Chemical Process Control and Chemical Modeling-Stephanopulos

TABLE OF CONTENTS '

PART I: THE CONTROL OF A CHEMICAL PROCESS: ITS CHARACTERISTICS AND THEASSOCIATED PROBLEMS . . . . . . . . . . , . . . . . . . . . . . .

Chapter 1. INCENTIVES FOR CHEMICAL PROCESS CONTROL . . . . . . . . . .

1.1 Suppress the Influence of External Disturbances . . . . . .

1.2 Ensure the Stability of a Process . . . . . . . . . . . . .

1.3 Optimize the Performance of a Chemical Process . . ,. . . . .

Chapter 2. DESIGN ASPECTS OF A PROCESS CONTROL SYSTEM . . . . . . . . .

2.1 Classification of the Variables in a Chemical Process . . .

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2.2 Elements of the Design of a Control System . . . . . . . . .

2.3 The Control Aspects of a Complete Chemical Plant . . . . . ..(

Chapter 3: HARDWARE FOR A PROCESS CONTROL SYSTEM . . . . . . . . . . .

'3.1 Hardware Elements of a Control System . . . . . . . . . . .

3.2 The Use of Digital Computers in Process Control . . . . . .

SUMMARY AND CONCLUDING REMARKS ON PART I . . . . , . . . . . . . . . . . I/t,

THINGS TO THINK ABOUT ,. . . . . . . . . . . . . . . . . . . . . . . . . ,-

R E F E R E N C E S . . . . . . . . . . , . . . . . . . . . . . . . . . . . . . .

P R O B L E M S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

PART II: MODELING THE DYNAMIC AND STATIC BEHAVIOR OF CHEMICAL PROCESSES ..

Chapter 4. THE DEVELOPMENT OF A MATHEMATICAL MODEL . . . . . . . . . .

4.1 Why Do We Need Mathematical Modelign for Process Control? .

4.2 State Variables and State Equations for a Chemical Process .

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4.3 Additional Elements of the Mathematical Models . . . . . . .

4.4 Dead-Time . . . . . . . . . . . . . . . . . . . . . . . . .

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4.5 Additional Examples of Mathematical Modeling . . . . . . .

4.6 Modeling Difficulties . . . . . . . . . . . . . . . . . .

SUMMARY AND CONCLUDING REMARKS . . . . . . . . . . . . . . . .

THINGS TO THINK ABOUT . . . . . . . . . . . . . . . . . . . . . .

Chapter 5. MODELING CONSIDERATIONS FOR CONTROL PURPOSES . . . . . . .

5.1 The Input-Output Model . . . . . . . . . . . . . . . . . .

5.2 Degrees of Freedom . . . . . . . . . . . . . . . . . . . .

5.3 Degrees of Freedom and Process Controllers . . . . . . . .

5.4 Formulating the Scope of Modeling for Process Control . .


/ THINGS TO THINK ABOUT . . . . . . . . . . . . . . . . . . . . .

REFERENCES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

P R O B L E M S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

PART III: ANALYSIS OF THE DYNAMIC BEHAVIOR OF CHEMICAL PROCESSES . . .

Chapter 6. COMPUTER SIMULATION AND THE LINEARIZATION OF NONLINEARSYSTEMS . . . . . . . . . . . . . . . . . . . . . . . . .

6.1 Computer Simulation of Process Dynamics . . . . . . . . .

6.2 Linearization of Systems With One Variable . . . . . . . .

6.3 Deviation Variables . . . , . . . . . . . . . . . . . . .

6.4 Linearization of Systems With Many Variables . . . . . . .


THINGS TO THINK ABOUT . . . . . . . . . . : . . . . . . . . . .

Chapter 7. LAPLACE TRANSFORMS . . . . . . . . . . . . . . . . . . . .

7.1 Definition of the Laplace Transform . . . . . . . . . . . .

7.2 The Laplace Transforms of Some Basic Functions . . . . . .

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7.3 Laplace Transforms of Derivatives . . . . . . . . . . . .

7.4 Laplace Transforms of Integrals . . . . . . . . . . . . .

7.5 The Final-Value Theorem . . . . . . . . . . . . . . . . .

7.6 The Initial-Value Theorem . . . . . . . . . . . . . . . .


THINGS TO THINK ABOUT . . . . . . . . . . . . . . . . . . . . .

Chapter 8. SOLUTION OF LINEAR DIFFERENTIAL EQUATIONS USING LAPLACET R A N S F O R M S . . . . . . . . . . . . . . . . . . . . . . . .

8.1 A Characteristic Example and the Solution Procedure . . .

8.2 Inversion of Laplace Transforms. Heaviside Expansion . . .

8.3 Examples on the Soiution of Linear DifferentiationEquations Using Laplace Transforms . . . . . . . . . . . .



Appendix 8.A. The General Solution of an n-th OrderDifferential

Appendix 8.B. The SolutionDifferential

Equation . . . . . . . . . . . . .

of a General System of LinearEquations . . . . . . . . . . . . .

Chapter 9. TRANSFER FUNCTIONS AND THE INPUT-OUTPUT MODELS . . . . . '.

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9.1 The Transfer Function of a Process with a Single Output .

9.2 The Transfer Function Matrix of a Process with Multipleoutputs ., . . . . . . . . . . . . . . . . . . . . . . .

9.3 The Poles and the Zeros of a Transfer Function . . . . . .

9.4 Qualitative Analysis of the Response of a System . . . . .



Chapter 10. THE DYNAMIC BEHAVIOR OF FIRST-ORDER SYSTEMS . . . . . . .

10.1 What is a First-Order System? . . . . . . . . . . . . . .

10.2 Processes Modeled as First-Order Systems . . . . . . . . .

10.3 The Dynamic Response of a Pure Capacitive Process . . . .

10.4 The Dynamic Response of a First-Order Lag System . . . .

10.5 First-Order Systems with Variable Time Constant andGain . . . . . . . . . . . . . . . . . . . . . . . . . .


THINGS TO THINK ABOUT . . . . . . . . . . . . . . . . . . . . . 17 :

Chapter 11. THE DYNAMIC BEHAVIOR OF SECOND-ORDER SYSTEMS . . . . . .

11.1 What is a Second-Order System? . . . . . . . . . . . . .

11.2 The Dynamic Response of

11.3 Multicapacity Processes

11.4 Inherently Second-Order

a Second-Order System . . . . . .

as Second-Order Systems . . . . .

Processes . . . . . . . . . . . .

11.5 Second-Order Systems Caused by the Presence ofControllers . . . . . . . . . . . . . . . . . . . . . . . .

SUMMARY AND CONCLUDING REMARKS . .., . . . . . . . . . . . . .

THINGS TO THINK ABOUT . . . . . . . . . . . . . . . . . : . . .

Appendix ll.A. Examples of Physical Systems with InherentSecond-Order Dynamics . . . . . . . . . . . . .

Chapter 12. THE DYNAMIC BEHAVIOR OF HIGHER-ORDER SYSTEMS . . . . . .

12.1 N Capacities in Series . . . . . . . . . . . . . . . . .

12.2 Dynamic Systems with Dead Time . . . . . . . . . . . . .

12.3 Dynamic Systems with Inverse Response . . . . . . . . . .

SUMMARY AND CONCLUDING REMARKS . . . . . . . . . . ; . . . . .

THINGS TO THINK ABOUT . . . . . . . . . . . . . . . . . . . . . .

R E F E R E N C E S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

PROBLEMS............... i...............

PART IV: ANALYSIS AND DESIGN OF FEEDBACK CONTROL SYSTEMS . . . . . . .

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Chapter 13. INTRODUCTION TO FEEDBACK CONTROL . . . . . . . . . . . .

13.1 The Concept of Feedback Control . . . . . . . . . . . . .

13.2 Types of Feedback Controllers . . . . . . . . . . . . . .

13.3 Measuring Devices (Sensors) . . . . . . . . . . . . . . .

13.4 Transmission Lines . . . . . . . . . . . . . . . . . . .

13.5 Final Control Elements . . . . . . . . . . . . . . . . .



Chapter 14. THE DYNAMIC BEHAVIOR OF FEEDBACK CONTROLLED PROCESSES . .

14.1 Block Diagram and the Closed-Loop Response . . . . . . .

14.2 The Effect of Proportional Control on the Response of aControlled Process . . . . . . . . . . . . . . . . . . .

14.3 The Effect of Integral Control Action . . . . . . . . . .

14.4 The Effect of Derivative Control Action . . . . . . . . .

14.5 The Effect of Composite Control Actions . . . . . . . . .



Chapter 15. STABILITY ANALYSIS OF FEEDBACK SYSTEMS . . . . . . . . .

15.1 The Notion of Stability . . . . . . . . . . . . . . . . .

15.2 The Characteristic Equation . . . . . . . . . . . . . . .

15.3 The Routh-Hurwitz Criterion for Stability . . . . . . . .

15.4 The Root-Locus Analysis . . . . . . . . . . . . . . . . .



Appendix 15.A. Rules for the Construction of Root-LocusDiagrams . . . . . . . . . . . . . . . . . . .

Chapter 16. DESIGN OF FEEDBACK CONTROLLERS . . . . . . . . . . . . .

16.1 Outline of the Design Problems . . . . . . . . . . . . .

16.2 Simple Performance Criteria . . . . . . . . . . . . . . .

16.3 Select the Type of Feedback Controllers . . . . . . . . .

16.4 Controller Tuning Techniques . . . . . . . . . . . . . .



Chapter 17. THE FREQUENCY RESPONSE ANALYSIS OF LINEAR PROCESSES . . .

17.1 The Response of a First-Order System to a.SinusoidalI n p u t . . . . . . . . . . . . . . . . . . . . . . . . . .

17.2 The Frequency Response Characteristics of a GeneralLinear System . . . . . . . . . . . . . . . . . . . . . .

17.3 Bode Diagrams . . . . . . . . . . . . . . . . . . . . . .

17.4 Nyquist Plots . . . . . . . . . . . . . . . . . . . . . . .

SUMMARY AND CONCLUDING REMARKS . . . . . . . . . . . . . . . . .


Chapter lg. DESIGN OF FEEDBACK CONTROL SYSTEMS USING FREQUENCYRESPONSE TECHNIQUES . . . . . . . . . . . . . . . . . . .

18.1 The Bode Stability Criterion . . . . . . . . . . . . . .

lg.2 Gain and Phase Margins . . . . . . . . . . . . . . . . .

lg.3 The Ziegler-Nichols Tuning Technique . . . . . . . . . .

18.4 The Nyquist Stability Criterion . . . . . . . . . . . . .


THINGS TO THINK ABOUT . . . . , . . , . . . . . . . . . . . . .

Appendix 18.A. Complex Mapping and the Nyquist Criterion forStability . . . . . . . . . . . . . . . . . . .

R E F E R E N C E S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

P R O B L E M S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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PART V: ANALYSIS AND DESIGN OF ADVANCED CONTROL SYSTEMS . . . . . . . .

Chapter 19. FEEDBACK CONTROL OF SYSTEMS WITH LARGE DEAD-TIME ORINVERSE RESPONSE . . . . . . . . . . . . . . . . . . . .

19.1 Processes with Large Dead-Time . . . . . . . . . . . . .

19.2 Dead-Time Compensation i . . . . . t . . . . . . . . . . .

19.3 Control of Systems with Inverse Response . . . . . . . .



Chapter 20. CONTROL SYSTEMS WITH MULTIPLE LOOPS . . . . . . . . . . .

20.1 Cascade Control . . . . . . . . . . . . . . . . . . . . .

20.2 Selective Control Systems . . . . . . . . . . . . . . . .

20.3 Split-Range Control . . . . . . . . . . . . . . . . . . .



Chapter 21. FEEDFORWARD AND RATIO CONTROL . . . . . . . . . . . . . .

21.1 The Logic of Feedforward Control . . . . . . . . . . . .

21.2 The Problem of Designing Feedforward Controllers . . . .

21.3 Practical Aspects on the Design of FeedforwardControllers . . . . . , . . . . . . . . . . . . . . . . .

21.4 Feedforward-Feedback Control . . . . . . . . . . , . . .

21.5 Ratio Control . . . . . . . . . . . . . . . . . . . . . .



Chapter 22. ADAPTIVE AND INFERENTIAL CONTROL SYSTEMS . . . . . . . .

22.1 The Concept of Adaptive Control . . . . . . ., . . . . . .

22.2 Self-Tuning Controller . . . . . . . . . . . . . , . . .

22.3 The Concept of Inferential Control . . . . . . . . . . .



Chapter 23. EXPERIMENTAL MODELING OF CHEMICAL PROCESSES . . . . . . .

23.1 Why Do We Need Experimental Identification of ProcessDynamics? . . . . . . . . . . . . . . . . . . . . . . . .

23.2 Least-Squares Regression for Linear and NonlinearSystems ................. . . . . . . . .

23.3 Pulse Testing . . . . . . . . . . . . . . . . . . . . . .



REFERENCES . . . . . . . . . . . . . . . . . . . . . . . . . . . ;. .

PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

PART VI: MULTIVARIABLE CONTROL SYSTEMS

Chapter 24. THE CONTROL OF PROCESSES

FOR COMPLEX PROCESSES . . . . .

WITH MULTIPLE INPUTS, MULTIPLEOUTPUTS (MIMO) . . . . . . . . . . . . . . . . . . . . .

24.1 Formulation of the Control Problems . . . . . . . . . . .

24.2 Degrees of Freedom . . . . . . . . . . . . . . . . . . .

24.3 Generation of Alternative Control Systems . . . . . . . .

24.4 Practical Guides for Screening the Alternatives . . . . .

SUMMARY AND CONCLUDINGREMARKS : . . . . . . . . . . . . . . .

THINGS TO THINK ABOUT

Chapter 25. INTERACTION AND

25.1 The Interaction

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DECOUPLING . . . . . . . . . . . . . . .

of Control Loops . . . . . . . . . . . .

25.2 Selecting the Loops. The Relative-Gain Array Method . .

25.3 Design of Non-Interacting Control Loops . . . . . . . . .

SUMMARY.AND CONCLUDINGREMARKS . . . . . . . . . . . . . . . .

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Chapter 26. CONTROL OF CHEMICAL PLANTS . . . . . . . . . .

26.1 The Characteristics of the Problem . . . . . .

26.2 Selecting Control Objectives and Manipulations

26.3 The Cause-and-Effect Diagram . . . . . . . . .

26.4 A Decomposition Strategy . . . . . . . . . . .

26.5 An Example . . . . . . . . . . . . . . . . . .

SUMMARY AND CONCLUDING REMARKS . . . . . . . . . . .

THINGS TO THINK ABOUT . . . . . . . . . . . . . . . .

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REFERENCES . . . . . : . . . . . . . . . . . . . .'. . . . . . . . . .

PROBLEMS.......: . . . . . . . . . . . . . . . . . . . . . . .

PART VII: PROCESS CONTROL USING DIGITAL COMPUTERS . . . . . . . . . . .

Chapter 27. THE DIGITAL COMPUTER CONTROL LOOP . . . . . . . . . . . .

27.1 The Hardware Elements . . . . . . . . . . . . . . . . . .

27.2 The Design Characteristics . . . . . . . . . . . . . . .

27.3 A Physical Example . . . . . . . . . . . . . . . . . . ;



Chapter 28. SAMPLING AND RECONSTRUCTING CONTINUOUS SIGNALS . . . . .

28.1 Sampling Continuous Signals. The Impulse Sampler . . . .

28.2 The Reconstruction of Continuous Signals . . . . , . . .

28.3 Types of Hold-Elements and Their Characteristics . . . .



Chapter 29. DISCRETE-TIME SYSTEMS AND THE Z-TRANSFORMS . . . . . . .

29.1 Converting Continuous to Discrete-Time Systems. TheDifference Equation . . . . . . . . . . . . . . . . . . .

29.2 The z-Transform and Its Properties . . . . . . . . . . .

29.3 The z-Transform of Some Basic Functions . . . . . . ; . .

29.4 The Inversion of z-Transforms . . . . . . . . . . . . . . .

29.5 The Relationship Between Laplace and z-Transforms . . . .


THINGS TO THINK ABOUT . . . . . . . . . . , . . . . , . . . . .

Chapter 30. THE DYNAMIC RESPONSE OF SAMPLED-DATA SYSTEMS . . . . . .

30.1 The Pulse Transfer Function of a Continuous DynamicSystem . . . . . . . . . . . . . . . . . . . . . . . . .

30.2 The Transfer Function of Discrete-Time Dynamic Systems .

30.3 The Equivalence Relationship between Continuous andDiscrete Time Dynamic Systems . . . . . . , . . . . . . .

SLWARY AND CONCLUDING REMARKS . . . . . . . . . . . . . , . .

THINGS TO THINK ABOUT . . . . . . . . , . . . . . . . . . . . .

Chapter 31. FEEDBACK CONTROL USING DIGITAL CO>fPUTERS . . . . . . . .

31.1 The Block Diagram and the Transfer Function of aClosed-Loop System . . . . . . . . . . . . . . . . . . .

31.2 The Response of a Closed-Loop System and ItsCharacteristics . . . . . . . . . . . , . . . . . . . . . .

SUWARY AND CONCLUDIXG REXARKS . . . . . . . . . . . . . . . .


Chapter 32. THE DESIGN OF SAMPLED-DATA, FEEDBACK CONTROL SYSTEMS . .

32.1 Conditions for Stability of Sampled-Data Systems . . . .

32.2 The Effect of Sampling on the Closed-Loop Response ofof Sampled-Data Systems . . , . . . . . . . . . . . . . .

32.3 The Design of Sampled-Data, Feedback Loops UsingFrequency Response Techniques . . . . . . . . . . . . . .

EXJNNARY AND CONCLUDING REMARKS , . . . . . . . . . . . . . . .

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aTHINGS TO THINK ABOUT . . . . . . . . . , . . . . . . . . . . .

Chapter 33.

33.1

33.2

33.3

33.4

THE DESIGN OF ADDITIONAL SAMPLED-DATA, CONTROLCONFIGURATIONS . . . . . . . . . . . . . . . . . . . . .

Feedforward Control and Ratio Control . . . . . . . . . .

Cascade Control . . . . . . . . . . . . . . . . . . . . .

Adaptive Control . . . . . . . . . . . . . . . . . . . .

Supervisory Control . . ,. . . . . . . . . . . . . . . . .

SUMMARY AND CONCLUDING REMARKS , . . . . . . . . . . . . . . .


REFERENCES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

PROBLEMS : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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PART I

THE CONTROL OF A CHEMICAL PROCESS: ITS CHARACTERISTICS AND THEASSOCIATED PROBLEMS

The purpose of the following three introductory chapters is:

- to define what we mean by chemical process control,

- to describe the needs and the incentives for controlling a

chemical process,

- to analyse the characteristics of a control system and to for-

mulate the problems that must be solved during the design of a

control system, and finally

- to provide the rationale for studying the material that follows

in the subsequent chapters.

In order to achieve the above objectives we will use a series

of examples taken from the chemical industry. ,These examples are

usually simplified and serve only to demonstrate the various

tlualit.i3tive points made.

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CHAPTER 1

INCENTIVES FOR CHEMICAL PROCESS CONTROL

A chemical plant is an arrangement of processing units (reactors, heat

exchangers, pumps, distillation columns, absorbers, evaporators, tanks, etc.),

integrated with each other in a systematic and rational manner. The plant's

overall objective is to convert certain raw materials (input feedstock) into

desired products using available sources of energy, in the most economic,way.

During its operation, a chemical plant must satisfy several requirements

imposed by its designers and the general technical, economic and social con-

ditions in the presence of ever-changing external influences (disturbances).

Among such requirements are the following:

- Safety: The safe operation of a chemical process is a primary requirement,

for the well being of the people in the plant and its continued contri-

bution to the economic development. Thus, the operating pressures,

temperatures, concentration of chemicals, etc. should always be within

allowable limits. For example, if a reactor has been designed to operate

at a pressure up to 100 psig, we should have a control system that will

maintain the pressure below

to avoid the development of

plant.

this value. As another example, we should try

explosive mixtures during the operation of a

- Production specifications: The plant should produce the desired amounts

and quality of the final products. For example, we may require the pro-

duction of two million pounds of ethylene per day, of 99.5% purity, from

an ethylene plant. Therefore, a control system is needed to ensure that

the production level (2 million pounds per day) and the purity specifi-

cations (99.5% ethylene) are satisfied.

- Environmental regulations: Various federal and state laws may specify

that the temperatures, concentrations of chemicals and flowrates of the

effluents from a plant be within certain limits. Such regulations for

example exist on the amounts of SO2 that a plant can eject to the atmos-

phere, and the quality of water returned to a river or a lake.

- Operational constraints: The various types of equipments used in a chemical

plant have constraints inherent to their operation. Such constraints should

be satisfied throughout the operation of a plant. For example, pumps must

maintain a certain net positive suction head; tanks should not overflow or

go dry; distillation columns should not be flooded: the temperature in a

catalytic reactor should not exceed an upper limit since the catalyst will

be destroyed. Control systems are needed to satisfy all these operational

constraints.

- Economics: The operation of a plant must conform with the market con-

ditions, i.e. the availability of raw materials and the demand of the

final products. Furthermore, it should be as economic as possible in its

utilization of raw materials, energy, capital and human labor. Thus, it

is required that the operating conditions are controlled at given optimum

levels of minimum operating cost, or maximum profit; etc.

All the above requirements dictate the need for a continuous monitoring

of the operation of a chemical plant and an external intervention (control) to

guarantee the satisfaction of the operational objectives. This is accomplished

through a rational arrangement of various equipment (measuring devices, valves,

controllers, computers) and human intervention (plant designers, plant

operators), which constitutes the control system.

There are three general classes of needs that a control system is called

to satisfy:

- Suppress the influence of external disturbances,

! - ensure the 'stability of a chemical process, and

- optimize the performance of a chemical process.

Let us examine these needs using various examples.

1.1 SUPPRESS THE INFLUENCE OF EXTERNAL DISTURBANCES.

Suppressing the influence of the external disturbances on a process is

the most common objective of a controller in a chemical plant. Such dis-

1_turbances denote the effect that the surroundings (external world) have on a

reactor,

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separator, heat exchanger, compressor, etc., and usually they are out

of the reach of the human operator.- Consequently, we need to introduce a

Icontrol mechanism that will make the proper changes on the process to cancel

the negative impact that such disturbances may have on the desired operation

p

of a chemical plant.

Example 1.1 - Controlling the Operation of a Stirred Tank Heater

Consider the tank heater system shown in Figure 1.1. A liquid enters the

tank with a flowrate Fi (ft3/min), and a temperature Ti (OF), where it is

heated with steam (having a flowrate Fs, lb/min). Let F and T be the

flowrate and temperature of the stream leaving the tank. The tank is con-

sidered to be well stirred, which implies that the temperature of the effluent

is equal to the temperature of the liquid in the tank.

The operational objectives of this heater are:

- Keep the effluent temperature T at a desired value Ts.

- Keep the volume of the liquid in the tank at a desired value Vs.

The operation of the heater is disturbed by external factors like changes in

the feed flowrate and temperature Fi and Ti' If nothing changed, then

after attaining T = Ts and V=Vs, we could leave the system alone without

any supervision and control. It is clear though that this cannot be true

since T i and Fi are subject to frequent changes. Consequently, some form

of control action is needed to alleviate the impact of the changing disturbances

and keep T and V at the desired values.

In Figure 1.2 we see such a control action to keep T = Ts when Ti or

Fi changes. A thermocouple measures the temperature T of the liquid in the

tank. Then T is compared with the desired value Ts yielding a deviation

e=T -T. The value of the deviation E is sent to a control mechanismS

which decides what must be done in order for the temperature T to return

back to the desired value Ts. If E > 0 which implies T < Ts, the con-

troller opens the steam valve so that more heat can be supplied. O n t h e

contrary, the controller closes the steam valve when e-c0 or T>Ts. It

is clear that when T = Ts, i.e. E. = 0 the controller does nothing. This

control system that measures the variable'of direct importance (T in this

case) after a disturbance had its effect on it, is called Feedback control

system. The desired value Ts is called the Set Point and is supplied

externally by the person in charge of production.

A similar configuration can be used if we want to keep the volume V, or

equivalently the liquid level h, at its set point hs when Fi changes. In'.this case we measure the level of the liquid in the tank and we.open or close

the valve that affects theaeffluent flowrate F or Fi (see Figure. 1.3). It is

clear that the control systems shown in Figure 1.3 are also feedback control

systems. All feedback systems shown in Figures 1.2 and 1.3 act post facto

(after the fact), i.e. after the effect of the disturbances has been felt by

the process.

Returning back to the tank heater example, we realize that we can use a

different control arrangement to maintain T = Ts when Ti changes. Measure'

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the temperature of the.inlet stream T i and open or close the steam valve to

provide more or less steam. Such control configuration is called Feedforward

control and is shown in Figure 1.4. We notice that the'feedforward control

does not wait until the effect of the disturbances has been felt by the sys-

tem, but acts appropriately before the external disturbance affects the system,

anticipating what its effect will be. The characteristics of the feedback and

feedforward control systems will be studied in detail in subsequent chapters.

The suppression of the impact that disturbances have on the operating

behavior of processing units is one of the main reasons for the usage of con-

trol in the chemical industry.

1.2 ENSURE THE STABILITY OF A PROCESS.

Consider the behavior of the variable x shown in Figure 1.5. Notice

that at time t = to the constant value of x is disturbed by some external

factors, but that as the time progresses the value of x returns to its

initial value to stay. If x is a process variable like temperature,

pressure, concentration, flowrate, etc., we say that the process is stable

self-regulating and needs no external intervention for its stabilization.

is clear that no control mechanism is needed to force x to return to its

initial value.

o r

It

In contrast to the above behavior, the variable y shown in Figure 1.6

does not return to,its initial value after it is disturbed by external

incluences. Processes whose variables follow the pattern indicated by y

in Figure 1.6 (curves a,b,c) are called unstable processes and require

external control for the stabilization of their behavior. The explosion of a

hydrocarbon duel with air is such an unstable system. Riding a bicycle is an

attempt to stabilize an unstable system and we attain that by pedaling,

steering and leaning our body right or left.

Example 1.2 - Controlling the Operation of an Unstable Reactor

i

-’

4

Consider a continuous stirred tank reactor (CSTR) where an irreversible

exothermic reaction A+B takes place. The reaction mixture is cooled by a

coolant medium that flows through a jacket around the reactor (Figure 1.7).

As it is known from the analysis of a CSTR system, the curve that describes

the amount of heat released by the exothermic reaction is a sigmoidal function

of the temperature T in the reactor (curve A in Figure 1.8). On the other

hand, the heat removed by the coolant is a linear function of the temperature

T (curve B in Figure 1.8). Consequently, when the CSTR is at steady state,

i.e. nothing is changing, the heat produced by the reaction should be equal to5I.

the heat removed by the coolant, thus yielding the steady states Pl, P2, Pg

at the'intersection of the curves A and B (Figure 1.8). The steady states

p1 and Pg are called stable while the P2 is unstable. To understand the

concept of stability let us consider the steady state P2.

Assume that we are able to start the reactor at the temperature T2, and

the concentration cA that corresponds to this temperature. Consider that2

the temperature of the feed Ti increases. This will cause an increase in

the temperature of the reacting mixture, say T;. At, T; the heat released

by the reaction (Q;) is mqre than the heat removed by the coolant, ,Q; (see

Figure 1.8) thus leading to higher temperatures in the reactor and consequently

to increased rates of reaction. Increased rates of reaction produce larger 1I

amounts of heat released by the exothermic reaction which in turn lead to

higher temperatures and so on. Therefore, we see that an increase in Ti

.?::E:i.jtakes the reactor temperature away from the steady state P2 and that the tem-

I:. . .

- -perature will eventually reach the value of the steady state P3 (Figure 1.9a).

1:

Similarly, if Ti were to decrease, the temperature of the reactor would take.:ti off from P 2 and end up at Pl (Figure 1.9b). By contrast, if we were

operating at the steady state P3 or Pl. and we perturbed the operation of

the reactor, it would return naturally back to the point P3 or Pl where

it started from (see Figures 1.5c,d). Note: The reader should verify this

assertion.

Sometimes we would like to operate the CSTR at the middle unstable steady

state for the following reasons: (i) the low temperature steady state P1

causes very low yields because the temperature Tl is very low. (ii) the

high temperature steady state P3 may be very high causing unsafe conditions,

destroying the catalyst for a catalytic reactor, or degrading the product B,

etc.

In such cases we need a controller which will ensure the stability of the

operation at the middle steady state. Question: The reader should suggest a

control mechanism to stabilize the operation of the reactor at the unstable

steady state P2. This example demonstrates very vividly the need for

stabilizing the operation of a system using some type of control in the

presence of external disturbances that tend to take the system away from the

desired point.

,\Bi.

:*: 1.3 OPTIMIZE THE PERFORMANCE OF A CHEMICAL PROCESS.,:4.:;- Safety and the satisfaction of the production specifications are the main

1/ ::t/ ::! Once these are achieved, the

12two operational objectives for a chemical plant.

. .Given the

I

next goal is how to make the operation of the plant more profitable.: :.,!.1

','i.:

1, ,$ :\ '.,

fact that the conditions that affect the operation of the plant do not remain

the same, it is clear that we would like to be able to change the operation of

the plant (flowrates, pressures, concentrations, temperatures) in such a way

that an economic objective (profit) is always maximized. This task is under-

taken by the automatic controllers of the plant and its human operators.

Let us now see an example from the chemical processing industry where the

controller is used to optimize the economic performance of a single unit.

Example 1.3 - Optimizing the Performance of a Batch Reactor

Consider a batch reactor where the following two consecutive reactions

take place:

A +B -t C1 2

Both reactions are assumed to be endothermic with first order kinetics. The

heat required for the reactions is supplied by steam which flows through the

jacket around the reactor (Figure 1.10). The desired product is B while C

is an undesired waste. The economic objective for the operation of the batch

reactor is to maximize the profit @ over a period of time tR, i.e.

tR

,Maximize @=I

{[Revenue from the sales of product B]- [cost of purchasing

0 A + cost of steam]] dt (1.1)

where .'

revenue from product B = p cB(t)

cost of raw material A = crcA(0)

cost of steam = Ch Q(t)

P = price per lb-mole of product B

C r = price per lb-mole of raw material A

8

8

8

8

8

8

8. .

8

I8

8

8

8

8

8

8

'h = cost per lb of steam

cA (0) = concentration of A at the beginning of the batch reaction

and

tR = the period of reaction.

The only variable that we can change freely to maximize the profit is the

steam flowrate Q(t) which can vary with time. The steam flowrate will

affect the temperature in the batch reactor and the temperature in turn will

affect the rates of the desired and undesired reactions, The question is how

should we vary Q(t) with time so that the profit @ is maximized. Let us

examine some special policies with respect to Q(t).

a. If Q(t) is given the largest value that we can for the whole reaction

period tR, then the temperature of the reacting mixture will take the

largest value that is possible. Initially, when CA is large, we will

have high yields of B but we will also pay more for the steam. As

time goes on and the concentration of B increases the yield of C

also increases. Consequently, towards the end of the reaction period

the temperature must decrease, necessitating decrease in the steam

flowrate.

b. If the steam flowrate is kept at its lowest value, i.e. Q(t) = 0,

for the entire reaction period tR, we will not have any steam cost,

but also we will not have any production of B.

We see clearly from the above two extreme cases that Q(t) will vary between

its lowest and highest values during the reaction period tRa How should it

vary in order to maximize the profit is not trivial and requires the

solution of the above optimization problem.

In Figure 1.11 we see a general trend that the steam flowrate must

follow in order to optimize the profit a. Therefore, a control system is

needed which will: (a) compute the best steam flowrate for every time during

the reaction period and (b) will adjust the valve (inserted in the steam line)

so that the steam flowrate takes its best value (computed above in (a)). Such

problems as the above are known as optimal control problems.

This example indicates that the control of the steam flowrate is

not used to ensure the stability of the reactor or to eliminate the effect of

external disturbances on the reactor but to optimize its economic performance.

CHAPTER 2

DESIGN ASPECTS OF A PROCESS CONTROL SYSTEM

2.1 CLASSIFICATION OF THE VARIABLES IN A CHEMICAL PROCESS.

The variables (flowrates, temperatures, pressures, concentrations, etc.)

associated with a chemical process are classified into:

a. Input variables, which denote the effect of the surroundings on a

chemical process, and

b. output variables, which denote the effect of the process on the

surroundings.

Example 2.1

For the CSTR reactor discussed in Example 1.2 (Figure 1.7) we have:

input variables: cA , Ti, Ti, Tc , Fe(F)i i

output variables: cA, T, F, Tc , V0

Notice that the effluent flowrate F can be considered either as input or

output. If there is a control valve on the effluent stream so that its flow-

rate can be manipulated by a controller, the variable F is an input, since

the opening of the valve is adjusted externally, otherwise F is an output

variable.

Example 2.2-.

For the tank heater discussed in Example 1.1 (Figure 1.1) we have:

input variables: Fi, Ti, Fs(F)

output variables: F, V, T

The input variables can be further classified into the following

categories:

.1. Manipulated (or adjustable) variables, if their values

:can be adjusted freely by the human operator or a con-

&rol mechanism and

ii. disturbances, if their values are not the result of

adjustment by an operator or a control system.

The output variables are also classified into the following categories:

i. Measured output variables, if their values are known

by directly measuring them, and

ii. unmeasured output variables,- if they are not or cannot

be measured directly.

Example 2.3-

Suppose that the inlet stream in the CSTK system (Figure 1.7) comes from

an upstream unit over which we have no control. Then, CA , Fi, Ti arei

1 disturbances. If the coolant flow-rate is controlled by a control valve,

then

FC

is a manipulated variable, while

T is a disturbance.ci

Also, if the flowrate of the effluent stream is controlled by a valve, then

F is a manipulated variable, otherwise it is an output variable.

With respect to the output variables we have the following: T, F, Tc ,0

V are measured outputs since their values can be known easily using thermo-

couples (T, Tc ), a venturi meter (F), and a differential pressure cell (V).0

The concentration CA can be's measured variable if an analyzer (gas

chromatograph, infrared spectrometer, etc.) is attached to the effluent

stream. In many industrial plants such analyzers are not available because

they are expensive and/or have low reliability (give poor measurements or

break down easily). Consequently, in such cases cA is an unmeasured output

variable.

Example 2.4

For the tank heater system (Figure l.l>, the inputs Fi and T. are dis-1

turbances, while FS

and F are manipulated inputs. The output variables

V and T can be measured easily and they are considered measured outputs.

According to their direct measurability or not the disturbances are

classified into two categories: the measured and the unmeasured disturbances.- - - -

2 . 5Example

The disturbances Fi and Ti of the stirred tank heater (Figure 1.1)

are easily measured; thus they are considered measured disturbances.

On the other hand, the feed composition for a distillation column,

extraction unit, reactors and the like, is not normally measured and conse-

quently is considered an unmeasured disturbance.

As we will see later on, unmeasured disturbances generate difficult con-

trol problems.

Figure 2.1 sununarizes all the classes of variables that we have around a

5 chemical process.

2.2 ELEMENTS OF THE DESIGN OF A CONTROL SYSTEM

Let us see now what are the basic questions that we must ask while

attempting to design a control system that will satisfy the control needs for

a chemical process.

A. Define Control Objectives

want

The central element in any control configuration is the process that we

to control. The first question that is raised by the control designer

is:

Question 1: "What are the operational objectives that a control

system is called to achieve?"

The answer to this question determines the so-called control objectives. They

may have to do with:

- Ensuring the stability of the process, or

- suppressing the influence of external disturbances, or

- optimizing the economic performance of a plant, or

' - combination of the above.

At the beginning the control objectives are 'defined qualitatively and sub-

sequently they are quantifi'ed, usually in terms of the output variables.

Example 2.6

For the CSTR system discussed in Example 1.2 (Figure 1.7), the control

objective (qualitatively defined) is to ensure the stability of the middle,

unstable steady state. But such a qualitative description of the control

objectives is not useful for the design of a control system and must be

quantified. A quantitative translation of the qualitative control objective

requires that the temperature (an output variable) does not deviate more than

5% from its nominal value at the unstable steady state. 8 L

Example 2.7

For the stirred tank heater of Example 1.1 the control objectives are to

maintain the temperature of the outlet (T) and the volume of the fluid in the

tank at desired values. For this example the quantification of the control

objectives is direct and straightforward, i.e. *

T = Ts

v = vs

where T and vS s are given, desired values.

Example 2.8

For the batch reactor of Example 1.3 the qualitative control objective

is the maxfmization of the profit. The quantitative description of this

objective is rather complex. It requires the solution of a maximization

problem, which will yield the value of the steam flowrate, Q(t), at each in-

stant during the reaction period.

B. Select Measurements

Whatever are our control objectives, we need some means to monitor the

performance of the chemical process. This is done by measuring the values of

certain processing variables (temperatures, pressures, concentrations, flow-

rates, etc.). The second question that arises is:

Question 2: "What variables should we measure in order to monitor

the operational performance of a plant?"

It is self-evident that we would like to monitor directly the variables that

represent our control objectives, and this is what is done whenever possible.

Such measurements are called primary measurements.

Example 2.9

For the tank heater system (Example 1.1) our control objectives are to

keep the volume and the temperature of the liquid in the tank at desired

levels, i.e.

keep T = T and VS

= vs.

Consequently, our first attempt is to install measuring devices that will

monitor T and V directly. For the present system this is simple by using

a thermocouple (for T) and a differential pressure cell (for V).

Sometimes it happens that our control objectives are not measurable

quantities, i.e. they belong to the class of unmeasured outputs. In such

cases we must measure other variables which can be measured easily and

rcl..Lably. Such supporting measurements are called secondary measurements.- - --

Then we develop mathematical relationships between the unmeasured outputs

and the secondary measurements, i.e.

unmeasured output = f (secondary measurements)

which allow us to determine the values of the unmeasured outputs (once the

values of the secondary measurements are,available). In a subsequent chapter

we will see that the above mathematical relationship between measured and

unmeasured outputs results from empirical, experimental or theoretical

considerations.

Example 2.10

Consider a simple distillation column separating a binary mixture of

pentane and hexane into two produce streams of pentane (distillate) and

hexane (bottoms). Our control objective'is to maintain the production of a

distillate stream with 95% by mole in pentane in the presence of changes in

the feed composition

It is clear that our first reaction is to use a composition analyzer to

measure the concentration of pentane in the distillate and tllcn using fcrtl-

back control to manipulate the reflux ratio, so that we can keep the

distillate 95% in pentane. This control scheme is shown in Figure 2.2a. An

alternative control system is to use a composition analyzer to monitor the

concentration of pentane in the feed. Then in a feedforward arrangement we

can change the reflux ratio to achieve our objective. This control scheme is

shown in Figure 2.2b. Both of the a'bove control systems depend on the compo-

sition analyzers. It is possible that such measuring devices are either very

costly or of very low reliability for an industrial environment (failing

quite often or not providing accurate measurements). In such cases we can

measure the temperature at various plates along the length of the column quite

reliably, using simple thermocouples. Then using the material and energy

balances around the plates of the column and the thermodynamic equilibrium

relationship between liquid and vapor streams, we can develop a.mathematical

relationship that gives us the composition of the distillate if the tem-

peratures of some selected trays are known. Figure 2.2~ shows such a control

scheme that uses temperature measurements (secondary measurements) to estimate

or infer the composition of pentane in the distillate, i.e. the value of the

control objective.

The third class of measurements that we can make to monitor the behavior

of a chemical process includes the direct measurement of the external dis-

turbances. Measuring the disturbances before they enter the process can be

Ghighly advantageous because it allows us to know a priori what the behavior

of the, chemical process will be and thus take remedial control action to

alleviate any undesired consequences. Feedforward control uses direct

measurements of the disturbances (see Figure 1.4).

c. Select Manipulated Variables

Once the control objectives have been specified and the various measure-

ments identified, the next question is how do we effect a change on the

process, i.e.

C&estion 3: "What are the manipulated variables to be used in order

to control a chemical process?"

Usually in a process we have a number of available input variables which

can be adjusted freely. Which ones we select' to use as manipulated variables

is a crucial question as the choice will affect the quality of the control

actions we take.

Example 2.11

In order to control the level of liquid in a tank we can either adjust._Ithe flowrate of the inlet stream (Figure 1.3b) or the flowrate of the outlet

stream (Figure 1.3a). Which one is better is an important question that we

will analyse later.

31

D. Select the Control Configuration

After the control objectives, the possible measurements, and the available

manipulated variables have been identified, the final problem to be solved is

that of defining the control configuration.

Before we define what a control configuration is, let us look at some

control systems with different control configurations.

The two feedback control systems in Figures 1.3a and 1.3b constitute two

different control configurations. Thus, the same information (measurement of

liquid level) flows to different manipulated variables, i.e. F (Figure 1.3a)

and Fi (Figure 1.3b). Similarly, the feedback control system (Figure 1.2)

and the feedforward control system (Figure 1.4) for the tank heater constitute

two distinctly different control configurations. FOP these two control sys-

tems we use the same manipulated variable, i.e. Fs but different measurements.

Thus, for the feedback system of Figure~l.2 we use the temperature of the

liquid in the tank, while for the feedforward system of Figure 1.4 we measure

the temperature of the inlet.

In the above examples we notice that two control configurations differ

either in:

- The information (measurement) flowing to the same manipulated variable or

- the manipulated variable where the information flows to.

Thus, for the two feedback control systems in Figures 1.3a and 1.3b we

use the same information (measurement of the liquid level) but different

manipulated variables (F or Fi). On the contrary, for the control systems

in Figures 1.2 and 1.4, we have different measurements (T or Ti) which are

used to adjust the value of the same manipulated variable (Fs).

Later on we will also study other types 0E control configurations, but

for the time being we can define a control configuration (or control---.- ~. . . __structure_)

as follows: :

Definition--II_-

Control configuration we will call the information structure which is

used to connect the available measurements to the available manipulated

variables.

It is clear from the previous examples that normally we will have many

different control configurations for a given'chemical process, which raises

the following question:

Question 4: "What is the best control configuration for a given

chemical process control situation?"

The answer to this question is very critical for the quality of the con-

trol system we are asked to design.

Depending on how many controlled outputs and manipulated inputs we have

in,a chemical process we can distinguish the control configurations into:

single-input, single-output (SISO) or

multiple-input, multiple-output (MIMO) control systems.

For example, for the tank heater system:

- If the control objective (controlled output) is to keep the liquid level

at a desired value by manipulating the effluent flowrate, then we have a

SISO system.

- On the contrary, if our control objectives are (more than one) to keep the

level and the temperature of the liquid at desired values, by manipulating

(more than one) the steam flowrate and the effluent flowrate, then we have

a MIMO system.

In the chemical industry most of the processing systems are mulitple-

input, multiple-output systems. Since the design of SISO systems is simpler

we will start first with them and progressively we will cover the design of

MIMO systems.

Let us close this paragraph by defining three general types of control

configurations.

a. Feedback control configuration: Uses direct measurements of the con-- - -

trolled variables to adjust the values of the manipulated variables (Figure

2.3). The objective is to keep the controlled variables at desired levels

(set points). We can see examples of feedback control in Figures 1.2 and 1.3.

b. Inferential control configzration: Uses secondary measurements, because

the controlled variables are not measured, to adjust the values of the mani-

pulated variables (Figure 2.4). The objective here is to keep the (unmeasured)

controlled variables at desired levels.

The estimator uses the values of the available measured outputs, along

with the material and energy balances that govern the process, to compute

mathematically (estimate) the values of the unmeasured controlled variables.

These estimates in turn are used by the controller to adjust the values of the

manipulated variables. An example of inferential control configuration can be

seen in Figure 2.2~.

C . Feedforward control configuration. Uses direct measurements of the dis-

turbances to adjust the values of the manipulated variables (Figure 2.5).

The objective here is to keep the values of the controlled output variables

at desired levels. An example of feedforward control configuration we can see

in Figure 1.4.

E. Design the Controller

In every control configuration, the controller is the active element that

receives the information from the measurements and takes appropriate control

actions to adjust the values of the manipulated variables. For the design of

the controller we must answer the following question:

Question 5: "How is the information taken from the measurements-

used to adjllst the val.ues of the manipulated variables?"

The answer to this question constitutes the control law, which is imple-_

mented automatically by the controller.

Example 2.12

Let us consider the problem of controlling the liquid level (h) in a

tank (Figure 2.6), in the presence of changes in the inlet flowrate Fi. Our

measurement will be the liquid level and the manipulated variable the outlet

flowrate. The feedback control configuration used is shown in Figure 2.6.

The question is: "How should F change with time to keep the liquid level

constant wlien 17 i changes?" In other words, we want to develop the control

law.

Let us assume that the heater has been operating for some time and that

1 its liquid level has been kept constant at hs while the liquid temperature

has remained constant at a value T We say that the heater has beens' I

operating, at a steady state (where nothing changes). Under these conditions

the material balance around the tank yields,

0 = F i,s - Fs (2.1)

where Fi,s

and Fs are the inlet and outlet flowrates at steady-state.

Let hs be the liquid level corresponding to steady state operation. Suppose

that the Fi increases suddenly as it is shown in Figure 2.7. If nothing is

done on F, the liquid level h will start rising with time. How h changes

with time will be given from the transient material balance around the tank,

i.e.

dVdt = Fi - F

or

Adh=F -Fdt i

where A is the cross sectional area of the tank.

(2.2)

Subtract eqn. (2.1) from (2.2) and take

A * zdt

(Fi - Fi,s)

- (F -u Fs)

Ad(h - hs>

dt = (Fi - Fi,s) - (F - Fs) (2.3)

since hS= const.

The variable h = h - hs denotes the error or deviation of the liquid- -

level from the desired value hs. We want to drive this error to zero by

manipulating the flowrate F.

The simplest control law is to require that the flowrate F increases

or decreases proportionally to the error h - h S ’i.e.

F = a(h - hs) + b (2.4)

This law is called Proportional Control law, and the parameter a is known as

Proportional Gain.- -

From equation (2.4) we notice that when h - hs = 0 then F = Fs and

consequently b = Fs. Thus the proportional control takes the form,

F = Fs + a(h - hs) (2.5)

If we substitute F given from equation (2.5) into equation (2.3) we

take,

d(hA

- $1dt + a(h - hs) - Vi - Fi,J (2.6)

This last differential equation is solved for (h - hs), and for various

values of the proportional gain a yields the solutions shown in Figure 2.8.

We notice that none of the solutions is satisfactory since h - hs # 0. Thus,

we conclude that the proportional control law-is not acceptable.

Considerable improvement in the quality of the resulting control can be

obtained if we use a different control law known as Integral Control.

According to this law the value of the manipulated variable F is proportional

to the time integral of the error (h - hs), i.e.

F = a' (h - hs)dt + b'

0

When we are at steady statd (h - hs)dt = 0 and F = Fs. Consequently,

b' = Fs. Thus, the 0 integral control law takes the form

F=F + a 'S

(h - hs)dt

0

Substituting F from eqn. (2.7) into eqn. (2.3) we take,

(2.7)

(2.8)

The solution of eqn. (2.8) for various values of the parameter a' is shown

in Figure 2.9. We notice that integral control is an acceptable control law

since it drives the error h - h to zero. We also notice that depending on1 S

the value of a' the error h - hs returns to zero faster or slower;

oscillates for a longer or shorter time, etc. In other words, the quality of

control depends on the value of a' in a very profound manner. Note: In

subsequent chapters we will see how to solve integrodifferential equations

like eqn. (2.8).

Combining the proportional control action with the integral control

action we have a new control law, known as Proportional-Integral Control.

According to this law the value of the outlet flowrate is given by,

F = Fs - a(h - hs> - a'I

(h - hs)dt

0

37

In subsequent chapters we will study the characteristics of various forms

of control laws, but it should be remembered that the selection of the

appropriate control law is a very important question to be decided by the

chemical engineer control designer.

2.3 THE CONTROL ASPECTS OF A COMPLETE CHEMICAL PUNT- -

The examples that we discussed in the previous sections were concerned

with the control of single units like a CSTR, a tank heater, and a batch

reactor. lt should be emphasized a,s early as possible that rarely if ever is

a chemical process composed of one unit only. On the contrary, a chemical

process is composed of a large number of units (reactors, separators, heat

exchangers, tanks, pumps, compressors,, etc.) which are interconnected with ,

each other through the flow of materials and energy. For such a process the

problem of designing a control system is not simple but it requires experience

and good chemical engineering background.

Without dwelling too much on the control problems of integrated chemical

processes, let us see some of their characteristic features which do not show

up in the control of single units,,

Example 2.13

Consider a simple chemical plant composed of two units: a CSTR and a

distillation column (Figure 2.10). The raw materials entering the reactor are

A and B with flowrates FA, FB and temperatures TA, TB respectively.

They react to yield C, i.e.

A + B - C

The reaction is endothermic and the heat is supplied by steam around the

jacket of the reactor. The mixture of C, plus unreacted A and R enters

the distillation column where A + B is separated from the top as the over-

head product and C is taken as the bottoms product.

The operational objectives for this simple plant are:

1. Product specifications:

- keep the flowrate of the desired product stream

FP

at the specified level, and

- keep the required purity of C in the product

stream.

ii. Operational constraints:

- do not overflow the CSTR, and

- do not flood the distillation column, or let it go dry.

iii. Economic considerations:

- Maximize the profit from the operation of this plant.

Since the flowrate and the composition of the product

stream are specified, maximizing the profit is

equivalent to minimizing the operating costs. It

should be noted that the operating cost involves the

cost for purchasing the raw materials, the cost of

steam used in the CSTR and the reboiler of the dis-

tillation column, as well as the cost of the cooling

water used in the condenser.

The disturbances that will affect the above operational objectives are:

- The flowrates, compositions, and temperatures of the streams of the two

raw materials.

- The pressure in the distillation column.

- The temperature of the coolant used in the condenser of the distillation

column. (For example, if the coolant is water it will have a different

temperature during the day time than during the night.)

At first glance the problem of designing a control system even for this

simple plant looks very complex. Indeed it is.

The basically new feature for the control design of such a system is the

interaction between the units (reactor, column). The output of the reactor

affects in a profound way the operation of the column and the overhead product

of the column influences the conversion in the CSTR. This tight interaction

between the two units complicates seriously the design of the control system

for the overall process.

Suppose that we want to control the composition of the bottoms product by

manipulating the steam in the reboiler. This control action will aEfect the

composition of the overhead product (A+B) which in turn will affect the

reaction conversion in the CSTR.

On the other hand in order to keep the conversion in the CSTR constant

at the desired level, we try to keep the ratio *A'53 = constant and the tem-

perature T in the CSTR constant. Any changes in FA/FB or T will affect

the conversion in the reactor and thus the composition of the feed in the

distillation column. A change in the feed composition of the column will

affect the purity of the two product streams.

The control of integrated processes is the basic objective for a chemical

engineer. Due to its complexity though, we will start by analyzing the cgn-

trol problems for single units and eventually we will treat the integrated

processes.

CHAPTER 3

HARDWARE FOR A PROCESS CONTROL SYSTEM

In the previous chapter we examined the various considerations that must be

taken into account during the design of a control system and the associated

problems that must be resolved. In this chapter we will discuss the physical

elements (hardware) constituting a control system as it is implemented in

practice for the control of real physical processes.

3.1 IIARDWARI:, ELRMMENTS OF A CONTROL SYSTEM.- - - -

In every control configuration we can distinguish the following hardware

elements:

a. The chemical process: It represents the material equipment together with

the physical or chemical operations that occur there,

b. The measuring instruments or sensors: Such instruments are used

to measure the disturbances, the controlled output variables or to measure

secondary variables, and are the main sources of information about what is going

on in the process. Characteristic examples are: ,

- thermocouples or resistance thermometers, for measuring the temperature,

- venturi meters, for measuring the flowrate,

- gas chromatographs, for measuring the composition of a stream, etc.

A mercury thermometer is not a good measuring device to be used for con-

trol since its measurement cannot be readily transmitted. On the other hand the

thermocouple is acceptable because it develops an electric voltage which can

be readily transmitted. Thus, transmission is a very crucial factor in selecting

the measuring devices.

Since good measurements are very crucial for good control, the measuring

devices should be rugged and reliable for an industrial environment.

C . Transducers or transmitters: Many measurements cannot be used for con-

trol until they are converted to physical quantities (like electric voltage

or current, or a pneumatic signal, i.e. compressed air or liquid) which can be

transmitted easily. The transducers or transmitters are used for that purpose.

For example, the Strain Gauges are metallic conductors which change their

resistance when subjected to mechanical strain. Thus, they can be used to

convert a pressure signal to an electric one.

d. Transmission lines: They are used to carry the measurement signal from

the measuring device to the controller. In the past the transmission lines

were pneumatic (compressed air or compressed liquids) but with the advent of

the electronic analog controllers and especially the expanding usage of

digital computers for control, the transmission lines carry electric signals.

Many times the measurement signal coming out from a measuring device is very

weak, and it cannot be transmitted over a long distance. In such cases the

transmission lines are equipped with amplifiers which raise the level of the

signal. For example, the output of a thermocouple is of the order of a few

mil 3. iv0I.t s . Before it is transmitted to the controller, it is amplified to

the level of a few volts.

e. The controller: This is the hardware element that has "intelligence".

It receives the information from the measuring devices and decides what

action should be taken. The older controllers were of limited ,"intelligence",

could perform very simple operations and implement simple control laws.

Today with the increasing usage of digital computers as controllers the

available machine intelligence has expanded tremendously, and very compli-

cated control laws can be implemented.

f. The final control element: This is the hardware element that implements

in real life the decision taken by the controller. For example, if the

controller “decides” that the flowrate of the outlet stream should be increased

(or decreased) in order to keep the liquid level in the tank at,the desired

value (see Example 1.1, Figure 1.3a), it is the valve (on the effluent stream):

that will implement this decision, opening (or closing) by the commanded‘,

amount .

The controi valve is the most frequently encountered final control element but

not the only one. Other typical final control elements for a chemical proces,s

are : -

- Relay switches, providing on-off control,- Relay switches, providing on-off control,,,

11 ‘I‘I,. *:‘~‘I,. *:‘~‘I ..- variable speed pumps,-- variable speed pumps,-

‘I, 1‘I, 1 ,. . .,. . . i ti t ..- variable speed compressors,- variable speed compressors,

.*.*ii I’I ’etc.etc.

ll

,?,?&.?’ ,’ .s ~!&.?’ ,’ .s ~! SiSi **

83.83. ‘Recording ‘elemen’ts :‘Recording ‘elemen’ts : These are used to provide. a visual demonstration ofThese are used to provide. a visual demonstration of) si) si

how the chemical firoce&‘behaves.how the chemical firoce&‘behaves.3:3: :: 11 :: i;i;

Usua%~y the variables recorded are theUsua%~y the variables recorded are theI “~._<.I “~._<. :: -1 a,.‘$-1 a,.‘$ 2.2.

variables which are directly Ameasured as a part of the control &s‘tw.variables which are directly Ameasured as a part of the control &s‘tw.2’:; ! ._ g:.2’:; ! ._ g:. . _,’ .) 3. _,’ .) 3II *,*, 1 .,.*1 .,.* .,.,

Various types-of ra~orders”,(te&erature, pre_ssure,“.flewrate; composition, etc.)Various types-of ra~orders”,(te&erature, pre_ssure,“.flewrate; composition, etc.).’. ’

can be seen in the control room of: a chtsmica), p&ant, monitor&q continuouslycan be seen in the control room of: a chtsmica), p&ant, monitor&q continuously,.....,.....

the behavior of the p&e&$” “.the behavior of the p&e&$” “.3.r” ,‘r..3.r” ,‘r.. ,.,: :,.,: : / ,.~J :, >;/ ,.~J :, >; *z*z ‘51 ”‘51 ”

The -recent introduction of the digital computersThe -recent introduction of the digital computers

in the processin the process44 j ;j ; // ., .(^., .(^ ‘,‘, ,.=,.=‘“T” ,,I‘“T” ,,I

d&trol has also expanded the rekording opportunities, throughd&trol has also expanded the rekording opportunities, through

the video display units ‘(‘VQU).the video display units ‘(‘VQU). ‘.‘- ”‘.‘- ” LL.G.G ,.,.

,, f.,, f. .fi.fiFigure 3.1 des&ib& the hhrdiare el.esien~ts used for the cqntrdl of theFigure 3.1 des&ib& the hhrdiare el.esien~ts used for the cqntrdl of the

: .:: .:stirred tank heater.stirred tank heater.

.-.-

3.2, THE USE OP. DTCITAL -COMPUTES TN : PROCRSS. ,CONTRO&.3.2, THE USE OP. DTCITAL -COMPUTES TN : PROCRSS. ,CONTRO&.,,

;; ::II

T h e r a p i d technologica+ @+opment of,d$gital.,.eomputers during t h e l a s t *The rapid technologica+ @velopment of,d$gital.,.eomputers during the last *,,

ten years,ten years, coupled with. s$gpifieant .reduction of their cost, had a very pro-coupled with. s$gpifieant .reduction of their cost, had a very pro-. ,. , ; . . .; . . .

found effect on how the ohemical-plants are controlled.*,‘The expected futurefound effect on how the ohemical-plants are controlled.*,‘The expected future

il’ :il’ ::: . .. .

,

,

ation of the control design tech& " ::improvements aqng with the growing sophistic:. . ' i

niquesmake the digital computer the centerpiece.for the development of a 9 .,;'-~'-+:f7;r.

control syitem for chemical processes.. , ..:' ,i 'i :I:,i -, ._ . .'

i Already large chemi'cal plants like petroleum refineries, ethylene pla&,.-. i,:a i. I I 8. ti ' / , 1~ Ji, O. .r

ammonia,plants and many others.

, are wnder digital computer control. The ::I * -<

effects have* been-very substantial,I' 4‘! ._: , .*vleading 'to better-control andreduced

i,operating costs. 1 r '.,.,

-'~. _ 7 '.‘ : .. ).. ' '_ '.,In the past the control laws that ceuld be imoloment& hv a-rontroller, ~ -..= ---_-__--- -, - --.

:.were very simple like the proportional or progotifional-integral control wa) i

discussed in'section'2.2. .The fu~. ,:i ,' 'i I '5 ," '. w

en& revolution introduced by,fhes., ". ,,

digital computer in the.prahtice'of process'Mntro~1 is the virtually unlimited,"' !,. ._,; '.

intelligence that can be exhibited;by such 'un,irs. .,'I&& phencmenon imilies .: >.';,'. _, :

that the control laws that can.be u,s‘ed are'&.u$ m&e comple?i ar&sophisficsted.., 7". ,~ ,::, ( ) .;.z,: .‘C i *.. ,:.T.

Furthermore, the'digital computer with i!--,eas,ily yropr d inherent *!I

intelligence ": ; . . J3 :, 1', ,~ 'II

can learn" as it receivks aieasur&en$s from-the proces,s,r

and it1 i ,', .,,*,. ,%' ;l;<:~, >: : 79; 1

can' change the control law that is imple&en&& in:tfie 3;1_ . 7:s: ,.-4, '. actual' :.operation of*-' 1 ,(

the plant.< .:. ; ' b ,c>.: ,:

The digita computers have found very~~diversi$&@- control applicat$+sI, ,, j I . 3 '~ * i ,". 2'.

in the process industry. In subsequent cha'&zrs we,pill.study both the. .';. j - !?'. ,::,c* 'p : -*, : .&',I',, ,*, '.theoretical and practical aspects associated with the us,e of,,digital computers

'** "j$ 7 J,( ‘&Afor process control. In thefollow%ng psragrophs~,$a i'the time being, we

./ #~,*":.; j : :. :.'r.",, ,J,( l,r *_. ,,'.*, '5::" .will diecues some applications characreristic'of'.the diverse usaBe gf the

'; 1~ '.digital computers.

;)*f.

',.l,.,a. , Direct Dig&al Control (DDC$i .'~R,.BUldhl86~~'i~~~~o~~~ the c&put& riceivesII <,+. : , ?

>_directly the measurements from'theSproces8' and b&&6'&~theicontrol law, which/ Iis @ready programmed and,resides %a ite:?eolrjrjr,~~alh~ata;Lb~the'valuos of the

,, :,manipulated variables, i These,dec&ons, are,nois- implemented direc$ly on the

process by the computer through the proper adjustment of the final control,

elements (valves, pumps, compressors, switches, etc.). This dfrect imple-

mentation of the control decisions gave rise to .the name direct digital con-

trol, or sim& DDC.' Figure 3.2 illustrates a typical DDC configuration.

The process can be a&of the units'we,have already considered such as; heaters, reac

separators, etc. The two interfaceslbefore and after' the &&pitter are hard-

ware element6 and they are used to create the interface 1 between the computer

and the' process.: In a let& cbapter"we.wiil diecuss the nature of these'

interfaces. Finally., the human operator c&"%nteradt~‘ititb 'the Computir and

affect the operation of the DDC'si

Today the chemical industry is moving more and more towards the DDC of

the plants. A typical system oYf DDC's for an ethylene plant can includei ' ,.*.Pbetween 300 and 400 control loops. aAl1 the compan&es which furnish the con-

'., i -,, ‘, r /: ,,.," 'trol systems for the chemical industry: rely 'more and more on.,DDC.

1 ' ,_ ad,,, '; L ,-. ! I " v.b . Supervisory computer control: -Aswe discussed earlier one of ;the

incentives for process control is the .opt&zat%on,of the‘ pla&'s economic.. i _ . I, i I I,a A.. .:r " .; i :

performance. Many times the human operator‘does not or cannot find the best ,.z *;,- *,'&! .'. :".r. ' :A. ' ; *$..j

operating policy for a plant which will minidil&th~ operating cost. This'y.. _.( t ?'S '/, I1 ,.. ; is. : ! p 'deficiency is due to the -enormous esmplexity of.a'typlcal chemical plant.'

,.. ; ," t,, j .*$~1 , j ; I,' : ( >ti L +..i \ I,:..In such cases we can use the,s#eed &d the progreamn;hd intelligence of a

:' 4; ..J.% .: /" * I..:,* 1 .: /,_.b. 5" :j: .%%. ,,. j : .,i'Idigital computer to analyze the sit&j&on and'to‘suggest the best policy... . , 2;s ," ,. j .: 1 _ I': I. , j

In doing so the computer coordinates the act$vit&es of the basjc DDC loops, :

(see Figure 3.3).

C* Sch,eduling computer control. Finally, the computer can be.used to-.I

schedule the operation kof a plant. For example, the conditions in the market

(demand, supply, prices) change with time, requiring the'management of the ):

chem&l plant to change its op.erational schedule like cutting:production to

avoid overstocking, increasing prodticti'on to meet the demand, changing-over

to a new production line, etc.

,

.,

,, !..,, !..These decision can be $made rationally with then aid &.a digital cpmputerThese decision can be $made rationally with then aid of.a digital computer

. . .’. . .' ii ,_.‘i, I,_.'i, I ~~ . . r. . r -3-3 77

which in, turn will communicate these ds$.sions to the supenrisptiy computerwhich in, turn will communicate these ds$.sions to the supenrisptiy computer./ .;./ .;

controllers.controllers.. .. . Pinally, these supervisory eontrollers will implement the,sePinally, these supervisory eontrollers will implement the,se.I.I ; / . ..'I; / . ..'I ,<,< .i".i",, 22 .,..'.,..'

decisions on the chemical plant through$the ;DDC'8+decisions on the chemical plant through$the ;DDC'8+-II-II !. .L!. .L.i.i ,, *,, * II

In subsequent shapters we ~i$l. +++Z .px!,edominantly~with &he DDC and aIIn subsequent shapters we ~i$l. +++Z .px!,edominantly~with &he DDC and aI;";" ',', // .s.s , !, ;‘, !, ;‘ -^-^ .'.' ,- .:,- .: ii

little with superyisory computer %ent&.~,wh~ile we..$lllittle with superyisory computer %ent&.~,wh~ile we..$ll not'conkern ourselves -': .I: .I ii not'conkern ourselves'.i'.i,'.t :,'.t :/I . ,./I . ,. 2.2.with 'the scheguling computer c*t$;ol $$ch $~,the subject u&pr qf ii dif- :with 'the scheguling computer c*t$;ol $$ch $~,the subject u&pr qf ii dif- :

I. :I. :ferent field.ferent field. ';';

i-i- ._' / "._' / " _, .I_, .I'_'_.<'.<' ,.,.

CONCLJDlNG R&g ON PART ICONCLJDlNG R&g ON PART Ir d,. _,r d,. _, / './ '. ..;' ,,..;' ,,

',',LL i.i.:(.:(. : .j: .j ',',

It is hoped that the reader now his a sk&chy outl$ne oftIt is hoped that the reader now his a sk&chy outl$ne oft,.*:. -;:..,.*:. -;:.. ..,Tf 6 _' , ~5:,Tf 6 _' , ~5: :,:' . i ,;, ;:,:' . i ,;, ; _ I_ I

I.- 'I%0 iced8 &id 'the incen~tives for' 'procees ~:onerol,:~I'~ .<I.- 'I%0 iced8 &id 'the incen~tives for' 'procees ~:onerol,:~I'~ .<I ,l.I ,l. ,,_,,_

St \,St \,,* /_ ‘:;;“‘,* /_ ‘:;;“‘ ,~ 1 > -i $ ',, ',,~ 1 > -i $ ',, ', .',..',. 1. .:1. .: ,;,.$;t "' :,;,.$;t "' : ;* :;* :

- the basic questions involved'during the design 6f'a control syst.en‘for a- the basic questions involved'during the design 6f'a control syst.en‘for a/ ** */ ** * : 3,.: 3,..,,r+ .' :- '.,,r+ .' :- ' ~~ : 14: 14/'l.l :/'l.l :

chemical process,chemical process, b ( * .,'b ( * .,' III.,. . II.,. . I , :a:$,' ', :a:$,' ' ,s , (' ),,s , (' ),,, .,, .- the‘hardware elements involved in a; &on& 8ys't&~;&,4- the‘hardware elements involved in a; &on& 8ys't&~;&,4

II :: 'j-k'* .;' 'fJ :" * ,?j;i -,z.f ,,'j-k'* .;' 'fJ :" * ,?j;i -,z.f ,, .:," ,'.:," ,' -F ..;i-F ..;i- the importance of the digital,comput&s~f& the pres.en.t"a~ future- the importance of the digital,comput&s~f& the pres.en.t"a~ future

,, '; .,, '; . * ,* ,~I~I ,z d$:'yje ; I,z d$:'yje ; I ‘‘implementation of advanced control'teohriiques.~ '~~~'~~~~implementation of advanced control'teohriiques.~ '~~~'~~~~

, :, :__I. jI. j I *:,rI *:,r ,,&:$;.! -1,. ,$;? : '. I (.: b,,&:$;.! -1,. ,$;? : '. I (.: b ""

In the remainingIn the remaining chapters we~will.~~s&~t'~ ay~tkmatic analysis &,thechapters we~will~~s&~t'~ ay~tkmatic analysis &,the

various questions raised in thi.s chept$F:, -wfW'the final,

design a rational control system f,or a':$i.v&n proce:.,. ,I ,) ,( ‘i :; (. :' _ " .',,. ;:.‘i$. i. ,.chapters will be leas chatty and more ‘&&xH$~‘~‘~-~’

*’

I,

REFERENCESi :I ( 1

. ..ChaRter 1: Numerous examples of the needs and %ncent%ves'for prowess control*'can bc'foun&'& the following booti:

(1) Techn-Jl&@ of Process .Cqmts@‘, by Fi S‘i 'Buckley, John Wiley &

Son&, Inc. , N&# T*yii (~9~4),V _ “', : c ',

(2) Process Control Systems, /2nd edition, by F. G. Shifnsky,.*

McGraiw-Hi&I, dew York -(19@]:;' ' ', 4;, I

More on the stability &haracter,istics of, &TR's tith'exokhermfq reactions,

can be found in: ' c i:b,. - /, ; ., I ..

(31 Ele

I& p. 2 (1976)‘ r.>> (The reader is encouraged to return to these articles later after he has become': ".,Ifamiliar w&h the terminology' Included in the above refere&ea,n .Chapter 3: Details on the chslracteristice and th6 design of the measuring+

devices, Oransducersi transmitters, controllers, final control elements and

recorders, can be found in: : d 4 $

1

L 4,

I

3%

(7) The Chemidal Engineer\8 Handbook; J. H. Perry (editor), 5th

edi$n, MqGrqyHill, New Vprk (1974). ,..* I

(8). Pro&a Instruments and,bontrdls X&dbqqk,, D. M. Considine,:

(editor.), ,2qd eaitiqn, l&&aw$iill;. New York (19Jk)..,.*I __, .:, An excellent reference for the cor&ter; contgo;;l ,& th&_ch+cel pro-

:.:cessea is the boa&: ?(~‘ SC.' ,i.' -I ;: ._c_ s. .

(9) Digital Computer Pn,,c&s Con&&, by,,C,:L, Smi&, Intext Educ,,

.Publ. ,. New York' (l!T$TZ$, ..I' ..', ,' ;,I ,>

Applications of computer contro4 qan be found in the following articles:-

(10) "Digida Cmtro~ oi a Dis.~~~Jatisn,Sye'~,l~.by, ~.~~.~~+~~ellano,* .:' ,I.C. A.,McCain'and F. W."fFblas, me+. J&. Erogr., -.:74(4), 56

.(197t3j. ,: ./ (1 i )L, : :I (3 (. _“ <i ',,

I-.' (11) "Ener@ Conser&ionl~ia Process Computer Cont.p3i,*!..by P. R.zs‘i..I , nI, . '_ ,,'

LaWuS,,, Chem., Fpg,; Pr0&~.,,;12(4) , 76*;.(19i@b ~' " _/ ‘ :'; ,,!>

7' (52) 'Qhnppr Co&l of Aqm+&&anfq~,ll,,:. .

!yL. C. Daig$e;III and.i.' ~,

.;;1. G. I(, Nieman, C&m. .&g, progri;',.~70(2)j..‘Hf.11974),,~ ": ' .',;- ;.:.i. _, '_ ,_ " " ,

(13) "Applying Cpn*ol, (Sompuzt&/ t&a'& X~t$y$ .'.,

.* ,,I.(Nisenfeld, Ch". Eng. Progr.;.,-

-- - -- - - -

, , :

.-b---b

.

..-----------------,

c..

i

,,'

Coolant Water

Controller I

COLUMN

Contrdller

I ”

I-----y----, .: ’

5-J

.

Feed I/

-4

--a

I

. c

Reflux Distillate '

_ ' 1 feedback (a)'; fet%-lforward (b); inferentia) (c)The three control schemes of the Example 2.8:

a

I-1

: I d

1

t

.

. kimat& o f t h e. ” Values zrf the

‘k-ti’ -4 VariabIesUnmeasured CX&rolled

f ’----w-a

‘Estimateso f t h e ’

,’, :

UnmrzrsuredControlhd ./

Variables i’

I. : a3,. :

.,

T h e g e n e r a l s t r u c t u r e o f t h e dnFetenfia1 c o n t r o l configurntim.

Measured,,

Oukputs-

.’

,U n m e a s u r e d

outputs

The gcnernl. s t r u c t u r e OC t h e CeodCorward control COIIC J:;+II-;:t i,,lI. ’

4: ” .,

C

,

,

. . . . C-...-w-. * -...a -. .-...L-w-.. . . ..-- . -.__- _ ,, ., ..-. ” . a.. .- ..d. 1

. .

-, . _ - . --- _ _. -. .-._ . . *. . . . . . .I . . - . . . . . ”5q ,. :.’

1 ir,- l----I-----g*

Disturbances

c..

-..+-+Measured

Unmeasured

y L Intertace \: L,- ,y- ..- -3-

..-- L 4 .-e-J - - - .A... _- ,-I 4. I-.. .+ -1-1 ,w-.. .- J..- .-A -- ,w-' .Y

c CHEMICAL PLtiT

>

-J

‘.

” “I

I

THINGS TO THINK ABOUT ‘ ‘

1. What is the control objective while you are riding a unicycle or a9

bicycle? What are the measurements that you instinctively make while

rid&g, and what are the manipulated variables at your’diaposal?rid&g, and what are the manipulated variables at your’diaposal?

2.2. While you are taking a morning shower, what $8 your control ob.jective,While you are taking a morning shower, what $8 your control ob.jective,

your measurement and the manipulated variab,les at your disposal? qyour measurement and the manipulated variab,les at your disposal? q++ i.i.

3.3. Compare a ,si”fple feedback to., a ,s’fmple f,yedf,orward control configurationCompare a ,si”fple feedback to., a ,s’fmple f,yedf,orward control configuration. . ’ 1 ‘) i. . ’ 1 ‘) i(Figure Q.I-1).(Figure Q.I-1). Which one would you trust to .perform better inWhich one would you trust to .perform better in II

88 _- ~-_- ~-a ch iev ing your contro l obj,ect$ve? Why? ‘:I,.j,.. ‘:a ch iev ing your contro l obj,ect$ve? Why? ‘:I,.j,.. ‘:;., ‘.’;., ‘.’ ,( *:’ “$2,( *:’ “$2 ,-,-,,

4.4. What factors >should you co&ider &n diter$?nir$gwhat variable? toWhat factors >should you co&ider &n diter$?nir$gwhat variable? to:: l,ll,l .a..- , J . ,-.a..- , J . ,-‘1‘1

measure for the control. ‘$f .a t$p$$$~~'~~~oces~~.~, $+wer qualitatively.measure for the control. ‘$f .a t$p$$$~~'~~~oces~~.~, $+wer qualitatively. --I I . 1. 1 _*_* .I.I .,., .ydl.l.i; ,:, ’.ydl.l.i; ,:, ’jj5 .5. When is: an -‘inferentia;l., ,contkol ‘configg$re&n ,neede@ ,:.$&at do you thinkWhen is: an -‘inferentia;l., ,contkol ‘configg$re&n ,neede@ ,:.$&at do you thinkI )”I )” .:‘ ” .&’.:‘ ” .&’ . .,. .,

primary weaknes$primary weaknes$..

is i tsis i ts “* f: ;“* f: ; Compare* *t to p,,s>mple feedback control cbn-Compare* *t to p,,s>mple feedback control cbn-&->,&->, I ,_I ,_figuration.figuration. Which one is prefhrabla?Which one is prefhrabla? *.,*.,

:. .:. . ,‘i :,,‘i :, 9;9; 7 q* ‘i7 q* ‘iii’ii’6.6 .Describe the steps that’*u ‘would go through in’order to dCsign*a con,-Describe the steps that'jmu ‘would go through in’order to dCsign*a con,-

,, /Lo,, /Lotrol system for maintaining* .the pH of the liquid in a stirred tanktrol system for maintaining* .the pH of the liquid in a stirred tank<<‘.“, ,: I,‘.“, ,: I, .<.,?..<.,?. .I i.I i “, i.’“, i.’ .i.i(see Figure Q.I-2) at a deetred value.(see Figure Q.I-2) at a deetred value.

ipip , f.., f.. _._. 44What questions mu+ you ryolve?What questions mu+ you ryolve?.,’.,’

Develop a feedback and a”feedfoiward control configuration ‘for thisDevelop a feedback and a”feedfoiward control configuration ‘for thisi, ’i, ’ ..* ‘ ,.L,..* ‘ ,.L, II ! -! -

system.system. ,,,I {”,I {” .,, II.,, II

7.7. What is a SISO and what is a MIMi system? Give examples gram theWhat is a SISO and what is a MIMi system? Give examples gram the 11

chemical engineering field for both.chemical engineering field for both. ‘‘~~

8.8. Define the term “control configuration” and develop three differentDefine the term “control configuration” and develop three different

control configurations for the pH control problem presented above incontrol configurations for the pH control problem presented above in

item 6.item 6.

9.9. In the tank system shown below (Figure Q.I-3),the flowrate F of the effluent sIn the tank system shown below (Figure Q.I-3),the flowrate F of the effluent s

is proportional to the square root of the”liquid level h in the,tank.

Show that such a system is self-regulating, i.e. if the inlet flowrate

,!J

‘.

t

increase or decrease by a unit, the tank will not overflow or empty.,’f,

completely. ‘3.* 1

. :8 . .. : I. :; iF ! ‘. ,‘:.

10. What is a differential.‘pr’&‘sure cell and how doee’ it measure the liquid ’

level in’& tank? )-” ’ . ’ ’1 I

’ i

11.

12..$’

14.

_

Is a Venturi meter a good ‘me&zing device for monitoring and trans-+ J ,. ’ ‘...

mitting the U&rate vaJ%e’of a i&earn?’i

d ,.I 8”,:,;! “,

Determine the hardivare‘elem~~‘tequired for the feedback co&&con-.. ..,( f. i ’

figuration of- the pli ‘in the &&red tank de&ibid in”‘qu&tion 5 above.{ 9. i ./ ;‘,,

If you were to use a digital computer ‘ita your controller in. the control., r *:;, : .

I_ i,’ ,.,configuration above (queer tion 8) ‘&at ‘&w hard&&‘ ‘element@ would’ ‘$0~

.need? ”

,_ “$-j :f .I; i”,..~ ,, I”. a . .. : 1 : J , ::..g I _-. ’._

i (i ~ ,.,What ire’ the,b&ic and moB‘<i.import!ant advantages offered by

t,.the’ digital, , i ‘$ - ‘; :- . _ .’ I

cornput ere in process control? Discuss the size, Cbp&ilities and ,thesj: . . : /,(I : 5, t :rP’ %. =f ‘:.<’

prices of the most recent’higital m~croprochaeore~ evailable in the , ’t.: ,a I’ 1

market. Do you realize the inexpensive potential, tha?. they offer for*, s

process Atroi?* t ‘<.,. ‘.

“ :,. ‘.., < ;: 2I

, 2,’

~: :“? .: I”.

i i.9. ,

,1 ” .I.

