATLAB plotting functions. There are many options for "systems" to simulate, including ;'statc space" models and s-functions which describe nonlinear systems. The lIscr is encouraged to experiment with state space models, and to read the SlMULINK User's Guide for more advanced usage.
EXERCISES Usc SIMULINK to perform the following. Show the SIMULINK diagram used for your 5i Illul ations.
1. Plot the step response of the following undcrdamped systcm
3
I:(s )ee
+ 5s +
when a step input of magnitude 2 is made at t;::; ), 2. Plot the step response of the following two transfer functions by performing a single simulat.ion
3(--3.1 I- I) g,(s)=·, .. 25s· + 5s -1- 1 I: (s) = 2
3(~3s I-
I
25.\' -t 5s I-
Usc a step input change of 2 at I::::: I. 3. Compare the following firsHH"der + timc-dclay system 3e----~s
I:,(s)
co
lOs
I-
I
with it's first-orderPaciC approximation
3( -- 22.5s II:,(s) = (2.5s I- I)(lOs for a step input change of I at f ;::: O.
I- I)
STIRRED TANK HEATERS
MODULE
5
After .':"tudying this module, the reader should be able to: Develop the Ilonlinear dynamic model of a perfectly mixed stirred lank heater Find the state-space and transfer function form of a linearized stirred lank heater Compare the dynamic responses of the nonlinear and linear models Usc MATLAB for linear and nonlinear simulations
Analyze phase-plane behavior crhc major seclions in this module are:
M5.1 M5.2 M5.3 M5.4 M5.5 M5.6 M5.7
M5.1
Introduction [)eye-loping the Dynamic Model
Steady-State Conditions State-Space Model
Laplace-Domain J\1.odcl Step Responses: Linear versus Nonlinear ivlodels Unforced Systcnl Responses: Perturbations in Initi:::ll Conditions
INTRODUCTION Mixing vessels arc used in Illany chemical processes. Often these mixing vessels are healed, either by a coil or a jacket surrounding the vessel. For example, a mixing vessel rnay serve as a chemical reactor, where two or more components arc reacted to produce 471
472
Stirred Tank Heaters
Module 5
Tank F;
~
nlet Ti ~
-
V
T
•
Tank Jacket Fj
Jae ket inlet
~'i
Tji
Ta nk outlet
Jacket
F T
outlet Tj
FH;(JU.E MS.l
Jacketed stirred tank healer.
Olle or more products. Often this reaction lllust occur at a certain temperature to achieve a desired yield. The temperature in the vessel is maintained by varying the fJowralc of a fluid through the jacket or coil. Consider a stirred lank heater as shown in Figure MS.l, where the tank inlet slream is received from another process unit. The objcCLive is 10 raise the temperature of the inlet stream to a desired value. A heat transfer fluid is circulated through a jacket to heat the fluid in the tank. In some processes steam is used as the heat transfer fluid and most of the energy transported is due to the phase change of steam to water. In other processes a heat transfer fluid is used where there is no phase change. In this module we assume thaI no change of phase occurs in either the tank fluid or the jacket fluid.
M5.2
DEVElOPING THE DYNAMIC MODEL In this section we write the dynamic modeling equations to find the tank and jacket tcm~ pcraturcs. 'fhis methodology was developed in ~~hapter 2. We make the following assumptions: The volullle and liquids are constant with constant density and heat capacity. Perfect mixing is assumed in both the tank and jacket. The tank inlet flowrate, jacket flowrate, tank inlet temperature. and jacket inlet temperature lllay change (these arc the inputs). The rate of heHt transfer from the jacket to the tank is governed hy the equation Q:::::; UAClj - 1), where U is the overall heat transfer coefficient and A is the area for heat transfer. The notation that we use is presented in Table MS.l.
Sec. M5.2
473
Developing the Dynamic Model
TABLE MS. 1
Notation
Variables
Subscripts
;\
area for heat transfer heat capacity (cncrgy/mass"'tcmp) yolumetric flowrate (volume/time) density (mass/volullle) temperature time rate of heat tnlllsfcr (energy/lime) heat transfer coefficient (cncrgyltimc"· arca$tcmp) volume
p
T
Q II V
inlet jacket jacket inlet reference stale steady-slale
j ji ref s
M5.2.1
Material Balance Around Tank
The first step is to
~ritc
a material balance around the tank, assuming constant density: dVp
.
dt
I
.
, = Fp - [op
Also, assuming constant volume (dVidt ;;::: 0), we find: F= F,
M5.2.2
Energy Balance Around Tank
The next step is to write an energy balance around the tank. accmllulation = in by now
~
out by /low + in by heat transfer + work done on system
c[. dn; = /.'P·'.,·-'I) '7'[' F '7.,,+ dt I
~
f)1
+ W"/'
The next step is to neglect the kinetic and potential energy and write the total work donc on the systcm as a combination of the shaft work and the energy added to the system to get the fluid into the tank and the energy that the system performs on the surroundings to force the fluid out. This allows liS to write the energy balance as (sec Chapter 2 for the details):
Stirred Tank Heaters
474
Module 5
dU = Fp (U + Pi) - Fp (U +'P) + Q + ~\' i
df
Pi
P
and since H:::: U + [lV, we can rewrite the energy balance as:
dff !It
!lpV,
,
.._, =h,[J-h,J!+Q+W, dt I
Note that dpV/dt:::: V dp/dt + P dVldt, and the volume is constant. Also, the mean pressure change can be neglected since the density is constant:
dJI I dt = Fp' ff-Ff>J!
+
f)
+
W
~~.I"
Neglecting the work done by the mixing impeller, and assuming single phase and a stant heat capacity, we find:
!IT V pc" !It =
Fpc"Cli -
COI1-
1) + Q
or
!IT til
P =V (1;- 1) +
Q VPC p
(MS. f)
We must also perform a material and energy balance around the jacket and use the connecting relationship for heat transfer between the jacket and the tank
M5.2.3
Material Balance Around the Jacket
The material balance around the jacket is (assuming constant density):
assuming constant volume (dV/dt ::::: 0), we find:
M5,2.4
Energy Balance Around the Jacket
Next, we write an energy balance around the jacket. Making the same assumptions as around the tank:
dTi = Fj (1" _ T) +
cit
VJ)I
)
Q V;PjcpJ
We also have the relationship for heat transfer from the jacket to the tank:
(MS.2)
Sec. M5.4
State-Space Model
475
Q
~
UA Uj- T)
(M5})
Substituting (MS.3) into (MS.I) and (M5.2) yields the two modeling equations for this system:
dT dt
~
F UA (T- T) (T - T) + ' . V Vpc p
(M5.4)
t
,17; _
(M5.5)
dt
M5.3
STEADY-STATE CONDITIONS Before linearizing the nonlinear model to find the state-space form, we must find the state variable values at steady-slale. The steady-state is obtained by solving the dynamic equations -for dxldl ;::: n. The stcady~stalc values of the system variables and some parameters for this process arc given below.
M5.3.1
Parameters and Steady-State Values It'
F,
~
1~s
= 50"1"
7~'is
~
1
111m
p Cp
T
20t)" I"
I
,
TJS
Btu
= 61.3 "I" I'l'
p/~pj
125"1"
V
150T
V
,
~
61.3
Btll T ftl
10ft3
I It'
Notice that the values or UA and Fis have not been specified. These values can be obtained by solving the two dynamic equations «M5.4) and (MS.S» at steady-state. From dTldl:= 0 at steady-state, solve (M5.4) to ol.1tain UA:= 183.9 I3tutF min I
M5.4
STATE-SPACE MODEL I-Icre we linearize the nonlinear modeling equations to find the state-space form:
xooAx+Hu y ~ ex where the state, input, and output vectors arc in deviation from:
476
Stirred Tank Heaters
XC-"
I II."~- I1'..
= state variables
1
J
Module 5
j.\
y,-II-I'I
output variables
=
. Ij-IjI
Recall that our two dynamic functional equations arc;
dl
(MS.4)
dt dl;
dt 'l,(T:Ij,F;,F:I;:Ii,,,)
FJ (T _ T) _ VA (TI) J .
Vj
V ' jf~'Cpj
Jil
The clements of the stale-space A matrix arc found by: A _,
ar ax
::.,-;.",1
II
J
=
a(1 - I,)
af;
=~f2
A 21
A
aX j
.= ill; = 2 . 2" oX
2
=
al;
F,
VA
aT
V
Vpc p
at;
VA
a(1; - '0,) ali ab
acr - T,)
at;
(-)( '1'j -
aT
~0.4
= 0.3
vpc p
VA
= 3,0
V;P/;pi
?Js V;
'r') 'js
lJA
-4,5
~'f~Cpj
The elements of the II matrix arc:
Nil =
8 12
-
Nn
..
lJ 14
ai,
ill,
JU I
iVi - 0,)
aj; = aFJ
0
ai, a(F - F,)
at;
I 1,1'
aF
ill; alI, af,
il/;
F
acr; - 1)
JI~
V
(Ju 4
acri, - 7jiJ
alji
al; dU 2
1,1"
ilf,
=0 =
v
-
0
'f~\'
--7,5 --
OJ
=0
(MS.5)
Sec. M5.5
Laplace Domain Model
477
ill, ill';
lJcilh~ "
' eli, 'j.,
Ii", "
ill; il",
ill;
ill;
il(F - FJ
ilF
'('/'i - '/'.) U H
ilt, ill;
ilUi, - Ij,J
'Ii.)
'[jis -
50
VI
-
0
~
0
0
~
0
ill;
FI.Y
J7ji
Vi
1.5
Since both states arc measured, the (' lnatrix is I. The numerical values for the matrices arc: A~
\ .... OA 3
03 -4..') 1 -7.5 0
0
B- 1 50
o
1.5
I
I(~ ~I
C
M5,5
0.1 0
LAPLACE DOMAIN MODEL Recall that a transfer function matrix relates the inputs and outputs:
y(s) •..• G(s) II(S) The input/output transfer function Inatrix is found from:
G(s)
C(sI-A)
'n
(M5.6)
Using the MATLA13 routine ss2 t f (sec appendix), we find:
15 G(s) ~
[
(50s + 20)
(7.5s- 33.75) (O.ls + OA5) - 22.5 0.3 s' + 4.9s + 0.9
OA5 (1.5s + 0.6)
I
The poles of the characteristic polynomial (8 2 + 4.9s + 0.9) arc -0.191 and -4.709. The transfer function relating output I (tank temperature) to input I (jacket f1owratc) is: ~
(,)
,II'
~
15 s2+4.9s+0.9
(MS.7)
Dividing (MS.7) hy 0.9, we find:
1.111
16.6667 + SA444s
+
(MS.X)
478
Stirred Tank Heaters
Module 5
(MS.S) can be factored into: ~
(\)
~
• II .
16.6667
+ 1)(5.23207.1' + 1)
(0.21236.1'
(M5.9)
which is the gain-time constant form. Any of these forms can be used, however \vc will generally LIse the gain-lime constant form. The transfer function relating output 2 (jacket temperature) to input I (jackel l1owralc) is:
50s +- 20
+ 4.9.1' +
(M5.10)
0.9
Similarly, lVe find (dividing (M5.10) hy 0.9):
1:" (.1') Also, (M5.] I) can be
l~lCtored
1:,,(.1')
M5.6
22.2 (2,.5.1' cO
+ I
(M5.11)
1.1 Ills' + 5.4444.1' " 1
into:
22.2 (2.5.1
+
1
(0.212361' 1- 1)(5.2307.1'
+
I)
(M5.12)
STEP RESPONSES: LINEAR VERSUS NONLINEAR MODELS In this section we will compare the step responses of the nonlinear model with those predicted by the linearized model. For cOlnparison purposes, we will plot the physical variables, r'or the linear model. \vhich is in deviation variable form, we convert to physicnl values hy realizing that:
x,(t) ~ T(I) - T,
X,(I)
c-
fj(t) -
Ii,
So, for the linear model:
'1'(1)
~
T,
+ .\,(1)
J;(t) ~
fj, + x,(t)
",(I)
F(t) - .FI" J
Also,
We will consider step changes of different magnitudes and differellt directions for this system.
M5.6.1
479
Step Responses: Linear Versus Nonlinear Models
Sec. M5.6
Small Increase in Jacket Temperature
Consider a step change of 0.1 (from the steady-stale value of 1.5 to 1.6 t'('I!min) in the jacket flowratc. 'fhe responses of the nonlinear and linear models arc shown in Figure M5.2. For hoth the tank temperature and the jacket temperature, the linear model closely approximates the nonlinear process.
M5.6.2
large Increase in Jacket Flowrate
Consider a step change of I J) (from 1.5 to 2.5 n3/min) in jacket f1owratc. '['he responses of the linear and nonlinear Illodels arc shown in Figure MS.3. The linear response is quali~ 127 linear LL
126.5
nonlinear
af D cL E
126
2
x
c ~ 125.5
5
10
15
20
25
30
time, min
a. Tank Temperature
152.5
linear
152
nonlinear
151.5
151 150.5 150
o
5
10
15
20
25
30
time, min
b. Jacket Temperature
FJ(;{nu: M5.2
Response to sinall incrca::;e in jacket [Iowratc (0.1 ftJimin).
480
Stirred Tank Heaters
Module 5
145 linear lL
140
g>
nonlinear
D
ci
E J!l
135
'"c
J'l
125
o
5
10
15
20
25
30
time, min
a. Tank Temperature
175 linear
170 165
nonlinear
160 155
150
o
5
10
15
20
25
30
time, min
b. Jacket Temperature
F1GlJRE MS.3
Reponsc to a large increase in jacket fJowratc (1.0 ftJ/min).
tativcly the same as the nonlinear response, hut the magnitude of change is different. 'fhe gain (change in olltpul/change in input) of the lincar model is greater than the gain of the nonlinear model.
M5.6.3
Large Decrease in Jacket Flowrate
Consider a step decrease of -- t J) (from 1.5 to 0.5 ft 3/min) in jacket flowrate. The responses of the Iincar and Ilonlincarmodcls arc shown in Figure M5.4. Notice that gain of the linear model is now less than the gain of the nonlinear modeL This is the opposite effect from what we observed for a large increase in the jacket f1owralc.
Sec. MS.6
Step Responses: Linear Versus Nonlinear Models
481
125 120 LL 0>
u
115
110
x c J'l
105
'"
j!)
linear
100
nonlinear
95 0
5
10
15
20
25
30
time, min
a. Tank Temperature 150
140 linear
130
120 nonlinear
110
o
5
10
15
20
25
30
time, min
b. Jacket Temperature
FIGlJRE M5.4
Response to a large decrease in jacket llowrate
(~1.0 f(3/ m in)
M5.6.4
Small Decrease in Jacket Flowrate
Consider a step change of ~O.l (from 1.5 to 1.4 ft3 /min) in jacket floWratc. The responses of the linear and nonlinear models arc shown in FigUfCM5.5. Since the the change in input was small, the linear model provides a good approximation to the nonlinear model.
M5.6.5
Remarks on Step Response Behavior
For the nonlinear model, the gain (change in oulputlchangc in input) varied as a function of both the input magnitude and direction. If small input changes were made, the gain did
482
Stirred Tank Heaters
Module 5
125 LL 0)
124.5
-15 0:
E
2 x
c
J'l
123.5
linear nonlinear
123
o
5
10
15
20
25
30
time, min
a. Tank Temperature
150 LL
'" '"
149.5
u
0:
149
E
2
"
x
148.5
0
."'-
148
linear nonlinear
147.5 0
5
10
15
20
25
30
time, min b. Jacket Temperature
FIGUn.E M5.5
Response to a small decrease in jacket flowralc (-0.1
ftJ/min).
not change much from the linear model case. For large input changes, the gain of the nonlinear system was less than the linear system for incrc;:lscs in jacket flowratc, but more than the linear system for decreases in jacket rlowralc. The response of the jackel temperature is faster than tbat of the tank temperature. This makes sense fro III a physical point of view, because the jacket volume is one-lenth of the heater vollllnc. Also, the jacket flowratc has a direct effect of jacket temperature and an indirect effcct on tank tcmperature. Noticc that the numerator timc constant partially "canccls" the slow denominator time constant for the transfer function relating jacket flowrate to jacket temperature.
Sec. M5.7
M5.7
Unforced System Responses: Pert Urbations in Initial Conditions
483
UNFORCED SYSTEM RESPONSES: PERTURBATIONS IN INITIAL CONDITIONS The eigenvalues of the A matrix in the state-space model provide information about stability and the relative speed of response. The eigenvectors provide information about the directional dependence of the speed of response, The MATLAB eig routine is used for eigenvalue/eigenvector calcuations, Rend I that a positive eigenvalue is unstable, while a negative eigenvalue is stable. A large magnitude eigenvalue is "faster" than a small magnitude eigenvalue. The eigenvalues for this system arc (sec appendix):
AI 1,2 ~
- 0.191 1 -4.7089
slow fast
The eigenvectors are:
-10.82071 0.5714 0695 c·' 0. 1 0.9976
12
I -
''',
I'·
slow fast
The second eigenvalue and eigenvector tell us that a perturbation in the initial condition of the second state variable will have a fast response. The first eigenvalue and cigcllvector tell us that a perturbation in thc direction of VI will have a slow response.
M5.7.1
Slow Response
Consider an initial perturbation in the slow direction. Let the perturbation be of Inagnitude 5:
.'. x () 0 = 5"'v)
-
~
4 035 y."1.0.82071 .~. 1 .1 0.5714
1
2.R57
The physical state variables (tank and jacket temperature) arc:
·1(0)·1 = Ilj(O)
1125j + l4.1 035 1 = 1129.10351 150 2.R57 152.857
'The responses for a perturhation in the slow direction arc shown in Figure M5.o.
M5.7.2
Fast Response
Consider an initial perturbation in the fast direction. Let the pertubation be:
x (0). =
.'. 5·"12
•.• 2
5·'"·I·-o.O()95j = 1.-0.34751 0.9976 4.9RR
484
Stirred Tank Heaters
Module 5
130 129
LL
en
'"
u 128
2
'"J'!c
127 126 125 0
5
10
15
20
25
30
25
30
time, min 3.
Tank Temperature
10
20
153 152.5
151 150.5 150
o
5
time, min b. Jacket Temperature
FIGURE M5.6
Initial perturbation in the slow direction.
The physical state variables (tank and jacket temperature) are:
T.(O)·j [.7;(0)
=
[1251 + 1.-0.3475] 150 4.988
=
[J24.6525.·] 154.988
The responses for a perturbation in the fast dlrection are shown in Figure MS.? (note time scale change from Figure M5.6). In Figures M5.6 and M5.? we have shown the effect of initial condition "difC(otir>ll This can be illustrated more completely by viewing the phase-plane behavior.
Sec. M5.7
Unforced System Responses: Pert Urbations in Initial Conditions
485
125
~ 124.9 w
D
ci E 124.8
2
~
:"'l
124.7
124.6 0
0.5
1.5
2
2.5
3
2.5
3
time, min a. Tank Temperature
155 LL
154
'"w
D
ci E
153
2 ill
152
'"
"
.<;!.
151 150 0
0.5
1.5
2
time, min
b. Jacket Temperature
FIGUUE wI5.7
M5.7.3
Initial perturbation in the fast direction.
Phase-Plane Behavior
The phase-plane behavior is shown in Figure M5.S, where the Tank Temperature is plotted on the x~axis and the Jacket Temperature is plotted on the .v-axis. Notice that initial conditions in the direction respond much more slowly than initial conditions in the
!i.1iii"ij
1.
--0.3475] 4.lJ88
direction
-.
486
Stirred Tank Heaters
Module 5
FIGURE 1\'15.8 Phase-plane behavior of the stirred tank heater
SUMMARY In tbis 11lOdu1c we have developed the dynamic modeling equations for a perfectly mixed stirred tank healer. We solved for the steady-state conditions, linearized to obtain the state-space model, and found the transfer function model. We compared the step responses of the lincar and nonlinear models. We also showed the importance of the '"direction" of initial conditions. PC11urbations in certain directions cause a fast response, while perturbations in other directions yield a slow response.
FURTHER READING The following text is a good reference for chemical process modeling, with a particularly nice presentation of energy balances: Denn, M.M. (1986). Process Modeling, New York: [,ongman.
STUDENT EXERCISES "I. In the heater example we performed stcp tests on the jacket ftowmte. Perform step tests on the other inputs: (i) Tank flownlte, (ii) tank inlettelnperature, (iii) jackel inlet temperature. l"ind the transfer functions relating the inputs and OlltputS. Discuss the linear versus nonlinear efTects.
2. Show that a decrease in jacket volmne will cause the difference in magnitude between the fast and slow poles to increase.
487
Appendix
3. [;iud the nonlinear stcady~stalc relationships between jacket flowrate and the two temperatures (tank and jacket). Plot the input (jacket flowratc) versus the outputs for the physical parameters given. Discuss how the gain changes with coolant fJowratc.
4. Develop 11 simple model with imperfect mixing on the jacket side. Assume that the jacket can be modeled as two compartments, where the flow is from one cOlllpart~ Illen! to another, as shown in the diagram below. Tank ~. inlet T;
~
Jacket inlet
f--
;--
T
V Tj2 Jacket
Fj
outlet
7]2
Tank
1j1 Tank outlet
F
~
T
APPENDIX 1.
Using MATLAB to convert from state-space to transfer function models-ss2tf
STATE SPACE MODEL a ~O
.4000 3.0000
0.3000 -4.5000
0 50.0000
~7.5000
b
0.1000 0
0
0 1.5000
c
1 0
0 1
0 0
(]
c1
0
0 0
0 0
Find transfer function polynomials for input 1:
»(numl,den] numl
ss2tf(a,b,c,d,l)
=
o o
0.0000 50.0000
15.0000 20.0000
1.0000
4.9000
0.9000
den
Find transfer function polynomials for input 2: »[num2,c1en] Dum2 =
ss2tf(a,b,c,d,2)
o o
-7.5000 -0.0000
-33.7500 -22.5000
1.0000
4.9000
0.9000
Find transfer function polynomials for input 3: »[num3, den] num3
ss2tf(a,b,c,d,3)
-
0 0
0.1000 0.0000
0.4500 0.3000
1.0000
4.9000
0.9000
dE-;Il
Find transfer function polynomials for input 4: »(num4,den] num4
ss2tf(a,b,c,d,4)
~
0 0
0.0000 1.5000
0.4500 0.6000
1.0000
4.9000
0.9000
den
Eigenvalues and eigenvectors »[v, lambda]
v
eig(a)
~
0.8207 0.5714 laml)(ia
-0.1911
o
0.0695 0.9976
o -4.7089
489
Appendix
2.
MATlAB function routine for nonlinear heater model function xdot
% %
~
heater(t,x);
Dynamics of a stirred tank heater (e) 1994 - B.W. Bequette 8 July 94
% % % % % %
x(l) x(2) de1F:j
% %
Tin
T
Tj
F
:5
'l'j i
%
V
%
Vj
% % % %
rhoep rhocpj
temperature in tank temperature in jacket change in jacket flowrate Tank flowrate Tank inlet temperature Jacket inlet temperature Tank volume Jacket volume density*heat capacity density*heat capacity,jacket fluid
parameter and steady-state variable values are:
% F
1·
Fjs Ti Tji
1.5; 50i
200; 10; V vj 1; rhocp 61. 3; rhocpjo::o 61.3 ; UA i83.9; %
delFj -0.1; Fj ~ Fjs + delFj; T~x(l);
Tj~
x(2);
%
% odes %
dTdt = (FjV)*(Ti - T) + UA*(Tj - T)j(V*rhoep); dTjdt = (FjjVj)*(Tji - Tj) - UA*(Tj - Tij(Vj*rhocpj); xdot(l) dTdt; xdot(2) = dTjdt;
FigureM5.7a is obtained using the following commands: xO : : : 124.6525 154.9880 »( t, xl :::;: ode45 ( 'heater I ,0,5, xO) ; »plot (t, x ( : , 1) )
ABSORPTION An Example of a Linear Equilibrium Stage System
MODULE
6
After studying this module, the reader should be able to: Understand step response behavior of absorption columns Use the MATLAB functions initial and step for state space simulations Use eigenvalue/eigenvector analysis to understand unforced system behavior Understand how zeros affect the step response hehavior oftranslcr functions The major sections of this module arc:
M6.1
M6.1
Background
M6.2
T'he Dynamic Model
M6.3 M6.4
Steady-State Analysis Step Responses
M6.5
Unforced Behavior
BACKGROUND Separations processes play an important role in most chemical manuracturing processes. Streams from chemical reactors often contain a number or components; sonlC of these components must be separated from the other components for sale as a final product, or for usc in another manufacturing process. A common example of a separation process js gas absorption, which is normally used to remove a dilute cOlnponent from a gas stream. 490
Sec. M6.2 Liquid Feed
The Dynamic Model L Xf
-.
I
491
v
Gas Product
v
Gas Feed
1
n
Liquid Product
L
t
x"
YIJ+l
FIGliRE M6.J Gas absorption column, II stages.
Consider the gas ahsorption column shown in I"igurc M6.1. COinponcnts that enter the boltom of the column in the gas feed stream arc absorbed by the liquid stream, so that the gas product stream (leaving the top of the column) is more "pure." An example is the usc of a heavy oil stream (liquid) to remove benzene from abellzene/air gas feed stream. Absorption columns often contain "trays" with a liquid layer flowing across the tray; these trays arc often lnodc'lcd as equilibriulll stages.
An objective of this module is to develop a dynamic (time dependent) model of a gas ahsorption colunm. We will :"ihnw how to solve [or the steady-state stage compositions llsing matrix algebra. We will then place the dynamic model in the lincar state-space form. Dynamic results for stage compositions will he shown for step changes in the inlet feed compositions.
M6.2
THE DYNAMIC MODEL M6.2.1
Basic Assumptions
We assume that the major comIJonent of the liquid stream is "inert" and docs not ahsorb into the gas stream. We also assurne that the Inajor component or the gas stream is "inert" and does not absorb into the liquid stream. It is assumed that each stage of the process is an equilibrium stage, that is, the vapor leaving a stage is in thermodynamic equilihriulll with the liquid on that stage.
M6.2.2
Definition of Variables
We use the following variable definitions:
Absorption
492
moles inert liquid
liquid molar flowratc
time
v M
w-
Module 6
:~nolcs iI~Et vapor
= vapor molar flowratc
time moles li9l!_~~~
= liquid molar holdup per stage
stage moles\,(~v()_r
vapor molar holdup per stage
stage moles solute
i)
mole inerlliquid (stage i)
y, M6.2.3
i)
moles solute
mole inert vapor (stage i)
Equilibrium Stage
The concept of an equilibrium stage is important for the development of a dynamic model oCthe absorption column. An equilibrium stage is represented schematically in Figure M6.2,
M6.2.4
Solute Balance Around Stage i
The lotal amount of solute on stage i is the sum of the solute in the liquid phase and the gas phase (that is, MX i + WyJ The rate of change of the amount of solute is then d(Mx i + Wy)/dt. The component material balance around stage i can now he written (accumulation = in - out):
il(M"
+ W y) _
- LX i _ l
ilt
+ VYil1 -
,
LXi~ VYi
(M6.1)
Since liquid is much more dense than vapor, we can assume that the major contribution to the accumulation term is the MX i tcrm. Equation (M6.1) can now be written:
il(Mx,) ilt
=
L
X i_ i
LXi__ j
+ V Yi+ I
~
f,
Xi
~
(M62)
VYi
VYi
FIGURE M6.2
A lypical gas
absorption stage.