I. :

pRoBLEMs ‘,’ .’ * 1’ ~$7 :; e.,

1. Consider;the heat exchangei: shown in Figure'P.I-1, Identify:“-

(a) The'control objectives fok this system,.; I

(b) All'the. extetial dis&bances'thdt will .affect the tieration-of ':], i~

(c) All the available $Xnipulated variables for the contm-1 of the(c) All the available $Xnipulated variables for the contm-1 of thei(i(

exchanger; in the presenc%"df disturbances. ,"exchanger; in the presenc%"df disturbances. ,"

2,2, For the same heat exchang& &kiwn in F%.gu3e P;I-'Q', considtir that'thie:For the same heat exchang& &kiwn in F%.gu3e P;I-'Q', considtir that'thie: ~,,~,,

temperature T2 k 190'F is our bakfc contrd$,objootive (i.e. maintaintemperature T2 k 190'F is our bakfc contrd$,objootive (i.e. maintain ",,,",,,,.,.*.*. ::

third temper$turk in the pr+ese%ce,:of ~d;tstiikbsnc&)~third temper$turk in the pr+ese%ce,:of ~d;tstiikbsnc&)~ QMstruct two',dif-QMstruct two',dif-,I.' ;,,.,,I.' ;,,., ,,

i ferent f&dbsck and txqo: diffeknt feed@&?& &Mrol~conf$$qations . .i ferent f&dbsck and txqo: diffeknt feed@&?& &Mrol~conf$$qations . .

/"/"

* I'* I'.; : ,'.; : ,' .'.' '.'. .. :,:,

that will satisfy the:co~:~~~~objec~~~~~i~-theipr~nqe of4disturbances.that will satisfy the:co~:~~~~objec~~~~~i~-theipr~nqe of4disturbances.,:,: j.4j.4 ,i:,i:. ..r..f.‘*.. ..r..f.‘*.

3.3. i A seke &rl>k&q$r~*~m a :'ki A seke &rl>k&q$r~*~m a :'klili ,; '7 I .,; '7 I . 11

!‘Wb&2 lo;ad cs$ Clml~~:r!‘Wb&2 lo;ad cs$ Clml~~:r..&&+&(pQp @+&&+&(pQp @+

,, ',', ..with'time; *$1$$%1 v&&ions -$n:'the-&aft speed i6fL~4h~~tu~bfne gre con- .with'time; *$1$$%1 v&&ions -$n:'the-&aft speed i6fL~4h~~tu~bfne gre con- .

.. _ a;_ a; ,:*,:* ,:,: ,'.I,:,: ,'.It$olledLthrough the use .of* a-flybsll Ofroad.-.,gove&qk.<' Fok'thitt$olledLthrough the use .of* a-flybsll sfroad~-.,~ove~~~~' Fok'thit system,.system,.

""""(a) : Identify '41.' th.e$&,ternal d&turb&tci&' an@+‘(a) : Identify '41.' th.e$&,ternal d&turb&tci&' an@+‘

\'\'I' ai .",(I' ai .",(

@),-L&&l -the avaiiab~~.~&&$at~ ~a&&&~"~ . . Lj:! ..:I I : t~~F?j@),-/h&l -the avaiiab~~.~&&$at&i va&&&:". . . Lj:! ..:I I : t~~:"j :,_ r:,_ r._'.._'.

.. ..Also $Ctermine :ttr~'~a8i~~~oon;trsl"object,i~~ and-ku&st~~a $&.d&& con-Also $Ctermine :ttr~'~a8i~~~oon;trsl"object,i~~ and-ku&st~~a $&.d&& con-

troller that would try to satisfy it. !.'troller that would try to satisfy it. !.''' ,,'),,') ;;i .;ii .;i

_I. i_I. i ,, II4.4. In Figure P,I-3 the dkWillation configukation fur these&zation~ofIn Figure P,I-3 the dkWillation configukation fur these@k?ation~of '.,,'.,,

;'.;'. .i.ibenzens fromitulqens is Biven.benzens fromitulqens is Biven. The feed &"the,distillat,&& icomesThe feed &"the,distillat,&& icomes11 _'_'

from the reactor where tolu&e hks been :deh$dxodeakk&ted &produce .,:'from the reactor where tolu&e hks been :deh$dxodeakk&ted &produce .,:'/'/'L'

<benzene, Ai;. e.<benzene, Ai;. e. 5 1:: 'I' *.5 1:: 'I' *. _ , j_ , j .i":. ,.:.i":. ,.: ,,L’

Toluene + H2 --+Toluene + H2 --+ Benzene + CH;Benzene + CH;.'.' IIrr '_'_

after the excess 'H2after the excess 'H2 and 'ihe'produced CH;and 'ihe'produced CH; have deenrem&ed in a ' Ihave deenrem&ed in a ' I.' .-.' .-

'flash unit.'flash unit. For the distillation system, 'For the distillation system, '

‘%I (a) Identify all the control objectives (make sure that you have ,,I,!I ‘.

,i

included all the operational object,ives) , - ’. , t. 3 : .;-.I

(b) identify all external ,disturbapces, and,, , ,, I

(c), all ,.t$e available ,measurementa, end ~mar$pulated variables. .

5. *For the distillation system of Figure P.I-3, t:1

(a) Wggest a feedfoxward ,&ntrolJ.er that ~$11 oontrol the’ operation‘I

of the column in tha presence of’ changes in the feed flowrate,+,

(b) .. suggest a if eedback. co@& co&&ration, to account.. for changes , *

in the feed Llowrate~:‘. I ,Xi ,,._.,, a

(e) if the control objective. is to; keep the purity of fhe..~@#h@, ,: ” “:,.r ~,: J:., _‘:

,prodpce (benzene) constant and.: the .use of, concencratiqn rn&$&?#g

.* dev ices (gas lchromatographs,z’ infrared;l,anaiyzerS,;le~~?)a~B:ao~ .; ~,,‘. *,” : .:,,‘recommended d% fo. their low relwility;,i sugge#;

+ control configuration.‘,

What secondary mea&em* _’31; ,,

..une?. How would you use them in principle.. to.Se$t,ima~+4,;& ,:‘i,i+ ,--x

6. (jonsider the air-heating system used to regu&&..the l~~~,~=~~~.~~. ; A‘.A . _,_ : :

a house (Figure P.I-4). The .heat is.,, e&&i& tOrom: ,&he., ,combust@n of G, i 1

fuel oil. !- ,’ :,i * .’ “.;, .; /I,, ” < -? :. ) _ j i,! ‘wrY,

Identify the contro’l ohjec&ive&&#e ‘%&i&b&s! m&suremente;(a)i ,c

man&pulated variable& What ‘are the &xtQbme% Gar :

?. fgg thfa a-,g&qjs sy&&, I I, ; , ‘$5 ’ ::..:+%,y ? ;T *

(b) Develop a feedback co.ntrol configuration to achieve your -control,>,

t objectives. ‘: 1 .‘-, ‘.

,*(

(c) Is a ‘f card&ward control conf igurat,iLan poaaible;*for achieving: ’“.‘, * /: 7 2‘ ‘” ‘>,

your control objectives? ~ + 3., .i

I:.

,,

I 1. . , _

I

7. Ftgure ,P.I-5 shows a system of ‘tw0 tanks which are used fur the temporary )

(Tank 1) and longer term (Tank.2), sborage of a liquid chemical product. ’.:” ’

I ‘he ‘d,hand d. s satisfied from the temporary Tatorage -tank, while Tank 2

lis used. :to a&uimA%te the l&@id. product in excess of the demand.

(a) ‘1, Idi?nt!Lfy: E%ternal diatur~arices, control objectives, mbasurements

J - \ ‘and- manPpuk&ted’ variables available to you. Is this a. SISO or a 1

:! i, MIMO systemT ,“ : : ‘i >’ ~‘<

(b) Develop alhernAtive feedback sand/or f e&f or-ward Control conf igu-

rat ions to ‘achieve your control ob j etitives . % i ‘, (i q

(c) Is the er any si$,uation, that ‘may arise durgng “which you&&notI

avoid’ overflotiing “the ‘storage tanks’3 !A” .:

8 . Consider ii s&em of two ~contitiuou~~~ stirred Gnkreactors in%&.&_. .f

x(Ffgure P. I-&) wh e r e t h e - foll&#ing’ &ndqth&mic Tre@tion takds pl’titie: V’‘ ,’ ? (..’ A + Cataiybt,t 4 B 1 t ‘, ,:‘:. ,~ , ;‘,,Z-:T i, i.l/ ‘,’ .j..:.I-. ,t

.:t (a)“ ‘fd’entffy t+ c,,&ro~-.odj$&&!!& xrbr ‘the ~pe+a&+ &;t’hg &o’1

CSTR’S.~ : G>‘“,. : / +‘*,’ * ), ’(1;) ’ Cl-&y ‘the ~va&.abl& of &c *syst& ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ “’

“P . 1’.* .‘“,. , ( 4,, . C I z.subsequent&y classify :the irtputs"in,~~‘"dist~~~~~~~~~~~~~~aaipulated 1. ti

,:: ,., ,/ 1 T', ',, : '.;

variabJ.. and '&& o&ji,ts &to measbked as ;~,&,~a#&~' &&,&&~ ,,, ,it _I

?I ('. ., "

Is this ,a SISO ‘or’ a’MIM0 system?+ ! *.d ) ~ J CI .!,i,.i i: --r (/ _-

aj ; .b ‘, :- ’.i‘*-!_,” :’(c) Develop .a feedback control &nf i&rat& “chat! iat”iif&&&ur

$ \., i f ..,.“; :c k.

: j i..1 . objectives using a c&iporitio’n ‘~naljio;‘cit4~~~‘exit it&m or ’

‘5 Lhe second CSTR.

‘,$ I I ,/ .’ :.,

(df Develop’an Inferential control’cdnfigurat~on that uses’.tempetatures

and flowrates only, assuming that a composition analyzer is not

available. !~ (,! 1

(e) Develop a feedforward control configuration whiti can also use

composition analyzers if they are needed. ;

(f) In your opinion, which syst@ is easier to .control, the two CSTR

system shown in Blgure P,lI+$~'or~an equltvaXent one CSTR,system'

2 that achieves the,,saple ~nve;r;aQu~? Explain, qualitatively why.

9. Con8ider.a tubular cataly,ti@ Ssactor where an endothermic reaction

A$B takes place (Figure P.I-?i). ,The reacting mixture is heeted'with,:.

steam flowing in a jac&t around the: tubular reac&orz. ,The stream of

the raw &&+a1 A 'includes s,lso a chemical C .which,poisons' thet'

* catalyatover ,a period of, five days.. As the catalyst decayswe can

increase the supply of heat to the,Yeaetor through;the steapl,thuss bincreasing -the temperature of thereacting mixtuke and:.Fonsequ&iy :

,'t,he.conversion to .B. &at us.,assuw &hat th+ reacrtor -is isothe,Jc i',, I"_. *along its lengrh.;. Pigure P.-I-'lb,:: shows thotempersture inthe reao%or -

during,the reaction pqriod, fl~#~ ‘imize+.the profit fromthe'operationI, 'of the tubular reactor. '(

1 c,' ,.;.

(a?.,. Fo,qm.d.,~t~ .the optijmiqsy+, ~~~blem~E~~~;~~~eld~~:the .temperat&-. : '..4

L pro$$I$ of F&g&e PUtlb ..as, its sqlu,9i411~ , ~- 7.._. .I 1 .*: d..:

(b) DweloR a .feedb,ack :co@r~l &y@em @ich.~ill,reguIate the tern-.^ ':..tC' . ._. .I_ .' .i.iperature of the rd[actor to ,t@t,.sh#k iqP~gur= P.I-?b..< whatT, /'" ,,I.'

Q? the kontfol $@hm 'bed 1.: j . ' , ,,. !‘,, l'. ..i, ,.i.: .-F ,., 'E:- “"y:. 1 <:t <'

I Cc) Draw a control system that uses ,&digital- computer for the.L.~ * : . ‘.&Se -'i,..I Iimplementation of the feedback system in (b), Include measuring

,>

I devices, transmissionlines,,:final con,trol elements and whatever,. * ',

else is necessary,.. , .,' p .,r: ;)Y*

10. TWO liquid streams with flowrates Fl, F2 ~j and temperatures T1, T2".

flow through two separate pipes which converge at a mixing functkon '.~

.*

/. :.r.: .,i

(Figure P.I-8). We want to maintain constant the ,flowrate .Fj and

the temperature .T3 of the liquid stream resulting from the mixing\

of th.9 fir& two.etreama. /z4..; :

*(a) [email protected] the control objectives, ,disturbances, available,P/.measurements and manipulased variables. Is this a SISO or a

MIM&'system? .‘

(b) Develop a control system that uses only feedback,controllers.

(c) Develop a cotitrol system that uses only feedforward controllers.

(d) Develop two different control systems that use both feedback andY.

.

I I vC

-T -

- -

S?ART IIS?ART II

MODELING THE DYNAMIC AND STATIC BEHAVI6R OF CR~EMICAL PROCESSESMODELING THE DYNAMIC AND STATIC BEHAVI6R OF CR~EMICAL PROCESSES$8. :$8. :

In order to analyze the behavior of a chemical..process and to ansyerIn order to analyze the behavior of a chemical..process and to ansyer

some of the questions raised in the previous chapters about its control,some of the questions raised in the previous chapters about its control,

we need a mathematical ,%epresentation of the physical andwe need a mathematical ,%epresentation of the physical and **

chemical phenom&a taking place in it.. Such a- mathematical representationchemical phenom&a taking place in it.. Such a- mathematical representation::

constitutes the model of the system, while the activities leading to theconstitutes the model of the system, while the activities leading to the::

construction, of the modei' will be, r&f&&d to .as modeping.construction, of the modei' will be r&f&rid to .as modeping.r ,r , ::Modeling a chemical process is $i very synthetic activity,requiring the ‘,Modeling a chemical process is $i very synthetic activity,requiring the ‘,

//use of all the basic principle&of the chemical.pn@eering science, such asuse of all the basic principle&of the chemical.pn@eering science, such as

thermodynamics ;*thermodynamics ;* kinetics, ,trane,port phenomena, etc. For the design of con-* .kinetics, ,trane,port phenomena, etc. For the design of con-* .

trollers for chemical processes,trollers for chemical processes, modeling is a very critical step,,.-It shouldmodeling is a very critical step,,.-It should,’,’ .a7.a7

be approached with care and’ th&htf.ulGese.be approached with care and’ th&htf.ulGese.

The purpose of the following two” Ichapters is :The purpose of the following two” Ichapters is :

- to explain why we need to deve,lop a iathematical description (model) of- to explain why we need to deve,lop a iathematical description (model) of

a chemical process as a pr,er&iei.te,to the design of its controller,a chemical process as a pr,er&iei.te,to the design of its controller,I ; L_I ; L_ ,,

- to describe a methodology for the modeling of a chemical process- to describe a methodology for the modeling of a chemical processss

using the balance equations and provide examples of its implementation,using the balance equations and provide examples of its implementation,

and finallyand finally

- to determine the scope and the difficulties of the mathematical modeling- to determine the scope and the difficulties of the mathematical modeling-.,-.,for process control purposes.for process control purposes.

??It should-‘be noted that the isubeequent chapters do; not constitute aIt should-‘be noted that the isubeequent chapters do; not constitute a

fificomplete treatment of ali the aspects ‘on mathematical modeling but it iscomplete treatment of ali the aspects ‘on mathematical modeling but it is**limited to those of interest for process control.limited to those of interest for process control.

JJ THE DEVELOPMENT OF A MATHEMATICAL MODEL ,.THE ~EVELOIWNT OF A MATHEMATICAL MODEL ,.

,_,_ .fd ‘_ !‘..fd '_ !'.Consider a general processing system with.its associated variables asConsider a general processing system with.its associated variables as

shown in Fig&e 2.1.'shown in Fig&e 2.1.'././

To investigate 'how a chemical process (i.eiTo investigate 'how a chemical process (i.ei its 'outprr~ts)i ts 'outprr~ts)_.._.. ,, ;.;. "3"3~_~_

changes:kth tiny@ under the~'ix&enca of the external disturbances’and mani-changes:kth time under the’inf’luenca of the external disturbances ‘and mani-..

pulated variables and consequently design an appropriate controller, we canpulated variables and consequently design an appropriate controller, we canr:r:

use two different approaches:use two different approaches:',‘, . ',. ‘,

-. Kxjjerimental approach: '-. Kxjjerimental approach: ’ ii this case the physical equipment.(s) of theii this case the physical equipment.(s) of the ’’.L.L , ;’, ;’

ch&&kl process. is available to us.ch&&kl process. is available to us.“i,’ i“i,’ i

Consequently, we change,deliberatelyConsequently, we change,deliberately_' ,_' , :: * , ,* , , I -:-';~..', i( :I -:-';~..', i( :

the values of various inputs -(disturbances,the values of various inputs -(disturbances,. .I.. .I. manipulated variables) andmanipulated variables) and, I ), I ) ;I* :,;I* :, 1 .f,1 .f, , _ “: ( ’ L :,, .,G ”, _ “: ( ’ L :,, .,G ”

through appropriate me&.&&g devices we” obser&e,how the outputs ‘item-through appropriate me&.&&g devices we” obser&e,how the outputs ‘item-$ ’$ ’ j_.j_. ww J.J.

peratures, -pressures, ,fl’o#rates, condentrations) of’ the’ chemkk processperatures, -pressures, ,fl’o#rates, condentrations) of’ the’ chemkk processI .qi,I .qi, ,,j.: f’ ‘Y’ 1 ;,’ ; I . ’ ’,,j.: f’ ‘Y’ 1 ;,’ ; I . ’ ’ *z :*z ::.:. uy;i’$i,uy;i’$i, ,A : , ,,A : , , ‘,‘,

change with time..change with time..i , .i , . S u c h pro&dure i s t ime and effort consuming as&“it is iS u c h pro&dure i s t ime and effort consuming an&“it is i. tJy. tJy I T‘ (I T‘ ( '- "i * : I-‘- “i * : I- ; *; * !<'.',!<‘.‘, : ,y-: L,, /Y ‘>! f. n": ,y-: L,, /Y ‘>! f. n” fj$ )~, +fj$ )~, +

usually quite costly because 8 large number of such experime&ts’ muqt.~;#e-:, ,.usually quite costly because 8 large number of such experime&ts’ muqt.~;#e-:, ,.:: : *: * ‘& ,) .‘& ,) . . .. .

/( $1/( $1 rr ,‘--k-,‘--k-h”h”i 4, . .i 4, . .‘. -.‘. -. . ( F. ( F 1’ ,.1’ ,. ,: p , .:,‘-jj,: p , .:,‘-jj

p e r f o r m e d . ,’p e r f o r m e d . ,’. ’. ’ff

I’,, .’I’,, .’- y&J&icalT ap;;ohc;;- y&J&icalT ap;;ohc;;

77 ;._>h; 1. ; ^. { ,-.iT, :;._>h; 1. ; ^. { ,-.iT, :. 1. ,‘I. 1. ,‘I

..*= : _ ii::! ;e&& ,~,:~~v..*= : _ ii::! ;e&& ,~,:~~v i&C.? t:’ .:.c ‘$i&C.? t:’ .:.c ‘$It:‘,$s quite -of ten the case: that we Bkve .‘t& .desXgnIt:‘,$s quite -of ten the case: that we Bkve .‘t& .desXgn 11

,. ’,. ’.:..:. . .. . G*&,>. :G*&,>. : , .* ; 2:;, .* ; 2:; k. ‘,k. ‘, i:: *;i:: *;<<t he c ont ro l sy&em 'ko$"a ch,emical process before,&'bkas,been c&nstruct&&the cont ro l sy&em 'ko$"a ch,emical process before,&'bkas,been c&nstruct&&'I'I

i. , .,i. , ., I 'I ''9 .'9 . ""In such a casa we cannot rely on the experimental procedure, and we needIn such a casa we cannot rely on the experimental procedure, and we need ''ILIL ,. ',.I,. ',.I ‘‘

'. a d i f f e r e n t re~~eeenra~~n'of!.;~~~~~~~~l'pr~ess im ordar,~~o.e~yit.~;~'. a d i f f e r e n t re~~eeenra~~n'of!.;~~~~~~~~l'pr~ess im ordar,~~o.e~yit.~;~(( "'"' . .. .. :. : ; d y n a m i c b&h&orr; d y n a m i c b&h&orr ~Thie~r&tired ropressnta$t9n is ~usiq&Q @A& kn twml~Thie~r&tired ropressnta$t9n is ~usiq&Q @A& kn twml

i :i : 33 _ .j> __ .j> _ '_'_

" I ' o,f'a se t o f ma.then%at&al squations (diffirkkQIS~ '&kg&&c) ‘wh.088:~" I ' o,f'a se t o f ma.then%at&al squations (diffirkkQIS~ '&kg&&c) ‘wh.088:~ ’’

solution yields the dynamic or2.stati.c behavior' ofa&& ehe~i&l.'procel~s *solution yields the dynamic or2.stati.c behavior' ofa&& ehe~i&l.'procel~s * ',&,',&,

under quest ion)under quest ion)) ‘..) ‘..~, ,.,~, ,., ffi ,i_i ,i_ ,. ; .‘.,. ; .‘. j ‘Ij ‘I

In the/present tex,t tie wiIl-~discuss both:aQproaches for%tW developmentIn the/present tex,t tie wiIl-~discuss both:aQproaches for%tW development

of a modelfor a, chemical process;of a modelfor a, chemical process; IQitialZy, we. will examine :the theoretikalIQitialZy, we. will examine:the theoretikal

approach.while leaving the experiment&l for a,subsequent &apt&r (Chapter ).approach.while leaving the experiment&l for a,subsequent &apt&r (Chapter ).

4.1 WHY DO WE NEED MATHENATICAL MODEIZNC FOR PROCESS CONTROL?4.1 WHY DO WE NEED MATHENATICAL MODEIZNC FOR PROCESS CONTROL?“(“(

Let us repeat that our’goal is to develop,s control system for a chemgcalLet us repeat Ghat our’goal is to develop,s control system for a chemgcal

process wh$ch will guarantee that the operational objectives of our processprocess wh$ch will guarantee that the operational objectives of our process., !., ! L :.L :. a*: ,a*: , 1, ‘<1, ‘<

are satisfied in the presence of ever dhangiag disturbances.are satisfied in the presence of ever dhangiag disturbances..‘. /.‘. /

/ .”/ .” . .. . ’ 1 ; (’ 1 ; ( ,,, ci “;;,,, ci “;;Then,whydoye .’Then,whydoye .’

2.’2.’ *t,,, ; :“:I ,I*t,,, ; :“:I ,I: 2, ;;.: 2, ;;.need to develop a mathematical description (model) for the process @z2.,want’,ztAa ’need to develop a mathematical description (model) for the process @z2.,want’,ztAa ’

.,!C r . :“4.,!C r . :“4 dd ‘: ‘: f‘: ‘: f ? ,,~.? ,,~. ‘.‘. ::iicontrol?control?

,:;, ‘;“; ” s,:;, ‘;“; ” s ;i.’;i.’ .. ~.,f&:, I-.. ~.,f&:, I- II‘7.‘7. I /I / 1s1s i *.,;;t’yi *.,;;t’yIn the introductory paragraphs of ‘the .previous page we noticed :&at, “In the introductory paragraphs of ‘the .previous page we noticed :&at, “,I” .~,I” .~ *_,*_, * ’* ’:: xx.I.I

often times the physical equipment of the c&mica1 process we want to controloften times the physical equipment of the c&mica1 process we want to control “,“,:: :: !:: !

have not been ‘constructedhave not been ‘constructed-.,e-.,e , ; L; ’, ; L; ’ ” /, L’..i hi--’” /, L’..i hi--’-.- . : b’: b’ .d’.d’

Consequently we cannot experiment to determineConsequently we cannot experiment to determine ’’.._._. .’.’ , ,:,k.‘”, ,:,k.‘” ‘,‘, ,’ *,’ *‘J “’‘J “’

how the process reacts to various inputs and therefore design the appropriate -’ _Ihow the process reacts to various inputs and therefore design the appropriate -’ _I~. “~. “ I, ‘1I, ‘1 ,-f r-fq,-f r-fq 1 a,,1 a,, *:. ,*:. ,** .:,. :.:,. :

control system,control system, But,. even if the process equipments a&available fop.But,. even if the process equipments a&available fop.. j. j ,a.:,,. *_*.,a.:,,. *_*. . . : .:.:. I. . : .:.:. I ‘,‘, .,< .’ ’.,< .’ ’ I/h rI/h r. .. . ‘ I”,‘ I”, ,“-e- “8,,“-e- “8, .P+‘ :.P+‘ :

experimentation, the procedure is usually very costly. Therefore, we’need aexperimentation, the procedure is usually very costly. Therefore, we’need a II‘. ::,.‘. ::,. .-.- . +.. +. ).. ; ii x .$).. ; ii x .$

- m ‘I.- m ‘I. I .:’I .:’ LqLq ‘i‘i .) i..) i.:x:x1 simple description of how the process reacts to various, .&iputs; ;a& this- issimple description of how the process reacts to various, .&iputs; ;a& this- is

a> :a> : . ,. , I ’I ’ ,: :,: : :’ ..&:’ ..& r;r; ,~i ;.; ,: ii,~i ;.; ,: ii &s$ .T-’ & _ .:Q,f;&s$ .T-’ & _ .:Q,f;‘‘ what the ‘mathematical models ,can provide to \&a control d&ig&&.what the ‘mathematical models ,can provide to \&a control d&ig&&. :, .’:, .’

;r;r ‘.,‘I /‘.,‘I / :,;, ;-:,;, ;- x: -~,) . , . , 3%. *. ; ‘, ;.x: -~,) . , . , 3%. *. ; ‘, ;. 1 ..‘1 i$‘1 i$ 1 ... .,4’ I. .,4’ Ii, f.’i, f.’ ;, 1.;, 1.Let us demonstrate nbw in terms of some -*I&es th&*‘nee& for’ theLet us demonstrate nbw in terms of some -*I&es th&*‘nee& for’ the ‘,. _‘,. _

:: __,’,’ ,.,,‘ ..,.,,‘ .. “,i’ r“,i’ r””development of ‘a mathematical model before we $&sign the control., system ,fo‘r *, I’development of ‘a mathematical model before we $&sign the control., system ,fo‘r *, I’

a’ .‘,a’ .‘, e.e..I-. _/‘I.I-. _/‘Ia’ chemical process.a’ chemical process.

ii .,I.,I :: :-;-I:-;-I .‘:,i~’.‘:,i~’ ., f.., f. j’:I;‘,j’:I;‘, .,,, ,.” ,_...,,, ,.” ,_.. ::.J 9.J 9 : .1: .1 I.“~,, e.. . iI.“~,, e.. . i . I. III :, ,z:, ,z 1,1,: ‘& ‘*: ‘& ‘*55 ii(( *.*. bb I”, .*)I”, .*) ‘1.. ?\,‘1.. ?\,i ,’i ,’ :‘” , ’:‘” , ’,.,. + :i,“+ :i,“ -I-I ,’,’ cc

s’s’,I,I ,’,’ ;: ‘, -, i).;: ‘, -, i). ‘d i .’‘d i .’ \\ 1’1’ ‘,‘‘,‘;. me /’ r : .;. me /’ r : . ; ’; ’ ,.,.

:: ’’

Example 4,l t Design && Integral &o~ro&&~‘t~ Ll,q~id Level. in e Tank I, ,Example 4,l t Design && Integral &o~ro&&~‘t~ Ll,q~id Level. in e Tank I, ,:::: . .,-. .,- , , I_, , I_

. Consrlder .the. problem. of ..con.trolling the IeveJ~of the Uquld 3n a tank,. Consrlder .the. problem. of ..con.trolling the IeveJ~of the Uquld 3n a tank,‘.‘,‘.‘, ;;%%

using integral control., <:@xample- $612) .1 i From F&@re 2.@.&~no%ic$ that’ theusing integral control., <:@xample- $612) .1 i From F&@re 2.@.&~no%ic$ that’ the+ ‘.+ ‘.“f . * .“f . * .

quality of the.,control depends$on the &&ue:o-f~ the pet&m&er: a’.quality of the.,control depends$on the &&ue:o-f~ the pet&m&er: a’.- I- I

B u t ! th&j:B u t ! th&j:

quas tion is :quas tion is : how does a’ affect the quagity of control and-what is it;‘how does a’ affect the quagity of control and-what is it;‘ : >: >II

best value?best value? To answer &his: question we:n&ed, to, know how-- the value-of theTo answer &his: question we:n&ed, to, know how-- the value-of the ..liquid,, level h 1s af f egted. by. changes!& the value, of.* the inlet$!lowrate 1 **liquid,, level h 1s af f egted. by. changes!& the value, of.* the itilet$!lowrate 1 **

.. ::Pi or the integral ,acltion’ of the controller;Pi or the integral ,acltion’ of the controller;(: ,,( : ,, This.,is given byfequation (2.g)This.,is given byfequation (2.g)

_’_’

, ., .ii

^ ,

which constitutes the mathematic’al model of the tank with integral control.

“.Example 4.2 - Design a Feedforward Controller for s Process L

‘< , ^ ::

In the feedforward control arrangement shown in Figure 4.1 we measure. , : ,..

the value of’th,e disturbance and we hnticipate what its effect will,be on the.‘, . _ _ “! I

output of the process that we want to control. In order to keap the value ofi.I2’. ‘<.& ) : : n ., __ : P”this output’ at the desired level,. we’need to change the value of the msni-

:’ > .- .,3 ‘i ‘,. ‘:,/-I I .,pulated ,vrriable by such ‘&~u&uni ‘ae t,o d&Rat& the impact that .iha

I’ .:

disturbance would have on the output. &e ques’(ion is:”.

how much should, we<.I .I r s‘; ..-

change the manipulated variable‘ in order., to,,; c&&cel’ thb effect of the dis-j :Lq,::’ ; I’ : :‘- ” ( _/ $ . ,I&

turbance? To answer this qupat$on tie must” kno; the following twoi; ,

‘ I */‘ .( .)

relationships,;’

;i _,*output - f l (disturba&%?) :*

; , , .’,,

- f2 (mani&ated v a r i a b l e ) ‘-f.“. 3 ?I,.. f,

output “.,) _I

: .t ‘. /.: : . , ~ ‘..i. ”. , I ::Q i’s :which are provided by a mathematical ,model of the process. ,Indeed, ‘if the. ,A-;,

: ‘ . . , . .+. 1 :I. ..’+: $ +<I; : 2’ ” s,+s; ., 14:; i-i ‘( f routput is’ to remain the same,, the manipulated variable &st. take’ such :W valu’&,

:f ,’ ;, ‘.’..,_’ *: ’ -‘.i “, .‘, ,.&7:< .*. ; p& ‘-*,,: j .-,,&A ‘, *

that .B . , ,. ; / ‘,. .

sfl (dis++&)““~~ f;i.:~m&ipulat& &&blG$ “: S

:;#.& ( ” -*$ A ‘, *, ,, “,f j :

“t I ‘.:’ 2 , I i.‘ ,‘.

This example demons&&s very vividly how important ‘is the mathematical.i . -, t.7, iI’ .j ~ ,.” _~ .,. <i” 1 , I ’ !

modeling for the ,deaign of a feedforward control system. In fact, iwithout.: /.

.good and accurate: mathematical modeling we .can&t desigti efficient feedfarward. 5, .~ ., *: ’control sys terns. ’

: . i; : ‘$.

Example 4.3 - The I&sign of an Inf.erential Contr:ol SystemP

In the inf ereptial control *s&me .shown in Figure 4.2 we ,measure the”

measured output and we try to regtilate the value of the unmeasuredcontrol

objective at al:desired value.objective at al:desired value. Since the control objective+3 not measuredSince the control objective+3 not measured.*.*

d:irectly, it can only be estimated from: the value of the measured output if ad:irectly, it can only be estimated from: the value of the measured output if a

relationship like the following is available,relationship like the following is available,9 .9 .

,.,. control objective -control objective - f (me&red output). .f (me&red output). . *,*,

Such a ‘relatio&hip ‘in turn Is‘ not poss$ble’3f we “do .not have ‘a’ mathematicalSuch a ‘relatio&hip ‘in turn Is‘ not poss$ble’3f we “do .not have ‘a’ mathematical, ;., ;..:..:. ~‘I’~‘I’ ,I’ , , i ; ) , ,,I’ , , i ; ) , ,

representat&“.of the process (mathematical model) ;. Once the value of therepresentat&“.of the process (mathematical model) ;. Once the value of the,,>- a,,>- a

control ob j e&e can be e&mated of tom- a &atXonship ‘like the above,control objective can be e&mated of tom- a &.atXonship ‘like the above, it canit can>> ‘iL, j‘iL, j : ‘,‘;j, _~: ‘,‘;j, _~

b; compared to’the desired-value (setb; compared to’the desired-value (set‘_ .)‘_ .)

po;nt)po;nt)‘-8‘-8 ,., f’,., f’

and,,sctivate the iontroller forand,,sctivate the iontroller fori: ,i: , .a.a , 6 .I$,”, 6 .I$,”.,, ‘I .- J _* $,,.,, ‘I .- J _* $,, ;. ,~ ” /;. ,~ ” /

an appropriate action like in the feedback control*an appropriate action like in the feedback control*.. II ,:.: /.,:.: /. ,‘I, *,‘I, * ,I,,I, ,..‘/ I! .,;,..‘/ I! .,;

I We notice’ therefore that the availability of a good mathematical mod&We notice’ therefore that the availability of a good mathematical mod&1.1. 33 : “I,: “I, .>,I** (, .‘:.. ‘.. ._: ~.>,I** (, .‘:.. ‘.. ._: ~ -:L“ *-:L“ *4 for the process is indispensable: for the d”t;arign of good inferential controlfor the process is indispensable: for the d”t;arign of good inferential control

I ,.I ,. ii - . .- . . \, ri\, ri !,‘r!,‘r .).) II ‘,y ‘,II ‘,y ‘, ‘. ,:,i‘. ,:,isystems. : ’systems. : ’

.’.’ A*A*** :‘:.:‘:. ii s1s1 ‘..,‘.., II:: .:. _. .,.:. _. .,*_a 3*_a 3I %- >“, -II %- >“, -I :,* ::,* :.,., ..,‘l/..,‘l/ ,.. i,.. i , rr, i, rr, iii

,, f’ ‘* , . i!f’ ‘* , . i! 33.-.- ,. *,. * .:. * ...:. * .. ,~,,~,** ’ 3’ 3.”.” /. ‘:/. ‘: . J:. J:I.I. . , . - >. , . - >The above‘ three exkmples” ind!lcate very ‘clearly ttit the mathemst$&lThe above‘ three exkmples” ind!lcate very ‘clearly ttit the mathemst$&l

.*.*“’“’77 _ :_ :f.f. ,1” a’:. “’,1” a’:. “’

modeling of a process. is of parsmount Gportance for ihe des%gn 02 ‘good andmodeling of a process. is of parsmount Gportance for ihe des%gn 02 ‘good and ~ ’~ ’tt ,*. :.,,*. :., f. :f. : ::

ef f iciest control‘ systems for a chemical process.ef f iciest control‘ systems for a chemical process., ‘., ‘. : 1 ‘.: 1 ‘.,In the follow& sections,In the follow& sections

‘!.‘!. ) _) _ 1111 f :’f :’ ‘+ rr j‘+ rr jwe will develop a mathodology for the’ con&e modeling of &&&al pr.ocesses.we will develop a mathodology for the’ con&e modeling of &&&al pr.ocesses.

‘.‘.

4.2 STATE VARIABLES AND STATE EQUATIONS..FQRA CHEMICAL PgOC&SS~,4.2 STATE VARIABLES AND STATE EQUATIONS..FQRA CHEMICAL PgOC&SS~,I,’ I ” e?--I,’ I ” e?--

In order to characterIzea? proceeelng dly$tem (tar& heater; ‘b,atch’* reactor,In order to characterIzea? proceeelng dly$tem (tar& heater; ‘b,atch’* reactor,i. 1.. .~i. 1.. .~ ; I ,’ i : ‘ ,; I ,’ i : ‘ , “5 ; ** ),,“5 ; ** ),, ,; i,; i ,” ‘*: ‘.,” ‘*: ‘.; ..:I; ..:I .I.I__

distillation column, heat exchanger,. etc.)., and its, behavAor we. ne+l:distillation column, heat exchanger,. etc.)., and its, behavAor we. ne+l:, *.a- ‘,, *.a- ‘, :‘f i:‘f i I .*.,I .*., QF.‘>QF.‘>

- A set of fundamental dependent quantities whose values .will describe the- A set of fundamental dependent quantities whose values .will describe the,‘.‘,,‘.‘, ii , , 7, c, , 7, c ,i,i1 ‘.1 ‘. SF, ,‘ I a :, *I :SF, ,‘ I a :, *I : ,(/ .,’,(/ .,’ “. ,_’“. ,_’

natural state of a’given system, and .:natural state of a’given system, and .: I ’I ’44 aa.. ..

- a set of equations in the above variables-which will describe how, the- a set of equations in the above variables-which will describe how, ther.r.

natural state of the given system changes with time.natural state of the given system changes with time.

For most of the processing system of interTat to a izhe&icalI’&ginesr thereFor most of the processing system of interTat to a izhe&icalI’&ginesr there::

are only three such fundamental quanti&es,,are only three such fundamental quanti&es,, i.e. ‘mass, energy, and’ momentum.~.i.e. ‘mass, energy, and’ momentum.~.i ,i , ,.,.. _. _ .’.’ .’.’

j ._:j ._:,, *..- J_.**..- J_.* ., ; a .., ; a .

:: _. * /

r ’

_’

Quite often though the fundamental dependent variables cannotbe~measured. I

dir eetlv and conv&ientlv. In such cases we select other variables which can

he measured conveniently.i and when grouped appropriately they determine- the_’ L’

value of the. fundamental variables. Thus, mass, energy, and momentum can be*

characteriped by va&riablea ouch ai danbity, roncentxation, t&mpsiature,

pressure, flowrate, etc. These characterizing variables are tailed state.-LI 6 ‘ivariable<s and their values define the’ state of a proces~‘~&“system.

_ f \.“F‘:‘,‘; ’ rf’; +The equattons :w&~ich relite the state mar&ah&& ‘(de@$dent variables) to

&!the varlouq, indgpendenti,,varLrtblee are ,der$ved from >the app~lication of the 1 .>;.;conservation principle on the fund,&ea;tal ti qugntikies and ,:z&re called Hate

,. I. Iequations.

<,. ._,’ , ::. ” j. <&! ; 8 ’;’“, :,

The principle 03 cons&vat&on of a quantity. S:; etatsr *h&t t:“’ ;:-:.” ,, ’ ,;, :,..*:‘..i :: , /

The quantity. S can be any of the following fundamental3 qu$ntities: ’ ji L /., a .“. , ) , : .y1

7- total mass, :.; * !1 “i

’ - mass of individual components,\ ., I,\ ., I i .i . ‘..i ))‘..i ))

j. ‘I 02,. , ,j. ‘I 02,. , , ;, ‘)”;, ‘)”ii

- total energy, and- total energy, and, ‘I, ‘I !_!_ :: $., ‘,$., ‘,

- .momentum.- .momentum. $ jl$ jl _’_’I.I. a 2.48a 2.48 . , : , I ‘ : .. , : , I ‘ : . IIRemark.Remark. It should be remembered that for the phy&al and chemical processesIt should be remembered that for the phy&al and chemical processesY? .’ .PY? .’ .P 4040 ,,. 7,,. 7 . .. .$$“’ 1 .“’ 1 . 5; _” i5; _” i .: , ;,_ .*a..: , ;,_ .*a.

‘L‘L rJe will be studying, the total mask and total energy can not berJe will be studying, the total mask and total energy can not be,:,:

generated from nothing neither they disappear.:’ .; J -._ l__l ‘. ,L,..; _I j<

Let us review now the most often used forms’for the balance equations.I ,’,,. ‘- dl.

Consider the sys tern shown in Figure 4.3. Then, we have:I, : . . . ,,.‘, ! ‘c

i

Total Mass Balance

Mass. Balance on a Component A r;,. , ?

d(q)‘. : :R ‘,: 1 F ” ,, : L’

‘,d(c,V) ‘,

UI. aL‘ii piet ‘cIA‘~i I,-

. ky s’, (4vTb)i

: j :;;“t;;; ci PJ ’ 4j ,,‘I

i ,‘_, ,? 7 ,.j, . L 5 ,P; Z i:,?,, 1’

P is; the’ .dens.ity of the matergal in ‘the’system; . 1 -.‘,,. I -: ./ r

pi .( is the de&y of the material in an inlet s.tream; I ‘Ll. .,-,P j . is the dens&ty sofa tha,materiial in an aut.k.ez: bstreami’ s *I ,’ c

v is the total volume of the system; “, ,I ,.: I.)1: “, #$ :;.’ \

is the v%lumetric flowrate of an .&E&i stream; :.I ’ .’ I:, ” ! _(Fi.

. ,,,’ ‘.

Fjis the v.olumetr$c flowrate ofi’anioutlet ‘~tr~eam; *r .% .-,‘

i ’ .! ,_, I.:: ..,1 ,. ,I_nA ,, is the number -of moles of colapanent A ‘in the system;. _ ., &A,

.‘. ;.: _:. ,: x

cA is the’.molar concentration (moleri;/vol.ume) of A in the’iystem;.y_ v ’ /.,‘A, , . ”is the molar concentration of A in an &&et;’ * ” ’ ’

‘. ,“I’ *

CAi” *;’

I ’ . ,: ‘/_ ,‘.Ais the molar concen tratlon of A in an outlet;

”CA

.I ) I’..,” 2.’ ; r : , 5.’

r is the reaction rate per unit volume ‘#or c0mponent.A in the ‘system;’ ? ‘I‘. : . _., ., < I

hi ia the specific enthalpy of the ma,t&ial in an inlet stream; : I,’ i:”

is the specj.fic enthalpy of the mat’erial in an outlet .s&eam;y; y::,,

hj :.: : * ‘. L.’ .,.I,U,K,P are the internal, kinetic and potential energies of the systea,

‘* ,,.‘L-’ f

l$‘,. ,, :. ’ . .: *._ . , ~ a**,-5respectively; ’ /.

‘.,:, ,,’ ? r: 8 ‘i.. ), )‘,, . ‘“!Q is the amount of heat exchanged between the system and its s&rQu.ndlngs

: . 7 . e jper unit t ime;

., .y,: . . ‘, 9:’ ., 1 :( A$+ ; ,.

wSis the shaft. work exchanged ‘between the system and its surroundings I

L

,

per unit time.

j/O-:

- The total mass in the tank is / ,

total mass - pV - pAh (4.2)j.

where p the density of liquid; V the volume of liquid; A the cross

sectional area, of the ‘tank and h the height of the liquid level.

- ‘The tohal energy of the liquid in the tank is

E = U+K+P

but since the tank does not move, dK/dt - dP/dt * 0 and dE/dt =‘, 1 _ t* ..# .: ,,

dU/dt. For liquid systems *,, ,_:.

dU/dt a .dM/dt

where H is the total enthalpy of the liquid in the tank. Furthermore,

H - pVcp (T - Tref) - pAhc (T - Tref j’, ”: p

(4-3),’. ,:_ ,’where 9.’ . t :.a,$ ‘I’)

.( L a, .. :, . .:.. ‘,Y:: L; a;.‘

cPis the heat capacity of. the, l,iquid in ‘the tank, and

I ‘+ ;i’ i -, I __,” ri ),;’ _ ..L.T-,c is the reference tsmper~fur&:~ where the sPe$if,lc en thalpy ofLtzI. ‘ ,::y $k> -i , . ;’

the liquid is assumed to b& %ero., . .the liquid is assumed to b& %ero., . . _-_-2 Lb,2 Lb,

FrmiFrmi eqne, ‘-eqne, ‘- (4 .2 ) i&f- (49) qii;i+i$adi t&at the eta&i &i$$+“~ for. a(4 .2 ) i&f- (49) qii;i+i$adi t&at the eta&i &i$$+“~ for. a, ‘ Y :, ‘ Y : ._._ (. .(. . ::::

the. stirred tank heate) t&& t~~~~~$&wln$; ‘, a’ : . ‘. ’ “~:, ‘. ‘,the. stirred tank heate) t&& t~~~~~$&wln$; ‘, a’ : . ‘. ’ “~:, ‘. ‘, 7’7’..’..’ I ’ 5 I”I’ 5 I”s,tate variablesi :. h, and -,?P .’s,tate variablesi :. h, and -,?P .’>:.t: I ”>:.t: I ” , _, _ :,i:,i .;.; .,” ,_’ , ‘;,.>.,” ,_’ , ‘;,.>

66*,,. ,,*,,. ,,

, ,, , IIwhile thewhile the 4\’ *’ . . ) .I4\’ *’ . . ) .I : ‘,: ‘, ‘.‘. .. 4 I, 1,.. 4 I, 1,

))constant parameters: p,constant parameters: p, A,.cA,.cp’; Trefp’; Tref **2.2.

are characteristic of the‘ tank systemi4,are characteristic of the‘ tank systemi4,

‘Note: It has been assumed that the d&si.ty , p ,. 2e.i independ&nt’ of: the ..,/I, i

temperature. ’ ” $ :*‘, .- : .” ‘, 1Let us proceed now to develop the state equations for the stirred tank

b

heater . We will apply the conservation principle on the two fundamental.,‘;

quantities, i.e. the total mass and total energy-.

(a) Total Mass Balance ,! _” , ,

.

,for

Accumulation of Input of. ,‘Output 0.f :total mass 3 total mass t o t a l mass

time .< t&me .’ j ,- t i m e

:*- ,, - A - j i

‘. fdW$! p piEi _ pF i ;I,,+” ( ‘c. .(&- ’‘. :..,qI

‘/ ..+:3where Fi and F are the volumetric f iowr’ates , i .e, volupe beti unit;‘of ‘time

( /- c?. . ;t: ’ “d : ke ‘(ft3/min, or m3]min), “for the inlet and outlet streams respectively. l Assuming

,, , i

s“$

constant‘” density (independent of temperature)/ .* 4:.; t .:i

eqn . (4.4) becomes: ’ _

.s P P. -‘: ‘. .;,I :..

u /i. la\ L

or

where 0 is the amount of heat supplied”‘b~.;hC’..c

can take the following simpler form Cas’ske ‘T.~ .,

Id:-‘. ,*Additkonal ‘ a lgebra ic mnipulations on. ‘eqn. (4.%$ y&e,-.

i 2’$

- ..FiTi - ’ FT.+.--q-.,:’ ‘7.pc- ’

p.4, / ,, : i,‘ !

‘i L .i ‘.’I T,!“

&dTdt = Fi (Ti' - T) + + .: ‘.: (4.5b’) :

p ,‘L J.

.’j

Summarliziag the above modeling steps we havel _ 5

i. t&Eat* auua~l~ns ‘ : : ‘,

L/, I i(4.4a)

Ah dTdt (, - Fi (Ti - ‘0 + $- .I

P” _, (4.5b)

The variables “in F4qns. (4.4a). .and “4.: 5b) can be classif ied a$ follows (see