Sec. M6.2
493
The Dynamic Model
Our next assumption is that the liquid InolaI' holdup (M) is constant, then we can write (Mo.2) as:
L
V
lW XiJ
L
V
+ M Yill -Mx;-M Yi
(Mo.3)
The solution to this problem will be easier if we can develop an explicit relationship between vapor phase composition and liquid phase composition. We will assume that the vapor on each stage is in equilibrium with the liquid on that stage.
M6.2.5
Equilibrium Relationships
The simplest relationship is a linear cquilibriuln relationship: (Mo.4)
where y is the gas phase composition (mole solute/mole inert liquid), x is the liquid phase composition (mole solute/mole inert vapor), ; represents the ith stage, and a is an equilib~ riuID parameter.
M6.2.6 The Modeling Equation for Stage i Substituting the vapor/liquid relationship from (MG.4) into the material balance (MG.]) yields:
dX i ill
_
(Mo.5)
which can be written in the following form:
+ Va)
dX i L Jt,-c-=M- ri + 1 -
M
xi
-!-
Va M
Xi + 1
(Mo.o)
It should be noted that (M6.6) will yield a matrix with a tridiagonal structure.
M6.2.7 Top Stage The balance around the top stage (stage I) yields:
dX i ill
-- (I. +Nt Va) x,
j-
Va M
x, +
(M6.?)
where XI' is known (liquid feed composition).
M6.2.8 Bottom Stage The balance around the bottom stage (stage n) yields:
ilx". L df -= M
(L Xn --- 1 -
+ Va) M
where YI/+[ is known (vapor feed composition).
XI/
+
V M
YIl11
(Mo.8)
494
Module 6
Absorption
----------------
EXAl\;IPLE j\iJ().J
Fivc~stagc
Absorption Column
The modeling equations (M6.6) -(M6.X) can he written in the following form:
+
dX I til L
tit tL"r J
r!t dX,j
tit dX 5
tit
Xl
M M
Xj
L M
(L X1 -
Va)
I
M
(L X :; -
Va M
.rl
Va)
I
X4
M
(L I Va) L t AI' 4 ,"vI
L
;Yf X 2 - /1,1/
x2
M
L M
+
-
Va
I
+
(stage I)
x/
);3
(Slage 2)
Va M
X4
(stagcJ)
Va M
);5
(stage 4)
V M Y()
X5
(slage 5)
'I'he state variables arc _\ (i;=o I to 5), and the input variables are xrC1iquid feed composition) and V6 (vapor feed composition). If is assl/llwd thaI the liquid and voporjlolFmtes are COIlSIOl/f. These equations can be written in matrix form as (notice the tridiagonal structure): (LI- Va) M
L M
I:"I ~
,Xs
0
Va M (L I Va) M L M
0
0
0
0
0
Va lvl
()
0
Va M
0
L
L M
()
+
x,
>1 X4
Va M
(L
I
_ X,;_
Va)
M
0
0 0
()
()
()
()
X2
(L I Va) M
M
L
so~called
()
(L I Va) M
M
which is the
0
0
I_Y6_I XI
V
M
statc-space form
x
Ax+Hu
(M6.IO)
M6.3
495
Steady-State Analysis
Sec. M6.3
STEADY·STATE ANALYSIS We have shown that the form of the dynamic equations for the absorption column arc:
x=Ax+Bu At
stcady~statc
the time derivatives arc zero, so:
o The
stcady~statc values
Ax+ HII
(M6.11)
of x can be found by solving (M6.ll) to find:
x
EXAMPLE M6.2
(M6.IO)
ccc
-A 'BII
(M6.12)
Parameters for the Fifth·stage Column
Lel the liquid feed flowrateL:::::; 80 kgmol inert oi1lhr, vapor feed flowratc V:::::; 100 kgmol air/hI', equilibrium parameter a:;;: 0.5, liquid feed composition xI';:::. 0.0 kgmol benzcne/kgmol inert oil, and vapor feed composition Y6 0::: O.l kgmol bcnzcnc/kgmol air. Let's operate with units of minutes, so L 0::: 4/3 kgmol inert oil/min and V:;;: 5/3 kgmoJ air/min. Assume that the liquid tnolar holdup for each stage is M :;;: 20/3 kgmol. The numerical values yield the follOWing matrices:
0.325 0.2
0.125 - 0.325
o
0.2
o
o o
A
o 0,2
o o
II
[
o 0.125 0.325 0.2
o
o o 0.125 0.325 0.2
o o o
(M6.13)
0.125 -0.325
~;
o
o o
o
025
(M6.14)
The steady-state Input is:
US
Solving for T at steady~state, we find
X"I·--=
=Y6S 1
X s "'"
x,
00 1JU. . ]
(M6.ls)
-Aln u,\"
~
1J.()076. 0.0198 O.1J392 [ Il.0704 (l.! 202
(M6.16)
496
Absorption
Module 6
which is simply a vector or the liquid phase compositiolls. The liquid product composition (leaving stage 5) is x'):;;:; 0.1202. The vapor product composition (leaving stage I) is obtained from the lincar equilibrium
I
___ rC~'h~\'~i(~)(~lS~h~ip~.~\·I~~_()~.5~.~rl~~_()~.5_«(~L~O(~I7~(~')~~_()~.()~O~3~X~I~b~m~O~1 ~b~CJ~ll~,CI~lC~/~lb~)(~)(~)t~'~\il~ ..
M6,4
~
~
STEP RESPONSES Here we will usc the MATLAB fUllction s Lcp to find the responses to a step change in feed compostioll. 'T'hc step function requires a linear state space model in deviation variable fonn. Ax~AAx +
M6.4.1
H Au
Step Change in Vapor Feed Composition
At time t ::: 5 minutes, the composition of the vapor stream entering the column changes ('rmn Y6 ::: n.] kgn101 bCl1zcl1c/kgmol air to )'6 ::: 0.15 kgmol bCIlZcllc/kgrnol air. Usc the MATLAB step function to find how the compositions of the liquid and vapor streams (x() and )'1) leaving the COIUllll1 change with time. How do the transient responses for all of the stage liquid compositions (XI through x:,,) diller? step is used to solve the following equations
A x = A Ax + H Au and
Ay
~
C Ax + D Au
where Lly is the output vector (deviation form). }·'Ol' this problem, we consider x'j (COl1lPO~ sition of liquid leaving the bottom of the eolulTm) and)'1 (composition of the vapor stream leaving the top of the column) as the outputs.
Therefore, the C and
J)
matrices arc:
C~I 0
0
O.S
D~IOo [y,x, tl
0
o o o o
01
0
step(A,B,C,D,u.2)
where the 2 in the final column indicates that the second input is being step changed. A comparison of 1
497
Step Responses
Sec. M6.4 0.07
;----~-_.-.
T-
...............•
0.06 0.05 0.04
"',I 0.03 0.02 0.01
o o
20
40
80
60
100
t(min)
FIGlJRE M6.3 Response of bottom composition feed composition of 0.05, at t :::: O.
10
a step increase in vapor
makes physical sense, because the disturbance must propagate through six stages (from the bottom to the lOp of the colmlln) to affect the top composition. figure M6.5 shows that the magnitude of the change in stage composition is greatest on the bottom stage. Deviation variables (xi(l) ~ xi(O)) have been llsed for case of comparison. 'fhe relative "speed of response" is faster the closer a stage is to the bottom of
1.5
0-
ro
>
,.: 0.5
___J,
0 0
20
-j--
40
60
80
100
t(min)
FIGURE M6.4 Response of top vapor composition to a step increase in vapor feed composition of 0.05, al t;;;:; O.
498
Absorption
Module 6
0.07
5
0.06 0.05 0.04 '"ill'"
,,'
0.03 0.02
3
--------- _--..
~::-;;;;-~
0.01 0
20
0
40
60
80
100
'(min)
FIGURE M(.,SRespollse of stage compositions to a step increase in vapor feed of 0.05, at I = O.
the column, as shown in l<'igurc M6.6. Scaled deviation variables (xJt)
~
x/O))I
xi(t;:;;;: 1(0) ~ x,{O) have been llsed for case of comparison.
M6.4.2
Step Change in Liquid Feedstream Composition
At time t ;:;;;: 0, the composition of the liquid stream entering the column changes from 0.0 kgmol hCllzcne/kgmol inert oil to "j;:;;;: OJ)25 kgmol henzcnc/kgmoJ inert oil. Find how the compositions of the liquid and vapor streams (.\:5 and .1'1) leaving the column
.\/~
5
0.8
D
0.6
'" u
,,'
0.4
0.2
40
60
80
100
'(min)
FIGlHU'-: M6.6 Response of stage compositions to a step incrc
----------------------~-~ .... .
!
--.........
Sec. M6.4
Step Responses
499 ----,~-----
0.01
.--_.__ --_. .....•
~-~
0.008
0.006
'"~I 0.004
0.002
o o
40
60
80
100
t(min)
FIGURE M6.7 Response of hottom composition to a step increase in liquid feed composition of 0.025 at t "" O.
change with time. How do the transient responses for all of the stage liquid compositions (xl through .r() differ? A comparison of Figures M6.7 and M6.8 shows that the bottom composition (x:J responds more slowly to the liquid feed change than the top vapor composition (VI)' This makes physical sense, because the disturbance Illust propagate through six stages (from the lop to the bottom of the column) to affect the bottom composition.
0.014 0.012 0.01 0.008
~
0.006 0.004 0.002 0
'-
0
--'--
20
~_~~~____
,
L. _
40
60
80
100
t(min)
FIGURE M(j.S Response of lOp vapor composition to a step increase in liquid feed composition of 0.025 at t :::0 O.
Absorption
500
Module 6
0.025
0.02
w g'
0.015
175 >(I
0.01
0005
-----'------
20
60
40
80
100
t(min)
FIGURE M6.9 Response of stage compositions to a step increase in liquid feed composition of 0.025 at t :;;: O.
Fi'igure M6.9 shows that the magnitude of the change in stage composition is greatest 011 the top stage. Deviation variables (A)t) ~ xi(O» have been used for case of comparison. The relative "speed of response" is faster the closer a stage is to the top of the column, as shown in Figure M6.1O. Scaled deviation variahles (x,(I) - x,(O))/(x,(I = 150) - x/O» have been used for case of comparison.
0.8
'0
0.6
w
w
~I
0.4
0.2
100
t(min)
FIGURE M6.1O Response of stage compositions to a step increase in liquid feed composition of OJ)25 at t "" 0 min. Normalized deviation variables arc plotted.
Sec. M6.5 M6.4.3
501
Unforced Behavior
Transfer Function Analysis
The reader should lise the MATLAB function ss2tf to find the transfer functions relating each input to each stale (student exercise I). Show that the order of the numerator polynomial gets smaller the farther that the state is from the input variable. Also, find the zeros and poles for each transfer function and explain why the speed of response is faster the closer a state is to the input, in terms of the trans!Cr functions.
M6.5
UNFORCED BEHAVIOR Eigenvalue/eigenvector analysis can he Llsed to understand the effect of perturbations in the initial conditions on the relative speeds of response. The eigenvalues span an order of magnitude, from ~O.05 min- 1 to -0.6 min-I, as shown by the MATLAB command: »[V , d]
-~eig
(arnat)
cl
--0.59886127875258 -0.05113872124742 -0.32500000000000 -0.16688611699158 -0.48311388300842 The first eigenvalue is the fastest, while the second is the slowest. The first and second eigenvectors arc associated with the fast and slow directions, respectively. v
Columns 1 throuqh 4 -0.16931748198570 O. )"109'':';601705416 -0.54181594235423 0.59352962728666 -0.433452"15388338
-0.16931748198570 -0.37095601705416 -0.54181594235423 -0.59352962728666 -0.43345275388338
-0.314446"15225915 0.00000000000000 0.50311480361464 -0.00000000000000 -0.80498368578343
0.2"/472527472527 0.34750303957894 -0.00000000000000 -0.55600486332631 -0.70329670329670
Column ') -0.27472527472528 0.34750303957894 0.00000000000000 -0.55600486332631 0.70329670329670
The reader should show (see student exercise 2) that the following perturbation in initial condition:
502
Absorption
XII
-
0.1693 03710 05418
Module 6
··0.025
O.s935 - OA335 yields it response tll,1! is u\'cr ten times faster than the follmving penurhation in inilial condition:
o I m, XII ~
IU710 0.5418 0.5935 OA335
'·0.025
SUMMARY In this module we developed a linear model for an absorption column and Llsed lhe i\ilATLAB step fUllction to study the erfect or input composition changes. T'hc magnillilk of composition change was greatest on the stages closest to the step input change, and the "speed" of composition change, \v{\s faster on the stages closest to the step disturbance. For example, for a step change in the vapor feed composition {\vhich enlers the bottom ()f the column) the crfect was greatest (magnitude largest and speed fastest) Oil the compositions of the stages close to the bottom of the colunlll,
FURTHER READING
The gas absorption example is a modification of an example presented by: LuJ'belL \V.L, & L.A. V'/eIlzcl. (I 9X8). Chemical Process Analysis: J\.4oss and ergy Ba/aJ/ces. Englewood Cliffs, NJ: Prentice {Iall.
Ei1~
EXERCISES I. Usc the t'vlATLAB function ss2 t: f to find the transfer functions rehlting each input to each state. Show that the order of the numerator polynornial gels smaller the farthe I' that the state is from the input variable. Find thc zcros and P(lIeS for cach transfer functioll. Explain why the speed of response is faster the closer a state is to the input- in terms of the transfer functions and the pole-zero values.
503
Appendix
2. Use the MATLAB initial function to show the effect of the "direction" of a perturbation in initial condition. Use plots to compare the "fast" and "slow" <.lircctions.
APPENDIX: MATLAB
absorp~ex.m
absorp C'x. m absorption example - Process Dynamics % orginal version - 29 June 93 % (c) B. Wayne Bequette % revised - 7 June 1997 96
5[,
% aeq gas/liquid equil coeff % ns :; : number of stages % liq :; : liquid molar flowrate % vap :; : vapor molar flowrate % mholc1 liquid molar holdup per stage % xfeed liquid feed composition % yfeed vapor feed composition %
% set parameters n~3
liq vap
.,
"
5; 80/60; 100/60;
mho1d - 2013; xfeed 0; yfeed 0.1; aeq :; : 0.5; amat bmat
zeros(nS,ns)i zeros(ns,2);
%
%
calculate ratios
%
-(liq+vap*aeq)/mhold; Ivratio lratio - liq/mhold; vratio - vap*aeq/mholdi amat(l,l) amat(1,2)
lvratioi vratio;
for i
2:ns-1; - Iratio; amat (i, i) lvratio; amat(i,i+l) vratio; end amat(i,i~l)
arnat(ns,ns-l) - lratioi Ivratioi amat: (ns, ns)
504
Absorption
bmi'J.t(l,l)
:::: lratio;
bmat(ns,2) %
vap/mhold;
calculate the steady-state xs ::::
-inv(amat) *bmat* [xfeed;yfeed];
%
perform step tests using step
'(;
,,;tep c11(:1.n90 thf" vapor feed composition at t
0
Clni'\.t zero:,:-; (2, n:::;) i cmaL(l,Ils) 1;
cmat(2,l) dmat
:::: aeq;
zo1'08(2,2);
'(,
[Y,x,L]=step(amat,bmat,cmaL,dmaL,2); % actual step magnitude was 0.05 "-(I y - y-kO.Ol:J; :::: x-kO.O:)i
x
plot the liquid composition leaving the absorber (deviation variables) ploL(L,y(o,l), 'w')
xlabel (' t (min) , ) ylabel ('x_liq')
%
pause ~)
plot. the vapor composi tion leaving the absorber plot (L,y( ° ,2), 'w')
xl abel ( , L (mi n) , ) ylabeJ ( 'y_vap ')
pause %
plot each stage liquid composition (deviation) plot (t , X, 'w') xlabEd ( t (min) ') yL:dx'l ( r X sL_a~Jc') pause F
plot- each std9(-;,
nt -
lC'nqth(t);
normalized by the final change
Module 6
Appendix
505
for i : :-, l:ns; xscale(:, i) end
x(:, i) Ix(nt, i);
%
%
plot (t,xscale, 'w') xlabel ( ' t (min) , ) ylabel ( 'x_scaled' ) pause now, do a step change in liquid feed comp [yl, xl, tl] =step (amat, bmat, cmat,
%
dmat~,
1) i
actual input change was 0.05 yl yl *0.025;
xl
xl*O.025;
% %
%
plot: the liquid composition leaving the absorber (deviation variables)
%
plot (tl,yl (:,1), 'w') xlabel ( , t (min) , ) ylabel ( 'x. 5 ' ) %
pause %
%
plot the vapor composition leaving t:he absorber plot(tl,ylj:,2))
%
pause
% plot each stage liquid composition (deviation) plot (tl,xl, 'w') xlabel ( ' t (min) , ) ylabel ( 'x_st<Jge' ) pause
%
plot scaled liquid compositions
%
ntl
Length (tl) ;
% for i = l:ns; xscalel (:, i) end
xl ( : , i) Ixl (ntl, i) ;
% plot(tl,xscalel, 'w') xlabel ( 't (min) , ) ylabel ( 'x_scaled' )
ISOTHERMAL CONTINUOUS STIRRED TANK CHEMICAL REACTORS
MODULE
7
After studying this module, the student should be able to: Determine the steady-state behavior of an isothermal CSTR (number of stcadystates) Determine the stability of a particular steady~statc
• Find the statc-space and transfer function form of a linearized CSTR model at a particular steady-state Show that steady-state gains calculated from the steady-state equations arc consistent with the transfer function gains The major sections of the module arc:
M7.1
M7.1
Introduction
M7.2 M7.3
First-Order Irreversible Reaction Van de Vusse Reaction
INTRODUCTION Chemical reactors arc generally the most important unit operations in a chemical Chemical reactors come in many forms, but two of the most common idealizations arc continuous stirred tank reactor (CSTR) and the plug flow reactor (PFR). These two serve as limiting hounds for the behavior of many operaling reactors. The CSTR is
506
Sec. M7.2
First~Order
507
Irreversible Reaction
F
v
CB
FIGURE 1\17.1 Continuous stirred tank reactor (CSTR).
used in dynamic modeling studies, because it call be modeled as a lumped parameter system. A dynamic model of a plug flow reactor consists of partial differential equations (also known as a distributed parameter system). Consider the continuous stirred lank reactor shown in Figure M7.1. In this module we will assume that the reactor is operating at a constant temperature (it is isothermal), so we do not need an energy balance (and can also assume that the reaction rate parameters are constant). In addition, we will assume that the volume is constant. The standard single irreversible reaction system is presented in Section M7.2. The Van de Vusse reactor, a re~ action scheme with a number of interesting characteristics is presented in Section 1\17.3.
M7.2
FIRST-ORDER IRREVERSIBLE REACTION Recall that a model of a CSTR for a first-order irreversihle reaction was developed in Chapter 2. Here we provide a quick review of the model ing equations, before presenting a detailed analysis. Consider a single irreversible reaction A ~ B. Assume that the rate of reaction per unit volume is first-order with respect to the concentration of A: molar rate of reaction of A per unit volume:::::: r A
:::::
kCA
Each mole of A reacted creates a mole of B, sO we have the following relationship for the rate of formation of B: molar rate of formation of B per unit volume:::::
M7.2.1
I"n:::::
kC;\
Component Material Balance on A
Our first step is to write the dynamic modeling equations for the reacting component, A:
rate of (rate) (rate) (rate) accumulation = in by flow - out by flow - out by reaction dVC A dt = F CA/- F C,l - V k C A
508
Isothermal Continuous Stirred Tank Chemical Reactors
Module 7
where k is the reaction rate constant. Since V is constant:
F .
VeA( which we can write as:
dC A dt M7.2.2
(M7.1)
Component Material Balance on B
It is natural to assume that there is no B in the fccdstrcam, which yields the following modeling equation:
dVCJ3
tit where again, k is the reaction rate constant. Since V is constant:
dC'/)
til
F
V Cli
+ Ie C,\
(M7.2)
The two dynamic modeling equations arc (M7.l) and (M7.2). Notice that the concentration of B docs not playa role in equation (M7.1), so equation (M?I) can be solved independently of (M7.2). Before analyzing the dynamics of this system, it is important to understand the
steady-state behavior.
M7.2.3
Steady-State Behavior
Assume that FIV is the input variable of interest. In the reaction engineering literature F/V is known as the "space velocity." Similarly, VIF is known as the "residence time" or "space time," that is, the amount of time that it takes for the reactor volume to be "swept out" by the flow. Equation (!'vI7.1) in steady-state form can be arranged to solve for the steady-state concentration (here we lise the subscript s to indicate steady-state value):
Fs(' V ~Afl' F S
(M7.J) _I k
V Notice in (M7.3) that CAs is a monotonic function of F/V. As F/V gets large (-----t 00), then CAs approaches C Als ' That is, the fluid is flowing so fast through the reactor thaI there is no time for allY cOllversion of A to B. As FJV gels very small (-----t 0), then C A .I. approaches 0, indicating complete conversion to B. This makes sense hecauscf:./V:= 0 corresponds to a batch reactor, and at steady~state we would expect complete conversion since we have an irreversible reaction.
Sec. M7.2
First-Order Irreversible Reaction
509
The sleady~s[atc gain between the input P/V and the output (e A ) is simply the derivative of the output with respect to the input. l1'fom (M7.3) we find:
(M7.4)
We call also find the steady-state gain between the input «('.v) and the output «('A) from (M7.3) by:
kV
dC/Is
(M75)
kV +k F,
DC(I/I'
Notice that we can rnakc a general plot of the solution of (M?3) by dividing by C"VI' to find
F kV F
·1 CA.JC~v:\,
kV
(M7.6)
+
can be considered the dimensionless concentration of A. The relationship in
(M7.6) is shown plotted in Figure M7.2. We can also plot the relationship in equation (M7.6) as shown in r;jgurc M7.3. Notice that, if we consider F to be the manipulated variable and C A to be the output variable. then we sec that the steady-slate gain (change in output I change in input) decreases as the r10wralc increases. That is, the gain is high and low nowrates and low at high rlowrates.
0.8
J
J
0.6 0.4 0.2
0 0
2
4
6
8
10
kVIF
F(;ORE M7.2
Dimensionless concentration as a function of kV/F.
Module 7
Isothermal Continuous Stirred Tank Chemical Reactors
510
0.8 ~
cJ
J
0.6 0.4
0.2 0 0
2
6
4
8
10
FlVk
FIGURE \\'17.3
Dimensionless concentration (equation (M7.4» as a function
or NkV. This result is consistent with (M7.4).Wc will find the same results when using the Laplace transfer function analysis. Notice from Figure M7.3 that to react most of A we need a large volume/feed ratio (residence time), which would require large vessels for a given nowratc. I'his indicates that economics must he used to guide the reactor design, and a trade-off must be made between the capital cost of the reactor versus the operating cost (which includes the difference in values of the reactants and products). The residence time is also related to the process time constant, which ultimately effects the quality of control that is possible. Solving (M7.2) for Ihe sicady-siare value of Ii we find:
k CA~'
(M7.7)
17\. V
subsriruring (M7.5) into (M7.7): k
(J)
I" V
elf,
V t k
k CeVi' I" +k
(M7S)
V
which can be written: C.lt>'
I" y kV
+
CArs
Notice in (M7.8) that the dimensionless concentration of B is also a monotonic function of the space velocity. As F./V gels large (-----) 00) then C'ns approaches O. That is, the fluid is
511
First-Order Irreversible Reaction
Sec. M7.2
0.8
."
0.6
J
0.4
<J
0.2 0 0
4
2
6
8
10
F/kV
Dimensionless concentration of B as a furH:tion
FIGURE M7.4
or reactor
f1owratc.
rtowing so fast through the reaclor that there is no time for any conversion of A to B. As F/V gets very small (~) 0) then CBs approaches C~W' indicating complete conversion of A to B (sec Figure M7.4). Notice that this figure is lIie mirror image ofFigurcM7.3.
M7.2.4
Dynamic Behavior
Let the state variables in deviation form be represented by: (:A -~ CAs
Xl '\:2
=
er; -
C'IJ.\'
(M7.9a) (M7.9b)
Also, let the input variables in deviation form be represented by:
P -PI'
Il] lt
z ::;:::
C'AI - C;1j~
(M7.9c) (M7.9d)
The state space model (by linearizing (M7.1) and (M7.2») is then:
dX J
(M7.IOa)
ell d''(2
(M7.IOb)
ell
or, dx]
(M7. lOa)
ell dX 1
ell
=
an
Xl
-+- an
X2
-+- b 2 !
llJ
(M7.IOb)
512
Isothermal Continuous Stirred Tank Chemical Reactors
Module 7
Writing in state-space matrix notation:
!J"II"II o '" we can see that the eigenvalues of A arc simply (J I J and ([22 (the reader should show this). Since the eigenvalues arc negative, the system is stahle. Since dx1/dt is not a function of _:1: 2 , we find (from the Laplace transfonn of (M7.IOa)): (M7.lla) which can be written in gain-time constant fortn as:
x,(s) .~
TI'"
+1
"I(S) + TjS
+
1 ",(s)
(M7.llb)
where
k"
/.'Jl. '= [CA!'oO~('AO'J F =
a
.\
"
+
k
V
k F ( V'+ k
(M7.12)
o
.
)
'
Notice that k ll is consistent with (M7.4).
T
I
a"
[
JI J Ie
(~)( 2"'11,:)
(M713)
'" ~ ~::: ··10~;1~ (2;;)
(M714)
Notice that k l2 is consistent with (M7.5). From (M7.1l) we sec that the process gain for output I decreases as the llowratc increases. We sec that T[ is always less than the reactor residence time from (M? J 2). Also notice that decreases as the reactor flowrate incrcases. From (M7.14) wc sec that k J2 is always less than L Taking the Laplace transform of (M7.lOa) and (M7.IOb), we find:
X2(.I')'~ b2l .l' + (b ll "21 .~ b 21 "1l) 1/,(.1')1 (.I' -
"ll)(S -
"n)
1/ (.1') (.I' - "JI)(s - an)
2
Sec. M7.2
513
First-Order Irreversible Reaction
which can he written in the form of:
ky) .1.) uIC') + (TIS + I )(~,s + I) 1l,(S) where: (M7.15)
(M7.16)
(M7.17)
kl)
h I2 {{:n
(M7.18)
(lll(ll1
Notice that Tn 0:::: 72' so there is pole-zero cancellation in gZI(s) and the transfer function becomes first-order. Also, since the eigenvalues are always negative (the time constants arc always positive), the system is always stable.
M7.2.5
Numerical Example
Consider a system with the following parameter and steady-state values:
/; V
k
~.
0.2 min
I
=
0.2 min
-I
gmol CAj, = 1.0 liter
which yields the following statc~spacc model:
[
clXll
cI/ - 0.4 clx,= 0.2 cit
I
II I I
UI oO.2h XI O.S O.21I + - O.S o 11., .1
The following figures compare responses of the nonlinear and linear systems
514
Isothermal Continuous Stirred Tank Chemical Reactors
Module 7
0.515 A
0.51 0.505 0
c
0.5
0 0
0.495 0.49
B
0.485
5
0
10
15
20
time, min
.FIGURE M7.5
Small (0.01) step change in FlV. Solid = nonlinear, dotted
-= linear.