j Section Z.1) ’ ’..

~~~-stke variabies:%: .h,T

,output ‘v&fabl:es : h,T ‘(both measured)

- d&sturbances:. Ti; Fii ’ .., ’ ‘ ” ‘,‘. ,_ .,:;f, ‘-J,,

- manipulated var iabiea : :Q, F ( f o r ‘feedbhck controls)

and the Parameters constitute the mathem&ttc

heater. We need only solve them in orqer to fin38 its7I *‘.,I : , .. .A”

b e h a v i o r .

Let us now rtudy the dynamic end atatie be&?&x of the‘ntf-ynd: tqtilc.‘.‘:.. ) “.,>

.a L i. .

.. he&er using the 8 tatq eqns *,,.‘. (‘4.4id c?nd (4&) g ,,.,‘Fw- ci, , .-.,‘

.

,i 1

initially the tank hea.ter is at steady state, i’.d.,]noth:: ,..’

situation is described by the state equation?, if ‘the’tate o

(ieft-hand s ides o f (4.4a) a n d (4,5b) i s . aet’,te zero, ‘$:e* II ‘<, ti , I ,..I

Fi,s - Fs - 0 \‘,I

r ’ *I , >,

Fi,i (Tg,s - 5) + g ” O ,,P1

rj., /

:

from the,golution of eqn. (4.5b) using & inj.t$al condition the.--.-, I

steady state value of ,T, +.e, Y3.1 ..' L. ‘i 8' ;.:

T (q$-ll ..= Ts. .t I.: ' ,_

.i i iFigure 4.4 indicates the static and dynamio'behavior of the'tank:,for * :~..'ai: I , !1' '. .,..1% $1. '. ~~ .,- ii i

this case. We observe that after a.~ert&~time the tank heater .has‘_1) ;. '5 : ,$ : ,;- ,,I;'$ :

.reachas again steady 'stste 'izondit~ons. I i : ii- i ",,,8 ., : ; “ '?,': ,j

r'.P," , "ST,. ;, : , . ,

ii.'*i ;~Consider tL;;t inirially ‘;hg*'g&k ,$-key t& I;;:;Ke&$;i;@ts '&& : Tr& ), ;: :

: ,,,' ., '.'

hs;>*

Ti,s' Qs* "i,s and %*':$: .!':> _, I ':I, .'. .- -.,; . .-I 1Thg& 'it' tim.&f q&g, .ihZ‘ $$& fiLrate r -1:

.?i

., :decreases by 10%. It is,:clear that&$& the id& and the temperature

'. ,,'

change

h

;I fi-1 \_'_," i'(4.5b) using as initial conditions,

(tco) p hsand '*'

' ^,v '_ .'.;T~!p'*) 6,' Ts*:' " I_,,

,,* ,,,4.5 summarizes the static &d dynamic behavior of the tankFigure

heater for this case.

.

The subscript :s ,denotes the steady state value of the, corresponding variable.

The system will be disturbed~from' the steady state situation df any of

the input variabfes changes value, Let: us examine the following two

situations: I' 6,j;.&

i. Consider that the inlet temperaturec

: T i decreases by 10%. 'from,- its ..: Q j ,,-~‘- I ,::,,,'.* p

k steady state value, Then value.:qf,~~he,l~~uid. leve!, wil,l remain thq . +i f/

same at t,he steady state value, hs,.since 'Ti does not influt&ee 1), '

.:~I' .:~& .+

the tot&mass,in the tank (see aJ,so equation (4.4a). ,$n t&con- :,.' .:,

trary, the temperature of tke,liqlr~d,wilZ,$lso.~tart decreasing with,

time. How the temperature T, changes with 'time will be detsrminedi*

By coriue&on, a”quantity’ is considered positive- if it fl&s in the system-

and negati&$if it flows out. j

The ‘statecequa,tiong with the aqk’ociated st8te variables constitute the

mathematicil model of a process whiqh gields the dynamic or static behavior

of the process. The ap$l&tion of th;e conservation $rinciple’as defined by

eqn . (4.1) will yield a set of d$f.ferantial equations wi,th the funda-

mental quantities as the dependent vari@les,, and time ~ ths,Jndependent I‘, 1.

time,, i-e8 :it wfll determine the. d.ynam&cihehsv$or of ,the prjcess. ,.:..

Consider the stirred tank. heater of tha &ample, l..l”:(F&& 1.1) r TheConsider the stirred tank. heater of tha &ample, l..l”:(F&& 1.1) r The::

fundamental quantities whose values, provfde every’ information about the: -Gfundamental quantities whose values, provfde every’ information about the: -G.,., : , .: , .

reactor are:reactor are: >‘,>‘, :: II

““- the total mass of the liquid in the tank, *- the total maas of the liquid in the tank, * ~ y ’~ y ’

- the total energy of the material in the tank, and- the total energy of the material in the tank, and ,‘.*.,‘.*.,,- its momentum.- its momentum.

The momentum of the heater rematns constdint even when tb,di&turb&nzes ChangeThe momentum of the heater rematns constdint even when tb,di&turb&nzes Change

value and will not be further c@sider?ed.

Let us now identify the state variables for the tank heater.

‘/’

Remark: It. is worth noticing that, after Fi. has changed,,. the level h .I) ,_ i

and the temperature. T reach their~‘new steady states with 1 z.: :’

. ‘.different speeds. In particular,, the level, h, achi&es its new-: . . -’. .

steady state fast,er than the.,.temperature. In a subsequent chapter : . ‘,; :,D -. ,II’we. will analyze the reasons fo>r such behavior:., s1 ‘,,,>.,- _,:I “, .-+ *: 1,‘~ 1

‘. :st _i i .‘.I> :

‘,, . .I II: ;$ : / i,ADDITIONAL’RLWRNTS OF TUZ,’ MATIiEMAqICAL MOD&$ ~‘_ ” ‘

, ” -I. . ,‘,4.3 i,r .. .A.:.!, -;I ‘.? ,I>.il, “,

* ., ; ..,

In additian’to the balance equationsA, ,;:

, we need’ other’ reldtion$hi$ to’; iy4 , . ‘;

G ;.:: < . ‘j. . 0 I. )‘. ”express thedodynamic equilibria, react.ion rates, t$ansp&t ‘ra&e for h&t,

1 I : d:’”5, p,“’ ,f i : ,, ” ;mass, momentum, etc. Such additional r$‘a&onshiRs~needed to comple& ‘the Y’

/ **‘,

where j_ .,* /

.* u i s t h e o v e r a l l . h e a t transfer; cbefficleat, T ‘9 : :c ., :. /,A t ,+s the total area of heat transfer, and . ,*.’ i( : -.

TS is the temperature of the steam. . : i I . : .c :, ,,

“heeded’ to”describe the rates of chem+cal reactions taking place in asyst& l :s: :

‘s&h equations are developed iti 3 course on Chemical Kinetics.

.

:

Example 4,6 L1 1 ) ,: i ;. ~., . *

The reaction rate of a first-orde? reaction taking #acf: in the C$3’lY$ is.‘.

given by ,ul i .js %,,h ., ’ I .-: : i ; ‘.,.

where

(i

.

Needed to describe the equilibriunt’ ‘t$ttia.tiona reackyd: king a ’ ah&i&l... !.,‘ ‘,j + t *,j 22,;:. ) .’ :l:‘reaction, or by mc) or more ‘phases. These ~elacionah& .‘&e dev@X6&ed.’ In ,x1 .,__ : : ,course8 on Th&nodynami&~ ” “I’ ’ ‘-

;‘5* ,,“,‘, .,*-: y.<.t ‘,‘d.‘,,$yy

Example 4.7

--------- - --1--- _____ -T---C-T-- -- -..- ---..=--‘~---- _T~ .Y I” - ““8..

_.,‘x _. d

p r e s s u r e yf fnd temp,Frature Tf . ’ I f the pres;ti;y pf is .l(lT~ .th&n, t h ej nx ;-. 7 .,‘>

bubble-point pressure of the liquid. ‘at t9periitur.e Tf, no vapor’ ‘phase $.ll be,,, ii” ’present. The liquid stream passes $h&&h a resjzriction (waive) and is

--.i , j, *“i , l!. r“flashed” in a drum. i.e. its oressure is reduced from D, to D fF*owp b.6) -

.‘I

\ - ...“.-- ‘., . . ,

i 7’I

1

‘

This abrupt expansion takes place under constant enthalpy. If the pressure’-_

b in the drum is smaller than tha bubble-point pressure:-of the liquid stream. , ; ; .) : : (at the ,.t,emperature .T,,f’ the ,tiqu,id will partially vaporize and two’ phases atI ~

equilibrium with each other wifl’.he present in the flash drum.I

The thermodynamic equilibrium .,betwee& ‘the vapor and liquid phases imposes. I - Is

certain r.estrictions on the ‘state variables ,:of the system which must’ be “.“;,* “. \

included in’ the mathematical model ‘of t&e. flas,h,drum;‘ if it ,‘is to. bk c9n~

sistent and Correct. These equilibrium relationships, as. it is known ~from’~

chemical thermodynamics, are: *

(1) ’ t&perature of liquid phase - temperature of vapar phase <’ ‘I, ,* ,,:. i(ii)l,’ pressure of liquid phase,t”; &essure of vapor ,&ase,: _ p , . _ : 2

(iii) chemical ptent,ial’ of, ‘con~opent, i; in the liquid phase =. . ‘L1chemical ‘pp tentiai of!: ;~rn~~~nent.,! ,, i &n1 th.i; vapor phase q

%I,The equilibrium Cela!:~~nsq~ps~.....int~roP~~~ a$Id$tional,l @@ations among the .., ,_,

‘+state variables of a system and reduce the numb& of,, them. Care mu& be* f “’exercised so that all the equilib,sium re&tion&ips h&& been accounted fhr,‘,. : 2: :_. ” i” I ii ’ ”Pp<,; “, i I , , 1 ‘6, ,:;II. I .;.(. ~.v- * i : I .’ ; a-*,*. _,, ci * .,:, , , ~. ,

_’i.: * ‘.,.__‘,, ,,.: ,. .i:,:.I jl .I’, ,.Ik p, :z$ ID. Equations of State 1’ : ‘4. ;<

(\ ,’ :, ,“;‘:‘& ,9 -I. ,i , ,;, $& ) ’ ‘: 4 Ii-p,, _I(.Needed: to describe ,,the relat&nsh<p ,a&+g the intensivi< var’iabl.es des-. $ * +* ..(” , _i ic&‘: _/ ‘:

cribing the thermodynamic s&te of ’ a sy.st,e&~‘~, The ideal gas law,.I. the Van d,er

” ^ Ia.:Waals equation, ’ are two ‘typic& e&ations :$f’state foe g&zous.-‘systems; I ,

,f r ’ ^ ;. I I,d ~ , . , :. _/,I). . . .!I,-,7: -, - . . :: 3’ ‘, ,A_,.b .:-i ,.’ k ‘“>

>_,.” ;,, ;: j f I ,,;, 2 ? ‘.,.: ” :. “, j ’ ” % : ” : ‘y ‘)Example 4 .g li . i:‘. . , .,i ‘, (. L

: . .a ‘_ I * _( ,I ‘*y* _ ! ? : ,.:./ /, x ,I /4‘:> Let us return to the flash drum ,ey,stem discussed above in Example 4.7.

:

1 t. .r .,_ ;,. SF’. f I/For the vapor phase from the ideal“gas law we have:

‘ ’ L.’ I. ,G I

-c i < *<. : j, >.i (moles of”Q + moles, pP R)aRT

‘. . i

. D%a$or (4.6) !‘/ I c>,,. ,,’ ,’ : !,i

~.

,’ 8,

.

or considering'that ,

Average Molecular Weight = yAMA' + ys% I". " "< " '-

_- . *,!~:, 1 i ?.we have

',

Pvapor (4.6a)

.?“;where yA9 yB are the molar fractions of componen& A and I‘.,'B and MA, MB

are the molecular we&h& oi ' ‘A-. and "k?. ."j ,.

-_ ,I,_

' liqyiid - IO,&,)

b. IIn all the modeling e@n$les discussed in the e&lCei sections'it has

,?J

been assumed thatfshenever a cha& takes &a&in&e of, the input variables‘.

(disturbances, manipulated variables), 2ts effect is instantaneously observed9

.in the state variables ana.the outputs.\ Thus, whenever the feed qomposition,*.

c. ,Ai-

or the feed knroerature. T,.I * 1'or the coolant temperature, T,, change ini - Y'c-. -.

the CSTR of Example 4.4, the effect of the change is fek: immediately and the

k temperature, T, or conc,entration, CA, o.f the outlet stream start changing;*, /,

ii I :themselves.

The above oversimplified picture is contrary to our physical experience,:, i ‘/ i>’

which dictates’that: whenever- &$n input variable of a system ‘changes, there

is a time interval (short or long) during which no effect is observed on. theI

system itself. : , This t&me interval ,j.s caller&dead time, or transportation lag, ’‘_ ‘.o r p u r e delay* o r dfstanceyelokity leg!. ‘,’‘.. , ‘. ; : _.

*4) : :’.,‘j_ II’ I :>,, I.

*Example 4.9 L,:,:A. . ,g.*’ , , .9- *r ‘f;:, : Pi’ ’ p ’

Consider the f:low of an imcompres$ble, nonreacting .llqutd,, through.,k#pe_. ‘f’., -(Figure 4.7a). If the pipe 3s completely thermally insu&ed and the -heat ,,i

y: : .I . ,‘C : *. dr’.1;,. :. . . i; : ‘. ,.+generated by the friction of the flowing. fluid’r .,;s: .’ is negligJble, i t is ‘+sy to

. ,:. j :. *,i .? r:, ,,’ I I’ “: Y( ,I

see that at steady state the temperature,Q;” i.* ToUe.,. % ._ ,f a ,.’ ,of the outlet ,stream will be.

equal to that of the inlet, Tin. AssuM now that starting at MO, theVtek+,I, a..: I_, ,“* ._ ,“‘I1perature of the inlet changes as shown by curve A in Figure-%%7b,’ It *is,‘,.* i

clear that the temperature of’ the outlet,: T out; L will remain the> same until‘/ if‘. . .1

the change reaches the end of &e pipe. ‘Then, ‘we will observe.the temperature, ” iI & I%>’ ,,!: > , F ( J *, ‘., ,+ _ i .., :I ’ ’

of the outlet changing, as shown b$ curve..B in Figure .4.7bi‘ - ‘We notice thatt I) ..’

the change of the outlet temperati&e~‘follows :the same pattern ,as the change I2,.of the Inlet temperature with a delay ‘of td’

/aeconda . td is the d~acJ,S,Gne”

a*and from physical considerationa it is raay ‘to see thsf,, ., , (.I

_ *c :!,/ 1 _* I< ;; “‘3” i.,; .r * , .,. -* ‘(i .I.,>2 :

td *volume of : the :p&pe

voldinetrfc”~fldw.rate,.Jy. L&t.. p e (&& ,:,

I , . ,‘.I I ’where ”

+ I ,&f~. ,.U av is the average-velocity of the f&id over the ‘croks sectional

area”of the pipe.if.,

‘.Fuuctionafly, we can relate '-Ti end To as follows: ,, .-

, ;', .F,

T : (4.7) ., .’4’ o;t( t3 - Tin(t - td)

/, ;. / .::’ rr,

.i :_ ‘! ,‘. _ 2.i. $- ‘l :’

“ ’ :-. I~ ,., .‘( ., ”

-;,t.” ,. 1 ,-$“;: .’The dead time is an impor,tant el;ement,fn the mathematical modeling of

I'3r..;'

chctnicnl processes and has aserious 'impact on the design of efrtectivc con- /

rrol%ers. As'~e.~will see in a later s-ectiou, the presence of deid":time can

dest&il~~-~,v~~~ easily the [email protected];jor aE,n,ayatem. /,:.; 'i.',I..' _1 ,. "'.' : ** I[, ; Ij ADDITrONhL 's&r,i~s .iOF ~~~~~~~~~~~.~~~~~~~~ “

- the mass of the chemical A "Yn,-the react&d mixture, and'-' L

- 'the total energy of'the'rea,&ing m$xture in the 't'&nk. " ' *

'Remarks :,; *

(1) The mass of component B can be fouti from sthe,total ma&,)?,* _ and the mass'of component A. Therefore, it ii‘not,.a‘ *:.. ((,.

- .,_ +fundamental quantity.:

(2) Themomentum of the'CSTR does no; 'change under any operating. I *- - -

-*’ : ‘<‘.<

conditions for the reactor and wiil be neglected in sub-

sequent 'ppragraphs;.If$. ?

Let us apply the conservation principle on the three fundamental

quantities:,, I : I/ _ ,

QFi - .P,F f .! 0 ;-:! .,> ;-i: :,' " i: (4.3.": / 8,_where pi and p represent the densities of &he ,.&let: and o&et-&reams," <'

*i and F represent t~e'volumeti-ic',fl~.~rgtes of thFinlet.@and,,,, .' .i ,'::., :‘ IL l,,:' p; , ,I

outlet str@~t$iS8.e. ft3/6in or m /min, &fid'_: .3- I ,A- ',

''A I ; . . ( :x (. i, :;*') I 3,. *.: : / ;; ;r - ,, : .,. I

V is the volume of the reacting"$ixture; ':,J" :.. . .i ,;'

(b) Mass Balance on Component A ;:-

' Accumulation Input .‘of AL-.--m df ;A“'time *.' - j fime - i' __j"< --r _ /

r represents the rate,of reaction-per.un@ $&me,:,. .i ii i

j *(.’ ;” ,,,,

j *(.’ ;” ”” ,d,d ._._

,

"A"A is the number .of mqlas of A,in.'&he,.;iris the number .of mqlas of A,in.'&he,.;ir_-a._-a. et&n& '&~yprq. ,, ,,, -'et&n& '&~yprq. ,, ,,, -' ..'.I..'.I :r .,:r .,,,

(c) TotaS‘Eng,rgy ,(c) TotaS‘Eng,rgy ,4 -'v-q4 -'v-q , i,, i,

BalanceBalance'( T‘,,.'( T‘,,. , I', .'&. I, I', .'&. I

. c . . .. c . . . $:;;j &c I, i , ) :$:;;j &c I, i , ) : -lz ,~? '-lz ,~? ' '.','.',f ,.f ,. “t "“t "

Accumulation ofAccumulation oftotal ‘energytotal 'energy

Input o f tqq$: _Input o f tqq$: _ output /of tbraloutput /of tbral Merge removedEnergy removed

timetime ,I,Ienergy .with f Bed’> enilsrgy. w:$eyutXet. _ by coota;;'energy.with fked’$ enilsrgy. w:$eoutXet- by coota;;

time ..‘. .b,b.“i, i , ,time ..‘. .b,b.“i, i , ,‘.‘.

1 * / ; ‘.< ‘$, I1 * / ; ‘.< ‘$, IIn the above'b&nce we have'neglected the &ft,work done byIn the above'b&nce we have'neglected the &ft,work done by. .._,, ,I.. .._,, ,I. ..f..fthethe impeller ofimpeller of

<< : .': .' ',, :.,',, :.,the stirring mechanism.the stirring mechanism. The total energy,df'hhe reacting mixture isThe total energy,df'hhe reacting mixture isP ,,f (-I IP ,,f (-I I ,fi,fi9;9;

where U is the internal energy, K the kinetic energy and P the potential

energy of the reacting mixture. Thefefore, assuming that the reactor does not,#:>f.-., : . > :I

move? i,e-. dK/dt - dP/dt - 0 , the left-hand side of the total energy balance

yields ,,.

dETE-

d(U + K + P) dUd t -xi

Since the system ii a liquid system, we -can‘make the following approximation, “:’-. 12 . ., .:

Accumulation of totalenergy of thematerial in- the CSTEper unit time *‘:““: * i :’ , ’ ._

Furthermore, I . i 4,:’I L ‘j2) *.a. I_.I “ t 4 ‘~:. *,: _:/ .. (Input of total energy-&th;feed’ per unrt,“t!ej ~$,p,~Fi lhi(Ti)‘; :“ ’,‘. -2

, :i ‘,?, 1 >’ 1. ” ~

and, :.. ..a ..,5. ‘, ’ ,; .i”

,\’ i ;‘T$, ,, _.’ *,. . .?’,?. i” ).(Output of total ‘o&ergy with the. ou,t&$$&$#n ,per’ .un@“~$@& A. pF h’(T)

:{,*’ -) .I ,’ ‘.: ,. ._ .$: :., ,, _ , _ ::’ 2 .: :,b,.: ._, i.1 ?:‘ : )where hi ‘is the specific enth&y (&Ghalpy per un$” maas)&&,‘:the fee& at$,eam>$ ,% I. 6. ,,i , -,:,L I .:” ..‘and h is the specific enthalpy of ‘dhe ‘cid.et % tF&rni

.z‘i (kqkytadti+, the ‘totalI

). ‘.energy balance leads to the foll~~i~~~~~~~uafr,~n,

;. @,’‘“i. ; ,’ I” .II‘: .Ih . . .:. *-: . i.

.,“r.’ r,z ‘;

;, ‘:- Q’rz

. 7 ._ I ‘ .-1 i/ ‘( - 1: ;,i‘,, j .‘e,. * : .‘_II . , . ‘ _ ,.

Q is the amount of. heat &moved by ‘the. coolaqt ,per unit: t&me.where., : ,.,. .::.:$ ,_(.I ,.‘a’ ;;, ‘_ ,;,: , , :; ; :i;* $ .Y : .a1;; ;I.

Equations (4.8)) (4.9) and (4.10) are %ot t&heir f&a! and most con-I ’ ,(,. :.“‘,~ ~” -! I< i : ,.:.. : ‘? . . “S1 y , i ,< .‘,

venieht form for process control design. studies.I’ To bring them to such form 1;,~*..?f~,;’ ,i ( , , :., . ,L i

I we need to identify the appropriate state v&tab&; ‘I ’qf4, : t $ ::

- Characterize total mass: We need the denCity of ‘the reacting’mixture, p,.+

;and its volume, V. ,,The density will be a functioh of- the co&&ntration

., WI

CA and=B

and of the temperature T. Quite often the dependence’ of p ,

o n ‘A’ ‘B and T Is weak and the density can be .roosidered constants as” I , , : :‘a. 8 :; ,.,

the reaction proceeds. Therefore, the left-hand side of eqn. (q.8) yields

Und,er &he ab&ve ,assumptLon ,4 ; : f? ,., V is the only state variable which is need&j itti..i. /‘8. .( -i-i:,, .<,i ‘. ., r I .:‘r....

characterize the..total masse. Then eqn. (4.8) becomes: j li ” ‘. *>.,, ‘- i: , . ‘, ._j ‘; ,&i’: .‘I‘#I i 3;

dV 2dt ‘; Fi ‘,L : F,

: , (4&)” , ~

- Characterize the ma& of compo&nt A.[ _ This is .simple.. From eqn.S.;(4.9),iwe, _ I’ ‘_’I r “fy ;; ‘, 1:. ‘- ” ~,.’*is

realize thiit the state variables needed are:,I ; . - :

, --,.”_c:

Algebraic’m&.$<ti&s on eqn. “(4.9) ‘lead to,

d(cAq? x’.‘&~ : d,fA : 7 -E/&T F ‘_ , ,

d t ‘- ~y4.zF ‘: .y - -dt _. - cAF LI- .ko” / “A’ . _.*‘ ’

I:,:, ,- 1 a I* -: i,or

.B ‘IdcA .” 1. , - . ./ ., i;”,

v-z- - ‘CA<FicAv ’ i ’

enthalpy of. a Z&quid .&Q&I is a fu

sition of .the liquid sy’mtebn, i.e. ..> _,

j , *. ,,_“: ,) :. ::

,,

H - H(Ts “A, .!$$.

‘3 ,J ,i -+ 1 , 1where n ‘.’A and -4 iy~‘~~~e m o d e s of A ‘.snd B ..,?n j&e CSTR.Differentiating

_ ““.the above expression we take, j- .,% ”

dH aH :dT + aH,dnA

- -dt - aT dt ?nA d t

+ ali d%’-as d t i (4 .11)

.‘< , . ‘,.’ :

il+l,t, ;‘;a , I “. i ;_. .-. ) L ‘: : 1: . ( , I . “., ~. (be>; :,. 2. : a’,aI4aT. - PVC *m

P ’ an, fiA(.T) , - fig(T) ,; I’ ’ ‘I

where cp: 3s the specific heat capacity of the reactihg miiture and “A,’

HB 1.the partial molar enthalpies of ’ A and B. Furthermore, from ‘eqn: ’

:(4.9) / ,

H n-Le sdn..dt

d(c.V)

P

and a similar balance on component B,

Substitut.e the above quantities in eqn. (4.11) and take:,‘

< _.,. i

dHdt * PV c,$ + $1 cA Fi - cAF -

,.g. ‘.

i ?? .(.,IrV1 + ,f& I-cBF + rV 1 :>, ,,.: I:, I * ;<. :.

Sub,stituting the dH!dt’ by. its equal., ‘in &he total’ energy balance ,JM &k&, ” ‘).t’i ., i ,)‘<; 1 ,. ‘__L -HA[cA~F~ - ‘AF,- ,tq -,, ii [-ch,F.. + r&J + p F, B i i h, &:- PF%-Q(4d?&),. ‘. ’.I:. _ , ,

Let us now notice that W

Consequently: eqn. (4.lOa) becomes 9 ’ : ‘, i ii.,

‘” ‘. *‘,

dT -iiAcAIpi + fiAcA + n,ri + &cBF, y, fiBrV ’PV cp dt - et ”:

or

i i .i Pi,(Ti

‘A A- FcB%. ‘: Q

:

that temperature T is the state variable2’. ,; .;_/ .,*.

!ne&y’of rthe. system. -:j’ ,..‘,‘T ” ’ ;, ‘ i) SE- : . . :‘.

y -: 7 - _..,_ ___ .e ‘steps in tha;mathemi’tical modeling of a CSTR tie; ;: b‘ ‘. , , .; ,: ‘, ii

have the following :‘ ,i‘ g ;’ /I ‘ *~,,: -

s tate var iables ; V , cA, T ‘, : >’ I.; i t .

state :equatAons : . ’ .* .._,

input variables: cA q R,,.‘$ii $.F (when feedback control ‘IS tied)i (“L : ., ‘, . . >A_ .

. .’ ,.:

Among the input variables the most ‘;~a&& ‘&&rub&&ss’ are : .-’ , l:cA , ,Pi, Ti ,., “. . . . . ! 1 ij :‘_ _s’ :,,

Idisturbances:

i’, ‘,.zIIz,.. ,.:,i: I

1’,*1-”- * ., )I

while the usual manipulated variables ’ atie: -\ _,.,; ‘fI‘ ~.

: manipulatgd variables:x

Q; F (oocasionally Fe o r T i )_ : I ‘, + ‘.

;iThe remaining (variables are parameters cha~acter&lc of the reactor system,

i.. e. ‘. Iconstant parameters : p, cp9 (-PII,) ,A ko, E, (activatYon energy), R. I’

i

,* _-

In the presence bi changes in the input variables, the state variables I

change. Integration of eqns. (4.8a), (4.9a) and (4.10b) yields the CA(t),. '

V(t) and' T;(t) as funrtions of' time. 'r‘: I; 1,; / : '.j. .Theieteady state behavior of the CSTR is given by eqns. (4,8a), (4.9a) and

(4.10b) Setheir left-hand sides are set equal to zero, i.e. from the'solution

of the following set of algebraic equations: a_ :.'

.,

Example 4.11 - 'J&e Mathematical Model Q& a Mixing Pro;cess : .I q (,

Two streams 1 and .2 are being m+ced~i&~a.well stlrred:tank, prtiducing a

product stream 3 (Figure 4.8‘). Each 'of the‘ltio feed streams“&+ composed?of

two components. A and ..B, with, moilar .&&entrat%ons c.-~.., c% - Bl.'

iandjcA2 ' cB2*

respectively. Let also ,F1 and F2 bra the volumetric flswrlrtes of the two

streams (ft3/min, m3/m%n> &?W‘-T 1, T2- their cprre&ondtig :&mperatures.

Finally, let C~ , cB , F3 ,and ;'T33 3

be the concentrati&,, fl&r,ate and tern-l

perature.of ,the product str~pm. A cqii. is also immerspd.,k the; li#iid of the.,

tank and it is used to SUDD~V or remove heat from the svstem with steam or

'cooling water,'cooling water, :,>, i:,>, i L ; , ', ,~ 3;:.L ; , ', ,~ 3;:. -I-I : 'ii .: 'ii .

The fundamental quantities neefled to deecribe#th$ mixing t,ank,are:The fundamental quantities neefled to deecribe#th$ mixing t,ank,are:__,,_,,,_,

.'.''L the total mass in the tank,:;'L the total mass in the tank,:; , >Z', >Z'

- the amounts of components A and B in the tank,.- the amounts of components A and B in the tank,.. ..lr. ..lr i: .:i: .: '..'..::

- the total energy, and- the total energy, and;v;v

- the momentum of the material in the tank. :- the momentum of the material in the tank. : ;.-;.- ,'/,'/

.'Remarks:"'.'Remarks:"' (l),,The momentum doas not change under, any op&rat$ng:condit$dns(l),,The momentum doas not change under, any op&rat$ng:condit$dns

,$nd*.it will.be neglicted.in further treatment,,$nd*.it will.be neglicted.in further treatment,

(2) We:"only need to constie; twofof the foiiowfng three quan-I

tities; ,total mass, mass of A, mass of B. The third can .

be computed from the&her two.

. ,. .':.. ',

u

I.-

Con,s%der nuw the balances on the -fundamental quantities: ,, /.:

: ‘i ,(a) ,% Total’ Mass Balance-i

,_ ,, ,

,(“~~,~~i:h”e”:a~~tal) (m;;;“;;qf+-Q;;.k) ( @Jtput.; of to.t@

t,im,e ,:Y in , ‘: time_ mass from the tank

>,

agLl blF1 + p2F2).I ,.2; !biF3 ‘ (4;15j.’ :

where %’ p2,.p3 are the densities ‘of the streams 1, 2, 3, respectively. : . .

Since the tank is well mixed the density of the’ product stream., p3” is equal ” ’ ’,

to the density of th,e material in t&e tank, p, i;e.’ ‘~3 * p. V is the volume . . .

of the material in the tank which Ia charaeterizM by the cross +ectional area ‘i.. i

of the tank, A;* and the height, h, of ‘$he liquid level, ire. .‘: ! ( . _>__’

V I A*h,. 4,: ._‘~ ‘. ;,a i

In general, the densities -p, pl and 6,” ,, depend on thq corresponding conce~

trations and tes@eratures, I, e.,, ‘1 ‘, .I : )

‘,, ,,ly: ’ ” ‘_: ,%..$ ‘. 2 ,,:‘*;r*

-i

P. = Pp = ~(CA~Y.~~YT~)Y PI + f(CA~,yA~,T1) iii ~~~~~~~~C~2~,~2~k&) li,.i!:f:I\ /

I ” .>::._i, ,i;Usually (but not always) the above de&&&n&s ‘&a’ti&ak wd %,, assume.!‘Lhat the :., ,,s: .”densities are independent of the concentrations and tem&atures.id‘. :

Ths,r&~$k.;; ‘.

we assunK? thbt,,‘, .‘,) ‘j* ’ **.. ,s: .i”& . ‘.J; : >c. *,.. (t ; ‘B .\ r : *’ ‘i” “, ,( , ;

Pl - P2 ‘= P3 - Q*,;*..

: ‘ . 2 ., : L *,,. ‘...* : [,‘. , : “t? ‘, <IThis transforms, eqn. (4.12) to the following:

)’ iL, :: i

, , dV dhdt = AZ - (Fl + F2) - F3 .i ,, )( . . 87 (4,12a)

>%..((b) Balance OB da‘mponent’ h L:; I’. ;, ‘> ’ * ;’ / :

Accumulation of ,’ Tota l in&t of ” Total.output “ofi component A )(component A in ) ( c o m p o n e n t A )

t h e t a n kL in; t h e t a n k _ ? f r o m t h e tanktime time : i time.< j :

o r, ‘ T., , ‘:

. .

‘, dcA. ,

V-dt +cfi.,:.i A dt = (cAIF1 + cA2F2) -i CA3F3 .,. (4.13),

Substituting dY/dt by fts equal from eqn. (4,lia) we have,

dcAY x +- cAflFl + F+ 1 i31 = (cAIF1 + cA2F2) ‘- cA3F3

i.and since CA =’ CA due to the we!@‘st$rring assumption, ,’ ’

i ‘;i.’. , .*

3 ,I . .

The total energy of the meterig& in the,+

E - U (internal) +, K ,(kinet$) $ P (pot) 2:

Since the tank is not movi,ng , dK/dt * *dP/dt *s- 0. 3%~ uip, dE’/dt - dU/dt and ‘for. .

(Heat. added or r.gmoved )with, ,the coil ‘1, .I :?‘

t&g, *, “‘i;

liquid sys terns,

dU -. dH,dt - dt

I +a,- <, : 1_. 7..I

where H is, the total enthalpy t3f thematerial. in the tank. Fuf t h e r m o r e ,., *.

Input of total .” %li. ,.with. feed stream v‘

per unit time4 F2h2) ”

‘1 ,, I.

and

I

4.

Output of total energywith product stream

1

) ” ,:.p,F3h3 ” i i <, , _: . =

per unit time ,‘,

where h1, h2, ,h3 are the specific enthalpies (enthalpy per unit mass of

streams 1, 2 and 3. Due to the perfect stirring assumption: the specific,:A, ‘,enthalpy of the material *in stream 3 ia ithe:, aa& aa ,$be specifi$ enthalpy of,,,- ,,. ‘.a 9 :the material in the tank, Thus, :f I

H - pV h3,i ‘)’

I i i _’ *\

Consequently, the total, energy balance’ yields,,’ ,, .?-y..t

d(pV h3)’ 4. :.:. , ..;.

dt - p(Flhl + F2h2) - .: FP3h3 +, Q ,: ’ : (4.?4);

.‘tThe question now Is’ how to characterize .,I hi, :h, and 63 in terms bf. ether

‘.’variables, i.e.. temperatures, concentrat’tons, etc. , Weiknow thao; . _ ,+

r ; ‘, di-” ,A. . ‘*. r

h3(T3$ 9 h3(To)-, t, 6 ‘%T .- T 1, ~‘;

:

. .

where ‘To is the reference temperaturer:,. ‘At this tkmpekature “I ”

J “.Ph3(To) (4.16a)

,. ,: $’ I :: ;,

&(T,) ,(. (4439)

.j .ph2’qbl =, cA2tiA +b$iw +; cA2A$2(To) , (4.16~)

.;, ’

where 8, and’ fiB are the molar’enthalpka ‘(en$halpy~ ‘*!.

per mole).of components

A and B at temperature To. “gs , A$ ,.’

A$‘3 ,.are t&heat -of solution for,;1 2 , : J ,“.

‘,

streams 1, 2 and 3 per moie of A at temperature To. Substituting eqns.

(4.16a,b,c) and (4.16a,b,c) into the totsi energy balance eqn. (4,14) we have:, . :,

i , dtI

= F,(CAfin+caJ . i$+~~Afi~)+pF c .(Tl-To)i nln D1 ,1 1 ~. 1 PI

.:4 F,(c. ii. +'cL-A A,

2 B2i$ + cA Ati ) + PF

2 s2,'c (T2 p2 2 - To)

\ 1- F3c"& + cB:$&.' cn,hti ) QF

3, ;3 3 s3' ,c

? -P3iT3 -'To) %,Q t

. j _A' ':.AT-“I

, .' 4 0 @alance on A)

)'+3f, + cA A*

3 s3Cp3@3 - To)1 :

.a

_- = Fp*,'"S, + PFlcP.J. J . -L

01 - Tb> + F2cA2AHs2 4 pF2cp2(T2 - To).' ,'. '. :

PF~c~,($~ - To)"I

A - F3CA3AHS3 - .a3 .ld ,/, . ’ ; ,’

and finally,

,

,

dTpc .t'$ -

p 3

3

cA F;[Aiig L 1'1 1 '

AI? ] ‘p':.+ CA P2[Ad !.g3 2, 32

- Ai&3

72

+ PF~~~~~,(T~ - To’ - cp3(T3-T$1 + F 11’ cp (T - To) - :c2 2 p3

(T3-To) f Q-:. ,,.

If we'aesume t&t c = c - c.*'.

pl p 2- c , we have: ,

q3 p. .

'.!

+ P1 cp(T1 - T3) + H2 cp(Ti - T31 * Q (4':14a).

Summarizing the, above -steps we have: 1! A' :-c. .1 ,'state variables: V, c 'T

A3' .3' / ,I~

', ',':t., -.,'

,lstate equations: ,' .'

'8 < '

.g, "- i,Tl +,F2)+ ,- F3j I . :,.-,' y

(4.12;) "., $i^ .; ' i '0' .p. II.

:, :dcA.' ,I' .: :- ‘.

,. y-2dt 'I @A

1- kA )F + PA

3l - 'A3)'2 (4.13a) ;2 \ ‘

,Ii

dT3.'I :, .I ., 1f 'I . .,, : I .,* ., .,:: '.

:% 'i ,.r'

f?cpvdt- c F fAii_: Al l

-Atis‘ ] "cA F2[A#s1 3 2 ‘ i s2

-AiiIs3

~+PP~c~(T~-T~)+PF;C~(~~-T~)~I~Q ' _-:+ : -

: ii,/". ' :,

_ '. ) (4.14a)

input variables: 'I!F1s cA1' T1s F22 cAiP T2,, ,3(for fa&bac~~~i$ntroL) I:<.I b'

output variablea: *V (,or equi;d;entQ-. ;*:' .?“ ,,,,,

theheight*,I.

oE ,liqaidl.eve~: h)';': i .;::?

:=A3 a n d T3i,

t .' .‘, 1% -.p,s,dfis ,, Aii;

., I“parameters (constant): p, c

_( t;l. s2'A& ~:, .'*. ;

r * . ?: _,

I/Remar lcs : (3) U&ally a mixing tank is equipped writh aeooiing or heating '

coil or jacket through,which flows a coolant (if heat is ' 1' 3 (* 4,,,'2,,. :

released,during the mixing of the two solutions) or a heating ' .: i\" . ,. c*,

t

'!- :1/0, ,, :

medium (if heat is~absorbed during mixing:) in an attempt to keep.I )

the mixing, isothermal. L( -.. .‘, ”~: _ *.

(4) ‘If the,heat of solution..sre @rang functions of concentration, i.e.?:

i f [hii ’%

- A6 ] and ‘[Afi,s3 ,! s2

- Afi, ] a re not smal l quant i t i es ,3,

” then from the total ‘energy balance eqn. (4.14a) we notice that+i I “: ‘7 : *, i.. ‘* 7:

jI tqgiderature. T3’. depends strongly ‘on the *concentrations of the‘., .) 7 ;

. -’ ;feed,strear& and their temperatures. If on -the o&r ha& ;

Consider. the shell-an&t

flows through the Snner tub~‘aird ,.it is being h~,~,ted~:by:+team ttfiieh flows outside/‘* , I :” ; ~ ~, ,_

the tube. The temperature of the .liquid. does c&t,‘only change.<wS& time ,but it.:‘. ” ,_,,.,y; ,i ,.also changes’ along the axial direction t :. ‘_.

~&NU thi :valua :I$. at the $ntrance“ L ~i I ‘i . . ;. I,. ,‘,

t o t h e v a l u e T2 a t the ex i t . We wiil &&I& t&t’ the” t~emp&ature does not;I

change, al.qng the radius. qf the pipe,. S&J.:- .we will+~ssitimp p$.& nlh. CO&. \ , ;.!J.sequently, we have two indep.endents variable& $..$. ‘,;ti.“!’ a$ni,‘;:‘t,: .me..stgt$ i _

1 ‘,,variable. of interest for t&e heat +&hanger i; .th&&peratt&. l(T) bf the’ ‘,;., ,, ‘., fheated liquid. Therefore, we need the energy baltitic.e;-foi;i..athe $b&&!briiation r,

of the temperature. To perform ‘this balance consid.er the element of length -1, . ’

> 1’ /AZ defined in Figure 4.9 by, the dotted lines. Fori ‘this syatcin, andl,over a :ii ’

a , 1: /r ,: .*_i . ” ‘;,r _. .*.:i period of time At’ we have:

.‘. ,, )I i i / c’$

Energy Balance . ‘C

p ypA*bz- [(T)

):A ‘.^: ,/(.’ ., /. .

I- ( T ) ] - p cpvA(T)‘:- *At -. p cpvA(T)

t?f*t 1 ./‘. eI ( I

Q*t +.’ 2 z+Az

Accumulation of ! Flow- .in of ’ F l o w o u t o fenthalpy during enthalpy during enthalpy duringthe t ime per iod i the time period the t ime periodAt’ At /.’ At

r.r ‘\ :, _ “‘8;

73

+ Q*Dt*(vrD*Az) (4.17)I’,, ,,I',, ,, .:,Enthalpy trans&rred

, from the steam ‘$0 the I . . . .liquid ,1 through the

. . !,!, 7%7%, :, :

wall ,. during s&e timeperiod, At. .‘I_

’’.’.’ tt IIII 1 I,1 I,wherewhere .d;.,.d;., ff __;yri _ ! ’ ’ *;yri _ ! ’ ’ * *i*i

Q is the amount of heat transferred. from the steam to the liquid perQ is the amount of heat transferred. from the steam to the liquid perunit of time. and unlit ,u$,heat tratisfer ,greo, -1.unit of time. and unlit ,u$,heat tratisfer ,greo, -1. :: ,:, .j,:, .j

1.1. ii‘1‘1

A is the $ross section& :ar,ea of .the ,inmr tube,. ’A is the $ross section& :ar,ea of .the ,inmr tube,. ’77

v is the velocity of. the liquid, an$v is the velocity of. the liquid, an$ .:;; L.:;; L ‘1‘1:, ‘ ”:, ‘ ”

D is the eFterna1 diameter of the inner tube.D is the eFterna1 diameter of the inner tube.:: ,,

Dividing both sides of eqn. (4.17) by Az*At ‘and letting A&& and AtA,Dividing both sides of eqn. (4.17) by Az*At ‘and letting A&& and AtA,, <, < ii

)r)r we take,we take,..

11 aTaTP cp AGEP cp AGE + p cp v+$ ,F$,* Q,:** I.).' \',, .:, :.+ p cp v+$ ,F$,* Q,:** I.).' \',, .:, :.”" ,:*,:* (4vW(4vW5, *.‘I5, *.'I

In eqn* (4*18) we can subst’it&&. Q.:!:‘gf~&& e*ai - ’In eqn* (4*18) we can subst'it&&. Q.:!:'gf~&& e*ai - ' :,:,_ i_ i

j’J 5:: . ..y. :,j'J 5:: . ..y. :,

Q - h(Ts I T) ~; :: - ; t y :::.a-Q - h(Ts I T) ~; :: - ; t y :::.a-,,

..‘u ,’..‘u ,'‘, “, 1 ~.,I i.’ , ’', ", 1 ~.,I i.' , '.; 0.; 0

:I:I

ttand take’ ’and take’ ’

“: .*:“: .*: 1 ‘,{ i;; “j1 ‘,{ i;; "j : ‘$i: '$i : .’: .' jj **j**j ::I, ,, 1I, ,, 1‘/ ”‘/ ” ,,Z$,,Z$

STST‘1”‘1”iT ,’ :iT ,’ : ‘I i ,j ) i ” i, ;*, : .‘I i ,j ) i ” i, ;*, : . ,*:A .$ :i”,*:A .$ :i” q ;.q ;.

::1 pcpAat,,1 pcpAat,, + P c V A - --h*D;(Ts-T)+ P c V A - --h*D;(Ts-T) , ,, ,

P J a?.>!,:,.::~ iP J a?.>!,:,.::~ i ”” : ‘;A,: ‘;A, ( 4 . 1 9 ) ’( 4 . 1 9 ) ’ii:‘.:‘.

This is. the equation of st.&te that models t.ha- behavior of the, liquid’51 tern-This is. the equation of st.&te that models t.ha- behavior of the, liquid’51 tern-

perature (state variable) along .thp~.~eng~~..of!-thre .exehanger.perature (state variable) along .thp~.~eng~~..of!-thre .exehanger..,.,

L,L, S&me e&‘.~~(.4.&S&me e&‘.~~(.4.&

is a partial differential .eq&ion: we say$+&at the -changer *has baen*modeledis a partial differential .eq&ion: we say$+&at the -changer *has baen*modeled,&S.,&S.vv

0~ 41 ~li8rrab~t,t~~ornnoter syatct,, ‘t*: .‘;0~ 41 ~li8rrab~t,t~~ornnoter syatct,, ‘t*: .‘; I ‘II ‘I.e ..“.e....mW”.I.e ..“.e....mW”.I ‘,i> “( ‘5 . , j ..,‘ :;‘,i> “( ‘5 . , j ..,‘ :; .i.i

:: s : ‘I’-,:s : ‘I’-,: i -i -

‘.‘.Example ‘4.13’.Example 4.13’.

,‘. t,‘. tThe Mathematical Model of~‘&~$&, Binary.,DSsti$J.ation CQ~U~The Mathematical Model of~‘&~$&, Binary.,DSsti$J.ation CQ~U~

II i:‘,. .,*,i:‘,. .,*,’’

“, ”“, ” ,;r .1:,;r .1:Consider a binary mixture of components A atid‘ B to be separated intoConsider a binary mixture of components A atid‘ B to be separated into ._._

; . 1. :; . 1. : .. “,“,two product streams using’conventional dlstillatian,two product streams using’conventional dlstillatian, ‘J‘J

The m+5ke is fed inThe rn+5ke is fed in11.. ;<.‘a .I, :,;<.‘a .I, :,the column as a saturated liquid, i.e. at its bubble .point, ‘onto the feed traythe column as a saturated liquid, i.e. at its bubble .point, ‘onto the feed tray

Nf (Figure 4..10): with a molar flowrate (,moles/mi&te)‘.FfNf (Figure 4..10): with a molar flowrate (,moles/mi&te)‘.Ff

-(. ‘--(. ‘-

ani a molar fraction“ani a molar fraction“,-,-

:: ?l?l: *, ’ ;‘g,

i

of c'omponent A, cf'

..'The overhead vapo r stream is cooled and completely con-I

dPnsed,band then it flows into the reflux drum. The cooling of the overheadr

vador is acoomdlished with cooling, water. The liquid from the'reflux drum is'L

partly pumped back in the column (top tray, N) with a molar flowrate Fk

(reflux stream) and partly is removed as the distillate product Qith a molar*.-

flowrate FI,. 'Let us call MRD the'1iqu.i.d holdup in th‘e reflux drum and xD

the molar fraction of component ‘A in theliquid of'the.re'flux'drum, It‘&

clear that xD./

is-the composition"Yd"r ,both 'the r&l& ,and dis'til.l~te &reams, I

At .the base of the distil,lation column, a'liquid &duct ,&'eirn (the' ',

bottoms product) is removed with 'a 'flo&a'tb'~'~b, and's com#osition xB (molar3>. '

fraction of A). -A liibid stream w$& ,& wlar flowrate 6s' i# "&&A &$,, fr<im,:

the bottom of the column and after it has been heated pith steam,“&t'retucns,.I

to the base of the column." The~'-Eompoeition of the recirculaddng',back to column'_. :

stream is xB* Let s, be the,S&iquid .h+dup at,the baee og:%the column.

The column contains N trays~nnmbered from thebottom o-f..the:,&lumn toA

the top. Let Mi be the liquid.holdup‘on the i-&h, tray, ~The vapor holdupon

each tray will be, assumed to be, neglig-ible. __ ,‘+, j ; .: '. ..'9

In Figure 4.11a we see the material flows in and out of the feed,,,hray,.

Figures 4.11b and.4.llc show the material flows for the tag (n-rth) and" bottom '

(1-st) trays. Figure 4.11d refers to any other tray. . ,',.J..,' I. II(,. ,

To simplify the system we are considering, w,ecwfll, make the following

assumptions: 1 : ."I

- Vapor holdup on each tray will be neglected. .;: j " * .,%

:,The molar heats of vaporization of bo,$h'components A and i i are approxi-'

mately equal., (i This means that one mole,of condensing vapor releases enough' ,A : .heat to vaporize one mole of liquid., 4 P!

1,- The heat losses from the column to th,e surroundings are assumed to be ,

negligible.

- The relativei,volatil&ty a of the t@ components remains constant through-J ~ : ,-. 1- The relative~,volatil*ty c(~ : ,-. of the t@ components remains constant through-J 1

!out ,the column. '.!out ,the column. '. 1 I1 I: , -": , -"<< . ,. , ."."LL- Each tray is assume< to be 100% efficient,- Each tray is assume< to be 100% efficient, i.e. the vapor leaving each trayi.e. the vapor leaving each tray

## 4'4' 5 \5 \ .'.' :: ff ~7~7

la in~.equil.&brium pith the liquid,,on the tray.la in~.equil.&brium pith the liquid,,on the tray.,, ..a ,",, :,, ..a ,",, :ii ..?.#..?.# '.'.

‘&e f iret’ thr& ‘arraumptioF,e .imply ,$hatT6e f*~stthr~~'agsumptions .imply,that.-,.-,.,.,

:/,- .:/,- . ,a,-> . , t,a,-> ., t ::’ : . . ”::' : . . " - , .-,. : t: t ‘.‘. ::

VV *,?.$ ,w“-v -.*,?.$ ,w“-v -..l & 2.l & 2z .'z .' '.'* L-Y'.'* L-Y

,* 'Y; ., 7:: VN ,;',* 'Y; ., 7:: VN ,;' jjtt ,,.'.', jl _.,, jl _., '' .i.i:*:*and,there is no,peed for! energy '~ala~cq~around each,tray..and,there is no,peed for! energy '~ala~cq~around each,tray..

,' I,' I :: .L :.L : a" +a" + '.,,, ,,'.,,, ,,

The last&m, assump~~qns~.~~~y"thaf'a.,Bampl~ vaporrliquid equ,ilibrium‘ ! ,The last&m, assump~~qns~.~~~y"thaf'a.,Bampl~ vaporrliquid equ,ilibrium‘ ! ,, h~-~ ,, h~-~ , LL .*.* . ,..). ,..),-,,-,./.j j./.j jrelationship c&n be used to rele;~~~~~,~~~~Olay.,.frs,ction ofrelationship c&n be used to rele;~~~~~,~~~~Olay.,frs,ction of

;r,;r, ai, ,"":;ai, ,"":; .. __.. __ A in the;,~yap~~~,A in the;,~yap~~~,,O,‘ ),O,‘ ) IIII''

leaving the i-th tray (yi) with the,molar~.f9~aci$on~ of A in the liquid leavingleaving the i-th tray (yi) with the,molar~.f9~aci$on~ of A in the liquid leavingII,a;,a; ::::the same tray (xi); i.e.the same tray (xi); i.e. 4,4,I. _ ,,j ,_ ,. *,; ;I. _ ,,j ,_ ,. *,; ;2"2" ,,i,,i .;.; ', ,,', ,,,, I"I" // --"I"I ;;"/, rr:"/, rr: :..:..J :J : axa*

11._ 2' ': :,;. 1,; _.._ 2' ': :,;. 1,; _. a: *a: * '1'1 * ,,. :* ,,. :

yi 7yi 7 ,1 + (a-lhi,1 + (a-lhi(4.20)(4.20).I ;.I ; *. .,*. ., f _<' :I.f _<' :I.

r‘r‘ "".i,.i, " ff!',$ I+" ff!',$ I+

is the relative vola~ility'.o~.th~'twc~.~ornponeqtsis the relative vola~ility'.o~.th~'twc~.~ornponeqtsI.I.

where awhere a AandB. i *!-' ,AandB. i *!-' ,e,7 'e,7 ' ',',.,.,

88 The final assumption$ that tie Wil'i :ni&e’e$kl”thd’ fo&mihg:The final assumption$ that tie ~~~i~riU?e'@&&hb~ fo&x#ihg: ‘*'* II

;:Neglect the dynamics of the cond'enser~%nrd 'the r'kboiler,;:Neglect the dynamics of the cond'enser~%nrd 'the r'kboiler, ''it :is clear: thatit :is clear: that::

these two units (heat exchangers) consfktute ~roces$ing'syefems ‘on theirthese two units (heat exchangers) consfktute ~roces$ing'syefems ‘on their ''

,rate of the liquid leaving each tray is related to the f;iquid,holdup of“ ". 'I. .!i ,,the tray through Francis weir foir&tA; -1’ $

,',''own right and ds such they have'ti'dynaniic behavior ‘(see &ax$le‘ 4.12).'own right and ds such they have'ti'dynaniic behavior ‘(see &ax$le‘ 4.12).

Therefore, any good'modeling should include she state'equations tihich des-Therefore, any good'modeling should include she state'equations tihich des- .,.,..

tribe their- dynamic behavior. .' 'tribe their- dynamic behavior. .' ' : ‘: ‘

- Negl.ect the momentum balance.& each tray and assume that t-tie molar flow-- Negl.ect the momentum balance.& each tray and assume that t-tie molar flow-

,rate of the liquid leaving each tray is related to the f;iquid,holdup of“ ". 'I. .!i ,,the tray through Francis weir foir&tA; -1’ $

Li - f(M$.**“‘i

wi - 1,2,***,f,***,N (4‘:21.) -

:I Let us now davelop the s‘tate equat$&which will detkiba'the dynanici

Li - f(M$.**“‘i

wi - 1,2,***,f,***,N (4‘:21.) -

:I Let us now davelop the s‘tate equat$&which will detkiba'the dynanici

behavior of a 'distillation column. The Fndamental quantities are total iaassi. i

and mass of component' ,A. But the 'question is:<i

behavior of a 'distillation column. The Fndamental quantities are total iaassi. i

and mass of component' ,A. But the 'question is:<i

__

"What is the system around which.we will make the'.balances?" F+om a"What is the system around which.we will make the'.balances?" F+om a++

pract;fcal point of view,pract;fcal point of view,:.* .I:.* .I

the bound,ary of the system of interest' is outlined bythe bound,ary of the system of interest' is outlined by,I. '.,I. '. ,,

dotted lines &-I Pigurt 4.10.dotted lines &-I Pigurt 4.10. Such's boundary clearly identifies the inputsSuch's boundary clearly identifies the inputs II

and outputs of practical significance for the overall system.and outputs of practical significance for the overall system. Tt -is alsoTt -is also

evident that, unless we can desctibe-how the concentrations and liquid holdups

on each tray-change with time, we cannot find how the variables of practical*rsignificance, like xD and x

B~khqti~e with time. Therefore, we are forced

to consider the'balances around each *tray. Thus, wehsve (see also Figure< .*> ,:. .4.11):‘ : ., f ':

,I i

Feed+'j"ay (ipf) it -, cr:,i. *.: . ::' ci ' ' CJ : .,i

T n t n l A&am. N’L.w r: .:: 1.I tr -1 -tr -m -1'R T VNwl - aaM - .WN - PR - UN '(4.23a) 1UL

" ," ..' .

Component A:d"!$& , -, .' ,:s-

dt. - FR*xD + $R-l*~~~l - L$$.'~~~ *i* Ec r '/ & 1 & ",,_

(4.23b),;

.j, ," .; <'I; i i._ Bottom Tray (i=l).,I-i

d(M,) ' '". ,' - ; i ::

- ‘n-L--,--L L..- ‘--1--1’,

- + .<i.,.

&“LcI.l. AIUZID.dt -.&I 2 - ? :

\9. apa)j r, i’*

dfM-x-l <’ ;. IUXll~UWNlL~A~ dt = L2x2 ? vyO - 'Llic&

-...' vlyI: :‘ ." i :! ; ,'(4; 24b)

1, ,\.,::; ., ; -: ',A;. I *. ; !c I.i-th 1tray (112,***,R-1 and if, :' :- '. i 1 ,

d(Mi) ~..Y ,: b._

Total Mass: .r = LieI - I.,~ 1 ; I '(4,2.59)

L

4

1

t

Reflux Drum

! i4’. .‘”. i D; (4.26a)

,I j’. ,’

d(fIRDxD) c’

i Com‘pon&t 8:dt

iNAYS - uR + FDjXD ‘- .I (4.26bj .l),I : . G , .

’ ” J.$’ “:.::.” : ,, ._

Column Base .i”

d($ ”j II, * .:i : i .

Total Mass: dt’- Ll<$. -. V - EB$ ~, *. ‘(4.27a).’ /) !

I’ . . :i d(MB.sj ! #j _/..‘“. . ‘-,,Component A:‘2.., dt - Llxl r Vu, L F& (4.27b)

( ;’11’ ,A11 the above equations are the, state eqwtions ‘and descrl;be the dynamic,/ 2

behavior of the distillation column. T#e stgte variable&, of the model are:‘AT$ r )~, * . .

Liquid’ holdups; M1)M2, l v ? ,Mf ,“I FI* ,M& G, and MB v* . . ,.G.’

Liquid concentrations; x1’x2.,‘;:*‘ixf5.* l l ;,x ‘;’ ‘$c ‘- ’ .’ j . ~ $;* I &. ;q 2!nd xl$ .., i,”

To complete the modeEing of the co&n, in addition ‘to. the state equations, ,’,t .’ ,.3 ) i

we need the following -relationships: I “. 1 ‘*. I,:: 1‘ I,/ ’ I : .i; _. “,

::_, .;, ,’ 6,. ‘.’ r

)I (a) Equilibrium relationships: “I i ‘, ,: . .,>I .& .)” ‘!

: “‘.’ : : :+, i ,’‘i,. -i’. ” ‘.,‘

a x i 6 ,I

Yi = 1 f (a-l)xi~ i~,Q,2, l l * , f ,‘a k&NiB- ‘, (4 .‘20)s* ,jr ii. ‘.V.’ .’) I’ “$ .:’

( b ) Hydraul ic re lat ionships (Francis we%+,, foimulg) ) ,, , ._ _. *r

Li = f(Mi) * i %.1,2,***‘if,***,N* I ( 4 . 2 1 )L1

When all the above modeling equations’ are solved wg find how the flowrates

and, concentrations of the two product s&e,ams (distillate, bottom) change with,’ i;time, in the presence of changes in thevarious input variables.

.& .;,.. . . /,The modeling steps outlines above indicate that the overall proceduri of

modeling a processing system may be tedius and fulilof simplifications. A t

times the resulting model is overwhelming in size ana the solution of the car-

responding equations may be cumbersome:I / . 5: ”

For the binary distillation column we \‘- :I

1;,/ .‘.’ ,’

‘,

1 x ’ ‘. .:

:

' have tcj s&lve~aikystem ,of.. ., 1 2;;L.

' 2N + 4 nonlinear diff&enti@ equations (state equatioT8). I,. i &. ,IL ': :' (L

and

.ZI$ +I; algebraic equations (equilibrium,'hydr&+c relgtionghips)

I

/‘(’ I : j.4.6 MODELING DIFFICULTIES

‘I. . ,! . 8, : '-. : <..

The modeling examples discussed in the previoue s.ections or,tgis ghapter ’. > 1 :., .I :.

should have alerted the reader to a series nf diff+rttltJna +lra* -ma k.-....enr-

in his efforts to develop a ~ea+ngful, and

of a chemical process,hi j

realist$$,mathematic

-' i : I'. ,'> ;_;i, ,' 1

Example 4.X4. .Considering the Mathematical Modeling.of.,.he CSTR (Example 4,lO)the Following Difficulties Arise: .,

- Determine with the desired accuracy the"values of'various parameters such‘

as preexponential kinetic constant, ko, the activation energy, E, and the.L

overall heat transfer'coefficient;'U:. ".

- Although the specific heat capacities; e, and z,c have been constdered.>.P 'Pi

constant, .they are in general functions of the temperature, T, and the

concentration, cA ' How do we decide that this dependence is weak (so that:,,wecanueeconstantvalues as 'in the e~~mple)~or &tiong (in which case theI ", . ,:modeling becomes very compli&ted)? The same '$&tions arise,for the

*de%sities, p and p

. ( iI Ii&i, and the hest'of r&ction,'(-AH;).

- During the operation of,the CSTR, scaling, fouling, etc.', wiil.alter the/ 6 ">

t:

value of the!overall heat tranafq coefficient. How can we scco~np for; '_ ithis effect in the mathemgticaf model?,

: '.a.,+-

- We have considered first-order kinetics to describe the reaction-rat&. IS,c'

this correct? .:$(' .I _. i

We can classify the difficulties encountered during the mathematical _I*

modeling of a process in three categories:

(a) 'those arising from poorly understood chemical or physical phenomena,

(b) those caused from inaccurate values of various parameters, and ,; "_

(c) those caused from the size and the complexity of the resulting model. 'I,,~

A . Poorly Understood Processes._ .,;

To understand completely the physical and chemical phenomena occurringI'

in a chemical process is,virtually impossible. Even <an aaceptable degree of

knowledge is at, times very difficult. Typical examples include:

- Multicomponent reaction systems with poorly known interactions among the

various components and imprecisely known kinetics.

6

4 ./'

1

- Vapor-liquid’,or ‘liquid-liquid thermodynamic equilibria for multicomponent

‘systeI5. :

- Heat and mass transfer interactions in distillation columns’with nonideal

multi@om@Xent mixtdres, .azeotropic mixtures, etc.

.:.

IZxample 4.15

Consider. the ffuidimd catalytic reactor shown in Figure 4.12. An oil_1‘,.? -.,

feed composed of heavy,*hydrocarbon molecules ia mix&d “gith catalyst ‘and enters

a fluidiz’ed bed reactor. ”* The long molecut+ react on the surface of’ the”. ,

catalyst and’ ‘they are ‘cracked” in&’ light&“product t&$ecules (like ga&&.ne)

regenerator where the,,materiLl‘ deposited on its surface 1s burned~~~ith “air;:.!“ be~I .I “,L~ c . ;

The regenerated catalyst‘ returns then to the reactor ,aft&” it.,,i;i’&x& W&h j

*?, ,, j, I J-’ ( I . i , ’f r e s h f e e d . I

I n o r d e r t o modek“the.,two u n i t s , t h e ’ foilowing ii&&mat&n must.‘be .“’ ” vI* ? -

available: ’ , .a 5 <* .,

- The reaction rate of the cracking ijraci?ss; .:i <. . ‘,

; , .’- The rate with which. carbon and heavy material are’ deposited %n the catalyst

( th i s w i l l de termine the ra te ’ o f catafyst~deactivatioh) ;““ “.“. ‘-I_ +,., ,.

- The dependence of the above two rates on the temperature of’ the reactor and‘.. . .

the quality of the ‘feed (light or heavy) ; ’ ’ ”

- The rate of combustion of the carbonaceous material’ deposited on ‘the

catalyst, in the regenerator , and its dependence on temperature.

All the above information is not only difficult to acquire but at times ‘it

leads to contradicting contentions. For example, in l?igure 4.13 we see two

,, .:.

‘. ._

/w

; ”

models that describe the ef f ec,t, of the ‘heavy ;;Jn?l$.l feed rite. on the reactor tem-i _‘ , 9 ,,. k.;”

pera ture. We notice that the qualitative behavior predicted by the two models‘“, 1)

is’ quite different. ,_ i ,2i:r * L

Finally, the two units (reactor ,reg.enerator) are fluidized beds and it is:well known how poorly~ understood are the fluid mechanical characteristics of

such units.

‘ E:4 ‘- ,11 I

B-. Imprecisely Known Parameters,, _/ -.: / I ,,* ‘ I

The availability of accurate va$ues for the p9rameter.s of a model is,:

indispensable for any quantitative. analysis of the behatior of a, process.‘. *

Unfortunately, this. is not always possible. Typical examples incl+de-. the ,- :

preexponential constant of a kinetic rate expression, : ,‘Y::

It should also be poi+ed out; that the yaJ,ues of the peametire doApt:!

rema*<n constant over long per&de. of time., Therefore EorV,:8ffective jnodeliryl i.‘t.

we’need not drily accurate values but alap some qu&sitasive description on how!

the parametric values change with time. Typical examples of changing,.$arameters

are the activity of a cafalyst, .a@ the iver&Ll; heat tr&sf,er, coefficient of

heat transfer systems (heat exchangers, jacketd reactors, etc.). _I , ,, n

The dead time is also a critical. parameter whose va1,u.e +$a usuelly; impre-

cisely known. .+ we will see in; a,,later sect$on.,,, the poor ,knowledge Sof thej_,

dead time can lead to .s+riou#~ stlility jroblems for the proce$s,lI

When no r&able, values. for the pa&u&a,. are available, we .-rwor t t o: . , c,.’

experiments tin the real process in an.,effort $.a estimate: some “goodj’~ valuesi ‘%’

for them. More’on’ the experimental. prqcedures will be dispussed in a sub-,;” 1.

sequent chapter. ,.,c ‘*’ , ‘i,,,

1’/

,~,:. ‘.

C . The Size and Complexity of a Mod&*$ ‘ ,>.' i: i

f;nan effort to develop as accurate and precise a mathemat+al,model as:, .:.: .:9.. ii : iIpossible!, th& size ano,the complexity pf the model increases,si.gnificantly.

Consider a-distillation column with 20 trays, a rebsiler and a-condeker.,. ;

The feed,.is;.~,two-&omp.onent mixture.. Then , as',wc have seen in Extim#le 4.13,*: :

'the mathematical.~model is ,composed of: .:/ v:..

2N + c, = 2(20) + 4 = 44 differential equations';'and :',

2N + 1 -"2,f20) f X,.- - 41: algebraic equatfone,‘ : I'

The size.of the model ,for such simple syetems‘is altady prohibit&e. Since',

the common distillation systems include feeds'with mo& tha&.two compt.k&n$r and,%.' ~ " (‘li.

possess larger numbers'af trays,',

it is clear that such .an~:~xten~iv~:'md;d~~ing

would lead to cumbersome and hard to use mod&. :: I..1

q 9.:, ,9:, 0. ,

;.:. $i "

iCare must be exercised too that the size.and the comglexity,,,of c:~odel do

I ,,ap_) , '; /

not exceed certain manageable levels,,geyond Gh$ch the model,:%$ses its vgfue:" :

and becomes less attractive. '. :I ,:

SUMMAk AND CONCLUDING REXARKS ' : "a I

i ‘J. '.FQr the math&atical modeling of processing systems $e need to:

I,-:j+. 9kL

:* a. -- identify the st&$,variables whose values charagterizs-the fundameztal

!: 'f !j

quantities of th&process, andI I. .." ).

.,!, 9'

The state equatjzons are developed from the application of the cgnservation .’:.

+ principle on fundamental quantities like. total’tiass mass of ‘various ~comp&&hts,. * .,tot41 energy, momentum.i. , _ :~ ’ :: t ?. ‘I. 1 I. i ’

To complete the .mathematioaLmodeUng of a process, ad&itional’ equations,, .;

are needed to Bescribe the:1 : . ._.. .( i.,

- reaction kinetic rates :’ : ,: :_‘s ‘-, *’ * ; i: , :

- (equilibrium conditions Yamong * various phases : ~~> ,a ./ :, ii . .”+ ‘. i

- thermodynamic equations of state-(40 descr,ib.e, the behavior&f varioug ‘:,.”1 i’;. _.

,materiale (gcfeas, l i q u i d s , ,mix#Cyyecq, @XL+). . , :. I., ’ _,~: ::. ,‘: 1 j i:- / < . ~ i V+z,.:‘,’ ,I, , , 3 “.

The modelgng of a process should a&aye consider t&: poss4~ility &! thi-; ..,,y’,3

(.’ ,I; r:’

presence of etgnificant dead time. The dead time Is i v&y %mportant f*ea,tur’e,t j

( : :; ,?1 : : .,J .zz, ,;~ I : ( Iand it’ plays a ,significant role”inW’tfie &sign of effective controllers -for a : : I u

,. .chemfcal process. Only when ‘the’ de&* t$@a is very ‘amall (since”i& priaciple

(. _/‘,_,\ it will never b,e zero) it can be neglected from the d&iopment”of ‘a mod+.

Several difficulties arise during .the modeling, of a &he&c& process.:: ’:‘t

j.’. .

.: -

*:‘farmulate the state equations whose s’olutibn depicts ,ho$‘the values of the#.

‘,sttite’variables, art? consequently the -uatural state of the process, changest ’ :. “,j .-we ;; -_ ! ’with ‘ t ime. .c

-

- rates of mass,. energy or momeritum tranifer . ‘a ,i I, ’ t‘, ;;.

- equilibrium conditions of reacting sy&eti ,,. ;:‘$ / ,,’ .*. :’,._I-

These difficulties can be classified int the following,, categories: ;.1’. q

- Difficulties arising from Amprecisely known chemical or.chemical phenomegi,/,. : 4:i ~ _ ;,: -..,.

affecting the,, eff ectiveness of the selected equilibrium nor rate reJ.at$onships.P .

- Difficulties arising from the inaccurate values of the various model

parameters an& how they change. :

- Practical difficulties caused by the large size and complexity of the

resulting model.

/

: . ,

q

’ ’ A-mode1 is considered acceptable when it can predict the dynamic behavior.?: j.

of a syste& with at? acceptable degree of accuracy. h The assumptions made by the

modeler will’ influence the, quality, of the resulting model.,’

‘Consequently, the ,

mathematical modeling is an- art not very well defined, where the ‘previousI* _, ‘,:5, :. rt :;, \ 1,’

educ’at’i&, ,practical experience and intuition play a very important role..:,, .’

Although ‘the modeling ‘procedure outlined in this chapter is in principle;-‘,: *ir * I .:_ 1’ :’

feasible’, additional practical considerationa motivated by the, ‘needs of process.

control dictate certain char&es’ that will be discussed in Chapter 6,.,8 ‘, (

*

ATHINGS TO THINK ABOUT ..(

1. What is a mathematical modgl.of’a physical process and what do we mean. . . :/’ . awhen we talk about mathematical modeling?

: ., ‘. _ i, 8 ‘. -. x ;’&“I2. In Figure 4.13 we see two dJLfferent curves.,which relate the ‘temperature

! ,’ ‘~,I’ 1,,,.. ,.z., ‘, _, .and the feed rate of the .reactorc in the, fluid catalytic cracking:‘?nPt,.:.: _’ idiscussed in Example 4.15,

(. : . JI@ the term “model” approp,f,iate for each of

:..these curves?

‘+ I j..?, i..’ ‘.,. i 4 (a%. : t;.

3. Let us recall that the Steam Tables give the temnerature .af whkh wnt&

liquid and water vapor are at equilibrium for a given ‘Gressure. They*i. - :a,

also give ‘the specific values for enthalpy, ,entropy, and, volume...of both I

liquid and vapor phases. Do ,these tables of values consti@ute a ”: A-:. 1 I

i1

mathematical model? i: : . ‘:i *,

4, Consider the graphs shown in the Figure Q.&l. These’ graphs were pro-*I .i”. ,’ ,

duced by measuring the concentratie,n of B in, the rea$ibn,~ ,A 4 B,; /’ ’

over time, and at various temperatures.’ Do these glraphii represent. a,i. :. , _mathematical model?

5. Why do you need to develop the mathematical model of a process you want,. ..:.,* s

to contro l?: *.

6.

,

, 7.

a.

. . _* i,t i :What are the strte variables and what are the state equations? What are

i. 1they used for? . .How many state variables do you need to describe a system which is com-

,, ? .( . . ,, + 8 ,, _/ I ‘) : ,*i’posed of ‘M phases and”N components?

, -’ ,‘~ +r ‘- *We know that when two’ phases are at a thermodynamic equilibrium, the

.j .) j,, ‘li-. <,chemical iotential of every compotient (i) in the phase I(u< $,l) 2b (- - 1 .equal to’ the chemical potential of ‘the same component in the phase II

(lJ i II), Le.,2

Q,i: ‘- .‘%,I1 i i lJ;;,.,N ’ - ‘.

_: ’Express the above equilibrium relationship ‘in. terms of the mole concen-

.- I ..“$. ,,tration of the ‘N components in the two phases. The answer to this

j. : ,question will denianstrate to you that we‘don’t need the ‘concentrations

“k, : ,/,Iof the N components in both .phaaea in order to describe the system.

,9.

I+

Write a relationship that will give you the molar or the specific.i.)

entlialtiy of a liquid at tempe&ure,t, b /I ‘

T, pressure, “p, with a composition

of N components known.~

10. RePeat question 8, ‘but with a”&# instead of a liquid.,

11. For the fash ‘drum example (Examples’ 4.7 and 4.8) develop an expression

for the density of the vapor phase , using the Van der~ Waals equation of*. .‘, .’

state. State also an expression for the d,ensity,of the liquid phase.

12, When is a system at’ steady state?’.I

13...i /What is the main reason for the presence of dead time iu’a process?

,, ,‘.14. Do you know of any systems which do not possess dead time?

15. How would you find the dead time of a system?

16. In the Figure 4.4-2 we see the behsvioi of the concentration at the

/)

“ :.. _ 1

18.

19.

20.

outlet of two processesoutlet of two processes , after the concentration at the inlets and at, after the concentration at the inlets and at

time t=O was incrjeased by 10%.time t=O was incrjeased by 10%. Which process possesses dead time?Which process possesses dead time? ..

What are the assumptions leading to equimolar vapor flow rates, i.e.What are the assumptions leading to equimolar vapor flow rates, i.e.

v1 = v2 - ,*** - VN - vv1 = v2 - ,*** - VN - v for a binary distillation column?for a binary distillation column?

Why have we neglected the energy galances for the binary ideal distillationWhy have we neglected the energy galances for the binary ideal distillation

column of Example 4.13?column of Example 4.13?

What are the assumptions leading to the equilibrium relationship (4.20)What are the assumptions leading to the equilibrium relationship (4.20)

and how is it derived?and how is it derived?

Could you have dead-time between the overhead vapor and ,the distillateCould you have dead-time between the overhead vapor and ,the distillate

p r o d u c t ? I f y e s , w h y ?p r o d u c t ? I f y e s , w h y ?

**

*a*a

.A..A.

h

1 (Clnmeurure

TOtd Steady Sfafe,-e-m---- --------we_

I

\

-err---- -------

Neti 5Yeady Sfufk. Y

n--w ---

- - - -- - --, xA& -_---a..&.- --i -.

f-.

,,;;I;’

c Y..i

/,Yrcxesr 45

,.

:<,:, . ,CHAPTER 5

MODELING CONSIDERATIONS FOR CONTROL PURPOSES

',The mathematical modeling of the,physical and chemical phenomena,c'. .'

encountered ,in processing systems of interest to chemical engineers is a form*,.5 _,;;. '.<-*. --, I ./ .3 ': '

of $&f&ic art. As every type of art it does not conform to neither does1 1'

it obey“vsry clearly' specified rulee and recipes. It is an expression of all, :... i ?'i'

the educations1 bk&gr&nd kud previous practical experience of the modeler._1

Therefore, if'"the prkious sections have generated more questions to theI / ,

reader than answers to their modeling problems, he should not despair. Good,1. .,,;.' i /

and efficient'modeling is acquired slowly with ever increasing ability for it.i

In this chapter we will attempt to focus the mathematic&l'&deling to the,I ;

control purposes and n&s.i _.. i(-:;: ;

Thus, we will examine the following issues:i' ? * ,',_.

- Starting from the state variables.model, how can one develop an input-output_, -: * e

model which is very cbnvenient for control purposes..c.-

8.- Using the mathemitidal model of a process',

: .how can one determine the degrees

.of freedom inherent in the process,'and consequently, identify the control

problem to be, solved. s' i- ... _".I :_

~ We will close~this chapter with soma general guidelines: which will,help

the control designer to formulatethe~scope of modeling fdir control pur&e~.

I,,, >.

5.1 THE INPUT-OUTPUT MODEL

every chemical process and its associated variables can be described pic-

torally as shown in Figure 5.1. The main block represents the process while

the arrows indicate the inputs and outputs of the process.

The mathematical model which is convenient and useful to a control systemI

designer should conform with the‘.above picture, i.e: be such that,. given the

values of the inputs it provides directly the values of the outputs. In6

particular, the ,model should have .I th-e- following general form for every output;,.;. , ;.”

output = f (Jwut ~ ~~f.~,+$p) ;;>” _, ;; .,:‘i c . . r / \!

Using Figure 5.1 the above relationshtp implies .’. . . .

y i - f(ml,m2,*‘** ,mh; dl,d2,‘;$* ;dQ). for i 2 1,2,***,m,;_I ,“. ,’ . .

Such a model describing directly the, relationshiP between the anPut andI: 2 ,F’ : 1output variables of a process is called input-output model. It is a very con-

’ -2,:. . .’venient form since it represents directly.the cause-and-effect relationship in

, I /,, j * .l,, 1~;. b , *processing systems and it is appealing to process engineersand control

8‘ : ,, s -. I j I :a-IIdesigners.

if ,/ .- 5 .L F;,( ‘.‘,’ ‘,The mathematical models we learned to Qevelop (in the’ previous,:chapter)

.; ’ .i -i; ,J $ i j :.using the state variables, are not of the dfrect input-output tyPe.,. , .:-’

Neyer the-: - .

less, they constitute the basis for the development of, an‘ input-ou,tPut model,_.. , . ’ 1 I ,

This is particularly easy and stralghtfopard when the state Variables roin$ide; I, ; ( j..completely with rtha output variables of a. process. ’In such a case vei can. ‘iintegrate the state model to produce the inpu’t-output model of the process.‘

: ‘. Pt. :

Example 5 .1 -The Input-output Model for the Stirred Tank,Heater ’

that Fi - F , which yields dV/dt = 0, leaving.tha~total enkgy balance as ,the

only equation of the state model,: . - ,i si, ‘,’

V g b Fi(Ti :, T) +, $ ,, ‘. \_ (47 5b)P

The amount of heat, Q, supplied by the steam is given by

T\Q = UAt(Ts -

.

where U. is the overall

transfer’and T, ‘is the

Q, eqn. (4.5b) becomes:

heat transfer coefficient, A, is the area of heat/x f :

temperature of the steam. Use the last equation for.’

. . /( .dT

"z + (Fi +UAt

FIT

* UAt

P= FiTi + pc TS

P

o r

dTdt f aT = + KTs (5.1)

where

, ' c1 " ',-;+L

Fi ' UA,a - LT and K - -T ,I vpCp :

Equation (5.1) is the'mathematical model of the stirred tank heater with T

the state variable and Ti and T, the input variables. Let us see how wer:

can develop the corresponding input-output model. I‘.,,

At steady state, eqn. (5.1) yieldsi'.'.. .,- */

L'? I. , ,, i' 7 z- .

fTO + aT(s) - r i(s) + KTs(s) _h ,c;5*2)7, , t

are the steady,state va&ues of Mie correspbndln$I

where T(8) ' Ti(s)) Ts(s>

variables. Subtract (5.2),from (5.1) and take,,,.''6:

d('r - Tts))

o r

dt + a(T - T(s)) - $ ('Q - Tits)) +: K(T, -, Ts(s))1 ir ,"~

I.1dT'dt + aT' = .k T'

T,# i+' KT; .;: . * (5.3)

:'

where

indicate

The

T' = T - T(s) , T; = Ti - Ti(s) and Ti * T.%,T Ts(s),.'

the deviations' from the correspdnding steady state values. Tk'

solution of (5.3) is '* I,: I !

,'d ,,:

T'(t) = cleVat +“. . ‘ a (5.4):

0 '

.

Assuming .that initially the heater is at steady state, i.e. at t=O, T'(t=O) ='.

0, we .find easily 'cl - 0. Therefore eqn. (5.4) gives

t

T'(tf = !' f$ T;+ &A dt (5.5)‘0

,'Equation (5.5) expresses ,the.relationship between the inputs (T;,Ti) and the

output (T'), and constitutes the INPUT-OUTPUTS MODEL for the tank heater. Thist:',relationship is also depicted p$ctorially in Figure 5.2,

‘. :

,/ : * ,Example 5.2 - The Input-Output Model for,a Hixing Process

; ,: 1Consider again the mixing of two siream&

::. 'process discussed in Example

'_ , .(I4.11 (Figure 4.8). Assume that:

:- Fl -k F2 - F3;' which implies dV/dt = d, i.e. "V'i constant..'

- The heats of‘solut&ons'are independent of the-‘concentration which,,z&mplie$:

rsl - Af?d - p82.-'Afis3] =' (.j ' ~ , i <,' ".'

*i

Then, the state equations are reduced to the following:

dCA1 _ .c : (5.6a)~I

-2+ ($++cA3 = &i; +' +cA2

* 'and

,t

dT3 Fl .,F2 " Fl ;. F2 '

dt + \v + $T3 - ?j-Tl'+ 7T2(5 6 ; '&. (, ,,,;iG,,

+ -& (5.6b)' , , ,' 7,‘~ '*' i 'I, ,;

At steady state eqns, . a ' .)' .'

: I' i '.

F1 ” ‘> F2 t:- V.CA

*!(5.7a)

U8)t”+CA ,:.

Z(s)

and

-- ,, ,;i't.j- j.: ,. F. F,F1 F2 F1 F2

0, +> Ctf. + -$T3(oj.. I T Tx(idd) + p T2(#j f -jQ(# f !I (5.7b).I p=p..i ;

Subtract ,$5,7a) from: (5.6a) and (S.fb) from (5.6b) and take:

(5.7b)

(5.8a)

F2,+ 7 Td *+.,'KQ'., ,: ( ; ,(4,8b),

where c' ,:.*c' ‘c' T', ,T'A1 Ai' A3' 1: 2, Ti 'and Q' ,are ,deviarjlon vhriablesdefined as

follows: L . ,; 31 :.i‘ .

*' - =A1 - Cilo ' =i; = 'A2 - CA2(sj ' %3($; :A3 - %3(s),. ; -. ‘

and ..I :. > ;i' ,I

Ti 9 T1 L Tl(s) , T; = T2 - T2(s) *, TJ .* T3 - T3(s+Q'+ Q.- Q(s)" /.. ,(

Also, "-(i I--.;;, s-z

,, ',

.,i.;,..I x

a &* F1 F2.v..+ v; and K.+-ql .:I:;' 3 ,,: ;: j .:.:

i li ): "

The solution of (5+8a) and (5.8bl) yields:,. '_,: ., ;:.

and

p2 *,,ci (t) - cleeal + '- +- c'3. : 1 :, ..,;i:- /_; " ; / ',

dt :'.A1 : _r P : A2 3 I; c,g,,:.

; '. (5,9a)_t, :

/ ,./I' , :i ':* 'i,'. .f,$.,"L

: I .i “ . .T;(t) = c2eBat + 'I'.:. ,(5.9b)

~,. ,$' ':‘0 ., :( : ,

If the system is iriitially, i.e.. at tA0, at.bEeady state, then ,

ci (VO) = 0' and' T;(t=O) -~ 0 '3

I Equations (5.9a) and (5.9b) represent the input-output,model for the

mixing :. process “s$ “lis shown schema tic-ally in Figure 5.3.-; c ;

R&arks : (‘1) In Examples 5.1 and 5.2 the;.output, variables coincide with the,’ : .

state variables of the two ‘processes. @&equently, in order-‘.

to develop the,, input-output* model’we need only’ solve the dif-‘; i’ :ferenti+ equati&s of ‘the: ‘mass and energy ,balances., ‘phi; is I,-

, . :

not always trued;. -,Taks as’ an example the binary distfllation,not always trued;. -,Taks as’ an example the binary distfllation, ‘.‘.tt

,$umn’ model (Examble 4.13;: and Figure 4.10).,$umn’ model (Examble 4.13;: and Figure 4.10). F o r t h i s systemi’ :’F o r t h i s systemi’ :’

wi have:wi have:> ”> ” :*:* :*:*“J“J

state variables:state variables:

Liquid holdups, Ml,M2,*“,;~f,***~~,‘~~Liquid holdups, Ml,M2,*“,;~f,***~~,‘~~ a n da n d,:,:

-M& .: ~-M& .: ~* /* / i ,i ,

Liquid concentrations, Xl,X2,**‘*,Xf,***Liquid concentrations, Xl,X2,**‘*,Xf,*** ,x,,. s ,ana I$,x,,. s ,ana I$,,,,

: ,: ,o u t p u t v a r i a b l e s :o u t p u t v a r i a b l e s :

.’. ’

D$stillate produc t flowrate ‘ind,, &m&si&on, TD, a n d 4.D$stillate produc t flowrate ‘ind,, &m&si&on, TD, a n d 4.:: ., , .., , .

1111.__ :.__ :

,,I,,I x.x.Bottoms product flowrate a&l tiomposltioni $‘R and XRBottoms product flowrate a&l tiomposltioni $‘R and XR “I t I“I t I

.‘..‘. j . ’j . ’‘<‘<

We notice that we have many more state.y$rlables thin outputs.We notice that we have many more state.y$rlables thin outputs. ::,, .,,.,,

For such systems,For such systems, theC.development of- the input-output’. model istheC.development of- the input-output’. model is

quite involved and difficult Equite involved and difficult E:::: Figure, 5 P 4 depidts pfc &allyFigure, 5 P 4 depidts pfc &ally,..,,..,

the input-output model thet we would like to develop’ for t&e,’the input-output model thet we would like to develop’ for t&e,’’’

binary Ideal dist,illatlon column,binary Ideal dist,illatlon column, :::’:’ L f ’L f ’.I.I

(2) In subsequent chapters we will study the method of Laplace”

transforms which aalows a much sinreler develonment of innut-

5.2 DEGREESOF FREEDOM '

.The degrees of freedom of a processing system are the independent

variables which must b? specified in order to define the process completely.

Consequently , ,the desired control of alprocess will be.achieved when and only

when all theidegrees-fofS;freedomrhave been specified.when all theidegrees-fofS;freedomrhave been specified.

A good understanding of how many degrees of freedom are inherent in aA good understanding of how many degrees of freedom are inherent in a

process and whick,:are :they,.is very crucial forth& design of‘effective con-process and whick,:are :they,.is very crucial forth& design of‘effective con- tt_'_' 2' .,.2' .,.trollers.trollers. For a specifiedsystem, its mathematicil model is the basis ofFor a specifiedsystem, its mathematicil model is the basis of

: ): ) ::

'finding the degrees'of fi;i4e‘dom unaer'both~dyn&ic 'and 'static conditions.'finding the degrees'of fi;i4e‘dom unaer'both~dyn&ic 'and 'static conditions. LetLet'. :'. :

us start with't'wo characteris& &amples‘; I" '.us start with't'wo characteris& &amples‘; I" '. ::

lh,,/lh,,/ 'i'i .. . ‘i,. ‘i, .$..$.Example 5.3Example 5.3 - Degrees of Free&m in,a Stirred Tank Heater'- Degrees of Free&m in,a Stirred Tank Heater'

. .. . .' c.' c :;:;The mathematical model of“a'$tirred tank heat&The mathematical model of“a'$tirred tank heat&

;:I'>;:I'>(Example 4.4) is given by,(Example 4.4) is given by,

A;$ p F ":a FA;$ p F ":a F" .'." .'. / i/ i

, ., .

A; j$A; j$ 9: "$(Ti9: "$(Ti_ ;jry + ,$ : ,;: _*/_ ;jry + ,$ : ,;: _*/ ;];] (4’: Sbj,,(4’: Sbj,,

p,p, .'.'.*..*. ::. .. . ; //; //;; :jy /. (:jy /. ( ,>,:.':)‘i2: ;;. f,>,:.':)‘i2: ;;. f

When eqns. (4.4a) and (4.5b) are solved, simultaneously,-we ca, :fi,d.how hWhen eqns. (4.4a) and (4.5b) are solved, simultaneously,-we ca, :fi,d.how h .,.,;,1 *: "',.iJ;,1 *: "',.iJ ' "' "

(liquid level) and T (liquid temperature) change with t&&e: taben the,inputs(liquid level) and T (liquid temperature) change with t&&e: taben the,inputsii

(Ti,FiiQ) change-.(Ti,FiiQ) change-.jljl ,%;. .:.,%;. .:.

Let us ask though' the following twoquestionst _Let us ask though' the following twoquestionst _'.'. ,,

I.I. II- Is the solution of the equations possible? ‘ ' :I::~- Is the solution of the equations possible? ‘ ' :I::~ ' ., 6:".' ., 6:". . .. '. .. '

- If the solution is possible,: how many satu'tiona *e&t? .r ‘;"~r- If the solution is possible,: how many satu'tiona *e&t? .r ‘;"~r '-:'.:' *:'-:'.:' *:StSt,/,/To answer the above questions let us count'equations -&d Lariabl~s. c :To answer the above questions let us count'equations -&d Lariabl~s. c :

.'.'

Number of equations = 2;; Equations (4.4a) and.'(4.@). "'Number of equations = 2;; Equations (4.4a) and.'(4.@). "' 12 :12 :i,i,

Number of variables = 6s h.; T, Fit F, Ti, and Q 'Number of variables = 6s h.; T, Fit F, Ti, and Q ' .i.Vt.i.Vt-a-a

We have assumed'that A, 6 and'."cWe have assumed'that A, 6 and'."cPP

are parameters with given constant values.are parameters with given constant values.

We notice that

^ ‘\ -z :‘,._‘ 3: 6 *:~.

: :‘1. il, ! , 1~:;‘I-

,: ‘j?’ ” \:; : ;,,

f :”)V. , , ‘,;

_-.* . ,-?; .

_-& . L./. ,,.&’

.,. If

Number of variables > 'I -' $mbe%iof equationsI .,; ,*-: .,

Consequently,.the answer to. ths-fir.&,;. I; 1:

q#@$tion.is~, .yI$S there exists at least: ;. ."; ,i-_.

one solution to;the equations mode&ing,. -hank. h*ater; With respect to the,.. I/ , .',-';,, ) i;

second question!we eagily+conclud$ that&here "is an' infinite number of solu-~ .t -i

tions since we can specify arbitrarily&e values+ four variables (4 - 6 - a. si-‘ I., 13 variabl'es.and solve eqna,.,(4.4a) and (I).Sb),.~fqr th$‘r+nainiqg twc

_' :The arbitrariiy,specifie@ vs$iables a& tke .degreea,,>of freedom and their,‘ ; .'

z_number is Siveni'by the:followinS ~~vioue.ie~a~~~nship,~::, ; ,~ J .'

,f - j.(Number of variables) - (I$umbqr of eq$$.onsj, -'

Suppose th?t we specify the values of $I$) folIowing, foyr va,niabl+s,:

% TYs I7and f$ 4:.

Then, we can integrate eqns. (4.4a) and <4.,5bjY anb find how h . and T change,('4:. ' '2 -', <

with time. If we give different value "'1 /:_ T+' I co ,-;pi a ?UX Ti or ,P or Q,.we,find .

j., :', ,, , ./I I ~,'

thst h and -T,' change d,iffe"re@tly than>befoke.‘ Consequently; 'if we want h'.' ,yr +and T to change in a prescribed manner we$ho&ld,,not ha& any d&reas of

,<: ;

Fri?edom, i.e. vclriabl.oswlrich can. t&e a&$&~,&$$&$,uee. 'Ttijk: lc?p&us to the'_ ,. _.I i. ,. 4 I -*'

conclusion that,,'in order to specify compIet& a'*procesh the number of,,degress* -I, : f "' /, * I;~

of fre'edom shoulh be. zero. ." % ,',( ! :: 't," :v I. 1'

I, /.~<3. ),, ~,, + * -" -z ..,). : .i: jr .: :. ,~ ; i 1%. :: ,., > '.

I 11 I, ,: ,--y, I,'I ,i::_;, ' '.,J. ‘;f,” : ,,:- /..A ,,: qg$.T, .'.i:

Example 5.4 - Degrees of I?ree&m'in -an $deal:Bin&ry Bistillat!~on‘@dl~mni

Consider t&e model for an idea18'Bin&y distillati~ca~~n'deire~~psd in,‘

Example 4.13. We have: : ,; ';.. 4~: .;' ", 7 ,-., ..' _I

Number of Equations ."a< ‘.L or&+&:

N-b1 : Equilibrium'reXb&&hips~ Xeqn. (4.20))

N -' llydrtluiic relatio&hips'" (eqn. (4.21).) '.;.

2 Balances on"feed'tray (eqns. (4.22a), (4*22b)).'I

2 Balanties o n ‘ t o p t r a y (aqns. (4.23a); (4.23b))

2 Bake&es on bottom tray I(eqna. ‘(4..24a), (4.24b))

2 (N-3) Balandes on i-th tray; i # l , N , f ( e q n s . (4.25a), (4.25b))

2 Balances on reflux drum (eqns. (4.26a);(4.26b))

2 ’Total = 4N + 5

Balances on column base (eqns. (4.27a), (4.27b))

i; I_

Number of Variables m.

N-I-2 xi i=1,,2****,f,***, N,D,B liquid compositions

s

1 .i :. , J i,Msl, yi # W,2,:***,f,*** ,N,B v a p o r com$&sit$zons ’

-,N+i ’ “,Mi fnl,.2,r**,f,*;*,k;I,~ l i q u i d h o l d u p s ”

.I ,. ,. ,j ..: . . :.)N Li “, ’i=l 2 l **,f,*** ,N l i q u i d f l o w s t

; .,

6Total = 4N f 11

Ff’ c,; FD” FB,’ fFR’ ,!! \j

The number of degrees of freedom For the ideal ‘bin&y distiliation column is

f - (4N C 11)_ (4N i $ &’ 6 ,’ ‘I .f* - :. ‘:: ‘.I;

( . . . z

i. e, we need to specify the values of alx variabWs“before kk c&. solve the ‘.

model of the b&ary distillation. $,I.~ : * :,:y * _‘,y‘

). “.I” ,”‘/’ _i. _ L/ : *I:

_.’ ,, . . ,. _; al..’ 8, ‘,.:$,,c :I '.gi.. I

,’The observations made and the conclusions drawn from the above two ‘&tamples

_“,S f. J., / :. “f *’ b :. ican now be generalized for any pradessing system deefribed by.,a s.et of E ‘,

* I’equations (differential and/or algebraic) containing V varisb$es. The num-,I. ., _b e r o f d e g r e e s o f f r e e d o m forsuch’a s y s t e m is given%:by, :, li,, ’ “.”

. ,“f i Y _ E ,1 _ / t _/ ” * ,I:‘.,” ;.’ ; :’ ‘,)

vAccardfng to the value of f ‘t we can have the falluking c@s&k: !, ‘, ” ”

Case 1. If f-O., then we have a system of equant&ons w$th,?equal numb&~~of .’i

variables. The sdlut,ion of the E equations ‘yi’elds udique~‘\ralued’

f o r the V var iab l es . In this ease we. say that: the process is

exactly specified.

,

1

Case 2. If f>O, then we have more variables, than equations, Multiple

soluti!yns result from the, E equat&~~ng since we can specify'e

arbitrarily L of the variables. ,In this case we say that the-:

process is underspecified by f equations, i.e. we need f,)

additional squationa in order to haves uniqua solution.

Case -3.

Remarks:

If f<O, then we have more squation s than variables and in general; >

ther

that

to r.

(1)

~.“’~.“’

e%s no solution to the E 'equations, 'In this case tie say 1e%s no solution to the E 'equations, 'In this case tie say 19.

the system is ovirs$ecified by f equations, i.e."we need I ,_the system is ovirs$ecified by f equations, i.e."ws need I ,_*.*. _ ::

em&e. f <equat,ions in order t;q have a solution for the &&am, 'em&e. f <equat,ions in order t;q have a solution for the &&am, " "

,It is clear .from- the above analysis that a sloppy.mopellng of,It is clear .from- the above analysis that a sloppy.mopellng ofr*r*

a* process may lead to a model which does not include all thea* process may lead to a model which does not include all the.Y.Y '"'"'n'n ,*,*relevantrelevant eouations .andeouations .and va&ables:o?"includesva&ables:o?"includes redundnti6redundnti6

:.

equations and variables. :In either'cise we have an"erroqeous,.- :e,

determination.,of;the:degr$e{'of ErFedom ,which'may ‘imply, ._ ,( , ',' I. I .; . I.i'incorrectly that we have an infinite number of solut.ions'or

>'I, .A' ."

no solution at all. .; :.i .,, ,* :' .:;, azi::; .,:h 'G:4 v q "> :. "/,,'(2) The presence of a cont'rol,;jloog in,~.~~~~al,,.proSeas. intro- '

duces an additiotial equation be$wedt+$the m(,(I/. :.*I.. _; 4,and the manioula&l vari&bles. thus reduoir

initial

fExample 5.5

The stirred tank heater is modeled 'hy two equations"conta&ng six

variables, thus:yielding foup:degreea oft freedom (Example 5,3); This is. true

if the effluent flowrate.,F ia detarmined by a pumpi:valve; ets.. ':-Let us

suppose that this is not the case and' that the‘liquid. flows out 'fromlthe tank' 1

I/I 3’

I,,

-, ‘:‘i.:. k+ ’

\. ’ % ; :

freely under the hydroa{atic pressure oE the liquid, ih the t&k, In tli38 cn8c.; *t ,,,.; ,: ! t

there is an additional equation relating F to h, e.g. F = $&, which, .? _* -i

reduces the numbyr of degrees of freedom by one. ” r, ’.‘I

Example 5.6 ' ., I

Consider’ again the stiired task heater, but noti &id& f.eed&ck control

( F i g u r e 5 . 5 ) .%ontr& Loop 1 inaintainlb the. l&&d level ‘ai a desired-'v&&&by, "

measuring the revel of ‘the liquid atid adj'ui3&lni &e va$e~of the,efflue&t

flowrate. Therefore, Control Loop 14ntroducei a'relationship bqtw&eri F and 1

h . SimilarI)', Contrd'l Loop 2 ma&&s the &p&i&r4 bf the liquid at the 1

desired value by*?n&ipulatirjg &ha,floG 02 ‘eta&$ a&I &I~“‘&$*‘&.& of’ heat 6..

Consequently,?&ntrol Loop 2 &trod&es'.& tela&&hip"'b&&en ' g and T.,

It is cliar ‘from th’e. above analysis ttiat th& two dofit& lopp.& introdtice

two additional equaitons, thus red&irig t&. d&$&s of fk’&do!m b$’ two.,

I*‘,<’ I% ‘. ). i : “; !_

5.3 DEGREES OF FREEDOM AND PROCESS CONTROLLERS 'T ,.". :: .,