SMALL STEP CHANGE IN F/V In Figure M7,5 we compare the responses of the linear and nonlinear models for a small (0.01) step change in FlV. Notice that the models yield virtually identical results.
LARGE POSITIVE STEP CHANGE IN FIV In Figure M7.6 we compare the rcspon~cs of the linear and nonlinear models for a large positive (0.2) step change in F/V. The nonlinear model has a lower "gain" than the linear model for bOlh states.
0.8 A
0.7 0.6 0
c
0
0.5
0
0.4
,,
0.3
B
0.2 0
5
10
15
20
time, min
FIGURE M7.6 = linear.
Large (0.2) slep change in FIV. Solid
=:
nonlinear, dotted
First~Order Irreversible
Sec. M7.2
Reaction
515
B
0.8
---
----_.~----
0.6 0
c 0
u
0.4 0.2 A
0 0
5
10
15
20
time, min
FlGUREM7.7
Large negative (-0.2) step change in F/V. Solid:;:;: nonlinear,
dotted:;:;: Ii near.
LARGE NEGATIVE STEP CHANGE IN F/V In Figure M7.7 we compare the responses of the linear and nonlinear model s for a large negative (-0.2) step change in FlY. The nonlinear model has a higher "gain" than the linear model for both states.
M7.2.6
Phase Plane Analysis
To determine the important directions for this system, we find the eigenvalues and eigenvectors of the A matrix: The slow eigenvalue is -0.2, which has the associated cigcnvcc~ tor ~l := I~J· The fast eigenvalue is ~O.4 with an eigenvector ~2 = I_;::~;:::;:!. These results indicate that an initial perturbation in the second state variable (concentration of B) will respond slowly, while an equal magnitude pertubation in both state variables will respond twice as fast.
M7.2.7
Summary of Behavior of the First-Order Irreversible Chemical Reaction
We can make some interesting observations about the behavior of this simple system. We notice that both reactor concentrations have first-order transfer functions with respect to the dilution rate, FIV. 'l'he concentration of A is also first-order with respect to (~\fi but a second-order transfer function relates eN to the concentration of B. The response ·of B to a change in the feed composition of A· is slower than the response of A for the same change.
Isothermal Continuous Stirred Tank Chemical Reactors
516
M7.3
Module 7
VAN DE VUSSE REACTION Often reaction schemes will exhibit a maximum in the concentration of product versus nowratc curve. Here we will consider such a reaction. It turns out this this type of system cail have significantly different input/output characteristics, depending on the operating condition chosen. Consider the reaction scheme consisting of the following irreversible reactions:
A 21\
--7
"
--7
B
--7
C
k2
D
"
This scheme was presented by Van de Vussc (1964). Engell and Klatt (1993) note that the production of cyciopcntinol from cycJopcntadiene is based on such a reaction scheme, where:
A ~ cyclopcnladdicnc B : : : cyclopcntcnol C:::::: cyclopcntancdio} f) ::::::
dicyetopcl1tadicnc
In the following development, we assume that the feedstream contalI1s only componeIlt A.
M7.3.1
Modeling Equations
OVERALL MATERIAL BALANCE Assuming constant density and constant volume:
dV dt
~
()
and
F =
F;
COMPONENT A BALANCE The balance on Ais:
accumulation
=
in - out by flow - out by reaction 'I -- out by reaction 2
d(VCA ) dt Since V is constant:
d(CA ) dt
(M7.19)
Sec. M7.3
517
Van de Vusse Reaction
COMPONENT B BALANCE Similarly, we can write: d( ell)
(M7.20)
dl COMPONENT C BALANCE Also, for component C:
d(C e )
(M7.2! )
dl COMPONENT 0 BALANCE And component D:
d(Cu) dl
(M7.22)
Look at equations (M7.!'!) through (M7.22). Notice that (M7.!'!) ami (M7.20) do not dc· pend on Cc or Cf). If we arc only concerned ahout CA and Cu' we only need to solve equations (M7.!9) and (M7.20).
d(CA)
(M7.!'!)
dl d( ('ilL
(M7.20)
dl Solving for the steady-slate for (M7.19), we find a quadratic equation in
Solving this quadratic (using the positive root), we find:
\
+
~ " + 4k, (r) ~ I( k, + r)' 2k,
and solving for Cns:
v + k2 These results lead to the input/output curves presented in Figure M7.8.
C~h:
Isothermal Continuous Stirred Tank Chemical Reactors
518
Module 7
.m so
E
o o
2
3
4
5
6
7
8
9
Fs1V,1/mJn a. Concentration of A as function of the space velocity.
1.4 1.2
.& ~
E
0.4 0.2 0 0
2
3
4
5
6
7
8
9
FsIV, 1/min b. Concentration of B as a function of the space velocity.
FIGURE M7.8
Steady-slate relationships for the Van de Vussc reactor.
Notice that there is a maximum value of the concentration of B. If the objective of the process is to maximize the production of B, then there exists an optimum residence time (FsI\!) of 1.292 min-- 1 (a function routine to find this is shown in the appendix).
STATE-SPACE MODEL Here we linearize the nonlinear modeling equations to find the
x=Ax+Bu y = ex
statc~spacc form:
Sec. M7.3
Van de Vusse Reaction
519
where the state, input, and output vectors arc, jn deviation form: CA --
C"'J
X
:0--;,,'
II
c
y
7~-1 C:
.
'"= slate vanables
I.C B ~ C/J.I'
[G -~I A
-
",en -
C:
AS
~ inplIt variable = output variables
,')
C B.1·,
Recall that our two dynamic functional equations arc:
The clements of the state-space A matrix arc found by: AU::::: ~t;/aXj A lI
A
I2
_ elI;
elI
eli; ~ ;);2
21
eli;
22
aj;
- - - '---",C'-' ax, a(CA - CA ,)
~
A
a;;
,
y-k l -2k'(A'
el( CII - C II,.) ~ JEll ~ 0
_ eli; _
A
P;
~ I -_) . .)( )" (XI (C A - '( A' oC,1
_ aj; _ k
-
~
aX 2
,.,
ae"
I
ai; il( C II - Cfi")
dC" =
all ------ii Dy ili; F
Dy and the state-space model is:
-
,
= C Afl' -
,
CAs
520
Isothermal Continuous Stirred Tank Chemical Reactors
Module 7
llcre we will consider several different steady-state
CASE 1 Consider a dilution rate of F/V = 4/7 min I, which is on the left side of the peak shown in «'igure M7.8b. The steady-state concentrations are:
mol 3 . liter
lOS 94
=
CHI'
alOl
1.117 .
liter
which yields the following stale-space model:
AB
-2.4048 r 0.8333
I
o
"2.2381
7.00001 1.1170
C"
The transfer function relating input I to output 2 is:
" 1.1170.1' + 3.1472 + 4.6429.1' + 5.3821
g2l(S) which has a zero in the right-half plane:
zero
=
3.1472 1.1170
~"
2.8175
The process transfer function can also be written in the form:
0.5848( 0.3549.1' O.1858.l + 0.8627
+1 .I' +
CASE 2 Consider now the following dilution rate F.JV:::: 2.8744 min-- I , which is on the right side of the peak in [
mol C A , = 6.0870 .
hter
105 94
1.117
11101
liter
Sec. M7.3
521
Van de Vusse Reaction
And the state-space model:
A
[-5.7367 0.K333
B
=
o
~4.5411
I
9130 I-1.1170 3. 1
The transfer function relating input I to output 2 is:
1.1170 s - 3.1 472 1O.277K s + 26.050K which has a zero in the left-half plane:
zero
= -
3.1472 ,= - 2B175 1.1170
Notice that cases 1 and 2 arc based on the same concentration of component B. Also note that case 1 has a right-half plane zero, while case 2 has a left-half plane zero of the same magnitude. This means that case I exhihits inverse response, while case 2 does not. The different types of behavior arc shown in the following figures. Figures M7.9 and M7.JO arc characteristic of a system with a positive gain and inverse response (right-half plane zero). I-'igure M7.9 shows that, for a small step change, the linear model is a good approximation to the nonlincar model. FigureM7.IO shows that the gain of the nonlinear model is significantly lower than the linear model, when a largc step input is made. Figures M7, 11 andM7 .12 are characteristic of a system with a negative gain. Figure M7.11 shows that, for a small stcp change, the linear model is a good approximation 1.123 linear
1.122
nonlinear
1.121
2 ~
E .0 0
1.12 1.119 1.118 1.117 1.116
3
2
0
4
time, min
FIGURE M7.9
Small step changc «WI) in space velocity, Case I conditions.
ESCOLA D~ ENGENHARIA BIBLIOTECA
5
Isothermal Continuous Stirred Tank Chemical Reactors
522
Module 7
linear
.~
,
"o
E .ci
,
, )
I
I
o
nonlinear
I I
I
time, min FIGIJRE M7.10 Response to a large step change (doubling. from 4/7 to 8/7 min-- I ) in space velocity. Case 1 conditions.
to the nonlinear model. Figure M7.12 shows that the gain of the nonlinear model is lower than the linear model, when a large step input is made. It is interesting to note that all operating points to the left of the peak in the input/output curvc-
1.1 1.1 1.114 ;~
~
E .ci 0
1.112 1.11 1.108 1.106 1.104 1.102 0 time, min
FIGURE M7.lt Response to a small step change in space velocity (0.1), case 2 conditions.
523
Van de Vusse Reaction
Sec. M7.3
1.2
1.1
2
~
E .ci 0
0.9 nonlinear
,,
0.8
0.7
\
linear
o
2
4
3
5
time, min
FIGURE M7.12 Response to a large step change (doubling of space velocity), case 2 conditions.
CASE 3 OPERATING AT THE PEAK "fhe optimum opcraUng point for the production of B is at the following dilution rate:
F -, V
= 1.2921 min
I
which yields the composition:
C,,,
mol
= 4.4949 .
C", = 1.2660
lIter
and the state-space model:
l
A
3.6237 O.R333
o
--2.95RR
mol liter
I
I ~:~~~{I) I The transfer function relating input I and output 2 is:
g.'l(s)-
-- 1.2660 S
- s' + 6.5R25 s + 10.7217
which has a zero at the origin.
Isothermal Continuous Stirred Tank Chemical Reactors
524 1.266
i
linear
Module 7
nonlinear
I 21.265 j
~
,
E
i
i
I
tf
j
1.265
1.264
a
2
3
4
5
time, min
FIGUREM7.13 point) conditions.
Response to a small step change, case 3 (peak operating
Notice that there is no steady-state change in the concentration of B Cor a small step change in space velocity, as shown in Figure M7.13; this behavior is consistent with the zero gain in the ,l;2J transfer function.
SUMMARY Two example isothermal reactor systems were presented in this chapter. We noted that V/F (residence lime; or equivalently, FlY, space velocity) is a common design parameter chemical reactors. We noted that a first-order reaction system has a monotonic rela""""", (no input or Olltput multiplicity) between output concentration and reactor nowrate. lower reactor nowrate leads to a higher gain with respect to the concentration of A. The Van de Vusse reactor had the interesting characteristics of input and right-half plane zeros (inverse response). At the maximum in the concentration nowrate curve the gain was zero.
REFERENCES AND FURTHER READING The following reference presents a more detailed study of an exothermic reactor that the same reaction scheme as the Van de Vusse reactor. Engell, S., & K.-U. Klatt. (1993). Nonlinear control of a nOIH11IIHlllllln-[lhasc CSTR. Proceedings (~fthe 1993 Arnerican Control Conference, 2041 --2045.
Appendixes
525
There arc a Ilumber or good reaction engineering texts. Most of the emphasis of these texts is on the stcady-stale behavior.
Fogler, H.S. (1992). Elements q{ Chemical Reacthm !-':ngineering, 2nd cd. Englewood Cliffs, NJ: Prentice-Hall. l"roment, G.P., & K.B. Bischoff. (1990). Chetl/feal Reactor Analysis and Design, 2nd cd. New York: Wiley. [,evcnspicl, O. (1972). Chemical Reaction/:;nghzeering, 2nd cd., New York: Wiley. Smith, 1M. (1981). Chemical Engineering Kinetics, 3rd cd. New York: McGraw-Hill.
STUDENT EXERCISES
t. Consider the A
----7 B example. Extend this to a reversible reaction A H B in a CSTR, where both the forward and reverse reactions are first-order, and where there is no B in the feed stream. a. Derive the state~space model. b. For the case where the reverse reaction rate constant is 1/2 the forward reaction rate constant (which is 0.1 Illite l ), find the value of F/V where thc conversion of A is SO%. c. [,'or a stcp increase in the feed concentration of A of IO/X), find the change in the A and B concentrations in the reactor.
2. Consider the yan de Vusse reactor. Include C and J) as stale variables and write the slale-space model. Find the transfer functions relating F/V and C c and CJ).
APPENDIXES A1
Generate Steady-State Input/Output Curves for the Van de Vusse Reaction
funct~ion
% %
[fsov, cbs, cas]
vandss(kl,k2,k3,cafs) ;
25 july 94 (c) b.w. bequette
%
% ~6
finds the steady~state input/ouLpuLcurves for the vand de vusse reactor
%
% % % % %
cbs cats
st:eady"-state space velocity, fs/v st:eac1y--state concentration of a steady~staLe concentration of b feed concentratioon of a
fsov
O;O.LIO;
fsov cas
526
Isothermal Continuous Stirred Tank Chemical Reactors
Module 7
% -(k1 I fsov)/(2'k3); casl cas2 - sqrt ((k1+fsov) ,A2 + 4*k3*fsov "leafs) / (2*k3) ; cas casl + cas2 i
%
% cas - O.Ol*ca£s:cafs/100:O.80*ca£s; % %
[sov:::
(k3.*cas.*cas + kl*cas) ./lcafs - cas);
%
cbs::: kl*cas./(fsov + k2);
A2
Van de Vusse Function File {vanmax.m} for Finding the "Peak" in the Input/Output Curve
l·'inding the maximum in the input/output curve for the Van de Vussc reaction:
Use [min fmin( 'vanmax'
I
0,2)
function cbsmin ::: vanmax(x) ; % %
25 july 94
%
(e) b.w. bequette
()6
9"
finds the maximum of cbs with the van de vusse reaction
r(~spect
to [soy for
9"
% % %
use mat lab function fmin which finds the minimum of a function take ~cbs as thp function Lo be minimized, which is equivalent to maxirnizins,J Cbf.:>
% )8
fsov
~)
cas
% %
cbs cafs
steady-state space velocity, fs/v st~oady-'-state concentration of a steady~·state concentration of b feed concentratioon of a
;';
k1 - 5/6; k2 ~ 5/3; k3 1/6;
cafs
10;
fsov
x·
% %
casl cas2 cas
-
-(kl + fsov) / (2*k3); sqrt((k1+fsov)"2 I 4*k3*fsov*cafs)/(2*k3);
casl + cas2;
% cbs
cbsrnin
k1*cas/(fsov + k2);
- cbs;
527
Appendixes
AJ
Van de Vusse Function File for Integration Using ode45, vanvusse.m function xdot :::: vanvusse{t,x);
% %
%
%
(e) b.w. bequette 25 July 94 revised 20 July 96
% °6
modE:~l
dynamic
for the van de vusse reaction
<[)
% % % % %
% % % %
uS(-~
to
this funct,ion file with ode45: ode45{'vanvusse',tO,tf,xO]
[tc,x]
where to is the initial time (usually 0) tf is the final time xO is the ini t~ial condition for t::.he states
x
states - concentration of A concentration of B x(2) time t
x(l)
% % 7>
%
fov ca cb caf
dilution rate concentration of a concentration of b feed concentrat::.ion of a
%
% % %
0,
the reaction scheme is A --> B
-~>
(kl)
% 2A
~~>
C
(k2)
D
(k3)
%
%
kl k2 k3 cat % % % %
~
~
5/6; 5/3; 1/6; 10;
% rate constant for A -> B (minA-l) % rate constant for B -> C (minA-l)
% rate constant for A + A ~> D (mol/(liter min) % feed concpntration of A, mol/liter
for step changes in fov, fov:::: 4/7 + delf;
use the following
use familiar notation for the states ea eb
xl]) ; x(2) ;
% % %
modelin~J
equations
dcadt - fov*(caf ~ ca) ~kl*ca ~ k3*ca*caj dcbdt ~fov*cb + kl*ca - k2*cb;
528
Isothermal Continuous Stirred Tank Chemical Reactors
%
derivatives placed in vector notation xdot (1) dcadt; xdot(2) - dcbdt;
A4
Various Cases for the Van de Vusse Reactor
al -2.4048 0.8333
»[vl,dl]
=
a -2.2381
eig(al)
vI ()
1.0000
0.1962 -0.9806
d1 ··2.2381
o
o -2.4048
,,~j.'l367
»a2
0;.8333
cl2
o
~).'j367
0.8333
-4.54111
~4.5411
»[v2,d2]= eig(a2) v2
o
1.0000
0.8204 -0.5718
d2
o
-4.5411
a
",13
-::i.7367
··3.6237 0; .8333 -2.9588J
a3 -3.6237 0.8333 »[v3.d3] v3 -
a -2.9588
- eig(a3)
a 1.0000
0.6237 -0.7817
c13 -
-2.
9~:i88
a
a -3.6237
Module 7
BIOCHEMICAL REACTORS
MODULE
8
After studying this module, the student should be able to
Develop the dynamic modeling equations for a
two~statc
biochemical reactor
Understand the concept of "washout" Understand the different types of steady-state and dynamic behavior exhibited by the MOl1od and Substrate Inhibition models
Find the numher of steady-state solutions and to determine the stability of each steady-stale The Inajor sections of this module arc: MS.I
M8.1
Background l~quatiolls
MS.2
Modeling
MS.3
Steady-state Solution
MS.4 MS.)
Linearization
MS.6
Phase-plane Analysis
MS.7
Understanding Multiple Steady-states
MS.S
Bifurcation Behavior
Dynamic Behavior
BACKGROUND Biochemical reactors arc used to produce a large number of intermediate and final procl~ ucts, including pharmaceuticals, food, and beverages. Biochemical reactor models arc similar to chemical reactor models, since the same type of material balances are per-
529
530
Biochemical Reactors
F
Module 8
.-
X1l
x"
X, F
x2
X,
V
FIGUH.E MS.l
X2
Biochemical reactor.
formed. In the simplest reaclor we consider two components: biomass and substrate. The biomass consists of cells that consume the substrate. Onc exmnplc would be a wastewater treatment system, where the biomass is lIsed to "cat" waste chemicals (substrate). Another example is fermentation, where cells consuille sugar and produce alcohol. Consider the schematic of a biochemical reactor shown in Figure Mg.l. In this module we aSSlllllC that the reactor is perfectly mixed and that the volume is constant. We usc the following notation: Xl
mass of cells
= biomass concentration
X2 =
volume
mass of substrate
substrate concentration
1'1
= rate ofceH generation
"2
-_'::0
volume
mass ofc~II~_g?ncrated vol lImc -time
mass of substrate consumed
rate of substrate consumption
F = volumetric f10wrate
volume- time
-= volume/timc
Now we can write the material btdances to describe the behavior of this systell1.
M8.2
MODELING EQUATIONS The dynamic model is developed by writing material balances on the biomass (cells) and the substrate (feed source for the cells). Biomass grows by reeding on the substrate.
M8.2.1
Biomass Material Balance
We write the biomass material balance as: rate of accumulation:;:: in by flow -- out by flow + generation
dVx, dl
,= ['\1/ - fit,
+
VI',
(MS.I)
Sec. M8.2
Modeling Equations
531
where ,x lf is the concentration of biomass in the feed stream and F is the volumetric flowratc.
M8.2.2
Substrate Material Balance
The substrate material balance is written: rate of accumulation::;;: in by now - out by flow
dVx, dt ,
.
=~
~
consumption
.
l'x)· .} ~ [-<x".. - Vr.,
(M8.2)
where x 2I is the concentration of substrate in the feed stream.
M8.2.3
Specific Growth Rate
The reaction ratc (mass of cells generatedlvolume time) is nonnally written in the following form:
r1
:::::
(M83)
fL
where /.L is the specific growth rate coefficient. We can think of /-l as being sitnilar (0 a first-order reaction rate constant; however, jJ. is not constanl-·-it is a fUllction of the substrate concentration as shown in Section M8.2.6. The units of /L arc time·- I ,
M8.2.4
Yield
There is a relationship between the rate of generation of biomass and the rate of COnS1l1l1p~ tion of substrate. Define Yas the yield, that is, the mass of cells produced per mass of substrate consumed: y~
mass of cells produced mass of substrate consumed
1',
(M8A)
I'"rom (MBA) we can write:
(M8.5) and substituting (M8.3) into (M8.5). we find: r
)
= P'~~J
Y
We assume in the subsequent analysis that Y is a constant.
M8.2.5
Dilution Rate
Assuming a constant volume reactor, we can write (M8. J) and (M8.2) as:
(M8.6)
532
Biochemical Reactors dX I
dl
F V
Module 8
(MS.7)
Xjf -
F
F
VX2f-
V
XZ - r2
(MS.S)
Defining F/Vas D, the dilution rate, and using the rate expressions in (MS.J) and (M8.6), we find: dX I
(MS.9)
dl
dx] dl
D x_21 - D..~tJ .\,. y
(MS.IO)
Generally, it is assumed that there is no biomass in the fecd stream, so xlJ':::;: O. The bioreactor modeling equations arc then norrnally written in the following form:
dX 1
dl
(MS.II)
(MS.12)
The dilution rate (D) is the same as the space velocity in the chemical reaction engineering literature. It is also the inver,",c of the reactor re.,;idence time and has units of time·-- l . The expressions for jJ- (specific growth rate) are developed in the following section.
M8.2.6
Growth Rate Expressions
The growth rate coefficient is usually not constant. A numbcr of functional relationships hetween the growth rate coefficient: and substrate concentration have been developed. The lllost common are (i) MOJ/od and (ii) Substrate inhibition.
MONaD The growth rate coefficient oftcn varies in a hyperholic fashion. The following forin was proposed by MOl1od in 1942. Notice that jJ- is first-order at low x2 and zcro order at high x2' (MS.I3) Notice lhat
jJ-
is firsl-order at low x 2 and zero order at: high x 2 . That is, when x2 is low:
Sec. M8.2
Modeling Equations
533
and when '-\"2 is high:
Since the reaction rate is: rl
/LY 1
this lncans that the Monod description is similar to a second-order (bimolecular) reaction when x 2 is low, since r
~_.
I
and to a first-order reaction when x 2 is high, since 1'1
=
!Lilla;; Xl
[~quation (MB.l3) is the same form as the Langmuir (u.Jsorption isotherm and the sl2uHJ;Jrd rate equation for enzyme-catalyzed reactions with a single substrate (Michaclis-l'viclllC'1l kinetics).
SUBSTRATE INHIBITION Sometimes the growth ra.le coefficient increases at low substrate concentration. but decreases at high substrate concentratioll. The physical reason may be that the suhstrate has a toxic effect on the biomass cells at a higher concentration. This effect is called suhstrate inhibition and is represented by the t()llowing cquation:
0.6 Monad
0.5 0.4
E
03 0.2 0.1 0 0
2
3
4
x2
FIGURE IV18.2 growth rate.
COIllparison of Monod and substratc inhihition models for
Biochemical Reactors
534
Module 8
TABLE MS.l I'aramctcrs for lVIonod and Substrate Inhibition lVlodcls Substrate Inhibition
MOllOd
0.53 hr l 0.12 g/litcr
fL max
0.53 hI
kill
0.12 g/Jitcr 0.4545 litcr/g
k,
y
0.4 4.0 g/Iitcr
fL~
0.4 4.0 g/li'c!
Y
kill + x 2
I
+ 1<1
(MBI4)
Notice that the Monod equation is a special case of (M8.14), with k 1 = O.
SPECIFIC GROWTH RATE RELATIONSHIPS The characteristic relationships between substrate (x 2 ) and specific growth rate (f.-l) arc quite different for Monnd and substrate inhibition. The curves for f.L as a function of _\:! for both models arc compared in I
M8.3
STEADY·STATE SOLUTION In this section, the MATLAB function fsolve will be used to solve for the steady-slate values or the biomass and substrale concentrations. The nUlllerical values used in our silll~ ulations are shown in Table MR.I. We \vill study the following cases: Case 1. Medium Dilution Rate, 1\::::: 0.3 hr" Case 2. Low Dilution Rale, D s ::::: 0.15 he l Case 3. High Dilution Rate, D, ~ 0.45 he'
EXAM PI ,E MS. I
Case I Results (Il
~
0.3)
The function file bio ss. m (Appendix 1) is set for Case! (D =: 0.3) and Ihe sub.'ilrale inhibition model (kl -co 0.4545). The l'vfATLAB function fsolve is used to solve for the steady-slatc values by entering tbe following in the comfnand window (with an initial guess of x (1) =: 1 andx(2) 1): 0;:
»
X =
f~301ve(
'bio
)3~:;'
,[1;1])
Sec. M8.4
Dynamic Behavior
535
The sh:ady-statc solution obtained is: x
= O.99 tj1 1.5122
Different initial guesses resuil in two other solutions for the substrate inhibition model. Also. the MOJlod model has two steady-slate solutions. "fhe reader should find the following results using [sol vc and bio_s~;, m, by entering different initial guesses.
Monod (2 steady-state solutions) Equilibriulll I Equilibriulll 2
0 'b = 1.5374
Xis::;;:
X2s::::;
4.0
x 2s :;:;:;
0./565
Substrate Inhibition (3 steady-state solutions) Equilibriulll I
Equilibrium 2 EquilibriullI J
0 0.9951 .\1.1-:;;;' 1.5302 X
Is
:=
Xis;;;::
X21::::;
4.0
.1 2.1'::::;1.5123
'2, = 0.1745
Notice that EC]uilibriUl,11 3 on the Sf model is ahnosl identical to Equilibrium 2 for the Monad model. In this section we have discussed case 1 results (D :;:: 0.3) only. Cases 2 and 3 will be discussed in Section M8.? In the next section we will analyze the dynamic behavior of this systeln, and in Section MS.7 we will show how multiple steady-state solutions arise.
M8.4
DYNAMIC BEHAVIOR In the previous section we found that the Monod and substrate inhibition models had t\V() and three steady-state solutions, respcctively, for the Case 1 paramcler values. In this section we perform simulations or the dynamic behavior of this system. A function file named bio. m is shown in Appendix 2.
M8.4.1
Case 1 (0 = 0.3), Substrate Inhibition Model
'rhe initial simulation is with the substrate inhibition parameters under Case I conditions 0.3). The simulations for two different initial conditions arc shown ill f,'igurc M8.3.
(D:;::
,,[tl,xlJ "
odc,45( 'bio' ,0,30, [1;lJ);
[t2, x2 J - ode45 ( 'bio' ,0,30, [0.75; 2 J) ;
536
Biochemical Reactors
Module 8
2
a a
5
10
15
20
25
30
25
30
time
4
"I
2
o
o
~---,
5
10
15
_
.... -
20
time FIGlJIU~ iVI8.J
Substrate inhibition, Case I. xO:::o II, I] (solid), xO
0:;;:
[0.75,2]
(dashed),
Although both initial conditions arc reasonably close to the Equilibriufll 2 solution found in section 3, one simulation converges to Equilibriulll I (dashed line) while the other converges to Equilibriulll :3 (solid line). We find in the next section that F~quiJih rillm 2is unstable: Furtber simulations will be performed and analyzed in the phasc-
plane (section 6),
M8.5
LINEARIZATION In this section we find the linear state-space and transfer function models. So that there is no confusion in notation, we will usc the following fonn:
i y
Az+BII ~
Cz
where: 21 =
x 2s
f) -
D,
x 2I --
Xl/i-
II, il Z
The st<:ltc-spacc matrices arc:
XI-XIs
2 2 = Xl ccc
Sec. M8.5
Linearization
537
where it is assumed that both states are output.s. The notation tL,; is used in the A matrix to represent the derivative of growth rate with respect to substrate concentration, evaluated at steady-state: aILs
rb: 2s For the MOl/od model:
dr<, (}X;:'I
IJomaJ(m
(kill
(MS. IS)
+ x::"J 2
and for the substrate inhibitioH model: aILs
V'11HI_~~~h (1
dX 2s
(kllJ + x 2.\"
P-_I~l,!X (km +
+
~1~J__' (2s)
+- {(I-ttY
+ k1xis) - J.LmllX X2\' (1 + 2k 1x 2s ) (kll/ + X2,y + k 1xiJ 2
(MS.16)
~ klxI,·) 1-
M8.5.1
k 1-'-2.1 ,_')2
Substrate Inhibition Model
Here we analyze the substrate inhibition model under Case I conditions. A MATLAB m-file, bio..._jac.m (Appendix 1), is used to generate the A matrix and the eigenvectors and eigenvalues.
EQUILIBRIUM POINT 1 The sleady-slale value (seclion 3) is (x".x,) = (0,4). The following command is entered: »(jac, evec, lambda]
:::; bio jac ([
a i 4]
)
538
Biochemical Reactors
Module 8
where j ac is the Jacobian (A matrix), evec is the eigenvector matrix and lambda arc the eigenvalues.
jac :; :; -0.1139 -0.4652 evec ;::;;
o
1.0000
o -0.3000 0.3714 -0.9285
lambda ~ -0.3000
o
o -0.1139
so,
A
~ r.~O.1l39 ~0.4652
"j
~ ~O.3
"2
~ ~ 0.1139
- OU.300]
~j ~ [~] "
0.3714 "I
~2 ~-O.9285 I
Since both eigenvalues are negative, the system is stahle at equilibrium point 1, verifying the simulation results shown in Section M8A.
EQUILIBRIUM POINT 2 The steady-state value is (xb,x,) »[jac,evec, lambda]
~
jac :; :; 0.0000 -0.7500
-0.0679 -0.1302
evec ;: ; 0.3714 -0.9285
0.2209 0.9753
lambda ~ 0.1698
o
~
(0.995 I, 1.5 I22).
bio~jac([0.9951;1.5122])
o -0.3000
The positive eigenvalue (0.1698) indicates that equilibrium 2 is unstable.
539
Phase~Plane Analysis
Sec. M8.6
EQUILIBRIUM POINT 3 The steady-state is (x b ,x2,) = (1.5302,0.1746).
»[jac,evec,lambda] = jac
bio~jac([1.5302;0.1746])
:::0
0.0000 -0.7500 evec
0.9048 -2.5619
=:::
0.9492 -0.3147
-0.3714 0.9285
lambda = -0.3000
o
o -2.2619
Both eigenvalues arc negative, indicating that equilibrium point 3 is stable.
M8.6
PHASE-PLANE ANALYSIS The m-filc bio~phas_gen.m (Appendix 2) was llsed to generate the following phaseplane plot for the substrate inhibition model under Case I conditions (sec Figure M8.4). Notice that all initial conditions converge to either the washout steady-state (trivial solu-
5
2 o 1 -
o o
0.2
0.4
0.6
1.2
0.8
1.4
1.6
x1
FiG-URE M8.4 Phase-plane plot for suhstrate inhihitioll model, Case I conditions (x:;;:: stable steady-state, 0 = unstable steady-state).
Biochemical Reactors
540
Module 8
lioll. equilibrium I) or equilibrium 3; while cqliilihriutn 2 is a saddle point (ullstable).
These results arc consistent with the stability analysis of Section M8.). A phase-plane plot for the Monad model was shown in Chapter 13.
MS.7
UNDERSTANDING MULTIPLE STEADY-STATES* In this section we find analytically the steady-state solutions for the biorcactor model and determine their stability. The stcady-slate solutions (djdt = d,,1dt = 0) of (MH.!!) and (MH.12) arc:
o= o=
(fL,·
OJ x"
f).I' (.t2j.i' -
x 2.1,)
(MH.I7I
---.y.-fLx"
(MHIHI
where the subscript s indicates steady-state. 'fhcrc arc two different types of solutions to (MS.I?) and (MS. IX). One is kno\vll as the trivial or ;'washout" solution. The other type is the nontrivial solution.
M8.7.1
Washout Condition
l:;'rorn (M8.I?) and (M8.IS) we can immediately sec one solution, llslli111y called llw trivial
solution.
x" x2_~
0 =
(MHIl)1
X2{I'
This is also known as the washout condition, since the reactor concentrations arc equal to the feed concentrations; that is, there is no "reaction." Since there is no biomass in the feed stream, then there is no biomass in the reactor under these conditions; all of the cells have been "washed out" of the reactor.
M8.7,2
Nontrivial Solutions
From (M8.17), assuming that .r ls *- 0, then: (ivlH.20)
fL., = D,
which indicates that the specific growth rate is equal to the dilution rate, at steady-state. From (MH.IH) we find thai:
y
(ivlH.211
and from (MH.20) and (MH.21): X 1.\
(M8.22J
"'This section contains a detailed analysis which the reader may wish to skip on a first reading.
Sec. MS.7
Understanding Multiple Steady-States
541
We can solve for x 2s ' by llsing the relationship for IJ. s as a function of x 2s (either Monod or substrate inhibition), since we kllow that IJ.- s =- D s (from (M8.20». Let !J..Jx2.\') represent this general functionality. Then, we tHust solve:
"(c)··D rs - 2.\ S
(M8.2.1)
for x 2s ' then substitute this value into (M8.22) to solve for .1: 1.1" The specific cases of Monoc! and substrate inhibition arc shown in the subsections below.
MONOD From (MS.13), the dilution rate at steady-state is I-LJ!laxXh'
kllJ + x 2s
(M8.24)
Solving (M8.24) for x2s' we find: (M8.25)
and since i-L s ;;;:;; D.I,
k ll!_ D.
I,
iJ-max --
D,,_
(M8.26)
For (M8.26) to be feasible, we note that D,I' < /L ma :c Actually, there is a more rigid requirement than that. 1'1'0111 (M8.22) we note that the highest value thal x2.1' can be is x 2f.i' otherwise Xis will he less than zero. Thc maximum /).1' in reality is thcn f.1,Jx 2 ,..,J, (Jr (from (M8.22), letting x2s ;;::; x 2JJ: . ILJn'IX ..\:2D:
kill
(Mollod)
+ x2(\'
(M8.27)
We also see from (M8.26) that there is a single solution for x2.\' as a function of D,I. 'rhis means that there is a total of two steady-state solutions for the Monod model, since there is also the washout (trivial) steady-state.
SUBSTRATE INHIBITION We found in the previous subsection that there are two possible steady-stales for the Monod model, for a given dilution rate. [11 this subsection we find the number or possible steady-states for the substrate inhibition mode1. From (M8.14) at steady-state: /-Lmilx j..L.\'
kill
+ x2. + 1,
ESCOLA Dc E~:C;'!I'li\IW\ BIBLlQI60/\
k JX~.I'
(M8.28)
542
Biochemical Reactors
Module 8
l'rom (M8.28), we find thai:
k,xl, +
(I - fL'''",) x" J.Ls
+ k",
~0
(M8.2'J)
()
(M8.30)
Since fL., = U, (M8.20), we snhslilutc into (M8.2'J) to find:
, , +
{(lXiI'
(1
fL"'''')
,,-,
D,
+ I(m
X 21' -
=
Since (M8.30) is a quadratic equation, there will he two solutions for x2S" This means that there arc three stcady~state solutions for substrate inhibition, since there is also the washout (trivial) steady-state. We sec from (M8.30) that for positive values of X2s the coefficient multiplying x 2.\ must be negative. The implication is that l.1 max must be greater than D s (the same result as the Monod equation). This implication can be seen more clearly frolll the solution of the quadratic formula for (M8.30):
_ fL"'''')' _ 4k I k D,
III
2k,
(M8.31)
So, for solutiolls with physical significance:
k ( 1 _fL""")' D, > 4k 11/1
(M8.32)
(M833)
and
Because of (M8.33), \vc know that the tefm inside the hrackets in (M8.32) is negative. For (M8.J2) to be satisfied, then we know:
< ( 1 _ fL'''''') lJ,
V4k 1/11 k
(M8.34)
which implies that: Ds < 1
_J-ll.!!
+ 2'v(,:k 1\ 1/11
(substrate inhibition)
(M8.3S)
We could have found the samc result from viewing Figure M8.2. Notice that there is a peak in the J..Ls curve, and again recall lhat D.I , ;::;: f.L r The steady-Slate dilution rate, D s canuot be abovc the peak in the x2s versus J..Ls curve. We can find the peak by finding ilfL,Idx2' ~ n. From (M8.16):
d/-1 s
dX 2s
(M8.36)
Sec. M8.7
Understanding Multiple Steady-States
543
We sec from (MS.36) that dfL./dX2' ~ () if: (MS.37) We can substitute this result into (M8,28) to find: fLlll(lx
ik;" l 1 \X
(MS.3S)
km +
!em
+ 1 +- k m
k
'"
fL.,
~I
__ }l..l!1~1-"' __
(MS.3,)
+ 2 VX,k,,,
sO the maximum dilution rate (for the nontrivial steady-state) is: f.111l
D,
1+2'r. -;Vk kl/(m
which is the same result as eMS.3)).
M8.7,3
Summary of Steady-State-Monod and Substrate Inhibition
WASHOUT (BOTH MONOD AND SUBSTRATE INHIBITION) Both Monod and substrate inhibition models have a washout (trivial) steady-state: XIs :~~
0
(MS. I')
NONTRIVIAL STEADY·STATE FOR MONOD The nontrivial steady-state solutions for substrate and biomass arc:
kI/lD,\.
(MS.26)
f.Lm
Y (x 2I, - -"') with the requirement that D s < IJ. max 'Y~f.Jklll + x~/.i' (that is D.I, < f.L.lr2j.J)
NONTRIVIAL STEADY-STATES FOR SUBSTRATE INHIBITION The two nontrivial steady-state solutions for substrate arc;
(MS.22)
544
Biochemical Reactors
1:'',' ' )' - 41
1(
\ .1 _ D
Module 8
\1 1(/II
2/(,
(M8.31)
unci the associated biomass concentration is:
x"
CC"
Y (X 2!, - x,,)
(M8.2S)
with the requirement for dilution rate:
(MX.32)
M8. 7.4
Stability of the Steady-States
The stability of each steady-stale solution is determined from the eigenvalues of the Jacobian matrix (matrix A in the state-space form), For a two-state system we know that the eigenvalues are found by: X' - lr(A) X + del(A) ~ ()
dCl(Al- A)
(M8AO)
j::"wm Chapter 13 we know that the following conditions mllst be satisfied for stability of a second-order system:
lr(il) < 0 del(A) > 0
(MXAI) (MXA2)
That is, the eigenvalues (A) will be negative if conditions (M8.41) and (M8.42) arc satisfied. The Jacobian of the biorcactor modeling equations (M8.11 andM8.12) is:
A C"
[1:'.' - D,
x "I:':
y,),"
,- f)" _ ..lJv'r ,I
-ij;
J
(M8A3)
where we have used the notation /-L,; to represent the derivative of growth rate with respect to substrate concenlration, evaluated al steady-stale:
i!/-L,\
(M8A4)
c)x:2y
The trace and determinant of Ii are: triA)
,c
(I:' - D) _
"
.,
J) _ I:';x"
.\
det(A) "" - (I:' _ D) (D
.\
.\
Y
.'
+
/L;
x,,) + X,dL;I:', Y
Y
(M8A5)
(MXA6)
We will usc (M8AS). (M8A6), and the eouditions shown in (M8AI) and (M8A2) 10 dctermine the stability of each steady-state.
Sec. MS.7
Understanding Multiple Steady·States
545
STABILITY OF WASHOUT STEADY·STATE Under washollt conditions: X2.\':::: '\.2/.'1' and xis::::: o. For stability of the washout steady-state, the following criteria then must be met. r"irsl, from the requirement that tr(A) < 0:
fL,-D,-D,
(MXA7)
and from the requirement that dct(A) > 0:
.. (fL, - DJ(D') > 0
(MXAX)
F"rom (M8.47) we sec that the requirement for stability is thell:
D > P'.I' .\ 2
(MXA9)
while from (M8.48) the requirement for stability is:
D s ::> /-Ls
(MX.50)
Notice that f..L s is evaluated at the substrate feed concentration for the washollt condition. Perhaps the expression I-L,lx2/) should be llsed to designate this relationship. Comparing (M8.49) and (MS.50), we sec that (MS.50) is the more rigorous requirement for stability of the washout steady-state. 'fhe growth rate expression for Monod kinetics is: (MX.51)
while for substrate inhibition kinetics: (MX.52)
Notice that IJ-.lr 2j:) is simply a shorthand expression for the specific growth rate evaluated at the suhstrate feed concentration. We must use (MS.50) along with either (M8.51) or (MS.52) to determinc the stability or the washout steady~state. Notice that the washout steady~state will only bc stable if D.I , is high enough. Wc can think of D.I , as a dynamic bi~ rurcation parameter, becausc the stability of the washout steady-state will depcnd on the valuc of the dilution ratc.
Stability of Washout Steady-State for Monad.
From (MX.50) and (M8.51 l. Iht
washout stcady-statc will be stable if:
and unstable if: (MX.54)
546
Biochemical Reactors
Stability of Washout Steady-State for Substrate Inhibition. and (MS.52), the washout sleady~state will he stable if:
Module 8
From (MS.50)
and unstable if: (MS.56)
NONTRIVIAL
STEADY~STATES
For the nontrivial steady-states, D s = fL.\.. The stability requirerncnts for the nontrivial steady-states arc then: II f
t:
- D _~·\~L>:
Y
s
<0
(MS57)
from the trCA) specification, and (MS.5S) from the det(A) specification. Since D.I. (and therefore 1J..), xis' and Yare positive, (MS.57)
and (MS.58)
rcdu~c
to the requirement that:
fL.: > 0
(MS. 59)
for slability.
Stability of Monod at the Nontrivial Steady-State.
[1; -_ ...• ILm
(k m + x 2 \)-
From (MS.24): (MS.60)
We see ilnmediatcly that fL,: is always positive for the Monod model at the nontrivial steady-state; therefore, the nontrivial steady-state is always stable. Recall that D.\. < fL.1.(X2!1.) for a nontrivial steady-state solution.
Stability of Substrate Inhibition at the Nontrivial Steady-States. We can tcll from the substrate inhibition eurvc in Figure M8.2 that a steady-state that is on the left side of the peak will be slable (since fL.:: > 0), while a steady-state on the right side < 0). will be unstable (since Numerically, from (MS.36):
f.<
~_!lF1X _(k'II=!~J~~3,~:2. _
(kll/ + .T2I. + k]xisf
(MS.61 )
Sec. MS.7
Understanding Multiple Steady-States
547
The x2s that is on the left side of the peak, and is therefore slable, is (from (MS.31»:
(fLO'' ' _1) - ~(I _iL",,,,r - 41< ,I<", (M8.62)
21<,
The x 2s that is on the right side of the peak, and is therefore unvtahle, is (from (MS.31 »:
X2~
=
Also, recall that Ds < fJ-ma/1 solution).
M8.7.5
(fL'D,' ' ' _I) + .
_fLO' D ' ' )' _41< I< L
.,
III
(M8.63)
+ 2Vk--I)~::: (which is equivalent to requiring a real
nontrivial
Case 1 (Ds = 0.3)
The reader should find the following results: M0I10d
Equilibrium I-washout Equilibrium 2-nontrivial
0 = 1.5374
Xis::::::
xl.>
X2,1':::::
4.0
x,., = 0.1565
unstable stable
Substrate Inhibition Equilibrium i-washout Equilibrium 2-l1ol1trivial Equilibrium 3-nontrivial
M8.7.6
.t Is =0
X2,
xL, = 0.9951
x2, x2,
xl.>
= 1.5302
= 4.0 = 1.5123 0.1745
=
stable unstable (saddle point) stahle
Case 2
For a steady-state dilution rate of D.).::::: 0.15, the reader should find the following results: Monod Equilibrium Equilihrium
l~washout
XLI'
L~nontrivial
Xli =
= 4.0 = 0.0474
=0 1.5811
X2.\'
=0 = -0.6104 = 1.5809
X2.\':::;:
x"
unstable stable
Snbstrate Inhihition Equilibrium I-~washout Equilihrium 2~nontrivial Equilibrium 3-nontrivial
XL,
Xli XL,
4,0 x2, = 5.5261 = 0,()478
x"
unstable not feasible stahle
Although there is a mathematical solution for cquilihrium 2, it is not physically feasible, since it corresponds to a negative biomass concentration.
548
M8.7.7 r'or a
Module 8
Biochemical Reactors
Case3
steady~statc
dilution rate of D s
:::::
0.6, the student should find the follmving results:
Mouod Equilibrium l~~washout Equilibriulll 2-nontrivial StC;:HJy~state
XIs::::
0
x",~2.0114
X2, ~
xl.,
4.0
~ ~ I J)286
stable not feasible
2 is not feasible. because it corresponds to a negative substrate concentration.
Substrate Inhihition EquilihriuI11 l,,--,washout Equilibrium 2-11ontl"ivial
Equilibrium 3-nontrivial
XIs:::::
0
x'" ~ 4.13 -
0.20j x,-,~4.13 +0.2(lj
X2., ~
4.0 -0.13 + 0.50j x,., ~ -0.13 - 0.50j
x2,
~
stable not feasible not feasible
The second and third steady-states (Irc not feasible because the concentrations for both the biomass and the StJbslratc are complex. There
LOW DILUTION HATE Case 2 had the lowest dilution rate (D,I' ::;:; 0.15). The Monod model has two steadyslates-the washout steady-state is unstable and the other (nontrivial) steady-state is stable. This means that any set of initial conditions will eventually converge to the nontrivial steady~state, for the Monod model. The substrate inhibition model has only two feasible steady-states-- the washout steady-state is unstable and the high conversion steady-state is stable. The interesting result is that at low dilution rates, the substrate inhibition model behaves like the Monod model.
MEDIUM DILUTION RATE Case I had the next highest dilution rate (D,I' ::::: 0.30). The Monod model has two steadystates----the washout steady-state is unstable and the other (nontrivial) steady-state is stable. 'fhis means that any set of initial conditions will eventually converge to the nontrivial steady-state, for the Monod model. I'he substrate inhibition model has three feasible steady··states. The w,-1shout (no conversion) steady-state is stable, the medium conversion steady-state is unstable and the high conversion (low x 2) sleady~state is stable. This means that any set of initial conditions will converge to one of the two stable steadystates. A phase-plane mllst he drawn to determine if a particular set of initial conditions will lead to washoul.
Sec M8.8
549
Bifurcation Behavior
HIGH DILUTION RATE Case 3 had the highest dilution rate (D\, ::: 0.(0). Both models had only one feasible steady-slate, the \vashout steady-state, and it was stahle. The student should he able to "sketch" phase planes for all three conditions for each of the models (Monad and substrate inhibition).
M8.8
BIFURCATION BEHAVIOR The conditions for stahility developed in Section MS.7 can be used to develop stcadystate input-output diagrams for the numerical example presented in the previous sections.
M8.8.1
Diagram for the Monad Model
The diagram for the Monod model is shown in Figure M8.5. As calculated, the Monod model has two steady-states for dilution rates that arc less than the specific growth rate under the Iced conditions, D.I, < l-L.Jx2J-). The nontrivial steady-state is stable under those conditions, while the washout steady-slate is unstable. For Ds > fL/:t2f.J there is a single ' steady-stale, lhe washoul steady-state, and it is stable.
M8.8.2
Diagram for the Substrate Inhibition Model
The diagram for the substrate inhibition model is shown in Figure M8.6. At low dilution rates, where D s < fL,JX 2f.i.) , there are two steady-slales (like the Monod model). The nontrivial steady-slate is stable under those conditions, while the washout steady~_~t~~!C is unstable. Por tbe intermediate dilution rate range, 1J-./or2j) < D s < /--L rna Jl + 2Vk 1k/ll'
Monad Model
5 unstable
stable
4r----------------....--j 3
2
o-------~ o 0.1 0.2 0.3 0.4
0.5
dilution rate
FIGURE M8.5
Input-output diagram for the Monad model,
0.6
550
Biochemical Reactors
Module 8
Substrate Inhibition Model
5
stablo \
3
,, ,,
2
,
'\. unstable
" "0'Lo
-
stable 0.1
0.2
"-
.->
0.3 dilution
FIGUHE M8.6
"-
Inplll~outpnt diagram
0.4
0.5
0.6
rate
for the substrate inhibition model.
there arc three stcady~statcs. Two of these arc stable, while one is unstable. The stable steady-slate that is attained will depend upon the initial conditions of the concentra~ lions, or on the way that the process is started lip. When the dilution rate meets the condition that D,I_ > f.1.Jx 2j ), there is a single steady-state, the washout steady-state, and it is stable.
M8.8.3
Hysteresis Behavior for the Substrate Inhibition Model
It is interesting to note that the way that the biorcactor is started up will determine the steady-state concentrations that the reactor achieves.L,ook at Figllre M8.6. NotiC\.,~ that if we start at a very low dilution rate we will have only one stable steady-state, so the reactor IIltl:;;! operate at that condition. If we slowly increasc tJ.l~~tilution ratc, we relnain on the lowcr curve of r'igure M8.6. When V,I' > ~ll1a/1 + XVk1kl/i (1\ : ;: 0.36126 for this example), the stable solution suddenly "leaps" to the upper stahle steady~state (washout cOIl(Ii~ tions). As we increase Ds Luther, we remain on the washout curve. Now, assume that we arc starting out at a high dilution rate along the upper curve, the washout conditions. As we slowly decrease the dilution rate, we remain on the washout curve until D,I' ::::: ~,Jx2Jj')' which is D,I' ::::: 0,186] for this example. The stahle steady-state then "jumps" down to the lower curve. As we continue to deCl'case the dilution rate further. we remain on thc lower curve. The type of behavior shown in Figure M8.6 is known as hysteresis and is exhibited by a number of processes, including exothermic chemical reactors and valves that "stick." The chemical reactor example is discussed further in Module 9, The student should be able to show how the phase-plane behavior changes as a function of dilution rate, for the example shown in Figure M8.6.
551
Student Exercises
SUMMARY The modeling equations for a biochemical reactor were developed for Monod and sub~ strate inhibition kinetics. We found that the Monod lnodel normally has two stcady~state solutions, while the substrate inhibition model normally has three steady-state solutions. At low dilution rates the substrate inhibition model behaves similarly to the Monod mode], with a single stable steady~state. Washout will not be a problem at the low dilution rates. At medium dilution rates the substrate inhibition model behaves quite differently from the Monod model. Depending on the initial conditions, the reactor will either con~ verge to a high conversion or to washout conditions for the substrate inhibition model. II has not been discussed thus far, but if we wish to operate at an intermediate (unstable) conversion level, then feedback control must be llsed. Notice that the Monod model still has only one stable point, and there is no danger of wash-out. At high dilution rates, both reactor models have only one feasible solution--washout. The flow is simply too high (residence tinle too low) for any cell growth.
FURTHER READING An excellent source for an introduction to biochemical engineering is: Bailey, .I.E., & D.F, Ollis. (1986), Biochemical Engineering Fundarnentals, 2nd ed, New York: McGraw-Hill.
STUDENT EXERCISES 1. In this module we developed the modeling equations assuming that no biomass is fed to the reactor, Analyze the system studied for the case where the biomass feed concentration is 2.5% of the substrate feed concentration (so xII::::: 0.1 for the numerical values used ill this module). Is there still the possibility of a washout steady-state? 2. Modify bio, phas_gen.rn and bio.rn to perform a phase-plane analysis for cases 2 and 3 with the substrate inhibition model. 3. Data for specific growth rate coefficient as a function substrate concentration for a biochemical reactor are shown below: fh, hrA-1
o
o
0.1
0.38 0.54
0.25
Biochemical Reactors
552
0.5 0.75 I 1.5
3 5
Module 8
0.63 0.66 0.68 0.70 0.73 0.74
a. Estimate the parameter values for a Monod model (kill' ~max) b. The production rate of cells (biomass) is DBxl. Find the steady-slale value of the dilution rate that maximi7.es the production rate of cells. 'fhe substrate fecd concentration is 5 gllitcr. c. Find the steady-state concentration of biomass and substrate at this dilution rate. d. Find the linear state-space model at this dilution rale, with dilution rate and substrate feed concentration as the input variables. Also find the transfer function relating dilution rate to biomass concentration. e. Simulate the responses (using the nonlinear dynamic model) or the conccnln.l" lions or biomass and substrate to step increases and decreases of IWit;, in the di~ lution rate (changes are from the dilution rate found in b.). Compare these re~ suits with those of the linear system (remelnher to convert deviation variables back to physical variables). 4. In this module we have analyzed how the biomass and suhstrate concentrations change depending on the dilution rate. If the purpose of a particular biochemical re~ actor is to produce cells, then we arc more concerned with the production rate of cells. The' production rate is mass of cells produced per unit time: steady-slale production rate of cells = D/'xls For both the Monod and substrate inhibition models present(~din this modale, find the dilution rate that maximizes the production rate of cells. Analyze the stability of the reactor under this condition.
5. Consider a biochemical reactor where the consumption of substrate (\2) promotes the growth of biomass (Xl) and formation of product (x 3 ). The three modeling lions arc:
eqlla~
dX I
lit
lil where the specific growth rate is a function of both the biomass concentration and the product concentration:
Appendixes
553
with the following parameter values:]
Variable
Value
Variable
y
0.4 gig
f'
0.2 hr-] 50 g/liter 0.04545Iitcr/g 0.202 hI' 5 glliter
"!J.
Pm
,
k, IJ -"1'-2
Value
2.2 gIg max
kill x21 xJ J.-:;
OAg lu·-- J 1.2 g/litcr 20 gil iter 6 gIl iter 19.14 g/liter
Compare and contrast this rnode! with that of the two-state model with substrate inhibition kinetics presented in this module. h. Verify that the steady-state values for Xl' x 2 , and x] presented in the table above are correct. For a steady-state input of f):::;: 0.202 (and all of the other parameters constant), arc there (my additional solutions for the states (for example, the trivial solution?). Analyze the stability of all steady~state solutions obtained. c. Perform dynamic simulations of the nonlinear model, with step changes of ± 10(#) in the dilution rate. Discuss the resuIts or your step changes (Le., docs an increase or ~lecrease in f) have a greater effect on the biomass concentralion?). Compare your results with linear simulations. H.