In general, a carefully modeled process will.,pqqsass oyV9,,oi more degree?.I , , i: ,, ? ,:.of freedom. Since for f>O L the &qxs~ +.ll have an @fini@ ntimber pf , ’.i “’

solutions the following question ar+es;t" t' '.'j:Q -I

"How do you reduce t& number -of d&grees of. freed4p to szaro :so that ydu can have a cr”omplately dpecified syst4m w&hunique behavior?" , ‘ .I . ,

It is clear that, for an unspecified system wit$ f degreeqof ,frF,edc?n, we;"need.i

to introduce f addtional equations to,make the Tyetem cpmpJ+@y, spe#ic?d,

There are two sources which ,prq,vid+ the,addifional equations$~' (a) the.

external world and (b) the control system. Let us examine them closer using:I

the: stirred tank heater as our example,

\’,” ,.

! ,’

. _

,’ IN ,I‘: *

Recall from Example 5.3 that the s&red tank heater powetises four‘I .

degrees of freedom. Therefore, we need four a?lditional’ relation&hi&, inde-.’pendent of the linodeling equations (eqns J (4.4a) , (4.5b)“. ‘These are provided

from the following considerations: ’ ’ ” ”

- The feed flowrate Fi and feed temperature Ti are the maiin two dis-

turbancae for the stirred tank heater ;and they are both speclf,ied by the

external worlds, elgl th,; unit that pre&ee# the tank, hea.ter.t* r ,’ A& though the

equations which specify Fi ( and Ti may not .be known te, us, nevertheless: ,. . . .,they exist .,and remov. two (2) degrees of freedom.( .’ Tbus,,we have. 4 -‘2 - 2‘

remaining ,+degrees of freedom.:! ; .,i ,.

t - The acceptable operation of the tank heater,;r<equires that!,,the liqu,$d level( 3:and liquid temperature in the tank heater are maintained at desired values..I ) ,I.. ), I I. . ..( /. / ,‘, .I a’ ,I 85%

These two b,contrql ob+t+ves can be, gc;fiieved with ,\the two. $ontEol looqs.i ,. ”.shown in Figure

\5.4 and discussed !n $.ample ,5: 6 i! . ~ r~ .* ,*. BQ$-~ : tp, $y$$+~tfont of

the two contry loops add,s ty, additiqnal ^equations‘r .. c ‘ (see Example ,5.6) thus,: ..‘,

removing the temaining two degrees.,of freedom.

Summarizing the above observations we co&lude the following: ”;.; ::

- The external ‘world by specifying the’ values’%f the.d$sturbances, :it removesI ’ >;+ -, 1 Ias’ many degrees of freedom 8s the number of, &turba&es’. ‘...,

L The control system required to achieve: the’ ‘iontrol objec&‘ve$, ‘it removes

as many degrees of freedom as the nurnber”~~:.‘cdn~r;;~ obf&%ie’s~.i

‘_

During the* reduction in the numberI,,of degrees of f?eedom ‘fir. a, chemical’ 1’ _: :t.;- . I .,,i

process, care must be exercised not to specify more control’ obj’ectives than it

is possible for’ the particular sys’tem.-’ I

Thus we can have tit most two control

objectives for’ the stir&d tank heater. Attemptfng to have three contro’i

objectives we are lead to an overspecif ied syst& &Yth fe0. ,,” ’ ”:; ‘/ i ;I.0

, ‘;‘,. ,L :,3 * . _

\

-_

(

Example 5.7 :L Re.duce the, Degrees of Freedom of an ideal’ Binary DistillationColumn , ; I; .“) ,Ir -,

Return’;0 the ideal binary distillation column (Figure 4.16). The system.’

poss&ases six degrees of freedom (see Example 5.4) which are specified as;.

follows : 1’1 ‘i

Specif &&on of the disturbances. Two are the main disturbances for thej ,, .i ,. .I

binary. disfil,lstlon column ; (I the feed flowrste Ff’, and the feed< ,.,composition cf.I,‘ The&r,values are speoified by the external world, e-.g. a

IL co ~mt\ I A 1 tllou~:ll, _ ;.reactor whose, affluent stream is :.the Feed to the JSs ti3lation

the equations specifying Ef ,and j ‘Q are not known to us j nevertheless they

exist and remove two degrees of freedom, leaving four for additional

specifications. .?, ., I’,i . , ::

Specification of the control objektives. We can have up to four control

objectives since there are four remain$ng degrees of freedom. The acceptabled.

operation of the binary col,umn requires that$he follo@ng variabZ& be mainc

tained at desired va2ues% j ,’ ‘7

(i) c o m p o s i t i o n o f the; distillltizr &ream@ Ei,;~~ : Ij b s:‘.’ ‘- <’ ”1

(ii) composition of the bottoms ‘stream,~XB; ri 1 : : : , ’

(iii) liquid holdup in the reflux drum, J$,; ’:

(iv) liquid holdup at the. base of the column, MB., .

Specification (i) and (ii) characterize the two product str8ams.i Speci f i - :

cations (iii) and (iv) are, required for opkrational, feas&blii.ty; i.e. -we ‘do . .{”I.

not want to flood or dry up the reflux drum or the base of the’rolumn for

safety purposes. Figure 5.6 shows the ,four control loo,ps,which aatlsfy the

above four ob j ectives .

We must note that the above four specifications of the control objectives

may differ, according to the particular operating objectives. For example, in

1

r-

L

‘

*

a different application we,may impose the following control objectives:;: * ,

"Keep at the desired values the distillate flowrate FD,, itscom$osition xl.), and the two"liquid holdupsi,,MD and MB !

or , t8' a' .i

.

"Keefi at the desired values the bottoms flowrate Fg, itscomlksition xB, and the two liquid holdups MD and' MB." .' '

'? 1 ,.Care must ‘be exercised noi'to%pecify &re'contro& objectives'than the

’ *

available number of degrees of fre.ed0m.i In such case‘ the.system becomes over-_I F'

specifiid'alta"i,f'ig'impo8aibra..tr &sign'a~control syktr& that satis&s~$ll‘:p

the desire&i ~c)ntrtil'bbjbctives?,':‘T~~s,! '.,'.,,,

it is:imbossible to design a'.,dontrol "

system for the"idea1 biniry;'drLitlPrstion column that can‘satisfy the following

six operational (control) objectives: '?. ""' -' , .> 1'

"Keep at the desired values &he FI), x~; Fg, xg; MD and' I$?

i 2 .:c;.. s I %, ,~':

< -,o ,. %@,) ‘! _'

Example 5.8:+ 6cgrees:Of.~reedom~pilla Mksing$%ocess.,,I:, : ,.'8. /

'Consider the non-isothermal mixingiof two str'ea~$!. diPcussed in Example

4.11 (Figure 4:'8). ‘The mathematic& model.& given by the equstions, ~

d c ;

dV 1’.

TEII (F1 + F*) _ .pj ' : ~ _ ": f<. , i d.. (4.12a). .I, ( '. ' .a / .>I

it,v -4 - (c, -

"F *1CA )t 'i

3l(4.13a)

.'! I,

dT,. .

~F~c~(T~-T~)~~P~~c~~~~-T~)~~~Q- -

,, i (4.14a)

Number of variables - 17: v,c A3'. T3, pi.7 'F2', Fj I "Al' CA; 9 'Ai'

f ’ Tl, T~,;Q, 6, cp, Ap, ) aii, , AtiS .12 3

Number of state equations - 3 .j '. ', ;.. I

Initial degrees of freedom - 17 - 3 = 14

‘ ‘:1.

6 Ii

_’

. . Further 1imLiation‘of the degrees of freedom.: m_ ,.

-“PhysP&l @rope&& of the.liquids are specified, i.e. p and=P’

- The heats of solution, Afi, , A$, , and AiS are functions of the corres-1 2 3

ponding concentrations, and the tefe?ence temperature To, i.e.,.‘i

Afi.a

s1- fl(cAl,cB1,To). AHs2 - f2(cA2,cB2,To), $3 - f3(CA3,cB3Jof

These three “equations reduce the degrees of freedom by 3; Consequently; i

a f ter the, abwe:spec%fications we have’.left,

14 - 5 - 9 degrees of freedom. ’ i ‘. ’

The eight degrees of freedom are now specified as foll&s: l:‘*,.; , i‘,

There , ak& pix’ uain- d4ifhAhmncee corni*.:.,

from the two feed styream,’ *tee., : , “. /r\ ‘$. *,: 1 , . ~1 ,:I. ” ’3; * _‘L ‘j , ‘_.’

1 ‘.Feed stream 1; Fl, Tl, and tii , _ :p&& stream .2; “%?2’~

1’ ‘,T2.’ And. cA .

, 2, ‘.“,‘,I. L: I,. ,,.: *I /

The values of the dist$rbances are’$pe$f&d by’ the extert&,world.I .

.‘. ,. .;‘..‘,,“.“ *: s .3,s: ,, >‘.I .‘; _:Specification of’,:rfie son&o& &bje&t$v&,~ a, Considering s&x disturbances

Q i + *specified by the external world,. we ha&$&y

i. ,.

9 ‘- ‘6 = 3’. ‘degrees o f .freedom* ,. ” .‘i I .,:,j -@..$ L,, ‘;‘....i; ;. *:,,:;‘- A,: j ‘j _, ’ :l e f t . Therefore, ye dan specify up to fhre$ control’objectives~ ‘I”* ese are:I. ; , ,.& “/& $

“Keep the volume (i) of “the mixture tn ‘ithe t&k ‘a& tie11 &II

, ,the temperature (T) a‘nd;;compositrtod’ (a& ) of‘! &e’~eff&uent 8 ,stream at d’esired values.“” - 3 ’ 3.

.,*.”‘, ‘2;: _:.i.

Figure 5.7 shows three possible feedbaa conttol* loops which satisfy ‘<the above: . ,‘. <” .:“., * , _ 3. j : 4,!:objectives. .

,_ < r,4, .; .”

It should be .clear by now that effi.c%ent modeling of B” chemical process:>

is a non-trivial task but, also-very cr’udfal. for the design: of a control system.

Before closing the’ present chapter let ‘us’emphasize’some of the factors which

will determine .the scope of modeJ.$,ng for .qontrol purposes,will determine .the scope of modeJ.$,ng for .qontrol purposes,: It”’: It”’. T. T””LL Before attem4pting to mode!., a* pro!esg, tie wuqt pose the, follow$ng questionsBefore attem4pting to mode!., a* pro!esg, tie wuqt pose the, follow$ng questions

tt ad try to understand their +mplicat&ns’ well:ad try to understand their +mplicat&ns’ well:

(1) What are the control ,objectives,we must oat&y?(1) What are the control ,objectives,we must oat&y? II

(2) What are t.he expected disturbances and their impact?(2) What are the expected disturbances and their impact?

(3) What are the dominant’ physical and chemical phenomena taking place in(3) What are the dominant’ physical and chemical phenomena taking place in

I,I, theprocess to be ,controlled?theprocess to be ,controlled? /, :’/, :’ ,,

Clear understanding of the above .questions and their, ansyera will, helpClear understanding of the above .questions and their, ansyera will, help

great ly to de f ine and s impl i fy ,great ly to de f ine and s impl i fy , _,,_,,

(a) the system-iwhich we will attempt to.model,(a) the system-iwhich we will attempt to.model, ,-,,-, / *.I ./ *.I .I.I. ,I,Is..s. .

(b)(b) the mass, knergy. and momentum balances that we sho@tLd ,-develop and,the mass, energy. and momentum balances that we sho@tLd ,-develop and,

(c) the additional equations that, will be needed: tb complete tha!,mathe-(c) the additional equations that, will be needed: tb complete tha!,mathe-

matical mod& ‘of. the process- i.e.matical mod& ‘of. the process- i.e. ‘transport and” kineti,crrate‘transport and” kineti,crrate ::. . ( ,. . ( ,

expressions, reaction and phase equilibria relationships)expressions, reaction and phase equilibria relationships) etc.etc.:, ;:, ; , , -, , -: _, ‘!., I: _, ‘!., I ! , .i . ..J! , .i . ..J

They will ‘also help to identify,They will ‘also help to identify,. .’. .’ 3.7 z -1”3.7 z -1”

’’ jj11 . i. i T.T. .6.6 ,i,i;.+:rp -:,‘I: > (“’ i, j (. . .,>, _,;.+:rp -:,‘I: > (“’ i, j (. . .,>, _, .,: ‘6..,: ‘6.. .. .,, I” ,.“.. 1I” ,.“.. 1

(11(11the stat; variables, , : ”the stat; variables, , : ” ~~~~ * - .“.‘,: . ,* - .“.‘,: . ,

, :, : titi _*_* / ?’/ ?’ a. ’a. ’ ,. ‘, ‘I,. ‘, ‘I ,’ I//,’ I// ‘. I :‘. I :the $put variables (manipu&ted &d disturbances) andFhe i,nput variables (manipu&ted kr@ disturbances) and

‘.‘.(ii)(ii) >I(>I(

> I >> I > ,, ;,, ; t.:.- .:t.:.- .: < :! . _< :! . _ ~:’~:’ 1>1> ::(iii) the output variables, ,,(iii) the outptt variables, ,, ‘. :‘. :

i, “; :i, “; :

./‘../‘. .... ;r;r i’l.i’l. -2:-2: ‘ ::‘ :: rr\*rr\*.:.: ,\,\that the mathe&tical model sSI&EdT Ynclu&. * ‘: .I;that the mathe&tical model sSI&EdT Ynclu&. * ‘: .I;1 - 1 ., .1 - 1 ., . jj

‘,‘, :,:,1,. , ;1,. , ; vv , , ‘., , ‘.

Let us now’ examine each of-. the above three questions and how they affectLet us now’ examine each of-. the above three questions and how they affect’’ .I *.I * i -,i -, ? :? :1( :,.1( :,. : *: * .‘f’ j ,.‘f’ j , : , , 1: , , 1 tt

the modeling of a process for control purposes through i’series of examples.the modeling of a process for control purposes through i’series of examples.

A. ’A. ’ Control ObjectivesControl Objectives

As ‘it has been discussed in Chapter’ 2 the,_objective.s. that a control systemAs ‘it has been discussed in Chapter’ 2 the,_objective.s. that a control system

is calted to satisfy may have to doi with’: ., ;:;is calted to satisfy may have to doi with’: ., ;:; ‘!‘! .,.,. ,... ,.. _,_,

- bnsuring the stability in the operetip)tl ,of a process, or l- knsuring thiat stabikl.ty in the oper%tip)tl ,of a proows, Qr l ‘.‘..:.:

- sup~,ressing the influs& of, ex~ar~i,‘8Ssturbancas, o r- sup~,ressing t h e influs& of, ex~ar~i,‘8Ssturbancas, o r ;.,,;.,,

- optimizing the economic performance of’ a plant, or usually , j- optimizing the economic performance 9;’ .+ plan!, or $ually , j ,-. i,-. i

- combination of the above.- combination of the above.

1.

All the abov’e’.dbj’ectiv& are translated in quantitative’ e2pr&sions in

terms of the floGrates; -‘temperatures, pressures, compositions, volumes, etc.,1’ ‘I,.

of the formb

v a r i a b l e x - d e s i r e d v a l u e o r

v a r i a b l e x 2 d e s i r e d v a l u e

where variable x = flowrate, temperature, pressure, volume; composition, etc.!,. j, . - I ,‘, 1 .- : _ . J.,, ,,

It is clear therefore that if we have identified the variables x which& ::- ;: “S

define quantitatively our, control objectives the mathematical model ~that wei: , ‘, . ,I. i” ‘, ‘_‘( ,

will develop must descrJ,be how the$e variables change with time.! ~.I ” :. .

i A l s o ;p.a .

will help ua determine what balances are. needed for the development of the ’+ ‘, ,*‘? ‘, , r:‘!. “,:.I

mathematical model. I’ : ; -: ,‘.. ” *‘_\i., t

.,’I ‘,.,. : ” 1 I‘ I _:. ,i

,

E x a m p l e 5 . 9 ‘-L/ jjrl_ <..’ !

:

Consider the stirred tank heater dlscussed .in Exam$le 4.4.Is- ,;*,,, ),.

A . If our control objective is to keep the -liquid level at a desired value,:i - ,

then the only state variable of interst “is the volume if the&quid in :< . ’ a I

the tank (or* ‘equivalently the 1 height of the liquid l&l> %nd.,‘con-,,,, .; ,

sequently we need only consider * the-‘total mass’ balance.” . I

The d$st&bance.

‘; ; i .: : sof interest-is the flowrate of the inlet stream, Pi, while the manib

” 6 / . ..I..~ _,pulated variables to be considered are the outlet flowrate

‘A_$ or the ,

i n l e t Fi.: ., ;.’ y

;F i 4>,. 1~ /B. If on the other hand our control objective fs’.‘to k& the.tem$ature’

;. ‘“‘of ‘the outlet stream, T, at a ‘desired value, “‘I

:. : 3’ ;: 1 :,‘pf y ,.,ii’ I/,them we r&et ‘.cons;Ld& ~both_:” .

: b,state variables, i.e. the temperature and the level t& the:%qu$d in ‘the. ..,~tank. This implies that we need write both total mass and lenergy

balance.l,.’ ,f

The disturbances of interest are the temperature and the

flowrate of’ the inlet stream; ‘while the available mani&ated.var&bles.’

are Fi, F and Q.

‘_ .h,

--

c. If, finallyr, OUT control objectives”are to keep the temperature of the

effluent s;t;re$m. and ,the: liquid, level, at desired values, we ,have a

situation similar to the case, B above. ’, L‘,‘,

# ’# ’:: :. :.: . :.,:*,:* .‘..‘.

,

A. If our control objective i&o” keep’ the concentration of the effluent

00 ‘,‘, ::Example 5.10Example 5.10 ’’ IIII

! ’ ‘.! ’ ‘.Consider the continuous m&id ‘proiess d&cussed in’ Example 4.11 (Figure :Consider the continuous m&id ‘proiess d&cussed in’ Example 4.11 (Figure :

‘,‘, *.*.We can distinguish the following control situations:4.8).4.8). We can distinguish the following control situations: ,i,i

:*’,

A . If our control objective i&o” keep’ the concentration of the effluentii /‘, ,/‘, , xx - I.- I.

stream in A at a desired value, t&n ‘the. state varsables of intereststream in A at a desired value, t&n ‘the. state varsables of interest.I I.I I

, are the vo#.ume o,f the mixture in the ta6k and’ its concentration in A., are the vo#.ume o,f the mixture in the ta6k and’ its concentration in A.

The relevant balances are thoaa‘,on total mass and on comionent A. ‘tie’The relevant balances are thoaa‘,on total mass and on comionent A. ‘tie’

dlsturbancfa of intereet are:dlsturbancfa of intereet are: cAcA i0ri0r11

CB 1, FlsCB 1, Fls11

cAcA22

( o r cB2) a n d F21( o r cB2) a n d F2;

: The available manipulated variables; are: Fl, F2, F3 or the ratio: The available manipulated variables; are: Fl, F2, F3 or the ratioii ‘6,‘6, ::.-.-

Fl,/F2. :Fl,/F2. : .,!.,! .:.:7 ‘I7 ‘I k4k4 ,.,. _Ij, j,_Ij, j,”” **

Ir .Ir . .Lf on the other hand our conttiol objectives are to’ keep“the composiion,..Lf on the other hand our conttiol objectives are to’ keep“the composiion,.I ,’ ~, .’I ,’ ~, .’ . .. . i -.i -..,., ‘17 ‘,‘17 ‘,

and the temperature of efflu?nt stream ,at .desired values, then we needand the temperature of efflu?nt stream ,at .desired values, then we needi .*i .* ii ” ,$i” ,$i :*-” ‘::*-” ‘: ‘-t ‘.(I, : ~‘-t ‘.(I, : ~

consider a.11 three state variabl.es (cconsider a.11 three state variabl.es (c,.,:,.,: ,‘V,T -) and ,formula ter-all thre,e,‘V,T -) and ,formula ter-all thre,eI _:I _:ll

;.A.3 .‘. 3 .i s :(A3 “. 3 .i * :

balances (&o.tal mass, component PI, total energy). .I” this, case thebalances (&o.tal mass, component PI, total energy). .I” this, case the: r: r

important disturbancesimportant disturbances are:are:!. ‘Al’.!. ‘Al’. Fp TIT !42 s F2 ant, T2 l s II TheFp TIT !42 s F2 ant, T2 l s II The

available manipulated variables are:.available manipulated variables are:.F13 .Fa. F3, the r+oF13 .Fa. F3, the r+oa...a... Fl/p2Fl/p2 a ia i_I. ‘ . ._I. ‘ . .( and Q.( and Q. 11

B.B. Expected Disturbances and Their ImpactExpected Disturbances and Their Impact// .’ ,i.’ ,i /, ‘..../, ‘.... / . ;, ,, ‘/ . ;, ,, ‘

The external disturbances which arei expected,.to appear and affect theThe external disturbances which arei expected,.to appear and affect the‘.r,‘‘.r,‘ .>.> , .:, .:

‘(‘(operation of a process will influence the mathematical model that we need>,,tooperation of a process will influence the mathematical model that we need>,,to,’,’.<.< 1 ,1 , ,:,:

__develop.develop. **

:: 1’1’

‘\ Fur thermor;‘\ Fur thermor;..

,, .disturbances with very small impact on the operation of the.disturbances with very small impact on the operation of the.::r.::r ./ I./ I I iI i

process ean bs’heplected while dirturbanoee with significant Impact on theprocess ean bs’heplected while dirturbanoee with significant Impact on theI.I..... ,:,, . ,,:,, . ,

IIII

_’

process must be included in the model. This will’ deter&e what ’ complexity 1-.,

model is needed, i.e. what balances andiwhat state variables should be‘,

included in the model. . t

State variables which are af,fected very little by the expected disturbances, _ it, ‘0‘Ccan be eliminated from the model and.&png with them the corresponding

balances. : : ) ., ~ ,

Example 5.11 ;.,

Let us return to the stirred tank heater (Example 4.4). If the feed

flowrate (disturbance) is not.expecte&to vary*significantly,.then.the volume ’‘+ \

of the liquid in the tank w$ll”remain almost constant. In this; case.,. ;_ * : 1 .jl >,A .’ ‘./

dV/dt - Adhjdt a 0. and we can uegledt the total:mass balance and the jr

associated state ‘variable hi&e-.;&~~aatical m;del .& in’te;;st ‘ior’ c*r;$rol

ipurposes is given by the total: energy balance alone’ ieqn; (4, 5bj)‘iwith tern;

paratura the o;ly && v&iabg* II .i .* Y, ‘T’:, i p 22, (“‘,! “‘

Note that if’ the feed tempera&e,’ Ti,;

Remark: ” ke “no&‘exp&ted .to qary sig-’ ‘;,, *_

nificant’iy but the feed floGrate, Fi, is expected” to! change

-substantially, then the mathetitical model can&$ b’e simpl&ied’as

above, but it will be given by’both eqnsi (4’:4;) andi’(4,%?$‘

Similar results as .‘above can’ b,e deduced “f-b+ the~“&i’&‘sys,&m~ ~l%~arnpie’ 4.10) ._

If the feed flowrate is not expected to change significantly, the model can be

s impl i f i ed , i . e . dV/dt = 0 and it is given by eqns. (4.9a) and (4.,lOb).f’ 1 ,,,:

. . 2’, . ’ :

C. The Physical-Chemical Phenomena in a Process l

r

A good understanding. ‘of the physical, chemical phenomena taking place in

a process can’ lead to signif icant ‘simplifications for control purposes. This

simplMicot.ion can be done by excluding from the balances (model) those terms

which have small contributions.j

‘*

Example 5.12

,

Let us return to’ the continuous mijring ,prodess dkussed in’Example 4 .ll.Let us return to’ the continuous mijring ,prodess dkussed in’Example 4 .ll.1.1. :.:.

Assume that for the particular combnents A and B of the mixture, tlieAssume that for the particular combnents A and B of the mixture, tlie

heat of solution do&s not depend significantly bti the.-composition of A andheat of solution do&s not depend significantly bti the.-composition of A and.;.; :.:.

B .B . In this case (see Example 4.11)In this case (see Example 4.11) :’: ’ ‘.‘.ff

aaAitCC

- Afi- Afislsl s3s3

zz AfiAfiI [I [

.. AfiAfi .I’.I’s3s3

z 0z 0

ii 3 ,~ j <i,.‘,.3 ,~ j <i,.‘,.]]11

and .the total knekgy ‘balance (eqn. (4. Ida)9 &n be simplif &d r 6 .the following:and .the to eal knekgy ‘balance (eqn. (4. Ida)9 &n be simplif &d r 6 .the following:,,

dT3 ~dT3 ~ aa:.:. ..(, ‘(..(, ‘( ::

:: PCpV.rdt -PCpV.rdt - %=p% : )%=p% : )- T3) + cPF2CP(T2 , . 3 __- T3) + cPF2CP(T2 , . 3 __-‘T) f Q-‘T) f Q

)) i-.i-.

In other words,In other words, we see that the qature of thk sr$xing phenomenon leacjs to awe see that the qature of thk sr$xing phenomenon leacjs to a. Q. Q I_ ._-I_ ._- : ,,: ,,

s&plif ication ,of the model.s&plif ication ,of the model. :.:.7.7. ,I,I 11 ; j i”; j i” __

Furthermore, assuming that from ali~possible disturbances,on\y the feedFurthermore, assuming that from ali~possible disturbances,on\y the feedII :,.II :,. II

compositionscompositions CA,CA, and cand c aye expected, to change significagtly whiJe the,aye expected, to change significagtly whiJe the,by,. ,,’by,. ,,’ ::

feed flowratesfeed flowratesF1-F1- a n d F2a n d F2 and feed temperatures. Tl, andand feed temperatures. Tl, and ‘i;‘i; are expectedare expectedii

to remain almoat the same w& can n+gleg&from t+e mathematical mbdel the totalto remain almoat the same w& can n+gleg&from t+e mathematical mbdel the total!! ‘,)!! ‘,) I jI j ‘,s‘,s

cucrgycucrgy balapce grid, from, fhs sezFr_of state; variables the temperature J’i,balapce grid, from, fhs sezFr_of state; variables the temperature J’i, TINIS )TINIS ).,:..,:. _*.:_*.:

the s impl i f i ed po$,el Ss &e,n o,n+y‘&ythe s impl i f i ed po$,el Ss &e,n o,n+y‘&y.. t;h,e; balegce in @nponent A ( e q n . 4.13a).t;h,e; balegce in @nponent A ( e q n . 4.13a)..L.LII7.7. “,tf .“,tf . B ‘”B ‘”

::.. .i; ).i; ) ;; ’;; ’ (’ -,(’ -,

Example 5.13Example 5.13

Consider agian the CSTR discussed ih Example 4.10.. If the heat ofConsider agian the CSTR discussed ih Example 4.10.. If the heat of

reaction for the particular reaction A-Breaction for the particular reaction A-B is :very Small. and the temper%tureis :very Small. and the temper%ture,_ _I,_ _I ‘i‘i . -.. -.

of the feed stream is. not’ expected to cpange signiflcpntly‘, the ,Wmperature ofof the feed stream is. not’ expected to cpange signiflcpntly‘, the ,Wmperature ofii

the reacting mixture will not change appreciably. In this case the reactor,. /,

can bei assumed iso thermal.can bei assumed iso thermal. We can ‘exclude-‘the total energy balance from theWe can ‘exclude-‘the total energy balance from theII

mathematical model and the temperature from the ,set of state variables;mathematical model and the temperature from the ,set of state variables;

., I., I _ ‘. (, ,.,’_ ‘. (, ,.,’ :: ,.,.

.... LLExamples ‘;‘.9,Examples ‘;‘.9,

‘5‘55.10, 5.11, 5.12 and 5.13 demonstrate very simply but also5.10, 5.11, 5.12 and 5.13 demonstrate very simply but also

->> ! ;! ;

)’)’vividly how’ the mathematical model o;f a process .can be simplified when we takevividly how’ the mathematical model o;f a process .can be simplified when we take: ,’: ,’ %. ,y%. ,y : ‘_: ‘_ ;; ‘,‘,

into account variousinto account various consider-ations related to the nature of the process andconsider-ations related to the nature of the process and:,t i‘i, 4 I:,t i‘i, 4 I LL

the characteristics of the control problems.the characteristics of the control problems.. _. _ ~~,~,’,~,’The control designer always looks out for such sThe control designer always looks out for such s f icatio.n~~.~ ! i( ., z1f icatio.n.j$ ! i( ., z1 ‘?I:‘?I:’’ (_,f :(_,f :i’i’ ‘-‘-

”” * ‘I* ‘I

SUMMARY biND CGNCLIJDING REMARKSSUMMARY biND CGNCLIJDING REMARKS,i!,i!

-<; , F[.x 8 5?’-<; , F[.x 8 5?’i 1..i 1.. * _** _* ,,,,.‘ _ i .*..‘ _ i .*.The mathematics,!: model .describingLi,the dygmic behayi&, of ,$: pqoc& isThe mathematica,l: model .describing,:i,the dynamic behav,i&, of ,a: proces& is

,”,”the result of a series ‘of assumptions made the modeler* : :For ;socessi &ntrol .-the result of a series ‘of assumptions made the modeler* : :For ;socessi &ntrol .-

,I,Ipurposes the !aasumptionsl made should: . , ,~ __purposes the !aasumptionsl made should: . , ,~ __8.8. ‘ .“i.‘ .“i.__ ,.L ”,.L ”(,,” . . b : :;,(,,” . . b : :;,

- identify and retain the %tipo~tant,~iqput (dieturl?gncesl.~~nipulated’ v&iabl.es) ,- identify and retain the irn~ortant,~input (dieturl?gncesl.~~nipulated’ v&ables) ,/ ./ .

output and state yF:iabl& onlyb,’output and state yF:iabl& onlyb,’ .,., :; 1:; 1 ;; “ (“ (;’ .::,t;’ .::,t‘3‘3 ,‘v (, ,.’,‘v (, ,.’”” .a,.a,- select the simplest type of, ,model thtl& describea tibe; prooess,w$thin- the- select the simplest t&b of, ,model thtl& desq$bee tibe; proqe&Iw$,thin- the

-.’-.’desired accuracy *.desired accuracy *. 11 “‘-“‘-

_’_’ .” ”.” ” 3 !..3 !.. ! : . . : 9! : . . : 9 ._ .;,.w t&J.‘, .it’.:y ’ “:._ .;,.w t&J.‘, .it’.:y ’ “:: /’: /’

The construction of a s&m@ ‘and effec;tiv~~.model.,,requ~~e$: a:;goo$., knowledge dfThe construction of a s&m@ ‘and effec;tiv~~.model.,,requ~~e$: a:;goo$+ knowledge dfii

II) . i, :). i, : ii : ‘,: ‘,*:*:

the physical a,nd chemical ptierromena under cons~~ezario~.~~nd..a~~;,~~a~a :‘under.ethe physical a,nd chemical ptierromena under cons~~ezario~.~~nd..a~~;,~~a~a :‘under.e,.,. ’’standing of the ‘control problems ,involved (eig, oQntrh1 dbjact~~~ir,~~~pactedstanding of the ‘control problems ,involved (eig. oQi&rtjl dbjact~~~ir,~~~pacted,d,d ‘. 1 ,,‘. 1 ,,disturbances and their magnitude, etc.):disturbances and their magnitude, etc.):

,I ‘.,I ‘..i;t ;‘. *..i;t ;‘. *. ,_I,_I_-_-

The input-output mo,del is a matheatical ~@#r~pti+ ,$aj$q directly ,The input-output model is a mathematical des&ipti+ ,rela$$g directly ,. ,. , ‘,’‘,’the inputs (disturbances ; matiipulated variables) to the outputs of a .,process .the inputs (disturbances ; matiipulated variables) to the outputs of a .,process .. *. * .dIt is preferred over a stat;e model because it is simpler and depictsnicely

the cause-and-effect logic in a physical system. The input-output model can,

be developed from the St-ate model through the integration of the ,state

equations when the outputs coincide with the state variables. In Part III we

It is preferred over a stat;e model because it is simpler and depictsnicely

the cause-and-effect logic in a physical system. The input-output model can,

be developed from the St-ate model through the integration of the ,state

equations when the outputs coincide with the state variables. In Part III we

/

’

,’

will study a simpler method to develop an input-output mode3 us+ng&he Laplace.; I ,,-:

transforms, :.2’Using the mathematical model of a process we can find the number of its

r:inherent degrees of freedom. A well modeled process should never be over-

.‘.specified. Usually it is”underspecif id’ and we need to provide .additional

equations to make it’ exactly speoified. “’>. ,, *l.

Such additional equations come from._ L I .,

the specifidation of the ~external’c&&ba,nces and of the ‘control objectives.*.,< .

For a system with f degrees of freedq‘ahd ’>/ : J)

d ,dleturbances we can. have no

more than (f-d)’ independent control objectives,./

‘. !‘

THINGS TO THINR ABOUT 1. _.;,. .’

1 . What is an input-output model’ and’.haw can you develop it.“from’a s&t=-, I, ~

model? When is this possible? +: :. ’ ’ .’ *. ” ’I. .

2. Describe a.: procedure which would alilow‘you to develop the ?&xit-output ’

model for an’ ‘ideal, binary’distillatiod do&m~.1 lii ,: :’

3. Define the:.concept of degrees of f.r&xiom”and &lat$it to the’ solut$on’.‘ ~I

of E equations with’ V &riat.il&$; ' ">' ~ :. ''$ : ‘i ""., c

4. How many degrees of freedom do you have in a system composed o’jE P’ ‘,

p&ses.. vi& C componen&? (.&&&~~;$&.,a $;i).; )’ ” ” .

5. How many degrees of freedom do..you’h&e in”~~$&!m composed ok P>.>

’ phases *w&h’ C components If ‘the mass of each’“phase^is gi&&, i.e.” ’ _

M1,M2,*** ,MP? (Recall Duhem’s ruler) ,. t.“I r. ‘. ; ^ ,

6. How does the number of *degrees of “&e&iom affect the number and the

selection of the control objectives in a, chem$eal :.,process?

7 . Why do we claim that d disturbances reduce the number of degries of

frPredom by d? , * ,_ l_. b .” *i ,;.

’ ‘:8, W h y c a n ’ t y o u deuld a control, syst&a f o r ‘an “tierirpecified procelic’i .’

I. ,, ‘.. !I :*., .

” .’” .’ J5J5**

.,..,.

’’II

9 .9 .Can you have the desired operation’ for an underspecified process? If yes,Can you have the desired operation’ for an underspecified process? If yes,

explain w h y .explain why. If no, explain how,can you lift the underspecification.If no, explain how,can you lift the underspecification. tt

10.10. ConszQler a system modeled by the following set of state equationsConsi@%- a system modeled by the following set of state equations

‘hi‘hi eedt:dt: fl(xl,x2,ml,m2,m3,dl’dZ)fl(xl,x2,ml,m2,m3,dl’dZ)

dX2dX2dt-dt- f2h1,x3al,d2) ,, ’f2h1,x3ayQ) ,, ’

dx3dx3dt-,dt-,f(x x x M m d d d)f(x x x m m d d d)3 1’ 2’ 3’ 2’ 3’ 1’ 2’ 33 1’ 2’ 3’ 2’ 3’ 1’ 2’ 3

wherewhere xx1 2 31 2 3, x , x, x , x are the state variables,are the state variables, mm 1' m2' m31' m2' m3 are tbe manipulatedare tbe manipulated

‘d‘dvariables,.and d l , 62,variables,.and d l , 62, 33 are the external disturbances.are the external disturbances.

(a) How many degrees of freedom does the system .possess?(a) How many degrees of freedom does the system .possess?

(b) How many control objectives can you specify at most? -;(b) How many control objectives can you specify at most? -;

(c) Consider the above system at steady state.(c) Consider the above system at steady state. How many degrees ofHow many degrees of., ”., ”

freedom does it possess?freedom does it possess? . .. .ii

11.11. A system is described by the follow$ng sbt of state equationsiA system is described by the follow$ng sbt of state equationsi‘,‘,

dXldXl..dx ,..dx ,

dr*dr* fl(ml,m2~dl,d2) ~ kd ,-J$ <A f2(ml,n$,dl)fl(ml,m2~dl,d2) ~ kd ,-J$ <A f2(ml,n$,dl)..

;i,,;i,,‘,‘, *,‘.*,‘.

Find the degrees’ of freed&n for the system at its dynamtc state &dFind the degrees’ of freed&n for the system at its dynamtc state &d: I.: I.steady state.steady state. Are they equal? If ‘ndt, why? What ‘are the implica$ionsAre they equal? If ‘ndt, why? What ‘are the implica$ions

on control in this case?on control in this case?zz__

12. What are the main control consideratsons which afiect t.he slope of12. What are the main control conslderatsons which afiect t.he slope of

mathematical modeling for a”chemica1 process?mathematical modeling for a”chenkl process? .’.’

13.13. In what sense do the control considerations gffect the~mathematicalIn what sense do the control considerations gffect the~mathematical

modeling of a chemical process?modeling of a chemical process?

14,14, What are the usual, general quantitative representations of the controlWhat are the usual, general quantitative representations of the controlI.I.

objectives?, In terms of what variables are they expressed?-objectives?, In terms of what variables are they expressed?-

. .

“,cV I

I

1I Ii ; fl\ #?N

I III,“,

Ts /I II h’

1Llnll-r - ‘)UfPUT ~0DE.t. I ,_’I

L - - w e - - - -------c---m-----------c--. I :

F;qtlre k.2~ *d d ;

r --------------------e--*-r.--‘--- III II

'I

I

I

‘a \-, fF, ,4(,-

a'&) j I OUTPUTZI1

INPUTS 1II

I~~PuT-OUTPUT EJIOPEL Il- --------.----c& - - - - - - - - - - - - -

ricjure 5 . 3

1..,--. _ __ I

.

,

1

*.

. . . .c

m1i ,,

‘\

A ‘). j

REFERENCES ""', ,, , .,. :

' ; ,i :

Chapter 4: ,Three' eiceptional references -with ,a large number of process,i:, ., ; " . .

modeliqg examples parer .~:, + 5, .' .,' .rl S,‘ a = " ,:,.Process Dynamics and Control.' -voi ;i j., I b$ *J, MI": Dousl&, $~eniidg-Hall,

Inc., E~glewoO;d Cliffs, N.d. (1972). ,] ,

.I .,:-'\ ', ,.* .,:I -f,i ,,1 ./ , .~ ",.

Process Modeling, Simulation and 'Control for Chemical %ngin&&, by'W. L. Luyben, McGraw-Hill Book C.o* p 'New York (1.7,3)i, .* :. ,,:,' t: .*I . .

(1)

(2)

(3) Dynamic Behavior of Processes, by J.;p .%i -'

C. Fri&ly, Prentice-Hall, It&.,Englewood Cliffs,G N.☺j. (1972) l , Iz �i . ,, II I_ f7Q,� + ,,~ ;

_ .�

15:+"H&w caia the ?Jnpact'of the disturbances simplify the model of a process?

Give an example other than that discussed in the text.: i

",:16. G&ve examples to demonstrate how you can'simpIify the model' of a process

', I

':~by'~Ef&regar$¶.ng'physicai and khemacal phdndmena with-sma&l impact on 1

the behav.fpr of .the process.-3 ,, * ,a::. _'- ;, ," ,..'17. Give'an outline of the steps tbat'yo$ should take .during the development

. .of the tnathema&al model f'or a 'cbemidai process,.

1 . ',..:I_ '\, )'

.;

For the development of the dynamic material and energy..;ktalanees, the rgader7

could also consult &he following book mhere Examples: 4.l@'.and 4.1-11,have been. . . .*' ,* j:" 1 '. < 1 a :' : i ;. /

adapted from: I.I. .

(4) Introduction, to Chemical Engineering ~~~~sis'~:'by,lj?'.:'IW;:Pir.,RueselJ. andM. M. Denn, J;.Wiley and Sons, Inc., New York (1972);:

/I' 53 o we,:: .‘Additional references for material'and,energy balanoer are: : '

, : ?\I* a

'j *y;; :,, . . .s,...i. '.(5) Basic Principles and Calculations,.in Chemical Engineering,'3rd Ed., by

D, M. Himmelblau, Prentice-Hall, Ina*, EngL&?00&C1@!f*,.N.J* (-l.974).L.

(6) Elementary Principles of Cbemieal~Pgdc~se~~~.by~~K. M...Feldsr* an&R. W,Rousseau, J. Wiley and Sons, Ink., New York,$l978),

,"f "/ '._ : ,. (.' 1' :"., is .'For the modeling of specific unit operations and reactors, there &fists a large

I'.number of textbooks that the reader could use. Not all models included'in

these books are convenient for process control purposes, but they could,help

to develop simplified and useful models..: Among all the available references. .'-, ::the following constitute a par.tial.fist‘: ,

(7)

For the modeling of ch,emical 'reactors:, , ,:

Chemical Reaction Engineering, by 0. Levenspiel, J. Wiley and Sons, Inc.,New York (1962).

(8)

(9)

(10)

.

(11)

(12)

(13)

An Introduction to Chemical Engineering kinetics and Reactor Design, byC. G. Hill, Jr.,i* .I, Wiley and Sons, New Yorks (1977).i _I

Elementary Chemi'cal reactor Analys$s, by R. Aris, ;P,rentice-Hall, Inc.,Englewood Cliffs, N.J. (1969).

Chemical and Catalytic Reaction Engineering, by J. J , * Carberry, McGraw-Hill, New York (1976).

For the modeling of transport pro&&es:"

TransportiPhenomena, by R. Bi Bird; W; E. Stewart and E. N;~Li&tfoot, 'vJ. Wiley and Sons, Inc,, Naw York (1960). t

Q, I._ ,h , , .,.~

'Mass-Transfer Operations, 2nd" Ed.;New York'"(1968).

by R.'k. Treybal, M&raw-Hill Book Co.,I.: ?.' 1 :. * .,

Heat and Mass Transfer, 2nd Ed.',.,: '.

byE, R, G'.'Eckart‘ and R; M. Drake, Jr.,McGraw-Hill Book Go., New, York (1959). , * > >.'. ._

For more on the reaction equilibria and'phase equilibria the reader could con-‘

sult the following books:~ ' ' '

(14) Introduction to 'Chemical ER‘B;in,e~~~~g".~ermod~lamics, 3rd Ed.', by J. M.Smith and.H. 6. Van Ness, McGraw-IiXlt Book Co., New York(1975).

:(15) Chsmlcal En inaarin Kinetics, 2ndE&.,.&;, 'NewJ& (197$, -'-

by J. M, Smith, McGraw-Hill Bookc‘ .P.;, : ,., _~:.L;.. , I; ty

++;F ," I : ..;,';:ir;:; i/ -..,,: i ?. ~ z, : ; : .‘V _ !r *

For an extensive discussion of the mathematica.1 modeling ok'am ideal.; binary/ ‘;. :. j. ;_ ' ,j. ;* .!a "i., :' f. I',$

distillation column and of a nonidaal multicomponent column, the reader can: ' ,.. : ", ;; j: , .

consult the books byi1T. ~~~'D~&gla~~~(Ref; 1); 'W,. L.'Luyb& (Ref.'2) a& J, C.

I Friedly (Ref. 3). An'interesting discussion of the difficulties encountered'‘ .% , :~:'

during the modeling of chemical processes 'can $e found in .<.I‘ ). t:

(16) "Critique-of Chemical Processi.iontrol Theory,"i..P' ,i. _:

2, p. 209 41973),'andby A., S. Foss, AIChE J.,

1 * '

(17) "Advanced Control Practice in the-chemical Process Industry: A View fromIndustry," by W. Lee and V. W. Weekman, Jr., AIChE J., 22, pa 27 (1976).

Chapter 5: For additonal study on the,degrees of freedom and thair impact on

the design of process control systemsS the reader should ref.er to the following

book:

(18) Automatic Control of Processez, by P,.:W. Murrill, International Textbookco., Scranton, PA (1967).

I& ;

PROBLEMS :. I / ; . .I'

1. Consider the two systems shownin Figure P.&-l. .,$y&m 1 differs;fromii ,

System 2 by the fact that the level of liquid.in t&k 2 does hot affect

the effluent flowrate,from tank 1 which is the case for System,e2. I~

(a). Develop the mathematical model for each of the two systems.'I

(b) What are the state vari&les,for e@i'syst;km,:and what type,of;,,'

balance. equa$ons %ave'you used? .;,,

(c) "Which mathemhtical model is elsier to a&ve~;~'0hat for System 1

' or that for System 2 andi,why8 i ! "._1:

Assume ,that the,fl&rate of an effluent stream from aGt&la pro-

portional to the hydrostatic liquid prelesure.jthat cau8es the flow-of

liquid. The cross sectional area of tank 1 is' A,$&t2 and of.'Itank 2d

'is A2" ft3 (for both systems). The flowratee :Fl, Pi, F3' are $ai

ft3/min.i

2. Do the eame work a8 in ProbUm (abova).for th&syetb:8h@n in Figure

P.II-2. All the flowrates are'volumetric, whi&a th&,cro8s.earctional ,,

areas.of thg three tanks are ;Al, A2, A3. in ft2, rerapectively. The2.

flowrate F5 i8..constantand does not depend on hg.a l*i:i d

; /Y

3 . Cons.ider t.he tvo ,grtirred ‘tank,,,heat,$rs, shov, ,in Figure P..II.q. :/ : 1. .:_ ‘(2

( a ) Ident i fy the state variabl,es $f,.the, system. .b*,,l J ,’ . ’

(b) :D,etermine what balances you, should, perform., ’ ” ‘,” . I

(c) ,Deve$op the state model .th$t $escribes the dgn.amic behavior!1

o f the system. t,

c

(d) How would you express the hea$s given by’ the ttio ,steam flbws I_,1

in terms of other variables?

The flowrates of the effluent etraa+ are assumed to be proportional‘ to,’

the liquid static pressure that. causes the fl&w of the liquid. T h eI

” cross i?ipaetipeX areas ‘of the. two tanks .are , . Al ‘and A2 i;n’ f.t? ,an&.iirhe. . is 1f lowW&s“&e l’volumetric , iNo vapor ia produced either in the first orI 1:-

nd+a*& AtX a n d At2 are the heat: .exchange areas for, the,.,

: _’:: , . h <‘,” x .,, ( ,‘^ ,I ,‘1cip ?robQm 3 C+pq$ for C&B s,t+yd tank h,qvs’I,

For the TankLi, the st,eam ia,.~,z&njocted

; & the Liquid fwater. rMater, vapor ..is produc*: ,in ,;the‘ .,e,ecc,nd’

are the cross sectional ar.aas .cf the &NJ tanks?,.d. i ,‘ . . .-

I’ Assume that :the ,effluent flowrates .are.:proportional,;ta,,4h~. l&&d, ! ,

static pressure that causes their flow, At is the heat transfer , , zr :,

a fea for the steam coi l . !*

5. Consider the mixing process taking;: place in., a two-tank system, (Figurei

P.II-.5). *’ \:’ !* : .:: V(*,‘.

(a) Identify the state variables of the system. /’ ,,! : j

(b) Determine what balances you should perform.. : ’

(c) Develop- the state model that describes the dynamic behavior. of

the process, assuming that the heats of solution are strong

functions of tha composition.I”‘. ..,

i ,, a..., ‘+.‘y. . . ;;

\

6,

7.

8,

weak functions of’ the com’pos’i’tib’n?” ’ ” ’ ” ”. ,f% ,‘.‘i.,‘, :,,, ai *

Assume that the flowrates are volumetric $nd ,the compositi&s ‘are in

moles/volume. ‘The effluent ’ f lowratea ’ are proportional ‘to the liquid

static pressure that causes their ‘flow. bl a n d A2are: the cross

sectional areas of the two tanks and At” is the-heat transfer area’ 1 ‘_’ :. t” .~ ,I“a . . /I !

for the steam coi l .,.i I. ; 1‘ $ ‘i CI’!(/ t

Develop then state model for .the’.b&chbemixing of two solutions (Figure,‘. )i : . i. _ .‘ s di * ,; 3~ ‘;; .I . ,P.II-6). Initially ‘the tank is “empty., The volume of the ‘Qank is,l,;cY ,’ r

’ a,‘_. -& i’ , . ‘ _’ “‘ -* : ,

(f2). The flowrates are volumet2i.c apcj”‘the ‘eohcentrat~ons ategilt ~,_ *: ).I”: ,‘i.‘ - i CL!.: j .‘,, imoles /vo lume. a’* ,t . .I . “’ :“. ; :d ‘( -.‘. ”,’ I(H) How ‘long does- it take to fill up”the tank? ’

.: ‘, ,i.,;;/;&

,(b) Show how would you”f And’, the

‘:‘; :v “;”mixture in the t~nk*;fu&$ t

I ;gif i l led upu ’

,Assume that the flowrates are volume

volume, and that the heat, oft solution depends on the oom@s,~tio~n.. - ‘3

Develon the state model for’ a ‘batch ‘reactor whese ehe F&&Y

reactions take place.* ‘. ‘1; i ‘. &j I ‘$ ‘2 :*. I, s Z’“‘., I...r .’

‘I

k, k,

A ‘,tB ’ +C ’:: 9 4c!*_ . :‘...‘.. .’

\kZ D,, 1

~; 9 ,‘, ,r: *,All reactions are endothermic and *have, f irst-ord.er k&n,et+s,:, :@he :::#‘i,,

‘, ;r..r;;<‘reacting mixture is heated by,steam of., lS&~ps,ig.w?&zh. f&$ws”‘through.-. 2a jacket around the reactor with a rate of : Q (lEr/at$n)-,. ?’ ‘~1

, .’ <$Consider the continuous stirred tank rea&tor, system shot+W’in,,,~FiguVe

P.II-7. Stream 1 is’a’aixture of A ..and,-,,B _I with composition ‘CA1

and c (moles/volume) and has a volumetric flowr$te Fl and aB1

temperature Stream 2 ia pure,.,R.,J The,reactions taking place21.i', : ; .'( :i : T1',*

are:I‘ f ?,$S : <, _.',

(Reaction '1): ~,kl, '. .

A:+ R - PI

1 .i" '.

:~ Li, :.'9 L

content with! 'At ' 'h&t.transfer a&&.t ,, : i ';, ..*--:

(d) ,Iiow can you 8;mplify‘"'f). 2 -I,

the state,model ir', -'t ;,, ak2 &lo, ki - . . j i ;l" ,;. L.'

'large range of~tqnpera&,res? ;i1 ": ', ',* i:“ ; .F1., : .> ~ : &I :;* .': i .'1(e) Define the assumptions that, s,kould be made in order to have:&': ".' _

> c ,,:‘.,: v_( I. . s, r.,

isothermal reactor. .' ,,..j ; .; ,/, . p- .I .1 ,, '\ :.; ;-. ::g;

9. Develop the ,state model for the twoCSTR.ayetem of Figure:P.II-8:. ,A,'

,-, Tr I . . .b'; I,. : ias,_ ".: .,_I''/ %

t simple reyction with 1st order kine,tics takes place: A.B. .:L :

Assume isothermal conditions. : .,,; -*

4, ' ' j O'_

10. Assuming plug flow conditions for a jacketed tubular, react& (Bigure

P.II-4) develop-its state mo$eL A olmple exoth.etrmic reaction,'A u+ R

with lst,o$rder kinetic8 takes~,$&&c~ Due ta the very largeheat:&;

reaction, malt* salt*, which flows gounterourrantly.to the ,reactiort

' mixture around,the tube of t&e reactor,, ia used to keep-the reaction '

temperature et acceptable levels, Aqsume:,con@tant temperature for the

malten salt along the lengthsof the reactor. The reaction takes place

* .,

in the gaseous, phase.,.

Tl&“~lowrates are .volumetr& and thk compos&ions

in moles/volume. The’internal diameter of the tube is d (in.) ano.,

i t s langth >,;,II ,’.j,.; .,L’ _ 6

(a) fs the s&tern ‘a lumped parameter (described by ordinary!..

differential equations) or a ‘distributed p%rameter? .

(b) ‘Does the realtor possess dead time between inputs ‘and outputs? ,/

‘&plain. -.’ ’ . , . I

11. F i g u r e P.,lI-10 s h o w s a s impl i f i ed reoresentntion nf ‘a clt-lllll bnf’trrr. ’.c . .

j - __ ‘.~ . . . . - . . -. . -

,z

Feed water-“enters the boiler with a flowrate Fl (mass/hr) ,and ‘a,, tern- ’?: :

perature TX and it is ‘heated by an amount ‘ofS’heat Q (Btu/hr); :./

which is supplied by burned fuel. The.:generateci s t e a m flows o u t ’‘”r .

from the top of the boiler,’ with a flowrate P2’ (&ss/hr), “and aJ 1: . < ;’ 4, /

p r e s s u r e p (psig) . , :- ( :,: , . , ,: ‘?,’ z1 i’

A sjmple feedbaclk control system has been..instaUed tlo keep the level

of the water in the drum boiler constant’ bv ,:man&l$tine- t&-@idre;t&

(a) What are the state variables deacribinn the.+1

(b) What balances are aonronriate for the drum boiler

(c) Develop the state model of the system,; For the Fe

trol system use a relationship of the form: F1 -- f (N“ -, . :’ ::

. .h desired) l

,I, ‘: *_

*

12. Consider a p.ipe of length L -(in.) with ‘an inter& h&k&.” d ”

(ilk). Water flows through the pipe with’s ~~ol&etri~ fle&ate* $. ’

Let.- P l be the pressure at the entranc’e.of ‘the pipe and ’ 62 “the’. ”I

pressure at the exit,. *‘ ^ ’\’ : ’ ‘i!. . . ”

’., % . ’ , ’ .‘> .: .*ir,. 6 r . I

i ,;., -4. . 1 (. , _ “, 1; b ” , ’ * , ,i 3,, -p, ,,.’ , x- I ,*TL

,

(a) Identify the appropriate state variables to describe the system.5’; , ./,‘, : -,

(b) What are the relevant balances, for the system?! , 1 3 1

(c) Develop the state model for this flow system. ,

13. Consider again the flow system described in Problem 12, (above), eat. “P

time t=O a stream containing pure component A is mixed with the

entering water, causing a concentration cA (moles/volume) a’s the .

water enters the pipe. Assuming the t A does not diffuse along the

l e n g t h o f t h e p i p e : .I

1.. : , : 1 .

(a) Develop the state model that describes how the concentration of ‘. .(<‘,

A changes with time and along the length of the, pipe,.1 : -.:I (b) Show that the system possess& dead time between input and.

output and compute the value of the dead time. I$A “.. /

14 . A liquid stream is’ a mixture of two components A and .,B and has a?‘,

v o l u m e t r i c (volume/hr) flowrate Ff, tempprature Tf and pressure* _

Pf. Let! cA and cB be the mole’fractions %f t. A j and B’ in the

,lkquid s t r e a m . It. is eesumed that’, the..,pte~.sura pf i s l a r g e r t h a n/ , I: : ‘,

the bubble point preasuri of the r&ture A and. D; so ‘that>.there :,.%,:‘ /-.is io v a p o r ~pnosent. 1 ’ ,i .(” 3‘T i)“. LI

i_. / a: _I’ / , - 1,’The liquid stream passes through an isenthalpic ,expans.ion valve and

&:. 1 I2 /.is “Flashed” into, a flash drum (Figure 416).

i:..The pressure p in

I” * .I _s

the drum ‘is assumed .to be lower than the bubble point pressure 04‘

the liquid mixture at ,Tf. As a result, two phases at equilibrium!>!

.with each other appear in the flash drum; a vapor phase with a compo-,f, _,._ ‘.

sition yA and yB (molar fractions) whic$ is drawn with a h,‘ .

volumetric flowrate Fv and a liquid with a composition ,xA and

xB (molar fractions) drawn with a .volumetric flowrate FE,, Let T

:’be the temperature of the two phases at equilibrium in the flash drum.

;

, ’ .:‘a_:. : i

,. .’(a) What are the fundamental ,dependen,t quantities whose* values .’

describe the,znatural state of the flash drum?, (See Section

A ‘!:4&2) ,:*r : ‘ci, ‘.* \

‘L(b)‘ ..‘What :are the boundarias,.of the system(s) around which::you will

perform the -various balances%perform the -various balances% . ‘._. ‘._ .%.%( .( .> -> -

(c) What: ari the relevant balances?(c) What: ari the relevant balances? jj ‘.,‘., ‘, :+ ‘,‘, :+ ‘,

(d)(d) Besides“ the balance ‘equations what additional aelationshipsBesides“ the balance ‘equations what additional aelationships: a: a

“’ do you need to, complete. the bta.te model for .&he flash drum?“’ do you need to, complete. the bta.te model for .&he flash drum? II

( e ) Ident&& t h e :state’:&riablss a”d the-:input v a r i a b l e s (manipu-( e ) Ident&& t h e :state’:&riablss a”d the-:input v a r i a b l e s (manipu-..”..”

lations, disturbances) of the system,, -’ *, 9lations, disturbances) of the system,, -’ *, 9 (...(.... ,. ,

( f ), Dev&op :the’compkeee st$te mode,,& o f t h e system>( f ), Dev&op :the’compkeee st$te mode,,& o f t h e system>

15. StaFting with the statg,models for the two ,%&tams of Probl& L4 : : .i % I ,. : , : ., ii i”.” .., (..._ :‘;, \ .‘L ::, _ ‘:: I” I_

(Figure P*II-f), feve@p the corresp9~~~g~ipput~plutp~t. models. Also,i

*i,, : i,::., .: * ,. l.. “.. ‘t.;. .,,( a ) F i n d t+’ d e g r e e s o f f r e e d o m f o r e a c h .&~&em,~:~~d.~,,U, ‘I ,’( a ) F i n d t+’ d e g r e e s o f f r e e d o m f o r e a c h .&~&em,~:~~d.~,,U, ‘I ,’

::.I..“^. ,.I..“^. , ,‘. : v!..f,$.t .”,‘. : v!..f,$.t .”

II(b) Specify how many control objectivee c& you h’#ve’for;each:(b) Specify how many control objectivee c& you h’#ve’for;each:,, i,, i ..--I ,’ : .i ,‘,:;..--I ,’ : .i ,‘,:;t3.j f:;r),t3.j f:;r),11 2%2% , ,, ,“,“,

systaQi!lsystaQi!l ‘.,\‘.,\::

-:>.r “~ ,:.-:>.r “~ ,:.

For each of the fallowing syst&e find?” ’For each of the fallowing syst&e find?” ’ “:’ ; ,: .’“:’ ; ,: .’

( a ) . Th? n u m b e r o f d e g r e e s ,df fre$ddm:.’ “ *, ”( a ) . Th? n u m b e r o f d e g r e e s ,df freedom:.’ “ *, ” ,.,.

(b)(b) The number of independent ko&rol joikj ectives you can specify,The number of independent ko&rol joikj ectives you can specify,

(c) The input-output model.(c) The input-output model. I’, :i c’ !:s ‘:‘_. ‘, 1.I’, :i c’ !:s ‘:‘_. ‘, 1.I.I. 1..1..System A:System A: The three-tank system of R&tire, ~PC.114 (Problem 2) .’The three-tank system of Figtire, iPC,II-% (Problem 2) .’

System B:System B: The: two stirred tani heater s&&em of Figqre P.Z’f-4The: two stirred tani heater s&&em of Figqre P.Z’f-4

(Problem 4) . . .,

system c: The two-tank mixing proce&s:of Figure P.IIi5

(Problem 5) .’ 1, j 1

‘

:’:’ , .$mpT&& (k’ ag$ f , . ^ ’ ‘., .$mpT&& (k’ ag$ f , . ^ ’ ‘. .;.;!i!i

ll

20.20. A simple chemical reaction, A + B; with first ord& kinetics,. tak&A simple chemical reaction, A + B; with first ord& kinetics,. tak&

place in a CSTR.place in a CSTR. The. &fluent of the reactor enteis an ideal binaryThe. &fluent of the reactor enteis an ideal binary

distillation column where the unre&ted A :sis taken as tha overheaddistillation column where the unre&ted A :sis taken as tha overhead\\ /./.

product with a composition yAproduct with a composition yA (molar frection)‘and ‘is recycled back(molar frection)‘and ‘is recycled back

to the reactor after it has been mixed wbth’fresh feed “(Figureto the reactor after it has been mixed wbth’fresh feed “(Figure

P.II-11).P.II-11). Assume that ‘the mSxing process and the CSTR are both iso-Assume that ‘the mSxing process and the CSTR are both iso-

thermal.thermal. I ,’I ,’

i)

17. For the CSTR system of Figure P.II-7 (Problem 8) -..find:.

(a) The number od degreea of freedom, and 1 : j.

(d) Identify Maat would you use of:conlrol objectivea.(d) Identify Maat would you use of:conlrol objectivea. ::. ‘,::. ‘,bb

(e) Develop. the linearized model.of the CSTR::around’ the steady ,a:(e) Develop. the linearized model.of the CSTR::around’ the steady ,a: ’ ‘, ’’ ‘, ’??

state and then show howyou. would ,form, the correspondingstate and then show howyou. would ,form, the corresponding ::

input-output model. ,input-output model. , ::

18 .18 . Do the same work as in- Problem 29 for the two .CSTR ,system of FigureDo the same work as in- Problem 29 for the two .CSTR ,system of Figure

P.II-8 ( P r o b l e m 9 ) .P.II-8 ( P r o b l e m 9 ) . ., i., i

19.19. For the f,lash drum system of Problem 14 (see also Figure 4.6) find:For the f,lash drum system of Problem 14 (see also Figure 4.6) find:‘_.‘_.

(a) The number of degrees of freedom, and(a) The number of degrees of freedom, and _’_’

(b)(b) The ifi&nber. of independent control obj&tives” you can specify. “’The ifi&nber. of independent control obj&tives” you can specify. “’i,$.i,$.

_-_- : , ‘, (: , ‘, ( a.:a.:( c ) Identify,a se t o f ’ c ont ro l ’ ob j e c t i ves bhich’are m$&ngfui ‘ “;( c ) Identify,a se t o f ’ c ont ro l ’ ob j e c t i ves bhich’are m$&ngfui ‘ “;

( ,.( ,..I.I

from a practical operation point of ‘view,from a practical operation point of ‘view, “” ’“” ’ ’’. t ; ’. t ; ’ ii

+,:+,:,,:, s,,:, s

(d)(d) Show how wouii you develop an, input-output’ model’ &r’ theShow how wouii you develop an, input-output’ model’ &r’ the~ $~ $

(b) The ;umber of independent control objectives you can specify.(b) The ;umber of independent control objectives you can specify.

(c) Fo’r the number control objectives you have specified, do you(c) Fo’r the number control objectives you have specified, do you

have’an equal number of manipulated variables softhat youhave’an equal number of manipulated variables softhat you

can achieve your control objectives? :,can achieve your control objectives? :, .‘,. ’.‘,. ’

flash drum.flash drum.

(a) Identify the. fundamental dependent variables which describe

the natural state of, the piant. ‘. ”

(b) What are the relevant balances and what‘are the boundaries

of the systems around which you wiil perform the balances?;.,

(c) Formulate all the relevaht balance equations.‘-/

(d) Identify the state variables ‘of the plant. ’ ’

(e) Deter’&& the number of degrees”of freedom for the plant...

(f) If (d,egrees of’ freedom)‘> O,‘.:how would you specify the._

‘;’_,

additional equations needed to render an exactly ep&&!ied ‘. ’ ” “‘..f ‘& ‘idegrees of fr-ed;mj. I +b: _I _ “I ., ‘,’ * ’ ‘I

SyBtem, ,. :, 1 rj ‘2;. (g)

‘.iHow many disturbance specifications do you have &d’ how”many

.i , j_ ..::control’objectives c a n y o u identify? ” ’ ’ ,!’

21. Consider the small plant described in Problem 20 (F$gure‘P’i%ll) .

“(a) Determine the number of’ degrees of freedom, for ‘the ‘plant.’ “I

(b) If (number of degrees bf’ freedom)‘ ; ‘C’* ‘h&would ‘youspecifyi’:. “_a 2i ‘t&t ,I’- , : !I$; i ,; ‘i-g*,

t h e add.itlonal e&atioh n&d& t o E&d& k-wta&&$ Eipecifk-ed: ‘” ,,.: P.&&& , :~

system, i.e. (number of degrees of ‘freedom) = O?‘.,,

’ <. i I.. ,,I’. .! i -j. s !. ,.(c) Iow many disturbance specffications do you’%ave and h&manyn

control objectives can you ‘iden’$!fy? ‘?‘* i l ’!;..r’L :. .i Q.

’‘. ;’

.:i,-s *J(d) Specify the control objectives which’ have’a,~.pracfical meaning

for the plant.’ * ,,..^b. e .

. I .’ ’

(e) Develop a s$nple~inpu&output “model for ‘“the ‘ plant ‘without‘_

p e r f o r m i n g e x t e n s i v e computations,~+ *a:;, t:a-.

22. Develop the state model for an ideal binary ‘l&&h d’istilla”t& column* ,

with N ideal plates (Figure P.II-12). At t=O,- the composition of;.

the in i t ia l mixture i s cA a n d

mass is M (moles).

(e) How many degrees of ..freedom dqes the system pos?ess? I. I ’ ’

A (f) How many control objectives can you specify?/ ,‘ Additional questions: I’

., :

(i) Does the’ vapor flowrate remain constant if the heat input Q remainsDoes the’ vapor flowrate remain constant if the heat input Q remains*I*I LL. .. .

constant with time!constant with time! ~‘1‘.~‘1‘. , ::, ::

What about Fk,What about Fk, does it remain constant? Under what conditions woulddoes it remain constant? Under what conditions would*’ :.. I*’ :.. Iyou change it?you change it? j ‘,j ‘, , : >, : > ;; : ,, ;:: ,, ;: a’a’ ..Develop the state model for a gas a&sorption column (J?igurq F,.XI-13)Develop the state model for a gas a&sorption column (J?igurq F,.XI-13)1,. ‘,”1,. ‘,” ,, ! ““>,i.! ““>,i.: 3 . :‘: 3 . :‘, , a*>, , a*> : ’ i :”: ’ i :” 2121with N ideal transfer units.with N ideal transfer units.,:- / )I’ i,:- / )I’ i Tha, f&x&ate &blhr) of the?%a, f&x&ate &blhr) of the

*.*. : .: ::r ; # ’ :: .: ::r ; # ’ :‘+,.‘+,. ppr,,:~ ,.Lppr,,:~ ,.L ,’,’entering gas stream isentering gas stream is’ rl’ rl : ,rg: ,rg

ttw,lth a oomqbsitioxa.~(m#ar fraction) yA ’w,lth a oomqbsitioxa.~(m#ar fraction) yA ’’’.,., r,>‘, r;ilr? , . :.,r,>‘, r;ilr? , . :., ip,?:\‘. itip,?:\‘. it

>:>: . *,. *,while the ;composition of t!he effluent gas stgeam should bewhile the ;composition of t!he effluent gas stgeam should be (mofar(mofar. . ,. . , yb,yb,

:i:i ftt. :ftt. :1’1’) :, jf i’<) :, jf i’< ,) i-*x‘,) i-*x‘

fraction) :’fraction) :’ The flowrate of the% liquti ‘absorbing stream is l?$ (moles/The flowrate of the% liquti ‘absorbing stream is l?$ (moles/,‘_(. ,, ‘. { . .),‘_(. ,, ‘. { . .) , : :(.. “:’, : :(.. “:’ ‘L. .,‘L. ., :, ‘>:, ‘> =.=.

Ia-) and it Is’ free of .t!lp;cd,myncnt, A. '.'Ia-) and it Is’ free of .t!lp;cd,myncnt, A. '.'f, ‘!f, ‘! ,;:;i, 1..-,;:;i, 1..-,, ,_ ) .l,, ,_ ) .l :: ", .", .‘. ,‘. ,

Develop the state model for &nul t&mponent: !$$a compone$s). nonideal. .Develop the state model for &nul t&mponent: !$$a compone$s). nonideal. .i: .’i: .’ ; I *t,i9;,,; I *t,i9;,, : ,.: ,.,i : .,i : .””distillation column with ‘13 ’ trays.distillation column with ‘13 ’ trays. Use the goneral noneqcl.aturaUse the goneral noneqcl.aturaI,I, ..;a ...;a .

developed in &ample 4.13 for the, ideal bin&y dittillation,, ,!developed in &ample 4.13 for the, ideal bin&y dittillation,, ,!; ,,I; ,,I ! .! .

(4 state your assumptions l !il: ,). ..,*;;,, -:(4 state your assumptions l !il: ,). ..,*;;,, -: // ,( , (/ .,,( , (/ .,

(b) Include the dynamScs of the condenser and,, reboilor ,cons%der&(b) Include the dynamScs of the condenser and,, reboilor ,cons%der& ,i,i* .?* .? ,.’,.’

them as perfectly stirred tank cooler and heater, respectively.them as perfectly stirred tank cooler and heater, respectively.I: ’I: ’(c) What relationships do you need in addition to the’balance(c) What relationships do you need in addition to the’balance

equations?equations? ‘:‘: ..‘I..‘I

(W

23.

24.

i.

(a) -List the assumpt;lans you will ‘make for modeling the system.,

(b ) Ident i fy the re levant balance,;equations.

(c) :,I”, addition to the balance eq&tions what other, relationships

do you need to complete the state umodel?. . . -.

(d) Show -‘how you can find the time when the composition, of the ). ~overhead product has reached a composition ;:, yA. (molar fract.ion) . , .’: .,

. , E. . .,,

12,

(d) How m&y degrees bf freedom does the system possesa?.

(e) How many control objectives can you specify?

. .

L‘

. _- -I J L

ioo*psi \

Q, (lblmin)

‘- -

Fiqure ‘P.p- 5

,

8

t

I I FR : cons-kud

Figyw ?‘.0-8

. :

CST’Rr~,FR :

- I I AL’ -..a- 1

I

FeLiquid S.ream

dy; ‘.**-:.- :.I 1. *

. . . . . . .. .. . --, . . . .. ’ :. *. ..- *.. . . : ‘. -.a..:. -Fiqure P.X -13

*. , a.v ‘.‘ * . :. : . .. .** *. ‘1 :‘. ..* . . *. .* .. ..*. --:., ..*.. . .- . ..‘.

Liquid t A

1

: ’

:. , ,

.,;-”

PART III*.

ANALYSIk OF THE DYNAkiIC BEHAVIOR OF CHEMICAL PROCESSES/

In Part XII we will study the ctynamlc: rind lwl~nvlor of SCVCViIl , f3 iniplo, ‘~

process%ng sys terns. Understanding the dynamics of such simple s)tstoms a.l.low61

us to analyze the behavior of more complex systems such as the chemical

processes. _'.I ,) :

,The analysis is limited, to linear dynamic sysbems; T@s may se& incom-

patible with the fact tha<'moost of the dhemical'eng&k&&ing processes are'.) _, _ /..' .~ ",, ,.. Imodeled by nonlinear equations,' Rowe$or, linear technique; ,$e'very v&luab&e. .:* . _,:I,.' ,/ : $,,j .;and of great practical importance for' the follov~hg reascns:. (a)..:There is no3 ';.general theory for the analytic sokution of nonlinear' dif'ferential e@nations,

, I " : .; :"."and consequently no compreh&sive analysis o,f nonlinear dynamic syst+ms.,. ,I : I...'-ip ,. _,. ,. j': s*,(b) A nonlinear sjfstem can be ad,~~tely,appr~~~~~i‘(?d by ~linoar system nearI' : ., 1%! ' I ,'some operating-conditions; (c) Significant 'kdvancei in .,the'~inearl,;eontrol

I ' '!..theory permit the synthesis and dsisign of.veri effective co&llers even for.'

., ., -: !",nonlinear processes. : ,) *

Fundamental, therefore, is the concept‘.of linoariMtGn.and. t&e&to;$

cedure for approximating nonlinear s&&s with linear &ea~$hich%ill be

t

studied in Chapter 6.:/ ', :'

The Laplace transforms, which we will discuss $n Chapter 7, ,offer a veryf.

simple and elegant procedure ,Wsolve linear differential. equations and con-

sequently analyze the behavior of linear systems.

In the remaining chapters of Part III we will cover the following items:

- the development of simpie input-output models for chemical,processes,, using

the Laplace transforms.

. .a ; :

- the dynamic analysis of various typical pr'ocesses and their stabilityb

characteristics..

. i

.f”,...__,

,/CCMPUTBR SIMULATION AND THE LINEARIZATION OF NONLINBAR SYSTEMS_ .-

‘ ‘Q ; , ;r:’ I

‘, InJorder- to,‘find the.dynamic behavior of a chemical process, we have to; : LG., ;. ” . : 1’‘integratet2the state,equations used to model the, process.;,. But, most of the,

.1,’processing systems that we will be interested in are modeled, by nonlinear dif-

ferential equations, and it is well known thatthere is. no, general,mathematicalrI

theory for the solution:o,F .~nlinear.eq~iations.,,J ‘. ,. Only -for lineat differentialI

equations are closed form,~. ‘analytic: solutions avail,able. ;,,’ :,’ ,

When confronted with dynami.c,analysis,of nonlinear syst,ems, there,are.II ,‘L’ $several things that,& Gan ,do ,such as: ii 3 t

(a) Simulate.‘the nonlig~~r..sys,t~,,osl,an a-nalog or digital .compu,ter and^ ‘. ‘..>.compute its soiution num+ically.

~.(b), Transform~the nonlinear;~,&%tem into a linear one by:an npproprinto. 1,; ,,’ :

transformation of its variables. ’., “;.I ’ ..,..-

r ‘(c) Develop a linear &de1 which approximates the dynawi& behavior of-a

.Inonlinear system in thesreighborhobd of:epsc~~isd’~~peratfnjg, ‘&nditions..

$ ,. ,’: The alternative.(b) @an be used in v+xy fW M~~~~i~~.ir~~~~nathee (a).- )

and.(c) are, in principle, .alwaya feasible,.I

In :thih &a&r wal‘-+$‘dieeuss

the computer simulation of nonlinear processes very briefly~be&&Wse iq is a’< ..j‘>

subject to be covered primarily in a course’on numerical an&lysia:~~PIar~e::+

emphasis will be given on the approximation of non&ear model’s by line& ones.J. 4 ) 1. ;It should be noted that all the-theory,‘tor the design of control sy.@&n,

available”from pa,st work, is based on linear systems, and,th&‘very small \t

I L_. .:, 1advances have been,made.f towards the divelopment of a control theory for non+

linear systems. ,.

/-

6.1 COMPUTER SIMULATION OF PROCESS DYNAMICSI - i,

Nonlinear differential and/or algebraic equations cannot, in general, be‘. 1 1, (1, I

. . : :I .,

solved analytically, and computer-aided numerical solutions are required.. .l Numerical solutions are also preferred for the equations which can be solved

analytically, but the analytic'sbl'utions are very cim@ex 'and provide little!

insight Wthe b&&v&r of's system; '. i.

Let us consider two pro&e&es de have already modeled; the 'c‘ontinuous,-

stirred tank reactor‘and the ~dear,"~~~ary-diStfr~a~~~on 'column. ', 1. _1

The model for the CSTR (see E‘iiampke'4.'10) 'is given by eqns. (4.8a), (419a)

and (4.lOb). :These cdnstitutb a‘ set of nonlinear,equations for 'which 'there,is

70 analytic solution available.‘. Therefore; iri order to study t!he dyna&: '

behavior of the CSTR, '& must solve the%mxk&ixig eqilat.~aire'n~~eri~a~l~ using ,a*. ,I' (.,.,fpmputer. . :; ,‘I .Y:;.,, ,. !

'. .. .,:. ., (

The model for the ideal, binary d&stkll&t$on: coltin'(eeeik$ca&$a~4. 13)‘As

composed of + ::?,:.1;

2N+4 nonlinear differential, equaftions, akdl~~ ,.<>:'

2N + 1 nonl&near.algebraic lequatlons.. :' ..: ', : ? /.i ”

I It is not 'only the aonlinear$ty of, the eqwtlons but alsothe sise of model

(24 differential and 21 algebraic equations fo!r.a-.,modest 10 tray cdlu&").~-that,i

necess$tates a numerical solution fin order-t& study'the dytiamgt: behsvTor,of

the column. I -/ i I .1

Today, computer simulation-is uhed~kctensively $6 aaPal?ze the~‘dynamki of

chemical processes to aid in thedesign 09 controllers and &tu'dy~:thei~~:. I

effectiveness in controlling a given ptb&ess. 'I Ankiog'.'and d'i&it$& computers

Ihave been used for this purpose'with'the emphasis hia;ving‘sjlif-t;?d'-almost

entirely in favor of the digital computers.

I

,; *

historically, analog computers were the first to be used to simulate the. .

dynamics of chemical processes with or without control. They iermitted a ’~

rapid solution of the modeling equations, thus providing useful insight as to+. ‘:. I ,.

how a process would react to eiternal disturbances ir how effective was the/,. -1, ,,

control of the process using various measurements,.:‘imanipulated variables and

1: :control configurations. The analog computers have several serious drawbacks:

‘, / Y .’( a ) Requ i re s i gn i f i cant t ime t o s e t up the pioblem’and get i t iunning.

,,:, , + $(b) The”need ‘of one hardware element per mathematical. oberatioh prohibits the

/siinulation of ‘large, complex systems.

i: 3 * , i 5 ,2,(c) The nonlinear terms are simulated

i _.by rather ‘expensive %ardiare’ ei’d;ents “(function generators)“bi.th limited

flexibility . -” ’(d) They do not possessmemory like’the digital cohputers. iheI “) I’,!’ : 3: ” * .:,.

subeequent revolution;“brought ‘ibout by’ the digital eorniuters’i ‘made ‘the ‘analog,s

computers obsolete,, y&;. & ‘a 3 ; ;*i iE$? i t Y,q<.’ ,; ,*‘, .<:: >

re sti?{I used’ in a”nme,Il ..$caFe and’.$mari.ly ’_ ,_ ., ’ ‘

to train operzitorq‘ on the ‘dynamic opetiation’ Sf chAic% kiints. ,’.I ,.’ I”.

* :,-, ’ . ..I*.’ . iThe computational power introduced with the digital computers,‘;~a$Song wtth ,

: T‘ jhas expanded tremer@ously,” the scope

! I,. ,. 9. -*<; r&hi: _ i ‘i I ! ; ‘c ‘.~ -2 4% “ : ..1.. * I ( : 55 f ” , ?_.j’and the practical significance ,of computer simulation for prcc4s@zdmamics

.* 91‘.,,,; )) 1 __ j s , . : ; “? ! ! :! : J I ?, 1” ,.!

and control. The availability of sophisticated equation goAGing foQtines for:., “i$.$, .s, : 1; $&, q, ,’ I

almost every digital computer sy.stem availabe has simpli.Eled the’required. , . ‘. 2; ,J, )j a I . ~ i c. i

groundwork for process simulation and has relieved the engineer from the needi’.: aI , \ j_ ,;, ! ..*1

to be an expert in numerical analysis.’ :,’ I . .‘,I . . . ,

Digital computer simulation of process dynamics involves :the solution of. .it. ,‘. ‘. ‘(,/, : \ I ‘. *s:’ ,

a set of’ differential and algebraic equations which describe the process. li :. ,

There are several categories of numerical methods which can be used to”;_ 4.’ I,

integrate differential equations and solve algebraic ones. Let us examine/

briefly the simplest and most populat hong them.

* : *

‘

A. Numerical Solution of Algebraic Equations. . _: I 4,“.

At steady state, the state equations turn to simple algebraic equations,,, .. / .;

since the rate of accumulation becomes zero. Therefore, in order to determine1 .$ ,’ .:I

the steady state behavior of a process under given conditions, we should be1. 1 .; : 4

able to solve sets of algebraic equatkons. All available methods use an,’ II .b,. 1’. ^ .:

iterative trial-and-error procedure, which approaches (hopefully) clo.ser and.’ :,;. . \i$.. i L jv .,

&oser to the solution with each iteration. The key question is to se1t ’ : .dPg the

appropriate method, which for the given set of equation: converges ra$idly toi :

the correct solution.i ,l‘.:,’ ‘1 ,/‘. “l”

Unfortunately,,‘*>‘* this is a very difficult task and. in all,. .$. : i. +. &:

but a few instances is impossible to know % priori how.;successful will be’s,a* , j j i ., -,: ‘*‘. i : : , : t L .j

method in finding the solution to a particular set .of equations.* j +, r; > Quite. of ten

I . *I: Ii ”a method will not ,converge to the solution, or.& in other ins,tan&es it app?oaches‘> ., * % ’ I .;; 1 , 1 ’ ‘. ‘3 _‘

the solution very. slowly./_ ,.. v: Among the most of ten ,uoed.‘.t, a f ..i “,.I;, <r$j!y! *‘., ‘, : ‘” :’ : ,hniques are the*- ,- ,_,. .ii

following : (4) interval halving) (b) &nxeeslve aubr &‘tutio?%+ drnd (c) Newto?-I_. ;‘#. ,..I .3 j , .” ” 2,: * _ i: “- ;’ .s !’

R a p h s o n ..X‘,

‘3:‘, . ‘.,,f‘ ”Wi;me&ai Integratidn o f Rif#.erential.

.$ : ) ..: : j: y i ,: \.’B’ . IZe uations

z. ’ , : . ; ‘\ T, .,,’ t .j : , i ,.,.\ :r’*, >.C’ : :.:

Here’ “aga$ ‘& have a .very large &mber of availab&e techniques. * Numerical,s-: j h t ‘1;i . i , . / .’ 1 .-A,. j’ ;r ,; c”,,* ,< ,:.I

integration +mplies an approximation of;. the. continuous dlff er‘ential equatio&; 2 ,‘. ‘1 ye,,,%!

with discretei,f inite-difference equations.*T. ~: ‘,.’ , , , * 7‘The: various integration methods

: ‘-3 ‘, :,, ;. ;: -j, ‘4.differ in the .way they implement this a&o&atlon. Thus, we have explicit

:i * , ’ *: : .,~, : i I I 1 ! e, ” ! * i..method,s which march on in time yielding the .sofution tn one. pass, or we have

J;‘; * 5 _/r i/ S’.T h e k e y q u e s t i o n s Eo’r.i:fn,Rlitiit merho$s with predictor-cc&rrectQr capabilities,

-, . ., r,: 3 ,i.;an integration technique, aie the itabiiity of! the procedure and the speed with

ha: .(: ,. ;* .!, ._,. , . ,:i ),-; :which .it reaches ,.the solution,

;.. .’But again, these are questions which, in

, .’ ,’ (r I,, ,,‘.‘$ ”general, cannot’ be, answered to our satisfaction ahead of time. ’ Among the most.

:pppular integration methods is’ the explicit. fourth-order Runge-Kutta which pro-

:.vides satisfactory accuracy and stability of computations as well as low cost.

:i

j‘,

I,.-t. .~ ;

,.’ .,.;::,j..+ s .

-

: Digital computer simulation of the chemical process dynamics .is used/

extensively’ at the present. It allows the engineer to anticipate the’behavior.:.,

of a process not or&. qualitatively but also’ quantitatively. It has helped to/.1 ,:.

design more complelt and sophisticated control systems. The‘major drawback of

computer simulation, is that “it only gives you,.numbers” and not a generalv

analytic solution in terms of arbitrafy, unspecified parameters which in turn;/j ,

you taylor to your’ particular prob&m. ’,Theref ore, the results of computeria

simulation are of ad hoc nature , and you will have to make several runs with.‘ : ‘_ : ,

different values for the input va&&es and parameter8 before “you can estab-i’i” ,j

lish a good understanding of the dynamics of the particular ‘procese..,j .%

: . 1

6.2 LINEARIZATION OF S,YSTRMS WITj;: ONR ‘;ARiAF&

v,., ,; ,” ‘,C

JL i n e a r i z a t i o n ia the process by’which’tbe approz&at!e nor&rear &ten%

with linear ones. It is widely used fn-the,s’tudy of procea&‘dyndmics and‘

design of control systems for the following &(ai8sons: , ,.I , : . . ~ ./ /.(a) We can have closed-farm, analytic. solutions for linear q&terns. ThUS,

we can have a comRleQ ,aqd general picture of ~a process *, behavior .j i

independently of the par titular .values’ &.&the par&ers and< input‘_

variables. This Is not possible for’ n&linear systems,: and comp~u.ter(I ,.

simulation. provides us only, with.$he ,behalt%or of the .system at specs- *

f ied values of inputs and, par&neters. >’I i: **. ,’ I

(b) All the significant deV$!&qprR~ntS towards the design of. ef&titive &ontrol‘~:j

systems have been limited ‘to;‘line$r epteqe. .,,:,

.:': _ ,i .:.,:;: I': :

First, we will study the linwhatioir of a-+'pot&Ln@p ikqudtion :wi.th.one

zvariable and then we will extend it, to,multivariable systems’. “1

Consider the following nonlinear differential equation, modeling a given

process :

d x7 .= f(x)dt : -'--' ( (6.1)

Expand the nenlinear function’ f(x) into a‘Taylor series around the point x.‘,,

and take: 1i :.,

‘. II. .:

+,.~ + (&f) (” :: xo)n +;;,*

dx” “o n!‘(6.2)

? .I ~.. ,

If we neglect all the terms of order two and higher we take the followingr‘ \i Iapproximation for the value of f(x):: ’ ,_t;, , , ,:.,. )

.:f(x) 7.z ‘f(Xo) ’ + ($jx (x (6.3)

,o-ox,) /’

‘7: “: ,, 1 ‘. ‘ 4

ai /

It is well known that the error ,introduced: in ths appkoxima$i& (6.3) i s o f,. __ 1:

the same order. of magnitude as the term ! j + j I :‘ f I I~2.,

Conscq uen t ly , t h e linear approxZm&tioti ‘ ( 6 . 3 ) ~~i“s~‘t’is’f~Ct’oiy ‘ o n l y tihen i is

very &lose to xo where the’ vai&ue of the t’&m I : is: very small. :::’

Tn Pigure’ 6.1 w e ten sco .tl\o nonlihaar $unct$on f ( x ) )-and its’ J&tear,

linear approximation depends on, the ?lbcktian “of the point xo “around- &ich we

make, the expansion into a Taylor series’, .~~@@~e .the’.+&r appro&m;l;ion bf/I ::I::

f(x) at the points x. and xl (Figure aC.1). ’ %z$ kpproxi@tion is’ ex& t only

at the point of linearization. .: ’ -L. I’. ‘ ‘E.

In eqn, (6.1) replace f(x) by its linear approximation given by &qn;0,(6.3) and take, I’ 1 i

i’

dxdt - f(xo) + (glx (x - xo) (6.5)

0 a. I. .

This last equation is the linearized approximation of the initial dynamic

system given by eqn. (6.1). In later chapters the design of the process con- I‘

troller will be based on such approximate linearized models.

,

Example 6.1

Consider the tank system shown in Figure 6.&, The total ma@+! balance,, ,,,

yi&&,.*

. .’dh /

A dt = Fi - F. (6.6)i ._., '.

where A is the cross-sectional.area of the'tank,and h'the height of the

liquid level. If the outlet flowrate F is a linear funqtion'of the liquikf's_,, 2. ;> : !

level, i.e. ..,*I '.>

F. - a h, where a - constant (I_' , .II

: .i. . ,a I

which is a linear differential equation (modeling a linear dj&&c system) and"' ! . _‘,

no approximation is needed. , i ,' ',:

If on the other hand; I/

F. = B

then the resulting total mass balance yields a nonlinear dynamic model,total mass balance yields a nonlinear dynamic model,_._. _* ,' :_* ,' :

VfbFi :VfbFi : '..' (6.7)'..' (6.7)

Let us develop the linearized approximation for the nonlinear model. The only.,

nonlinear term in eqn. (6.7) is 13 v%. Take the Taylor series expansion of

this term around a point ho:

pig- i A0 + k (6 qh h +-ho> + $ (B ☺I;>�

as0

.,[ I: ( h - h )2( h - h )2210 + l �**lo + l �*

h=hh=h 0 ?0 ? ::

-sdi-sdit: o + ‘A-(h ‘- hi) + :,“’ fst: o + ‘.-dt-d(h ‘- hi) + :,“’ fs (h - ho)2� + l .;(h - ho)2� + l .;

Neglecting the terms of order two and higher, we takeNeglecting the terms of order two and higher, we take

h =&PO + m-k-’ ih - h o ) :

2%I, 2 , I,

which, if introduced in the nonlinear dyn8mic system (6.7) yields the following: , P

I, 2 , I,which, if introduced in the nonlinear dgn8mi.c system (6.7) yields the following

: ,j

Qnearized approximate model:Qnearized approximate model:"”

Adh+.di-hsF _Adh+.di-hsF _dtdt 2%2% ii vii-vii-

j 3 i O >p" :_ . ,#,, >.$ '.j 3 i O >p” :_ . ,#,. 2.s ‘.(6.8)(6.8)

.. !'!’

Ifl.i Let”us compare‘ the linearized, approximate’ m&h& given by eqn. ” (6:g) to3. P

the nonlinear one, given by eqn. (6.7).. Assume that tha tank is at steady

state with a liquid level1’ *,~j

Then at time t=O, & ‘stop the-supply of liquidh0 �

fl.i Let”us compare‘ the linearized,

Iapproximate’ m&h& given by eqn. ” (6:g) to3. P

the nonlinear one, given by eqn. (6.7).. Assume that tha tank is at steady

state With a liquid level h1’ *,~j

Then at time t=O, & ‘stop the-supply of liquid0’

to the tank, while we allow the liquid to flow out,.;

Th&,’ a t r-O th,o l i q u i dto the tank, while we allow the liquid to flow out,.;

Th&,’ a t t-o th,o l i q u i d

level is at the steady state value, i.e,” h(t=O) -‘ho;;: Curwe ‘(A) .$n Figure;

6.2(b) is the solution of eqn. (6.8) and,“curve (B) in the:: s&e figure ,is the,a> : ./ _’ : ,:i~ ,* 9,. ~solution of eqn. (6.7). We notice that the two curves are very close to each

.: ‘.iother for a significant period of time. This indicates that the linear$zed

I ,.

model approxim8 tes at the. beginning very well the nonlinear model.

1

level is at the steady state value, i.e,” h(t=O) -‘ho:! Curwe ‘(A) .$n Figure;

6.2(b) is the solution of eqn. (6.8) and,"curve (B) in the$s&e figure ,is the,a> : ./ _' : ,:i~ ,* 9,. ~

solution of eqn. (6.7). We notice that the two curves are very close to each.: ‘.i

other for a significant period of time. This indicates that the linear$zedI ,.

model approxim8 tes at the. beginning very well the nonlinear model.

1As the time increases and the liqudd,level continues to fall, fts value

,r;:i I / . ..h 1 deviates m&i’ and more’ ;rom the~~&ri&al value ho

‘, .,; - .around .which the

As the time increases and the liqudd,level continues to fall, fts value,r;:i I / . ..

h 1 deviates m&i’ and more’ ;rom the~~&ri&al value ho‘, .,; - .around .which the

linearized model was developed. Figure ,6.2(b) indicates very ‘&early that as

the differ enca ‘ho - h +ncrea8es the. l%:neolriaed appro)cimation becomeo ;pro-

gr’oesively less: accurate, 8s ?a0 expected. __s

linearized model was developed. Figure ,6.2(b) indicates very ‘&early that as

the difference 'ho - h +ncrea8es the. l%:neolriaed appro)cimation becomeo ;pro-