APPENDIXES
1
Steady-State Biochemical Reactor Model, bio~ss. m
function f :::,
bio~ss
(x)
%
% b.w. bequette % (e)
16 Nov 92
% revised 18 July 96 %
% find steady-states of bioreactor, using fsolve: %
%
x::::: fsolve( 'bio' ,xOl
IThis model is from Chapter 4 of the following Illonograph:Henson, M.A., & D.E. Seborg (cd.). (1997). Nonlinear Process Conlrol. Upper Saddle River, NJ: Prentice-Hall.
Biochemical Reactors
554
Module 8
% where xO is a vector of initial guessef3 biomas,;.>
% x(l) ~8 x (2) %
substrate
% biomass CI,
D
% y 96 mu 'l, mumax ~f, km "6 kl 'r, sf
("bugs") consumes the substrate dilution rate (F/V, timeA-l) yield biomass/substrate specific DrowLh rate param(:!Ler (both Monad and Substrate Tnhibi tion) parameter (both Monad and Substral~e Inhibition)
parameter (Substrate:; Inhibition only, subf3trate f(x~d concentration
kl : : : () for Monod)
'?6
% the function vector consists of 2 equations % f 96
~
zeros (2, 1) ;
parameter values D ~ 0.3; mumax 0:: 0.53;
Y ~ 0.4; km 0.12; sf 4.0; kl
%
0.4545;
Substrate Inhibition expression for specific growth rate
% InU
~
mumax k x(2)/(km+x(2)+kl*x(2)*x(2));
90
% %
steady-state equations [solve varies xll) and x(2)
f(l) f(2)
2
(mu
to drive f(l)
and f(2)
to Zero
D)*x(l);
(sf - x(2) )'D - mu*x(l) /Y;
bio.m, Function File for Dynamic Simulation Using ode45
function xdot = bio(t,x} % b.w. bequett.e % (c) 18 ~July 96 % 5(, dynamic equations for biorcactor, (i; usiuq the following command % [t,x]
= ode4.5( 'bio' ,t.O,tf,xO)
integrated using ode4.5,
555
Appendixes
% where to is the initial time (usually 0), tf is % the final time and xO is the initial condition vector % xO (1) biomass initial conclition % xO(2) substrate initial condition % 96 biomass ("bugs") consumeS the substrate '6 % state variables
% % x(l) 't x(2}
biomass substrate
q.
"
- dil.ution rate (P/V, timeA-l) yield biomass/substrate specific growth rate mu mumax - parameter (both Monad and Substrate Inhibition) parameter (both Manod and Substrate InhiJyit.1.on) km parameter (Substrate Inhibition only, kl :::; 0 for Monad) % kl 96 sf substrate feed concentration
90
D
% % % %
y
%
% the function vector consists of 2 equations I % % %
zeros(2,1);
param(~t~(,~r
values
D O . 3;
mumax 0.53; Y = 0.4; km
0.12;
sf - 4.0; kl % %
0.4545;
Substrate Inhibit_ion expression for specific growth rate mu :::; mumax-k x(2) / (km+x(2) +k1-k x(2) *x(2));
1-s 90
dynami c equa t i ODS xdOL (1) xdot(2)
3
{mu
D)'x(l);
(sI
x(2))*0 - mu*x(l)jY;
Phase-Plane Plot for Biochemical Reactor, bio---phas_gen.m
:6 'A b.w. bequett.e '6 (e) 19 July 96 %
% generates phase-plane plots for the bioreactor
Biochemical Reactors
556
Module 8
%
% set-up the axis limits %
axis{[O 1.75 0 5]); %
% stable and unstable points for substrate % inhibition model % xlu x2u
[0.9951]; [1.5122];
xls x2 s
[0;1.'j302J; [4; 0 . 1'146] ;
%
% place an 'x' on stable points % place a 'a' on unstable points %
plot{xlu,x2u, 'wo' ,xls,x2s, 'wx')
hold on %
% select different initial conditions % xl ranges from 0.1 to 1.5 (every 0.35)
a
:!, x2 ranges from
to 5
(every 1.25)
% total of 16 initial conditions %
xlinit x2 ini t xlinita x2 lni ta
=
[0.1 0.45 0.8 1.15 1.5 0.1 0.45 0.8 1.15 1.5]; [0 0 0 0 0 5 5 5 5 5] ; [1.5 1.5 1.5 0.1 0.1 O.lJ; [1.25 2. 5 3.75 1.25 2. 5 3.75] i
%
xG = [xlinit xlinita;x2init x2inita] ; %
% ncol
number of initial conditions
=
%
[mrow,ncol]
size(xO)
i
% run simulations for each initial condition %
for i
;:c.
1:nc01;
% [t,xJ
"ode45('bio',0,30,[xO(:,i)]);
%
plot (x ( : ,1) ,x ( : ,2)
I
'w' )
% end %
xlabel ( . xl ' ) ylabel ('x2')
hold off
______________.L !
Appendixes
4
557
Direct Calculation of the Eigenvalues
Here we calculate the eigenvalues of the nontrivial solution:
= A2 - (r(A)A + dc(A) = 0
del(A/- A)
(MXAO)
where the trace and determinant arc
(MXA5) (MXA6) For the nontrivial solution,
11-.1'':;:::: f),I':
(r(A) c= -D, -
11-,;'X 1,1'
Y
(MX.AI)
X Is 11-,~. 11-.1'. (cl I (A) ,-
(MX.A2)
Y
The rools of (MXAO) arc: - 4 de( A
A-
(MX.A3)
2
(MX.A4)
2
+
+
2
y
2
4 .r,_ fL,;' /.1", Y
(MX.A5)
2 x.,_ /-L,;' /Ls y (MX.A6)
± A=
and our roots are:
x';:r
2 y
A=
~(=;L, -
± 2
(I~'_
(MX.A7)
x'p:) (MX.AX)
558
Biochemical Reactors
A.[ = --/1-)'
and
Module 8
(MX.A9)
and since /1-.1' ::::: D,I" we are assured that one pole will always he negative. The second root will only be positive if /1-.: is negative. Since /1-.: is positive for the Monod model, the nontrivial solution is stahle as long as the solution is feasible (D.I, < /1-./x2f~.)). The fL; can be either positive or negative for the substrate inhibition model, so a J1()[]trivial slcadystate may either be slable or unstable.
DIABATIC CONTINUOUS STIRRED TANK REACTORS
MODULE
9
The purpose of this module. is to understand the steady-state and dynamic behavior of jacketed continuous stirred tank reactors (CSI'Rs), also referred to as diabatic CSTRs. After reviewing this rnodulc the student should be able 10 Understand the assumptions made in developing the classic CSTR model Understand the possihle steady-state behavior based on an analysis or the heat gellcrated by reaction ~lI1d removed through the cooling jacket Understand what is meant by multiple steady-states. Develop the hysteresis (1~1l1 lion/extinction behavior) input-output diagrams based on analysis of the heat gener-
ation and rClllOval curves. [(clate these to saddlc-llOdc and hysteresis bifurcations from Chapter 15 Understand the dynamic behavior by linearization of nonlinear dynamic equation:.; followed by eigenvalue analysis Use phase-plane analysis to understand which steady-state a given initial condilinll will converge to Realize the possihility or limit cycle behavior (llopfhifurcations) 'rhe l11i:\jor sections inthis module are: M9.1
Background [~quatiolls
M9.2
The Modeling
M9.3
Steady-State Solution
M9.4
Dynamic Behavior
M9.5
Linearizali(m of Dynamic
M9.6
Phase-Plane Analysis
[~qllations
559
Diabatic Continuous Stirred Tank Reactors
560
Module 9
F r-
~
r-
TjlrJ
Fi
LU CA
T
F
T,
FIGUH.E M9.1
M9.7
UlldcrstandingMulliple Steady-State Behavior
M9.S
Further Complexities [)illlcnslonlcss Model
M9.9
M9.1
Continuous stirred tank reactor with cooling jacket.
BACKGROUND The most important unit operation in a chclnical process is generally a chemical reactor. Chemical reactions arc either exothermic (release energy) or endothermic (require energ) input) and therefore require that energy either be removed or added to the reactor for d constant temperature to he maintained. Exothermic reactions arc the most interesting systems to study because or potential safety prolJ]cll1s (rapid increases in tcrllpcraturc, S(HllCtimes called "ignition" behavior) and the possibility of exotic behavior such as Illultiple steady-states (for the same value of the input variable there may be several possible val~ ues of the output variable). In this Illodule we consider a perfectly mixed, continuously stirred tank reactor (CSl'I<-),l shown in Iiigure M9.1. The case of a single, first-order exothermic irreversible reaction,;1 -> 11, will be studied. We will find that very interesting behavior can arise in such a simple system. In I<'igure M9.1 we see that a fluid stream is continuously fed to the reactor and another fluid stream is continuously removed from the reactor. Since the reactor is perfectly mixed, the exit stream has the same concentration and temperature as the reactor fluid. Notice that a jacket surrounding the reactor also has feed and exit streams. The jacket is assumed to be perfectly mixed and at a lower temperature than the l'C
ISomctimes this type or reactor is called "non-adiabatic'" The term adiabatic means "no he,ll loss," so the term nOli-adiabatic cOllstitu{es a double negative (This point has been disclissed h) Barduhn, Chelll. Eng. Education, 19, J 71 (19X5)). We prefer to usc- diabatic to describe this reactor.
Sec. M9.2
The Modeling Equations
561
There are many examples of reactors in industry similar to this olle. Examples include various types of polymerization reactors, which produce polymers that arc lIsed in plastic products such as llOlystyrcnc coolers or plastic bottles. The industrial reactors typically have more complicated kinetics than we study in this module, but the characteristic behavior is similar.
M9.2
THE MODELING EQUATIONS For sitnplicity we assume that the coolingjackct temperature call be directly manipulated, so that an energy balance around the jacket is not required. We also make the following assumptions: Perfect mixing (product strcanl values arc the same as the bulk reactor fluid) Constant volume COnsl,lI1( parameter values
The constant volume and parameter value assumptions can easily be relaxed by the reader for further study.
M9.2.1
Parameters and Variables
The parameters and variables that will appear in the modeling equations arc listed below for convenience.
A
Area for heat exchange
(~\
Concentration of A in reactor
C~V
Concentration of A in feed stream
cp
Heat capacity (cnergy/mass*tcmperature)
F
Volumetric f10wrate (vohltne/time)
ko
Pre-exponential factor (timc-- J )
R
Ideal gas constant (energy/mol*tcmperature)
r
Rate of reaction per unit volume (mol/volume'; time) Time
T
Reactor temperature
Ii
Feed temperature
T.I
Jacket temperature
I~'(l U
Overall heat transfer coefficicnt (cnergy/(timc*area*tcmpcrature))
Reference temperature
\I
Reactor volume
ali
Activation energy (energy/mol)
(-a H)
Heat of reaction (energy/mol)
II
Density (mass/volume)
ESCOLA Dc ENGEN HARIA 616LIOTECA
562
M9.2.2
Diabatic Continuous Stirred Tank Reactors
Module 9
Overall Material Balance
The rate of accumulation of material in the reactor is equal to the rate of rnatcrial in by flow - the material out by flow.
Assuming a constant amount ofrnatcrial in the reactor (dVp/dt:::::: 0), we find that:
If \VC ;:I1so assume that the density remains constant,2 then:
F
M9.2.3
and
dV/dt
~
0
Balance on Component A
The halance on component A, assuming a constant volume reactor, is:
V
(11191)
dt
where r is the rate of reaction per unit volume.
M9.2.4
Energy Balance
The energy balance, assuming constant
Vpc"
dT dt
~
VOIUlllC,
heat capacity and density, is:
FjlC,,("/"t- T) + (-t:.!IWr- UA("/"- "Ii)
(1119.2)
where (-liB) VI' is the rate of energy contributed by the exothermic reaction.
M9.2.5
State Variable Form of Dynamic Equations
We can write (M9.1) and (M9.2) in the following state variable form:
t(C;\' T) ~ ,1
dCII dl
i/1' dt
F V (CAl
IV" (1;- T)
(1119.1 a)
+ (...- !1II) r - UA (1' 7i) pc!,
Vpc p
(1119.2,,)
2It should he noted that the dellsity of all streams docs not need to remain constant for the modeling equations to holtL F'-or example, DCllJl (1986) shows that as long as the density is a lincar function of concentratioll, the final modeling equations are correct.
Sec. M9.3
Steady-State Solution
563
The reaction rale per unit volume (Arrhenius expression) is:
r
= k o exp!, i C" (- !1F)
(M9.3)
where we have assumed that the reaction is firshOlrdcr.
M9.3
STEADY-STATE SOLUTION The steady-state solution is obtained when dC,//dl ::: 0 and dlldl :::: 0, that is:
(M9.1 s)
VA (T- T)
Vpc p
("19.2s)
I
To solve these two equations, all parameters and variables except for two «(~\ and T) must be specified. Given numerical values for all of the parameters and variables we can lise Newton',,; method (Chapter 3) to solve for the steady~sLatc values of C~l and T. l<'or convenience, we usc an "s" subscript to denote a steady-state value (so we solve for CAs and [,J. In this module we will study the set of parameters showll for Case 2 conditions in 'Table M9.1. Cases I and j arc left as exercises for the reader.
M9.3.1
Solution for Case 2 Parameters Using fsolve and cstr_ss.m
The function nl~file for the steady~state equations is cstr_ss . m and is shown in dix I. The command to run this file is X
:=
Appen~
fsolve( 'cstr_ss' ,xO);
where xO is a vector or the initial guesses and x is the solution. lkfore issuing this com~ mand the reactor parameters must be entered in the global parameter vector CS'I'R PAR. TABLE M9.l
Real'tor Parameters
Parameter
FlV. hr ko, hr l
Case 1
1
(~-!J.H), kcallkgrno! E, kcal/kgmol pC'f" kcaJ/(m:'\QC)
I"(~J'"Ckgltlollm
J
UA/V, kcal/(m:\oC hr)
'Ij, "C
I 14,825*3600 5215 11,:-\43
500 25 10 250
25
Case 2
Case :3
I
I
9,703"'3600 5%0 11,843 500
18,194*'3600
25 10 150 25
Xl95
11,843 500 25 10 750
25
564
Diabatic Continuous StIrred Tank Reactors
Module 9
We find that different initial guesses for the concentration and temperature lead to diffcr~ cnt solutions. When choosing initial guesses for a numerical algorithm, it is important to usc physical insight about the possible range of solutions. For example, since the feed concentration of A is 10 kgmollm 3 and the only reaction consumes A, the possible range for the concentration of A is 0 < Ct\ < 10. Also, it is easy to show that a lower bound for temperature is 298 K, which would occur if there was no reaction at all, since the feed and jacket temperatures arc 298 K. Notice also that there should be a correlation betwc-en concentration and temperature. If the concentration of A is high, this means that not much reaction has occurred so Iittlc energy has been released by reaction and therefore the temperature will not be rtlllch different than the feed and jacket temperaturcs.
GUESS 1 l1if.:h cOll(..entratioll (1011' conversion), Ienv temperature. lIciT we consider an initial guess of (, =') and T = 300 K. X
:::::
fsolve{ 'cstr ss', [9;300]);
x
8.5636 311.1710
so the steady-state solution for guess 1 -is I-C'r,-,,] conversion) and low-temperaturc.
GUESS 2 Intermediale concentration and temperature. X
::::
fsol ve ( 'cstr ss ' , [5; _350] )
x 5.5179 339.0971.
so the steady~state solution for guess 2 is [(;-_:'-1
GUESS 3 Lmv concentration and high temperature. x
fsolve('cstr ss', [1;450J)
x 2.3589 368.0629
I-~._,112' )(,.:1('1 that is-, hioh concentration <.10\1/ l:>
Sec. M9.4
565
Dynamic Behavior
TABLE Mt).2
Guesses and SOIUtiOIlS[Jsillg fsol ve (,uess I
Guess and Solution xO ( 1 ) , C~\ guessed xO (2 ) , T guessed x ( 1 ) , C~\ solution x (2), T solution
CJUCSS
2
(ILlCSS
9
5
I
300
350
450
8.564 311.2
5.518 339.1
2.359 368.1
3
l(r"1
so the stcady-stale solution for guess 3 is l,;;,~~~I, that is, low concentration (high conversion) and high temperature. ' The results arc summarized in TablcM9.2, Other initial guesses do not lead to any other solutions, so we sec that there arc three possible solutions for this set of parameters. In Section M9.7 we show how to use physical insight to determine the Humber of steady-state solutions for this problem.
M9.4
DYNAMIC BEHAVIOR We noted in the previous section that were three different steady-state solutions lo the Case 2 parameter set. Here we wish to study the dynamic behavior under this same parameter set. Recall that nlilIlerical integration techniques were presented in Chapter 4. The m-file to" integrate the modeling equations is cstr_ dyn.m, shown in Appendix 2. The command to integrate the equations is [L,x]
=
ode45( 'cstr__ dyn', to,tf,xO);
where to is the initial time (usually 0), tf is the final time, xO is the initial condition veClor, t is the time vector, and x is Ihe slate variable solution vector. lJcfore performing the inlegration it is necessary 10 define the global parameter veclor CSTl{ PAR. '1'0 plot only concentration or temperature as a function of time, use plot (L, x ( : ,1)) and plot (t, x ( : 2) ), respectively. J
M9.4.1
Initial Condition 1
Here we use initial conditions that arc close to the low temperature steady-state, The initial condition vector is [conc, temp] :;:: [9,300]. The curves plotted in r,'igure M9.2 show that the state variables converge lo the low temperature steady-stale.
M9.4.2
Initial Condition 2
Here we usc initial conditions that arc close to the intermediate temperature steady-state. The initial condition veclor ror the solid curve in Figure M9.3 is [conc, temp] :;;: [5,350], which converges to the high temperature steady-slale. The initial condition vector for the
Diabatic Continuous Stirred Tank Reactors
566
Module 9
9.5
"c " 0
9 8.5 0
2
4
6
8
10
6
8
10
time, hr
"'"
320
ID
"0:
310
'"-
300
E
0
2
4
time, hr
FIGURE l\iI9.2
State variable responses with initial condition xO::::: [9;300j.
dotted curve in Figurc M9.3 is leonc, tcmpl 0::::: fS,32Sl, which converges to the low temperature steady-state. If we perform lllallY simulations with initial conditions close to the intermediate temperature steady-state, we find that the telnpcraturc always converges to either the low temperature or high temperature steady-states, but not the intennediate temperature steady-slate. This· indicates to us that the intermediate tcmperature steady-state IS lIflstable. This will be shown clearly by the stahility analysis in Section M9.5.
10
" " c 0
5 0 0
2
4
6
8
10
6
8
10
time, hr
"'"
400
"'"0:
350
'"-
300
E
0
2
4 time, hr
FIG1JRE 1\19.3 Stale variable responses with initial condition (solid) and xO:;;: [5;325] (dashed).
~\"O
:;: 15;3501
Sec. M9.5
567
Linearization of Dynamic Equations 4
u
c
2
0
u
0 4
2
0
6
8
10
6
8
10
time, hr
" '"'"0:
400
2
360
'0
E
380
4
2
0
time, hr
FIGURE 1\'19.4
M9.4.3
State variable responses with initial condition xO ::00 [I ;400].
Initial Condition 3
Ilere we usc initial conditions that arc close to the high temperature steady-state. The initial condition vector is [COile, temp-l -= f 1,4001. The curves plotted in P'igureM9A show that the state variables converge to the high temperature stcady~statc. In this section we have pcrfonncd several simulations and presented several plots. In Section M9.6 we will show how these solutions can be compared on the saille "phascplane" plot.
M9.5
LINEARIZATION OF DYNAMIC EQUATIONS The stability of the nonlinear equations can be dClcnnincd by finding the following statc-
space form:
x=Ax+BII
(M9.11)
and dctcrmining the eigenvalues of thc A (state-space) matrix. The nonlinear dynamic state cquations (M9.la) and (M9,2a) arc:
dC., dt dT dt
(M9.1 a)
( p;',;I) k C ~ CT. el
UII
vpe"
T+
UII T VIIC I
"
+ FT V
(M9.2a) f
568
Diabatic Continuous Stirred Tank Reactors
Module 9
L,ct the state and input variables be defined in dcviittiou variable form:
CA'J
x '" I. c" T- T "
II
M9.5.1
=-~
r
J TTJS
l
C:'ilf - CilI~'
Ii - Ii,
Stability Analysis
Performing the linearization, we obtain the following clemellts for A:
il/;
il/, ile.,
ax, aI, at, ax} aT ai, a/;
All
.11 12
/\21
An
-
F
V ..
- ks
- C A /< (- 6./1)
ks
ax!
aCA
pCI'
a/;
il/;
F
(fA
V
VpC~)
aT
ax')
+
(6.11) pCp
CA,lk:
where we define the following parameters for more compact representation:
k,
.=
k exp () ak,
aT
( RT6.F). · 1
(- 6.1') (6.E ih'? )
k" exp . fiT,
or,
From the analysis presented above, the state-space A matrix is:
F
V ·-k\
A =
[
(-6.11), ' ( p,p
(M9.12)
J(\
The stability characteristics arc determined by the eigenvalues of A, which are obtained by solving del('\' I - A) ;::; o.
Sec. M9.5
Linearization of Dynamic Equations
A1- A
det()" 1- A)
I,..
569
A-A" -An
()" - A,,)(A - A 22 ) -- ),,2_
-
A"A 2I
(A" + A 22 »)" + A"A n - A"A"
- ),,'-(trA»)"
+ del (A)
'file eigenvalues are the solution to the second-order polynomial:
A' - (lr A) A + det (A)- 0
(M9.13)
'fhe stability of a particular operating point is determined by finding the A matrix for that particular operating point, and finding the eigenvalues of the A matrix.
Here we show the eigenvalue..., for each of the three case 2 steady-state operating points. We usc the function routine cstx_"amat. ill shown in appendix 3 to find the A matrices.
OPERATING POINT 1 The concentration and temperature arc 8.564 kgmol/m 3 and 31 J.2 K, respectively. »
[ama t. , 1 ambc1a ]
amat = -1.1680 2.0030
cstr_amatI8.564,311.21
-0.0886 -0.2443
lambda : : ; -0.8957 '-0.5166
Both of the eigenvalues arc negative, indicating that the point is stable, whieh is consistent with the results ol'Figure M9.2.
OPERATING POINT 2 The concentration and temperature arc 5.51 Hand 339.1, respectively. »[amat,lambda] amat = --1.8124 9.6837 lambda = -0.8369 0.4942
=
cstr amat(5.518,339.1)
-0.2324 1. 4697
Diabatic Continuous Stirred Tank Reactors
570
Module 9
One of tile eigenvalues is positive, indicating that the point is ullstable. This is consistent with the responses presented in Figure M9.3.
OPERATING POINT 3 »[amat,lambda]
~
cstr_amat(2.359,368.1)
amat :::; ~4.2445
~O.3367
38.6748
2.7132
lambda :::; ~O.7657
+
~O.7657
~
O.9584i O.9584i
The rcal portion of each eigenvalue is negative, indicating that the point is stable; again, this is consistent with the responses in Figure M9.4.
M9.5.2
Input/Output Transfer Function Analysis
The input-output transfer functions can be found from:
G(s) = C(sI - A) 'B where the clements of the H matrix corresponding to the first input (lIl =
iJj; =
all,
aj;
1~.
- 7jJ arc:
=0
a'lj
a/2 = aI,
= -_
dU l
(M9.14)
aTj
=
VA VpCp
The reader should find the elemenls of the B matrix that correspond to the second and third input variables (sec studcnt cxercise 8). llerc wc show only the transfer functions for the low temperature steady~state for Case 2. Thc input/output transfer function relating jacket tcmperature to reactor concentration (state I) is:
-0.0266
+ l,4123s + 004627
- 0.0575 (1.I165s + 1)(l,9357s
+ 1)
and the input/output transfer function relating jacket temperature to reactor temperature (state 2) is
g2l(S)
O.3s + 0.3504 + 104123.1' + 004627
07573(0 856s + I] (1.1165.1' + 1)(1.9357.1' + I)
Notice that the transfer function for concentration is a pure second-order system (no numerator polynomial) while the transfer function for temperature has a first-order nutllcra-
Sec. M9.6
Phase-Plane Analysis
571
tor and second-order denominator. This indicates that there is a greater "Jag" bCl\vCen jacket temperature and concentration than between jacket temperature and reactor temperature. This Inakcs physical sense, because a change in jackel temperature must first affect the reactor temperature before atTccling the reactor concentration.
M9.6
PHASE-PLANE ANALYSIS In Section M9.4 we provided the results of a fc\v dynamic simulations, noting that diffcr~ cnl initial conditions caused the system to converge to different steady-state operating points. In this section we construct a phase-plane plot by performing simulations for a large nUlubcr of initial conclitions. The phase-plane plot shown in Figure M9,5 was generated using cstr_]un. m and cstr.m [rom the appendix. Three steady-state values arc clearly shown; two are stable (the high and low temperature steady-states, shown as '0'), while one is unstable (the intennediate temperature steady-state, shown as '-t'). Notice that initial conditions of low concentration (0.5 kgmolhn-1) and relatively low-to-intermediate temperatures (300 to 365 K) all converge to the low-telnperature steady-state, When the initial temper~ ature is increased above 365 K, convergence to the high-temperature steady-state is achieved. Now, consider initial conditions witb a high concentration (9.5 kglllollm 3) and 1m\' temperature (300 to 325 K); these converge to the low-temperature steady-state. Once the initial temperature' is increased to above 325 K, convergence to the high-temperature steady-state is achieved. Also notice that, once the initial temperature is increased to around 340 K, a very high overshoot to above 425 K occur;.;, before the system settles down to the high-temperature steady-state. AHhough not shown on this phase-plane plot, higher initial temperatures can have overshoot to over 500 K before settling to the hightemperature steady-state, This could cause potential safety prohlenls if, for example, secondary decomposition reactions occur at high temperatures. The phase-plane analysis then, is able to point out problem initial conditions. Also notice that no initial conditions have converged to the intermediate telnpcrature steady-state, since it is unstable. The reader should perform an eigenvalue/eigenvector analysis for the A matrix at each steady-state (low, intermediate, and high temperature) (sec student exercise 3). You will find that the low, intermediate, and high temperature steady-states have stable node, saddle point (unstable), and stahle focus behavior (sec Chapter 13), respectively. It should be noted that fcedback control can be used to operate at the unstable intermediate temperature steady~state. The feedback controller would measure the reactor temperature and manipulate the cooling jacket tcmperature (or f1owrate) to m8intain the intermediat.e temperature steady-state. Also, a feedback controller could be used to make certain that the large overshoot to high temperatures docs not occLir from certain initial conditiclIls.
Diabatic Continuous Stirred Tank Reactors
572
Module 9
cstr with multiple steady+states
420
400
'"
380
'" ID
"0
*~ 360 iiiID "-
~
- 340
320
300
o
2
4
6
8
10
concentration, kgmol/ml\3
FIGURE M9.5 arc marked with
M9.7
Phase-plane plot for Case 2. Stable points (conditions I and:1) The unstable point (condition 2) is marked with '+'.
'0'.
UNDERSTANDING MULTIPLE STEADY-STATE BEHAVIOR In previ()us sections we found that there were three steady-state solutions for Case 2 parameters. The objective of this section is to determine how multiple steady-slates might arise. Also, we show how (0 generate steady-state input-olltput curves that show, for example, how the steady-Slate reactor temperature varies as a function of the steady-state jacket temperature.
M9.7.1
Heat Generation and Heat Removal Curves
In Section M9.3 we used numerical methods to solve for the steady-states, hy solving two equations with two unknowns. In this section we show that it is casy to reduce the two equations in two unknowns to a single equation with one unknown, This will give us physical insight abollt the possible occurrence of Illultiple steady-states.
Sec. M9.7
573
Understanding Multiple Steady-State Behavior
SOLVING FOR CONCENTRATION OF A AS A FUNCTION OF TEMPERATURE 'fhe steady-slate concentration solution (dCidt) ::: 0) for concentration is: (M9.15)
We can rearrange this equation to rind the steady-state concentration t{)r any given steady-state reactor temperature, 7:1,:
(M9.16)
SOLVING FOR TEMPERATURE The steady-state temperature solution (dT/dt :::: 0) is:
UA
-(11) --- k ("'f-~(~~)
F
II (. ( , p ""
((j.{') - C R J~'
ex') """. 0
1
As
(M917)
The tcnns in (M9.17) arc related to the energy removed and generated. IT we ll111hiply (M9.17) by VpC/J. we find that:
Fe I- UA(F.\ ]- . F.) P - fI (F.\ - F) ',' Js 1
~.
(M9.18)
Q[':ell
I-~Ilcrgy
removed by flow and heat exchange
Heat generated by reaction
Note the form of Ql'em (M9.19)
Notice that this is an equation for a line, \vhcrc the independent variable is reactor temperature (T\). The slope or the line is I UA +- FpCpl and the intercept is [-UA Tj.\. Fre,/I:,]' Changes in jacket or feed tempen.llllre shirt the intercept, but not the slope. Changes ill [fA or F affect both the slope and intercept. Now, consider the Qgen term: (M921l)
Substituting (M9.l6) inlo (M9.20), we find that:
Diabatic Continuous Stirred Tank Reactors
574
k"
cx p ( /:/) -Mf V '
Module 9
(& C,v,)
(M9.21)
k Cxp( RT -!J.F)
I' + V ()
j
[~qLlalion
(M9.21) has a characteristic "S·, shape for Q"ell as a function of reactor temperature. From equation (M9.18) we sec that a stcadY~~late solution exists when there is all intersection of the QIClIl and Qgcll curves.
M9.7.2
Effect of Design Parameters
In FigureM9.6 we show different possible intersections of the heat removal and heat generation curves. If the slope of the heat removal curve is greater than the maximum slope of the heat generation curve, there is only Olle possible intersection (sec Figure M9.6al. As the jacket or feed temperature is changed, the heat removal lines shifts to the len or right, so the intersection can be at a high mlow temperature depending on the value of jacket or feed temperature. Notice that as long as the slope of the heat removal curve is less than the maxinlllm slope of the heat generation curve, there will always be the possibility of three intersections (see l<'igure M9.6b) with proper adjustment of the jacket or feed temperature (intercept). If the jacket or feed temperature is changed, the removal line shifts to the right or left, where only one intersection occurs (either low or high temperature). This case is analyzed in more detail in Section M9.7.3.
M9.7.3
Multiple Steady-State Behavior
In Figure M9.7 we superimpose several possible linear heat removal curves with the S-shaped heat generation curve. Curve A intersects the heal generation curve at a low temperature; curve n intersects at a low temperature and is tangent at a high temperature:
Q
Q
Reactor Temperature (Tr)
a. FIGURE 1\119.6
curves.
Reactor Temperature (Tr)
b. Possible interscctions of heat gener::ltioll and heat rcmoval
Sec. M9.7
575
Understanding Multiple Steady-State Behavior E 9
Q
Temperature
FIGURE !\/19.7 CSTR energy generated and energy removed as a function 01 reactor temperature.
curve C intersects at Imv, intermediate, and high lclnpcraturcs; curve D is tangent to a hl\\ temperature and intersects at a high temperature; curveE has only a high temperature in tersection. Curves A, B, C, D, and E arc all based on the same system parameters. except that the jacket temperature increases as we move from curve A to E (from equatioll (M9.7) we sec that changing the jacket temperature changes the intercept but not the slope of the heat removal curve). We can usc Figure M9.7 to construct the steady-state inputoutput diagram shown in !-<"igurc M9.8, where jacket temperature is the input and rC
'Also knowll as
saddle-node bifurcation point. Sec Chapter 15.
576
Diabatic Continuous Stirred Tank Reactors
c
B
A
o
Module 9
E
9 8
7
6
T
5
4
3 2
Jacket Temperature
FIGlJREM9.8
Reactor temperature a~ a function of jacket temperature.
one starts at a high ja~ket temperature (point 9) there is a single high reactor temperature, which decreases as the jacket temperature is decreased (points 8 and 7). As we move slightly lower than the high temperature limit point (point 6), the reactor temperature drops (also known as extinction) to a low temperature (point 2). Further decreases in jacket temperature lead to small decreases in reactor temperature. The hysteresis behavior discllssed above is also known as ignition-extinctioll behavior, for obvious reasons. Notice that region he tween points 4 and 6 appears to be unstable, because the reactor docs not appear to operate in this region (at least in a steady-state sense). Physical reasoning for stability is disclissed in the following section.
M9.7.4
Multiple Steady-States: Stability Considerations (Steady-State Analysis)
Consider the system described by the energy removal curve C in Figure M9.7. Notice that the steady-state energy balance is satisfied for the operating points 3, 5, and 7, that is, there are three possible steady-slates. What can we observe about the stability of each of the possible operating points, solely froTH physical reas(Hling?
LOWER STEADY-STATE (OPERATING POINT 3) Consider a perturbation from operating point that is colder than point 3, say T3 - 01'. At this point we arc generating more heat than can be removed by the reactor, so the lempcr-
Sec. M9.7
Understanding Multiple Steady-State Behavior
577
alllte begills to risc··---cvenlually moving to 1:-". If we start at T] + 0[, more energy is being rcrnovcd than we are generating, so the temperature hegins to dccrcasc~cvcntually Inoving to 7'-3'
The lower lcmperalut'c intersection,
TIt
may be a stahle operating point.
'fhe open-loop stability cannot be assured until an eigenvalue analysis is pcrl<mncd (sec Section M9.5).
MIDDLE STEADY-STATE (OPERATING POINT 5) If we start at 7:') - oT, we arc generating less energy than is being removed, causing T to decrease, eventually causing the temperature to go to 1:1 , [f we slart at '1:") + 0'1', more energy is being generated than is being removed, causing the temperature to risc, increasing until T, is reached.
The middle temperaturc intersection, 7'5' is an IIflswble operating point.
Here the steady-state analysis is enough to determine that point 5 is an unstable operating point.
UPPER STEADY-STATE (OPERATING POINT 7) Using the same arguments as
1:
1,
we find:
The high tcmperature intersection, 1 7, may be a stahle operating point.
Again, the stability or the upper steady~state can only be determined by eigenvalue analysis (SectionM9.5). We h;:IVC used physical reasoning to determine the open.. loop characteristics solely frOill steady-state analysis. Recall that when we studied the Case 2 conditions in SeLlion MY.S, we used eigellvalue analysis to find that the lower and upper temperature steadystates were stable, while the intennediate steady-state was unstable,
578
M9.7.5
Diabatic Continuous Stirred Tank Reactors
Module 9
Generating Steady-State Input-Output Curves
There arc two different ways to find how the steady-state rcaclOt temperature varies with the jacket temperature. One way is to solve the nonlinear algcbn1ic (steady-state) equations for a large number of different jacket temperatures and construct the curve rcl;1ling jacket to reaclor temperature. It turns out that there is another, simpler, way for this particular problem. lIcre we can easily solve for concentration as a function of reactor temperature. \Vc can also directly solve for the jacket tcmpcrallirc required for a particular reaCl(lr temperature. Then we simply plot reactor temperature versus jacket temperature. We show the calculations in detail below. As in Section M9.7.1, we can find the steady-state concentration for any steadystate reactor temperature, 7~\, (frotH the material balance equation for reactant A):
F V C'Ah
F + I< exp (-111\) -- --V () RT\ Prom the steady-state temperature solution (dTldt jacket temperature, Z:.s:
Fpc/T, T,
+
"liJ - (-
=:
0) we can solve for the steady-state
I11I)VI<" (fll
Similarly, if we assume that the jacket temperature remains constant, we can solve for the required feed temperature:
+ IUII(T, - -(.J - (- /:'11) VI<" eXI{~?~~\)
T \'
eeL,]
Fpcp
The function file csU','_io.m shown in Appendix 4 is used to generate the input-output curves for the Case 2 parameters shown in l<'igures M9.9 and M9.IO. »'I'empvec 300 :2 . 5 : 380 ; »[Tjs,fl'fs,conc] ::c: cstr,.io(Tempvec); »plot(Tjs,TC"mpvec) »plot (Tfs, 'rempveC')
M9.7.6
Cusp "Catastrophe"
In Chapter 15 we presented a simple example of a cusp catastrophe that occurs when the input-output behavior Illoves from monotonic through a hysteresis bifurcation to multiple steady-state behavior. This can be shown clearly in Figure M9.11 where we find thaL as we increase PIV, we move from monotonic to multiple steady-state behavior.
Sec. M9.7
Understanding Multiple Steady-State Behavior
579
380 370
"
0>
360
~
"0
1"
~ ~
350 340
0-
E
2
330
~
320
g ~
310 300
275
280
285
290
295
300
305
310
315
320
jacket temperature, dog K
FIGURE LVI?? Stcady~state reactor temperature as a function of jacket tcmpe-ralurc, Case 2 parameters.
380
"
370
"0
360
0> ~
~
" ];;
350
~
0-
E
340
~
330
~
320
2
~
310 300 290
292
294
296
298
300
302
304
feed temperature, dog K
FIGlJRE M9.10 Steady-state reactor temperature as a function of feed temperature, Case 2 paranlctcrs.
306
580
Diabatic Continuous Stirred Tank Reactors
Module 9
400 380 360 ~
I
0.
E
340
w
f-
320 300 280 0.2 0.4
0.6 0.8 FIV
Temp..1
FIGURE 1\/19.11 Cusp catastrophe for the diabatic CSTR, Case 2 conditions. The input/olltput (jacket tcmpcraturc/rcaclor temperature) behavior changes as a functioh of the space velocity (NV).
M9.8
FURTHER COMPLEXITIES In this module we have only been able to cover only a slllall part of the interesting behavior that can occur in CSTRs due to nonlinear dynamics. In this section we provide a COIIcise overview of other interesting topics in CSTR analysis. First (Section M9.8.1), we will draw a connection between our physical-based approach (heat removal and generation curves) for analyzing Illultiplicity behavior and the more mathematical bifurcation theory. 'rhen (Section M9.8.2), we will discuss the cffect of cooling jacket flowrate 011 the behavior of a CSTR. l<'inally (Section M9.8.3), we discuss limit cycle behavior and HopI' bifurcations.
M9.8.1
Connecting Bifurcation Theory with Physical Reasoning
In Chapter i5 we presented mathematical conditions for bifurcation behavior; we have focused on physical arguments in this module. The goal of this section is to draw a connection between the mathematical conditions and the physical conditions for multiple stcady~ stale behavior.
Sec. M9.8
Further Complexities
581
r"irsi, recall the lnathcmatical conditions for a bifurcation. Let x represent the sLate variahle, p., a bifurcation parameter, and gCq..l) a single nonlinear algebraic equation thai must be satisfied at steady~state. If the following bifurcation cOJ1(litions arc satisfied:
iJ'g
dx::' --- ... and
then there exists k steady-state solutions in the vicinity of the bifurcation point. In our case, we were ahle to write a single nonlinear algebraic equation that must he satisfied for a steady-stale solution:
(.('1' -T) + UI(T -'1')1- ( Ml)VJ I, F1'", I' I, I' ,~ '"
- !1E) exp ( Iii' .\
F V
\I F
+
k
()
C
'A/,
Mo') CXI1'(-
-
HI',
j
()
_
f"or simplicity, we write this equation as:
where 7~\, is the slate variable (steady-stale reactor temperature) and /-L is used to represent the vector of physical parameters that can possibly be varied. The ~teady-slalc solution is obtained by solving:
or,
which is simply the intersection of the heat relllOvaJ and heat generation curves. 'rhc first derivative condition:
is simply
iJQ",,,U:,JL) iJT,
iJQ""U:,JL) iJ'I',
These conditions arc satisfied hy curves Band J) in FigureM9.12 below. The slope conditions arc satisfied at a high temperature for curve B anc! a low temperature for curve D, and we directly see how a "limit point" on the input-output curve (Figure M9.12b) corrc-
Diabatic Continuous Stirred Tank Reactors
582
B 0
B
Q
Module 9
o
Tr
Jacket Temperature
Reactor Temperalure( Tt)
a. Heat Removed and Generated
FIGURE 1\'19.12
b. Input/Output
Illustration of saddle-node (limit point) bifurcation.
sponds to the heat generation and removal curves following conclition also holds: J2g(r,.I~)
aJ}
(f~'igurc
M9.12a). We also note that the
I ()
Since, although the sc.cond derivative of the heat rClllovallinc is always zero,
a2 QJl,m(T ,/.L)
en}
1
the heat generation curve is not at an inflection point: ()! Q"c.,J·(.iL)
DT\2
t ()
and we see that the conditions for saddle-node behavior arc satisfied. Now we illustrate the conditions for a hysteresis bifurcation, which are: (J)g(T"iL)
aT}
()
All of these conditions arc satisfied in f
Sec. M9.8
Further Complexities
583
Q
Tr
Jacket Temperature
Reactor Temperature (Tf)
a.
Q
Tr
Jacket Temperature
Reactor Temperature (Tf)
b.
Q
Tr
Jacket Temperature
Reactor Temperature (Tf)
c.
FIGURE M9.13
lIlustration of a hysteresis bifurcation. Bifurcation conditions are satisfied in b.
584
Diabatic Continuous Stirred Tank Reactors
Module 9
0.8
d
0.6
~ 0.4
I-
0.2
o 1
0.5
--------.......
~---~~0-6--~O.8 UA
o
o
0.2
1
.
Temp-J
FIGURE M9.l4
M9.9
0.4
Cusp diagram.
DIMENSIONLESS MODEL "rhe CSTR modeling equations have a large number of parameters. It is clear that these parameters afC not independent. For cxarnplc, the behavior is dependent on the ratio FIV, which call be considered a single parameter. Most of the analysis of CSTR bchavi~)r has been performed on dimensionless equations, with a minimum number of dimensionless parameters involved. One possible set of dimensionless parameters is shown in student exercise 6. For a more complete discussion of the development and lise of dilncnsionlcss equations, sec Aris (1994).
SUMMARY Diabatic CS'l'Rs afC perhaps the most studicd unit opcration in thc chcmical process dynamics literaturc. They can exhibit exotic and interesting steady-state and dynamic behavior, including: steady-state multiplicity, HopI' bifurcations, periodic behavior, and chaos. In this module we have pnwided an elementary introduction to the steady-state and dynamic behavior of these systellls. The reader is encouraged to usc MATLAB to "play with" the model (particularly for the parameter sets for Cases I and 3) presented in this chapter.
585
Student Exercises
REFERENCES AND FURTHER READING A number of undergraduate textbooks on chemical reaction enginecring present the ideas of multiplicity behavior in adiabatic CSTR. These include the following books by Aris and Fogler: Aris, R. (1989). E'lementar.v Chemical Reactor Analysis. Boston: Butterworths. This is a reprint of the book originally published by McGraw Hill (J (69). Fogler, H.S. (1992). 1'.:lem('l/{S (~l Chemical Reactioll Engineering, 2nd cd. Englewood Clirrs, NJ: Prentice lIall. The following book docs an excellent job of developing the modeling equations for a diabat;e CSTR: Denn, M.M. (1986). Process Modeling, New York: Longrnan. [)iabatic CSTRs have been analyzed for complex behavior (multiple steady-slates, llopf bifurcations, quasi-periodic, chaos) more than ;:my other chemical process. An excelleot review of the literature on these topics is presented in the following paper: Razon, LJ'., &R.A. Schmitz. (1987). Multiplicities and instabilities in chemically reacting systellls~a review. Chem. Rng. S'd., 42(5):1005---1047. The following book provides an excellent discussion of the lise of dimensionless models, with adiabatic CSTR as a 1l1;:~jor example. Ads, R. (1994). Mathematical Modelling Techniques. New York: Dover. This is a reprint of the work originally published by Pitman, London (1978).
STUDENT EXERCISES 1. Consider the Case I parameters shown in Table M9. I. 3. Find the steady-state using fsol ve. h. Determine the stability of the steady-state using lincarization and eigenvalue analysis. c. Simulate the dynamic behavior by perturbing the steady~state concenlration and temperature values by ± 10 9(). Usc ode4 5 for the simulations. 2. Consider the Case 3 parameters shown in Table M9.1. a. Find the steady-state using fsol ve. b. Determinc the stability of the steady-state using linearization and eigellvaluc analysis.
Oiabatic Continuous Stirred Tank Reactors
586
Module 9
c. Simulate the dynamic behavior by pcrtm:bing the steady-state concentration and temperature values by ± J0 %. Usc ode4 5 for the simulations. 3. a. For the Case 2 conditions, sketch the phase-plane plots for the process linearized (A matrices) at each of the steady-states (operating points 1,2 and 3) in Section M9.5.1. Usc the eigenvalues and eigenvectors of the A matrices to sketch your plots by hand. b. Generate the phase-plane plot (Figure M9.5) for the nonlinear system of Section
M9.5.I. Usc cs tr run. TO and cs tr . ill from the Appendix. 4. a. Modify the parameters in the file cstr_.arnat.m (from the Appendix) to generate the state-space A matrix for the steady-state operating point for Case I. Use the eigenvector/eigenvalue analysis of the A matrix to sketch the phase-plane. h. Modify cstr__xun.m (change initial conditions) and cstr. In (change cstI' parameters) to perform the phase-plane calculation for the nonlinear systelll for Case I. Discuss the phase-plane results. 5. a.Modify the parameters in the file cstr_~_amat.m (from the Appendix) to generate the slate space A matrix for the steady-state operating point for Case 3. Usc the eigenvector/eigenvalue analysis of the A matrix to sketch the phase-plane. h. Modify cstr run. ill (change initial conditions) and cstI.'.m (change cstI' parameters) to perform the phase-plane calculation for the nonlinear system for Case 3. Discuss the phase-plane results. 6. The CSTR modeling equations have a large number of pararnetcrs. It is clear that these parameters arc not independent. For example. the behavior is dependent on the ratio F/V, which can he considered a single parameter. Define the dimensionless variahles as:
f3 I)
(- !:llJ)C Af ~~
7
-
pC[l7j'o
UA ~
pC/li)
4,
F;) V V
~
l~;)
1
'I
koe--Y
q
TX2
Xt
XII'
~
_~_!~L C'Arn
XlI'
1'1
'Ij'o
eNG
~
'Tf -
Fro
lio
xJ
'I
where F o' CAIO' and 'Ij'o are the nominal steady~state values. a. Show that the diniensionless modeling equations arc: dx)
,IT dX 2
(IT
~
E
IU,,, F
-~
F;} ~
7/0
'I
Student Exercises
587
K(X,) ..
where
cc
CX P (
1
and if the inputs arc at their steady-state values, the modeling equations arc: dX I
d'T dr) {!'T 7. Consider a system with the following parameters. Show that steady-state temperature is 350 K, and the corresponding steady-state concentration is 0.5 gmolliiter. Determine the number of steady-state solutions and their stability characteristics. Generate a phase-plane diagnun. Consider the response to ± 5 K changes in the jacket temperature. 1"01' one direction you should find oscillatory behavior, while the other direction is stable. This indicates thal a Hopf bifurcation (Chapter 16) occllrsfor some value of Tj between 295 K and 305 K. Find the value of Tj where this occurs (recall that a IJopf bifurcation occur whell the complex eigenvalues have zero real portion. F \I
T
I
. __ J
ko' 111111 (cD.H), J/gll1o] EIR. K
J 00 liters/min ]00 liters ]00 K 7.2 x 10 10
50000 8750
rei" JIIiter K
239
if.
]50
K
CAl' kgmol/m:1 UA, J/min K
10
50000
8. C\msider the input-output transfer function matrix for a CSTR, G(s), from Section M9.5.2. Find the clements of the lJ matrix corresponding to inputs 2 and 3, and the resulting transfer functions for each of the Case 2 operating conditions. For the sta~ hie steady-states (low and high temperature) compare the step responses of the transfer functions with those of the nonlinear SyStCIll. Use input changes of ± 0.1 molllll J and ± I deg K in inputs 2 (feed concentration) and 3 (feed tempcrature), respectively. Remclnbcr that for comparison Oil the same plots you must convert from deviation to physical variables ror the transfer function responses. 9. Orten cOllcentration measurements arc not available in chemical reactors. Show how a steady~state energy balance around the reactor can be llsed to infer the rate of reactioll, and therefore the "heat flow" of the reaction, til":::: rV(-6Jf). Think of two ways: (a) assuming UA is known and the jacket outlet temperature is available, and (b) assuming UA is unknown, but the jacket flowrate and jacket inlet and outlet temperatures arc available.
Diabatic Continuous Stirred Tank Reactors
588
Module 9
APPENDIXES 1
Function m-fi1e for Steady-State CSTR Solution
function £ = cstr_ss(x) "o % 2-state diabatic CSTR model % Use [solve to find st
temperature '[, 'l'hn cal]. statement from the command window is: fsolve ('
x
cstr~ss ' f
xO) ;
% where x is the solution vector (x % and xO is an initial guess vector
[conCi temp])
% ~f; define the following global parameter so that values % can be entered in the conunand window % global CSTR_PAR
% parameters (case 2) % CSTR __ PAR{l) frequency factor % CSTR_PAR(4)
(9703-k 3600) heat of react.-ion (5960) activation E;~nen;.JY (11843) densit:y*hl"at cap_ (1500)
% CSTR_PAR(5)
heat trans caeff * area (150)
% CS'l'R..YAR(2) ,?, CS rI'R.PAR(3)
'6
CSTR~PAR(6)
ideal gas constant
% CSTR_PAR(7)
reactor volume (1)
% CSTR_PAR (8)
feed flowralce
(1.987)
I l)
% CSTR PAR(9) feed concentration (10) % CSTR_PAR(10): feed temperature (298) % CSTR~PAR(ll): jacket temperature (298) % % ::,tates
'r, ca ,,~ Temp
and 2 - concentration of A, kgmal/m A ) reactor temperature, deg K
% 16 July 96 - (c) B.W. Bequette % revised 2 LJanuary 199'7 - added CS'I'R_PAR %
f
=
zeros (2,1) ;
%
global CSTR]AR %
% below we use common notation for the states to '15 improve code readibili ty % ca
xiI) ;
Temp
x(2) ;
%
% parameter values, %
case 2 shown in parentheses:
589
Appendixes
kG
Ij
% frequency factor (9703*3600) % heat of reaction (5960) % activation energy (11843) % density*heat cap. (500) % ht trans coeff k area (150) eSTRPAH (6) ; % gas constant (1.987) cs'm PAR (7) ; % reactor volume (1) eSTR PAR(8); ~~ feed flowrate (1) eSTR_PAR I ')) ; ;6 feed concentration (10) C:STRPAR (1 G) ; % fc;~ed temp(?rature (298) es'm_pAR (11) ; % jacket temperature (298) eSTRPAR
) ;
H_rxn - eS1'RJ'AH (2) ; E_act eS'I'R_PAR (3) ; eSTR_PAR (4) ; rhocp eSTR]AR(5) ; UA R
V F
caf Tf cI'j %
% ratios used in the equations % fov F/V; UAoV - UA/V; %
% modeling equations: %
rate:::: kO*exp{--E_ act/(R*Temp)))'ca; dcadt dTdt
fov*(caf - cal - rate; fov-k (Tf - Temp) -J (H __ Txn/rhocp) *rate (UAoV Irhoc~:p) * (Temp - '1'j); =
~5 below we convert back to function notation. fsolve % seeks values of x(l) and x(2) to drive f(l) and £(2) to _ zc~ro
%
f (1) f (2)
2
dcadt; -
dTdt;
Function m-file for Integration of Dynamic Equations
[unction xdoL
=
cstr_dyn(t,x)
%
% 2-state diabatic CSTR model 9) (JS(~ ode45 to find concentration and temperature % '2;
The call statement from the corrnnand ",lindow is:
% % %
[t,X]
ode45 ( cstr dyn I
I,
to, tf, xO) ;
'i; where t is the tirnc~ vector ;t, x is the solution vector (x = [conc;temp]) % to is the initial time % tf is the final time % xO is an initial condition vector % deL-ine the followin9 global paramet.er so that valueE3 % can be entered in the cOlrunand window
%
global CSTR PAR
Diabatic Continuous Stirred Tank Reactors
590 9(,
paramet,ors
% CS'l'H PAR (1) 'b CSTI' PAI'(2)
lease 2) frequency factor
19703*36(0)
"ci CSTRpAR(4)
heat of reaction (5960) activation energy (11843) density*heat cap. (500)
% CS'l'R_PAR (5)
heat trans coctf * area (ISO)
'b CSl'R_PAR ( 6 ) 'b CSTR_PAR (7) "C, ('STR PAR (B) "C, ('STR_ PAR (9) : % CS'rI(J'AR(lO) 1, CS'l'R PAR I 1] ) :
ideal gas constant (1.987) reactor volume (1) feed flowrate (1) teed concentration (10) feed temperature (298) jacket temperaturE~ (298)
% CSTR_PAR(3)
Module 9
states 1 and 2 == concentration of A, kgmol/m A 3 c.l; Temp reactor temperature, d.eg K
'I;
~I:;
Cd
% 16 July 96 - (e) B.W. Bequette ~~) revised 2 ,January 1997 -- added CSTRPAR qlobal CSTR- PAR \\ below we, use conunon not:':-JLion for the staLes to % improve code readibility %
~i5 96
Cd
x Ill;
Temp
x (2) ;
parame ter values
-
rhocp UA R V F
celf Tf '1'j
case 2 shown _in parentheses:
('STR PARI.] ) ; CS'I'RPAR(2) ; CS'l'R pAR(3) ; CSTR_PAH.(4) ; CSTR .. PAR ( ') ) ; CS'I'R PAR (6) ; CS'l'H ___YAR (7) ; CSTICPAR (8) ; CSTR_PAR(9) ; C.s'l'R PARII0) ; CS'l'RPARIll) ;
kO Il ---- rxn
E (Jct
I
-
'.', b ~i)
?~
"--;') '(, ~t
96
%
'r, 96
%
frequency factor 19703*3600) heat of reaction (5960) activation energy (11843) dens.ity-kheat cap. ( ',0 () ) hL trans coeff * arc~a I l'iO) ga~, constant II 987) reactor volume (1 ) feed flowrate 11) feE~d conc'entration 110 ) f(~ed temperature (298) jacket temperature 1298 )
% ~f,
ratios used in the equat_ions
- F/V;
fov UAoV 'f,
UA/V;
model Lnq
rate
=
equat_ion~:-;:
kO*exp(--E_.dct/ (R*'1'emp) )-"ca;
Appendixes
591
dcadt fov* (cdf - cal - ratei d'I'dt fov* ('l'f ~ rl'emp) + (I-I _rxn/:chocp) -krate (UAoV I rhocp)* (Temp - 'I'j); % ~6
lx~low
we convert back to ode45 function notation.