gr’oesively less: accurate, 8s ?a0 expected. __s

6.3 DEVIATIO?j VARIABLES6.3 DEVIATIO?j VARIABLES

Let us now introduce the concept of the deviation variable that we willLet us now introduce the concept of the deviation variable that we will

find very helpful $I% later chapters for the control of processing systems.find very helpful $I% later chapters for the control of processing systems.

Suppose that xsSuppose that xs is the steady dtate'value of x describing the 'initialis the steady dtate'value of x describing the 'initial

dynamic system (6.1). Then,dynamic system (6.1). Then, ,, :: '.'.

-, . dx&'-, . dx&' III,I, .;.; ,,

(dt) - ! - f(x$) : .(dt) - ! - f(x$) : . '/ ;'/ ; (6.9)(6.9) ,

Consider xsConsider xs the point of linearization for eqn. ,(6.1), .i.e. x. E xs. Then,the point of linearization for eqn. ,(6.1), .i.e. x. E xs. Then,

eqn. (6.9) yields the following linearized model,eqn. (6.9) yields the following linearized model, "*"*,i,', ,,i,', ,dxdx d fd fqq x -'xx -'x= f'x#K) ,+ \dx'xg: ( . . $= f'x#K) ,+ \dx'xg: ( . . $ )) (6.10)(6.10). .. . ; :.; :.

Subtract eqn.Subtract eqn. '(6.9)?from,,(6.10~.~and't~~~' ''(6.9)?from,,(6.10~.~and't~~~' ',, _.,, _.

d(x - xs)' .L :" .d(x - xs)' .L :" .(df, ')(df, ')

:.i. : / a:.i. : / a .* ' *.* ' * ii'.'. i,2 ,.i,2 ,.mm

dtdt dx xsdx xs(( x - xs)x - xs) ., * 3., * 3 (6,. 11).(6,. 11).

If we define the deviation variable x' as,If we define the deviation variable x' as, ,, ,,,,, ,,, :<,.:<,.

j('- = xt c xj('- = xt c x 3.3. ,,SS ..*..* ii

then eqn. (6.11) takes the following 'form:then eqn. (6.11) takes the following 'form: "'"' '.*' i): " "'.*' i): " "

dx' ="(qdx' ="(qdtdt dxxsd -' S.&A *dxxsd -' S.&A * iiii .y (6.12).y (6.12)

-- .u’.u’i ::. :i ::. :~<~< 55 r ,‘.(r ,‘.(

Equation (6.12) is the linearized approximation of the nonljnear‘idynamic sys-Equation (6.12) is the linearized approximation of the nonljnear‘idynamic sys-. >,. >,t,. ,,'.'Pt,. ,,'.'P " " ;" " ;I :.I :. (.(.'St: -'St: - j .'j .'

tern (6.1), expressed in terms of the deviation variabletern (6.1), expressed in terms of the deviation variable x'.‘x'.‘' ,'' ,' I 'kj! *I 'kj! *.‘.‘

The notion of the.deviation variable is'vesy useful in' procesd,control.The notion of the.deviation variable is'vesy useful in' procesd,control.'?'?

Usually we will be concerned with maintaining the value of a process variableUsually we will be concerned with maintaining the value of a process variable‘.i .'‘.i .' .' ;.' ; .' a, T.' a, T I:,I:, !! :I:I 1

(temperature,(temperature, concentration, pressure, flowrate, volume, etc.) at some desiredconcentration, pressure, flowrate, volume, etc.) at some desired

steady state. Consequently, the steady'state becomes a natural candidate',

point around which to develop'the approximate llne&&ed &del. in such cases

the deviation variable describes directly the magnitu,de of the &l&cation of

1

steady state. Consequently, the steady'state becomes a natural candidate',

point around which to develop'the approximate llne&&ed &del. in such cases

the deviation variable describes directly the magnitu,de of the &l&cation of

a system from the desired level of operation. Furthermore, if the controllera system from the desired level of operation. Furthermore, if the controller

/c,

,of.the given process has been designed well, it will not allow the,process.

variable to move far away from the desired steady state value.. Consequently,

the approximate, linealieed model exp,ressed in. terms of, daviat$on variables,.G

will be satisfaqtory to describa the dynamic behavior of the .process near tha

steady atate. :r I.

In the subsequent chapters we will make extensive use of'thelfnearized

fdrlr5i Of dll:Furl!ntl,ill cqilcltions, in terms of deviation variables: ' "

r'! ,-

3;,Exumpta 6.2 al

:: I’_/:

Consider the linearized model of ‘the; t&k &is tem (given by eqn’. ,.(6.8), of’ ’_., ._/

I

1

Example 6.1. Let hs be the steady st,ate vakua of the.llq@l level,fcw a,, (,,,.’ .,,._,given value, Fi,s:* of the inlet flowrate, Fi.: Then, the linearlzed.model ^,

around ,,hs, i.e. h0

5 hs, gives

*-J&g + s ,J'd t 2 ~~ - Fi -: 2

"'.$ i ,,.. ,; V3)

zi G‘. 7 ./i

At steady state'from eqn. (6.7) we also'yhave. ".-igr ,: ,~ 'e, : , .,'F ,.

.a..

A-$+Bq* Fis 51 _',

0 9 - !

Subtract eqn. (6.14) from (6.13),* :' '.

2% ,(6.14)'

',i :* * 5 ii . 14, , .,; ;

d(h w,he)A ,df c + ---@--(h-h,):-':.

2’1m;;Ft. :- a,;, ) ,,. :

1Defining the deviation variables

.I ,.; . . 1,:" .

,, " )' 6 ';ii ish-hs ^' ~and @i,.' - Fi.( : t Firs

/we take the following linearized form interms of deviation variables

(6.15)

,

6.4 LINEAJ.UZATION OF SYSTEMS WITH MANY VARIABLES

In the previous sections we developed the linearized approximation of a

nonlinear dynamic system that had only one variable. Let us noG extend that

approach to systems with more than-one varf-able. ’ . .‘( . ‘-

Consider the ,following dynamic system,

‘Fround th’e poind (k, n ,x,h) _ and take

a&f2+c 1axlax ~~~~~~~~~ (x 1 -x )(x,-x2()) +*-**lo

Neglect the terms of order two and higher and take the'following approximations:

aflaflfl(XiSX*) T fl(Xl~‘X20) f (-)fl(XiSX2) T fl(Xl~‘X20) f (-)axI (Xax1 (X

“4 ’ (X - X“4 ’ (X - Xaflafl

1(p*$ 1 10~(p2~) 1 10) + (-) + (-ax2+x ‘ Xax2+x ‘ X2#x2 - x20)20)(x2 - x20)

**1o,1o,

,.,.and ’ -,and ’ -, C’C’

,1,1 ><>< ‘!‘!.I.I ;; . .. .

af2af2 .a+,. ,:.a+,. ,:f295) tf2(x+2()) +~~)(xlo’x~o)~~~-~~~~ +(;i;;;J(xf295) tf2(x1(92()) +~~)(xlo’x~o)~~~-~~~~ +(;i;;;)(x

,,’ ‘CL.,,’ ‘CL.10;2c~o) (x2 - x2*),10;2c~o) (x2 - x2*),

,:~,:~.,” .( :.,” .( : .. _-_-Sub,stitute t h e a b o v e l i n e a r a p p r o x i m a t i o n s o f f,b(x1,x2) a n d f2(x1,x2) in&Sub,stitute t h e a b o v e l i n e a r a p p r o x i m a t i o n s o f f,b(x1,x2) a n d f2(x1,x2) in&

11the eqns. (6.16) and (6.17) af ,the init& nonlinear dynamic system andthe eqns. (6.16) and (6.17) af ,the init& nonlinear dynamic system and take:take:

dXldXlx=f “(Xx=f “(X1 ~~,X2~)f~~)(xl~~x20)(Xl-Xl~)+~~)(x,~~,x~~)(.i‘~2~)(6*l8)1 ~~,X2~)f~~)(xl~~x20)(Xl-Xl~)+~~)(x,~~,x~~)(.i‘~2~)(6*l8)

;- dx2 ,+ ’;- dx2 ,+ ’- = f&&-J- = f&&-J

af; .,. +,:I.,af; .,. +,:I.,

dtdt:) + ;(---):) + ;(---) ’’ (x “X(x “X

:af2 ; ‘, ,’ . i:af2 ; ‘, ,’ . i

ax, (x?-y20) 1,ax, (x?-y20) 1, 10,10,) + ‘ax,l(x,) + ‘ax,l(x, (x(x,:- pyre?.. 2,:- pyre?.. 2- ~20) (6.19)- ~20) (6.19)

**These last two equations are linear,differential equatiod .and constitute theThese last two equations are linear,differential equatiod .and constitute the

linearized, approxamate model of the initial: nonlinear sytftrzim des&ibed’ bylinearized, approxamate model of the initial: nonlinear sytftrzim des&ibed’ by,‘> (,‘> (

eqns, (6.16) and (6.17).eqns, (6.16) and (6.17)._’_’

,-,-ii

The comments made earlier for the one-dimensf,$nal .&a& ?&ly also here,The comments made earlier for the one-dimensf,$nal .&a& ?&ly also here,, +, +

: .:;: .:; ‘1‘1‘I.‘I. ,i,ii . e .i . e . ;;

,,

- the approximation deteriorates as the point- the approximation deteriorates as the point (x(x x’) moves away from‘thex’) moves away from‘the1’ 21’ 2

po in t (x10,x2$ o f l i n e a r i z a t i o n , a n d Ipo in t (x10,x2$ o f l i n e a r i z a t i o n , a n d I

- the linearized approximate mod&depends on the point (x10,x29)- the linearized approximate mod&depends on the point (x10,x29) aroundaround

which we make the,.Taylor series expans&;which we make the,.Taylor series expans&; LL, :\’, :\’ )kL) ‘1 ^.)kL) ‘1 ^.** kk

Let us now express the linearized sF;tem in terms of deviation variables.Let us now express the linearized sF;tem in terms of deviation variables.“.“. II ,.,.

Select the steady stateSelect the steady state (x(x‘.<i‘.<i l,s ‘~2~s) as, the point around which you will makel,s ‘~2~s) as, the point around which you will make

iithe linearization, 1. e,, in eq&. (6.18) a&j (6.129) putthe linearization, 1. e,, in eq&. (6.18) a&j (6.129) put ‘x‘x .: E x.: E x

1O1O 1 , s and1 , s and

x20x20 = x2,s’ - := x2,s’ - :At the steady statem eqns. (6.16) and (6.17) yields:At the steady statem eqns. (6.16) and (6.17) yields:

0. -0. - fl(xl,s,x2,s) .fl(xl,s,x2,s) . ;.;. (6.20)(6.20)

0 -0 - f2(xl,s,x2,s)f2(xl,s,x2,s) - t- t.;.; i :;i :; (6.21)(6.21)

99

Subtract eqn’, (6.20) from (6.18) and (6.21) ‘from (6.19) and take:‘. ,., ~ r, , hrx ) -‘~.’ , ':"j' af

I~$&&& (1)axi (x l,~,x2,*)(x1-~l,a) ~,~~~~,l,*,x2,a)(“2-,x*,~)~~;22)

a n dc

l,s,x2,s)(x1-xl,s) + (~)~~l,s’X2,s)(XZ-.X2.a)(6*23)

D e f i n i n g the d$v+ytion var,$ablea b y ,* 1_/*

Xi 3 ,xl - x~,~ and *; = .*2 - XQg

eqns. (6.22) and (6,23) take the following form,in terms of’devlation variables:

-dXi

dt

, i %,,dx;

dt

where

hi

- i!' :, .

3 aiixi + a12x; *‘,L

3; :~ .: ‘- ‘I ., .+ ‘_:

‘akxi + “22x; : ’s, j_.

‘: . ~. c :.p ,.-I, *es;i

af- ' af.

” a2l “=. ‘axl)(xl s,x2,;) ,’ a22 - (q(x; s,x2;aj* :9 9

; .;i A final comment is in order. IN the previoue andiG&e.p%&ent oection,s we,, /! x. 1‘, 1’considered the presence of state variables only in the nonlinear &nctions.

Thus, for systems with one variable we had oniy the stateir ‘ii8.. ‘ill, .

.x’>‘&d for systems

a n d X2.j 9 ,. ‘,T :.;‘, f, <+’ ‘

with two variables we had only states xl The above formulation

should‘not be ‘perceived as ‘restrictive,.:t;i

*but it is’~~~ily,,expanded’-to’include ”,’

the presence of other ‘inpu@iariablis, like.the tna~~p~l~t~d’v~riablsa, and the? ,

disturbances. j1

Example 6.3 ,., ‘: ,,L’;

., Consider a dynamic system desciibed by two state irariables xla n d x2

and the following state eq’w$itino;, 1‘_

dxl,’

- z fl(x1sx2~yn2sdl) Id t

dx2 i= ,_, '. . 9

dt'.I

fZT”L,x2;ml,m2’d2)

w h e r e m1 .a n d m2 are two &nipulritedi variables aAd

distukbances, affea$ing the ,syst&m.’ ’ .‘I’ :

Linearizatidn* of,. tlqs above .&&ions around the nominal valu&..r

(x10sx20sm10s~20sd10s~20~ wi l l y i e ld : 8 ‘.,.!

dxldt

afl , afl= f1(x10sx20sm10s~20rd10) + (T&~ lx,- xl01 + $qo (x,- x2$‘**

t.. .

,

f (<). (ml L ml01 + ($flo (m 7 m20) + ($,n,v(d, -n.I. L, - - -- - dl(+

1(6.$)

i iand _1 3, .,

.i

dx,-g’f(X

!a_c

2 10’X20sm10Sm20~d20) + (a~_ n2) (X, Lap--

‘3. ” )rA” ’

1

wh-@l-e all the derivatives have been camntr~d a+ +h- n~4..+ -e ,.,-r--.t--ll--

(denoted by the- subscript 0).

Assuming t h a t t h e p o i n t o f linearization ia +hp n+a,Ar ara~a L-L-.-J-- -c

the system, we can define the deviation varfahlem a,**

xi. *; Xl - Xl0 mi p ml’ - ml0 di = dl -’ did

Xi 3.’ X2 -’ Xi0ml,

‘=di d2 - d20- --

Introducing theie deviation vari&les in’l.eqns. (6.24) and (6.25) we take ’

‘. *t ‘_(- * sallXi f .~;ZX; + bllmi + bl2m; + Cldi . (6.26)

. , and ‘7 :,.”

_ dxi”’ . ‘.c ,t xi ,f_: tal$ + a22xi + bzlmi f bZ2mi + ~~$1 5. .C6.27)

.*, ,;+where the cons tante a.i-j ’ bij’ =i are the appropriate derivatfves in the

eqns. ‘- (6.24) .annd (6.25)) i.e.

and

I’ * .’ ’form that we would like to have for ‘process control purposes, i;e. ‘linearized,.approximation of the nonlinear state equations, in terms of deviation

,! .:_

var iab 1 es. \’ 1. I, * .

Linearization of a Nor&o thermal CS~TR,

Example 6.4. ”I I ,$!:: ~ r : ’> y>

tations for a CSTR were given in Example 441O'by eqns. '5The modeling eqt

(4.ga), (4.9aj and (4.10b). Assume that the volume V of, the reactlpg ‘i ‘I-‘,mixture remains constant. Then,. ‘the dynamic :model of the ,reactok is !relktced

‘_ I_. “to the following : ,

dCAdt is.= $ cc - CA)’ - k, e-E/RT

CAAi : .O” /( 6 . 2 8 3

I,

“a n d

,

,

‘ dTdt’ - f

L (Ti -T) $ ’ J l’o .-E!P;lTv Q - --$(‘p -,‘jj .qL 29)P I ‘,’ . I ? .? .~

This model is nonlinear due to the ‘presence of the no&near term .e-& a:

. ’ -- cAS.

while all the o.ther terms are linear. ‘*Thus, in order to linearize ,-n?qns.

(6.28) and (6.29) we need only to rinearize the above”nonlinear term around

some point (ci ,To).0

-E/RTo ale-E/RTe-E/RTc ‘, e

’ Ql

A 'A +( aT * )To' CA (T - Toi0 b 0 :* 0

~

.-1 .Id[*-E’RTc ]

+,(, 1 ac A)A

! ‘(CA-CA :, To,Q .O

i t.. -E/RTo kcA + (- e-E/RTo ”

-eRTo2 i

CA ) (&A To) f ‘(8-E/RT

O)(cA-CA 10 0 *. i : ; , * : 0,

,. ‘\I 1 -: : : ~.

SubetitutXng the above approximation into‘ eqns. (6 .‘J&y-‘&d (6,;‘;;) “;le i&e the

following linearized mod;1 for a nonisothermal CS:;i’R:,

dc.. -E/RT. E -E/R%’ -E/RT;. ,-*-+ (cAi,-~A) -toe -.: o.#kS,-koL;iZ,,e

: RTof!!!$+ To,.-&;& ‘$y- cA )

0 / 0,. ) : :: ‘j I.

(6.630)_,) - ( -):, : ,; i '& . 4 :: -,i. j

$+ (Ti-T)+ Jko: : “R >-E/RTo ’ :’ .CA’, .+ (- e

0 R$ .:, i 0CA )(T-To)+(e

+&To)

(6.31)P _

I

$e can proceed a stlp further co develop a more convenient form for eqns.’

(6.30) and (6.31) using the deviation variab’les. Assume,that To and CA‘^ 0

.

are the steady-state cgnditions for the CSTR and for given input conditions.J.' , "_

cAio, Tie, Tco. Then from eqns. (6.28) and (6.29) we take,

0 = $ (;;4"' - cA ),- ki e-E/RTo

.-cAO

(6.32)i0 0",.

I

-E/RT_ .;.. . I0 - + 'Ti

.t.-To)+ 8

QJko e

_ ., 'I .,/ ;a- o 'A0 ‘.- 5 (To-~Tc ) (6.33)

. . ,.p :‘o j

Subtract eqns-. (6,32) arrd (6.$3)> from (6.30) and:(6.31); respectively, and

:- i

\ '-R/RT_"(T-T~)-~~ e u(~A- CA )

,A : ,) '! !':.- 0.~.

. (6.34)

<take: _' ,,,

-d'[(.cdcadt T Ai

-c

,::

;' / 1' 4 .Ljl s ,, ,. : i‘i '

dTdt:?' [(Ti- Tie) - (T,- To,It

-E/RT-. ')(T+To)

.I-E/RT+ (e O)(cA-,CA) I -3

,., i[(T-To) -- (Tc-Tc )I. 1I e(6.35)I

0 P 0

:, *Define the following deviation variables:

IThen, eqns. (6.34) and (6.35) take the following form,. in terms'of the deviation

: !-variables:

dCA f . i , ; i ‘i . ‘%

.[, I

koE -E:/RT -E/RT--7 Cc& - cl> -, ~~~ !+‘. T - hoe, ‘1~:d t (6.361,.

:’ 3’0

I; . ..1

: ,,*- m 5 (T;‘- +') + r JkdT'dt l ci

I---g CT'-T;)

P

(, (6.37)_ '<-.

1 SUMMARY AND CCNCLDDING:RRMARRS.,. iir, 1.

Most of the chemical engineer&rig syst& ire modeled with nonlinear

equations .(dhfferential, algebraic);: .Slnestherk is no general theory for the;

solution of such equations, digital computer simulation is used widely to study

the dynamics of:,chemical processing systems,' Computer simulation is nothing‘ :;,

else but the numerical sofution.of the 'eq&ations describing the behavior of aL

process. *

Linearization is the approtimation of a nonlinear model by a linear one: :

in the neighborhood of 'an operating point.' It is~'baas,ed~'on'the Taylor se&8:

expansion of nonlinear functions around a certain point and the retention of

only first order terms, while second and higher order terms areineglected.,: -i$, j: :"

The lineaiized model provides very 'good',~Bprol;tnate'de~.~ripsidn of a ,process

only near the point of linearization.':

For process control purposes the steady sta& operation of,, &stem is

taken as the point of. linearization. This allows the-jatroductionbf the.9 I' deviation variables whose values indicate how much a system is removed fromii

L the desired steady state operation. The deviation variables are very useful,

.i, "

quantitiss ip control and in subsequent chapters we will model a2'

prOc888 alWay 'in tqm8 a$ deviation ~Varfidib?!S~~ .;, : tI’ ‘,

?I

i (y

I.’ . I .-_ ,I

‘. , , . . ii’ ; _ ,._’ ,) ;?f . ’ “. “‘,. :;.::A . ,

THINGS' TO" THINK 'ABOUT

1 . What-i&n computer ‘simulation ifnd what is it used for?

2. Discuss the methods of interval halving, successive substitution and

.Newton-Raphson for solving nonlinear algebraic equations. What are

their relative advantages and disadvantages?

1 . What-i&n computer ‘simulation ifnd what is it used for?

2. Discuss the methods of interval halving, successive substitution and

.Newton-Raphson for solving nonlinear algebraic equations. What are

their relative advantages and disadvantages?

3. Do the bame with Euler's and the fourth-order Runge-Kutta integration

methods. b /'

4. What is linearization,? I*/i

5. Why are.the linearized, ap&%imate models useful for process control:

purposes? : i

6. When is the linearized model more accurate, near or far from the point.1

3. Do the bame with Euler's and the fourth-order Runge-Kutta integration

methods. b /'

4. What is linearization,? I*/i

5. Why are.the linearized, ap&%imate models useful for process control:

purposes? : i

6. When is the linearized model more accurate, near or far from the point.1

of linearization, and why?,..

7. What is the most attractive‘pdint of linearization for control purposes

and whyi?

8. Which linearieation is rnoi~~ useful, the one around the point A or the,. ‘

of linearization, and why?,..

7. What is the most attractive‘pdint of linearization for control purposes

and whyi?

8. Which linearieation ir mote uskful, the one erouad the point A arc the,. ‘

one around,'the point B, and why? (See Figure 6.4-1). \’ .one around,'the point B, and why? (See Figure 6.4-1). \’ .

9. What are the deviation variables? What is the point of linearization r

in order to define the deviation variables that wiil be- useful for

9. What are the deviation variables? What is the point of linearization '

in order to define the deviation variables that will be useful for

process control purposes? ,. .., ,* I

10. Consider the tank system discussed in Example 5.6 (Figure 5.4)'where

the flowrate of the outlet stream is proportional to the square root of

process control purposes? ,. .., ,* I

10. Consider the tank system discussed in Example 5.6 (Figure 5.4)'where

the flowrate of the outlet stream is proportional to the square root of

the height of, the liquid level. Show that we should relinearize thethe height of, the liquid level. Show that we should rekearize the

balance equation-every time that $e change the desired l&&d level at-..--- - - .I -_--_e-...- .I--

s t a t e .steady

.’

.-

11. The following differential equations provide the mathematical modelst

‘ for several processes.. Which of them $$e"linear and which nonlinear?

Process I: +Apx - lot +'5 '

'1. .' b

dx,Process II: a- ml(t) t,d;(t)

1 dxl

I bl-z-+ b2

dxlProcess III: al= + xw* 1

I //,!. B-’ 14

J’‘< ,

5,’ 1

I CH,APT?ER 7',;: *I l.

? *it LAPLACE TRANSFORMS

,,:; IThe use of L&place transforms offers a very simple and elegant method to

I, & i '

solve the linear ar linearized differential equations.which result from the,#*; ,\S , .j,mathematical modeling of .chemical processes.

:", :. j ' 6 ::,

The T,aplaee transforms also allow;1 ~ : < .*, :: “), i"

.

v a r i o u s external.influences. '... .,,ld i. !&.i 3

It is for all,',the above reasons' that the Laplace trans,forms have been,- :

included in a Process Control bo$', although they constitute a,purely;,_ \,rY' k, -i. imathematical subject.; : : !:

7.1 DEFINITIQN OF THE LAPLACETRANSFQRM ;, ; .,

Consider the function ,f(t)r;G,The',Laplrce, tr&or?n" ?(s):of the function.

f(t) is defined as follows; , j, ;.‘ 8

&f(t)] E ?(S) m'.I-

f(tl,e-st'& " :d' j (7.1)

.;o ,: ?, . I, />iI

Ranarks: (1) A more rigorous def,inftion'of the Laplace transform is given

by eqn. (7.la). "7: r

&[f(t)] - T(e) - l id

T"f(t)e

Y$;t 4 .. "'(7.fa)

E-d .LT-m E'

'3

If the function f(t) is piece-wise"ti&tinuous' and defined for

every value,of time from t=G to tm', %h& the .rigorous

definition (7.la) reduces to that of (7.1). For almost all

I

/<

4

the problems that we w%lL be concerned with in this book,

the simpler def initiqn given by (7.1) will suffice.

(2) From the definition (7.1) or (7.1a) we notice that the

Laplace transformatiun is.a transformation of.‘a filnction .

from the time domain .,&here time is the independent variable).‘

to ‘the s-domain (with “8.‘. the independent variable). s ‘is.I.,,i

a variable def iried in the: tmmpiex plane )’ i.e. s’h a + jb.

(3)’ From the ‘dafini&n: (i; i) zr &“: @. 18j i $8 &i&’ tfiihfi’$he :, “.

“.:

&aplscy ~r~aaf~~m~~f tde”$&& f(t); @xi@& ip;& : 1 t ” *j :: . . 6

integral iW’ f(t) e-*t ht”“ / takes a kr$te valuel;’ i”l e’.., ‘rem&is

bounded. C o n s i d e r t h e f u n c t i o n f ( t ) -: eat w h e r e ~0’. % ’ .’. , : ,. ^

men,‘*. I1 1. . i;.,: .,>‘,*:r .:*I. . , ‘_) ,-

1 [eat] o eatemst.dt sI- J=

e(a-s)‘t dt ! ~~ (7. 2);. :.,’ I .G

0 0 .

’ f-%iNow, if a -

’ ,’s > 0 o r s<a ‘then&e i&g&” i n ’ ( 7 . 2 ) ,?‘.

, “.: ., becomes unbounded. ‘Consequently, tire iapia&’ transform of

at is defined only for ’ .:’

e s>a, which yield finite values ”.,

f-or the integral in, (7.2).,;: AI,& the. fu$ctions that, we wi’ll ,;@’ *

be concerned with in this Ibook &ll possess Laplaca trans-. !

forms so Mat we +ll nob”‘test their, exis’teac’e every ,time . ‘.

time we need them,

(4) The Ladlace trax$fdrmation is a linear operation, i.e.$,l 7 ._’ I , _.:[alfl(t) + a2f2(t)J - ,ala’,-*Ifl(t)j + a2 [f,(t)] ‘(7.3)

. >

w h e r e a% _.‘a& a2 are con*tant param,eters ? <me p r o o f i s:*

straightfoeard, i.e,,,. ; ’ ’ + ,j& *. -.

I/ :

‘; S ,’

, 3.:’ :: .&~jL& ‘, .‘, .,:. 1 _ -“‘ ,a:‘.;. &“’ ..:A$

If,(t) I.

Note: A bar on top,of a variable<will signify the Laplace fransform.of that

variable.variable. .This conventioii.'will,be used throughout this text..This conventioii.'will,be used throughout this text.II:II: **

7.2 THE LAPLACE &UWFORMS OF f@4E BASIC F&CTIONS "7.2 THE LAPLACE &UWFORMS OF f@4E BASIC F&CTIONS " ""_! I ,_! I ,.q.q 6.'6.'

Let US now apply the Laplace transformation on some basic-.&ncfions that weLet US now apply the Laplace transformation on some basic-.&ncfions that we

will use repeatedly in the folloi;i5pg'chapters.will use repeatedly in the folloi;i5pg'chapters.II .~.~

A .A . Exponential Function : IExponential Function : I tt..',>..',> %, .1%, .1 ,.'? I,.'? I

This function is' def$ned as! ,_This function is' def$ned as! ,_~.~.//

,.,. . . .. . .-at '-at 'f(t) -. e,.f(t) =. e,. foifoi t?Q -'t?Q -'

From (7.4) 'it is clear thatFrom (7.4) 'it is clear that __

&eat] - &&eat] - & (7.5)(7.5)

I

o--CL

,... . ”

B . Ramp Functions (Figure 7.la) L ‘

This functian'is defined as: ' ,S? 1f(t) = at for tzo with a - ,-constant. _., . .

Then

Pro0 f :‘;

&c[at] - &=-St dt ;, ; .*.0

. ‘etIntegratffig by. parts, where t and e 1 ‘are ,Ghe two’ hnct%ons’, we take :

0 i .’!’ . ‘.: , , ‘P ._ 0 TJ’ .,

: a.,

Consider the sinusoidal function i’(t)‘,= ‘$&(ut) .- Then, .‘-

’,y.;

_(. .i7,7)

Proof:

-

-- -*: - -. .I-;

:.

“

^” .:r ,

f . , ,. ’ ”

ii” + w’

.’

: ‘, . .

. 1

Similarly, it can be proved that,,Similarly, it can be proved that,, .j.j<,<,,.,.

>’ i>’ i

i *’i *’

) ,’) ,’ :.:. ::.:. :b’b’

~[cos(wt)] - s2+42~[cos(wt)] - s2+42ri’ri’ ,^ p) L.,^ p) L.

nn

Note:Note: In the above proof we have us&d the Euler ,identityIn the above proof we have us&d the Euler ,identity ’ ’’ ’;~;~11

sina : -sina : -e3a _ g-Ne3a _ g-N

aa44

-_-_ .i I.i I‘X‘X

:_:_ ‘-.:‘-.: ,,I’,,I’

d

For the proof of (7.8) use

D.

e3a + Jacosa -

2

and

gtep,Functi,on, (Figure’7.lb) ‘I -’ ”

This funct%on is defined by,,]

f(t) - A for 00 r

-0 1 f o r tco

its Laplace transform is6

i

d[step f u n c t i o n o f s i e e A ] = sA (7.9), L ,.

. 9 .‘y,Proof: * 1 . ’a

We notice, that a.:discontinuitv in the value of- the function’ exists at

t=O, such that f (t=O) is undefined. The definition of Laplace transform.: 4 1i

from eqn. (7 .l) requires’ the knowledge‘ef the function at ‘t=O. *“The drawbackfrom eqn. (7 .l) requires’ the knowledge‘ef the function at ‘t=O. *“The drawback

is overcome if we consider the mere precisa ,mathematical definition of the_”

is overcome if we consider the mere precisa ,mathematical definition of the_”

Laplace transformation from eqn. (7.la)Laplace transformation from eqn. (71’I.a)

T‘ I.T‘ ‘I

&f(t)] *” ,,lsl;;, &e-St’&&f(t)] *” ,,lsl;;, f(t)e-st’dtII

(.(.&I c&I cT-MOT-MO

T h u s ’ f o r t h e s t e p f u n c t i o n , t h e u p p e r lfmit i’s, T*m b u t .the lrjwer l imit i&’T h u s ’ f o r t h e s t e p f u n c t i o n , t h e u p p e r lfmit i’s, T-m b u t .the lcwer limit i&’

t=Oi-, i.e. a very small but finite positive time, instead of’. t&O,t=Ot, i.e. a very small but finite positive time, instead of’. t&O, Hince,Hince,

for step function we .havefor step function we .have I li .:I li .: ’ .h”’ .h”

E .E . Translated FunctionsTranslated Functions ,’,’ ..

Consider the function, f(t) shown ‘in Figure 7.2a, sf .this’funct;ion is4

delayed by to eeconds we take the function shown in Figura.7;2b, and if it

i s a d v a n c e d b y to seconds then we have the .,curve of, Figure 7.2~. The

relationship among the three curves is:

f(t + to) = f(t) = f(t - to)

curve in curve in c u r v e i nFig. .;7.2c F i g . 7.2a Fi;g. 7,2b

L e t S!Z[f(t)] )-= f(s) be -the Laplace transform of, f (‘t) ;’ .Then .’ .’ ’t

‘. I’

$[f ;t1

- toI1 i e-sto I!(s) ” :. (7110)i

,; .-_

and‘.

$.[f(t + t )]0

= esto T(s) ‘, (7 .ll)_.

Proof:

,,$[f(t - to) ] =I-f (t - to)emst .dt .f e.

-a0

r

-SWto). <;‘f(t.- to).e , , : d(t- to)

0 .O .s?. . rs L ,.“.I I; ” ‘_ ,_ i

S ince d t = d ( t - to). Let ty to=ri t h e n ::.,.‘, ., .” ;;.r:. I * ,,’c j

-st OD0

If(t=to)e

-s(t-.to) :’ ., ., . ” ?qt*., Q)

I

../ “ _ ~: ;,.e d(tL t ) * e ._

I 0.f(r)e-*’ dr

0i”~.

z;‘to I: -., “, :

1*’’4

‘I-St0 *

- eI ,

f(T) $-ST dT,. 6: e ,o ?(s) ! : 1.4 ,(. .

0 ‘I .’ 1

Notice that in the last. equality,we ‘rep$aced tha lower., bound -to wPthv ,O.. (*

This will not change the value ,of the.Jntagra& since f<(t) 4 9 -for t&k d 0.

19 Equation (7 .lO) will be pakt$cularly useful in the computation of

,‘ ;;:tiplace transforms of, systems wiq,*,dead, time. V,

‘.,

i.!’ 1: %

; I>

/

Example 7;l 1’ 4,:. :;. -1

Let .us recall ‘the flow of an 3ncompressibla liquid through’& ‘pipe ”

(Example 4.9 and.Flgure 4.7a) .I’ From eqn. (4.7) we have‘

T*ut( t, - T& - t,) (4.7)t,

w h e r e Tout is the temperature of the liquid flowing out of the pipe and Tin

is .the temperature of the fluid flowing .in the pipe. The temperature of the

outlet is equal to the temperature of the inlet but’delayed by td .where td

is the dead-time (transportation lag), i.e. the time required for a change in

the inlet to reach the outlet of the pipe.

I f ,?&Tin( t ) ] - Tin(s)

J.

then, using (7.10) we have:

-stTout(k) = dfl[Tout(t)] - ,$%in(t - td)] - e d f,,(S). .

. .

,:

F./ I i

.

Unit Pulse Function

Consider the. function shown in Figure 7,3a., The height is l/A and the,‘: ‘i ,: ? s’.

width A. Thus, the area under the curve, is‘I

1/ if,’ ., :>L

area - X’”1 r’l 1 i. _

1 This function is called unit pulse function of duration A and. 58 defined by:

I

0

1: /

f o r ‘t<O‘,/I s -f 1 0,

“ii(t) i l/A f o r &ct<A; . *i .

I0’ ‘+ f o r t>A 1 ,

.* :It can also be described as the difference of two step functions of equal size

Ir..;1I.’

l /A . The first step fuktion occurs at time j t=O whiie the second is delayed’‘,t _’ .;

b y A uni t s o f t ime . Thus , i f

first step function: ,

t<o

second step function:

f,(t) = {l;A ;

then,

6A(t) - unit pulse of duration A = f l( t) -- f,(t)

3 ,’ ,-f.(t) - f.6A)

T h e Laplace t r a n s f o r m o f t h e u n i t p u l s e functibn o f &ration A isi;’

&s,(t)] I $ 1 - GSAa (7.12)

‘ . i

Proof:

;ei$(t) 1 - &fl(t) - fl(tiA)]I

‘ 4‘ 4 .~,.~,

G;G; Unit Impulse FunctionUnit Impulse Function:: II_ I_ I 3: ‘;3: ‘;

Consider that the duratidn A of ‘a unit pulse function& allowed toConsider that the duratidn A of ‘a unit pulse function& allowed to, *., *.

slirink,slirink,i. .; ,r)i. .; ,r)

approaching’ z e r o , w h i l e t h e heiiht l / A . ~ appr&ches i n f i n i t y . T h eapproaching’ z e r o , w h i l e t h e heiiht l / A . ~ appr&ches i n f i n i t y . T h eL /L /

‘4:‘4:area under the:curve remains alwaysarea under the:curve remains always ,’,’

A +“I I l 1A +“I I l 1 :’:’II ‘,,; 1 : : r j f’ ‘i‘,,; 1 : : r j f’ ‘i

/ ./ . ,,As A+0 we take the function shown in Figure’7.jb. This function is called. ”As A+0 we take the function shown in Figure’7.jb. This function is called. ”

i!i! 11 i’i’I~!L~,-Q!!~)~I~~ or I).Lrnc function and it Is usually represented hy

a(t) l I t i s

i-

defined as equal to zero for all times except for t=O. Since the area under). ;;-,

l.hc unit pulse remains equal to 1,“,i’I : .’ _-

tlihIj, c .!

i t in c l e a r tl& ia true for the unit.-I r., ,- ; ’ !,. .) :

i m p u l s e , i . e . L

I

m

G(t)dt - 1-cn

The Laplace transform

;erw1 - 1

Proof:

Since a(t) - lim SAft) ,,Ad-l

Using L'Hospital's iule

IIn Table 7.1 the,Lapla&i tra&&ms of..&&'d&ic'; functiqns have p&en

'tabulated.>; b,.

;7 i " ':' -:' 4 ": ,. . ., _: "

Remark: It ii Amp&tant'to nO&z& tba$'Xhe" ,i ;.a '.baeie fundtiona exam3n;ed-P$.3zhis.:se

shown in T&e 7 ,.l,ara ra$$& of twb p6ly&tuiajs~j.n e~~.~,,The only ~,. -..‘ .?.\, *;j. .,‘k -2 -, :-exceptions ~ ire the Laplaca’;tleiansfc32’ma,,,<rf fkn&bioiia &$wIa&d in ’

.’ ,_,’+og -*,time, which include the elrponential ie,, e,. . .!Bh+efo$e, for..

any function f(t) (not Including a tink translated ‘-teGrn) we willI .’ _. ;; ._

“Cd4 1 ‘. II -j :%(’ have ’ .\ _v’.

where qy(s) and q2(s) are two polytaomial$,in 8, i.e.,:*. ,.

: '! %.' g,:')<I

,

1

41 (.s)m-l- k,s" + k*ls + l ** + kls + ko " 5

. /

Example 7.2

i rf f(t) = cos(wt) then f(s) = 8 91w

8 + to22 = 42(s) w-?-y -13th q&ii) - i4t 0

and .q7(s)'L l-s2 + w?. From Table 7.1 if f<t) T emat cos(tiit) t h e n' _ ::“

1‘?(s.): - s+a &". s&Q

(s+a)2 + w2, q.2 ~~~,.* .’: “; . I . ‘:./ -,,‘.i

with Q,(S) = i-s + a a&I i:a,f85 88 1.9~ 2:‘ j x+ : “2 ,'

-J.- _ ~.. "FZ'"-, -7 - 2a’ 8 9’.(ig~ ,-r- tlJ2):

/ .'

0 ”

.,.~ ( 7 . 1 6 )‘

* From eqns. (?.14j. (7.15) and*$7.161 k~‘&tr

j '.i ,.-- .

I-, ~_ ----ce that in orde?.‘to ffnd the

Laplace transform of any derivative,,w$ n&$d,to have a number of initial con-

ditions, To find the Laplace transform of an n-th,order de?ivative we need n

initial conditions,

f(O) , f'(O) , f"(O) , l ** ) f("--l)(O) .l

7.4 LAPLACE TRANSFORMS OF INTEGRALS

t

Pi f(t)dt] = " (7.17)0

where. _

%3> -

Proof: .!' ?,I

- . ‘,

&[f(tIl ‘: ,., ;’ ,

Inte&rate by. par&i j Put

. . I

and dv

Now

t

f(t)dt]e-*tdt i -~-$J ,J0 0

” 0

7.5 THE FINAL-VALUE THEOREM ' .,I .

lim f(t) = lim[sf(s)J ':, (7.18)t-- s+‘o :.L 5;; + 1‘,...where

Proof:

Using the. Laplace transform'of .a derivative (eqn;' (7.14)) I"

“-‘dfdtt’ .--St dt p &cS) _ f(o) .- ‘,’ . ’0

‘,

Take the limit of both sides as s-9.-: I5,

01 "limS-+O

@&$ e-" dt - lim [s?(s) - f(O)]s-a 8.

0 +

Since variable s is independent‘of time t, we take'

lim"dflf) cost dt - lim [s?(s) -,f(O)I1

!, P;

Example 7.3

Let Z(s) = s+ls(s-l)(sf2)(

the final-value theorem we heve. i

limf(t) = Mm [ST(s)1 4 lim 8“t

: s+j.

t- S+O s+o s(s-l)(s+2)(s+&q

= limc

84-1 1 '_ & ~ ',( ; '~ ., ': ,. \.,S-+O

s-l)(s+2)(s+3) = 6:i t

I :* !: .The final-value theorem allows.us to compute the value that a function

approaches as t* when its Laplace transform is known.

A note is in order. The final-va$e theorem applies gnly if the limit

lim f(t)t-

.

a.. _,_. _

;

is‘ bounded. Therefore, the limitpi_

! lim [sF(s>l ‘ ~s-m

: .: i i $ .‘,must be bounded.

7.6 -TIN lCNITTAL,-VALUE ,THJOR@'l

Z(B) =.r-

i .-E(t)e-=dt - Laplece transform of f(t) .

'( .I ‘ o"; - ( )*? > _, / ":> I, "i‘ ', :" .:. ,

The proof ~follows the* $nme patteen &i:kdr the final-&alua theoreti,. '

‘; ‘. : ^ ,;r 1: : /^

, ‘.-,,.I ;, :*s, ;:*i

Example 7.4,*;.' :.. .; .,_ 1; ,. &I. ._ ;

Let Z(s)bwe have,

lim f(t) - lim [s?(s)] - limt-a e-gao )'

m lim L + -2s + lo 1 = l.+ llm -2s + lo =i _

~~ L s2 - 8 - 12J -:--___

2- -

Se S-+-B -s-I2

As was the case with the final-value theorem, similafly for the initial-,

value theorem, the limits.

lim f(t) "and l i m [s?(s)]mtlst be bounded ,.t+o Es+- :

i.e. to have finite values.,

I

F .__

*, c*, c , ’, ’

SUMMARY AND CONCLUDING REMARKSSUMMARY AND CONCLUDING REMARKS ;JI- , ;‘i i. ‘.Y?, ‘7 +a;JI- , ;‘i i. ‘.Y?, ‘7 +a*I*I

The Laplace transform of a functionThe Laplace transform of a function f(t) ,-defin&,by eqn;,~.:~(!.l), is af(t) ,-defined, by eqn;,t, ,(T. 1) , is a, ., .

linear transformation of a function from the’time “domain to the s domainlinear transformation of a function from the’time “domain to the s domain* ,:.* ,:.

wherewhere 8 is a complex variable.8 is a complex variable.. .

The important feature of the Laplace transform is that it always leads toI> h 1

a function’& t h e 13 d o m a i n , w h i c h -is t h e rgtl,pof two,polynomials In s,I

(with the only exception the tranelrted Sn .Qrn~ ,fufz@ionr) t in$apendantlyI ,! *

of the type of the initial function f’(t) which can be: constant, linear

in time (ramp), trigonometricb

, exponential, or linear combination of the ”

above, etc.. .

li ’ ‘, c 3,:I..‘, . I.,In addition, the Laplace transform <converts the derivatives and the integrals

into ratios of two. pplynomials. This .imRortant feature allows .tFie con-

version of integrodifferential equations into simple algebraic equations,

as we will see in the following chapter..

The final and initial-value theorems,:will be employed in,order to compute“‘<, 7 ‘* I 1

the unknown final or initial value of the function f(tj’,when Its L&lace

transform T(s) is known.


1. If T,(s) - ;Il'[fl(t)] and z,(s)'* i<[f,(t)] can we find the

[fl(t)f2(t)] fo: general functions fl(t) and f*(t)?

2. l)OeS tllt2 .tUtIC!EiOlI fl = ;-‘-i possess a Laplace transform?

3. What is the Laplace transform of the function f(t).= 5 cos(4t) f

e+ I- 5t?

4. What is the Laplace transform of the vector function

.

[ % ~' -'.ms ,',

asint '+ belCt

gt> = a + b '1

cos(t) + b sin(t- td),

'.

5. Using Euler's identity1

.'

e j "+e -jacosa = 2

show that

[cos(wt)] - *s2 + u2

6.:

Show that

$?Qf(t + t,)] = esed Z(s)

where

?(s) = iAf(t) 1

7 . Starting from the equation yielding the Laplace transform of a derivative,'a

i.e. 4

rdf(t) d-" dt - s?(s) - f(O)dt

0

prove the initial value theorem.

8. What functions have Laplace transforms which cannot be cast as ratios

of two polynomials in 6. i

,.‘% WI, /!/ ,,“’ 8’:.‘x !,’ 3ter>i <..’ ,,

_’ I-.-La., , .- ‘7 1 6) I, I J,

Tnb-le 7.1. Laplace 'Transforms .of Various Functions

Time Futiktion ftz0) Laplace Tranwfdrm

Unit Impulse, 6(to)

Unit Step

Ramp, f(t) = 1

t” n.!xia I ‘I

I /

~ TabLe 7.1 (continued)

Time Function (tz0) Lapldce Transform ~. '.' : ._A

cosh(wt)

t8

2s -U2, "-I,

I -.*

:. . . .

Table 7.1 (continued)

Time Fynction (tz0) Laplace T r a n s f o r m i 1

eaatsin(wtI w

I CHAPTER 8

,a., SOLUTION OF LINEAR DIFFERENTIAL EQUATIONSJ

'k’,; 1.

USING LAPiACE TRANSFORMS1 .‘, _ j.

,,,As it"was mentioned, earlier, the primsry:use,,,fpr the Laplace transforms5‘ '7.I;; to solve linear differential equations or systems of linear (or linearized

/I +nonlinear) differentialf equatiorik w!i,th'constant coefficients. The procedure

I: _I : '..was developed by theEnglish engineer Oliver Reaviside and it enables us to.;.

solve many problems without go&p .through the ttoub1.e of findiflg;$he comple- '

mentary and the particular solutions for l&near ddfferential equations. The:z "/ L_ : I - &t;,c" * ! 4 ; A i .,?'< ; r;'.' h y: L h &; :; I 1 *same procedure'can,,~ie'exte~deb"to 's%ple oti”q$!tems of paftial diiferential

_ ./’ , ‘..ir ,‘g ,$’ i .i $ “,I’ tequations and to iritegral equation&. ,. I., ”

_*

$f + aT’- $ Ti+ ,,iT* > j,, .* ,‘.,j- / .‘: (5.1)

I ’.’Equation (5.1) can be expressed in terms of deviation vkri+bl&, "" ?,

' dT' ;'+ ,aT' =' 5 T'i' + '^ KT'

.p .* ,! ‘< j. :: ',,!" 1

dt ,I: % ' (5.3)," ',71 : *i.. ,". ,' .,;:

where <; . _.,j

T' - T - T(s) , T; = Ti - Tiis) , T; '= Ts - Ts'(s) *I .

are the deviation variables from the steady state defined by the values<. . . ,*J _ ,'

Tb) a Tib) and Ts(s)'c r

Assume that the heater is initially at steady state: i.e. T'(O) - b.,'

At &O, the temperature of the inlet'stream increases by a step of 10°F from '‘ .I

its steady state value and remains at this new level. Thus, T'(t) = 10°F for/ I ', T ,~ L

t>o. The temperature of the h~uib’in tile tank will start increasing and we

I’

want to know how it changes with time, In other words, we must solve eqn..:

,* .

Equation (5.3) i3 a linear equation with constant coefficients. We can. . ‘. 2 , i.

use Laplace transforms to’solve it. Let us examine the solution procedure.

or. ,

.

T;(s) - 0.T;(s) - 0. Then eqn.’ (8.1) beg,ome$c )’Then eqn.’ (8.1) beg,ome$c )’ ‘:‘:‘.., *‘.., * -1 ,‘.-1 ,‘. . . . .. . . . \\..

tt .-.-T’(s) P $ . & . $’T’(s) P $ . & . $’

..:..:

.; ‘..; ‘. “,.I.“.“,.I.“.(8.2)(8.2)” _” _ i i , .-i i , .-

The function , T’(t),,w.bose ,Lapla.ce transgorm is+: given by the. right ha,nd<;rideThe function , T’(t),,w.bose ,Lapla.ce trap@&% 19: given by the. right hs,pd<;ride ::i: l’l.i: l’l.

o f eqn,o f eqn, (8.J) 5,s our”so$u~tion.(8.J) 5,s our”so$u~tion.I. :I. : It ,is easy to ~hhowthat, SC : 1 .:, ,,It ,is easy tijr ~hhokqxthat, SC : 1 .:, ,,“.“.

T’(s) - + : -& ? +y I s [$ - AL]T’(s) - + : -& ? +y I s [$ - AL] ; : ,b , ”; : ,b , ” ( 8 . 3 )( 8 . 3 ). I. I * .* .

From T a b l e 7.1.,we find,easily that;: :.From T a b l e 7.1.,we find,easily that;: :.I ,; _I ,; _

d,d, .:. : r ..:. : r . / I/ I** / .;.a/ .;.aII ,+,+

- ,the function Cith Laplaca transform IIs 1s. a unit. st+ function,- ,the function Cith Laplaca transform IIs 1s. a unit. st+ function, andand .,.,:: i .,:$i .,:$ . . .e,* .,. . .e,* .,

- the function. with Laplace; traneform”-l/,(e+a) is- the function. with Laplace; traneform”-l/,(e+a) is -at-at.$.$

ee .. ‘.‘..I.I

Therefore, from eqn. (8.3) we find,Therefore, from eqn. (8.3) we find, ;; .’.’ ‘/‘/T ’ ( t ) ‘ -T ’ ( t ) ‘ - g ( 1 - eeat)g ( 1 - eeat) ( 8 . 4 )( 8 . 4 )

,, _ i_ iT’(t) given by eqn. (8.4) is the solution to our Initial dlfferentlal eqn.

(i.3) . , Indeed, taking the Laplace of eqn. .(8.4), it yields eqn, (8.3j. ‘i$e< .‘- ~ ~l.‘

procedure by. wh%ch we find the time function when its, Laplace transform is‘. ‘ I

known is called Inverse Laplace Transformation and is the most critical step/’ ,s*. -I1, .’

while solving linear differential equations using Laplace transforms. To

h

*

..

sum~~~M.~e the solution procedure described in the above example, w’e cansumm&3.?e the solution procedure described in the above example, w’e can2’2’

hdentify, the following steps:hdentify, the following steps: *’*’ .s.s,, ( ,i .: ~ ‘.,, ( ,i .: ~ ‘.

A.,A., TakeTake the Laplacethe Laplace3 -its’i ‘3:: 1 fj I , :3 -its’i ‘3:: 1 fj I , : transform of bot,h’sid&. of the different&l equation.transform of bot,h’sid&. of the different&l equation.,,

,U& eqni., (7;14), ( 7 . 1 5 ),U& eqni., (7;14), ( 7 . 1 5 ) and (7.16) to develop; the La@aee , &ran&ormsand (7.16) to develop; the La@aee , &ran&orms

of thy vnr.tbus derivatives.of thy vnr.tbus derivatives. llle init tat cond-1.tions giucn ~Fffr .tlW tllf-llle init tat cond-1.tions giucn ~Fffr .tlW tllf-II

.ferential equation are incorporated in this step with the transforms.ferential equation are incorporated in this step with the transforms,:g “:,:g “: I” ‘I” ‘o f t h e deri&t$vtBs*.o f t h e deri&t$vtBs*.

;; L ^L ^.,.,

B .B . So,lve the resul,tYng.algebra;Lc equation 2n terms. of the Laplace transformSo,lve the resul,tYng.algebra;Lc equation 2n terms. of the Laplace transform

of the unknown function.of the unknown function. :.:.

c .c . Find the ,t+me function. #$ch bs ,& $ts Lap&ace tr&Morm, the. rightFind the ,t+me function. #$ch bs ,& $ts *place trstis*orm, the. rightII * :* : : ’: ’hand side. of ,tBe,,~quat~on~Rb,taiqed,, instep It,? This functiqn is thehand side of ,tBe,,~quat~on~Rb,taiqed,, instep It,? *is functiqn is the .;.;: _.: _. ,s, 8 i,s, 8 i . .. .

desired solution, ‘since it satisfies .&he differ&tiat equatlan and thedesired solution, ‘since it satisfies .&he differ&tiat equatlan and theIIin&tier1 conditions.in&tier1 conditions.

66

Step C is the most tedious,,Step C is the most tedious,, .Ghm Q general expression’ l$.ke,.Ghm Q general expression’ l$.ke, **~~ j_j_ (.(.

Z(s) -Z(s) -(s2 + als, + bl)-te + cl)(s2 + als, + bl)-te + cl)

’’ ;s(s3 t a2s,’ bss + c2);s(s3 t a2s,’ bss + c2)/ ,i/ ,i

it is not obvious at all what is the function jt( t) whi-ch has the” above’; I’it is not obvious at all what is the function jt( t) whi-ch has the” above’; I’” i” i

Laplace ‘transform. : In Section 8.2’we .wcili study a particular &ethodo$gy forLaplace ‘transform. : In Section 8.4’we .wcill study a particular &ethodo$gy for

the inversion of Laplace transforms by .P,art&a$.-Fractions &pa&on: ” *. *’ ’the inversion of Laplace transforms by .P,art&a$.-Fractions &pa&on: ” *. *’ ’0:0: ‘/ ?’‘/ ?’

8.2 INVEIiiON OF LA&ACE TRANSFORMS.8.2 INVEIiiON OF LA&ACE TBANSFOEMS.II

mw1~1m3 EEPAEsIoNmw1~1m3 EEPAEsIoN,. ; :,. ; : -- : i <: i <,, i:,,i:,,‘ .‘ .

As it was pointed out above,As it was pointed out above, the critical point in finding the solutionthe critical point in finding the solution‘$‘$ / y; Q, .:/ y; Q, .:

to a differential equation. using Laplace transforms is the inversion of’theto a differential equation. using Laplace transforms is the inversion of’thew .

Laplace transforms. In this section we: will study a method developed by :.I.,.t

Heaviside for the inversion of Laplace transforms known as J-leaviside. or?,.

Partial-Fractions Expansion. __, . I ., / :.I _,

Assume that the Laplace transform of an unknown function x(t) is given

where Q(s) and P(s) are polynomials in s of o&&r m and n respectively, ,.1 where m<n. The inversion: of Laplace”‘transf or&s’ using the expansion to partial

,fractions is compoeard of ‘the following three steps:‘- ’

A . Expand the Q(s)/P($) into a series of-fractions’, i.e. ’ .)

.tj;(s) P 90 a -c1 c2 c ,’n

P(s) rlb) + r,(e), + 0.0 +rJa) (8.6)

where rlW, r2’W,*** ,rn(is) are low order ‘polyno&ls like first,

second order, etc. . ;:’

B. Compute the values of the constants Cl, C2,*;*,Cn from eqn. (8.6).

c. Find the inverse Laplace transform’of every pirtial fraction; Then; .the

unknown function x(t) . Is giveh by ’ .** 1,’ : ,6

r

.,,,, = k-l@$j + X~[~~ + ii*-+,, &.?j]

z1f.

. wh.ere - symbolizes the inverse L&lace transfcirm of the expression within

the brackets. The inversion’of each fr:action ean be done rather easily b,y‘..,inspection using tables of Lap1ao.e transforms for typical funat,ions ‘like.

Tables 7.1 and, 8.1,. * -$ 1 ! , .1.IWhen Z(s) is given as the ratio of two polynomials (eqn. (8.&) its

..:, ,. , .-expansion into a series of fractions is governed by the form and the roots of

: . hthe polynomial in the denominator P(s), In general we will distinguish two

i _’ ‘,’ j. .!cases;

. i : . ?, ’i /’- polynomial P(s) has n distinct (all different) roots, real”orcomplex, or

- polynomial P(s) ‘has multiple ‘roots : ’,: -_

.IWe will examine each case separately using characte&tic examples: “- .’

.I

‘j ‘r -4 : I_-*

’i

\ I..!, ,’

given by

(8.7)

P(s) in- s3 - 2s2 - ,s -t 2 :..

and has t&se roots , ,’

Pl’ ; 1 *P,y* = - 1 *and . . pj &.:

Therefort%;

P(s) m ..s3,-2sL,- s+ 2

and eqn. (8,7.) become@’

Expand (8.8) into partial fractick aqd te

where “$9 cp GJ are unknown qocstar&s, to be &valuated. ’9

From eqn. (8.9) it is clear that . X ,:’

d-l& + L 9-2‘I -1 5,.I I.iland using Table 7 .l we find that

x(t) 3 Clel’t + C2ew1yt + C3e2*t ‘. ” L, (8.10)

which is the inve,rse Laplace ‘transfo& of t& expiession, in (8:7). I

Let us see now how we can compute the constants Cl, C2, C3..,

3

- Compute Cl: MultWe both sides of,(8,9) by (s-f),,.,t. (,'

I 'j ~qg> " qs-1) ! i

= c1+-+SC1 s-2 (8.11).. ff

Equation (8.11) holds for all values of .s:: Set s - 1 - 0, i.e. s=l.Equation (8.11) holds for all values of .s:: Set s - 1 - 0, i.e. s=l..i.i

The hast two terms in. the rip[‘ht ha& side of (8.10) become zero and.The hast two terms in. the rip[‘ht ha& side of (8.10) become zero and.

we take,we take,

- Compute C2: Multiply both sides of (8.9) by‘(s+l), ;'. 4 "'1i

2 9(8 )i.@!-D(s-;)~(s-*)

C1(s+l)- s-l ,:- + c2

c&+1) '*,+ �yg-- l ,. i

Set 6 + 1 = 0, i.e. 8 - -1, c , .,, _I ,. :

- Compute C3: Multiply both si+rq of @3..9) by (.&2);:,"; i , :, s

Set 6 - 2 - 0, i.e. 8-2,

,‘I.,’ f

II. Dlstlnct Complex Roots of tha'Polynomia1 P(s)-'. .

IConsider the following Laplace'transform ‘ '

s-i-1Z(s) f - - - - -S2 -2s+5

The polynomial P(s) is of second‘order,and has ,two distinct roots! which s ,'

not real (as in the previous case) but complex conjug!ates, i,e. .,ri ! , 'g

p1 -l-t 2j a n d P2 - 1 - 23,: <'

Therefore, _I','

P(s) - 82 - 2s + 5 - [s - (1+ 2j)][a - (1 - 2j) ,,. .!_

Expansion into partial fractjons yields:, ,' -. ,

fi(p = a+1 ,s+l ,cl 5'i &- 2s-f. 5’ = [s-(1+2j)].~.s-(b-2j~] - s-(1+23) +8_(l_zj),@*12)

and using the",transforms, of Table:7.1 we find '. +I A ' i' I.

x(t) * c .(1+2j)t '. '(l-2j)t1 ,? C2e (8.13)

The constants Cl and., C2 *are qnputed-as fn Case I. :

- Compute Cl: Multiply both sides of'(8.12)'by (s-(1+2j)], ,

(s+l)[a - (1 : a.La-h-c+-r-psfl[s'- z 2j)]' - c1 '+

03 Es -.(I + WI,8 - (1 -;,2j) P

-. i ,. .' ,Set [s - (1 + 2j)J = 0, i.e. s - 1 f 2j and take, .

cl -9 i J

- Compute C2: Multiply both sides of (8.12) by [e T (1 - 2j)J end then set

s - (1 - 2j) - 0, i.e. s - 1 -'23, to find ..:" : " I "' . */

c2 -A$i '/

Notice that the coefficients C.l

and C2 &re complex conjugates of each

other.

.

P u t t h e v a l u e s o f C l a n d C2 i n ( 8 . 1 3 ) a n d f i n d :1 ,i . .’ : j

x(t) = l-+ ,(1+2j)t + ,a ,&2j.)t. ,, * ,’. 2

or

Let us recall Euler’l;1 &entity‘< :ri ,-, ; &‘I $1

’‘, ;. :. ,.

(8 ! ‘e.ta’;

1 = cosa + joina (8.15)

Then we have:P

2jt. . J

e = cos(2t) + j s i n ( 2 5 )

pd ’

‘,. I ! ,<.’ Lt*::

.-2jt’ - cos(*2t) + jsin(h2t)” ‘A ‘~z?os(Lt) i jsin(2t) ’h

ig -AL. ‘! iiI n e q n . ‘(8.14), r ep lace eLJr by’ti leir .squal:’ f$bm’th& above ‘eq&~tions! a&$

8’t *_

take, .’ :,: . ? :p

., I’. ’* , , . . I . .

x(t) - g {(l-j) [coi3(;2t) lY .jsin(tLt) ]?+ “(‘l+j) &as(&) ?.: JL &&(2t) 1,): -.- - - *

,: . Is : b -5 ~5. 1, ‘. _rlor : 1

\

x(t) -x(t) - et[ooi(2t> + sin(2t) ] ‘-, ,.. .,et[ooi(2t> + sin(2t) ] ‘-, ,.. ., :: (8.16)(8.16)/ i’/ i’ ,,

Recall the trlgonomatrlc identityRecall the trlgonomatrlc identity

alcdsb + a2sinbalcdsb + a2sinb - egcrlo(b+~)- egcrlo(b+~)”” .>’ ,.>’ ,

(8.17)(8.17)

wherewhere ‘I‘I,,** cc

;y;y .’ .:.’ .: ii., .., . 11 , “:;, “:; 88 \. ;\. ; ._ ,,._ ,,II !! a3 .=a3 .= rr2 22 2

al + aal + a 22 ‘and.‘and. :. $I:. $I -. tan-l(al/a2) s-. tan-l(al/a2) s

A p p l y ( 8 . 1 7 ) t o e q n . (8,.16). a n d f i n d :A p p l y ( 8 . 1 7 ) t o e q n . (8,.16). a n d f i n d :

x(tj - Pt. &*sin(JZt + I$) ‘j *<-,, “’ *

Ahere 4 = tan-‘(l/l):.

= 45”.:

.

-..-..

,*” .I’,’,*” .I’,’

Remark:Remark: yhenever the polynomial P(s)-has complex roots:yhenever the polynomial P(s)-has complex roots: i'i'---YI I---YI I

"(1). they'will be always in'complex pairs, _r"(1). they'will be always in'complex pairs, _r .*-.*-I)I)

(2) the coefficients of the cirresponding terms(2) the coefficients of the cirresponding terms

',,',,

in the partialin the partial"T"T

fractions e&pansion will also be complex conjugates of eachfractions e&pansion will also be complex conjugates of each:. . : ;‘ ):. . : ;‘ )

other, andother, and.:.:II

(3) they will give rise'to a:p&iodic ter$ (e.g. sinusoidcil wave).(3) they will give rise'to a:p&iodic ter$ (e.g. sinusoidcil wave).

III.III. Multiple,Roots of the Polynomial P(s)Multiple,Roots of the Polynomial P(s)88 : , ,"I .:~ ': , ,"I .:~ ' // !!

The expansion into'p~rtial'fractions and the computation of theThe expansion into'p~rtial'fractions and the computation of the.',., %*..',., %*. ii

coefficients change when the polynomialcoefficients change when the polynomial P(s) has multiple roots. ConsiderP(s) has multiple roots. Consider: ;: ;..‘" '*..‘" '*the follpwing Laplace t&kirformthe follpwing Laplace t&kirform '',,,:,,,:

*;.+g:* **;.+g:* * ) *I ") *I "7.:7.: ../../

ii(s) -ii(s) --* ". 1. / .: (,,-* ". 1. / .: (,,**&~,3<~+~) n ' Ir, j'&~,3<~+~) n ' Ir, j'

!:!: , .:<" i, .:<" i (8.'18)(8.'18)*.*...'!'! ;i'I;i'I '( r0'( r0,,_,', )' $;'; j,,,_,', )' $;'; j, 1 ;:'1 ;:' **

The polynomialThe polynomial P(s) has three"&& e&al and the fourth different, i.'e.P(s) has three"&& e&al and the fourth different, i.'e.:: '$ ., . _‘'$ ., . _‘ *,g.;. ,; .+, ,'" ,, ? ‘.*,g.;. ,; .+, ,'" ,, ? ‘. :: ".".11

pl - p2 - p3 r -1 iand p4 -.: -2p1 - p2 - p3 r -1 iand p4 -.: -2>> )“ j)“ j

Rxpand (8?19) into pTrtia1 fractiops: h' “,I

From Tables 7.1 and 8.1 we find that .cI

/,t-,;, (

t

.-pc; ,-1- C2teWt and z4 C3 'C3 2

(s+l,j 2= Tte

-t

8. ' :. , * .&w3 I, . . i. r._Consequently, the inveroe Laplace tranlfosm 03 ' (8.19) ir easily foutid tq trek'

x(t) - Clewt + C2tP: cg ,.g I,+ 2 -t e

+.*,' p i .(8.20)

.+

Let us see then how can we compute the constants Cl, C2,1 C3*. and Ci.' ,

- Compute CA: This constant corresponds to the distinct root'and can be

computed using the procedure,described earlier. Thus, :raultiply both sides- Z.' *.i

of (8.19) by (s+Z) and they set sS2 - 0, i.e. s - -2 and find:

L

I

c4 = -1 .._ 3 .'

- Compute C3: Use the familiar procedure1,i,e. multiply both sides of (8.19)

.l-' - C1(s+l)Ls+2 + C2(sfl) +* c3 + -*+2 (8.21)I.

‘ iSet (s+I)~ - 0, i.e. B m -1 and f&d;i

1., , 1

.cgi'-l':, '/ : . :*$', :n ii

- Compute C2: The familiar procedure used ibove cbnnot be -employed for thei L..; ,) :" n

computation of C2. Thus, if we multiply ,both sides of (8,19) by (~+l)~',! '/

we take,

(s+&s+3) * e,(s+l) + ci + C3 i Cq(s+l)

iG ;( +; --GE- :

Then, setting s - -1, the term invo$vin$ C3 becomes infinite. The same_' i &r i"* ,,i.l,.'

problem ;8' encountered if we try to cpmpute Cl,.; i ‘ a, ^_Therefore, dn'alternite

:, * _'. "procedure is needed to compute C2 and Cl. ' '

ilDifferentiate both sides of (8.21) wtd respec@&' s, &d take,"r;.I <. I -:.

1>a 2 '/hLr\ /3an\

A 'F 2C,(s+l)‘ k '2, &> c,4

Set s, - -1 and find :' '_ -3 .I 1 3''_

- Compute Cl: To obtain t&e value of, ,C1, diffetentiatd'g8,22) once more and.:Ilita@, ..a ': * i'." ; '$ ,k‘-.. 1 ? , ) .,_ . '. ; ,!,

I I:,2

(e+2)3- " 2c1 + %;'c& ):i

.'

Set 8 - -li"and find “",,'., ,".. ,\ * ,; .‘-

9 : /> ' : .:.i Cl .- -1, :' : : '

~:'Substitute ihe values of Cl,.( .,

Cp*,C3t and‘ Ch in (g.20) knd find,._ i.'I/

,i ‘ I

Ii x(t) - -e,-t(l+t +.:f t*) ‘+ e-2t

T ‘If the polyiomial

.’ ’Remai k : P(s) has multiple roots, then the denominator

‘*of X(s) has a term (s - ,pi)m where pi is the multiple root-.2,--:” which is repeated m times. In such case the partial functionsa ~

‘.. expa&on p&duces terms ‘such as”.,?,’ ,r<

..’ Cl“..c* :( ‘. * ,cm-1 ‘:

(?- P,) + (sy$ *.+ l ** ++

” cm

( : ! (8 - pi)m-A (8 - Pi)m,

From Table 7.5.$a know that.;

I

htne-at1 p nln+l .,‘.. : ,‘,,’ , ./ ’

,(gcg), :

Therefore the terms of the 4bove expansion lead .to..the following inverse Laplace

transform:

* Cl[.

T3 ,2 : ’ cm-1-F .yt + 21 t .+ a*. +*f’tm-2 +c “m &l p.t

(m-l)t: t Iei

,: I .“1The constant Ck can be computed:*in bhe usual manner by multiplying both

,sides of the expansion .~ith (t

,.2 pi)m.‘-Yand setting” saXY-, pi. I The remaining,,,

constants Cm-l’l **,C2,Cl are computed’by su&ssive’differentiations~of theI,j,

equation resulting from the amltip&a&n .of :tRe expacl-;aionl by (8 - pi)m.< .’

8~3 EXAMPLES ON THE SOLUTION OF LINEAR DIFj?ER;ENTIAL RQ&IONS USING LAPLACRTRANSFOREis ?.’ \ a%il ” . , .: .‘3.

In this section we will give twa characteristic ex&~+.es ~.oP solving linear

differential equations using the’ Laplace transforms,Y, .The first example is the* I,

solution of a second-order differential equat’ien, while’ln the’ second example

I * we find the solution to a system of i two differential ‘equations,,

‘The solution of..any other Unear 1differ’ential:-iequation or of, a system of

linear differential ,,equations ; ‘twill f aU,Qw j the same general p&tern outlilned in

the two examples. For the solution of ghe general n-th order linear

.

differential equ?tion and of a general system of linear differential equations;:;.

the interested‘ reader is encouraged’ to consult ip&ndic& A anti‘ B at the end.of this clmaptey; j .’ . .I II

,c, P ,i _, ;

‘, * ,r,

Example 8.1 - The ‘Solution of a Second-Order Differential Equation: ‘<. -.. :PConsider the following seeond-order dif f &e&al equation\.

I ,d2*

:’ : _ .-,

a2 Jd x+& al dt f aox d f(t) (8.23)

I * :. ,;

where x(t) .is ,considered to be in the form of B deviation &iabale iith** : ’

initial conditions :. >

x(O) . ! - (&o. - 0 * ‘- . ! ,‘b & /’ I ,!(8.24),

Take the Laplace transform of (8;23) ’ .‘. if I

a2[s21(s) - >sx(O) - (2) ] + a [d(e) -0 i(O)] +‘ap - f ( s )dt F-O ,.-_ 1

or ,I“ _. i’:“,’ :2J ; .j,;, ; j’ I, : .: .“-‘I

L(s) - .’ I(s). I’ + +f=W> ,+ QC$!&~~ .+ alx(0)

a*s2 + y T 80 ’ :; : ‘.,

( iii 25)

” ‘, >: +.p2 ” ’+as..+aO I”:, J, < : : ‘y -:‘

Let us assume that f&t) is d unit .&i-fans&ion, ‘gtving * ‘., r (,’

Z(s) - L/s

e q n . ( 8 . 2 5 ) becomag.f,t ‘:.:: ;. -[.., is .,, : ; :..:;

S(s) -1 , -.\ ,._ I. ~,

2‘,(&&)

i e(apeJ., + a&.y.ao) ,, :“: * r’ < :.‘< s . *,L;

The polynomial P(s). 5’. ais + ale’.+ aiD,. is,aalled the characc&ristie polynon&al

of a second order equatiaai In ;order tie’ invert <the right hand chide of (8.26)r. . ’:swe need to knew .&e rooes of the poZyaomia1 P*(s) i Depeading .on. the values

:,* /‘

of the constants a2, a

::

,., .'. 1' and ao we can distinguish three cases:

...f',T .'. .;'Case.1: .;a2 - 4a2ao * 0. Then, we have two distinct real roots, .

e.g., 'let a1,:,y,4, a2 - 1, a, - 3 then ai ;.! I"

'.4a2ao * 16 -1294>0 and

sl =, -1 and.:: s2:'= '-3._., ,_ ,: L, ., %

. . ;1 " - ; 1 'Z;. i. I Cl e2 C3

s(a2s2 + als + a,) *(is2 -I- 48 + 3) &. s(s+3)(8+1) - -ii- +‘s+3’+s+l (8.27)

* t..a”: ..-

Multiply (8.27) by s and set '~0. Find‘i : ,

* i/3 , ':c1 _'

Multi& (8.27) by e+3 bnd'lset a.4 -3. Find

c2*: -l/G : I:, “,

*; : 1

Multiply (8.27) by' s+l and set B - -1.. Find:

,

Case2: a:- 4a2ao - 0. $hen, we,have two equal roots.

S1 = T2 = -al/2a2 I".l

1.,.

Let al - 2, a2 ~'1, a0 - 1. a: - 4a2ao - 4 -.) :

4*1*1 - 02,'',' '. .,,,. 1 '"

s1 - s2 - -1 *

:;

1 0' 1 '-. 1 V;%; C3.s(a2s2 + al8 + ao) s(s2 + 28 f 1) s(s+1)2 8 83-1 4*+$

(8.?28),

i

i Multiply (8.28) by s and set ~0. Find

c'.

Differentiate (8.29)'with respect to .s T' ' .‘. 'I :'. I_. :

set s - -1 and find,

c2 - -1

2Case3: al- 4a2ao <

Let a, - 2, a3 = 1, a, * 2, a? -$ b&a- - 4

roots are:

b

1' $2 ,$a,c3*-+-----+-(8.m)

s - k$U s - +J.I, ._ 3..' , -. t ,.>A‘. r ,v

Multiply (8.30) by s, set s-O, and find, _p:. "2..

- 112i :

c1L

. Multiply (8.30) by (s - q), set s = -Iti and find,

2(-1+j)(-l-j)(-1+j; = -l+j

Multiply (8.30) by (s - =$J-), set s = y,' and find,t

c.3' #ii 2 r 2(-lrj) ,I)-1+j (-1+j)(-l-j) - c -l-j -I'

Consequently, .

1 . _) :

x(t) e 4: y-l 1L t _ EYI + (-1+i) z$_-

-q 1; ,1s- . -- - _ .w,i-l. . . + (-k _ -yl

, .,;, ? i :

,,’ !o r if

- d ,..a *. _L -'a-. - _.

‘1 ,Is ’ ty<,’

x(t)’ - 1 .+ (-,,j)e*yw3G +,~~,,(,l-j)e;l/z(ll'j)~. ) (8.31)

Recall &k?i-ls;idemity'

,

4 - ta*-l(i/1) = tan-l(l): - ii:50 i :, i*

The use of Laplace transforms is not limited to the solution of simple

differential equations, like the seconf-order .equation of. Example, 8-1. It- -,’ ,e a j’

extends to the solution of sets of differential. equations. , . Coti~,i&r for!.

example the following system of linefirr differential equations;

dxl

t

dx2dt

P

3

allXl +

‘a.21Xf .+

.I

I - bllfl(t)

with initial conditions” xl(O) - ~~(0) ‘IL: 0.with initial conditions” xl(O) - ~~(0) ‘IL: 0. Taking the Laplace’ transforms ofTaking the Laplace’ transforms of.‘,.‘, .,.,,,

the above equations and after appropriate grouping- we find,the above equations and after appropriate grouping- we find,,,(8 - a(8 - alp$f’)lp$f’) - ‘l+$(‘)- ‘l+$(‘) - bllW + b&(s): ‘i,;,- bllW + b&(s): ‘i,;,

-ia21Gl(s) + (8 - a )G~(s)-ia21Gl(s) + (8 - a )G~(s)T iT i ii,2? ;ii,2? ; - b&p:. :’ <“~~p~p);‘-~- b&p:. :’ <“~b&p);‘-~/ ,’ ?:+/ ,’ ?:+ _,I,. ,; ‘a, ,,?_,I,. ,; ‘a, ,,?

The last two equations,for a set of two linear a,lgebtaie eq&&$ns~~~th ?l(s)The last two equations,for a set of two linear a,lgebtaie eq&&$ns~~~th ?l(s)‘.‘. ee ‘.‘. JJ

and Z2( 8) as the two unknown variables,and Z2( 8) as the two unknown variables, and can, be solved easily ‘using, forand can, be solved easily ‘using, for44 I ‘I ‘.<.<

example, Cramer ‘8 rule. 1 Thus; we’ find:example, Cramer ‘8 rule. 1 Thus; we’ find: .. ‘,>‘,>

j-p) -j-p) -&f1(8) + b12~2(e)~4 (a -.;az2) + al2[b2l~,l(s) ,+ b22f2(s)-1 i&f1(8) + b12~2(e)~4 (a -.;az2) + al2[b2l~,l(s) ,+ b22f2(s)-1 i

;.;. (8.35)(8.35)

I~,,~,(s) + b,&,fs) 3(@~2(8) -

- ai%>’ + az21[bll?l(s) + b12Z2(s) 1

82.. - (al1 + az2j8 - aa2a21 i(8.36)

j*

j ‘(c’-The above expressions can now be inverted using thepartial-fractions expansion,tas it was deecribed in ‘Section g. 2; to ‘find the u&own sol&ion xl(t) andas it was deecribed in ‘Section g. 2; to ‘find the u&own sol&ion xl(t) and

x,(t).x,(t). ..’’ :f:f && 44‘‘ ..

. The solution procedure described above’ can be extended to ‘larger systems. The solution procedure described above’ can be extended to ‘larger systems

of equations, but it is computationally more cumbersome, A short compactof equations, but it is computationally more cumbersome, A short compact

description is given in Appendix B at the end of this chapter.

Let us now discuss the details of the solution'procedure, in terms of an

example.

/I*

Example 8.2 - The Solution of a'Set of Linear Differential Equations s'

Find the solution ofthe following set of equatiotis: \

dxl*1 dt= 2xl + rx, + 1 with Xl(O) - 0

dx2- - 2xl -t x2 + etdt with

‘I ‘I’.’

Take the Laplace transforms and after rearrangement find.1I

Using Cramer"s rule to solve the above system of linear alg&afc equations in. ‘. : -z& :

?$(s), Z,(s), we find: : ' _'; .''_ ;..r.i ‘ -,

: : " & _( .o r

:, (3 9,s'fs+l :':. r.2, /

\ +) - s(s-l)(s-4)(s+l).and s2(s). i -..,; s:

s(s-l)(s-4)(s+l)

Expand into partial fractiotis,, 1, ' d i .~

* ., i .x r ; : ‘

q(s) =s2*s+1 cl c2

s(s-l)(s-4)(&#+1) -y +x :.;' )I :'.

Tz2w ="~2'-i 2 . '. Dl ,D2 /'*.

1 ' s(s-l)(s-4)(s+l) -,7,+x +tsi4 : ':+

. ,.1<, / :. ..L

Taking the inverses we finally have:, I i"..). I::.:

:,.;r ;' _ 41

SUMMARY AND CONCLUDING RRMARKS. \ !. " : :, ;.f; a . '\ s, __ ,:t+,< , ,> :

In the previous sections it was 'shown that the golution of lineqr dif-,t .,:.ferential equations (single or syeti?Ofj

* , .I :;, .'. ,- ''be&omes ~"sii&le'ilg$braic problem

'when Laplace transforms are used. The propedure,'~- ,. .p @,as follows:'i.b ; .", #

- Take the Laplace transforms ofkboth &ides of' the diffj3rentia$~,9 e$uations.',. :,' ,-yi i .; ., I , .s /",. 7 I: ,_ 3' s,'

- Solve for the Laplace transforms'ofjthe, tinknown fun<tions on%<: left hand<l I'8. ! */side, while'keeping the Laplace'itransforms of. &e known forcing functions;on /

. .: .L ,.$ *:-r,' ,.the right hand side of the equahio&, i : " ."' ],, Ie "5' : ,.:.:i& yy q l"yx < -.

- Find the'inverse Lap%a"ce tran:forms of") ' 3

the.'kn~~quantities'bnI

the right).* ,., %_

hand sidk of the equations. Tlie$te am\.& ;olutions to the &;rentinlt

&:I‘>"e quations. f& '_' i,'

To find the inverseLa'pl@ce tr+nsfo?m'of a gig&n expre$pt++,use .the i.. . . . L .I: i.,,

Heavside (or Partial Fraction) expans$on. This leady,to a suvtion of simple,. Y, ' _ L

termu.ln the s-domain, whose Inversa paplace transforms are easily found,_ * i

using Tables 7.1 and 8.1. :r

Whilocomputing the constants o$cthe terms in the Heavislde expansion,I ,'. ..'*, '.

it should be remembered that the multiplicity'of' \Lthe roots of the character-

. ;.;.

; ?

istic polynomial is a very“~important3f&tor. li ',

,APPENDIXg.A,,~LDTIQN OF {AN n~th~XR.LINEAR DIFFERENTIAL EJQIJATION‘+,' I '"5. 9.:" , ,. ,..<:. yet ) " J, r ,., i ..' ..:8,. 'I,3 '";"-:: "'7 / !"I ,,. , d:‘~;~,~'~'.Consider the following,n-th order, linear‘differential equation, .'b .c* / 1 ,, .,r,'.* : s,i' :., ir

& +,'n-l Y "

: 'n dt"a .dn-l',dt~~l,~ + ,.aox gi ' f&) (8A.l)

L' *'(i / ,,. 1,:'. :

:._with the following initial conditions: _ _:.

. ; '$3.'. ' 8

~(0) = ko, (dxjdt tm() t' kl'

A(8A.2)

'<..' ;: _ITake the Laplace transforms of both sides of (&A,l)'using the initial conditions:

/(8A.3), and after:algebraic rearrangements find,. ,, ,! I"iI ;:- _. i ')i, 1 I L_r ", ,

+a**+ als +-\a0 ansn + an-s'�: . +* i : ., \*,c ,). :.i< '& : .,ii .i I 'where -:. *

.i ','Z

bty LIi

n-l =ak b 'n 0 ' n-2 = a,kl + an$ko,***, I ,I _. . : / .;

bl*ankn.2+8 -',a kn n-l.-t-an-lkn,+~**+alko.'.il. s ", , . . : .::1: :'. _(' ,,> '7 i * .': ').f '* : A , _ ii! : : .;

Define the ch&&ze&t& polynomial,for eqn. (BA.l)"as follows; ,<L :f 'dj * ,

.,-" ii(s) -

and

>:,,-x(t) -

I .‘. ’ ..I’? k.:. J )I\. : .; .) /I’! :a! I) ;‘I / ,$ . i ,;,c ; ,$

If f(t) - C (no external influence) we have,the solution to the n-th,i '. .': i i“.._ ' iI f.f 2:'i

. . ,.: -' .+I! j :i 1 I *i' ,‘ : *> !'* '*.,. 1,

.I.x(t) - x2(t) -

I, q*]~: ‘* _ ":.. : ;__ j. i' ,.,' b.

which is the complementary solutioi arid depends’only on the initial conditions! i ‘., I y.

(see definition of q(s) arid of bn, bnLIS*** ,bc),. The. other part of the

solution‘,‘*

.

?gF) 1 = a-Q&] ; ;. J : ,. .*: * I, .-:,

is the particular solution and -depends % the f&m of the’ ,fcrcihg function. ‘2

j -!f(t). ,’ b: J _

I .-’_I -, . .iFor process control, purposes,~ the”:&yn&mics of a ptocess are ,~described in

: *terms of ,deviation variables. ,’ ,In such case, if the system is initially~.at<,” I

1(steady state, then

~’ .,r’; .‘; ‘“.F ..’,’ ;k. t= kl -’ l p l (*. k-i :i, 0 1,: , ~ i ?

# .A ‘, ,i.... ‘.C* , .‘z.. I” .’ I, :.{~li. *..and the solution to the n-th ‘order eqtiatioq, is given by the particular solution

,” ;. t.,. * ‘.;.*.i .,_‘/.‘. jonly* I :-

” .A . .‘: ji__4 Y i

/ .; : ‘IIf any of the roots’ of P*(a) isO’located to the’r#jht of’%& inagi&ry

‘I 1 1:. ,,;axis; the system desciibed by (8A.l) is unstable. In order to’be stable all

roots of the characteristic polynomial.. P*(s) should -lie i to the left of thei./. ’ I ‘.- .’iiimaginary axis. : .

.I, ,.r : ,. I’.iI& ‘/ .t.\

‘. . , A*‘i ,,:.

:. : k1.r: i / i,

.: +,4 .r.;

I .a.’ ,:q _‘, ,. : ” .. f, ... .L,, .

L.’ :, I ’ ,,A,-, ‘. . J,(’ . . .

- ‘.

APPENDIX8,.B4'HE SOLUTION OF A GENERAL SYSTEM OF,LINEAR DIFFERENTIAL EGUATIONS. I ; .- . I ,-,

Consider the following system of n linear differential equations:: .'

' dX15 - - allxl + al2x2 + l ** + alnxn + bllfl(t) + l ** + blmfm(t)dt .

i ' i 'i ~, i...,.! L,dki

:.iv 1. , _ , y; v

--$,= a2lxl -I- a22x2 + l ** ,f a2nXn f bqlflW + •*;~ +:b2nfm(t)

. . . . . . .: v: ‘:

dxn// ! /

-qTa x<,l 1.

+ a22x2 + '0. + bnmfm(t)8 '. ;, :

l ** f a2nxn + b,lfl(t) + :! ' !- j ,;* A I"‘ \ t 'A :5,,

I 1'with initial conditions xl(G), - ~~(0) - l ** I x,(O) - 0. In matrix: form the

* .jabove. equations yield,-

I/ j.,. ' )..'i'fl.

, ', " '; I a

di I" :-,., / ,',".‘&B‘:: . ;;, i .t* 1 ,. ‘.:~, .> 711 .,1:dt - & + Bf .. (8B.l),-j, . 3

where:

Taking the Laplrce transforms of .both sides of eqn. (8B.l) we have,s .: * ,

<s& - 8) g(s) '= @s,.

Or

gs) = (si - +g-l g(s) (8B.2)s/,

where: I =I [ 1111 O. = identity matrix; g(s) and f(s) the Laplace transforms

0 “1

of the vectors x(t) and f(t), respectively.

From eqn. (8B.2) by taking the inverse Laplace transform we have,

$t) - ~-lr.(sJ - g-l g(B)] (8~~3)

Let us recall that thk ii.ivrCr& of a matrix ‘# ite &@I ,$y+ ;; : ‘1. ” :*. . <

P :, ,(!!I : 2 : ,..: .,l I, ,:’ ” 1 ;, !” ”9

, : 1 : Lwhere adj($)x *a the adjoint ‘of m&+x S~U- (isee Appendix B)r:’ ,Thctefore, eqn. j ,

i8~.3),~g~es,., .i ir. ’ I ,.+ ,j ,? .! i , ,.. +’ ;I: 1

$t) *:t PA < /

The inversioq will be done using $a&ia$. ftactidns expansion. The polpnomfal,‘,P*;s) = 1s: - 4ii - sn + G-lC& +.;;Yp+ + ..�. l + c,B T cp

k. �& * .

is the characteristic polyn6miaJ of the fystem of equations and Its rootsi : y .: ‘,

Al, Q,***,Xnts i.e.’

P*(s) - Is& - kg = (8 - h$(s - AZ), “, (s :,+q ,) I**. 6” pi, .’j 1 (I.(

?, (, ,I ..:

are called the’ characteristic values or+ eigenvalues ofi.,tI$ aiatrk ,of : S.I 1. - :‘,: ,‘,{&i ,; ! id

coefficients &, which characterises the homogeneous’ iystk’af e&+ohs. . ,,.

.“$’ ! ‘,>

The eigdnvalues also determine ‘the forth of fh& solutiq6 ‘:z(t)‘. ‘,‘j “’I:/ v

II;);) : :: : *. . >; : / L i.*.. >; : / L i.

4%’4%’1.1. ,I;Jhat is 'the characteristic polynomial for a Pzi,r%t-order and +I socontl-,I;Jhat is 'the characteristic polynomial for a Pzi,r%t-order and +I socontl-

.. ic,.. ic,oz%jer eystem? Find i ts root&,oz%jer eystem? Find i ts roots,

: ,: ,‘,I‘,I

2.. Why are w%interested in-the roots of the charaeterlstic p’ol~no&~l of2.. Why are w%interested in-the roots of the charaeterlstic p’ol~no&~l of..’..’ : _/: _/ ..an.n-th order linear differential squation, or ‘a system of linear d.$f-an.n-th order linear differential squation, or ‘a system of linear d.$f-

,,I,,If erential equations?f erential equations? :: .’.’** ii dIdI

3.3. Ho~'doeg the prbcedure to’: $&@ute othe &netants of the terms resultingHo~'doeg the prbcedure to’: $&@ute othe &netants of the terms resulting

from t~e”pgre!ili’ fraotldns qpansibn v&y,ifrom t~e”pgrt!ili’ fraotldns qpansibn V&y;IjIj

I ’ $ 1I ' $ 1 _,:_ , :i”n ,thQ’preaence of, multiple ’i”n ,thQ’preaence of, multiple ‘

,.,. _,_,ro$qsi “, :; ;, :,, ,>* :: : L :,ro$qsi “, :; ;, :,, ,>* :: : L :,

’’1 (I1 (I .’.’ I.I.: ,,: ,,-. _,-. _,,.’,.’ .h.h a.,a.,

4.4. What ‘2s the complement&y eolutioni, and what; i$’ the particular eolutlonWhat ‘2s the complement&y eolutioni, and what; i$’ the particular eolutlon(. ., ,,. :’(. ., ,,. :’ j ; I. Ij ; I. I1’1’ $ &-, ’$ &-, ’ 3%. .’ 13%. .’ 1 I ./’ /I ./’ /* I.* I.for (a) an n-th order iin+ar differential equation, and (b) a 2x2 sys-for (a) an n-th order iin+ar differential equation, and (b) a 2x2 sys-

:: i! .’i! .’(.(. II ?? i.i.tem of lfnear diffaren~~al,8quatiansi ’ %WhatL do these eolutions mean?tem of lfnear diffaren~~al,8quatiansi ’ %WhatL do these eolutions mean?1.;1.; $. ‘.’$. ‘.’;:, ,;:, ,,,What fact&e’ deiermfne ‘.the@What fact&e’ deiermfne ‘.the@

‘, ,.,‘, ,.,,i. ‘; :,i. ‘; : ‘$,’‘$,’ 6. (6. (// ‘f .,‘f .,

::’’ :: _’_’5.5. Consider the foglowing ‘e$st+ of aimdtaneoue ifnear differential ”Consider the foglowing ‘e$st+ of aimdtaneoue ifnear differential ” ..

22equations .equations . :: : I: I ,(, (

dxl : I /dxl : I / * ‘_* ‘_I l.,I l., 1’ I. ~ ,<,.I i1’ I. ~ ,<,.I i

dt:dt: 811X1811X1 + a12xie ,” fg(t) with+ a12xie ,” fg(t) with Xl(O) - ~0Xl(O) - ~0::

1 dx21 dx2dt .-dt .- C2PlC2Pl t a22x2’ ‘j- f,(tjt a22x2’ ‘j- f,(tj withwith x2(0) - 0x2(0) - 0

--Show that this system ksn be converted to’ the following equivalentShow that this system ksn be converted to’ the following equivalent

sys tern:sys tern:

dX1dX1- -- -dtdt allXlallXl - h(t)- h(t)

and I

dx2bl 2

dx2+- .b2dt + b3x3 - g(t)

where bl, b2, b3 depend on ,all, a12, a22; h(t) and g(t) depend

on f,(t), ,f2(t) and their derivatives. Notice that the modified

. . . .

C’

System can b6 s~lv& seauentiallv and thlla mnre ma41v'+&&++kd.

1:,r, ! ‘_ ,

Table 8.1. Inverse Laplace Transforms of Selec!ted Expressions

Laplace Trans,form: I(s) xirn+ Function’: f(tI-J

4

,

r1. (*+a)l(B+b)

L.

3.

3(s+a) h:b) (f3+c) !

-ate e’Pt -ct(b-a>(c-a) + (c-b)(a-b) ” (a-c”) (b-c)

‘s+a

(Mb) (s+c) - ‘.’ ,-726T [ (a-b)‘dbt - (a-c)e’ct]

a \n at lcbt ’4.

5.

(s+b)’

a(s+bj3 ‘. /_

- - -

3 $ ,-btA -

r.i

.

6.-_

(a+b)“+’ ) ,. ‘i, ;, ::

3 t” ekbt

1 > ,$’ _,.. >.

7 .^.,._.

s(as + 1) “..:. 1I‘ _ ply

;j:I: ‘; a+t -t/a

.; ‘2.

>_: 7 -y- c. ,. ” ,;:

8. a

‘s(as + l).* .:,

2s.

* ., 9.

+ 11~0~ +, u2jl,

, .' -rwt

*2(s21 +?e, z;>p

10. 1(l+as)(s +w2 :? 2

- -(1 + il"wS)

12. 1 ’ e-at2 2(a+a) [(sfb) + w ] (a&j’ -ty w2 ’

te - ein(cJr- 4)w[(a-b)’ +.k2]1e

w h e r e cp - taa - (--“-1a-b.

_.’ ,

4-d

CHAPTER 9

TRANSFER FUNCTIONS AND THE INPUT-OUTPUT MODELS

The use of Laplace transforms allows us to form a very simple, convenient

and meaningful representation of chemical process dynamics. It is simple

because it uses only algebraic equations (not differential equations as we

have seen in Part II of this text). It is convenient because it allows a

quick analysis of process dynamics and finally, it is meaningful because it

provides directly the relationship between the inputs (disturbances, mani-

pulated variables) and the outputs (controlled variables) of a process.

9.1 THE TRANSFER FUNCTION OF A PROCESS WITH A SINGLE OUTPUT

Consider a simple processing system with a single input and a single

output (Figure 9.la). The dynamic behavior of the process is described by an

n-th order linear (or linearized nonlinear) differential equation;

a &+a d

n-l‘+

n dtn0.0 + al s + a,y = bf(t)n-l dtn-l . (9.1)

where y(t) and f(t) are the input and output of the process, respectively.

Both are expressed in terms of deviation variables.

that the process is described by one state variable

output variable].

[Note: We have assumed

which coincides with its

Assume that the system is initially at steady state. Then,

2Y(O) = (%I t=O = 84

dn-lL>dt2 t=o = l *. = cdtn-l tz() = O (9.2)

After taking the Laplace transform of both sides of (9.1) and using the initial

conditions (9.1) we find,

m - G(s) e b

m a,s" + an 1sn-l + l ** + als + a0(9.3)

G(s) is called the transfer function of the above system, and in a simple I.-

I

algebraic form it relates the output of a process to its input (Figure 9.lb).I

The‘diagram of Figure 9.lb is also known as the block diapram for the system.

If the process has two inputs, fl(t) and f2(t) as shown in Figure 9.2a,1

then its dynamic model is

a &+an-l

l ** + al% + a,y1

d y+n dtn n-l dtn-l = blfl(t) + b2f2(t) (9.4) _

Iwith the same initial conditions (9.2). From (9.4) we take,

Y(s) =bl b2 I

ansn + a n-l f,(s) +- n n-l Z,(s)

n-lS+ 0.0 + als + a0 ans +an-lS +***+as+al o

or equivalently,

Y(s) = G1(s)fl(s) + G2(s>7,b) (9.5)

with

G1(s) Ebl

and G2(s) Kb2

ansn+a sn-ln-l +***+as+al o ansn+a n-l

n-lS +***+as+al o

G1(s) and G2(s) are the two transfer functions which relate the output of the

process to each one of its two inputs. Thus, Gl(s) relates the y(s) to the

first input z,(s), and G2(s) relates j?(s) to the other input z,(s). These

relationships are shown by the block diagram of Figure 9.2b. A similar pro-

cedure can be applied to any system with one output and several inputs.

Figure 9.3 shows the block diagram for such a system.

Summarizing all the above, we can define the transfer function between an

input and an output as follows:

Transfer functionzG(s)= Laplace transform of the output, in deviation formLaplace transform of the input, in deviation form

(9.7)

hnilrk~. (1) The transfer function allows the development of a simpler

input-output model than that discussed in Section 5.1.

(2) It describes completely the dynamic behavior of the output

when the corresponding input changes. Thus, for a particular

variation of the input f(t) we can find its transform

F(s), and from (9.7) we see that the response of the system

is

Y(s) = G(s)ii(s)

Take the inverse Laplace transform of G(s)f(s) and you have

the response y(t) in the time domain.

(3) To find the transfer function for a nonlinear system, it

must first be linearized around a steady state and be

expressed in terms of deviation variables.

Example 9.1 - The Transfer Functions of a Stirred Tank Heater

The mathematical model of the stirred tank heater in terms of deviation

variables was developed in Example 5.1 and it is given by the equation

dT'- + aT' =dt t Tf + KT; (5.3)

where T', T;, T'S

are deviation variables, and

UAta = $+K,==L Fi/V and K = rP P

Take the Laplace transforms of both sides of (5.3);

(s+a)T'(s) = $ !i’f(s) + KTAb)

or

T’(s) = g T;(s) + & Q(s) (9.8)

Define the two transfer functions

Gl(4 = T'(s)/T;(s) and G2W = T'(s) /F(s)

Then,

T'(s) = G,(s)T;(s) + G2(s)T;(s) (9.8a)

and Figure 9.4 shows the block diagram for the tank heater. Gl(s) relates

the temperature of the liquid in the tank to that of the inlet stream, while

G2(s) relates the temperature of the liquid in the tank to that of the steam.

Remark. Compare the input-output model given by (9.8) and Figure 9.4 to-___

the more complex developed in Example 5.1 (eqn. 5.5 and Figure 5.1).

9.2 THE TRANSFER FUNCTION MATRIX OF A PROCESS WITH MULTIPLE OUTPUTS

Consider a process (Figure 9.5a) with two inputs, fl(t> and f,(t), and

two outputs, yl(t) and y2(t). Let its mathematical model be given by the

following two linear differential equations, with all the variables in

deviation form;

dyl- = ally1dt + alp2 + bllfl( t> + bl$2(t)

dy2-=dt a21yl + a22y2 + b21fl(t) + b22f2(t)

(9.9a)

(9.9b)

The initial conditions are

Yp = y2(0) = 0

[Note: Here again we have assumed that the process is described by two state

variables, which coincide with the two outputs yl and ~~-1

Take the Laplace transforms of both sides of the two eqns. (9.9a) and

(9.9b) and solve with respect to s,(s) and y,(s). [For the details of this

procedure see Section 8.3 and Example 8.21. Then,

[(s - a22)bll + a12b21l [(s -J;,(s) = 2 f,(s) + 2

a22)b12 + a12b22l?2(s)(9.lOa)

S - (all +a22)S -a12a21S - (a11+a22)s-a12a21

[(s - aY,(s) =

ll)b21 - a21bl11

.s2 - (a 11+a22)s-a12a21

ys) +[(s - all)b22 + a21b121

s2 - (a11+a22)S- a12a21

f2(s)(9.10b)

or

$4 = Gll(s)~l(s) + G12(s)f2(s)

Y2W = G21(5)+) + G22(4~2W

(9.11a)

(9.11b)

where the transfer functions G.,l, G12, G21 and G22 are defined as follows

(from eqns. (9.10a) and (9.10b));

s + (a12b21 - a22b11) s + (a12b22 - a22b12),

- (all +a22)S-a12a21 G12(S) E s2 - (all + a22) s - qp21

s + (a21b11 - a11b21) s + (a21b12 - a11b22)GZ1(4 - 2 , G22(4 - 2

S - (all + a22)S - a12a21 S - (all+a22)S - a12a21

The block diagram of the system

Remarks. (1) Eqns. (9.11a) and

matrix notation;

is shown in Figure 9.5b.

(9.11b) can be written as follows in a

The matrix of the transfer functions is called transfer

function matrix.

(2) For a system with two inputs and two outputs, like the

one discussed above, we have 2x2 = 4 transfer functions

to relate ail outputs to all inputs. For a general

process with M inputs and N outputs we will have

MxN transfer functions or a transfer function matrix

with N rows (number of outputs) and M columns

(number of inputs).

Example 9.2 - The Transfer Function Matrix of a CSTR

In Example 6.4 we developed the linearized model of a continuous stirred

tank reactor in terms of deviation variables, given by eqns. (6.36) and (6.37).

After rearranging the terms in these equations we take:

dcl; r- l2dt + L+ koe -E'RTjcL + E ;E'RTjT' = + cii

-E/RTot

Simplify the notation by defining

' =.i! T' +-UA T''A r i pcpV c

-E/RT0

all 'c=L+koe ,

-E/RT0

a21 =Jkoe ,

and

bl = l/-r ,

kOE-E/RTo

a12 = z e0

1 JkoE -E/RTo

a22 =-r -2 e+ uA

PCpV0

b2 = UA/bcJ)

Then, eqns. (9.13a) and (9.13b) become!

dcL- + all CL + al2 T' = blciidt

dT'- + a21 CA + a22 T' = blTi + b2TAdt

The initial conditions are:

(9.13a)

(9.13b)

(9.14a)

(9.14b)

c;(o) = T'(C) = C

Take the Laplace transforms of (9.14a) and (9.14b):

(s + all)?;(s) + al2 T'(s) = bl Ei (s)i

a21 CL(s) + (s + a22)T'(s) = blTf(s) + b2TA(s)

Solve for E:(s) and T'(s) and take:

qs> =bl(s + a22) a12bl a12b2

P(s) Ei (s)i

- ~.';'s' - p T;(s)

a21blWs) =- p(s> z; (s) +

bl(s + allI b2(s + all)

i P(s)P(s) +

P(s) T;(s)

(9.15a)

(9.15b)

2where P(s) E s + (all + a22)s + (alla22 - a12a21).

In a matrix form, eqns. (9.15a) and (9.15b) are written as follows:

cp

H

=f’(s)

Q) G12W G13W

G2#d G22b) G2+4-

In Table 9.1 we see the six transfer functions corresponding to the CSTR. These

can be derived easily from eqns. (9.15a) and (9.15b). The transfer function

matrix is nonsquare since the number of inputs is not equal to the number of

outputs.

Transfer function matrix = E(s) =

Figure 9.6 shows the input-output model for the CSTR in a block-diagram form.