%
xdotll1 - dcadt; xdot 121 dTdt;
3
cstramat.m function
[amat, lambda]
-;c:
cstr_amat (ca, 'I'emp)
% ,/, 16 July 96 'l; (c) B.1/,J. Bequette
\" fi.nd the A matrix for the 2-state CSTR to determine the % stability by eigenvalue analysis: %
% amat A matrix 7,; larnba - vector of eigenvalues %
%
paramc~ters
frequC-:-DCY factor, hrA-l F/V. hr iwat of rXTl, kcal/kqmo:l activation energy, kcal/kgmo:l densi ty-kheat capaci ty, kcal Im":3 clefJ K feed temperature, deg- K feed concentration of A, heat trans. coeff.*area/volume, kcal/m A 3 deg 1< hr
kG 7, fov ~6
(~
A
-]
I-l rxn
% E_dct % rho(:,p % 'l'f % caf '6 UAoV %
% manipulated input ?o Tj =: jack(,:,t temperature, deq K (Ii
°6
states
(b ca ~6
Tc'mp
-
% %
pararnet~er
conCE~ntration of A, kqffiol/m"3 reactor temperature, deq K
values
2: fov -;c: 1j
cas(~
kO - 9703*3600; ~)960j H_rxn E __ act 11843 ; rhocp '1'f
caf UAoV R
1'j
ks
500i 298; - 10; 150; 1.98'7; 298;
kO*exp(-E_act/IR*Temp));
592
Diabatic Continuous Stirred Tank Reactors
ksprime -
%
(E_dct/ (R*Temp":Temp)) ;
amat(l,l) - -fov - ks; amat(1,2) - -ca*ksprime; amat:(2,1) ks'kH_rxn/rhocp; amat(2,2) ~ -fov - UAoV/rhocp +
lambda
4
kSk
Module 9
(H_rxn/rhocp)*ca*ksprime;
ei<;J (amat) ;
cstr.run.m (Script File to Generate Phase-Plane Plot)
'6 cstr__run. ill % b.w. bequette >6 2 Dec 96
% revised - 4 Dec 96 - more documentation %
% generate phase-plane plot for nonlinear cstr (cstr.m) %
% set of initial conditions 96
xOla=- [0.:3;305]; xOl [0.5;3151; x02 [0.5;325J; xO:3 [0 . 5; 3.3:)] ; x04 [(L~)i34S]; x05 - [0.15;355J; x06 [O.5;365}; x07 [0.5-;375]; x08 [0.5;385J; x09 - [0.5;395]; x010= [0.'i;405]; xOll= [0.5;415J; x012a=[9.5;305] ,. x012 [9.5;31'i]; x013= [9.5;325J; x014 [9.5;335]; x:Ol~5 [9.5;345J; x016 [9.:5;35::iJ; x017 [9.5;365];
x018
[9.5;375J;
x019 x020 x021 x022
[9.5;385J; [9.5;395]; [9.5;4051;
[9.5;415J;
"o
% use ode45 for numerical integration [tla,xlaJ = ode45('csLr' ,O,10,xOla); [U,x1] ode45('csLr',O/10,xOl); [t2,x2J = ode45('cstr' ,0,10,x02);
593
Appendixes
[t3,x3J - ode4S('cstr',O,lO,x03); [t4,x4j oclo4S( 'cstr' ,0,lO,x04); [t.5, x5] ode45 ( , C~3 tr ' , 0, 10, xO 5) ; [t6,x6] ocle45 ('cstr' ,0, lO,x06); [t'l,x'l] ode4S( 'cstr' ,0,lO,xO'l); [tS,x8] ode4S('estr' ,O,lO,xOS); [t9, x9] odc,4S (' cstr' ,0, 10,x09) ; [tl0,xlO] ocle4'i('cstr',0,10,xOl0); [tlI,xlI] odo4'i ('estr' ,0, 10,xOlI); [lU,xl2] - ode4S( 'cstr' ,0, 10,x012); [L12a,x12a] ::= ode45( 'cstr', O,lO,x012a); [tU,xU] ocle45 ( 'estr' ,O,lO,x013); lt14,x14] ocle45('cstr',0,10,x014); [t15,x15] - ode4~)( 'cstr' ,0, 10,x015); [U6,xl6] ocle4S('cstcr',0,lO,xOl6); [tl'l,x17] ode4S('est:r' ,0,JO,x017); [tIS,xlS] ocle4S( 'estr', O,lO,xOlS); [tl9,x19] ode45( 'estr', 0,10,x019); [t20,x20] ocle4S('estr',0,10,x020); [t21,x2l] - ode4S('estr',0,10,x02J); [t22,x22] - ocle4S( 'estr', 0, 10,x022); x040 =
[0.5;3'10];
x041
[0.5;372.5];
x049 [9.S;320]; xOSO = [9.S;326.2S]; xOSl [9.'5;32'7.S]; x052 [9.5;330]; xOS3 - [9.si332.S]; [t40,x40] = ode4S( 'estr', 0,10,x(40); [t41,x4l] ode4S( 'estr', 0, JO,x04l); [t49,x49] ode4S( 'estr' ,0, 10,x049); [tSO,x50] ode45 ( 'estr', O,10,xOSO); [t51,x51] ode45( 'cst,r' ,O,lO,x051}; [t~52,x52] - ode45 ('cstr' ,O,lO,x052); [t53, x53] ode45 ( 'cstr' ,0,10, x053) ;
% phase-plane plot plot (xl ( ; , 1) ,xl ( ; ,2) , 'w' ) hold on plot (xla ( ; ,1) ,xla ( ; ,2) , 'w' ) plot (x2 (;,1) ,x2 (;,2), 'w') plot (x3 ( ; ,1) ,x3 ( ; ,2) , 'w' ) plot (x4 ( ; , I) ,x4 ( ; ,2) , 'w' ) plot (xS ( ; ,1) ,x5 ( ; ,2) , 'w' ) plot. (x6 ( ; ,1) ,x6 ( ; ,2) , 'w' ) plot(x'l (; ,1) ,x'l (; ,2), 'w') plot (x8 ( ; , I) ,x8 ( ; ,2) , 'w' ) p1ot(x9 (;,1) ,x9(; ,2), 'w') plot(x10 (;, J) ,xlO(; ,2), 'w')
Diabatic Continuous Stirred Tank Reactors
594
Module 9
plot(xll(:,ll ,xll(:,21, 'w') p]ot(x12 (:,1) ,xU (: ,21, 'w') plot (x12a( : ,II ,x12a(: ,21, 'w' I plot (x13 ( : ,1) ,xU ( : ,2) , 'w' ) plot (x14 ( : , 1 I ,x14 ( : ,21 , 'w' I plot (xIS ( : ,1) ,x15 ( : ,21 , 'w' ) plot (x16 ( : ,1) ,x16 ( : ,2) , 'w' ) plot(x1'7 (: ,1) ,x17 (: ,2), 'w') plot(x18(:, 1) ,xI8(: ,21, 'w') plot(xI9(:,I) ,xI9(:,21, 'w') plot (x2 0 ( : , 1) ,x2 0 ( : , 2 I , 'w' ) plot(x21(:,I),x21(:,2), 'w') plot.(x22(: ,1) ,x22(: ,2), 'w') % 'to mark the stable and unstable steady-states =
(0 =
stable,
unstable)
plot ( [8 . 564] , [31] .2] , 'wo' , [S. 518] , [339. 1] , 'w+ ' , [2 .3 S 9 ] , [368.1], 'wo') %
%
some initial conditions noL used
1; %
x030 x031 =
[5.768;339.1]; [S.368;339.1l;
% %
x032 xG33
(5.518;338.5]; [5.518;339.71;
'I)
[t30,x30J [t_3l,x31] [t32,x32] [t33,x33]"
ode4~)(
plot (x4 0 ( : , 1) plot (x41 ( : ,1) plot(x49 (: ,1) plot(xSO(:, 1) plot(x'51(:, 1) plot(x52.(: ,1) plot (x53 ( : ,1)
,x4 0 ( : ,x41 ( : ,x49 (: ,xSO(: ,xSl(: ,x52 (: x53 ( :
9(,
% % %
'cf-;tr' ,0, lO,x030); ode45('cstr' ,O,IO/x(31) i ode4[j{ rest.r' ,O,lO,x032) ; ode45('csLr',O,lO,x033);
J
,2I, ,2) , ,2), ,2), ,21, ,2) ,2) , I
'w' ) ; 'w' ) ; 'w'); 'w'); 'w'); 'w');
'w
I
)
;
% plot{x30 (: ,1) ,x30 (: ,2), 'w'); 'f, plot(x31(:,1),x31{:,2), 'w'); ~) 96
plot(x32(:,ll,x32(:,2), 'w'); plot(x33(:,l),X33(:,2), 'w');
%
% set limits aD plot, and label axes and provide plot title % axis ([0 10 300 425]) xlabel (' cODcent~nJtion, kgmol/m 3') ylabel ( , temperature, deg K Lit~le{'cstr wit.h multiple steady-states') A
J
)
-+
Appendixes
function
595
[Tjs,Tfs,concs)
=::
cstr__ io(Tempvec)
%
% Generate the steady-state i/o curve for the diabatic CSTR. % rrempvec is a vector of reactor temperatures generated % before this routine is called. '5 This routine solvE:?s tor vector of jacke1~ temperatures 96 that would yield 1:11E;" required reactor temperature vector 8; (ar3suming a constant teed tJ~mperature).
% In addition, this routine solves for the vector of % feed temperat.ures that would yield the required reactor % temperature vector (assuming constant jacket temperature) %
% For consistency with cstr ss.m and cstr__dyn.m, we use 96 a global parameter vector. The elements of this vector % should be set in the command window before this routine ~-; is run. % global CSTR_PAR % parameters (case 2) % CS'I'RPAR(ll frequency factor (9703*3600) % CSTR_PAR(2) heat of reaction (5960) 'Ii CSTR_PAR(3) activation energy (11843) % CSTR PAR(4) density*heat cap. (500) % CS'I'K_PAR(5) heat trans coef£ * area (150) % CS'l'R.YAR (6) ideal gas constant (1.987) ~; CSTR_PAR (7): reactor volume (I) % CS'l'R_PAR(81: feed flowrate (1) % CSTR_PAR(9): feed concentration (10) % CS'I'RPAR(lO) feed temperature (298) " CSTR_PAR(ll): jacket temperature (2981 'r, % 16 July 96 - (c) B.W. Bequette 96 revise~cl 2 ,January 1997 ~ acld(?d CSTR PAR %
global CS'I'RPAR --. %
% parameter values, case 2 shown in parentheses: %
kO H_rxn
CSTRPAR (I 1 ; CS'l'R.PAR (21 ; E~_act CSTR_PAR (31 ; rhacp CSTR Pi\R(4); - CSTR PAR (5) ; VA CSTR PAR(6); R CS'I'R.PAR (7) ; V F CS'I'R.YAR (B) ; cat CSTR_PAR (9) ; - CSTR PAR(lO); Tf
% % % % % % % % % %
frequency factor (9703*3600) heat of reaction (5960) activation energy (11843) density*heat cap_ (:500) ht trans coeff * area (150) gas constant (1.987) reactor volume (1) feed flowrate (1) fE:,(~d concentration (10) feed temperature (298)
596
Diabatic Continuous Stirred Tank Reactors
'1'j
CSTR_PAR(11);
% jacket temperature
Module 9
(298)
~)
% ratios used in the equations %
fov UAoV
F/V; UA/V;
% for i -= 1:1ength(Tempvec); 'remp 'I'empvec ( i) ; k kO*exp(-E_aet/(R*Temp)); ca fov*caf/ (fov -+- k); rate kO*exp (-E dct/ (R*'I'emp)) *ca; stuffl - fov-krhocp* (Tnmp - Tt); stuff2 - H rxn ).-ra te; Tjs (il 'I'empvec(i) + (stuffl - sLuff21/UAoV; conc5(i) .: : : ca; stuff3 UAoV* (Temp Tj) ; stuff4 H~rxn*rate; Tfs (il Tempvec(i) + (stuff3 - stuff41 / (fov*rhocpl; end;
6
cstr~cusp
.m-Cusp Catastrophe
% cst.r cusp.m % generates the cusp diagram, based on case 2 parameters % case2; % clear 'I'empr Tj s 'rfs cones; Tempr ~ 280:5:400; 'is vary UA or F npts c= length (Tempr) ; CSTR]AR(81 ~ 0.1; F--pI t = ones (npts) *CS~'ICPAR(8 I ; [T] S 'ris cones J : : : c8tr io (Tempr) ; p1ot3(F_plt,Tjs,Tempr, 'w-'I I
hold on for kindex
I
=
1:9;
CSTR_PAR(8) 0.1+ O.l*kindex; F_plt = ones (nptsl *CSTR_PAR(8);
[Tjs,Tfs,concs]
~
cstr_io{Tempr);
plot3 (F_Vlt,Tjs,Tempr, 'w-') end hold off % axis([0.25 1.375 260 360 300 380JI axis([O.l 1.0 220 400 280 400J) view(60,20) xl abel ('F/V') ylabc,I ( , Temp~j , ) zlabe1 ( 'Temp_r' I
IDEAL BINARY DISTILLATION
MODULE
10
After studying this module, the student should be able to Develop the dynamic modeling equations for ideal hinary distillation Solve for the steady-slale Linearize and find the slale space model Understand the dynmnic behavior of distillation columns Use MATLAB for steady-state and dynamic simulation
'fhe major sections of this module are: iVI 10.1 iVI I 0.2
M10.1
Background
Conceptual Description
(If l)isti Ilation
iVI 10.3
Dynamic Material Balances
iVI 10.4
Solving the Stc
iVI I0.5 iVI 10,6
State-Space l,inear Distillation Models
iVI 10,7
Multipliei tyBehavior
Solving the Nonlinear l)ynamic Equations
BACKGROUND Distillation is a common separation technique for liquid slre-anlS cOlltaining two or more components and is one of the more important unit operations in chl~mical manufacturing processes. Design and control of distillation is important in order to produce product streams of required purity, either for sale or for lise in other chemical processes.
597
Ideal Binary Distillation
598
Module 10
vapor
y
x liquid
FIGlJH:E !VItO. I Closed system \vith liquid and vapor in cquiJibrililn.
I)istillation is based on the separation of components of a liquid mixture by virtue or the difTcrcnccs in boiling points of thccomponcnts. For illustration purposes, \\le will base our discussion on the separation of liquid streams containing two cOinponents (I)inary mixture). We will refer to the pure component that boils at a lower temperature as the light component and the pure component that boils at a higher temperature as the heavy component. For example, in a mixture of benzene and toluene, benzene is the light cOlnponcnl and toluene is the heavy component. A saturated liquid mixture of two components at a given concentration is in equilibrium with a vapor phase that has a higher concentration of the light component than the liquid phase. Let x represcnt the mole fraction of the light component in the liquid phase and y represent the mole fraction of thc light component in the vapor phase. Consider r·'igme rvIlO. I as a conceptual representation of phase (vapor/liquid) equilibrium. 'fhe salu~ rated liquid is in equilibrium with a saturalcd vapor. The concentration of the Iighl COtll~ pOllcnt will be larger in the vapor phase than the liquid phase. Figure M 10.2 is an example of an equilibrium diagram that rcprescnts the relationship bctvvecll the liquid and vapor phase cOITlpositions (mole fraction). For example, if the liquid composition is 0.5 mole fraction of the light component, we rind frOID Figure M! 0.2 that the vapor composition is 0.7.
O.B
w
00
'"
0.6
"'"
OA
cQ
0,
>
:';;
02
0.5 x, liqUid phase light component
FIGURE i\:IlO.2 Vapor/liquid equilibrium diagram.
Sec. Ml0.2
Conceptual Description of Distillation
599
For ideal mixtures, it is cOlllmon to model the phase equilibriulll relationship based on constant relative volatility
o:x }'=
.
I
(MIO.I)
+ (o:-I)x
where (~is known as the relative volatility. FigurcMIO.2 was generated based tion (M 10.1) with (X ~ 2.5.
M10.2
011
equa-
CONCEPTUAL DESCRIPTION OF DISTILLATION Thcft.JlIowing is a conceptual description of the operation of a binary (two-component) distillation colunm. The feed typically enters close to the rniddlc of the column (above the feed stage), as shown in f'igurcMIO.3. Vapor flows from stage to stage lip the colullm, while liquid flows from stage to stage down the column. The vapor from the top tray is condensed to liquid in the overhead condenser and a portion of that liquid is returned as rcflux. Thc rest of that vapor is withdrawn as the overhead product stream; this overhead rroduct stream cOlltains a concentrated amount of the light componcnt A portion of thc liquidatthc bottom of thc column is withdrawn as a bol.loms product (containing a con-
Condenser (Heat removed from condenser)
Overhead receiver
_...-'::=r=::.-=-----... Reflux Distillate Product Feed Stream
-----1>1
NF
Vapor Boil-up NS
(Heat added to Reboller)
Bottoms Product
FIGURE MlO.3
Schematic diagram for a distillation column.
600
Ideal Binary Distillation
Module 10
W
E
8c
, o
o
FIGUIU: 1\'110.4 for
it
Schematic diagraol
distillation C()lumn tray.
centratcd amount of the heavy component), while the rest is vaporized in the rcboiler and returned to the column. 'file liquid from one tray goes over a weir and cascades down to the next tray through a downcomcr. As the liquid moves across a tray, it comes in contact with the vapor from the tray below. The schematic diagram for a sieve tray is shown in Figure MUlA. Generally, as the vapor from the tray below comeS in contact with the liquid, turbulellt mixing is promoted. Assuming that the mixing is perfect, allows one (0 IlHldcl the stage as a lumped parameter system, as shown in !·'igurc M 10.5. Notice that the vapor fronl stage i is modeled as a single stream with molar flowrate Vi and light component vapor composition (mole fraction) Yi' The liquid leaving stage i through the downcomer is modeled as a single stream with molar fJowrate Li and light component liquid composition (mole haction) xi' The conceptual diagram for the feed stage is shown in Figure MIO.6. It differs from Figure M 10.5 in that an additional input to the stage is from the feed to the column.
Li__ l
Xil
Lixi
Lnfxnf
ViYi
Vi-t-l Yill
VIlf11
Ynf-t-l
FIGURE 1\'110.5 Conceptual material balance diagram for a typical stage.
FIGlJRE i\-1l0.6 Conceptual rnatcrial balance diagram for the feed stage.
Sec. Ml0.3
M10.3
601
Dynamic Material Balances
DYNAMIC MATERIAL BALANCES M10.3.1
All Stages Except Feed, Condenser, and Reboiler
The component balance for the liquid phase of a lypical stage as shown in Figure MIO.S is: accumulation liquid frolll
tray above
vapor from liquid vapor tray below leaving leaving
dNlrt i dl
(M 10.2)
where Mi is the liquid molar holdup on stage i. [.'01' this simple binary distillation model, we will make the COllllllon assumption of cquimolal ovcrllow (King, 1980). For any stage except the feed stage, we assume that the vapor tlowratc from one stage is equal to the vapor molar fJowratc of the stage below: (MI0.3)
and that the liquid leaving the stage is equal to the liquid flowing from olle stage ahove: (MIOA)
M10.3.2
Feed Stage
Let {{F' represent the quality of the fccdstrcam. If the feed is a saturated liquid, thcn qF =: (, while 'iF ~ 0 fbr a saturated vnpor. The vapor molar f10wratc leaVing the feed stage is (where NF =: number of the feed stage) (M 10.5)
Similarly, the liquid molar nowrate of the slrcanl leaving the feed stage is: (M I0.6)
M10.3.3
Condenser
A total condenser removes cnergy from the overhead vapor, resulling in a saturated liquid. Assuming a constant molar holdup in the distillate receiver, the total liquid nowrate from the distillate receiver (reflux + distillate flows) is equal to the nowrate of the vapor frotH the top tray:
V2 where Lv and D represent thc reflux and distillate molar nowratcs, respectively.
(MI0.7)
602
Ideal Binary Distillation
M10.3.4
Module 10
Reboiler
A total material balance around the rcboilcr yields:
Ii = L NS _ 1 where
Vreboilcr
M10.3.5
Vrehoi1er
(MIO.8)
is the reboiler molar llowrate and B is the bottoms product molar flow rate.
Summary of the Modeling Equations
The rectifying section (top section of column, above the feed stage) liquid molar nowrates arc:
L R = L{)
(MIO.9)
The stripping section (bottom section of column, below the feed stage) liquid molar
flowratcs are (MIO.IO) The stripping section vapor molar flow rates are:
(MIO.II) The recti fying section vapor molar flowrates arc:
Vii = V, + FCI - 'lr)
(MIO.12)
In the following ~e assume a constant liquid phase molar holdup (dM/dt) = 0).
The overhead receiver component balance is:
(MIO.l3) The rectifying section component balance is (from i = 2 to NF~l):
dX j
"I
=
(MIOI4)
The feed stage balance is:
dX NF dl
(MIO.IS)
The rectifying section component balance is (from i = NF+l to NS-I):
1 [ dt = M Lx 'S i r
"Xi
(MIO.16)
And the rehoiIer component baLance is:
dxNS' =
"I
1 M [L\"NS_l - BxNS - V\YNS]
(MJO.17)
Ii
c
Sec. M10.4
Solving the Steady-State Equations
603
EQUILIBRIUM RELATIONSHIP It is assumed that the vapor leaving a stage is in equilibrium with the liquid on the stage. The relationship between the liquid and vapor phase concentrations on a particular stage can be calculated using the constant relative volatility expression: y-= , I
M10.4
O'X i
(MIO.18)
+ (IX - I )x,
SOLVING THE STEADY-STATE EQUATIONS To obtain the steady-state concentrations we must solve the system of equations, f(.\') :;:: O. From the overhead recci vcr component balance:
I,
~ y,~X, ~
0
(M 10.19)
}
+
VRYi+ 1 ~" LW'C i ~ VRYi =
(M 10.20)
0
From thefced stage balance: j~'F =
LgxN/' -J + VSYNr+ I + FZ r
~ LS'''-NF - VUYNI
'-co
0
(M 10.21)
From the stripping section component balance (i ::::: NF+ I to NS-l):
f;
:0'::'
LSX i _ 1 +
V.SYill -
LSx i
-
VSYi
=
0
(MIO.22)
And from the reboiler component balance:
f:vs
= LSX NS _ 1 -
Bx NS
-
VSYNS
= ()
(MI023)
where B = L, - V,.
We must realize that (M 10.19)--(M I 0.23) constitute a set of nonlinear algebraic equations, since the relative volatility relationship (M I 0.18) is nonlinear in the state variable. Equations (M IO.19)~(M 10.2:1) arc N.S' equations in NS' unknowns. A Newton-based technique will be used to solve the equations.
EXAlV1PLE MIO.l
Stead'y~State Operation
of a 41-Stage Column
Consider a 41-stage column with the overhead condenser as stage 1, the feed tray as stage 21 and the reboiler as stage 41. The following parameters and inputs apply ex F
1.5
::: 1 mol/min zr 0.5 mole fraction of light component R 2.706 mol/min D "'" 0.5 mol/min qF 1 (sat'd liquid feed)
Module 10
Ideal Binary Distillation
604
r:rom an overall material balance, the hottoms product f10wratc is;
B
=
F - D = 1 --- 0,5 mol/min
the stripping section flowr:ltc is:
Ls
=
R
-j
Fqr
2.706
=
+ I
=
3.706 mol/min
c=.
3.206 mol/min
and a balance around the reboiler yields:
V\' ,."'" L s -- B ,=
3.706 -- 0.5
'[ he ll1-tJle d 1 st C,b, Tn (shown in the Appendix) is used to sol ve for the steady-state compositions.
x
fsolve(
'di~-oL
S,3' ,xO)
The resulting compositions are shown in Figure M 10.7. Notice the strong sensitivity to rellux flowrate. alpha
0
1.5. R
0
2.706.
V
0
3206
_-J
[_~
08
__:_
nominal
~11 O~O,,
_
0.0.6
E
8
:c
£>04
0.2
o
____1...
5
__-------'__
_
10
15
20 stage
_
.-----L.-
25
l.,_ ...
30
-~
_
35
40
FIGURE MJO.7 Liquid phase composition (mol fraction of light component) as a function of stage number. Solid := nominal reflux, dashed;::: +1II(, reflux, dOlled:::o. --1% rellux. 'Che overhead composition (stage I) is 0.90 and the bolJOllls composition (stage 41) is 0.0 I for tile nominal rcllux rate (2.706mollmin).
Steady-State Input-Output Relationships The scnsitvity to rdlux rate is also shown by the plot in Figure MIO.X. The steady-stale gain (change in oUlput/dmnge in input) for distillatc composition is large whcn reflux is less than 2.7,
Sec. Ml0.5
Solving the Nonlinear Dynamic Equations
605
but small when the reflux is greater than 2.71 1l1OlImin. 'T'hc opposite relatiollship holds for bottoms composition, where the gain is small when reflux is Jess than 2.7 mol/min, but large for n.> flux greater than 2.71 mol/min. This sensitivity has important nnnificati{)HS for control system design.
~ 0.95
2.66
2.68
2.7
2.72
2.74
reflux
0.06 ~ 0.04
0.02
o ::--::.-----
_~J:_~_'::-
2.66
2.68
2.7
1_________ 2.72
_ ----L _ 2.74
reflux
FIGURE i\i110.8 Steady-state input (reflux)- -olltput (distillate or boUoms composilion) relationship.
M10.5
SOLVING THE NONLINEAR DYNAMIC EQUATIONS Equations (M IO.I3)--(M 10.17) arc a set of initial value ordinary equations, which can he solved using numerical integration techniques. 'rile next example L1ses the variable step size MATLAB routine ode4 5 to perform the intcgratiOlL
EXAMPLE lV110.2
Dynamic Response
Consider now the previolls problem, with the initial condiLions of Lhe stage compositions equal to the steady-state solution or Example M 1O. L The ~\dditional parameters needed for the dymunic simulation are the molar holdups on each stage. Here we use the following parameters:
MJ=M o tv}]
=
M 3 :::: 5
overhead receiver molar holdup feed tray molar holdup hottoms (reboiler) molar holdup
:::::;5 mol = 0.5 mol
=5 mol
606
Ideal Binary Distillation
Module 10
To illustrate the nonlinear behavior we compan.' the results of J-, I (j{ step changes in the reflux rate at time t:;:: 5 minutes. [t,x]o::ode45( 'dist_dyn' IO,400,xO)
Note that the currenl version of ode4 5 docs 1101 allow model p,lnullClerS to be passed through the argument list, so global parameters afC defined in the m-filc dis L_dyn. m shown in the Appendix. The following results arc shown in l;'igurc MIO.9. A positive lr;::- step change in the reflux rate yields a snlal! increase in the distillate composition; tbis makes since because the maximum possible increase in distillate purity is (J.()J (the composition cannot he greater than I mole hac lion) while it call decrease much more than that. A negative J (J{ step change ill reflux causes a larger change in the distillate purity. 'fhe opposite effects arc observed l~)r bottoms composition, where a positive reflux change yields a large bottoms composition change. A negative reflux change yields a small bottoms composition change.
0.98
~ 0.96 0.94
o
50
100
150
200
250
300
350
400
250
300
350
400
time, min
0.06 0.04
? 0.02
o o
50
100
150
200 time, min
FIGURE M10.9 Illustration in Ilonlinear response or distillate and bottoms compositions to :'>tcp changes in reflux. Solid line c::: +l(k, Dashed line;:::: -19().