Table 9.1. The Components of the Transfer Function Matrix for the CSTR Ioutput Input G

-ij Element Transfer Function

am cd, (s)i G1l. bl(s+a22)/P(s)

'i's) G12 -a12bl/P(s)

y(s) G13 -a12b2/P(s)

T’ (s) Eb, (s)i G21

G22

G23

bl(s+ all)/P(s)

b2(s+all)/P(s)

III

9.3 THE POLES AND THE ZEROS OF A TRANSFER FUNCTION

According to the definition of a transfer function we have

34 = G(s) IZ(s)

In general, the transfer function G(s) will be the ratio of two polynomials,I

G ( s ) = $$

The only exception are system with time delays which introduce exponential

terms (see Section 7.2(E)). For physically realizable systems, the polynomial I

Q(s) will always be of lower order than the polynomial P(s). The reasons will

become clear in subsequent chapters. For the time being, all the examples we 1have covered satisfy this restriction.

I1The roots of the polynomial Q(s) are called the zeros of the transfer

function, or the zeros of the system whose dynamics are described by the

transfer function G(s). When the variable s takes on as values the zeros

of G(s), the transfer function becomes zero.

The roots of the polynomial P(s) are called the poles of the transfer

function, or equivalently, the poles of the system. At the poles of a system

the transfer function becomes infinity.

The poles and the zeros of a system play an important role in the

dynamic analysis of processing systems and the design of effective controllers.

As we proceed along in the text, their usefulness will become clearer.

Example 9.3 - Poles and Zeros of the Stirred Tank Heater

The input-output model of the tank heater was developed in Example 9.1 and

it is given by:

i'(s) = G1(s) Tf(s) + G2(s)T;(s) (9.8a)

The transfer function Gl(s) is

l/TGlb) = s+a

and has no zeros and one pole at s = -a. Similarly, the transfer function

G2(s) which is given by

G2 = sta

has no zeros and one pole at s = -a. Notice that the two transfer functions

have a common pole.

Example 9.4 - Poles and Zeros in a CSTR

The transfer functions corresponding to the CSTR were developed in

Example 9.2 and are summarized in Table 9.1. All six transfer functions have

common denominator,

P(S) 5 'S2 + (all + a22)s + (alla22 - a12a21)

23

and therefore common poles. Since P(s) is a 2nd order polynomial, the system

has two poles which are given by:

-

-(all + a221 + (all - a221 2 + 4a12a21

p1,2 = 2

With respect to the zeros, the six transfer functions differ.

G12(s) and G13(s) have no zeros

G22(~) and G23(~) have one common zero at s = -all

Gll(s) has one zero at s = -a22.

9.4 QUALITATIVE ANALYSIS OF THE RESPONSE OF A SYSTEM

The dynamic response of an output y is given by

Y(s) = G(s) F(s)

For given input f(t) we can find easily its Laplace transform Z(s), while

the transfer function G(s) is known for the partiuclar system. Therefore,

the response y(t) in the time domain can be found if we invert the term

G(s)%).

Furthermore, in general,

G(s) = s

while the Laplace transform of all common inputs can also be expressed as the

ratio of two polynomials (see examples in Chapters 7 and 8 as well as Tables

7.1 and 8.1);

Pi(S)f(s) = -

P2(S)

Consequently,

P,(S)f(s) = $g -

P2(S)(9.16)

To invert the right hand side of (9.16) using the method of partial fractions

we need to know the roots of the polynomial P(s), i.e. the poles of the sys-

tem, and the roots of the polynomial p,(s). The terms resulting from the

inversion by partial fractions are uniquely characterized by the poles of the

system and the roots of Pz(s). Therefore, if we know where the poles of a

system are located we can determine the qualitative characteristics of the

system's response to a particular input, without additional computations.

Let us use the following general example to clarify the above statement.

Suppose that the transfer function of a system is given by,

G(s) =s = Q(s) (9.17)(s-Pl)(s-P~)(s-P3)m(s-P4)(s-P~)(s-P5)

where pl, p2, p3, p4, p2 and p5 are the roots of P(s), i.e. the poles of

the system located at various points of the complex plane (see Figure 9.7).

The partial-fractions expansion of G(s) will yield the following terms:

c1 c2G(s) = s- + s-

Pl P2

+ '4 + '2 '5y+-

s-p4 s-p4 S--P5

The following observations can be made for the location of the poles:

A. Real, Distinct Poles, like pl and p2, are located on the real axis

(Figure 9.7). During the inversion, give rise to exponential terms like

C epit p2t

1 andC2e

Since p1 < 0, Cle pit decays exponentially to zero as t -f 00 (Figure

9.8a). Also, because p2 > 0, C2eP2t grows exponentially to infinity

with time (Figure 9.8b). Therefore, distinct poles on the negative

axis produce terms which decay to zero with time, while real positive

poles make the response of the system grow towards infinity with time.

B. Multiple, Real Poles, like p3 which is repeated m times. Such poles

give rise to terms like

(c31 + J$ t + A$ t2 + . . . + (1:;) , tm-llep3t. . .

The term within the parenthesis gorws towards infinity with time. The

behavior of the exponential term depends on the value of the pole p3;

- if P3 ' 0p3tthen e + ~0 'as t + w

- if P3 < 0p3tthen e + 0 as t -f 00 and

-if p3=0p3t

then e = 1 for all times.

Therefore, a real, multiple pole gives rise to terms which either grow.

to infinity, if the pole is positive, or decay to zero if the pole is

negative.

C. Complex Conjugate Poles, like the p4, pt. We should emphasize that

complex poles appear always in conjugate pairs and never alone. Let,

P4 = a + jf3 and p$ = a - jS

In Section 8.2 we have seen that conjugate pairs of complex roots give

rise to terms like

eat sin(Bt + $) .

The sin(Bt + $) is a periodic, oscillating function, while the behavior

of eat depends on the value of the real part a. Thus,

- If a>0 then eat + m as t+m,and eat sin(St + $) grows to

infinity in an oscillating manner (Figure 9.9a).

- If a<0 then e + 0 as t + w, and eat sin(Bt + +I) decays toat

zero in an oscillating manner with ever decreasing amplitude (Figure

9.9b).

- If a=0 then eat = 1 for all times, and eat sin(bt + 4) = sin(St+$) /

which oscillates continuously (Figure 9.9c).

Therefore, a pair of complex conjugate poles gives rise to oscillatory

behavior, whose amplitude may grow continuously if the real part of the

complex poles is positive, decay to zero if it is negative, or remain

unchanged, if the real part of the poles is zero.

D. Poles at the Origin. Pole p5 is located at the origin of the complex

plane, i.e. p5 = 0 + j-0. Therefore, C5/s - p5 = C5/s and after

inversion it gives a constant term C5.

Remarks. (1) The above observations are general and can be applied to any

system. Thus, we can find the qualitative characteristics of

system's response if we know where are the poles of the cor-

responding transfer function located. It is obvious that

for a particular input, f(t), we should consider the

additional roots introduced by the denominator of f(s),

before we can have the complete picture of the qualitative

response of a system.

(2) Poles to the right of the imaginary axis grow exponentially

to infinity with time. Such systems with unbounded behavior

are called unstable. Therefore, a system will be stable (i.e.

with bounded behavior) if all the poles of its transfer

function are located to the left of the imaginary

9.7). In subsequent chapters we will define more

the stability of a system.

SUMMARY AND CONCLUDING REMARKS

axis (Figure

precisely

The use of Laplace transforms allows us to develop a very simple and con-

venient to use input-output model of chemical process. This model is based on

the concept of the transfer function, which is defined in the s-domain (complex

plane).

The transfer function between specific input and output is defined as the

ratio of the Laplace transform of the given output, over the Laplace transform

of the given input, provided that both have been expressed in deviation form.

For a process with N outputs and M inputs we have NxM transfer

functions, which compose an NxM transfer functions, which compose an NxM

transfer function matrix.

The block diagram is a very illustrative representation of the interactions

between the various inputs and outputs of a process. From the block diagram

we can identify very quickly; (a) what input affects what output, and (b) by

how much.

The zeros of a system are the values of s which make its transfer

function zero, while the poles of a system are those values of s which give

infinite value to the transfer function.

The location of the poles of a system will determine the qualitative

response of a system to external inputs. If the real part of any pole is

positive, the system is unstable.

Now, we are ready to analyze the dynamic behavior of various processes,

which are modeled with certain common forms of transfer functions, In

Chapter 10 we will study the dynamic response of the first-order systems,

leaving the higher order systems for Chapter 11.

2 .:

1 .

2.

3.

4.

5 .

6 .

7 .

8.

9 .


Define the transfer function. Why is it useful?

For a process with four inputs (disturbances and manipulated variables)

and three measured outputs, how many transfer functions should you

formulate and why? What is the corresponding transfer function matrix?

In Section 5.1 we developed a different type of input-output model.

Would you prefer that over the input-output model based on the transfer

function concept? Elaborate on your answer.

What is the block diagram of a process? What type of information does

it convey?

Equations (4.4a) and (4.jb) constitute the complete mathematical model

of a stirred tank heater. Develop the input-output model for the

process, by formulating the necessary transfer functions. Draw the

corresponding block diagram. Analyze the interactions among inputs and

outputs. What do you observe? (Hint. Start by linearizing the

modeling equations and expressing the variables in deviation form.)

Draw the block diagram of the distillation column shown in Figure 4.10.

Can you develop analytically the transfer functions among the various

inputs and outputs? If yes, explain how, but don't do it.

The stirred tank heater of Example 9.1, is it a stable system or not,

and why? For what values of the parameters a, r and K is it stable?

Can it become unstable? k

Does the location of the zeros of a system affect its response to

external inputs? Elaborate on your answer.

Repeat question 8 above, but taking the location of the poles of a system

into account.

10. Show that the poles of a 2x2 system are also the eigenvalues of the

,.. ., ,.r;<’I

matrix of constant coefficients in the dynamic model of the system.

11. Under what conditions can the CSTR of Example 9.2 become unstable?

IIIIIIIIIII

.

r------------,

I Fiqure 9.2 I

!L-e-,--,-- ,,,J

W

I 3Fi we 9.4I

..-

c

b)

I Figure 9.51

I 1 I

I

I ,II

? G&ll

IIIII

I

II

II

IIII

I

I,- --_____ _____ __ ____- _ _____ !.

I

I

I

I

ii,.*P4

(P>3

- --

,, 3 Real Axis

CHAPTER 10

THE DYNAMIC BEHAVIOR OF FIRST-ORDER SYSTEMS

The previous chapters of Part III have provided us with all the pre-

requisite tools we need in order to analyze the dynamic behavior of typical

processing systems, when their inputs change in some fashion, e.g. step, ramp,

impulse, sinusoid, etc. In this section we will examine the so-called first-

order systems. In particular, we will study:

- what is a first-order system and what physical phenomena give rise to

first-order systems,

- what are its characteristic parameters, and

- how does it respond to the various changes in the input variables

(disturbances and/or manipulated variables).

10.1 WHAT IS A FIRST-ORDER SYSTEM?

A first-order system is one whose output, y(t), is modeled by a first-

order, linear differential equation

Liz?!al dt + 'soy = bf(t)

where f(t) is the input (forcing function). If a0 # 0, then eqn. (10.1)

yields,

k!Y + y =

a0 dt $- f(t)0

Define

al-=TP '

and b

a0= Ka

0 P

4.Y'p dt + y = K f(t)

and take

(10.2)

I TPis known as the time constant of the process and K

Pis called the

steady state gain or static gain or simply the gain of the process. Their

physical meaning will become clear in the next three sections.

If y(t) and f(t) are in terms of deviation variables around a steady

state, then the initial conditions are:

y(O) = 0 and f(0) = 0

From eqn. (10.2), it is easily found that the transfer function of a first-

order process is given by;

G(s) = ti = ; :p+ 1w4 P

(10.3)

A first-order process with a transfer function given by eqn. (10.3) is also

known as: first-order lag, linear lag, exponential transfer lag.

If on the other hand, a0 = 0, then from eqn. (10.1) we take

a,dt 2. f(t)

al= K; f(t)

which gives a transfer function

G(s) =

F(s)S

(10.4)

In such case the process is called purely capacitive or pure integrator.

10.2 PROCESSES MODELLED AS FIRST-ORDER SYSTEMS

The first-order processes are characterized by:

(a) their capacity to store material, energy or momentum, and

(b) the resistance associated with the flow of mass, energy or momentum in

reaching the capacity.

Thus, the dynamic response of tanks which have the capacity to store liquids

or gases can be modeled as first-order. The resistance is associated with the

pumps, valves, weirs, pipes which are attached to the inflowing or outflowing

liquids or gases. Similarly, the response of solid, liquid, or gaseous systems

IIIIIIII

III

which can store thermal energy (thermal capacity, cp) is modeled as first-order.

For such systems the resistance is associated with the transfer of heat through

walls, liquids or gases. In other words, a process which possesses a capacity

to store mass or energy and thus act as a buffer between inflowing and out-

flowing streams will be modeled as a first-order system. The stirred tank

heater of Example 4.4 and the mixing processes of Example 4.11, are typical

examples of first-order processes.

It is clear from the above that the first-order lags should be the most

common class of dynamic components in a chemical plant, with the capacity to

store primarily mass and energy.

Let us examine now some typical capacity processes modeled as first-order

lags.

Example 10.1 - A First-Order System with a Capacity for Mass Storage

Consider the tank shown in FigurelO.la. The volumetric (volume/time)

flow in is Fi and the outlet volumetric flowrate is Fo. In the outlet

stream there is a resistance to flow such as a pipe, valve, wire, etc. Assume

that the effluent flowrate F. is related linearly to the hydrostatic pressure

of the liquid level h, through the resistance R, i.e.

F =$= Driving Force for Flow0 Resistance to Flow

(10.5)

At any time point, the tank has the capacity to store mass. The total mass

balance gives:

Adh = F _ F = hdt i 0 Fi - it

or

AR$+h=RF i (10.6)

where A is the cross sectional area of the tank. At steady state

hw = R Fi(s)and from (10.6) and (10.6a) we take the following equation in terms of

deviation variables:

(10.6a)

AR% + h' = RF!1 (10.7)

where h'=h-h(s)

and F;=Fi-F.l(S) l

Let

TP= m = time constant of the process, and

KP=R = the steady state gain of the process

then, the transfer function is:

i; ’ 6s) KG(s) = - = P

p;(s)TpS + 1

(10.8)

Certain notes are in order.

(1) The cross sectional area of the tank, A, is a measure of its capacitance

to store mass. Thus, the larger the value of A the larger the storage

capacity of the tank.

(2) Since ~~ = AR we can say that for the tank we have

(time constant) = (storage capacitance)x(resistance to flow) (10.9)

Example 10.2 - A First-Order System with a Capacity for Energy Storage

The liquid of a tank is heated with saturated steam, which flows through

a coil immersed in the liquid (Figure 10.2). The energy balance for the sys- I

tern yields:

Vpc dT = QP dt

= UAt(Ts - T) . (10.10)

where

V = volume of liquid in the tank

P,Cp = its density and heat capacity

U = the overall heat transfer coefficient between steam and liquid

At = the total heat transfer area and

Ts = the temperature of the saturated steam.

The steady state is given by

0 = UA(Tt s(s) - T(s)) (10.11)

Subtract (10.11) from (10.10) and take the following equation in terms of

deviation variables,

vpcp $ = UAt(T; - T') (10.12)

where T'=T-T(s)

and Ti = T, - T(s) *

The Laplace transforms of (10.12)

will yield the following transfer function

T’(s) UAt KG(s) I ~ = = P

TpTpS + 1

(10.13)

where

TP= time constant of the process = Vpcp/UAt

KP

= steady state gain = UAt

Remarks. (1) Eqn. (10.13) demonstrates clearly that this is a first-order

lag system.

(2) The system possesses the capacity to store thermal energy

and a resistance to the flow of heat characterized by U.

(3) The capacity to store thermal energy is measured by the

value of the term Vpc . The resistance to the flow ofP

heat from the steam to the liquid is expressed by the term

l/(UA$. Therefore, we notice that the time constant of

this system is given by the same equation as that of the tank

system in Example 10.1, i.e.

VPC,(time constant) = ~~ =J=(storagecapacitance)x(resistanceto flow).uA

t 1

Example 10.3 - A Pure Capacitive System

Consider the tank discussed in Example 10.1 with the following difference, I

"The effluent flowrate :Fo is determined by a constant-displace-

ment pump and not by the hydrostatic pressure of the liquid

level h (Figure 10,lb)"

In such case the total mass balance around the tank yields

dhA dt = Fi - F.

At steady state

(10.14)

0 = Fi(s) - F0

(10.15)

Subtract eqn. (10.15) from (10.14) and take the following equation in terms of

deviation variables:

which yields the following transfer function

P(s) l/AG(s) = - r - (10.16)F;(s)

S

Remark. A pure capacitive process possesses the capacitance to store mass,

energy, or momentum but there is no resistance associated with the

flow of mass, energy or momentum in reaching the.capacitor. The

lack of such resistance is not encountered often in physical

phenomena, and consequently the purely capacitive processes are

rather rare.

10.3 THE DYNAMIC RESPONSE OF A PURE CAPACITIVE PROCESS

The transfer function for such process is given by eqn. (10.4)

K’G(s) = f(s> = +m

Let us examine how y(t) changes with time, when f(t) undergoes a unit step

change, i.e.

f(t) = 1 for t>O

We know that for a unit step change

P(s) = f

Therefore, eqn. (10.4) yields

Y(s) = K;/s2

and after inversion we find (see Table 7.1)

y(t) = K;'

We notice that the output grows l-inearly with time in an unbounded fashion.

Thus,

I

y(t) -+ Q) as t-00 (Figure 10.3)

Such behavior, characteristic of a pure capacitive process, lead to the name

I pure integrator, because it behaves as if there were an integrator between its

input and output.

IA pure capacitive process has no steady state, i.e. a state of natural

equilibrium.

I

Its presence in a chemical plant will cause serious control

problems, because it cannot balance itself.

I In the tank of Example 10.3, we can adjust manually the speed of the

constant-displacement pump, so as to balance the flow coming in and thus keep

Ithe level constant. But, any small change in the flowrate of the inlet stream

will make the tank flood or run dry (empty). This attribute is known as

I non-self-regulation.

Processes with integrating action are quite common in a chemical process.

The most often encountered are tanks with liquids, vessels with gases,

inventory systems for raw materials or products, etc.

10.4 THE DYNAMIC RESPONSE OF A FIRST-ORDER LAG SYSTEM

The transfer function for such systems is given by eqn. (10.3).

G(s) = y(s) = KP

mrps+l

(10.3)

Let us examine how it responds to a unit step change in' f(t). Since f(s) =

l/s, from eqn. (10.3) we take

KY(s) = P Kp--

s(rps+l) = sKPrps+l

(10.17)

Inverting eqn. (10.17) we take,

-t/TY(t) = Kp(l - e ') (10.18)

If the step change in f(t) were of magnitude A, then the response would be

-t/Ty(t) = AKp(l - e ') (10.19)

Figure 10.4 shows how y(t) changes with time. The plot is in terms of the

dimensionless coordinates

y(t)/AK p vs. t/T P '

and as such can be used to determine the response of any typical first-order

system, independently of the particular values of A, KP

andTP'

Several features of the plot of Figure 10.4 are characteristic of the

response of first-order systems and thus worth remembering. These features

are:

(1) A first-order lag process is self-regulating. Unlike a purely capacitive

process, it reaches a new steady state. In terms of the tank system in

the Example 10.1, when the inlet flowrate increases by unit step, the

liquid level goes up. As the liquid level goes up, the hydrostatic

pressure increases, which in turn increases the flowrate F. of the

effluent stream (see eqn. (10.5)). This action works towards the

restoration of an equilibrium state (steady state).

(2) The slope of the response at t=O is equal to 1.

d[W/AKpl -t/rdt It=0 =

(e P)t,o = 1

This implies that if the initial rate of change of y(t) were to be

maintained, the response would reach its final value in one time

constant (see dotted line of Figure 10.4). The c llary conclusions

are:

The smaller the value of the time constant T the steeper theY

initial response of the system.

Equivalently,

The time constant T ^ of a process is a measure of the time necessaryY

for the process to adjust to a change in its input.

(3) The value of the response y(t) reaches the 63.2% of its final value

when the time elapsed is equal to one time constant, T .P

Subsequently,

we have:

Time elapsed2TP 3TP 4sP

y(t) as percentage of its ultimate value 86.5 95 98

Thus, after four time constants, the response has essentially reached.

its ultimate value.

(4) The ultimate value of the response, i.e. its value at the new steady

state is equal to K p for a unit step change in the input, or AKP

for a step of size A.

This characteristic explains the name steady state or static gain given to the

parameter KP'

since for any step change in the input

K =A(output)s

P A(input)s(10.20)

where A(output)s = change in the steady state values of the output caused by

A(input)s = change in the steady state values of the input. Equation (10.20)

also tells us by how much should we change the value of the input in order to

achieve a desired change in the output, for a process with given gain, K .P

Thus, in order to effect the same change in the output, we need

l a small change in the input if KP

is large (very sensitive systems),

and

l a large change in the input if KP

is small.

Example 10.4 - The Effect of Parameters on the Response of a First-Order System

Consider the tank system of Example 10.1. It possesses two parameters;

- the cross sectional area of the tank, A and

- the resistance to the flow of the liquid, R,

or from another but equivalent point of view,

- the time constant of the process T andP

- the static gain, K .P

Consider two tanks with different cross sectional areas Al and A2 where

A1 > A2 and the same resistance, R. From eqn. (10.9) we find that T > TPl p2'

i.e. the tank with the larger capacity has a larger time constant, while the

static gains remain the same. When we subject the two tanks to the same unit

step changes in the inlet flowrates, the liquid level in each tank responds

according to eqn. (10. ) and its behavior is shown in Figure 10.5a. We

notice that the tank with the smaller cross sectional area responds faster at

the beginning, but ultimately, both levels reach the same steady state values.

This is in agreement with our physical experience. Suppose now that both tanks

have different cross sectional areas A1 and A2 and different flow

resistances R1 and R2, such that

A1 R2

AZ =q(10.21)

Equation (10.21) yields:

Tpl= AlRl = A2R2 = rP2

But, since Al > A2 then from eqn. (10.21) R2 > Rl which implies that

K >Kp2 Pl'

Figure 10.5b shows the responses of the two tanks to a unit step

change in the input. Since both tanks have the same time constant, they have

the same initial speed of response. But, as the time goes on, the tank with

the larger resistance R2 allows less liquid out of the tank. Thus, the

liquid level grows more in this tank and its ultimate value is larger than

the value of the level in the tank with resistance Rl. This again agrees

with our physical experience

the static gain of a process

for the same input change.

and also demonstrates the fact that the larger

the larger the steady state value of its output

10.5 FIRST-ORDER SYSTEMS WITH VARIABLE TIME CONSTANT AND GAIN

In the previous sections we assumed that the coefficients of the first-

order differential equation (see eqn. (10.1)) were constant. This lead to the

conclusion that the time constantTP

and steady state gain KP

of the pro-

cess were constant. But, this is not true for a large number of components in

a chemical process. As a matter of fact, in a chemical plant, we will encounter

more often processes with variable time constants and gains than not.

Let us examine two characteristic examples:

Example 10.5 - A Tank System with Variable Time Constant and Gain

For the tank system discussed in Example 10.1, assume that the effluent

flowrate, Fo, is not a linear function of the liquid level, but it is given

by the following relationship (which holds for turbulent flow);

F. = BJi; , [3 = constant

Then, the material balance yields the following nonlinear equation:

Linearize this equation around a steady state and put it in terms of deviation

variables (this problem was solved in Examples 6.1 and 6.2);

Adh’+ f3 h’=dt F;

or

dh’ + h’rp dt = KF!

PI

where

rP= 2A$7O and K

P = 24733

We notice that both the time constantrP

and the steady state gain KP

depend on the steady state value of the liquid levelhw *

Since'we can vary

the value of h (s) by varying the steady state value of the inlet flowrate

Fi(s) 2 we conclude that the system has variable time constant and static gain.

Example 10.6 - A Heater with Variable Time Constant and Gain.

Let us return to the heater system discussed in Example 10.2. The time

constant and the static gain for the heater were found to be:

TP and KP

= UAt

The overall heat transfer coefficient, U, does not remain the same for a long

period of operation. Corrosion, dirt, various other solids deposited on the

internal or external surfaces of the heating coil, result in a gradual decrease

of the heat transfer coefficient. This in turn will cause the time constant

and static gain of the system to vary. This example is characteristic of what

can happen to even simple first-order systems.

The question then arises as to how one handles first-order systems with

variable time constants and static gains in order to find the dynamic response

of such systems. There are two possible solutions:

- First; we can use the analytical solutions which are available for first-

order differential equations with variable coefficients. Such solutions

are quite complicated and of very little value to us for process control

purposes.

- Second; we can assume that such systems possess constant time constants

and static gains for a certain limited period of time only. At the end of

such period we will change the values of T and K and consider that weP P

have a new first-order system with new but constant r and KP P'

which

will be changed again at the end of the period. Such an adaptive procedure

can be used successfully if the time constant and the stagic gain of a

process change slowly, in which case the time period of relatively constant

values is rather long.

1 SUMMARY AND CONCLUDING REMARKSI

First-order is a process whose dynamic behavior is governed by a first-

order, linear differential equation. All such processes are characterized by

their capacity to store material, energy or momentum, and constitute the most

common components in a chemical plant. Therefore, the majority of simple

input-output configurations that we will encounter in a chemical plant will be

represented by first-order dynamics with a transfer function

KG(s) = +

P

The two constants which characterize a first-order system are; the time

constant of the process rP

and the static gain K . The first is a measureP

of how long a process takes to adjust itself to the new value of an input,

while the second indicates the size of the change in the steady state value

of an output resulting from a unit input. Therefore, the time constant is-

associated with the dynamic behavior of a system, while the static gain is

related to its steady state behavior.

A purely capacitive process arises from a process which has only capacity

and no resistance to the flow of mass, energy, in reaching the capacitor. It

is a non-self-regulating process leading to serious control problems, unlike

the first-order lags which are self-regulating, thus causing milder control

problems.

In Chap ter 11 we will study the dynamic behavior of systems with higher

order dynamics. Particular attention will be given to the 2nd order systems.

Furthermore, we will examine how several simple capacity processes with first-

order dynamics, combined with each other in various ways, give rise to

higher-order systems.

I1IIIIIIIIII1III


1.

2.

What is a first-order s'stem and how do you derive the transfer functions

of a first-order lag or of a purely capacitive process?

What is the principal characteristic of the first-order processes and

what causes the appearance of a purely capacitive process?

3. In Examples 10.1 and 10.2 it was found that for a first-order process

4.

5.

6.

7 .

8.

9 .

(time constant) = (storage capacity)x(resistance to flow)

Is this appropriate for an isothermal, constant volume CSTR, where a

simple, irreversible reaction, A-tB, takes place?

Show that a tank with variable cross-sectional area along its height

also has variable time constant and static gain.

Discuss a system which stores momentum and exhibits first-order dynamics.

How would you regulate the purely capacitive process of the tank in

Example 10.3 so that it does not flood or run dry?

Consider a closed vessel with air flowing in it. Is this a pure

capacitive or a first-order lag system? Answer the same question if

the vessel is supplied also with an exit for the air.

Study the response of a first-order lag to a unit impulse input.

(Recall that for a unit impulse F(s) = 1).

Study the response of a first-order lag to a sinusoidal input. What

do you observe in its behavior after a long time, i.e. as t*?

257

0 1234I t/q

FJI’ ure m4

CHAPTER 11

THE DYNAMIC BEHAVIOR OF SECOND-ORDER SYSTEMS

Systems with first-order dynamic behavior are not the only ones encountered

in a chemical process. An output may change under the influence of an input,

in a drastically different way than that of a first-order system, following

higher-order dynamics. In this chapter we will analyze, (a) the physical

origin of systems with second-order dynamics, and (b) their dynamic

characteristics. The analysis of systems with higher than second-order

dynamics will be left for Chapter 12.

11.1 WHAT IS A SECOND-ORDER SYSTEM?

A second-order system is one whose output, y(t), is described by the

solution of a second-order differential equation. For example, the following

equation describes a second-order system:

&a2 dt2

&?L+ al dt + ao = bf(t)

If a0 # 0, then eqn. (11.1) yields

2 d2y-c-dt2

+ 25~ g + y = Kpf(t)

(11.1)

(11.2)

where

T2 = a2/ao 25~ = al/a0 and KP= b/a0

Equation (11.2) is in the standard form of a second-order system where

T = the natural period of oscillation of the system

?l = the damping factor, and

K =P

the steady state, or static, or simply gain of the system.

The physical meaning of the parameters r and 5 will become clear in the

next two sections, while KP

has the same significance as for the first-order

systems.

If eqn. (11.2) is in terms of deviation variables, the initial conditions

are zero and the Laplace transformation of eqn. (11.2) yields the following

standard transfer function for a second-order system;

G(s) = 'j(s> =KP /

f(s) T2S2 + 25TS + 1(11.3)

Systems with second or higher-order dynamics can arise from several physical

situations. These can be classified into three categories:

(i> Multicapacity Processes, i.e. processes which consist of two or more

capacities (first-order systems) in series, through which material

or energy must flow. In Section 11.3 we will discuss the character-

istics of such systems.

(ii> Inherently Second-Order Systems, like the fluid or mechanical solid

components of a process which possess inertia and are subjected to

acceleration. Such systems are rare in chemical processes. They

will be discussed briefly in Section 11.4 and two examples are given

in the Appendix ll.A at the end of this chapter.

(iii) A Processing System With Its Controller, may exhibit second or higher

order dynamics. In such cases, the controller which has been

installed on a processing unit introduces additional dynamics which,

when together with the dynamics of the unit, give rise to second or

higher-order behavior. An example in Section 11.5 will demonstrate

this point.

The very large majority of the second or higher order systems encountered

in a chemical plant come from multicapacity processes or the effect of process

control systems. Very rarely we will find systems with appreciable, inherent

second or higher order dynamics.

11.2 THE DYNAMIC RESPONSE OF A SECOND-ORDER SYSTEM

Before we proceed to examine the physical origin of second and higher-

order systems, let us analyze the dynamic response of a second-order system to

a unit step input. Such analysis will provide us with all the fundamental

dynamic features of a second-order system.

For a unit step change in the input, f(t), eqn. (11.3) yields:

Kf(s) = P

S(T2S2 + 2<TS + 1)(11.4)

The two poles of the second-order transfer function are given by the roots of

the characteristic polynomial, i.e.

T2S2 + 25TS + 1 = 0

and they are

/--c2-1p1=-$+ T

and

(11.5a)

J52_1p2=-$- T

Therefore, eqn. (11.4) becomes

(11.5b)

Y(s) = KPs(s - Pl)(S - P2)

(11.6)

and the form of the response y(t) will depend on the location of the two

poles, p1 and p2' in the complex plane (see Section 9.4). Thus, we can

distinguish three cases:

Case A - when ~-1 we have two distinct and real poles

Case B - when c=l we have two equal poles (multiple pole)

Case C - when 5~1 we have two complex conjugate poles.

Let us examine each case separately.

CASE A. Overdamped Response, when <>l.

In this case, the inversion of eqn. (11.6) by partial-fractions expansion

yields,

y(t) = Kp

where cash(0) and sinh(*) are the hyberbolic functions defined by

l- e-5t'r(cosh&$+l

f-c2-1-

(11.7)

a -asinh(a) = e -2 e and cash(a) = e' + e-'

2

The response has been plotted in Figure ll.la for various values of 5, <>l.

It is known as overdamped response and resembles a little the response of a

first-order system to a unit step input. But when compared to a first-order

response we notice that the system initially delays to respond and then it's

response is rather sluggish. It becomes more sluggish as 5 increases, i.e.

the system becomes more heavily overdamped. Finally, we notice that as the

time goes on, the response approaches its ultimate value asymptotically. As

it was the case with first-order system, the gain is given by,

K = A(output steady state)P A(input steady state)

Overdamped are the responses of multicapacity processes, which result from the

combination of first-order systems in series, as we will see in Section 11.3.

CASE B. Critically Damped Respone, when 5~1.

In this case, the inversion of eqn. (11.6) gives the result,

y(t) = Kp[l - (1 + $)e-t/Tl (11.8)

The response is also shown in Figure ll.la. We notice that a second-order

system with critical damping a-proaches its ultimate value, faster than an

overdamped system.

CASE C. Underdamped Response, when <<l.

The inversion of eqn. (11.6) in this case yields,

e-Tt/.r1 sin(wt + $)

I-l-c2 1where

/-l-c2oJ"-T

(11.9)

(11.10)

and

4 tan-l l-C2I-=5

The response has been plotted in Figure ll.lb for various values of the damping

factor, 5. From the plots we can observe the following:

- The underdamped response is faster than the critically damped or overdamped

responses, which are characterized as sluggish.

- Although the underdamped response is faster and reaches its ultimate value

quickly, it does not stay there but it starts ocillating with progressively

decreasing amplitude. This oscillatory behavior makes an underdamped

response quite distinct from all previous ones.

- The oscillatory behavior becomes more pronounced with smaller values of the

damping factor, 5.

It must be emphasized that almost all the underdamped responses in a

chemical plant are caused by the interaction of the controllers with the process

units they control. Therefore, it is a type of response that we will encounter

very often, and it is wise to become well acquainted with its characteristics.

Characteristics of an Underdamped Response

Let us use as reference the underdamped response shown in Figure 11.2, in

order to define the terms used to describe an underdamped response.

1. Overshoot: Is the ratio A/B, where B is the ultimate value of the

response and A is the maximum amount by which the response exceeds

its ultimate value. The overshoot is a function of r, and it can be

shown that is given by the following expression:

OVERSHOOT = exp(x) (11.11)

/-l-c2

Figure 11.3 shows the plot of overshoot vs. 5 given by eqn. (11.11).

We notice that the overshoot increases with decreasing 5, while as 5

approaches 1 the overshoot approaches zero (criticaly damped response).

2. Decay Ratio: Is the ratio C/A, i.e. the ratio of the amounts above

the ultimate value of two successive peaks. The decay ratio can be

shown to be related to the damping factor 5 through the equation

DECAY RATIO = exp(----2'r; ) = (OVERSH~~T)~ (11.12)

I-l-c2

Equation (11.12) has been also plotted in Figure 11.3.

3. Period of Oscillation: From eqn. (11.10) we see that the radian

frequency (radians/time) of the oscillations of an underdamped response

is given by

f-w = l-G2radian frequency = ___T (11.10)

To find the period of the oscillation, T, i.e. the time elapsed between

two successive peaks, use the well known relationships w = 21-rf and

f = l/T where f = cyclical frequency. Thus,

T 2x-r=-

I--l-c2

(11.13)

4 . Natural Period of Oscillation: A second-order system with <=O, is a

system free of any damping. Its transfer function is

KP

KG(s) = P= (11.14)

T2S2 + 1 (s-j +)(s+j +)

i.e. it has two purely imaginary poles (on the imaginary axis) and

according to the analysis of Section 9.4 it will oscillate continuously

with a constant amplitdue, and a natural frequency (see eqn. (11.14))

wn = l/T (11.15)

The corresponding cyclical period, Tn, is given by

Tn = 271-c (11.16)

It is this property of the parameter T that gave its name.

5. Response Time: The response of an underdamped system will reach its

ultimate value in an asymptotic manner, as t-toJ. For practical purposes,

it has been agreed to consider that the response reached its final

value, when it came +5X of its final value and stayed there. The time

needed for the response to reach this situation is known as the response

time, and it is also shown in Figure 11.2.

6, Rise Time: This term is used in order to characterize the speed with

which an underdamped system responds. It is defined as the time

required for the response to reach its final value for the first time.I

From Figure 11.1 we notice that the smaller the value of 5, the shorter/

/ the rise time, i.e. the faster the response of the system, but at the

I same time the larger the value of the overshoot.II

Remark: In subsequent chapters (Part IV), our objective during the design of

I a controller will be the proper selection of the corresponding 5

and T values, so that the overshoot is small, the rise time short,

r

the decay ratio small, and the response time short. We will realize

that it will not be possible to achieve all these objectives for the

same values of 5 and r, and that an acceptable compromise should

be defined. Good understanding of the underdamped behavior of a

second-order system will help tremendously in the design of efficient

controllers.

11.3 MULTICAPACITY PROCESSES AS SECOND-ORDER SYSTEMS

When material or energy flows through a single capacity, we get first-

order system. If on the other hand, mass or energy flows through a series of

two capacities, as it moves from the input to the output variable, the behavior

of the system is described by second-order dynamics. Two multicapacity systems

are shown in Figure 11.4 with two mass capacities (the two tanks) each.

Examine the two systems of Figure 11.4 more closely to identify a sig-

nificant qualitative difference between them. In System 1 (Figure 11.4a),

Tank 1 feeds Tank 2 and thus it affects its dynamic behavior, while the

opposite is not true. Such system is characteristic of a large class of the

so-called, non-interacting capacities, or non-interacting first-order systems

in series. On the contrary, in System 2, Tank 1 affects the dynamic behavior

of Tank 2, and vice-versa, because the flowrate Fl depends on the difference

between the liquid levels hl and h2. This system represents the so-called

interacting capacities, or interacting first-order systems in series.

Multicapacity processes do not have to involve more than one physical

processing unit. It is quite possible that all capacities are associated with

the same processing unit. For example, the stirred tank heater is a multi-

capacity process with capacity to store mass and energy. A distillation

column is another example of a multicapacity process. Every tray has a mass

storage capacity (liquid holdup), which in turn allows for thermal energy

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

Let us now see how multicapacity processes result in second-order systems.

We will start with the non-interacting capacities.

A. NON-INTERACTING CAPACITIES

When a system is composed of two non-interacting capacities, then it is

described by a set of two differential equations of the following general form:

dylrP1 dt- + y1 = K

p1f,(t) first capacity

dy2rp2 dt-+y2 = Kp2 y1(t) second capacity

(11.17a)

(11.17b)

In other words, the first system affects the second by its output, but it is

not affected by it (Figure 11.5a). Equation (11.17a) can be solved first and

then we can solve eqn. (11.17b). This sequential solution is characteristic

of non-interacting capacities in series. The corresponding transfer functions

are:

Y,(s) KPlGl(s) = ~ =

fl( s) rplS+l

The overall transfer function between tba external input, f,(t) and Y2(t) is:

Y,(s) y,(s) y,(s) KplK

Go(s) = - = - *-= G1(s)G2(s) =Pl

q(s) Y,(s) f,(s) rp1s+l * r

P*s+l

or

K'GO(s) = P

(Tt)2s2 + 25'T'S + 1

I where

(Ty = TPlTP2

25'T' = T TPl p2

and K'P=K K

Pl p2

(11.18)

(11.19)

Equation (11.19) indicates very clearly that the overall response of the system

is second-order.

overall transfer

Pl =

From eqn. (11.18) we also notice that the two poles of the

function are real and distinct, i.e.

1/TPl

and p2 = l/Tp2

If the time constants r andPl Tp2

are equal, then we have two equal poles.

Therefore, non-interacting capacities always result in an overdamped or

critically damped second-order system and never in an underdamped. The response

of two non-interacting capacities to a unit step change in the input will be

given by eqn. (11.7) f or the overdamped case, or eqn. (11.8) for the critically

damped. Instead of eqn. (11.7) we can use the following equivalent form for

the response

-t/T -t/.r

(-r p2e p1 - T e >

I

(11.20)Pl p2

where Ki = K K .Pl p2

Equation (11.20) can be derived easily by simple inversion

of eqn. (11.18) where fl(s) = l/s.

For the case of N non-interacting capacities (Figure 11.5b) it is easy

to show that the overall transfer function is given by

K K *..K

Go(s)'1 '2 'N

= G1(S)G2(S)“‘GN(S) = Gs+~)(~Pl p2

s+l)...(TPN

ss-1)(11.21)

Example 11.1 - T WO Non-Interacting Material Capacities in Series

System 1 in Figure 11.4a is such a system. The transfer functions for

the two tanks are:

and

where, according to Example 10.1 we have,

KPl

=R1 , Kp2

=R2 ,rp1

= AIR1 andrp2

= A2R2 .

and also variables hi, hi, F;, Fi are in deviation form. Since,

we can easily find that the overall transfer function is

q(s)K

Go(s) = - = p2

F;(s)(T s+l)(T s+l)

p1 p2

,

(11.22)

Equation (11.22) indicates that the relationship between the extewrnal input,

Fi( t), and the final output, h2(t)' is that of an overdamped second-order

system. Using eqn. (11.20) for the response of two non-interacting capacities

with r # Tp1 p2

, we find:

c -t/Tp1 - T e

-t/Th;(t) = K 1 ' (.c e p2

p2 rp2-rpl p1 p2

Figure 11.6 shows the qualitative features of the response, which are the

same as those of an overdamped system (see for example Figure ll.la with 511) *

A comparison with the first-order response would be instructive. Thus, from

Figure 11.6 we notice that:

- The response of the overdamped multicapacity system to step input change

is S-shaped, i.e. initially changes slowly and then it picks up speed.

I This is in contrast to a first-order response which has the largest rate

of change at the beginning. This sluggishness or delay is known also as

transfer la& and is characteristic of multicapacity systems.

- As the number of capacities in series increases, the delay in the initial

response becomes more pronounced. See also Figure 11.6.

B. INTERACTING CAPACITIES

In order to analyze the characteristics of such system, we will use the

two capacity System 2 of Figure 11.4b. The mass balances yield:

dhlAlx=F i - F1 Tank 1 (11.23a)

dh2A2dt = F1 - F2 Tank 2 (11.23b)

Assume linear resistances to flow, i.e.

F1 = (hl - h2)/R1 and F2 = h2/R2

Then, eqns. (11.23a) and (11.23b) become:

A R dhl1 l dt + hl - h2 = RIFi

dh2A2R2 dt

R2- + (l+R)h -R2

1 2 "1"' = O

(11.24a)

(11.24b)

We notice that eqns. (11.24a) and (11.24b) must be solved simultaneously. This

is the central characteristic of interacting capacities and indicates the

mutual effect of the two capacities.

The steady state equivalents of eqns. (11.24a) and (11.24b) are:

hw - h2(s) = RIFi(s)

R2 R2(l +ir;)h2&) - q Qs) = 0

(11.25a)

(11.25b)

Subtract (11.25a) from (11.24a) and (11.25b) from (11.24b) and after intro-

ducing the deviation variables take,

dhiAlRl dt + hi - hi = RlFf (11.26a)

dh;

A2R2 dtR2- + (l+~)h' -l2

(11.26b)

where

h’ =1 hl - hi(s) , hi = h2 - h2(s) and F! = Fi - Fi(s) .

1

Take the Laplace transforms of eqns. (11.26a) and (11.26b) and find

(AIRIS + l)Ci(S) - F;;(S) = RlFf(s)

-

R2 -,

- R1 hl(s) +

R2+ (1 + 5) I;;(s)1

= 0

Solve these algebraic equations with respect to El(s) and g2(s) and find:

CT p R1b + CR1 + R2)

r;;(s) = 12 (11.27a)

T T sPl p2

+('I +TPl p2

+ A1R2)s+1

i;;(s) = R2

2+(T +T(11.27b)

rP1TP2S Pl p2+ A1R2)s+1

whererp1 = AIRl and r = A2R2 are the time constants of the two tanks.

p2Equations (11.27a) and (11.27b) indicate that the responses of both tanks

follow second-order dynamics. Compare eqn. (11.27b) for the interacting tanks

with eqn. (11.22) which corresponds to the non-interacting tanks. We notice

that they differ only in the coefficient of s in the denominator by the

term, A1R2. This term may be thought of as the interaction factor and indi-

cates the degree of interaction between the two tanks. The larger the value

of AIR,, the larger the interaction between the two tanks.J. L

Remarks: (1) From eqn. (11.27b) it

the transfer function

is easily found that the two poles of

are given by:

I

p1,2 =

-(TV, +T~ +A1R2) I!7

4~ +TPl p2

+A1R2)% TPl p2

2r TPl p2

But

(T +T + A1R2)2 - 4 T T >o

Pl p2 Pl p2

(11.28)

Therefore, p1 and p2 are distinct and real poles. Consequently,

the response of interacting capacities is always overdamped.

(2) Since the response is overdamped with poles p1 and p2 given

by eqn. (11.28), eqn. (11.27b) can be written as follows:

qcs, R2 R2- =F;(s) (S--Pi) b-P,> = (rls+l)(T2s+l) (11.29)

where

'I1 = -l/P1 and r2 = -l/p2

Equation (11.29) implies that two interacting capacities can be

viewed as non-interacting capacities but with modified effective- -

time constants. Thus, while initially the two interacting tanks

had effective time constants

rp1and

=P2 *

when they are veiwed as non-interacting, they have different time

constants

r1 and

(3) Assume that the two tanks have the same time constants, i.e.

TP1 = rp2 =T. Then, from eqn. (11.28) we take,

7

VT2 = P2/P1 =-(2-c + A1R2> + A1R2 + 4-cA1R

# 1

-(2T + A1R2) - JA:R; + ~TA$X

Thus, we see that the effect of interaction is to change the ratio

of the effective time constants for the two tanks, i.e. one tank

becomes faster in its response and the other slower. Since the

overall response of h2(t) is affected by both tanks, the slower

tank becomes the controlling and the overall response becomes more

sluggish due to the interaction. Therefore, interacting capacities

are more sluggish than the non-interacting.

Example 11.2 - The Dynamics of Two Interacting Tanks

Consider two interacting tanks like those of Figure 11.4b. Let Al = A2

and Rl = R2/2. Then, 'c = TPl p2

/2=~. From eqn. (11.27b) we take:

~;(s> R2 R2~ =

F;(s) T2S2(11.30)

+ 5-rs + 1 = (0.21~s + 1)(4.8~s + 1)

For a unit step change in F;(t), i.e. for Pi(s) = l/s, eqn. (11.30) after

inversion yields

h;(t) = R2[1 + 0.014e -t/0.21T _ 5 2 e-t/4.8.r. 1

or

F;(t) = 1 + 0.014 e-t/0.21T _ 5.2 e-t/4.8T

If the two tanks were non-interacting, then the transfer function of the

system would be given by eqn. (11.22), i.e.

q<s> R2 R2- =

F;(s) (T s+1)(rp2s+1)p1

= (U?+1)(2Ts+l)

which by inversion yields

h;(t) = R2[1 + e-t/-c - 4e-t/2T]

or

F;(t) =l+ e-t/T _ 4e-t/2T

Let us compare the responses of the two systems:

(1) They are both overdamped. As such they have the characteristics discussed

in Section 11.2, i.e. they are S-shaped and with no oscillations.

(2) For the system of the two non-interacting tanks the time constants

are:

T and 2T

For the case of the interacting tanks the effective time constants

have become

0.21r and 4.8T

i.e. one was decreased and the other was increased. Their ratio

from l/2 changed to 0.21/4.8 = 0.044.

(3) As a result of the change in the effective time constants, the response

of the interacting tanks is more sluggish, or more damped than the

response of the non-interacting tanks. Figure 11.6 dramatizes this

result.

Example 11.3 - The CSTR as a System With Two Interacting Capacities

The linearized mass and energy balances for a constant volume holdup

CSTR are given by eqns. (9.14a) and (9.14b) in Example 9.2.

dci

dt + allci + a12T' = blcii (9.14a)

dT'- + a21ci + a22T' = blT; + b2Tidt (9.14b)

Both equations, being first-order differential equations, denote systems with

capacity. In particular, eqn. (9.14a) characterizes the capacity of the CSTR

for storing component A, while eqn. (9.14b) denotes the capacity of the CSTR

for storing thermal energy. From the form of the equations we realize that

they are interacting. Thus, the term a12T' denotes the effect of the thermal

capacity on the amount of component A. Similarly, the term a21ci denotes

the effect of the amount of component A in the CSTR on the thermal content of

the reacting mixture.

Consider now that the CSTR is at steady state when one of the following

variables changes by a unit step;

- feed concentration cd, , ori

- feed temperature T!, or1

- coolant temperature T'.C

According to the analysis made above we conclude that CL(t) and T'(t) will

respond to the input change like second-order, overdamped systems.

[Note: In the above example it has been assumed that the linearization of the

CSTR was made around a stable steady state (see Example 1.2) and that unit

step changes do not move the system far from this steady state.]

11.4 INHERENTLY SECOND-ORDER PROCESSES

Such process can exhibit underdamped behavior and consequently they cannot

be decomposed into two first-order systems in series (interacting or non-

interacting) with physical significance, like the

systems we examined in the previous sections. They occur rather rarely in a

chemical process, and they are associated with the motion of liquid masses

or the mechanical translation of solid parts possessing; (a) inertia to motion,

(b) resistance to motion and (c) capacitance to store mechanical energy. Since

resistance and capacitance are characteristic and of the first-order systems,

we conclude that the inherently second-order systems are characterized by

their inertia to motion. The three examples in Appendix ll.A clearly

demonstrate this feature.

Newton's Law applied on a given system yields

Balance of forceson the system

= (mass of system)x(acceleration) (11.31)

Since,

acceleration = d(velcoity)dt

and

velocity = & (spatial displacement)

we conclude that,

Balance of forces d2on the system = (mass of system) x----

dt2(spatial displacement) (11.32)

The second term of the right-hand side gives rise to the second-order behavior

of the system. Equation (11.31) or its equivalent (11.32) is the starting

point for the examples of Appendix ll.A.

11.5 SECOND-ORDER SYSTEMS CAUSED BY THE PRESENCE OF CONTROLLERS

The presence of a control system in a chemical process can change the

order of the process and produce a dynamic behavior which the process cannot

exhibit without the presence of the controller. In the chapters of Part IV

we will have the opportunity to examine many such situations. For the time

being let us consider a simple example.

Example 11.4 - A First-Order Process With Second-Order Dynamics Due to thePresence of a Control System

Consider the tank shown in Figure 10.1 (Example 10.1). This is a

simple first-order system with a transfer function given by eqn. (10.8). We

would like to control the liquid level at a desired value when the inlet

flowrate Fi undergoes step changes. In order to do that we use a feedback

control system (see Section 2.2) shown in Figure 11.7a. This control system

measures the liquid level and compares it with the desired steady state value.

If the level is higher than the desired value, it increases the effluent

flowrate Fo by opening the control valve V, while it closes the valve when

the level is lower than the desired value. Let us now see how the presence

of this controller changes the order of the dynamic behavior of the tank from

first to second order.

The dynamic mass balance around the tank gives,

A dhdt = Fi - F. (11.33)

while at the desired steady state we have

0 = Fi(S) - Fe(S)

Subtract (11.34) from (11.33) and take

Adh’ =dt F; - F;

(11.34)

(11.35)

where the deviation variables are defined by,

h' = h(s) - h 9 F; = F~(S) - Fi and I?: = Fe(s) - F. .

When the liquid level is not at the desired value, h' # 0, The measuring

device measures h and this value is compared to the desired value h(s).

The deviation (error) h' is used by the controller to increase or decrease

the effluent flowrate according to the relationship

(11.36)

where KC

and 'I are constant parameters with positive values. According

to (11.36j:

- When h' = 0 then F = Fe(s) and the valve V stays where it is.

- When h' > 0, i.e. the level goes down, then from eqn. (11.36) gives

.F. < Foes), 1-e. the effluent rate goes down and the level starts increasing.

I- When h' < 0, i.e. level goes up, then from eqn. (11.36) we find F. >

Fo(s) '

i.e. the effluent rate increases and the level decreases.I

The control action described by eqn. (11.36) is called Proportional - Integral

Control, because the value of the manipulated variable is determined by two

terms, one of which is proportional to the error h', and the other proportional

to the time integral of the error.

In eqn. (11.35) replace FA with its equal given by (11.36) and take:

A dh'- + Kch' + ydth'dt = F;

The Laplace transform of (11.37) gives

KA s K'(s) + K$'(s) + c 1 F;ys>

=I s= F;(s)

or

[

2 s2 + TIS + 1 E'(s) = K P;(s)I

=IK

C C

(11.37)

From eqn. (11.38) we find that the transfer function between the external

input F;(s) and the output 6'(s) is that of a second-order system and given

by;

m= KP

F;(s) T2S2 + 25r.s + 1

where

T2 =A~I/K~ , 25-r = TI and Kp = TI/K= .

From the above equations we find that

Depending on the values of the control parameters Kc and -cI we may have

the following cases:

-qp2. The, <cl and the response g'(s) to a step input in P;(s)

is that of an underdamped system.

- JK&A = 2. Then, -c=l and the response is critically damped.

- Finally, w > 2. Then, <>l and we have an overdamped response.

In Figure 11.7b we can see the dynamic response of the liquid level to a step

change in the inlet flowrate, with and without control.

The above example demonstrates very clearly how the simple first-order

dynamic behavior of a tank can change to that of a second-order, when a

Proportional-Integral Controller is added to the process. Also, it indicates

that the control parameters Kc and rI can have a very profound effect on

the dynamic behavior of the system which can range from an underdamped to an

overdamped response.


Chemical processes may exhibit second-order dynamics, which are: (i)

inherent, or (ii) come from first-order systems in series, or (iii) come as a

result of the control action.

The second-order systems are characterized by two parameters; the damping

factor 5 and the natural period of the system, '5. Depending on the value of

zj we can have; overdamped response (Al), critically damped (c=l) or under-

damed response (<cl). 'Ihe following observations can be made:

- The underdamped is a quick but oscillatory response and it is usually

exhibited by controlled processes.

- The overdamped or critically damped responses are sluggish, and are usually

exhibited by first-order systems in series. The sluggishness increases

with increasing value of the damping factor, and with the degree of inter-

i action between first-order systems.


1 .'

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

What is a second-order system? Write the differential equation describing

its behavior in the time domain and give its transfer function.

Explain the physical significance of the two parameters T and 3 of a

second-order system. Consult Reference 10 (Section 10.11) and Reference

11.

Identify the three classes of second-order systems and give one repre-

sentative example for each class. What is the origin of the most second-

order systems in chemical processes?

Discuss the overdamped, critically damped and underdamped responses of a

second-order system. Identify their distinguishing characteristics.

Describe the characteristics of an underdamped response.

Develop the expressions for the overshoot and the decay ratio (eqns.

(11.11) and (11.12)).

How do you understand the interaction or non-interaction of multicapacity

processes? Give ghe general set of two differential equations describing;

(a) two non-interacting capacities and (b) two interacting capacities.

Explain why two interacting capacities have more sluggish response than

two equivalent but non-interacting capacities.

Show that as the number of non-interacting first-order systems in series

increases the response of the system becomes more sluggish.

Develop the equations giving the response of a second-order system to a

unit impulse input for ~1, 5=1 and <cl.

Prove eqn. (11.20) for two non-interacting capacities.

A drum boiler (Figure P-11-10) has a capacity to store material and

thermal energy. Are these capacities interacting or not?

13. What is the origin of the most common systems with inherent second-order

dynamics? Describe an example. You can use References 10 and 11.

14. In Example 11.4 if you use Proportional Control only would you change

the order of the tank's dynamic behavior?

Il 6,.

“30Kp

IL

0

- - - A - - - - - - -

;z4

-- -

>

i.0

0.4

APPENDIX ll.A. EXAMPLES OF PHYSICAL SYSTEMS WITH INHERENT SECOND-

ORDER DYNAMICS.

Systems with inherent second-order dynamics can exhibit oscilla-

tory (underdamped) behavior but are rather rare in chemical processes.

In this appendix we will present three simple units which can be en-

countered in chemical plants and which possess second-order dynamics.

1. Simple Manometers and Externally Mounted Level Indicators.

Consider the simple U-tube manometer shown in Figure ll.A-la.

When the pressures at the top of the two legs are equal, the two

liquid levels are at rest at the same horizontal plane. Let us assume

that suddenly a pressure difference Ap = pl - p2 is imposed on the two

legs of the manometer. We like to know what is the dynamic response of

the levels in the two legs.

Let us apply Newton's law given by equation (11.31), on the mano-

meter. We take,

(forcepduEnt;ePgr;ssure) _ (forcepduEnt;e;rqssure)

1 2

force due to liquid force due to- ( level difference ) - (fluid friction)in the two legs

=(mass of liquidin the tube > x (acceleration)

or

PIAlii A (2h) _

- p2A2 - p gc 2( force due to )fluid friction

= p$ (ll.A-1)C

where,

pl'p2= pressures at the top of legs 1 and 2 respectively.

Al,A2 = cross sectional areas of legs 1 and 2 respectively.

Typically Al = A2 = A.

P = density of liquid in manometer.

g = acceleration gravity.

gC= conversion constant.

m = mass of liquid in the manometer = PAL

V = average velocity of the liquid in the tube. We have

assumed that the velocity profile in the tube is

flat (plug flow).

h = deviation of liquid level from the initial plane of

rest.

L = length of liquid in the manometer tubes.

Poiseuille's equation for laminar flow in a pipe can be used to relate

the force due to fluid friction with the flow velocity. Thus, we have

(Poiseuille's equation)

dh rR4 AP(volumetric flowrate) = A dt = --811 L

(ll.A-2)

where; R = radius of the pipe through which liquid flows.

u = viscosity of the flowing liquid.

L = length of the pipe.

AP = pressure drop due to fluid friction along the tube of

length L.

Therefore, applying Poiseuille's equation to the flow of liquid in the

manometer, we take:

(ll.A-3)

Recall also that the fluid velocity and acceleration are given by,

v = dhldt and dv/dt = d2h/dt2 (ll.A-4)

Put equation (ll.A-3) and (ll.A-4) in equation (ll.A-1) and take:

Ap*A -yh 8pL.A dh pAL d2h---=--C R2gc dt gc dt2

Finally, after dividing both sides by 2pgA/gc we take,

+ 41.1L dh gCTx+h=2pgApP@

Define,

T2 = Ll2g , 26~ = 4uL/psR2 and Kp = gc/20g

(ll.A-5)

and take,

T2 d2h dh- + 2r;T dtdt2 + h = KP'Ap

(ll.A-6)

Therefore, the transfer function between h and Ap is

k(s)-KP

*i(s) T2S2 + 25TS + 1(11 .A-71

Both equations (ll.A-6) and (11-A-7) indicate the inherent second-order

dynamics of the manometer.

For the measurement of liquid levels quite often we use the

externally mounted displacement-type transmitter, which is shown in

Figure ll.A-lb. We notice that the system of the tank-displacer

chamber, has many similarities with the manometer. The cross sectional

areas of the two legs are unequal and the Ap (external) pressure dif-

ference is caused by a change in the liquid level of the main tank.

Therefore, we expect that the response of the level in the displacer

chamber (h,) will follow second-order dynamics with respect to a change

in the liquid level of the tank (h), i.e.

h,(s)K'rn-=

a) ,;s2 + 26,T,S + 1

2 . Variable Capacitance Differential Pressure Transducer.

(ll.A-8)

The variable capacitance differential pressure transducer is a

very popular device which is used to sense and transmit pressure dif-

ferences. Figure ll.A-2 shows a schematic of such a device. Pressure

differences cause small displacements of the sensing diaphragm. The

position of the sensing diaphragm is detected by capacitor plates on

both sides of the diaphragm. A change in pressure pl in the process

(such as a change in the pressure of a vessel, or a change in liquid

level in a tank, etc.) will make the pressure p2 at the end of the

capillary tube to change.

A force balance around the capillary will yield;

Force due to the Force due to the

(pressure pl of the process) _ (pressure p2 exercised)

exercised at the end 1 at the end 2 of theof the capillary. capillary

= (mass) x (acceleration)

or

plAALP d2x

- p2A = (7) -C dt2

(ll.A-9)

where,

A = cross-sectional area of the capillary.

L = length of the capillary tube.

P = density of the liquid in the capillary tube.

x = fluid displacement in the capillary tube

= displacement of diaphragm.

The force p2A at the end of the capillary is balanced by two forces,

i.e.

resistance exerted byP# = ( the diaphragm )+(

viscous friction forceexercised by the fluid >

which acts like a spring

dx=Kx+cz (ll.A-10)

where,

K = Hooke's constant for the diaphragm, and

C = damping coefficient of the viscous liquid in front

of the diaphragm.

Substitute p2A in equation (ll.A-9) by its equal given by equation

(ll.A-10) and take:

(ll.A-11)

Equation (ll.A-11) clearly indicates that the response of the device

(i.e. the diaphragm displacement, x) when the process pressure, p1 ,

changes, follows second-order dynamics. If we define,

T2 = A-u/g c ' 26T = C/K and K = A/KP

we take the following transfer function,

24 _LFl (s) T2S2 + 2FTS + 1

3. Pneumatic Valve.

The pneumatic valve is the most commonly used final control ele-

ment. It is a system that exhibits inherent second-order dynamics.

Consider a typical pneumatic valve like that of Figure ll.A-3.

The position of the stem (or equivalently of the plug at the end of

the stem) will determine the size of the opening for flow and conse-

quently the size of the flow (flowrate). The position of the stem is

determined by the balance of all forces acting on it. These forces

are:

- PA = force exerted by the compressed air at the top of the dia-

phragm. Pressure p is the signal that opens or closes the

valve and A is the area of the diaphragm. This force acts

downwards.

- K x = force exerted by the spring attached to the stem and the

diaphragm. K is the Hooke's constant for the spring and x

is the displacement. It acts upward.

dx- Cx= frictional force exerted upward and resulting from the close

contact of the stem with valve packing. C is the friction

coefficient between stem and packing.

Apply Newton's law and take:

M d2xpA-Kx-Cz=r~c dt

or,

Let,

T2 = M/Kgc ' XT = C/K , K = A/KP

and take,

=2 d2x dx-+ 2<r dt+ x = Kp*pdt2

The last equation indicates that the stem position x follows inherent

second-order dynamics, when p changes. The transfer function is

ad _ A/K

P(s) (g-)s2 + ; s + 1C

(ll.A-12)

Usually, M << Kgc and as a result the dynamics of a pneumatic valve

can be approximated by first-order.

k - _ -_ -.:r+--c t-L L- -..- --. L _._ . -. --. c- _cc _-- L- -.-_ -_ -- _. - -. ._ __ - _ ,

CHAPTER 12

THE DYNAMIC BEHAVIOR OF HIGHER-ORDER SYSTEMS

Systems with higher than second-order dynamics are not uncommon in chemical

processes. Three are the most often encountered classes of higher-order

systems:

- N first-order processes in series (multicapacity processes)

- Processes with dead-time

- Processes with inverse response.

In this chapter we will analyze their typical dynamic characteristics.

12.1 N CAPACITIES IN SERIES

In Section 11.3 we found that two capacities in series, interacting or

non-interacting, give rise to a second-order system. If we extend the same

procedure to N capacities (first-order systems) in series, we find that the

overall response is of n-th order, i.e. the denominator of the overall trans-

fer function is an n-th order polynomial,

ans n + anls n-l+ l ** + als + a0

If the N capacities are non-interacting, then the overall transfer

function is given by eqn. (11.21)

K K **OK

Go(s)=G1(s)G2(s)"'GN(~) = (Tp1 '2 'N

p1s+l)(T s+l)***(-r s+l)

p2 PN

(11.21)

where Gl(s), G2(s),***, NG (s) are the transfer function of the N capacities.

For interacting capacities the overall transfer function is more complex.

In Section 11.3 we studied the basic dynamic characteristics of two

capacities in series when the input is changed by a step. Similar analysis is

possible for N capacities in series. The following general conclusions can

be easily drawn from the discussion in Section 11.3.

A. NON-INTERACTING N CAPACITIES IN SERIES

- The response has the characteristics of an overdamped system, i.e. it is not

oscillatory and very sluggish.

- Increasing the number of capacities in series increases the sluggishness of

the response.

B. INTERACTING N CAPACITIES IN SERIES

- Interaction increases the sluggishness of the overall response.

It is clear therefore, that a process with N capacities in series will

necessitate a controller which will not only keep the final output at a desired

value but will also try to improve the speed of the system's response.

Let us now examine some typical examples of processes with N capacities

in series.

Example 12.1 - Jacketed Reactors as Multicapacity Processes

Consider the batch reactor shown in Figure 12.la. The reaction is exo-

thermic and the content of the reactor is cooled by constant flow of cold

water circulating through the jacket. We can identify the following three

capacities in series:

- Heat capacity of the mixture in the reactor.

- Heat capacity of the reactor's wall.

- Heat capacity of the coolant in the jeacket.

It is easy to show that the three capacities interact.

For the jacketed CSTR of Figure 12.lb we have more interacting capacities,

i.e.

- total material capacity.of the tank,

- tank's capacity for component A,

29

- heat capacity of the reactor's content,

- heat capacity of the reactor's wall, and

- heat capacity of tht cold water in the jacket.

Again, all five capacities are interacting.

According to what we have said above, we expect that the response of the

reactors to input changes will be rather slow.

Example 12.2 - Staged Processes as Multicapacity Systems

Distillation and gas absorption columns are very often encountered in

chemical processes for the separation of a mixture into its components. Both

systems have a number of trays. Each tray has material and heat capacities.

Therefore, each column with N trays can be considered as a system with 2N

capacities in series. From the physics of distillation and absorption it is

easy to see that the 2N capacities interact.

Therefore, a step change in the liquid flowrate of the solvent at the

top of the absorption column produces a very delayed, sluggish response for the

content of solvent in the valuable component A (see Figure P.II-13). This is

because the input change has to travel through a large number of interacting

capacities in series.

Similarly, a step change in the reflux ratio of a distillation column

(see Figure 4.10) will have quickly an effect on the composition of the over-

head product while the composition of the bottoms stream will respond very

sluggishly (delayed and slow).

Finally, a step change in the steam flowrate of the reboiler will have

almost an immediate effect on the composition of the bottoms stream since very

few trays intervene between the returned to the column stream V and the

bottoms product. On the contrary, the effect on the composition of the over-

head product will be delayed and very slow.

12.2 DYNAMIC SYSTEMS WITH DEAD TIME

For all the systems we examined in Chapters 10, 11 and Section 12.1, we

have assumed that there is no dead-time between an input and the output, i.e.

whenever a change took place in the input variable, its effect was instan-

taneously observed in the behavior of the output variable. This is not true

and contrary to our physical experience. Virtually all physical processes

will involve some time delay between the input and the output.

Consider a first-order system with a dead time td between the output

y(t) and the input (forcing function) f(t). We can represent such system by

a series of two systems as shown in Figure 12.2a, i.e. a first-order system in

series with a dead time. For the first-order system we have the following

transfer function,

while for the dead time we have (see Section 7.2, eqn. (7.10))

[Yet - td)l

[y(t) 1-t&i

= e

Therefore, the transfer function between the input f(t) and the delayed output I

y(t - td> is given by (Figure 12.2b)I

[Ytt - t,)l

[f(t) 1

Similarly, the transfer

by,

-tds

= KP'e

rPs + l

function for a second-order system with delay is given

[y(t - t$l KP e

-tds

[f(t)1 = T2S2 + 25TS + 1(12.2)

Remarks: (1) Figure 12.3 shows the response of first and second-order sys-

tems with dead time to a step change in the input.

(2) Quite often the exponential term is approximated by the first

or second order Padi: approximations

1 td

ewtds - 2 ':: first-order approximation (12.3a)&

-tds (td)2s2 - 6tds + 12e ^ ^ second-order approximation

(td)'sL + 6tds + 12. _

(12.3b)

(3) Processes with dead time are difficult to control because the

output does not contain information about current events.

12.3 DYNAMIC SYSTEMS WITH INVERSE RESPONSE

The dynamic behavior of certain processes deviates drastically from what

we have seen so far. Figures 12.4b and 12.5b show the response of such sys-

terns to a step change in the input. We notice that initially the response is

in the opposite direction to where it eventually ends up. Such behavior is

called Inverse Response or Nonminimum Phase Response and it is exhibited by a

small number of processing units.

Example 12.3 - The Inverse Response of the Liquid Level in a Boiler System

Consider the simple drum boiler shown in Figure P.II-10. If the flowrate

of the cold feedwater is increased by a step, the total volume of the boiling-.

water and consequently the liquid level will be decreased for a short period

and then it will start increasing, as shown by the response in Figure 12.4b.

Such behavior is the net result of two opposing effects and can be explained

as follows:

- The cold feedwater causes a temperature drop which decreases the volume of

the entrained vapor bubbles. This leads to a decrease of the liquid level

of the boiling water, following first-order behavior (curve 1 in Figure

12,4b), i.e. -K /(rp1 p1

s+l).

- With constant heat supply, the steam production remains constant and con-

sequently the liquid level of the boiling water will start increasing in

an integral form (pure capacity) leading to a pure capacitive response,

i.e. Kp2

/s (curve 2 in Figure 12.2a).

- The result of the two opposing effects is given by

K Kp2 p1 =

(K 'cp2 p1

- K )s+Kp1 p2- -

S

rpls+l S(T s+l)

p1

(12.4)

and for

K-r <Kp2 Pl p1

the second term -K /(Tp1 p1

s+l) dominates initially and we take the inverse

response. If the above condition is not satisfied we do not have inverse

response.

[Note: When K 'c <K then from eqn. (12.4) we notice that the transferp2 p1 p1

function has a positive zero.]

The above example demonstrates that the inverse response is the result of two

opposing effects. Table 12.1 shows several such opposing effects between first

or second-order systems. In all cases we notice that when the system possesses

an inverse response, its transfer function has a positive zero. In general,

the transfer function of a system with inverse response is given by

G(s) =bmsm + b,-lsm-' + l ** + bls + b.

ansn + anels n-l + 0.0 + als + a0

where one of the roots of the numerator, i.e. one of the zeros of the transfer

function has positive real part.

Systems with inverse response are particularly difficult to control and

require special attention.

Example 12.4 - Inverse Response from Two Opposing First-Order Systems

Figure 12.5a shows another possibility of inverse response. Two opposing

effects result from two different first-order processes, yielding an overall

response equal to

K K

Y(s) = (p1 p2

Tpls+l - T s+lms)

p2

or

(K r -K Tp1 p2 p2 p1

)s+(K -K >

Y(s) =p1 p2

(T s+l)(,rp1 p2

SSl)

We have inverse response when

- initially PROCESS 1 reacts slower than PROCESS 2, i.e. T > Tp1 p2'

but_. _

- ultimately PROCESS 1 reaches a higher steady state value than PROCESS 2,

i.e. K > Kp1 p2'

Figure 12.5b shows the inverse response of the overall system.

[Note: When 'I >'c ,K >Kp1 p2 p1 p2

and K 'I < K 'cp1 p2 p2 p1

we find that the

system's transfer function has a positive zero, i.e.

K - Kp1 p2z = -

K T -K Tp1 p2 p2 p1


Chemical processing systems may exhibit higher order response. The most

common are; (a) N capacities in series, (b) systems with de-d time and (c)

systems with inverse response. N capacities in series yield delayed and

sluggish response. The sluggishness increases with the number of capacities

and if the capacities are interacting. Virtually all processes possess dead

time. First-order with dead time covers the large majority of dynamic elements

in a chemical process. The inverse response is the result of two opposing

effects, usually the difference of the responses between (a) first-order

systems, (b) second-order systems, (c) first and second-order systems, with or

without time delay. Systems with significant dead time and inverse response

are difficult to control.

IIIIII3IIIIII

THINGS.TO THINK ABOUT

1.

2.

3.

4.

5.

6.

7.

8.

9.

I 10

I

I

I

I

How would you define a higher-order system?

Using the above definition why is a system with dead time a higher-order

system? (Hint: See item 4 below).

Show that as the number of non-interacting or interacting capacities in

series increases, the response of the system becomes more sluggish.

Consider N identical non-interacting capacities in series, with gain

KP

and time constant ~~ for each capacity. Show that as N + 00)

the response of the system approaches the response of a system with

dead time TP

and overall gain K .P

In an ideal binary distillation column the dynamics of each tray can be

described by first-order systems. Are these capacities interacting or

not. What general type of responses would you expect for the overhead

and bottoms compositions to a step change in the feed composition?

How many capacities can you identify in the mixing process of Example

4.11? Are th-ey interacting or not?

What is the most common transfer function encountered in chemical pro-

cesses? why?

What is an inverse response aud what causes it?

Shw qualitatively ,tbat the response of the bottoms composition of a

distillation column to a step change in the vapor boilup, V, can exhibit

inverse behavior. (rOnsuIt References I , 1.

Why do you think-a system with inverse response is difficult to control?

REFERENCES

Chapter 6: Two very good references on computer simulation (digital or analog)

are the books by Luyben and Franks.

(1) Process Modeling, Simulation and Control for Chemical Engineers byW. L. Luyben, McGraw-Hill Book Co., New York (1973).

(2) Modeling and Simulation in Chemical Engineering by R. G. E. Franks,J. Wiley and Sons, New York (1972)

Both books provide a series of examples drawn from the area of chemical engi-

neering and demonstrate how digital computer simulation of chemical processes

can enhance our ability to understand the dynamics and develop better con-

trollers for such systems. Computer programs in FORTRAN for typical systems

are also included. For more details on the numerical techniques for the

solution of algebraic or differential equations the reader is encouraged to

consult the following two classic books:

(3) Digital Computation for Chemical Engineers, by L. Lapidus, McGraw-HillBook Co., New York (1962).

(4) Applied Numerical Methods, by B. Carnahan, H. R. Luther and J. D.Wilkes, J. Wiley and Sons, New York (1969).

The notion and the characteristics of the Taylor series expansion as well as

the linear approximation of nonlinear systems can be found in all the

standard texts on calculus.

In Section 6.1 of his book Douglas [Ref. 51 discusses a procedure that

allows us to ascertain the range of values around the point of linearization

for which the linearized model is acceptable.

(5) Process Dynamics and Control, Vol. 1, by J. M.,Douglas, Prentice-Hall,Inc., Englewood Cliffs, N.J. (1972).

Example 6.1 was motivated by the physical system analyzed in Section 2.3 of the

book by Russell and Denn where the reader can find more information.

(6) Introduction to Chemical Engineering Analysis, by T. W. F. Russell andM. M. Denn, J. Wiley and Sons, New York (1972).

Chapters 7 and 8: The Laplace transformation has been the object of a large

body of mathematical research. For more details on the theoretical aspects

of Laplace transforms the reader will find useful the following book:

(7) Operational Mathematics, 2nd edition, by R. V. Churchill, McGraw-HillBook Co., New York (1958).

For the use of Laplace transforms to the solution of differential equations

(ordinary, partial or sets of) the book by Jenson and Jeffreys can be very

valuable.

(8) Mathematical Methods in Chemical Engineering, by V. G. Jenson and G. V.Jeffreys, Academic Press London (1963).

In the following two references the reader can find tables with the Laplace

transformation of a large number of functions:

(9) Feedback and Control Systems, by J. J. DiStefano, III, A. R. Stubberudand J. J. Williams, Schaums Outline Series, McGraw-Hill Book Co., NewYork (1967).

(10) Handbook of Mathematical Functions, by . . Abramowity and . .Stegun,

Chapter 10: The book by Weber [Ref. 111 is an excellent reference for the

dynamics of first-order systems. The interested reader will find (Chapters

8 and 9) an extensive coverage of first-order systems based on mass, energy

and momentum balances, with a large number of examples. It provides also a

valuable physical interpretation of the notion of capacity for various pro-

cessing systems.

(11) An Introduction to Process Dynamics and Control, by T. W. Weber, J. Wileyand Sons, New York (1973).

In the books by Douglas [Ref. 51 and Coughanowr and Koppell [Ref. 121 the

reader can find the response of first-order systems to impulse or sinusoidal

I inputs. The response of a capacity process to a sinusoidal input is also

given in Chapter 12 of this text.

(12) Process Systems Analysis and Control, by D. R. Coughanowr and L. B.Koppell, McGraw-Hill Book Co., New York (1965).

Chapter 11: The book by Weber [Ref. 111 is also an excellent reference for the

development and physical interpretation of second-order systems (Chapter 10).

It contains examples of inherently second-order systems which the reader will

find quite useful. In the books by Coughanowr and Koppell [Ref. 121 and

Douglas [Ref. 51 the reader can find the response of second-orcer systems to

impulse and sinusoidal inputs. For more information on the externally

mounted level measuring systems, or the manometers, and their dynamic second-

order characteristics, the reader can consult the References 18 (Chapter 18),

11 (Chapter 10) or the book by Shiskey [Ref. 13, Chapter 31.

(13) Process Control Systems, 2nd edition, by F. G. Shinskey, McGraw-HillBook Co., New York (1979).

The following two references can be consulted for further details on the

variable capacitance differential pressure transducer and the pneumatic

control valve.

(14) Process Dynamics. Part 2: Proces Control Loops, by J. C. Guy,Chem. Engng., Aug. 24, p. 111 (1981).

(15) Measurements and Control Applications for Practicing Engineers, byJ. 0. Hougen, Cahners Books, Boston (1972).

Chapter 12: Luyber [Ref. l] has a good discussion on the inverse response of

the bottoms composition of a distillation column to a change in the vapor

boilup (Section 11-5). More details on this system can be f;ound in the

following paper:

(16) "by W. L. Luyben, Inst. Chem. Eng. (London), Symp. Ser. No. 32, p. 6(1969)."

Iinoya and Altpeter [Ref. 171 discuss the characteristics of the systems which

exhibit inverse response and give a table of the most common physical

situations (transfer functions) which give rise to inverse response.

(17) "Inverse Response in Process Control," by K. Iinoya and R. J. Altpeter,Ind. Eng. Chem., Vol. 54, No. 7, p. 39 (1962).

In the book by Shinskey [Ref. 131 the reader can find further discussion on the

inverse response of a drum boiler.

A-B

, ‘J I-

I Figwe .12 31

,!%.S

Input

+) ’

k .t,st4

(. )a

C-+.’

PsoCEsS 1

output>cc )S

PROCESS 2?j (9

*<-- ----

Cb)

Table 12.1. Systems With Inverse Response

1.

2.

3.

4.

5.

I 6.

I

Pure capacitive minus first-order response (Figure 11.11)

G(s) =

KP2

Kp1

S Tpl s+l=

(K T - K )s+KP2 p1 p1 P2

S(T s+l)p1

for K T < KP2 p1 p1

zero = -K /(K Tp2

-K)>Op2 p1 p1

Difference between two first-order responses (Figure 11.12)

K Kp1 p2

(K Tp1 p2

-K Tp2 p1

)s+(K -K >G(s) = p1 p2

Tpls+l - T s+l =

P2CT s+l)(T

p1 p2s+l)

for K T < K 'Ip1 p2 P2 p1

zero = (K - Kp1 p2

)/(K -rp1 p2

-K Tp2 Pl

)

Difference between two first-order responses with dead time

G(s) =

-T S

Kd2

l e

P2

s+lfor

Second-order minus first-order response

Kp1

KG(s) = - - p2

T2S2 + 25TS + 1 s+lfor

K >Kp1 p2

and T >'Idl d2

L 0

K >KPl p2

Difference between two second-order responses

Kp1

KP2

2 KG(s) =

r;s2 + 2C2T2S + 1

T2 p1 > 1for -7j K

Tl p2

Difference between two second-order responses with dead time

-T s -T S

K dll e K

d2l e

G(s) =p1 P2

T;s2 + 2yp + 1 T;s2 + 2c2T2S + 1for K >K

p1 p2and

‘dl ’ ‘Id2 2 0 .

PART IV

ANALYSIS AND DESIGN OF FEEDBACK CONTROL SYSTEMS

In Part III we studied the dynamic behavior of various typical processing

systems under the influence of changes in the input variables (disturbances or

manipulated variables). In doing so, we were not concerned about having the

system to respond in a specific manner: in other words, we were not interested

in controlling the behavior of the process.

Starting with Part IV, our main concern will be; how can we control a

process in order to exhibit a certain desired response in the presence of

input changes. We will start with the most common control configuration known

as FFEDBACK, which we touched upon very briefly in Chapter 2. In Parts V and

VI we will discuss additional control configurations such as; feedforward,

cascade, ratio, override, split range, multivariable, etc.

Thus, in the subsequent eight chapters of Part IV, we will do the following:

- Discuss the notion of the feedback loop and present its hardware elements.

- Identify the types of feedback controllers which are available for process

control and examine their effect on the response of a chemical process.

- Analyze the stability characteristics of a feedback control system and

learn how to design the appropriate feedback system to control a given

process.

- Solve some special problems which are encountered during the design of

feedback controllers.

CHAPTER 13

INTRODUCTION TO FEEDBACK CONTROL

In Chapter 1 we introduced the notion of a feedback control system. In

this chapter we will expand the discussion by introducing the hardware elements

of a feedback system and the types of the available controllers.

13.1 THE CONCEPT OF FEEDBACK CONTROL

Consider the generalized process shown in Figure 13.la. It has an output

y, a potential disturbance d and an available manipulated variable m. The

disturbance d (also known as load or process load) changes in an unpredictable

manner and our control objective is to keep the value of the output y at

desired levels. A feedback control action takes the following steps:

- Measures the value of the output (flow, pressure, liquid level, temperature,

composition) using the appropriate measuring device. Let yrn be the value

indicated by the measuring sensor.

- Compares the indicated value y, to the desired value ySP (set point) of

the output. Let the deviation (error) be E = yd - y,.

- The value of the deviation E is supplied to the main controller. Theicontroller in turn changes the value of the manipulated variable m in

such a way as to reduce the magnitude of the deviation 8. Usually, the

controller does not affect the manipulated variable directly but through

another device (usually a control valve) which is known as the final control

element.

Figure 13.lb summarizes pictorially the above three steps.1

The system in Figure 13.la is known as open loop in contrast to the feed-/I back controlled system of Figure 13.lb which is called closed loop. Also, when

Ithe value of d changes. the response of the first is called open-loop response

while that of the second is the closed-loop response. The origin of the term

closed-loop is evident from the Figure 13.lb.

Example 13.1 - Feedback Control Systems

The following represent some typical feedback control systems which are

often encountered in chemical processes.

(4

(b)

(cl

Cd)

Cd

Flow control. Two feedback systems are shown in Figures 13.2a and 13.2b,

controlling the flowrate F at the desired value Fd.

Pressure Control. The feedback system in Figure 13.2~ controls the

pressure of the gases in the tank, at the desired pressure pd.

Liquid Level Control. Figures 13.2d and 13.2e show two feedback systems

used for the control of the liquid levels at the bottom of a distillation

column and its condenser accumulation tank.

Temperature Control. The system in Figure 13.2f controls the temperature

of the exiting hot stream at the desired value Td.

Composition Control. Composition is the controlled variable in the

blending system of Figure 13.2g. The desired value is Cd.

Remark: To simplify the presentation of a feedback control system, we will

usually replace the diagrammatic details, e.g. measuring device,

comparator, controller, with a simple circle carrying the one of

the following characterizations:

FC for flow control,

PC for pressure control,

LC for liquid level control,

TC for temperature and

CC, for composition control.

Also, little squares with the characterizations LT, TT, PT, FT, CT are used to

indicate level, temperature, pressure, flow, and concentration measurements and

transmitters. Figures 13.3a and 13.3b are equivalent to Figures 13.2b and

13.2d, respectively.

All the above examples indicate that the basic hardware components of a

feedback control loop are the following:

1.

2 .

3.

4.

5 .

/

/

Process, the material equipment where the physical or chemical operations

take place (tanks, heat exchangers, reactors, separators, etc.).

Measuring instruments or sensors, like thermocouples (for temperature),

bellows or diaphragms (for pressure or liquid level), orifice plates

(for flow), gas chromatographs or various types of spectroscopic

analyzers (for composition), etc.

Transmission lines, used to carry the measurement signal from the sensor

to the controller and the control signal from the controller to the

final control element. These lines can be either pneumatic (compressed

air or liquid) or electrical.

Controller, it also includes the function of the comparator. This is

the unit with logic that decides by how much to change the value of the

manipulated variable. It requires the specification of the desired

value (set point).

Final control element, usually a control valve or a variable-speed

metering pump. This is the device that receives the control signal

from the controller and implements it by physically adjusting the value

of the manipulated variable.

Each of the above elements should be viewed as a physical system with an

input and an output. Consequently, their behavior can be described by a

different equation or equivalently by a transfer function. In the following

sections of this chapter we will take a closer look at the dynamics of these

hardware elements.

13.2 TYPES OF FEEDBACK CONTROLLERS

Between the measuring device and the final control elements comes the

controller (Figure 13.lb). Its function is to receive the measured output

signal y,(t) and after comparing it with the set point yd to produce the

actuating signal c(t) in such a way as to return the output to the desired

value YSP' Therefore, the input to the controller is the error E(t) =

ysp - y,(t) while its output is c(t). The various types of continuous

feedback controllers differ on the way they relate E(t) to c(t).

The output signal of a feedback controller depends on its construction

and may be a pneumatic signal (compressed air) for pneumatic controllers or

an electrical one for electronic controllers.

There are three basic types of feedback controllers; (a) proportional,

(b) proportional-integral and (c) proportional-integral-derivative. The

details of construction may differ among the various manufacturers but their

functions are essentially the same. Let us study each one separately.

A. Proporational Controller. (or P Controller)

Its actuating output is proportional to the error, i.e.

c(t) = Kcc(t) + cs (13.1)

where Kc = controller proportional gain and cs = controller bias signal,

i.e. its actuating signal when E=O.

A proportional controller is described by the value of its proportional

gain Kc or equivalently by its proportional band, PB

PB = loo/Kc

The proportional band characterizes the range over which the error must change

in order to drive the actuating signal of the controller over its full range.

Usually,

1 6 PB < 500

It is clear that,

"the larger the &ain Kc or the smaller the proportional band the

smaller the controller's actuating signal will be."

Define the deviation c(t) of the actuating signal by

c(t) = c(t) - cs

and take,

c(t) = K$t) (13.2)

The last equation yields the following transfer function for a proportional

controller

Gc(s> = Kc (13.3)

B. Proportional-Integral Controller (or PI Controller)

Most commonly it is known as proportional-plus-reset controller. Its

actuating signal is related to the error by the equation,

c(t) =KC

KCW + -y-f

c(t)dt + cs (13.4)I

where r1 is the integral time or reset time in minutes. The reset time is

an adjustable parameter and is sometimes referred to as minutes per repeat.

Usually it varies in the range

0.1 I TI 5 50 minutes

Some manufacturers do not calibrate their controllers in terms of TI but in

terms of its reciprocal l/rI (repeats per minute), which is known as reset

rate.

At this point it is instructive to examine the origin of the term "reset".

Consider that the error changes by a step of magnitude E. Figure 13.4 shows

the response of the output of a controller as it is computed from eqn. (13.4).

3f3,

We observe that initially the controller output is Kc~ (the contribution of

the integral term is zero). After a period of rI minutes the contribution

of the integral term Is

Kc(t)dt = 2 EATS = Kc~

r1

i.e. the integral control action has "repeated" the response of the proportional

action. This repetition takes place every TI minutes and has lent the name

to the reset time. Therefore,

"reset time is the time needed by the controller to repeat the

initial proportional action change in its output."

The integral action causes the controller output c(t) to change as long

as an error exists in the process output. Therefore, such a controller can

eliminate even small errors.

From eqn. (13.4) it is easy to show that the transfer function of a

proportional-integral (or proportional-plus-reset) controller is given by

Gc(s> = Kc(l +-+-)I

(13.5)

C. ProportionalrIntegral-Derivative (or PID Controller)

In the industrial practice it is commonly known as proportional-glus-

reset-plus-rate controller.

The output of the controller is given by,

c(t) = KcK$t) + y dc-I

E(t)dt + Kc 'D dt (13.6)

where TD is the derivative time in minutes.

With the presence of the derivative term, Kc d&/dt, the PID controller

anticipates what the error will be in the immediate future and applies a control

action which is proportional to the rate of change in the error. Due to this

property, the derivative control action is sometimes referred to as "anticipatory

control".3

The major drawbacks of the derivative control action are the following:

- For a response with constant non-zero error it gives no control action

since dc/dt = 0.

- For a noisy response with almost zero error it can compute large derivatives

and thus yield large control action, although it is not needed.

From eqn. (13.6) we can easily derive the transfer function of a PID con-

troller,

Gc(s> = Kc(l+ -&I

+ TDs) (13.7)

13.3 m~ME;AsuRING DEVICES (SENSORS)

The successful operation of any feedback control system depends in a very

critical manner upon the good measurement of the controlled outupt and the

uncorrupted transmission of the measurement to the controller. The first

requirement implies the need for an accurate measuring device while the second

necessitates good and effective transmission lines.

There is a large number of commercial sensors. They differ either in the

basic measuring principle they employ or their constructional characteristics.

For more details the reader can consult the various references at the end of

Part IV or the technical booklets circulated by the various manufacturers.

Let us look more closely at the various typical sensors used to measure

the most common process outputs.

A. Flow Sensors

The flow sensors most commonly employed in the industrial practice are

those which measure the pressure gradient developed across a constriction.

The, using the well known (from fluid mechanics) equation of Bernoulli, we can

compute the flow-rate. Such sensors can be used for both gases and liquids.

The orifice plate (Figure 13.5a), Venturi tube (Figure 13.5b) and Dal1 flow

tube are typical examples of sensors based on the above principle. The first

is more popular due to its simplicity and low cost. The last two are more

expensive but also more accurate.

A different sensor is the turbine flow meter which uses the number of

turbine revolutions to compute the flowrate of liquids quite accurately.

Flow sensors have very fast dynamics and they are usually modeled by

simple algebraic equations, i.e.

Flow = a & - (13.8)

where c1 is a constant determined by the construction characteristics of

the flow sensor, and Ap is the pressure difference between the flow con-

striction and a point with fully developed flow.

B. Pressure or Pressure Actuated Sensors

Such sensors are used to measure the pressure of a process or the

pressure difference which is employed to compute a liquid level or a flowrate

(orifice plate, Venturi tube). The variable capacitance differential pressure

transducer has become very popular. Figure ll.A-2 shows a schematic of such

device. Pressure differences cause small displacements of the sensing

diaphragm, Thepositionof the sensing diaphragm is detected by capacitor

plates on both sides of the diaphragm. The differential capacitance between

the sensing diaphragm and the capacitor plates is converted into a d-c voltage.

A force balance around the sensing diaphragm leads to the following second-

order model

2 d2z dz'c-dt2

+ 25~ dt + z = Kp*Ap (13.9)

where

Z is the displacement of the sensing diaphragm,

AP the actuating pressure difference, and

T,C,K Pare the +.hree parameters of a 2nd order system defined in this

case by the constructional characteristics of the device.

For details on the development of eqn. (13.9) see Appendix ll.A at the end of

Chapter 11. Various other types of sensors, all of them measuring the dis-

placement of a mechanical part under the influence of Ap, are also in use.

C. Temperature Sensors

The most common are thermocouples, resistance bulb thermometers and

thermistors. All provide the measurement in terms of electrical signals.

Independently of their constructional differences their basic dynamic behavior

can be examined in terms of the temperature profiles in Figure 13.6a and 13.6b.

The temperature sensing element is always inside a thermowell, Figure 13.7.

In the first case (Figure 13.6a) we assume that the major resistance to heat

transfer is located outside the thermowell casing. In such case we have a

single capacity with resistance and as we know from Chapter 10, it is modeled

by a first-order system, i.e.

dTm=P dt-+ Tm = T (13.10)

In the second case (Figure 13.6b) we have major heat transfer film resistances

inside and outside the thermowell casing. This is equivalent to two capacities

in series and as we know from Chapter 11 the thermocouple reading will exhibit

second-order (overdamped) behavior, i.e.

dT+ 25~ -$ + Tm = T (13.11)

The parameters r and 5 depend on the constructional and material character-

istics of the temperature sensing device (i.e. thermocouple, casing, materials

of construction). It is clear that the response of a thermocouple modeled by

eqn. (13.11) is slower than that of a thermocouple modeled by (13.10)(Figure 13.8).

D. Composition Analyzers

Typical examples of such sensors are: gas chromatographs and various

types of spectroscopiL analyzers. They are used to measure the composition

of liquids or gases in terms of one or two key components or in terms of all

components present in a process stream.

The dominant dynamic feature of composition analyzers is the time delay

(dead time) in their response, which can be quite large. Thus, for a

chromatographic column, the time required by the sample to travel from the

process stream to the column, plus the time required to travel through the

column, plus the time needed by the detector at the end of the column to

respond, can be quite large. Such long time delays result in ineffective

control.

Other features characteristic of composition analyzers are; (a) their low

operational reliability (easy breakdown) and (b) their relatively high cost.

13.4 TRANSMISSION LINES

These are used to carry the measurement signal to the controller and the

control signal to the final control element. There are two types of trans-

mission lines; the pneumatic (compressed air, liquids) and the electrical.

Unless the process changes very fast or the transmission lines are very

long, the dynamic behavior of a pneumatic transmission line can be neglected

from consideration. When the above assumptions do not hold, it has been

found that the following transfer function correlates successfully the pressure

at the outlet (PO> to the pressure at the inlet (Pi) of the pneumatic line,

p,(s)-T sd

e-=-

pirps+l

with r /Td P

s 0.25.

[Note: IN the subsequent chapters, as a rule, we will neglect the dynamics of

pneumatic transmissioLl lines.]

13.5 FINAL CONTROL ELEMENTS

These are the hardware components of the control loops which implement

the control action. They receive the output of a controller (actuating signal)

and adjust accordingly the value of the manipulated variable.

The most common final control element is a pneumatic valve (Figure ll.A-3).

This is an air operated valve which controls the flow through an orifice by

positioning appropriately a plug. The plug is attached at the end of a stem

which at the other end is supported on a diaphragm. As the air pressure (con-

troller output) above the diaphragm increases, the stem moves down and

consequently the plug restricts the flow through the orifice. Such valve is

known as an "air-to-close" valve (Figure 13.9a). If the air supply above the

diaphragm is lost, the valve will "fail open" since the spring would push the

stem and the plug upward. There are pneumatic valves with opposite actions,

i.e. "air-to-open" which 'Ifail closed" (Figure 13.9b). The most commercial

valves move from fully open to fully closed as the air pressure at the top of

the diaphragm changes from 13 to 15 psig.

In Appendix 11-A we developed the mathematical model which describes the

dynamic behavior of a pneumatic control valve. This was shown to be of second-

order. But, the response to changes of most small or medium size valves is so

fast that the dynamics can be neglected. In such case only a constant gain

term will remain which relates the output from the controller (air pressure

signal) to the fluid flow through the valve.

For non-flushing liquids the flow through the valve is given by,

F =

where

AP = the pressure drop across the valve,

K = a constant which depends on the valve size,

P = specific gravity of the flowing liquid and

f(x) = valve flow characteristic curve.

The valve flow characteristic curve, f(x), depends on the geometrical

shape of the plug's surface. Figure 13.10 shows the most common types of

plugs and the corresponding f(x). Figure 13.11 shows the flow capacity

characteristics for the various valves.

Other final control elements include relays to start or stop various

equipment, variable-drive motors for fans or pumps, heavy load electrohydraulic

actuators, etc.


Feedback is the most common configuration for the control of chemical

processes. Its basic idea is to measure the controlled variable and use its

deviation from a desired value to activate the controller, which in turn will

adjust the value of the manipulated variable through the final control element.

The controller's action is such that the output variable is returned to the

desired set point.

Every feedback loop is composed of the following hardware components;

(a) process, (b) measuring sensor, (c) controller, (d) final control element

and (e) transmission lines. Each of these elements should be considered as a

physical system with input and output. Their dynamic behavior can be modeled

using the same principles as in Part II, and as we will see in Chapter 14, it

is very crucial for the closed-loop response of the controlled process.

The three types of feedback control are; Proportional (P), Proportional-

Integral (PI or Proportional-plus-Reset) and Proportional-Integral-Derivative

(PID or Proportional-plus-Reset-plus-Rate). Their effect on the closed-loop

response varies and will be extensively analyzed in Chapter 14. In the same

chapter we will also study their relative advantages and disadvantages.

There is a great variety of measuring devices (sensors) and final control

elements. The selection of the appropriate element is important but it

requires information which the reader can find in technical manuals and hand-

books and are beyond the scope of this text.

THINGS 'TO THINK ABOUT

1.

2.

3.

4.

5.

6 .

7 .

8.

From all that you know so far, what are the strengths and weaknesses of

a feedback control system?

Describe one example of; (a) flow control, (b) pressure control, (c)

liquid level control, (d) temperature control, (e) composition control,

and which are not the same as the examples covered in this chapter.

Draw the appropriate diagrams.

Define an open-loop and closed-loop system. Why do we use the term;

open, closed, loop? Also define an open-loop or closed-loop response.

What are the basic hardware components of a feedback control loop?

Identify the hardware elements present in a feedback loop for the

temperature control of a stirred tank heater.

Write Bernoulli's equation for two points of a Venturi tube and show

how you can compute the flowrate through the tube by measuring the

pressure difference between the two points, i.e. prove the essence of

eqn. (13.8).

The model for a variable capacitance pressure transducer was developed

in Appendix ll.A and is given by eqn. (13.9). It shows that the system

is inherently second-order and can exhibit underdamped response. What

does this mean for the applicability of such device?

Is it possible to have an oscillatory behavior by the indicated tem-

perature (T,) of a thermocouple, if the measured temperature (T)

changes by a step?

Discuss some of the factors you should take into account, before deciding

if you would use an air-to-close or air-to-open pneumatic control valve.

9 . Compute the response of a PD (Proportional-Derivative) controller to a

ramp change in the error E, i.e. E = at with 01 = constant. Sketch

the contributions of the proportional and derivative actions separately.

On the basis of this example discuss the anticipatory nature of the

derivative control term.

10. Consult References 6 (Chapter 15) and 7 (Chapter 10) and discuss the

factors which affect the selection of the valve type, i.e. linear,

square root, equal percentage and hyperbolic.

11. When an error E(t) persists for a long time the value of the integral

Idt)dt increases significantly and may lead the output of a PI con-

troller to its maximum allowable value. We say that the controller

has saturated and in physical terms it means that the valve is fully

open or closed before the control action has been completed, i.e.

before the error has been driven to zero. This situation is also known

as reset windup. How would you handle such a situation? You can con-

sult References 7 and 15.

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CHAPTER 14

THE DYNAMIC BEHAVIOR OF FEEDBACK CONTROLLED PROCESSES

In the previous chapter we defined the basic notion of a feedback control

system and we discussed its hardware components. In this chapter we will

analyze the dynamic behavior of a process which is controlled by a feedback

control system when the values of the disturbance (load) d or of the set point

yd change. This analysis will be done by considering various types of feed-

back controllers, i.e. P or PI or PID.

14.1 BLOCK DIAGRAM AND THE CLOSED-LOOP RESPONSE

Consider the generalized closed-loop system shown in Figure 13.lb. For

each of its four components; process, measuring device, controller mechanism

and final control element, we can write the corresponding transfer function

relating its output to its inputs. In particular, if we neglect the dynamics

tmdsof the transmission l-es, we have:

Process

y<s> = Gp(s) iii(s) + Gd(S) a(s)

Measuring Device

Y,(s) = G,(s) y(s) (14.2)

Controller Mechanism

E(S) = Y,,b> - Ym<s>

C(s) = Gc(s) E(S)

comparator

control action

(14.1)

(14.3a)

(14.3b)

Final Control Element

G(s) = Gf(s) Z(s) (14.4)

where GP' Gd, Gm, Gc, Gf are the transfer functions between the corresponding

inputs and outputs.

Figure 14.1 shows the block diagram for the generalized closed-loop system

and it is nothing else but a pictorial representation of eqns. (14.1), (14.2),

(14.3a) and (14.3b) hnd (14.4).

the block diagram of Figure 14.1

Notice the direct correspondence between

and the schematic of Figure 13.lb.

The series of block between the comparator and the controlled output, i.e.

GC ’ Gf and G

Pconstitutes the forward path, while the block G is on them

feedback path between the controlled output and the comparator, If G=

GcGfGp then Figure 14.2a shows a simplified but equivalent version of the

block diagram.

Algebraic manipulation of the above equations yields:

m(s) = Gf (s)C(s) = Gf WGc(s>3s>

= Gf(s)Gc(s> [j;,,(s) - T,(s) 1

(14.3b)

(14.3a)

= Gf(s)Gc(S)[~Sp(s) - G,(4Tb)l (14.2)

Put the last expression in eqn. (14.1)

3s) = Gp(s)(Gf (s)G=(s) [y,,(s) - G,(s)y(s) 1) + Gd(s);i(s)

and after readjustment take;

GpbNf (s)G,(s) Gd(S)‘(‘) = 4+Gp(s)Gf(s)Gc(s)Gm(s) ‘S+) +4+Gp(s)Gf(s)G,(s)Gm(s) ‘(‘)

(14.5)

Equation (14.5) gives the closed-loop response of the process. We notice

that it is composed of two terms. The first term shows the effect on the

output of a change in the set point while the second constitutes the effect

on the output of a change in the load (disturbance). The corresponding transfer

functions are known as closed-loop transfer functions. In particular

GpGfGc Gl+GGGGp f c m ' l+GG = GSPm

(14.6)

is the closed-loop transfer function for a change in the set point and

Gd Gdl+GGGG

p f c m' 1 + GGm = GLOAD (14.7)

is the closed-loop transfer function for a change in the load. Figure 14.2b

shows a block diagram equivalent to that of Figure 14.2a but further simplified.

For every feedback control system we can distinguish two types of control

problems; the servo or the regulator problem.

- Servo Problem; the disturbance does not change, i.e. j(s) = 0 while the

set point undergoes a change. The feedback controller acts in such a way

as to keep y close to the changing yd. In such case,

T(s) A GspWs;sp(s) (14.8)

- Regulator Problem; the set point remains the same, i.e. &+> = 0 while

the load changes. Then

Y(s) = GLOAD(S) ii(s) (14.9)

and the feedback controller tries to eliminate the impact of the load

changes and keep y at the desired set point.

Remark: The closed-loop overall transfer functions GSP and GLOAD depend

on the dynamics of the measuring sensors, controllers and final

control elements. Consequently, the closed-loop response for set

point or load changes depends on the dynamics of these elements

whose selection becomes therefore critical.

Example 14.1 - The Closed-Loop Response of the Liquid Level in a Tank

Consider the liquid level control system for the tank of Figure 14.3a.

The level h is the controlled output while Fi is the load (disturbance)

and F. the manipulated variable. The transfer functions for each component

of the feedback loop are:

Process. The material balance around the tank gives

Adh=F -Fdt i 0

and easily we find that

K(s) = -& Fi(S) - & Fob> (14.10)

Measuring Device. This can be a variable capacitance differential pressure

transducer (section 13.3) measuring the pressure of a liquid column of height

h. The dynamic response of the sensor is given by eqn. (13.9). Let Ap = a*h

where a is a constant. Then take,

T2 d2z-+2&$+2=dt2

Kp*Ap = Kp*a*h

where z - hm, i.e. the value indicated by the measuring device. Therefore,

the transfer function for the sensor is:

Eim(s> =Kpoa

T2S2 + 2qLs + 1i;(s)

Controller. Let hSP be the set point. Then,

Z(s) = $pw - Grnw

and for a PI controller (eqn. (13.5))

F(s) = Kc(l +

(14.11)

(14.12)

Control Valve. Let us assume that for the control valve of this system, the

response is that of a first-order system, i.e.

Fo(“) = %qTi c(s) (14.13)

Figure 14.3b shows the block diagram for the closed-loop system, with the

transfer functions for each component of the loop. The closed-loop response

of the liquid level will be given by eqn. (14.5) where the transfer functions

GP' Gd, Gm, Gc and Gf are shown in Figure 14.3b. The servo problem arises

when the inlet flowrate Fi remains constant and we change the desired set

point. In this case the controller acts in such a way as to keep the liquid

level h, close to the changing desired value hSF. On the other hand, for

the regulator problem the set point hSF remains the same and the feedback

controller acts in such a way as to eliminate the impact of the changing load

and keep h at the desired value hSP.

Example 14.2 - The Closed-Loop Temperature Response of a Tank Heater

Consider the temperature control system for the heater of Figure 14.4a.

The temperature T is the controlled output while the inlet temperature Ti

is the load and the steam temperature is the manipulated variable. The

transfer functions for each component of the feedback loop are:

Process. If T, Ti and Ts are deviation variables, then from eqn. (9.8)

the response of the process is given by

T(s) = g T,(s) + & Tp (14.14)

The parameters r, a and K have been defined in Examples 5.1 and 9.1.

Temperature Sensor (thermocouple). Assume that the response of the thermo-

couple is very fast and its dynamics can be neglected. Thus,

T,(s) = K,T(s) (14.15)

Controller. Let Td be the set point. Then,

-(s) = T,,(s) - Tm(s) (14.16a)

and for a proportional controller the actuating output is given by,

E(s) = Kc?(s) (14.16b)

Control Valve. Assume first-order dynamics, i.e.

f(s) = & C(s)V

(14.17)

Figure 14.4b shows the block diagram for the closed-loop system with the

transfer functions for each component of the loop. The closed loop response

is easily found to be,

T(s) = Gsp(s>Tsp(s) + GLOAD(s) Q(s)

where the closed-loop transfer functions, GSP and GLOAD' are defined as

follows:

Gsp(4 =&I [Kc 1 [-&I

1+ [--s;al [Km1 [Kc1 [-q&lV

and

GLoAD(4 =rg1

1+[-.;a1 [Km1 [Kc1 &IV

Remark: To expedite the construction of the overall closed-loop transfer

functions for any feedback control loop use the following rules:

(1) The denominator of the overall transfer functions for both

the load and the set point changes is the same. It is

given by

1 + Product of the transfer functions in the loop

i.e.

l+GGGGp m c f '

(2) The numerator of an overall closed-loop transfer function

is the product of the transfer functions on the forward

path between the set point or the load and the controlled

output. Thus;

- The transfer functions on the forward path between the set

point ysp and output y are: Gc, Gf and Gp. Therefore

the numerator is

G;Gf l Gp

- The transfer functions on the forward path between the

load d and the output is only Gd. Thus, the corresponding

numerator is:

Verify these two rules with the overall closed-loop transfer functions GSP

andGLoAD ( =w l

(14.6) and (14.7)). Also, these rules can be used to

formulate the closed-loop transfer function between an input anywhere in the

loop and the output.

14.2 THE EFFECT OF PROPORTIONAL CONTROL ON THE RESPONSE OF A CONTROLLED PROCESS

Let us now examine how the response of a normal, uncontrolled process is

changed when a simple proportional, integral or derivative feedback controller

is incorporated. In this section we will consider only the proportional con-

troller and its effect on the most commonly encountered first- and second-order

systems. The effects of integral and derivative control actions will be

studied in the following two sections.

The closed-loop response of a process is given by eqn. (14.5). To

simplify the analysis assume that

G,(s) = 1 and Gf(s) = 1

Also, for a proportional controller

GJs) = KC

and eqn. (14.5) yields:

‘y(S) =Gp(s) *Kc G&d

1 + Gp(s)*Kc Y,,(s) + 1 + Gp(s)*Kc &s>

A. First-Order Systems

For first-order systems

L-!Y=p dt + y = Kpm + kdd with y(O)=m(O)=d(O)=O ,

which gives

K3s) = *ii(s) +

P+ &s)

Thus, for the uncontrolled system we have:

- Time constant, 'cP'

- Static gains; Kp for the manipulation and Kd for the load.

Put

KGp(s) = A

KdTps+l and Gdb) = 1:

P

in eqn. (14.18) and take the closed-loop response,

j;(s) =KpKC

~~s+l + K K Y,pb> +Kd

=P"+l + K Ka(s)

PC PC

Rearrange the last equation and take,

K'j;(s) = -q-- 7rps+l SP (s) + & a(s)

where

% -i&t-PC

K KK' =P 1 +PKCK

PC

(14.18)

(14.19)

(14.20a)

(14.2Ob)

and

K; = Kdl+KK

PC(14.20~)

The parameters K' andP "ii are known as closed-loop static gains.

From eqn. (14.19) we conclude that the closed-loop response of a first-

order system has the following characteristics:

(i> It remains first-order with respect to load and set point.

(ii) The time constant has been reduced, i.e. r' < -c which means thatP P

the closed loop response has become faster with respect to changes

in the set point and the load (see Section 10.4 for the effect of the

time constant).

(iii) The static gains have been decreased.

To gain a better insight on the effect of the proportional controller,

consider unit-step changes in the set point (servo problem) and the load

(regulator problem) and examine the resulting closed-loop responses. For the

servo problem, S;,,(s) = l/s and a(s) = 0. Then, eqn. (14.19) yields

K'sits1 = r�;+l l i

P

and after inversion we find

-t/T ’y(t) = Ki(l - e ') (14.21)

Figure 14.5a shows the response of the closed-loop system to a unit-step

change in the set point. We notice that:

"The ultimate response, after t*, never reaches the desired new

set point. There is always a discrepancy called offset which is

equal to

offset = (new set point)-(ultimate value of the response)

K K= 1 - K' 3 1 _ 1 +PKCK = 1

P l+KK *'IPC PC

3 “1The offset is characteristic of the proportaional control. It decreases as

Kc becomes larger and theoretically

offset _f 0 when KC-+'.

For the regulator problem, y,,(s) = 0. Consider a unit-step change in the

load, i.e. a(s) = l/s. Then, eqn. (13.19) yields

and after inversion

-t/T 'y(t) = Ki(l - e ')

Figure 14.5b shows this response to a unit-step change in the load. We notice

agains that the proportional controller cannot keep the response at the desired

set point but instead it exhibits an offset;

offset = (old set point) - (ultimate value of response)

0 - % = -Kd=

l+KKPC

The benefit of the proportional control in the presence of load changes can be

seen from Figure 14.5b. Although it cannot keep the process response at the

desired set point and introduces an offset, the response is much closer to the

desired set point than would have been with no control at all. Furthermore,

as we increase the gain Kc the offset decreases and theoretically,

offset -4 0 when Kc-=.

Remarks: (1) Although the offset tends to zero as Kc + ~0, we will never

use extremely large values of Kc for proportional con-

trol. The reason will become very clear in the next chapter

where we will study the stability of closed-loop systems.

337

(2) If Gm = Km and Gf = Kf then it is easy to show that

the offsets become:

For set point unit-step changes

K K Koffset = 1 - Pcf

l+KKKKp c f m

For load unit-step changes

offset = - Kdl+KKKKp c f m

Remark (1) above still holds.

(3) In the subsequent sections we will examine only the

response for the servo problem assuming that the reader has

gained already the facility to repeat a similar analysis

for the regulator problem.

(4) Processes having the term l/s in their transfer function,

when they are controlled with proportional controller, do

not exhibit offset for set point changes but they do for

sustained load changes, e.g. step changes. Let us demon-

strate this important feature for the liquid level control

system shown in Figure 14.6a. The output F. is constant

and the level is controlled by manipulating the inlet

flowrate Fi. The load (disturbance) is the flowrate Fd.

In terms of deviation variables the mass balance around the

tank yields,

*dh’ =dt F; + F;

and in the Laplace domain,

K’(s) = & F;(s) + & F;(s)

Therefore,

Gp(s) = l/As

Consider proportional control and for simplicity, Gm =

Gf = 1. The closed-loop block diagram is shown in Figure

14.7b and gives:

P(s) = ILl/Kc

$ s+li;;,(s> + A

y- s+lC C

For a unit step change in the set point, i&(s) = l/s and

F:(s) = 0. Then,

Ii’(s) = l ;+ s+l

C

From the final value theorem

h'(t*) = lim [s L'(s)] = 1S-4

Therefore,

offset = h& - h'(t*) = 1 - 1 = 0

For,a load unit step change,

l/Kg(S) 3 A c

K s+l. +

C

h'(t*) = lim [sii’(s) 1 = l/Kcs-+0

Therefore,

offset = 0 - l/K =C

-l/Kc # 0

For liquid level control systems like the one of Figure 14.7a, usually we

are not interested in maintaining the liquid level exactly at the desired value

but with a

acceptable

certain range. In such case the value of the offset l/Kc may be

for reasonably large Kc. Therefore,

"liquid level can be controlled effectively with proportional control."

Similar conclusion can be reached for gas pressure systems whose transfer

function also includes the term l/s.

B. Second-Order Systems (servo Problem)

The transfer function for a second-order process is

Y(s) =K

Gp(s) = - P - -

m(s)2 2

T s t 2STS + 1

Put this expression in eqn. (14.18) and recalling that for the servo problem

d(s) = 0 we take,

K'Y(s) = P

(q2s2 + 23'~'~ + 1- L,,(s)

where

T’

5’

K’P

T=

J~+KKPC

5=v'ltKK

PC

KpKc= l+KK

PC

(14.22)

(14.23a)

(14.23b)

(14.23~)

From the above we notice that the closed-loop response of a second-order

system with proportional control has the following characteristics:

- It remains second order.

- The static gain decreases.

- Both the natural period and damping factor decrease. This implies that

an overdamped process may become underdamped (oscillatory) with proportional

control and appropriate value of K .

Consider a unit-step change in the set point, i.e. yd(s) = l/s. Then,

K'Ji(s> = (,,)2s2 p

+ Z<�T�S + 1

l :

Depending on the value of 5' the inverse of the above expression may be

given by,

- eqn. (11.7) for the overdamped case, <' > 1, or

- eqn. (11.8) for the critically damped case, 5' = 1, or

- eqn. (11.9) for the underdamped case, <' I 1.

Independently though of the particular value of r,', the ultimate value of

y(t) is given by the final value theorem (Section 7.5). Thus,

K KY(t-> = lim [s y(s)] = K; =

s-to1 +PKCK

P C‘

Consequently, we notice again the presence of offset

offset = (new set point) - (ultimate value of response)

K K= l- PC 1

l+KK = l+KKPC PC

Again, offset -+ 0 for K --+ 0~.C

Remarks: (1) Depending on the value of the damping factor 5 for the

uncontrolled second-order system, eqn. (14.23b) shows

that 5' 5 1. If 5'>1 the overdamped response of the

closed-loop system is very sluggish. Therefore, we prefer

to increase the value of Kc and make r' < 1. Then, the

closed-loop response reacts faster but it becomes oscillatory.

Also, by increasing Kc the offset decreases.

(2) The increase in the speed of system's response and the

decrease in the offset,both vary desirable features, come

at the expense of higher overshoots (maximum errors) and

longer oscillating responses. Thus, as Kc increases

causing 5' to decrease

- from eqn. (11.11) we see that the overshoot increases

while

- eqn. (11.12) shows that the decay ratio also increases.

Finally, eqn. (11.13) shows that the period of oscillation

for the closed-loop response decreases as r;' decreases.

All the above features are demonstrated in Figure 14.7.

14.3 THE EFFECT OF INTEGRAL CONTROL ACTION

In this section we will repeat a similar analysis to that of the previous

section but using integral instead of proportional controller. Not to over-

whelm the reader with the repetition of algebraic manipulations we will limit

our attention to first-order systems and for the servo problem only.

Recall that for the servo problem, d(s) = 0, and eqn. (14.18) yields:

G G GS(s) = i-+ :pif:cGm 'd(')

(14.24)

Let for simplicity

Gm = Gf = 1

For a first-order process we have

KG z-f-..-.P Tps+l

while simple integral control action,

Gc = Kc&I

Substitute Gm, GP'

Gc, Gf in eqn. (14.24) by their equals and take:

B(s)

(--IF_TKS+PK 1)P

c TIS=

1+ (---TK:+l)(K A)

. Y,,(s)

A -, .,. ,

or

Y(s) =1

T2S2 + 25TS + 1l �j,,(s) (14.25)

where

T =PC

(14.26a)

51=J

?

-z TKKPPC

(14.26b)

Equation (14.25) indicates in important effect of the integral control action,

"it increases the order of dynamics for the closed-loop response"

Thus, for a first-order uncontrolled process, the response of the closed-loop

becomes second order and consequently it may have drastically different

dynamic characteristics. Furthermore, as we have seen in Section 11.3 and

12.1 that by increasing the order of a system its response becomes more

sluggish. Thus ,

"integral control action alone is ecpected to make the response of

the closed-loop system more sluggish."

Let us examine the dynamic behavior of the closed-loop system when the

set point changes by a unit step. From eqn. (14.25) we take:

T(s) = 1 1. -T2S2 + 2STS + 1 s

The shape of the response y(t) depends on the value of 5 (overdamped,

critically damped, or underdamped) but the ultimate value of the response can

be found from the final-value theorem (Section 7.5), i.e.

y(t-> = lim [s y(s)] = lim

[

1 1 = 1s-to s-to r2s2+ 25TS + 11

Therefore,

offset = 1 - 1 = 0

This indicates the most characteristic effect of integral action, i.e.

"integral control action eliminates any offset."

The reader can verify easily that for the regulator problem the integral con-

trol action produces a second-order closed-loop response and leads again to

zero offset.

Remarks: (1) Equation (14.26b) indicates that the form of the closed-

loop response (i.e. overdamped, critically damped, underdamped)

depends on the values of the controller gain Kc and restt

time rI. Therefore, tunign the integral control action for

the appropriate values of KC

a n d -cI is an important question

and will be discussed in Chapters 15 and 18.

(2) From eqn. (14.26b) we observe that as Kc increases the

damping factor < decreases. The consequences of decreasing

5 are:

- The response moves in general from sluggish overdamped

to faster but oscillatory underdamped behavior.

- The overshoot and the decay ratio of the closed-loop

response both increase (see eqns. (11.11) and (11.12) and

Figure 11.3).

Therefore, we conclude that we can improve the speed of the

closed-loop response at the expense of higher deviations

and longer oscillations. Figure 14.8 summarized the above

characteristics for set point changes.

(3) From eqn. (14.26b) we also observe that as -CT decreases,

5 decreases too. The consequences of decreasing ‘I on

the closed-loop response will be as above in Remark 2,

i.e. increased spped comes at the expense of higher overshoots

and long oscillations. Figure 14.9 demonstrates these

effects very clearly.

(4) The conclusions drawn by Remarks 1 and 2 above can be

restated as follows:

"increasing the integral control action, i.e. increasing

Q and decreasing rI, the response of the closed-

loop system becomes more sensitive."

In the next chapter we will see that such trends lead to instability of the

closed-loop response.

14.4 THE EFFECT OF DERIVATIVE CONTROL ACTION

For derivative control action alone we have

GC

= K/TDS

Assuming again for simplicity, Gm = Gf = 1, the closed-loop response of a

first-order system with derivative control action is given by

KI ?

___ l Kc(y)

T s+1

U(s) =

PK Ysp(s)

1+-J--*y+l KckDs)

or

Y(s) = (=KpKc’DS+KK-c)s+l Y,,(s)

P pcD(14.27)

Equation (14.27) leads to the following observations on the effects that the

derivative control action has on the closed-loop response of a system:

(5) The derivative control does not change the order of the response. In

the above example it has remained first order.

., ..,,.

TaQ .4..j L _*

(ii) For a unit-step change in the set point eqn. (14.27) yields,

KK-rsj;(s) = pcD

(TV + KpKc D)s + 1 * :

with an ultimate value,

Y ( t-m> = limS O

and

r[s Y(s)1 = lim l$lKcTDS

s-to+ K K ~ >S

pcD i= 0

offset = l- 0 =l.

This is an important result and demonstrates that,

Ifderivative control action does not affect the final steady

state the system approaches, i.e. it does not reduce to

offset".

(iii) From eqn. (14.27) it is clear that the effective time constant of the

closed-loop response is (rp + K K T ),PCD

i.e. larger than r . ThisP

means that the response of the controlled process is slower than

that of the original first-order process. Furthermore, as Kc

increases the effective time constant increases and the response

becomes progressively slower.

Remarks: (1) It is very instructive to examine the effect of the derivative

control action on the response of a second-order system.

Assuming again that Gm = Gf = 1, the closed-loop response

for the servo problem is,

KP

2 2 + 25TS + 1l Kc~D~

y(s) = T s K ~ - K&9

l+ 22P

T s + 25TS + 1l Kc~D~

or

f(s) = 2 2KpKcTDS

l Y,,(s)

T s + (25~ + K K -cpcD

)s + 1

From the last equation we observe that,

- the natural period of the closed-loop response remains

the same while,

- the new damping factor 5' is given by

Z<'T = 25-c + K K rpcD

i.e.

T' > 5

Therefore, the closed-loop response is more damped and the

damping increases as KC or -CD increase. This character-

istic produces more robust behavior by the controlled

process.

(2) The decrease in the speed of the response and the increase

in the damping demonstrate that the derivative control

action,

"produces more stable and robust behavior by the

controlled process."

14.5 THE EFFECT OF COMPOSITE CONTROL ACTIONS

Although proportional control can be used alone, this is almost never the

case for integral or derivative control actions. Instead, proportional-

integral (PI) and proportional-integral-derivative (PID) are the usual con-

trollers employing integral and derivative modes of control.

A. The Effect of PI Control

Combination of proportional and integral control modes lead to the

following effects on the response of a closed-loop system:

- The order of the response increases (effect of integral mode).

- The offset is eliminated (effect of integral mode).

- A s RC

increases the response becomes faster (effect of proportional

mode, see eqn. (4.20a) and more oscillatory to set point changes,, i.e.

the overshoot and decay ratio increase (effect of integral mode). Large

values of Kc create a very sensitive response and may lead to instability

(see Chapter 15).

-As -rI decreases, for constant Kc, the response becomes faster but more

oscillatory with higher overshoots and decay ratios (effect of integral

mode).

B. The Effect of PID Control

Combination of the three control modes leads to a closed-loop response

which has in general the same qualitative dynamic characteristics as those

resulting from PI control alone. Let us now describe the main benefit

introduced by the derivative control action.

We have seen that the presence of integral control slows down the closed-

loop response of a process. To increase the speed of the closed-loop we

increase the value of the controller gain Kc. But increasing enough Kc in

order to have acceptable speeds, the response becomes more oscillatory and may

lead to instability. The introduction of the derivative mode brings a

stabilizing effect to the system. Thus, we can achieve acceptable response

speed by selecting an appropriate value for the gain KC’

while maintaining

moderate oversh-ots and decay ratios.

Figure 14.10 summarizes the effect of a PID controller on the response

of a controlled process. Notice that although increasing Kc leads to faster

responses, the overshoot remains almost the same and the settling time is

shorter. Both are results of the derivative control action.


The dynamic response of a feedback controlled process is affected by the

dynamic behavior of all the elements in the loop, i.e. process, measuring

sensor, controller and final control element. The form of the closed-loop,

overall transfer function for set point or load changes, i.e. GSP (eqn. (14.6))

and GLO~D (eqn. (14.7)), indicate this very clearly.

With respect to the effect that the various types of feedback controllers

have on the dynamic response of a closed-loop system, we can observe the

following:

- The proportional control action, (a) does not change the order of the closed-

loop dynamic response while (b) it makes it faster by decreasing the dominant

time constant. The most serious drawback of proportional control alone is

its inability to reject completely the effect of a disturbance or track very

closely changes in the set point. This is demonstrated by the presence of

an offset between the desired set point and the closed-loop ultimate response.

- The integral control action, (a) changes the order of the closed-loop

dynamic response and (b) eliminates always the offset. The first character-

istic is of particular importance since it slows down the response while

processes with simple first-order dynamics may acquire oscillatory behavior

and even become unstable. The second feature indicates that integral control

action is needed wherever even small deviations (errors) are undesirable.

- The derivative control action is the only one that anticipates future errors

and takes appropriate corrective action. With respect to its effect on the

closed-loop response we notice that, (a) it does not change the order of

closed-loop dynamics and (b) allows for higher proportional gain due to its

stabilizing effect.

THINGS TO THINK BOUT

1.

2.

3 .

4.

5 .

Develop the block diagram of a generalized feedback control system with

one disturbance, incorporating in each block the appropriate -ransfer

function and on each stream the appropriate variable.

Develop the closed-loop responses for set point and load changes.

Repeat items 1 and 2 above for a process with two disturbances. Can the

feedback controller handle simultaneous changes in both loads?

Define in physical terms the servo and regulator control problems.

The following block diagram (Figure Q.14-1) corresponds to a control

system with two loops. .-

I c* -- r" _ -.. ._ -.. cws -..

, .I ;

_I--.--I-____Figure 4.14-l

Reduce the above block diagram to a simpler one like the following (Figure

Q.14-2) by identifying the appropriate transier functions G1, G2 and G3.

Figure 0.14-2

6 .

7 .

8.

9.

10.

11.

12.

13.

14.

15. Repeat item 14 above but for derivative control action.

What are the relative advantages and disadvantages of the proportional,

integral and derivative control actions? What are their characteristic

effects on the closed-loop response of a process?

The proportional control leads to a lower static gain for the closed-

loop response compared to the gain of the uncontrolled process (see

eqns. (14.20b and (14.20~)). Is a lower gain more favorable or less

for the controlled process? Recall the definition of the static gain

from Section 10.4.

What is the order of the closed-loop dynamic response for a second-order

process with PI control? Can the PI control destabilize such a process?

Discuss the effects of Kc and TI on the closed-loop response of a

process controlled with PI.

Discuss the effects of Kc, TV and TV on the closed-loop response of

a process controlled with PID.

Consider a first-order process. Could you have almost the same closed-

loop responses with PI and PID controllers and appropriate values of

their adjustable parameters?

Repeat item 11 above but for a second-order process.

Which one of the three controllers, P, PI, PID, would give more robust

closed-loop response to an underdamped second-order system?

Integral control action makes a process, (a) faster or slower, (b) more

oscillatory or less, (c) with larger deviations from the set point or

smaller? Explain your answers.

do)PROCESS

I 2-Lr ---- --- - ----_,

I

CONTROLLERCoN-rRO~ L-------- _ _ _ _ _ _ 1

ELEMENT

ym CSI Gm, WMEASURlr\lGDEVICE

F i

L-s-)

i,- _ _ yd(+)---a------_ L--3

4

+Kc '

I Fig we 14.51

W L

. .,. .

I F’i g ure 14-r 1

L---------

- ----

CHAPTER 15

STABILITY ANALYSIS OF FEEDBACK SYSTEMS

In Chapter 14 we examined the dynamic characteristics of the response of

closed-loop systems and developed the closed-loop transfer functions which

determine the dynamics of such systems. It is important to emphasize again

that the presence of measuring devices, controllers and final control elements-________- -----._--___.

change the dynamic characteristics Thk~s , non-- - - of an uncontrolled process._-----_.---.---__

oscillatory first-order processes may acquire oscillatory behavior with PI

control. Oscillatory second-order processes may become unstable with a Pl

controller and unfortunate selection of K andC TI'

Xhile designing a feedback control system, i.e. selecting its components

and tuning its controller, we are seriously concerned about its stability

characteristics. Therefore, before we proceed with the particular details of

designing a feedback control loop, we will study the notion of stabilitv and

analyze the stability characteristics of closed-loop systems.

15.1 THE NOTION 0.F STABILITY

In Section 1.2 we introduced a simple-minded notion of stability. A sys-

tem was considered unstable if, after it had been disturbed by an input change,

its output “took off” and did not return to a state of rest. Figure 1.6 shows

typical outputs for unstable processes. Example 1.2 also described the

unstable operation of a CSTR.

How do we define a stable or unstable system? There are different ways,

depending on the mathematical rigorousness of the definition and its practical

utility for realistic applications. In this text we will employ the following

definition which is often known as the bounded input, bounded output stabilitv..-- ---.-.L

"A dynamic system i.s considered to be stable if for every bounded input

it produces a 1~o11nc~~d wt.put , regardless of its initial state."

E7ery systc‘m wllicki is not stnhlr according tr? t11t alcove drl'inition wi1.1. be

f:n'l..led unstable. T0 complete the ti~finition, consider that.

- bounded is an input wl~ich always remains between an upper and a lower Iknit

( 12 I g . si.nitsoidal , step but not the ramp), and that

~.&2)o~ln+.if~il 01.1 tptitc; exist. c-n1 y in tl;t:c~r-y and nr?t in pr:3~tice b~~.a~lse all.

/,*5ys i :x1 rlilantit i<": 31'e !.imited. Thwefort:, the t crm unl~oiindecl means y:er!’

? xr F e ”

Accord-in:: t:!) tllc above def j.nition, a system with response like those of

F'jgure 1ij.l~ ic: A stabtt while Figure 15.lb shows the rpspnnses of unstable

'T-f '3 ! "Ills *

1.-t 11.5 consjder a tiv2ami7 system with input m and output y- 7.'h en )

j tr c!priamiI* bci~nvior can be described by a transfer function C(sj,

)I(sj = c(sj fiiis)

Tn Section 9.4 we concluded tbnt if G(s) has a pole with positive real part

then it gives rise to a term c1 ePt which grows continuously with time, thus

producing an unstable system. The transfer function G(s) can correspond to

an uncontrolled process or it can be the closed-loop transfer function of a

controlled system, e.g. %P Or GLOAD' Therefore. the stability analysis of

a system can be treated in a unified way independently if it is controlled or

llncont rolled.

The location of the poles of a transfer function gjves us the first

criterion for checking the stability of a system:

"If the transfer function of a dynamic system has even one pole with

positive real part, the system is unstable."

Therefore, al.1 poJes of a transfer function must be in the left-hand part of a

complex plane.

Example 15.1 - Stabilization of an Unstable Process With P Control

Consider a process with the following response,

Y(s) = ; iii(s) + & mClearly, this process is unstable because its transfer function possesses a

pole at s = 1> 0. Figure 15.2a shows the response of the uncontrolled

system to a unit-step change in the load d which verifies its unstable

character. Let us introduce a feedback control system with proportional

control only. Assume that for the measuring sensor and the final control

element

Gm = Gf =l

Figure 15.3 shows the block diagram of the closed-loop system.

The closed-loop response of the system is given by eqn. (15.5) which for

our system becomes:

lO*Kc3(s) = s _ (1 - 10K ) %P(') +

5*Kcd(s)

Cs - (1 - 10Kc)

From the last equation we conclude that the closed-loop transfer functions

10-K 5-KG C

SP = s - (1 - ;O*Kc) ' GLOAD = s - (1 - lo-Kc)

have negative pole if Kc > l/10. Therefore, the original system can be

stabilized with simple proportional control. Figure 15.2b shows the dynamic

response of the controlled system to a unit-step change in the load for Kc=l.

Compare it to the behavior of the uncontrolled system and realize the stabilizing

effect of the controller.

Example 15.2 - Destabilization of a Stable Process with PI Control

Consider a second-order process with the following transfer function

Gp(d =1

S2 +2s+2

The system has two complex poles with negative real parts

Pl = -1 + j and P2 = -1 - j

Therefore, according to our criterion the system is stable. Indeed, if we make

a unit-step change in the input, the response of the system is as shown in

Figure 15.4a. Introduce a PI controller. Let the measuring element and the

final control element have the following transfer functions:

G,(s) = Gf (s) = 1 .

The closed-loop response to set point changes is given by,

G GP(s) = 1 +PGCG l j,,(s) = Gsp-Ysp(s>

PC

To examine the stability of the closed-loop response we have to find where

the poles of GSP are located.

, T s+l.L

s2+2s+2. Kc m-i!-

?s

GSP =GpGc

KC(71s+1)/71

l+GG= rIS+l =pcl+ ' .K-

K

,2+2s+2 c -rIss3 + 2s2 + (2+Kc)s -I- 2

=I

Let

K = 100 and "I = 0.1C

Then, the poles of GSP and the roots of the polynomial

S3 + 2s2 + (2 + 100)s + g.

which are found to be

P1 = -7.185 p2 = 2.59 + j(ll.5) and p3 = 2.59 - j(11.5)

Example 15.1 - Stabilization of an Unstable Process With P Control

Consider a process with the following response,

Y(s) = $+ ii(s) +

Clearly, this process is unstable because its transfer function possesses a

pole at s = 1> 0. Figure 15.2a shows the response of the uncontrolled

system to a unit-step change in the load d which verifies its unstable

character. Let us introduce a feedback control system with proportional

control only. Assume that for the measuring sensor and the final control

element

Gm = Gf = 1

Figure 15.3 shows the block diagram of the closed-loop system.

The closed-loop response of the system is given by eqn. (15.5) which for

our system becomes:

Y(s) =lO.Ke

s - (1 - 10Kc) %P(') +

5’KC

s - (1 - 10Kc) '(')

From the last equation we conclude that the closed-loop transfer functions

lO*K 5*KGSP = s - (1 - ;O*Kc) ' GLOAD = s - (1 -clO*Kc)

have negative pole if Kc > l/10. Therefore, the original system can be

stabilized with simple proportional control. Figure 15.2b shows the dynamic

response of the controlled system to a unit-step change in the load for Kc=l.

Compare it to the behavior of the uncontrolled system and realize the stabilizing

effect of the controller.

Example 15.2 - Destabilization of a Stable Process with PI Control

Consider a second-order process with the following transfer function

GpW =1

s2 +2s+2

The system has two complex poles with negative real parts

Pl = -1 + j and p2 = -1 - j

Therefore, according to our criterion the system is stable. Indeed, if we make

a unit-step change in the input, the response of the system is as shown in

Figure 15.4a. Introduce a PI controller. Let the measuring element and the

final control element have the following transfer functions:

G,(s) = Gf(s) = 1 .

The closed-loop response to set point changes is given by,

G GP(s) = 1 +PGCG - s;,,(s> = Gsp*YspW

PC

To examine the stability of the closed-loop response we have to find where

the poles of GSP are located.

1 rIS+ll K -

GpGc ,2+2s+2 c vG

K$y+l) hI

SP = l+GG==

PC 1+ lT s+l

s2+2s+2'Kc&- s3

K

I+ 2s' + (2+Kc)s +c

-?

Let

Kc = 100 and r1 = 0.1

Then, the poles of GSP and the roots of the polynomial

S3 + 2s2 + (2 + 100)s + g.

IIIIIIIIIIII

which are found to be

pl = -7.185 p2 = 2.59 + j(11.5) and p3 = 2.59 - j(11.5)

We notice that p2 and p3 have positive real parts. Therefore, according

to our criterion the closed-loop response is unstable, when the values of

Kc = 100 and rI = 0.1 have been used. Figure 15.4b shows the response of

the system to a unit-step change of the set point. Compare it to the

response of the uncontrolled system and notice the destabilizing effect of the

PI controller. For different values of Kc and TI the response becomes

stable. Indeed, lowering the gain to Kc = 10 and increasing TI = 0.5 we

find that all the poles of GSP have negative real parts, i.e. the closed-loop

system is stable.

15.2 THE CHARACTERISTIC EQUATION

Examples 15.1 and 15.2 dramatized the effect a feedback control loop may

have on the stability characteristics of a process. In this section we will

organize and systematize our analysis, introducing and defining some

appropriate terms.

Consider the generalized feedback control system shown in Figure 14.1.

The closed-loop response for such system is given by eqn. (14.5)

G G G7(s) = Pfc

1 + GDGfGcGm 'SF(') + 1 + GIEfGcGrn '(')(14.5)

or equivalently

Y(s) = GSP Ysp(s) + GLOAD ;i(s> .

The stability characteristics of the closed-loop response will be determined

by the poles of the transfer functions GSP and GLOAD. These poles are

common for both transfer functions because they have common denominator and

are given by the solution of the following equation

l+GGGG = 0p f c m (15.1)

Equation (15.1) is called the characteristi equation for the generalized

feedback system of Figure 14.1

Let pl,p2,"*,p, be the n roots of the characteristic eqn. (15.1),

i.e.

1 + G G G G = (s - pl)(s - p2)***(s - p,)p f c m

then we can state the following criterion for the stability of a closed-loop

system:

"A feedback control system is stable if all the roots of its

characteristic equation have negative real parts, i.e. are to the

left of the imaginary axis."

If any root of the characteristic equation is on or to the right of the

imaginary axis, i.e. it has real part zero or positive, the feedback system

is unstable.

Remarks; (1) The stability criterion stated above secures stable response

of a feedback sytem independently if the input changes are

in the set-point or the -oad. The reason is that the roots

of the characteristic equation are the common poles of the

two transfer functions, GSP and GLOAD ’ which determine the

stability of the closed-loop with respect to changes in the

set point and the load, respectively.

(2) The product

Go, = GpGfGcGm

will be called open-loop transfer function because it relates

the measurement indication ym to the set point yd if the

feedback loop is broken just before the comparator, i.e.

Y,(s) = GOLW *Y,,(s)

Therefore, the characteristic equation can be written as

follows,

1 + GOL = 0

and we notice that it depends only on the transfer functions-

of the elements in the loop.

(3) The roots of the characteristic equation are also the poles

of the closed-loop transfer functions, GSF and GLOAD'

They are often called closed-loop poles.

Example 15.3 - Stability Analysis of Two Feedback Loops

In Example 15.1 we have:

G10=-

P s-l ' Gf=l , Gm=l and Gc = KC

Therefore, the corresponding characteristic equation is,

1 + GGGGp f c m =l+S* l*Kc*l = 0

which has the following root,

p = 1 - 10Kc

and the system is stable if p<O, i.e. Kc > l/10.

For the system of Example 15.2 we have:

Gp= ’

s2+2s+2, Gf=l , Gm=l and Gc = Kc(l+&)

I

The corresponding transfer function is:

l+GGGG =l+ 'p f c m s2+2s+2

l l*Kc(l + -+)*l = 0I

For KC= 100 and -cI = 0.1 the above equation yields

s3 + 2s2 + 102s + 1000 = 0

with roots, -7.185, 2.59 + j(11.5) and 2.59 - j(ll.5). The closed-loop sys-

tem is unstable because two roots of the characteristic equation have

positive real parts.

- ._. _ . . ./ . , .I%.3&t

I

I

I

I

I

I

15.3 THE ROUTH-HURWITZ CRITERION FOR STABILITY

The criterion of stability for closed-loop systems does not require the

calculation of the actual values of the roots of the characteristic polynomial.