M10.6
STATE-SPACE LINEAR DISTILLATION MODELS Linear state space models arc useful for slability analysis and control system design. Here we develop models of the (orm:
x ,- A x' t
y' ;: : : C x'
n u'
(M 10.24)
(MI0.25)
Sec. M10.6
State-Space Linear Distillation Models
where! is used to represent the deviation variables. indicates the steady-state values). Defining I( =
,
av.' =
ilx;
Xl :;:;: X -- x~,
0'.
(I+«I-I)x)'
607
u' ;:;;: u -- u.\ (the subscript
S
(MIO.26)
and linearizing the dynamic equations (M IO.13)--(M IO.I?) (M 1(27)
V/JC A11) .;01'
(M 1(28)
i = 2 to NF-] :
ill; JX i 1
(M 10.29)
M[
(M 10.30)
Au I
ill; I
=
-) (-t'i+ 1
VnJ(; I'vf/
I
(MI031)
F'or the feed stage:
illy/
(MI032)
ANF,NF I JXA'V_I
ill"
(MIO.33)
()x NP A,V/
ill,,·
V)<',/
M,
dX,vF' I
(M I 0.34)
Fori=NF+1 toNS-I:
(If; ax;
1
M[
(ivlIO.35)
(M 10.36)
A iJI and for the rcboiler (stage iV,,')
ill;
V/
()Xi i I
iVf/
(ivlI0.37)
608
Ideal Binary Distillation
A NS,NS-l
at; ."C:_-
O.li_ I
(M 10.38)
Mil
_(13 + V,K
ANS,NS
N ,)
(M 10.39)
MlJ
Now, for the derivatives with respect to the inputs;
Module 10
ttl ;;;;:
LR
;;;;:
L! and
aj,
B,.2
Ill::::
Vii':::::
(M IOAO)
0
au?
VrehoJler:
For i = I to NS-I:
Mr
B'.2
.
aj;
Yil
-
Mr
OllZ
(MIOAI)
anel for the bottom stage:
B
aj;
N.'),
ill,
.....
I ~--
au]
M NS
!!"\2 '. ~ 1'_,. ~ aU 2
- Yvs
MNS
(M I OA2)
If the output variables arc the overhead and bottoms compositions, tIlcn:
C u " 1, while C?NS
M10.6.1
= 1, while
eli .c. C'!.i -::0'
0 for i
1
(M I 0.43)
0 for i ;f!
(M I OA4)
j.
Transforming the State Space Linear Models to Transfer Function Form
The matrix transfer function is: G(s) ~ (,(sf-A)'!!
(M IOA5)
It is easy to generate MATLAB m-files to calculate each of the state~sIxlcc matrices (A,B,C) for a particular set of parameters (and steady-state comrJositions). For a colunm of reasonable sile (say the 41-stagc example) the denominator polynomial ill (M 10.45) would be quite large (say 4pt order). \Vhat is often more llseful is to be able to directly calculate the steady-state gain matrix, as shown below. The steady-state gain matrix is: (M IOA6)
where, again, it is easy to generate a MATLAB m-file to perform this calculation.
M10.7
MULTIPLICITY BEHAVIOR Even simple ideal binary distillation columns have been shown recently to have interesting steady~state and dynamic behavior, including multiple steady-states. Nice cxalnples are shown by Jacobsen and Skogestad (1991, 1994). The key assumption that must be made for this behavior to occur is that mass flows, rather than molar flows, arc manipu-
609
References and Further Reading
lated. The reader is encouraged to read these papers and modify the MArLAB m-files presented in this chapter to illustrate the behavior shown by Jacobsen and Skogestad.
SUMMARY In this chapter wc have dcveloped modeling equations to dcscribe tile steady-state and dynamic behavior of ideal, binary distillation COIUlllllS. The 41-stagc column example shows that stcady-statc distillate and bottom compositions arc a nonlinear function of the manipulated inputs (distillate and vapor boil~up flows). Also, the dynarnic responses or these COll1positions depends on the magnitude
REFERENCES AND FURTHER READING The following undergraduate chelnical engineering texts develop the steady-state modeling equations ror ideal binary distillation: King, C,J. (1980). 5,'cparatiolls Processes. 2nd cd. New York: McCiraw-l-lill. McCabe, W.L., & J.e. Smith. (1976). Un;f Operations (?F Chemlt.-ol EJlgineering, 3rd cd. New York: McGraw-Hill. The dynamic lnodcling equations for distillation are prescIlted by: L.uyben, W.L. (J 990). Process A1odding, Simulation 01/(1 COlltrol.!iJr Chemical h,'l1gineers, 2nd cd. New York: McGraw-HilI. More advanced treatments of stcady-st::lte and dynamic distillation models arc prescnted by: Holland, C.D. (1981). FUlldamenfals (~l Multicompoflent Distillation. New York: McGraw-Hili. Holland, CD., & A.l. Liapis. (1983). Computer Methods'for Solving Dynamic .)'eparatioll Problems. Ncw York: McGraw-HilI. The panlllleters for Example M I0.1 are presented in the following two references: Skogestad, S., & M. Morari. (1988). Understanding the dynamic hehavior of disljl~ lation columns. /ild. 6·lIg. Choll. Res., 27( 10): 1848-1862. Morari,M., & E. Zafirioll. (1988). Robust Process Control. Englewood Cliffs, NJ: Prentice-Hall. The possibility of multiplc steady-state behavior in ideal binary distillation is presented by: Jacobsen, E.W., & S. Skogestad. (199 J). Multiple steady~stales in ideal two~product distillation. A/ChE.I., 37(4): 499-511. Jacobsen, E.W., & S. Skogcstad. (1994). Instability or distillation columns. A/ChE .1.,40(9): 1466-1478
610
Ideal Binary Distillation
Module 10
STUDENT EXERCISES 1. Consider a simple I tray (3 stage) column with the overhead condenser as stage I, the feed tray as stage 2 and the reboilcr as stage 3. Usc the following parameters and inputs:
R = ifF
F lJ Zp
= = =
5 3 mol/min I 1 mol/min 0.5 mol/min 0.5 mole fraction of light component
Find the bottoms product Jlowratc, the stripping section flowratc and the vapor boilup rate (stripping section vapor Ilowrate). Usc fsolve and dist_ss.m to find the resulting compositions:
x =
0.703] [ distillate composition ] 0.486 = composition of stagc 2 (the feed tray) [ 0.297 bottoms product composition
Consider now the dynamic behavior, with the initial conditions of the stage COITlPO~ sitions equal to the slcady~state solution. The additional paramcters needed for the dynamic simulation are the molar holdups on each stage. Usc:
M, M2 M,
overhead receiver molar holdup feed tray molar holdup
kIn
= =
5
=
5 mol 0.5 mol.
=
bottoms (reboilcr) molar holdup
5 11101
At lime zero, the reflux is changed from 3.0 mol/min to 3.2 mol/min. Usc ode45 and dist_dyn. m to simulate the dynamic behavior shown in the figures below.
0.78
0.08
0.76
(J)
~
0.06
·x
~ 0.74
.:- 004 ~ 0.02
0.72
0.7
S time
is
10 time
The Reflux is step changed from 3.0 to 3.2 at 1 :::: I minute.
Appendix
611
APPENDIX
function f
;::: dist_ss(x);
%
% solve for the steady-state stage compositons in an ideal % binary distillation column using fsolve. %
% (c) 1993 B. Wayne Bequette - 21 june 93 % revised 31 Dec 96 %
% All flowrates are molar quantities. Stages are numbered % from the top down. A total condenser is assumed. % ThE~ overhead receiver is stage .1. The partial reboiler % is stage ns (the number of equilibrium "trays" is then % Ds-I). The column parameters should be specified in the % DIST_PAR array. % % to use this function, enter the following in the con@and % window, or from a script tile (after defining parameters % in the DIST_PAR array: % % x ;::: fsolve (' dist_ss' %
I
xO)
~~ wher(~ xO is a vector of initial guesses for the liquid % phase stage compositions (length(xO) :c: DS)
%
%
%
DIST~PAR
is a vector of distillation column parameters used by both dist sS.m and dist~dyn.m
% %
< 8; disp ( not enough parametlC':rs given in
if 1engthIDIST_PAR) I
disp ('
DIST~PAR ')
')
disp (' check to SQe that global DIS'I' PAR ha;" bCE"n defined ') return end % alpha
DIST_PAR(l) ; DIST_PAR(2) ;
ns nf
DIST~PAR(3) ;
feed zfeed
DIS'r_PAR (4) ; DIST_PAR('j) ; IJIST_PAR I 6) ;
~
qf %
reflux vapor
DIST PAR(7) -
;
D151'.. PAR(8) ;
96
relative volatility 12. OJ)
% total number of stages (3 ) % feed stage (2 ) % feed flowrate (1 ) % feed cc)mposition, Liqht comp (0.5) sat'd liqd, % teed quality ( l 0 = sal'd vapor) (1) % reflux flowrate (3 ) ~) rE'-"boiler vapor flowrat.e ( 3 . 'j )
%
% DIS'I'_PAR(9:19) used by dist_dyn.m (distillation dynamics) % dist = distillate product flowrate
612 ~t,
f (i)
°6
Ibot II' Is
% %
Ideal Binary Distillation
Module 10
ith comp mat bal equation bottoms product flowrate liquid flow in rectifying section (top) liquid flow in stripping section (bottom)
%
VI'
% %
VS
~
x(i)
::0:-
% %
y(i)
vapor flow
~
rectifying sec
(::::: vapor + feed* (l-qf)
vapor flow ~ stripping section (~ vapor) mole frae light component on stage i, liq mole frae light component OD stage i, yap
% rectifying and stripping section liquid flowrates %
reflux; refl_ux + feed*qf;
lr Is %
% rectifying and stripping section vapor flowrates %
vs
vapor; vs + feed*(l-qf);
vr %
% distillate and bottoms rates %
dist_
Ibot
reflux; Is - vs;
VI'
%
if dist < 0 disp('error in specifications, distillate flow < 0')
return end if Ibot < 0 disp{'error in specifications, stripping section ') disp('
')
disp ( liquid flowrate is negative') return end I
1;
% zero the function vector %
f :::;: zeros(ns,l)
i
% % calculate the equilibrium vapor compositions %
for i=l:nsi y(i)=(alpha*x(i))/(l.+(alpha-l.l*x(i)); end 9"
% material balances % % overhead receiver % f (1) = (vr*y(2) - (dist+xeflux) *x(I));
%
% rectifying (top)
section
Appendix
613
'r, for i";2:nf-l;
flil=lr*xli-ll+vr*yli+ll-1r*x(il-vr*yli) ; end %
% feed stage
lr*x{nf-l)+vs*y{nf+ll-ls*x(nflvr*y(nf)+feed*zfeed;
f Infl %
% stripping (bottom) section ~;
for i=n£+1:n5-1; flil ls*x(i-l)+vs*Yli+l)-ls*xli)-vs*y(il; end % % reboiler %
f(ns)=(ls*xlns-l)-lbot*x(ns)-vs*y(ns));
dist dyn.m function xdot.
==
dist~ __dyn (t
J
x) ;
o
o
% solve for the transient ;-Jtage compositions in an ideal % binary (listillat~on column using ode45. %
% (e) 1997 B. Wayne Bequette - 24 Jan 1997 % revised 31 Dec 96 % ~t, All flowrates art::" molar quant~ities. Stages are numbered % from the top down. A total condenser is assumed. % The overhead receiver if3 stage 1. The partial reboiler % is stage DS (tIle number of equilibrium "trays" is then ~f, n8-1). Th(~ column parametE~rs should be specifi(~d in the % DISrr PAR array.
% to use this function, enter the fol1ow,ing in the command % window, or from a script file (after defining parameters % in the DIST_PAR array: % 96
[t, xl
ode451 'dist_dyn' ,to,tf,xO)
% where xO is a vector of initial values for the liquid % phase stage compositions (length(xO) ~ DS) %
7; DIST _PAR is a vector of distillation column parameters % used by both dist 88.m and dist_dyn.m
614
Ideal Binary Distillation
Module 10
if 1ength(DISTPAR) < 11; di~3P ( not enough parameters given in DTST~PAH_') disp (' ') disp ( 'check to sec that global DIST_PAR has been defined') I
return end %
alpha n;:;
n£ [cedi
zfeedi
-
DIST_PAR(I) DIS'r__ PAH (2) DIST PAR(3 ) DIS'I" PAR (4) DIS'T' -- PAR ( ~) )
% relative ~olatility (1.5) '). total number 0 f s Laq(-'-;s (41) "-,:i" feed stage' (21) % initial feed flowrate (1) % initiaJ feed composiLiol1, light comp
;
; ; ; ;
(0.5)
qf
DIST PAR(6) ;
rc;£luxl
DIST .. PAR(7) ; DIST_PAH(S) ;
% feed quality (1 = sat:' d liqd, o ~ saL'd vapor) (1) 90 initial reflux flowrate (2.706) % initial reboiler vapor £lowrate
~~
vapori mel
mb ml-
(3.206 ) d.1.stillatE' molar
~6
DIST_PAR(9) ; DIS'I' PAR (10) ; DIS'l' PAR (] 1) ;
hold~up
(5)
% bottoms molar hold-up (5) 'x sLaqe molar hold-up (D.5)
%
if length (DIST_Pt\R) stepr U:,:;tepr
==
19;
DIST_PAH(12);
st~cpv
rnS'l'. PAR(14);
tstepv stepzf tstepzf stepf
DIS'1' YAR(15) i DIST_PAR(16) i
Lst(~pf
else stepr st:epzf end
~~-;
DIST__ PAR (13); % % % 96
DIST_PAR(17)i % DIST]AR (IS); IHST PAR(19) i
';, %
magnit_udn step in reflux (O) t~ime of reflux step change (0) magnitude stc':'};) in vapor (0) time of vapor step change (0) magnitude of fc(~d comp change (0) time of feed camp change (0) magnitude of feed flow change (0) time of foed flow change (0)
0; tstepr = 0 i stepv = 0; tstepv = 0 i 0 i tstepzf = 0; stept = 0; tst.epf = 0 i
%
:(; % %
DI~_::T_PAR{9:19)
dist Ibot
LI' % %
Is vr vs x (i) xdot(i) y (i)
used by di~3t_dyn.m (cJistillat:ion dynamics) disLillate product flO1.,vrate bottoms product flowrate liquid flow in rectifying section (top) liquid flow in stripping section (bottom) vapor flow - rectifyin9 sec {= vapor + feed* (l-qf) vapor flow - stripping section (= vapor) mole frac light component on stage i liq light component ith stage mat bal equation molc.:> frac light componenl- on stagei, vap I
~6
Vs <J(,
check disturbanc0f; in rt-'lflux, and feed flowrate
%
if t < tstepr;
vapor boil-up,
feed composit::.ion
Appendix
reflux else refLux
615
reflux,i;
refluxi
+
sLepr;
('Ond
if t < tstepv; vapor i;
vapor else
sLepv;
vapori
ViJ.por
end if t: < tsLepzf; zfeed. - zf(:,-cdi;
else zfeed - zfeedi end
+
stepzf;
%
if t < tstepf; feec1i; feed else feec1.i
feed
t-
SLOpr
end (I, rectifying and strippi.nq section liquid flowrates
reflux; reflux!
lr 1s 9,5
rectifying and
feed*qf;
st~.ripp_ing
section vapor flowrates
%
vapor; vs + feed*(l-qfl;
vs VI'
% distil.late and bottoms rates %
vr -- reflux;
dist. Ibc) t:
Is -
V~;;
if di;:3t: < 0
disp('error in specifications, distillate flow < D') return end if lbat < 0
disp{'error in specifications, strippin9 section ') disp (' 'I disp ( 'liquid flowrate is
return end
% zero the function vector %
zeros (ns, 1) ; %
Df:~gative ')
616
Ideal Binary Distillation
Module 10
% calculate the equilibrium vapor compositions %
for i==l:ns; y(i)=(aIpha*x(i)I/(I.+(alpha-I.)*x(i)l; end
%
% material balances %
% overhead receiver % xdot(II=(I/md) * (vr*y(2) -(dist+reUux) *x(II); %
% rectifying (top)
section
%
for i=2:nf-l; xdot (i I = (I/mt) end
* (Ir*x (i-l I +vr*y(itl) -lr*x (i) -vr-'y( i) ) ;
%
% feed stage %
xdot (nf)
(lImt) * (Ir*x (nf-l I +vs *y (nf+ 1) ·-ls'x (nf 1-
r*y(nf)+feed*zfeed); %
% stripping (bottom) section %
for i==0£+1:os-1; xdot(il=(l/mt)*(ls*x(i-ll+vs*y(i+l)-ls*x(il-vs*y(i)) ;
end %
% reboiler % xdot (ns 1= (l/mb) * (18 *x (ns-l) -Ibot *x (ns) -vs *y (ns) I ;
INDEX
Absorption, 53, 490 Algebraic Equations linear, 52 nonlinear, 54 direct substitution, 55 false position, 60 interval halving (bisection), 58 Newton's method, 60, 66 Analysis, II Balance equations, 17 integral, 18 instantaneous (differential), 20 Batch reactor, 143 linear regression, 458 semi-batch, 155 Bifurcation continuous systems, 361 Hopi; 388 pitchfork, 362 saddle-node, 366 transcritica1,368 discrete systerns, 342 Biochemical reactor, 128,250,529 bifurcation behavior, 549
dynamic behavior, 535 linearization, 536 model, 530 Monad kinetics, 532 multiple steady-states, 540 phase-plane, 320, 539 steady-state, 534 substrate inhibition kinetics, 533 Block diagrams, 262 SIMULINK, 278, 467 Boundary value differential cquati(l!ls, 283 Chaos, 337, 396 Characteristic cqu<1tion, 145, 216, 289 Characteristic polynomial, 444, 445 Cobweb diagrams, 339 Cofactor, 440 Complex variables, 421 Constitutive relations, 27 equilibrium, 28 heat transfer, 29 kinetics, 27 gas law, 27 valves, 29
617
Index
618 Continuous stirrcc!lank reactor (CSTR) isothennal, 21, 39, 200, 20 1,293,506 diaharic, 559 Convergence, 55 Cramer's rule, 441 Critically damped, 219 Cusp catastrophe, 374, 578 Damping factor, 216 Decay ratio, 221 Decision variables, 454 Determinant. 440 conditions for stability of two-stale systems, 3 J 2 Deviation variables, J 10, 288 Diabatic reactor. 559 cusp catastrophe, 578 dynamic behavior, S(lS linearization. 567 phasc~plane, 571 motkl,561 multiple steady-states, 572 stcady~statc 563 Dimensionless models, 35 Direct sub,<.;titution, 55
l)iscretl;>timc models, 290 stale space, 291 transfer function, 291 Distributed parameter system, 6, 8, 34 Distillation, 597 dynamic behavior, 605 linearization, 606 model. 601 multiple sleady~s{ates, 608 steady-state, 603 Eigenvalues, 443 nth-order differential equations, 146 state-space models, 120 phase~plane analysis, 311 Eigenvectors, 44:i state-space models, 120 phase-plane analysis, 311 Energy, 31 kinetic, 31 internal. 31 potential, 31 total,31
Enthalpy, 31 Equilibrium point, 52 Equilibrium stage, 492, ()Ol Exothermic, 560 heal of reaction, 561, 562 Estimation discrete models, 295 first-order ITanskr functions, 195 linear regression, 452 Euler integration, XI [~xtinction, 576 liaise position, 60 ricedback, 273 Feigenbaum's nUlnber, 353 r:inal value theorem. 177 Pixed point, 52, 344 Heated mixing tank, 32, 471 Heterogeneous differential equations, 152, 289 llo!llogelleous differential equations, 145,289 Hopi' bifurcation, 3X8 diabatie reactor, 584, 587 Lorenz equations, 401 llysteresis, 371 biochernical reactor, 550 diabatic CSTH., 582 Identity matrix, 439 Ignition, 576 ImpUlse responses first-order systems, 198 MATLAB,236 second-order systems, 222 Initial conditions, 95,120,146 Initial value theorem, 177 Input variable, 37 Integrating systems, 206 Integration methods, 80 Euler, explicit, 81 r~lller, iniplicit, 82 Runge-KuHa, 88 step size, 82 Inverse matrix, 53, 441 Laplace transform, 169
Index
619 Inverse response, 227, 229 parallel processes, 271 van de V usse reactor, 520 Jacohian matrix, 169 Laplace transforms, 168 applications, 178 definition. 169 derivatives, 173 exponential, 170 integrals, 173 Initial value thcorem, 177 final valuc theorcm, 177 irupulse, 176 pulse, 174 ramp, 174 step, 17 I table, 186 time delay, 172 Lead-lag transfer functions, 208, 264 Least squares, 452 Levcllocrg-M<1Il1Ui11dt,7() Limit cycle, 323, 384 Linear regression, 452 hatch reactor, 458 decision variables. 454 least squares objective function, 454 MIITLAB,45R solution, 455 Linear systems analysis, II Linearll,ation, 109, 117, 2f<,7 Logistic equation, 334 Lorenz equations, 195 introduction, 396 model equations, 397 stability, 399 stcady~state, 398 Lumped paranlCtcr systems, 6 MIITLAB,415 background, 416 commonly used cOllllnands, 431 commonly llsed functions, 432 complex variables, 421 diary, 430 entering matrices, 417 if-then, 427
for-loops, 427 function routines, 429 MATHWORKS,431 multiplication, 421 plotting, 422 script files, 428 toolhoxes, 430 transpose,418 workspace, 419 MATLAB, functions and routines conv, 269 det, 447 eig, 446 expm, 124 feedback, 276 [solve, 70 fze:ro, 63 impulse, 236 initial, 127 ode23, 95 ode45, 95 par,:;d.lel, 272 plot, 422 polyf it, 459 polyval, 460 roots, 64,275 ~;eries, 269 ss2tf, 252 "tel', 127,234 Lf2ss.234 trace, 447 Matrix operations and properties, 436 addition, 436 co-factor, 440 determinant. 440 diagonal, 439 eigenvalue, 443 eigenvector, 443 identity, 439 inverse, 53, 441 luinors, 440 multiplication, 437 trace, 440 transpose, 4:~8 vectOr norm, 448 Matrix transfer-functions, 248 Mean value theorem, 19 Minor, 440
620
Index Mixing tank, J 91 Model definitions, 4, 5 dimensionless, 35 mathematical, 16-40 uses, 17 control system design, 17 operator training, 17 process design, 17 safety, 17 Multiplicity (multiple steady-states), II biochemical reactors, 535, 540 diabatic CSTR, 564,572 distillation, 608 input Illultiplicity, 524 outpullllultiplicity, 574
Newton's method single equation, ClO multiple equations, 66 quasi-,69 Norm Vector,44X Nth order ordinary differential equations homogeneous, 145,289 non-homogcnC(lliS (heterogeneous), 152,2:)9
time-varying parameters, 154 Numerator dynamics, 226 Objective function, 454 Optimi/.ation,454 Orbit diagrams, 342 Orthogonality, 448 ()verdmnpcd,2IX Overshoot, 221 Padc approximation, 230 Parameter, 23, 37, 38 Parameter estimation discrete models, 295 first-order transfer functions; 195 linear regression, 452 Parlial fraction expansion, 179 complex roots, 183 distinct roots, 181 repeated roots, 182 Periodic behavior, 335, 348 Phase-plane analysis, 303
biochemical reactor, 321 diabatic reactor, 572 heated mixing tank, 486 linear, 304, 311 nonlinear, 316 Pitchfork bifurcation, 363 Poles, 145, 228 Pole~l.cro cancellation, 263 Population growth, 332 Process modeL 5 developing equations, l6 Quadratic map, 334 Rank. 443 Reaction-dilfusion. 35 Recycle, 273 Rcguli falsi, 60 Re-Jati vc order, 219 Relative volatility, 599 Regression analysis, 452 Rise time, 221 ROLlth stability criterion. 157.276 Saddle point, 305 Semi-batch reactor, 155 Shooting method. 2'cl7 Sine inputs, 224 Similarity transform, 120,447 Simulation, 10 SingUlar matrices, 443 SIMULlNK, 278, 464 Space time, 508 Space velocity, 508 Stability discrete-time models, 291 fixed-point solutions, 344 numerical methods, 78, 84 state space models, 130, 311 transfer function models, 229 State-space models, ] 06 absorption, 494 biochemical reactor, 536 diahatic reactor, 567 distillation, 606 heated mixing tank, 475 isothermal CS'rR, 5 { I van de Vussc reactor. SIR
Index
621 State variables, 23, 37 Step response absorption, 496 distillation, 606
effect of poleshcros, 228 first-order SystCrllS, 193 integrating system, 20X isothermal CSTR, 203, 5J4 lead-lag systems, 209 MATLAB, 127,234 second-order systems, 2] 7 second-order with numerator dynamks, 227 statc space models, 127 stirred tank heater, 478 time-deJay systems, 231 van de Vussc reactor, 521 Surge drums gas, 26 liquid, 106, 112. 115 Symbolic analysts, 4 System, 10
Trace conditions for stability of two-stale systems, 312 Ilmtrix, 44() Transfer functions first-order. 191 first-order plus time-delay, 204 integrating, 206 lead-lag, 208 matrix, 248 second-order, 2 J 6 series, 267 cI'ranspose,438 Tutori~tl reviews linear regression, 452 MATLAB,415 matrix algebra, 435 SIMULlNK,464 vall de Vusse model linear, 293, 518 nonlinear, 516 Washout, 540
Taylor series, 60, J 09 'fhcrmodynamics, 31 Tolerance, 55 absolute, 55 relative, 55
Zeros algebraic equation, 63 of a transfer function, 22R righ{~half plane, 229
ESCOLA CC c' ::~':HARIA BIBLI01t:CA
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I CHEMICAL ENGINEERING I
Process Dynamics Modeling, Analysis, and Simulation B. , Wayne Bequette In today's highly integrated chemical process plants it is important to consider process dynamics at the process design stage. This innovative book applies the power of MATLAB and SIMULINK software to the dynamic behavior of chemical processes. Computer simulation techniques allow the application of multiple analytical strategies, both linear and nonlinear, to create accurate, adaptable models for real-life chemical processes. With an emphasis on numerical methods and the dynamic nature of chemical processes, Process Dynamics covers: • Process Modeling: material and energy balances, lumped parameter systems. scaling, and constitutive relationships • Numerical Techniques: algebraic equations, numerical integration, associated MATLAB routines • Linear and Nonlinear Systems Analysis: Laplace transforms, transfer function analysis, block diagrams, bifurcation, the quadratic map, chaos theory Special review and learning modules further integrate the software applications with the key text concepts. End-of-chapter exercises complement the more extensive case studies at the end of the book. Practical, comprehensive, and easy to use, Process Dynamics: provides a thorough treatment of this complex subject.
About the Author B. WAYNE BEQUElTE is an Associate Professor of Chemical Engineering at Rensselaer Polytechnic Institute. He holds a B.s. from the University of Arkansas and a Ph.D. from the University of Texas at Austin. He has published numerous articles on process dynamics and control, with an emphasis on nonlinear systems. He also serves an Associate Editor of Automatica, a journal of the International Federation of Automatic Control. This book is based on seven years of experience In teaching Process Dynamics and ChemiCAll Process Control. '
ISBN 0-13-206889-3 90000
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9780132 068895