It only requires to know if any root is to the right of the imaginary axis.

The Ruth-Hurwitz procedure allows us to test if any root is to the right of

imaginary axis and thus reach quickly a conclusion as to the stability of the

closed-loop system without computing the actual values of the roots.

Expand the characteristic equation into the following polynomial form,

1tGGGG z aosn t als n-lp f c m + l ** t anmls t an = 0 .

L e t a0 be positive. If it is negative then multiply both sides of the above

equation by -1.

First test. If any of the coefficients al,a2,***,an,l,an is negative, then

there is at least one root of the characteristic equation which has positive

real part and the corresponding system is unstable. No further analysis is

needed.

Second test. If all coefficients ao,al,a2,***,an-l,a n are positive, the

first test cannot conclude anything about the location of the roots. Form

the following array (known as Routh array):

Row 1 a0 a2 a4 a6 ***

2 al a3 a5 a7 ” - *

3 Al A2 A3 l

. . 0

4 B1 B2 B3 l *�**

5 cl c2 c3 ’ -*-. . . . . . . . . . . . .

where

Al = ala2 - aoa3 ,

al

B1 = Ala3 - alA

A1>

c1 = B1A2 - A1B2 3

El

A2 _ ala4 - aoa5, A3 = “laba; aoa7 , . . .

al

B2 = A1A5 - alA ~ . .A1 ’

B1A3c2 = - - A1B3 ...B1 ’

etc.

Examine the elements of the first column of the above array, i.e.

a()> al' Al, B1, C1,-*J1

- If any of these elements is negative then we have at least one root to the

right of the imaginary axis and the system is unstable.

- The number of sign changes in the elements of the first column is equal

to the number of roots to the right.of the imaginary axis.

Therefore, a system is stable if all the elements in the first column of the

Routh array are positive.

Example 15.4 - Stability Analysis With the Routh-Hurwitz Criterion

Consider the feedback control system of Example 15.2. The characteristic

equation is

3 KCS + 2s2 + (2 + Kc)s + -r = 0

I

The corresponding Routh array can now be formed:

Row 1

2

1

2

K2(2+Kc) -$

3 I2

2+KC

Kc

?

0

The elements of the first column are

Kc2(2+Kc) -y--

IK

2 1C

' T-11

All are always positive except the third which can be positive or negative

dpending on the values of Kc and -cl.

- I f Kc = 100 and -cI = 0.1 the third element becomes -398 < 0, which

means that the system is unstable. We have two sign changes in the

elements of the first column. Therefore, we have two roots with positive

real parts (see Example 15.2).

t2- If Kc = 10 and rI = 0.5 the third element is equal to $1 > 0, and

the system is stable since all the elements of the first column are

positive.

- In general, the system is stable if KC and TI satisfy the condition

2(2 + Kc) > 5 ,r1

Example 15.5 - Critical Stability Conditions for a Feedback Loop

Return to Example 15.4 and let rI = 0.1. Then, the third element of the

first column in the Routh array becomes,

2(2+Kc) - 10Kc

2 .

The value of Kc that makes the third element zero is

KC

= 0.5

and constitutes the critical conditions for a feedback control system.

Therefore, according to the Routh-Hurwitz test we have:

- I f K/0.5 all the elements of the first column in the Routh array are

positive and the system is stable, i.e. all the roots of the characteristic

equation are located to the left of the imaginary axis.

- If Kc > 0.5 the third element of the first column of the Routh array

becomes negative. We have two sign changes in the elements of the first

column, therefore we have two roots of the characteristic equation located

to the right of the imaginary axis.

It is clear therefore that as Kc increases two roots of the characteristic

equation move towards the imaginary axis and when Kc = 0.5 then we have two

roots on the imaginary axis (pure imaginary) which give rise to sustained

sinusoidal term.

Remark: The two purely imaginary roots can be found from the equation

K2s2 + .-A = 0

r1

i.e.

2s2 + g = 0.

and they are:

+ j(2.5)

The coefficients 2 and Kc/r1 are the elements of the row in

the Routh array just before the element of the first column which

is zero, i.e. the elements of the second row.

15.4 THE ROOT LOCUS ANALYSIS

The preoceding examples have demonstrated very vividly that the stability

characteristics of a closed-loop system depend on the value of the gain Kc.

Thus, in Example 15.1 we notice that the closed-loop system becomes stable

when Kc > l/10. Also, in Example 15.4, the system is stable when

Kc2(2+Kc) > FI

which for =I = 0.1 yields

0 < K < 0.5 .C

The root loci are merely the plots in the complex plane of the roots of

the characteristic equation as the gain Kc is varied from zero to infinity.

As such they are very useful in determining the stability characteristics of a

closed-loop system as the gain K changes. Let us examine the constructionC

of the root locus using a specific example.

Example 15.6 - The Root Locus of Two Capacities in Series with P Control

The two capacities in series may be two stirred tanks, two heaters, etc.

and have a transfer function

KGpW = P

(T1s+1)(T2s+1)

Let,

Gm = Gf =l and

Then, the characteristic equation is:

Gc = Kc .

1+ KP l K = 0(Tls+1)(?2s+1) c

or

(~~s+l)(y+l) + K = 0 where K = K K .PC

Consider K as the changing parameter instead of the gain Kc, and make the

following observations:

(0 When K=O (i.e. Kc = 0) the characteristic equation has as its roots

the poles of the process, i.e.

Pl = -l/r1 and p2 = -l/T2

(ii) As K increases from the zero value, the roots of the characteristic

equation are given by,

-(T1+T2) t &+T2)2 - 4y2(1+K)

P1,2 =2TlT2

.

They are distinct real and negative as long as

2

K <(T1+T2)

4=lT2

_ 1

i.e. as long as Kc satisfies the inequality

K c1C

%

(iii) When

then, we have two equal roots

(15.2)

(15.3)

T +-r1 2

Pl = P2 = - -

2=1r2

(iv) For

K+ (15.4)P

we have again two distinct roots which are complex conjugates of

each other, i.e.

-(T~+T~) 2 j J 4'rclr2(l+K) - (TV+ 2)2

P1,2 = 2T T1 2

Notice that the real part is equal to

=1+=2-~2T1’2

and independent of K, while the imaginary part tends to infinity

as K -f a.

Using the above information we can construct the root locus of the system as

follows:

(a) The beginning of the root locus corresponds to Kc = 0 and is given

by the points A(-l/rl, 0) and B(-1/~~,0), see Figure 15.5.

(b) As long as Kc satisfies inequality (15.2) we have two distinct real

and negative roots. Therefore, the root locus is given by two distinct

curves which emanate from points A and B and remain on the real

axis. Furthermore, the two curves move towards each other and meet

at the point C (Figure 15.5). At this point, Kc i\as the value given

by eqn. (15.3) and we have a double root.

(c) For larger values of Kc satisfying inequality (15.4), we have again

two distinct curves of the root locus because we have distinct, com-

plex conjugate roots. Since the real part of the complex roots is

,...

constant, the two branches of the root locus are perpendicular to the

real axis and extend to infinity as Kc + 00.

The complete root locus is given in Figure 15.5 and since all its branches are

LQGated tQ the Left af the inagtnaary a?sis-, More CQRC~NdC tkat tke ClQsed- b2~

system is stable for any value of K . Furthermore. we conclude that for KC

satisfying inequality (15.2) the response of the system to a step input is not

oscillatory. It becomes oscillatory for Kc satisfying inequality (15.4).

Example 15.6 demonstrated that the root locus of a system does not only

provide information about the stability of a closed-loop system but informs

us about its general dynamic response characteristics as KC

changes.

Therefore, the root locus analysis can be the basis of a feedback control

loop design methodology, whereby the movement of the closed-loop poles (i.e.

the roots of the characteristic equation) due to the change of the proportional

controller gain can be clearly displayed.

The construction of the root locus for the system of Example 15.6 was

rather simple. For higher order systems to find the exact location of the

root locus branches we need a computer program that can find the roots of a

high order polynomial. Such programs are available in any large computer

system and the interested reader can find one in Reference 7.

Quite often though we are not interested in the exact location of the

root locus branches and simple but qualitatively correct graphs will suffice

to draw the general conclusions about the dynamic behavior of a closed-loop

system. Appendix 15.A gives a set of general rules which can be used to draw

the approximate root locus of any given system.

Let us close this chapter with one more example on the construction of

the root locus for a reactor system and its use for the analysis of the

system's dynamic response.

Example 15.7 - The Root Locus for a Reactor With Proportional Control- -

Douglas [Ref. 121 has developed the model for the reactor shown in

Figure 15.6. The control objective is to keep the concentration of the

desired

despite

control

tration

product C as close as possible to a given steady state value

the upsets in the inputs of the reactor. He attempts to achieve the

objective with a proportional controller which measures the concen-

of C and manipulates the flowrate of the reactant A.

The transfer function for the process is,

Gp(s) = 3s) = 2.98(s+2.25)

G(s) (s+l.45)(~+2.85)~(~+4.35)

Assuming instantaneous responses with gain 1 for the measuring device and

the valve that controls the flowrate of A, i.e.

G =m Gf = 1

we have the following characteristic equation for the closed-loop system:

1+ 2.98(s+2.25)2

-K = 0(s+1.45)(s+2.85) (s+4.35) C

(15.5)

When Kc = 0, it is easy to find that the roots of eqn. (15.5) are:

p1 = -1.45 , p2 = p3 = -2.85 and p4= -4.35

As Kc increases, we need an iterative, trial-and-error, numerical procedure

to find the roots of the characteristic equation. Such solution is feasible

through the use of a digital computer. Table 15.1 shows how the locations

of the four roots change with the value of Kc. These results have been

Table 15.1. The Roots of the Characteristic Equation-for the System of Example 15.7.

5-0

1

5

20

50

100

p1 p2 p3 p4- -

-1.45 -2.85 -2.85 -4.35

-1.71 -2.30 + j(O.9) -2.30 - j(O.9) -4.74

-1.98 -1.71 + j(1.83) -1.71 - j(l.83) -5.87

-2.15 -1.09 + j(3.12) -1.09 - j(3.12) -7.20

-2.20 -0.48 + j(4.35) -0.48 - j(4.35) -8.61

-2.24 -0.35 + j(5.40) -0.35 - j(5.40) -9.75

transferred in Figure 15.7 which displays the four branches of the root locus

for the closed-loop reactor system..

Let us examine the root locus branches of Figure 15.7 and draw some con-

clusions on the dynamic response of the closed-loop reactor system as the

proportional gain KC

changes from zero to infinity.

(i> The system is stable for gain values up to 50 because all the roots

are located to the left of the imaginary axis. For a gain value

between 50 and 100 the root locus crosses the imaginary axis and

moves to the right of the imaginary axis. Therefore, there is a

critical value between 50 and 100 for which the closed-loop response

of the reactor becomes unstable.

(ii) For any value of Kc > 0 until the critical value there are two

complex conjugate roots with negative real parts. They imply that

the response of the reactor to an input step change will be a

decaying oscillation.

(iii) For Kc larger than the critical value (where the system becomes

unstable) the roots that cause the instability are complex con-

jugates with positive real parts. Consequently, the unstable

response of the closed-loop system to an input step change will be

oscillatory with growing amplitude.

Remark: A very satisfactory approximate sketch of the root locus for the

reactor system can be constructed using the graphical rules of

Appendix 15-A.


Almost all processing units employed in a chemical plant are inherently

stable. Under feed back control, though, they can become unstable. Therefore,

the stability characteristics of a closed-loop system are of uppermost importance

during its design. They depend on the dynamics of all the components in the

control loop, i.e. process, sensors, controller and final control element.

The notion of stability that suffices for our purposes is that of the

bounded input, boudned output. According to this, a system is stable if it

produces a bounded response to any bounded input. The criterion that we will

use to examine the stability of a system is related to the location of the

poles of its transfer function. Thus,

- For uncontrolled systems the poles of the process' transfer function must

be located to the left of the imaginary axis for the system to be stable.

- For closed-loop systems the closed-loop poles (the roots of the character-

istic equation) must be located to the left of the imaginary axis for the

closed-loop response to be stable.

The Routh-Hurwitz tests allow us to examine if any pole is located to

the right of the imaginary axis, without computing the actual values of the

poles. This procedure should be used whenever we need to know only if the

system is stable or not.

The stability characteristics of a closed-loop depend on the dynamic of

the controller used which are affected by the values of the parameters Kc,

TI and ~~~ The root locus analysis permits us to examine the location of

the closed-loop poles as Kc changes from zero to infinity. From the location

of these poles we can draw conclusions on:

- the stability of the feedback loop and

- the general characteristics of the system's response for various values

of KC’

The root locus has been historically a popular method for the design of feed-

back loops. It has several drawbacks like its instability to handle

efficiently systems with dead time, and today it has been replaced by more

powerful design techniques like those based on the frequency response of a

system. Consequently, we will spend no more space and time expanding on its

use for control design.

3,


1.

2 .

3 .

4.

5 .

6 .

7 .

8.

9 .

Define what is known as bounded input, bounded output stability.

Based on the above definition examine if a system with a pole at s=O

is stable or not.

Define the terms: open-loop transfer function, characteristic equation,

closed-loop poles.

If a closed-loop response is stable with respect to changes in the set

point, is it stable to changes in the load? If yes, why?

How does the pole location determine the stability of an uncontrolled

or controlled process?

Does location of the zeros of a transfer function affect the response

of an uncontrolled process?

What is the major advantage of the Routh-Hurwitz criterion for examining

the stability of a system?

What conclusions can be drawn if one element in the first column of

the Routh array is zero? Consult References 13, 14.

The root locus analysis cannot handle easily systems with dead time.

why? Show how systems with dead time could be handled with root locus

analysis.

10. Examples 15.6 and 15.7 indicate that the root locus has as many

branches as the number of poles of the open-loop transfer function.

Thus, in Example 15.6 the open-loop transfer function has two poles and

the root locus two branches., while for Example 15.7 we have four poles

and four branches. Is this true for any closed-loop system? Explain.

t3

376Y

- - - -

t t

c?

A+R - ?I%+a-, cC4R”r-DD - l - - R - E

I Figure i5.4

-6

I6’Ik .;I00

4

\\

4 .,,,.... . ,-” ,,,. .,

. . .~.

APPENDIX 15.A. RULES FOR THE CONSTRUCTION OF ROOT-LOCUS DIAGRAMS.

Consider the characteristic equation for a general closed-loop

system,

l+GGGG = opmcf

Let the open-loop transfer function G G G G . be written as the ratiopmct

of two polynomials of order m and n with m < n. Then, the characteris-

tic equation becomes:

( s - z ) ( s - z ) l -* (s-z )

l+K 1 2c (s-p,)(s-p,) l a* (s-pr) = O (15.A-1)

Z1’ Z2’ -*-, zm are the m zeros of the open-loop transfer function,__-

while pl, p2' a*-, Pn are the poles of the open-loop transfer function.

As Kc changes from 0 to 03 we find different roots for the characteris-

tic equation (15.A-1). The geometric locus of all the roots consti-

tutes the Root-Locus Diagram for the corresponding closed-loop system,

The Root-Locus for the general closed-loop system with the char-

acteristic equation (15.A-l), can be approximately constructed, using

the following rules:

1. Number of Curves. The number of branches (loci) composing the

Root-Locus is equal to the number of open-loop poles, n.

2. Origin of Curves. Each branch (locus) originates from a pole,

whenK =O.C

From multiple poles originate multiple branches,

Thus, if p =1 -1 is a double pole, we will have two loci originat-

ing from the point (-l,O).

3 . Termination of Curves. As Kc + 00 the n branches (loci) of the

Root-Locus terminate at either

- the m zeros of the open-loop transfer function, or

- approach (n-m) zeros at infinity along specified asymptotes.

4. Loci on the Real Axis. The real axis is part of a branch (locus)

if the sum of the number of poles and zeros to the right of any

point on the real axis is odd. A double or triple pole or zero

should be counted twice, three times, etc.

5. Location of Asymptotes. The (n-m) asymptotes approached by (n-m)

branches of the Root-Locus emanate from the center of gravity of

the poles and zeros of the open-loop transfer function. The center

of gravity is given by

n m

c1 = i"l 'i - j~l 'i

n - m (15.A-2)

The asymptotes make angles of

n[(2k+l)/n-m)] k = 0, 1, 2, . . . . (n-m-l)

with the real axis. This means that they are equally spaced at

angles of

2T/(n-m)

with each other.

6. Breakaway Point. Whenever two branches emerging from two adjacent

real poles intersect, they leave the real axis at a point which is

determined by the equation,

m1 n

c c 1=j=l s - z.

J i=l ' - Pi(15.h-3)

The two branches leave the real axis at right angles.

7 . Entering Point. Whenever two branches coming from two different

poles move towards two adjacent real zeros, they enter the real

axis at right angles and at a point satisfying equation (15.A-3).

8. Angle of Departure. The angle of departure of a branch from a

simple pole on the real axis is 0 or TT. For multiple real poles of

k-th order, the k branches of the Root-Locus leave the pole, at

angles given by

Eli =+ (2i+l)v + T&, (pq-zj> + T& (pq-pj)j=l j=l !

7%

for i = 0, 1, 2, . . . . k-l (15.A-4)

where pq

is the k-th order pole.

9 . Angle of Approach. The angle of approach

zero is either 0 or IT. For multiple real

k branches of the Root-Locus approach the

given by

of a branch to a simple

zeros of k-th order, the

multiple pole at angles

oi = i

L(2i+l)n + Y& (zq-zj)

n

j=l+ ' ~ ('q-pj)

j=l !7%

for i = 0, 1, 2, . . . . k-l (15.A-5)

where z is the k-th order zero.4

10. Symmetry of the Root-Locus. Since the complex roots of the char-

acteristic equation (15.A-1) always appear as conjugate pairs, the

Root-Locus is symmetric with respect to the real axis.

Example. Consider again the Example 15.7. The open-loop transfer

function has four poles and one zero; pI = -1.45, p2 = p3 = -2.85,

P4 = -4.35 and z1 = -2.25. Therefore, the Root-Locus is composed of

four branches (loci) which emerge from the four poles. One of the

branches (branch 1, see Figure 15.7) terminates at the zero, while the

other three (4-l) approach zeros at infinity along the three asymptotes.

The center of gravity of the poles and zeros is given by equation

(15.A-2)

c1 = (-1.45) + 2(-2.85) + (-4.35) - (-2.25)(4-l)

= -3.08

The asymptotes make angles of

-1~[(=+1)/(4-1) 1 k = 0, 1, 2

with the real axis; i.e. 60", 180" and 300". The center of gravity

and the location of the three asymptotes are shown in Figure 15.7.

The following segments of the real axis are part of the Root-Locus:

- Between -1.45 and -2.25. One pole to the right of every point of

this segment.

- Between and --m. Four poles and one zero to the right of

every point of this segment.

4 ,. .” I. . . .

The segment from -1.45 to +m has no poles or zeros to its right, while

the segment from -2.85 to -4.35 has three poles (pI,p2,p3) and one

zero to its right. Both these segments cannot be part of the Root-

Locus, according to the Rule 4. The same is true for the segment from

-2.25 to -2.85.

From the double pole of -2.85 emanate two loci with angles of

departure given by equation (15.A-4).

Figure 15.7 shows the approximate Root-Locus for the reactor of

Example 15.7.

CHAPTER 16

DESIGN OF FEEDBACK CONTROLLERS

In this chapter we will confront the critical questions: how do we select

the type of the feedback controller, i.e. P, PI or PID, and how do we

adjust the parameters of the selected controller (i.e. Kc, TI, rD) in order

to achieve an "optimum" response for the controlled process? The answers to

both questions lead to the synthesis of the control system , and the analysis

of closed-loop behavior studied in the previous three chapters of Part IV.

16.1 OUTLINE OF THE DESIGN PROBLEMS

Consider the block diagram of a general closed-loop system shown in

Figure 14.1.

When the load or the set point change, the response of the process

deviates and the controller tries to bring the output again close to the

desired set point. Figure 16.1 shows the response of the controlled process

to a unit step change in the load, when different types of controllers have

been used. We notice that different controllers have different effects on

the response of the controlled process. Thus, the first design question arises:

Question 1: What type of feedback controller should be used to

control a given process?

Given that we have decided somehow to use PI control, we still need to select

the value of the gain Kc and the reset time r1* Figures 16.2a and 16.2b

demonstrate very clearly that these two parameters have an important effect

on the response of the controlled process. Thus, the second design question:

Question 2; How do we select the best values for the adjustable

parameters of a feedback controller?

This is known as the controller tuning problem.

1* ._. -.. . .

"keep the maximum deviation as small as possible,"

or

"return to the desired level of operation and stay close to it

in the shortest time,"

then, we would have selected the other controller yielding the closed-loop

response of type B. Similar dilemmas will be encountered quite often during

the design of a controller.

For every process control application we can distinguish,

- steady state, and

- dynamic response performance criteria.

The usual steady state performance criterion is zero error at steady

state. We have seen already that in most situations, the proportional controller

cannot achieve zero steady-state error, while a PI controller can. Also, we

know that for proportional control the steady state error (offset) tends to

zero as Kc -+ 00. No further discussion is needed on the steady state

performance criteria.

The evaluation of the dynamic performance of a closed loop system is

based on two types of commonly used criteria:

- Criteria which use only a few points of the response. They are simpler

but only approximate.

- Criteria which use the entire closed-loop response from time t=O until

t = very large. These are more precise but also more cumbersome to use.

In the remaining of this section we will deal with the first category of

simple performance criteria and we will leave the more complicated criteria

for the next section.

The simple performance criteria are based on some characteristic features

of the closed-loop response of a system. The most often quoted are (see

Figure 11.2):

- overshoot,

- rise time, i.e. time needed for the response to reach the desired value

for the first time,

- settling time, i.e. time needed for the response to settle within 25% of

the desired value,

- decay ratio,

- frequency of oscillation of the transient.

Every one of the above characteristics could be used by the designer as the

basic criterion for selecting the controller and the -values of its adjusted

parameters. Thus, we could design the controller in order to have: minimum

overshoot or minimum settling time, etc. It must be emphasized though that

one simple characteristic does not suffice to describe the desired dynamic

response. Usually, we require that more objectives be satisfied, i.e.

minimize overshoot and minimize settling time, etc. Unfortunately, controller

designs based on multiple criteria lead to conflicting response character-

istics. For example, Figure 16.4 shows that by decreasing the value of the

overshoot we increase the settling time. Scuh conflicts will always arise

while using simple design criteria as the above, The control designer must

intervene and subjectively balance the conflicting characteristics.

From all the above performance criteria, the decay ratio has been the

most popular by the practicing engineers. Specifically, experience has shown

that a decay ratio (see Figure 11.2)

c/A = l/4

is a reasonable trade-off between a fast rise time and a reasonable settling

time. This criterion is usually known as the one-quarter decay ratio

criterion.

Example 16.1 - Controller Tuning With the One-Quarter Decay Ratio Criterion

Consider the servo control problem of a first-order process with PI

controller. The closed-loop response was developed in Section 15.3 and is

given by eqn. (15.23) when Gm = Gf = 1

'*(s) =rIs+l

T2S2 + 25rs + 1Ysp (s)

where

(15.23)

T = JYpK KPC

and

r, = 1? J

r1T K K (' + KpK~)PPC

(15.24a)

(15.24b)

We notice that the closed loop response is second-order.

For the selection of the "best" values for Kc and 'II we will use

simple criteria stemming from the underdamped response of a second-order sys-

tem. Select the one-quarter decay ratio criterion. From eqn. (11.12) we

know that

Decay Ratio = exp( -2lT3--->

f-l-r2

Therefore, for our problem we have,

exp

After algebraic simplifications we take:

/ T--2r J4rpKp;c _ l (1 + KpKc) = an(1’4) (16.1)

Equation (16.1) has two unknowns; Kc, rI. Therefore, we will have several

controller settingg which satisfy the one-quarter decay ratio criterion.

Let KP=l and -cp=lO. Then, we find the following solutions

K =lC

Kc = 10 Kc = 30 Kc = 50 Kc = 100

r1= 0.49 7I = 0.16 ,rI = 0.06 ~~ = 0.04 ~~ = 0.02

etc. The question is which one to select. Usually, we select first the

proportional gain Kc so that the controller has the necessary "strength"

to push the response back to the desired set point and then we choose the

corresponding TI value so that the one-quarter decay ratio is satisfied.

16.3 TIME-INTEGRAL PERFORMANCE CRITERIA

The shape of the complete closed-loop response from time t=O until

steady state has been reached could be used for the formulation of a dynamic

performance criterion. Unlike the simple criteria which use only isolated

characteristics (e.g. decay ratio, settling time) of the dynamic response, the

criteria of this category are based on the entire response of the process.

The most often used are:

1. Integral of the Square Error (ISE), where

co

ISE = e(t)dt

0

2 . Integral of the Absolute Value of the Error (IAE), where

IAE =.P

Idt> Idt

0

(16.2a)

(16.2b)

AC L .,...,L ,, . .,

. . .-

34

3 . Integral of the Time-Weighted Absolute Error (ITAE), where

I

cm

ITAE = tlE(t) Idt (16.k)

0

where E(t) = yd(t) - y(t), i.e. the deviation (error) of the response from

the desired set point.

The problem of designing the "best" controller can now be formulated as

follows:

"Select the type of the controller and the values of its adjusted

parameters in such a way as to minimize the ISE, IAE or ITAE of the

system's response."

Which one of the above three criteria we will use depends on the character-

istics of the system we want to control and some additional requirements we

impose on the controlled response of the process. The following are some

general guidelines:

- If we want strongly to suppress large errors, ISE is better than IAE

because the errors are squarred and thus contribute more to the value of

the integral.

- For the supporession of small errors, IAE is better than ISE because when

we square small numbers (smaller than one) they become even smaller.

- To suppress errors which persist for long times, the ITAE criterion will

tune the controllers better because the presence of large t amplifies

the effect of even small errors in the value of the integral.

Figure 16.5 demonstrates in a qualitative manner the shape of the closed-loop

responses. When the controller parameters have been tuned using ISE, IAE and

ITAE performance criteria, two points are very important to emphasize:

- Different criteria lead to different controller designs.

- For the same time integral criterion, different input changes lead to

different designs.

Let us analyze these two statements on the basis of the following example.

Example 16.2 - Controller Tuning Using Time-Integral Criteria

Consider the feedback system shown in Figure 16.6. The closed-loop

response is:

L(s) =-rIs+l

Y,,(s) +5s- ;i(s)

r1 220Kc --Q s+l' +5(l+20Kc

or

TIs+l T sI+4 = r2s2 + 2cTs + 1 %P(s) +,282 + 2T.s + 1 d(s)

where

T =

and

(16.4a)

= 15 2 J

=I20K (1 + 2OICJ

C

(16.4b)

In order to select the best values for Kc and TI we can use one of the

three criteria: ISE, IAE or ITAE. Furthermore, we can consider changes

either in the load or the set point. Finally, even if we select set point

changes we still need to decide what kind of changes we will consider, i.e.

step, sinusoidal, impulse, etc. Let us say that we select ISE as the

criterion and unit-step changes in the set point. From eqn. (16.3) we have:

Y(s) = 2 2.rIs+l 1

. -

T s + 25TS + 1 s

Invert the last equation and find (if 5~1):

r 7

y(t)=l+e -rtlT 1 TII sin( f- l-5 2 5) - sin(/- l-5 2t -1 I- l-c2

I-

7 + t a n ___

l-C21r )

(16.5)

Then solve the following optimization problem

I03

"Minimize ISE = [ysp - y(t)12dt by selecting the values of

0

T and 5, where y(t) is given by eqn. (16.5)."

The optimal values of r and < are given by the solution of the following

equations (conditions for optimality):

a(ISE)a7

= a(Iw = (-Jar

Let -r* and <* be the optimal values. Then, from eqns. (16.4a) and (16.4b)

we can find the corresponding optimal values for the controller parameters TI

and KC’

If the criterion was the ITAE then we would have to solve the following

problem:

f

co

"Minimize ITAE = dYsp - y(t) Idt by selecting the values of

0

T and 5 where y(t) is given by eqn. (16.5."

The solution r* and c* is given by the equations

a(ITAE)/ar = a(ITAE)/ar, = 0

and in turn, from eqns. ((16.3a) and (16.3b)) we can find the optimal Kc and

It is clear that the solutions of the two problems with different

criteria will be in general different.

Let us consider now unit-step changes in the load. Equation (16.3)

yields;

T sY(s) =

I 1.-T2S2 + 2<TS + 1 s

and inversion,

*I e-?t/T

y(t) = f-sin( l-5 2 t

T l-c2/--

--) (16.6)

We can find the optimal values of Kc and TI following a similar procedure

as previously. Since the response y(t) is now different than it was for a

unit step change in the set point (compare eqns. (16.6) with (16.5), we expect

that the optimal settings of Kc and 2 will be different, even if we use

the same criterion, i.e. ISE or ITAE.

16.4 SELECT THE TYPE OF FEEDBACK CONTROLLER

Which one of the three popular feedback controllers should be used to

control a given process? The question can be answered in a very systematic

manner as follows:

- Define an appropriate performance criterion, e.g. ISE, IAE or ITAE.

- Compute the value of the performance criterion using P or PI or PID

controller with the best settings for the adjusted parameters Kc, -cI,

- Select that controller which gives the "best" value for the performance

criterion.

This procedure although mathematically rigorous has several serious

drawbacks like:

- it is very tedius,

- it relies on modes1 (transfer functions) for the process, sensor and final

control element, which may not be known exactly,

- it incorporates certain ambiguities as to which is the most appropriate

criterion and what input changes to consider.

Fortunately, we can select the most appropriate type of a feedback con-

troller using only general qualitative considerations stemming from the analysis

. “.

.

in Chapter 14. There we had examined the effect of the proportional, integral

and derivative control modes on the response of a system. In summary: the con-

clusions were as follows:

Proportional Control

- Accelerates the response of a controlled process.

- Produces an offset, i.e. non-zero steady state error for all processes

except those with terms l/s (integrators) in their transfer function, like

the liquid level in a tank or the gas pressure in a vessel (see Remark in

Section 14.2).

Integral Control

- Eliminates any offset.

- The elimination of the offset comes usually at the expense of high maximum

devications.

- Produces sluggish, long oscillating responses.

- If we increase the gain Kc to produce faster response the system becomes

more oscillatory and may lead to instability.

Derivative Control

- Anticipates future errors and introduces appropriate action.

- Introduces a stabilizing effect on the closed-loop response of a process.

Figure 16.1 reflects in a very simple way all the above characteristics which

should be used in a general manner.

It is clear from the above that a three-mode PID controller should be

the best. This is true in the sense that it offers the highest flexibility

to achieve the desired controlled response by having three adjustable

parameters. At the same time it introduces a more complex tuning problem

because we have to adjust three parameters. To balance the quality of the

desired response against the tuning difficulty we can adopt the following

rules to select the controller.

1. If possible use simple proportional controller.

Simple proportional controller can be used if: (a) we can achieve

acceptable offset with moderate values of Kc or (b) the process

has an integrating action, i.e. a term l/s in its transfer function

for which the P control does not exhibit offset. Therefore, for gas

pressure or liquid level control we can use only P controller.-

2 . If simple P controller is unacceptable use a PI.

?I controller should be used when proportional control alone cannot

provide sufficiently small steady state errors (offsets). Therefore,

PI will be used seldomly in liquid level or gas pressure control sys-

tems but very often (almost always) for flow control. The response of

. a flow system is rather fast. Consequently, the speed of the closed-

loop system, despite the slow down caused by the integral control

mode, remains satisfactory.

3. Use PID controller to increase the speed of the closed-loop response.

The PI eliminates the offset but reduces the speed of the closed-loop

response. For a multicapacity process whose response is very sluggish,

the addition of a PI controller makes it even more sluggish. In such

casestheadditionof thederivativecontrolactionwith its stabilizing

effect allows the use of higher gains which produce faster responses

without excessive oscillations. Therefore, derivative action is

recommended for temperature and composition control where we have

sluggish, multicapacity processes.

Example 16.3 - Selecting the Type of Controller for Various Processes

Let us discuss various processes which are to be controlled by feedback

control systems. We will address primarily the question of selecting the

appropriate type of feedback controller.

(a) Liquid level control. Consider the two liquid level control systems for

the bottom of a distillation column and its condenser's accumulation

drum (Figure 16.7). Our control objective is to keep each liquid level

within a certain range around the desired set point and not to meet it

exactly. This allowance for non-zero offsets dictates that proportional

control alone is satisfactory.

(b) Gas pressure control. Our objective is to regulate the pressure p in

the tank of Figure 16.7c, when the inlet pressure p1 or the pressure

p2 in a downstream process change. Usually, we want to maintain p

within a certain range around a desired value, thus making a proportional

controller satisfactory for our purpose.

Cc> Vapor pressure control. Here we can have loops which react quite fast

or are relatively slow. Consider for example the two configurations

shown in Figure 16.8. The loop in Figure 16.8a measures the pressure

and manipulates the flow of vapor, thus affecting directly and quickly

the vapor pressure in the process. For such systems with fast response,

a PI controller is satisfactory. It eliminates any undesirable offset

while maintaining acceptable speed of the response (despite some slowdown

caused by the integral mode of control). For the system in Figure

16.8b the situation is different. Here, the vapor pressure is controlled

indirectly by the flow of cooling water which affects the amount of

vapor condensed. Such systems may be used for controlling the pressure

in a distillation column. The slow dynamics of the heat transfer

Ic- 1 _ . .U”. ” -. 1,

. . .

process are introduced in the control loop. We expect that the response

of the system will be rather slow. A PI controller will make it even

slower and if we attempted to use high gains to speed up the response,

we may get an unstable system. Therefore, a PID controller should be

selected which will provide enough speed and robustness.

(d) Flow control. Consider the two flow control systems shown in Figure

16.9. Both respond quite fast. Therefore, a PI controller is

satisfactory because it eliminates offsets and retains acceptable speed

of response.

(e) Temperature control. Consider the temperature control system shown in

Figure 16.10. Our objective is to keep the temperature of the reacting

mixture at a desired value. Since the reaction is endothermic, this

is accomplished by manipulating the flow of steam in the jacket around

the reactor. Between the measured temperature and the control effect

we have two rather slow processes: (i) heat transfer between the

reacting mixture and the temperature sensor (see Section 13.3) and

(ii) heat transfer from steam to the reacting mixture. We expect,

therefore, that the overall response will be rather sluggish and a PI

controller will make it even more so. Consequently, for such systems

a PID controller would be the most appropriate because it can allow

high gains for faster response without undermining the stability of

the system.

(f) Composition control. Here we have a similar situation to that of

temperature control, i.e. very slow response caused by slow composition

sensors. Therefore, a PID controller should be the most appropriate.

397

16.5 CONTROLLER TUNING TECHNIQUES-

After the type of a feedback controller has been selected, we still have

the problem of deciding what values to use for its adjusted parameters. This

is known as the controller tuning problem. There are three general approaches

we can use for tuning a controller:

- Use simple criteria like the one-quarter decay ratio (see Example 16.1),

minimum settling time, minimum largest error, etc. Such an approach is

simple and easily implementable on an actual process. Usually, it provides

multiple solutions (see Example 16.1). Additional specifications on the

closed-loop performance will then be needed to break the multiplicity and

select a single set of values for the adjusted parameters.

- Use time integral performance criteria like ISE, IAE or ITAE (see Example

16.2). This approach is rather cumbersome and relies heavily on the

mathematical model (transfer function) of the process.Applied experimentally

on an actual process, it is time consuming.

- Use semi-empirical rules which have been proven in practice.

In this section we will discuss the most popular of the empirical tuning

methods known as the Process Reaction Curve Method which was developed by

Cohen and Coon.

Consider the control system of Figure 16.11 which has been "opened" by

disconnecting the controller from the final control element. Introduce a

step change of magnitude A in the variable C which actuates the final

control element. In the case of a valve, C is the stem position. Record

the value of the output with respect to time. The curve y,(t) is called

Process Reaction Curve. Between ym and C we have the following transfer

function (see Figure 16.10)

YmWGpRC(s) = ~ = Gf(s)*Gp(d-Gm(s> (16.7)

C(s)

The last equation shows that the process reaction curve is affected not only

by the dynamics of the main process but also by the dynamics of the measuring

sensor and final control element.

Cohen and Coon observed that the response of most processing units to

an input change as the above had a sigmoidal shape (see Figure 16.12a) which

can be adequately approximated by the response of a first-order system with

dead time (Figure 16.12b, i.e.

-tdsYmW

GpRC(s) 2 ___ = K e,rs+l (16.8)

C(s)

which has three parameters: static gain K, dead time td and time constant

T . From the response of Figure 16.12a it is easy to approximate the values

of the three parameters. Thus,

K = Output (at steady state) BInput (at steady state) = x '

T = B/S where S is the slope of the sigmoidal response at

the point of inflexion

td = time elapsed until the system responded

Cohen and Coon used the approximate model of eqn. (16.8) and estimated

the values of the parameters K, td and T as indicated above. Then, they

derived theoretical expressions for the "best" controller settings using load

changes and various performance criteria like:

- one-quarter decay ratio,

- minimum offset,

- minimum integral square error (ISE), etc.

The results of their analysis are summarized below.

. . .

(3)

whose response has the general overdamped shape of Figure

16,12a(seealsoFipures 10.4, ll.la and 11.6). The

oscillatory underdamped behavior is produced mainly by the

presence of feedback controllers. Therefore, when we 'iopen"

the loop (Figure16.ll)and thus disconnect the controller,

the response takes the sigmoidal shape of an overdamped

system.

From eqns. (16.9), (16.1Oa) and (16.11a) which give the value

of the proportional gain Kc for the three controllers we

notice that:

- The gain of the PI controller is lower than that of the

P controller. This is due to the fact that the integral

control mode makes the system more sensitive (may even lead

to instability) and thus the gain value needs to be more

conservative.

- The stabilizing effect of the derivative control mode

allows the use of higher gains in the PID controller

(higher than the gain for P or PI controllers).

Example 16.4 - Tuning Feedback Controllers Through Reaction Curves- - - -

In this example we will examine how the dynamics of various typical

processes influence the tuning results recommended by Cohen and Coon.

(0 Processes with very shosrt time delay (dead time). When td is very

small (almost zero) the process reaction curve (Figure 16.12a) reminds

the response of a simple first-order system. The Cohen and Coon

settings dictate an extremely large value for the proportional gain

Kc (see eqns. (16.8), (16.9a) and (16.10a)). In real practice we

IIIIIIIIIIIIIIIi

will use the largest possible gain to reduce the offset if a pro-

portional controller is employed. If a PI controller is used, then

the value of gain will be determined by stability requirements.

(ii) Multicapacity processes. These constitute the large majority of real

processes. Consider two first.-order systems in series with

KG = PP CT

Pls+l) (T

p2s+l)

Let the measuring device and the control valve (final control element)

have first-order dynamics, i.e.

G Knl Kf=- andm -cms+l Gf = rfs+l

Then, the transfer function between the control actuating variable

C and the recorded measurement of the output y, is given by

(see eqn. (16.7)):

K K KG = G G G = f pmPRC f pm (y+l) (T

Pls+l)(-r s+l)(Tms+l>

P2

(16.8)

Equation (16.8) indicates that the process reaction curve has the

same dynamic characteristics as the response of a system composed

of four first-order systems in series, i.e. it is slgmoidal curve.

Figure 16.13 shows the process reaction curve for the following

values:

KP

= 1.0 , Km = 1.0 , Kf = 1.0 and

TPl = 5 7 rP2 =2 , Tf = 0.00 ) Tm = 10.0

Draw the tangent at the inflexion point and find;

- S = slope at the inflexion point = 0.05

- B = ultimate response = 1.0

- -c = effective time constant = B/s = l.O/O.OS = 20.

For

For

- td = dead time = 2.5

- K = gain = B/A = l.O/l.O = 1.0.

Therefore, the process reaction curve can be approximated by the

response of the following first-order with dead time system:

GPRC1.0 e-2*5s

2Os+l

The approximate response is also shown in Figure 16.13. We notice

that the approximation is satisfactory until the response has

reached the 40% of its final value.

Using the Cohen-Coon suggested settings we find:

proportional controller:

KC= 8.0

proportional-integral controller:

KC= 7.2 and TI = 5.85 and rD = 0.89

Figure 16.14 shows the closed-loop responses with the above settings for

set point (Figure 16.14a) and load changes (Figure 16.14b). We notice that

the Cohen and Coon settings produce underdamped behavior with rather good

decay ratio.

Example 16.5 - Controller Tuning for Poorly Known Processes

The methodology of controller tuning using process reaction curves is

particularly appealing if the dynamics of the main process or the measuring

sensor are poorly known, i.e. we do not know exactly the order of dynamics or

the values of the parameters. In such case the process reaction curve reveals

the effects of all the dynamic components, i.e. process, sensor and final con-

trol element and provides an experimental, approximate model for the overall

process.

c-- ._., A,...

Take as example the temperture control system for the reactor of Figure

16.10. It is quite a complex system and we may not know with satisfactory

precision all or a few of the following:

- the reaction kinetics,

- the heat of reaction,

- the mixing characteristics in the tank,

- the heat capacity of the reacting mixture,

- the overall heat transfer coefficient between steam and reacting mixture,

- the effective order of the thermocouple's dynamics,

- the gain and time constant of the thermocouple,

- the characteristics of the steam valve, etc.

The process reaction curve for this system provides us with an experimental

model of the overall process which we can use to tune the controller without

requiring detailed knowledge of the dynamics for the reactor, heating jacket,

thermocouple and control valve.

SLIMHARY AND CONCLUDING REWBKS

To design a feedback controller it means: (a) to select the type of the

controller (P, PI, PID) and (b) to choose the values of the adjusted parameters

for the selected controller. There are two main classes of dynamic performance

criteria we can use to evaluate alternative controller designs, The first

includes simple isolated dynamic characteristics of a system's response

(overshoot, decay ratio, rise time, settling time), while the second includes

criteria which are based on the time integrals of various functions of the

errors (e.g. of the square of the errors, or of the absolute errors, or of the

product of time with the absolute errors). The first class of criteria,

1- are simpler to use,

- can be used even with poorly known processes, but

- lead to multiple solutions and need additional specifications to produce a

single solution.

The time integral criteria,

- are based on the mathematical model (transfer function) of a process,

- necessitate the solution of an optimization problem, and

- lead to unique solutions, since they depend on the entire dynamic response

of a closed-loop system.

In order to select the type of a feedback controller we can use the following

general heuristic rules:

- Use proportional control only if small offsets can be tolerated, e.g. liquid

level, gas pressure control.

- Use PI controllers to eliminate undesirable offsets if the response of the

open-loop system has satisfactory speed, e.g. flow control.

- Use PID controllers to speed up the response of sluggish multicapacity

open-loop systems, while maintaining satisfactory robustness.

For controller tuning the most common methods employ,

- the one-quarter decay ratio rule, or

- time integral criteria (ISE, IAE, ITAE)., or

- the Cohen and Coon settings based on the process reaction curves.

The first and last can be used with poorly known processes, while the second

necessitates mathematical models for all components of a closed-loop.

In the next two chapters we will discuss a methodology for the design of

feedback controllers, which is quite different from everything we have seen so

far. It is known as the Frequency Response Analysis and permits us to look at

the same design problems from a new perspective.

I


1.

2.

3.

4.

5.

6.

7.

a.

9.

10.

11.

What are the main questions arising during the design of a feedback con-

troller? Discuss them on the basis of a physical example.

What is meant by controller tuning?

Discuss the two classes of dynamic performance criteria. Give physical

examples and demonstrate how different criteria lead to different con-

troller designs.

Can you design a controller which minimizes the rise and settling times,

simultaneously? Explain.

Can you design a controller which minimizes the overshoot and settling

time, simultaneously? Explain.

What are the relative advantages and disadvantages of the three time-

integral criteria, i.e. ISE, IAE and ITAE? How would you select the

most appropriate for a particular application?

Why do simple criteria like minimum overshoot, minimum settling time,

one-quarter decay ratio, etc. lead to multiple solutions? How do you

break the multiplicity and come up with a single solution?

Why do the time-integral criteria lead to unique solutions?

Discuss a set of simple heuristic rules you could use to select the

most appropriate type of feedback controller for a particular system.

Discuss the philosophy of the methodology that leads to the Cohen and

Coon settings for feedback controllers.

How do you understand the "opening" of the control loop shown in

Figure 16.11? Explain in practical terms how one tunes a feedback

controller for an existing process in a chemical plant.

aa. ,. A . 1 .. .

12. Why do most of the process reaction curves have an overdamped sigmoidal

shape? Can you develop a physically meaningful system which has a

reaction curve with an underdamped, oscillatory shape?

13. Are the Cohen and Coon settings reliable for all processes? Explain.

14. What is the value of the proportional gain Kc for a pure dead-time

system according to the Cohen and Coon settings? Is it reasonable?

Explain.

15. If the dynamics of the process or measuring sensor are not well known,

what tuning techniques would you use? Discuss your answer.

L4, _ . ..- A,..

uNc~~TROLCED

R E S P O N S E

4 a.7

desired level of operaLion

1i 6.

lyue 1G.d-.--__ -_ I

-

/I

.r-- pc --:I u I

a

IFig u-e 4 6 . 8 1

Cls)=A/S

2jnJs)- - G,

. .

?/

1.0

0.75

0.5

0.25

0

/44slope =s

id!b)

t

-7‘t

1 A

I . .

CHAPTER 17

THE FEQUENCY RESPONSE ANALYSIS OF LINEAR PROCESSES

In Chapters 17 and 18 we will study a new technique which is often used

in designing feedback controllers. It is quite different from everything we

have seen so far and it is called Frequency Response Analysis.

When a linear system is subjected to a sinusoidal input, its ultimate

response (after a long time) is also a sustained sinusoidal wave. This

characteristic, which will be proved in Section 17.2, constitutes the basis of

the Frequency Response Analysis.

With the frequency response analysis we are primarily interested to find

how the features of the output sinusoidal wave (amplitude, phase shift) change

with the frequency of the input sinusoid. In this chapter we will deal only

with basic premises of the frequency response analysis, while leaving its

usage for controller design in the next chapter.

17.1 THE RESPOME OF A FIRST-ORDER SYSTEM TO A SINUSOIDAL INPUT

Consider a simple first-order system with the transfer function

Y(s) = KPG(s) = ~Us) TPS+l

(17.1)

Let f(t) be a sinusoidal input with amplitude A and frequency w, i.e.

f(t) = A sinwt

Then

T(s) = Aws* + u*Substitute F(s) from eqn. (17.2) into eqn. (17.1) and take:

(17.2)

KY(s) = * l Aw

P s2 + cd2

Expand into partial fractions and find

5Y(s) = s + l,T +

c2 + c3

Ps + jw s - jw (17.3)

Compute the constants Cl, C2 and C3 and find the inverse Laplace trans-

forms of the three terms in eqn. (17.3):

K Aw-ry(t) = 2p2

-tl ppe -

KpAo-r K AP

rPw +IT202 -I- 1 cos t + r2U2p + 1 sinwt

P P

-t/-cAs ta the e '+O and the first term disappears. Thus, after a long

time the response of a first-order system to a sinusoidal input is given by:

Y,,W = - KPAwr P,2w2 + 1 coswt +

KPA,2,2 + 1 sinwt (17.4)

P P

Use the following trigonometric identify;

a cosb1 + a2sinb = a3sin(b+$)

where

a3 = m and

Then, eqn. (17.4) yields:

Y,,(t) = P

OF

sin(wt + $I)

+1P

where

I$ = tan-'(al/a2)

(17.5)

4 = tan-1 (-o.rp) (17.6)

From eqns. (17.5) and (17.6) we observe that;

- The ultimate response (also referred as steady state) of a first-order system

to a sinusoidal input is also a sinusoidal wave with the same frequency W.

A ,,.I ..I

. ..“.

9/3

- The ratio of the output amplitude to the input amplitude is called amplitude

ratio and is a function of the frequency, i.e.

AR = amplitude ratio = P

P

(17.7)

- The output wave lags behind (phase lag) the input wave by an angle ]$I,

which is also a function of the frequency w (see eqn. (17.6)). Figure

17.1 shows the ultimate response of the system and its relationship to the

input wave.

The above three observations do not hold only for first-order systems but are

expandable to any order linear system. Before we proceed with the

generalization of the above results let us make the following remarks related

to the algebra of complex numbers.

Remarks: (1) Consider a complex number W defined by

w = a + jb

where a = real part of W = Re(W) and b = imaginary part

of W = Im(W). Define the following terms:

- Modulus or absolute value or magnitude of W is repre-

sented by IW] and defined by

IWI = J[Re(W)12 + [Im(W)12

- Phase angle or argument of W is represented by $W

or arg(W) and defined by

+W = tan-1 Im(W)L- I~ =Re(W)

9

From Figure 17.2 it is clear that

a = lwlcose and b = /WlsinO

(17.8)

(17.9)

and

w = (WI ~0.~0 + j[Wlsine

Recall also that

cosa =ejfJ + e-je Jo _ e-je

2 and sin8 =2j

Then

w = /WIejO + .-jO ej 8 _ .--jO

2 + j/WI 2j

= lWleje

(2) Put s = jw in eqn. (17.1) and take

G(j > = jutp+l = juFp+l-jw-iP+l

P P -jOrp+l

(17.10)

or

K K WTG(jw) = P P P

T21112 + 1 - j T2U2 + 1P P

G(jw) is a complex number. Therefore according to Remark

(1) above,

Modulus of G(jw) = L=amplitude ratio (see eqn. (17.7))

P

and

Argument of Gm =tan--I(- wrp) = phase lag (see eqn. (17.6))

The last two relationships indicate that the amplitude ratio

and phase lag for the frequency response of a first-order

system are equal to the modulus and argument respectively

of its transfer function when s = jw.

This is an important result which we will generalize in the next section for

any linear system.

17.2 THE FREQUENCY RESPONSE CHARACTERISTICS OF A GENERAL LINEAR SYSTEM

Consider a general linear system with the transfer function

G(s) = (17.11)

where Q(s) and P(s) are polynomials of orders m and n respectively,

with m<n. We will prove that:

- The ultimate response of this system to a sinusoidal input is also a

sinusoidal wave.

- The ratio of the output amplitude to the input amplitude is a function of

the frequency w and it is given by the modulus of G(s) if we put

s = ju, i.e.

AR = modulus of G(jw)

- The output wave is shifted with respect to the input wave by an angle 4

which is a function of the frequency w given by

$ = argument of G(jw)

PROOF

For a sinusoidal input f(t) = Asinwt we have f(s) = Am/s2 + w2) and

eqn. (17.11) yields:

Y(s) = G(s) l 2Aw

S + cd2

Expand the last equation into partial fractions:

y(s) = G(s) Aw as2 + w2=s + jw + b + 5 + 52 + . . . + 5-l

s - j w s-p1 s-p2 s - 'n

The terms

c1 c2 %. . .s-p ' s-p ' ,

1 2 s-p n

give rise to exponential terms

eplt p2t pnt, e . . ., , e

If the poles pl, p2,*=*,pn have negative real parts, all the above terms

decay to zero as t-tco (see Section 9.4). Therefore, the ultimate response is

given by

Yss(s> = G(s) Aw = a + s -b jws2 + cd2 s + ju (17.12)

Compute constants a and b as discussed in Section 8.2 and find

a = AG(-jw)-2j and b=h!t$+!d

Therefore, eqn. (17.12) gives

AG(-jw) 1Y&4 = - zj s + jw

or

AG(-jw)y,,(t) = - zj

.-jut + AG(jw) ejwt2j

(17.13)

Use eqn. (17.10) to express the complex numbers G(-jw) and G(jw) in polar

form, i.e.

G(-jw) = IG(-jw) Ie-j' = IG(jw) [e-j'

and

G(jd = IG(jw) Iej'

where 4 = argument of G(jw). Substitute the values of G(-jw) and G(jw) in

eqn. (17.13)

yss

ct) = - &k%ikAd e-j(at+4) I AIG(jw) 1 ej(wt+$)2j 2j

= *IG(jw) 1ej Cut+@) _ e-j (cot+@)

2j

or

Y,,(t) = AIG(jw)l sin(wt+$)

The last equation proves what we set out to prove, i.e.;

- the ultimate response as t+= is sinusoidal with frequency w,

- the amplitude ratio is

AR = AIG(jo)l/A = IG<jw)/

- the output sinusoidal wave has been shifted by the angle

+ = argument of G(jw)

Example 17.1 - The Frequency Response of a Pure Capacitive Process

The transfer function is

G(s) = !kS

Put s = jw and take

KG(jw) = ' =jw

Consequently, for the ultimate response;

- the amplitude ratio is

AR = IG(jw)l = Kp/" (17.15)

(17.14a)

(17.14b)

- the phase shift is

4 = tan-l(-02) = -90” (17.16)

i.e. the ultimate sinuosidal response of the system lags behind the input wave

by 90".

Example 17.2 - The Frequency Response of N Non-Interacting Capacities in Series

The transfer function is (see eqn. (11.21), Section 11.3)

K K

G(s) = G1(s)G2(s) ,*-,GN(s) = p1 . p2 . . . KpNT s+l

rp2s+l ' ' T s+l

p1 PN

Put s = jw and take

G(jw) = Gl(jw)G2(ju) ,-'-,GN(j~)

But, according to eqn. (17.10)

(17.17)

Gl(jw) = IGl(ju) lej+,

, G2(ja) = IG2(ja) lej+,

,***,GN(ju) = bN(j") tej+,

where +1,42,***,4N are the arguments of Gl(jw), G2(jW),***,GN(jW). Then,

eqn. (17.17) becomes

G(jw) = 1 IGl(jW) llG2(ju)l - ,-*- ,- IGN(ju) II l ej(41+$2+,***,+4N)

Consequently, the response has the following characteristics;

- amplitude ratio,

AR = IG( jd I = IGl(ju) I*IG2(jw) 1 9--*9 IGN(jd

or

K -K *,**-,-Kp1 p2 PN

AR = ~--qyK~~,"~, /L--q?1 p2

- phase shift,

or

4 = tan-'(-wrpl) + tan-'(-UTP2

)+,**a,+ tan-'(-wrPN

)

Since +<O the response lags behind the input.

Example 17.3 - The Frequency Response of a Second-Order System

For a second-order system the transfer function is

(17.18a)

(17.19)

(17.19a)

KG(s) =

r2s2+L+l

.

A/!‘”

Put s = jo and take

K K 2 2G(jw) = P = P . (-T w +l)-j25rw

(-T2u2+1)+j2<Tu (-T2u2+1)+j2cTu (-.c2u2+1) -j 2y-rw

or

G(jw) =Kp(l--rLuL) Kp*2T-rlri

(1-*2fA2)2+(257w) 2 -9(1-r202)2+(2r7U) 2

Therefore, the ultimate response has the following characteristics;

- amplitude ratio given by

AR = IG(jw) 1 = P

+ (25TW)2

- phase shift is

+ = argument of G(jo) = tan-I(- 25?W

1 - 92w2)

which is phase lag since 4~0.

Example 17.4 - The Frequency Response of a Pure Dead Time Process


--T sG(s) = e d

Put s = jw and take

G(jw) = e-j Tdw

Clearly;

- amplitude ratio = IG(jo)l = 1

- phase shift = argument of G(jo) = -dew

i.e. a phase lag, since 4~0.

(17.20)

(17.21)

(17.22)

(17.23)

Example 17.5 - The Frequency Response of Feedback Controllers

Let us now shift our attention to the various types of feedback controllers.

(i> Proportional Controller


Gc(s) = KC

Therefore,

AR = Kc and $=O.

(ii) Proportional-Integral Controller


GcW = Kc(l +$-)I

Therefore

AR = IGc(jw)l = Kcm(UT >I

$I = argGc(jw) = tan-l(-l/wrI) < 0

(iii) Proportional-Derivative Controller


GcW = K,(l + -rDs)

Therefore,

(17.24)

(17.25)

AR = IGc(jw)/ = Kc x + rkw2 (17.26)

$ = argGc(jw) = tan-1 (~Dw) > 0 (17.27)

The positive phase shift is called phase lead and is the consequency

of the derivative control mode and another manifestation of its

anticipatory control nature.

(iv> Proportional-Integral-Derivative Controller


GcW = Kc(1 +-&I

+ -rDS)

and it is easy to show that

AR = lGc(jw) 1 = J D(T w - A)2 + 1I

4 = tan-l(rDw - +-)I

(17.28)

(17.29)

Notice that;

- AR is always larger or equal to 1 and

- depending on the values of 'rD and TI and the frequency w

we may have $>O (phase lead) or 4~0 (phase lag).

17.3 BODE DIAGRAMS

The Bode diagrams (in honor of H. W. Bode) constitute a convenient way to

represent the frequency response of a system. As we can see from eqns. (17.14a)

and (17.14b), the amplitude ratio and the phase shift of the response of a

system are functions of the frequency w. The Bode diagrams consist of a pair

of plots showing;

- how the logarithm of the amplitude ratio varies with frequency and

- how the phase shift varies with frequency.

In order to cover large range of frequencies we use a logarithmic scale for

the frequencies.

Let us now examine the Bode diagrams of some simple dynamic systems that

we have encountered in the previous chapters.

A. First-Order System

For a first-order system we have seen that

Kamplitude ratio = m = P

I-1 + r202P

(17.7)

phase lag = 0 = tan-'(-rw) (17.6)

For simplification, let Kp = 1. Then, from eqn. (17.7) we find that,

1ogAR = - + log(1 + T;w2) (17.30)

For convenience, since 'IP

is constant, regard -rpw as the independent

variable instead of w. The plot of log(AR) vs.- log(rpw) is shown in Figure

17.3a (solid line) and can be constructed from eqn. (17.30) for various values

of the frequency w. Instead of the very elaborate numerical work needed to

plot this graph, we can give an approximate sketch by considering its

asymptotic behavior as w-t0 (low-frequency asymptote) and as o- (high-

frequency asymptote). Thus, we have:

- As w-t0 then -rp~+-O and from eqn. (17.30) logARM or AR-tl.

This is the low-frequency asymptote shown by a dotted line in Figure 17.3a.

It is a horizontal line passing through the point AR=l.

-As w-, then -rpw- and from eqn. (17.30) 1ogAR = -log(-rpw)

This is the high-frequency asymptote shown also by a dotted line in

Figure 17.3a. It is a line with a slope of -1 passing through the point

AR=1 for -rpw=l. The frequency w = l/-r is known as the cornerP

frequency. At the corner frequency, as it can be seen from Figure 17.3a,

the deviation of the true value of AR from the asymptotes is maximum.

The plot of phase shift=. ( rpu) is shown in Figure 17.3b. It can be

constructed from eqn. (17.6). We can easily verify the following character-

istics of this plot:

- As w-+0 then ++O

-As w- then @tan-1(-m) = -90"

- At w = l/rp(corner frequency), $ = tan-l(q) = -450

Remark: If Kp # 1 then as it can be seen from eqn. (17.7) the low fre-

quency asymptote shifts vertically by the value logKp. Equation

(17.6) shows that Kp has no effect on the phase shift.

B. Pure Capacitive Process

For such processes we know that (see Example 17.1)

AR = Kp/w and 4 = -90"

The Bode plots are easily constructed and shown in Figure 17.4.

C. Second-Order System

In Example 17.3 we found that

KAR=...-.-- P - and I$ = tan-5

-2??w>

J(l-T2U2)2+(25Tw)21 - T202

The two plots are shown in Figure 17.5 for various values of < when Kp =

The two asymptotes for the plot AR vs. TW are determined as follows:-

- As w-fo, then logAR+O or AI+1 (low-frequency asymptote)

-As w-, then logAR+-2log(rw). This is the high-frequency asymptote. It

is a straight line with a slope of -2 passing through the point

AR =l,rw =l

From Figure 17.5 we notice that for underdamped systems, i.e. 5~1, the

amplitude ratio can exceed significantly the value of K .P

When Kp # 1

the low frequency asymptote shifts vertically by the value 1ogK .P

D. Pure Dead Time System

From Example 17.4 we have that,

AR = 1 and $J = -TdW

1.

The Bode plots for this system are easily constructed and shown in Figure 17.6.

E. Systems in Series

Consider N systems in series with individual tranfer functions

Gl(s),G2(s),...,GN(s)

The overall transfer function is,

G(s) = Gl(s)G2(s),-,GN(s)

put s = jw and take

G(jw) = Gl(jw>G2(jw>,**o,GN(jw)

or

G(ju) = IGl(ju)

or finally

jw> lej@,

jG(jw) 1 ej' = IG,(jw>/ IG2(jw)l*l'**,*/GN(jo)lej ($1++2+,** l ,++,I

The last equation yields

IG(jw) 1 = IGl(ju) I* IG2(ju) 1 l ,***,-IGN(j~) I (17.31)

and

@ = $1 + $2 +,“‘,+ $N

From (17.29) we have

AR = (AR)l*(AR)2*,~*~,*(AR)N

or

(17.32)

log(AR) = ~o~(AR)~ + l~g(AR)~+,---,+log(~~)~ (17.33)

where

(AR>,,(AR>2,***,(~>N

are the amplitude ratios for the individual systems in series. Equations

(17.31) and (17.32) are very important and indicate the following rules for

the construction of the Bode diagrams:

If the transfer function of a system can be factored into the product of

N transfer functions of simpler systems, then

(1) The logarithm of the overall amplitude ratio is equal to the sum of

the logarithms of the amplitude ratios of the individual systems,

(2) The overall phase shift is equal to the sum of the phase shifts of the

individual systems,

(3) The presence of a constant in the overall transfer function will move the

entire AR curve vertically by a constant amount. It has no effect on the

phase shift.

Example 17.6 - Bode Diagrams for Two Systems in Series

Consider the following two systems in series;

GIW = & and G2W = &

The overall transfer function is

6G(s) = & l -

5s+l

Then,

or

1ogAR = log6 + log(AR)l + log(AR)2 (17.34)

where (AR)l and (AR)2 are the amplitude ratios of the individual systems,

when their gains are 1. Figure 17.7a shows the amplitude ratios of the two

systems as functions of w. The addition of these two curves plus the factor

log6 will yield the amplitude ratio of the overall system versus the frequency

w. The overall curve is also shown in Figure 17.7a. From this curve we

A .“. -

notice three distinct frequency regions. JCJe slope of the asymptote in each

region is the algebraic sum of the slopes of the asymptotes for the two systems

in the corresponding region. Thus, we have:

(2) Region 1. From w =O to w = l/5

Slope of the overall asymptote = 0 + 0, i.e. horizontal, going through

the point AR = 6.

(ii) Region 2. From w = l/5 to w = l/2

Slope of the overall asymptote = 0 +(-1) = -1 going through the

point AR = 6 o = l/5

(iii) Region 3. For w > l/2

Slope of the overall asymptote = (-1) + (-1) = 02

Figure 17.7b shows the phase shift for the overall system versus the frequency,

as the algebraic sum of the phase shifts of the two individual systems, i.e.

9 = $1 + $2 = tan-l(-2w) + tan-'(-5w)

It is clear that;

- when w-to, @l-to, 42%) and $+O

- when w-, y-90°, ($J2+-900 and (p-t-180'

F. Feedback Controllers

The Bode diagrams for various types of feedback controllers can be con-

structed easily using the results of Example 17.5.

(i> Proportional Controller

The Bode plots are trivial. The AR and 0 stay constant at the

values KC

and 0" for all frequencies.

(ii) Proportiaonal-Integral Controller

From eqns. (17.24) and (17.25) we take;

and 0 = tan-l(-l/wrI)

As ~0 1 ,>jl , then log@) = -log(w~I)(WTI) L RC

and the low frequency asymptote is a straight 1

- High frequency asymptote:

ine with s

Therefore:

- Low frequency asymptote:

1A s w+-= - -2 +O and ARi.e. - =

(UT >log& = 0 , K

I C C

lope -1.

1 .

The high frequency asumptote is a horizontal line at the value

AR/Kc = 1.

The (AR/Kc) vs. (UT,) is shown in Figure 17.8a. For the phase-

shift we have the following:

as w-to p-90" and

as w+-= 4-f 0"

The 0 ys-. (wrI) is shown in Figure 17.8b.

(iii) Proportional-Derivative Controller

The AR and $I are given by eqns. (17.26) and (17.27). The Bode

plots can be easily constructed and are shown in Figures 17.9a and

17.9b.

(iv) Proportional-Integral-Derivative Controller

The AR and 9 are given by eqns. (17.28) and (17.29), respectively.

The Bode plots are easily constructed and they are shown in Figures

17.10a and 17.10b.

Example 17.7 - Bode Plots for an Open-Loop System

Consider the feedback control system shown in Figure 17.11. The open-

loop transfer function is (see Remark 2 in Section 15.2)

GOL = Gc*Gf*Gp*Gm

or

GOL = 100Kc*(l + -+)* lI O.ls+l

with ? = 0.25 and Kc= 4. We notice

product of six transfer functions, i.e.

l (2s+&+1) l

1 -0.2s0.5s-U' e

that the GOL can be factored into a

1 1 1 -0.2s- -2s+l ' s+l ' 0.5s+l ' cl+&), l

I O.ls+1 ' e

with the following corner frequencies (in the same order)

9 = l/2 = 0.5, w2 = l/l = 1, w3=1/0.5=2, w4=l/0.25=4, w5=1/0.1=10.

The Bode plots of the individual transfer functions are easily constructed and

they are shown in Figures 17.12a and 17.12b. The Bode plots for the overall

system can be constructed following the rules discussed in paragraph E of the

present section.

- First; we identify the following six regions on the frequency scale:

0 2 w < ol, w2 ,( w c w2, w2 5 w < w3, w3 d o < w4, w4 I w < w5 and

w5 5 w < a.

- Second; for the AR vs. w diagram, the slope of the overall asymptote-

is equal to the algebraic sum of the slopes of the asymptotes of the

individual transfer functions (Table 17.1). The overall asymptote is shown

in Figure 17.12a.

- Third; the overall phase shift is equal to the algebraic sum of the phase

shifts for each individual transfer function and is shown in Figure 17.12b.

III

IIII

-..

+Q Y

17.4 NYQUIST PLOTS

A Nyquist plot is an alternate way to represent the frequency response

characteristics of a dynamic system. It uses the Im[G(j,)] as ordinate and

Re[G(jw)] as abscissa. Figure 17.13 shows the form of a Nyquist plot.

A specific value of the frequency w defines a point on this plot. Thus,

at the point 1 (Figure 17.13) the frequency has a value ml and we observe

the following:

- The distance of the point 1 from the origin (0,O) is the amplitude ratio

at the frequency wl, i.e.

distance = hRe[G(ju1)]12 + [Im[G(ju1) II2 = IGkiw)( = AR

- The angle $ with the real axis is the phase shift at the frequency wl,

i.e.

$I = tan-l[Im[G(jwl)]/Re[G(jwl)]] = argument G(jw) = phase shift

Thus, as the frequency varies from 0 to ~0 we trace the whole length of the

Nyquist plot and we find the corresponding values for the amplitude ratio

and phase shift. The shape and location of a Nyquist plot are character-

istic for the particular system.

The Nyquist plot contains the same information as the pair of Bode plots

for the same system. Therefore, its construction is rather easily given the

corresponding Bode plots. Let us now construct the Nyquist plots of some

typical systems using their Bode plots developed in the previous section.

A. First-Order System

The corresponding Bode plots are given in Figure 17.3. We observe that:

(0 When w=O, then AR = 1 and $=O. Therefore, the beginning of the

Nyquist plot is on the real axis where $=O and at a distance from

the origin (0,O) equal to 1 (see point A in Figure 17.14a).

1 .93 0

I

(ii) When u-)co then AR+0 and ++--90". Therefore, the end of the Nyquist

plot is at the origin where the distance from it is zero (point C in

Figure 17.14a).

(iii) Since for every intermediate frequency

0 < AR ~1 and ' -90" < c$ < 0

the Nyquist plot will be inside a unit circle and will never leave

the first quadrant. Its complete shape and location are shown in

Figure 17.14a.

B. Second-Order System

The corresponding Bode plots are shown in Figure 17.5. Notice that:

(i> When w=O then AR = 1 and Cp=O. Thus, the beginning of the Nyquist

plot is on the real axis at a distance equal to 1 from the origin.

(ii) When wxo then AR-+0 and (p-+-180', i.e. the Nyquist plot will end

at the origin and will approach it from the second quadrant.

(iii) When ~1 then AR51 and the Nyquist plot stays within a unit

circle. When 5~1 then AR becomes larger than 1 for a range of

frequencies. Thus, the Nyquist plot goes outside the unit circle

for a certain range of frequencies . Figure 17.14b shows the Nyquist

plot for a second-order system.

C. Third-Order System


G(s) = 1$s+l) (3s+2) $s+U +9,3 real and positive

It is easy to show that:

- When w=O, then AR = 1 and $=O.

- When w-tco, then AR = 0 and +-+-270'.

Therefore, the Nyquist plot starts from the real axis at a distance 1 from the

origin and endsatthe origin, going throughthethird quadrant (Figure 17.14~).

D. Pure Dead Time

From the corresponding Bode plots (Figure 17.6) we notice that:

AR = 1 for every frequency

and

$= d-T w

Therefore, the Nyquist plot for this system is a circle of radius 1 and

encircles the origina an infinite number of times (Figure 17.14d).

E. Pure Capacitive Process

From the corresponding Bode plots (Figure 17.4) we notice that:

.- When w-to, AR*, while

- When w*, ARM).

The phase lag remains constant at -90" for every frequency. Therefore, the

Nyquist plot coincides with the negative part of the imaginary axis (Figure

17.14e).

F. Feedback Controllers

In a similar manner as above we can construct the Nyquist plots for P,

PI, PD and PID controllers. They are shown in Figures 17.15a, 17.15b, 17.15~

and 17.15d, respectively.


The ultimate response (also called sometimes steady state) of a linear

system to a sinusoidal input has the following characteristics:

- Is a sinusoidal wave with the same frequency as the input.

- The ratio of the output amplitude to the input amplitude is a function of

the input frequency w and equal to the modulus IG(jw) I.

- The phase of the sinusoidal response is shifted by an angle 4~ with respect

to the input. We have a phase lag when @O and phase lead when (p>O. The

phase shift 9 is equal to the argument of G(jw) .

Frequency response analysis is the study of the ultimate response of a

linear system subjected to a sustained sinusoidal input. Since this response

varies with the frequency of the input wave, Bode diagrams and Nyquist plots

are used to represent the frequency response characteristics of a system. The

Bode diagrams consist of a pair of plots showing how the amplitude ratio and

phase shift vary with the frequency. The Nyquist plot is a polar plot with

Re[G(jw)] and Im[g(jw)] as coordinates. Both contain the same information

and can be constructed from each other.

Frequency response analysis and the Bode diagrams or Nyquist plots will

be used extensively to design effective controllers and identify the dynamics

(transfer function) of unknown systems.

8 ,4.3


1 . What are the characteristics of the ultimate response of a linear system

with a transfer function G(s) to a sustained sinusoidal input?

2 . Define the frequency response analysis.

3 . What means could you use to represent the results of the frequency

response analysis for a dynamic system?

4. Define the Bode diagrams and Nyquist plots. Do you have any personal

preference on one of them over the other? If yes, why?

5 . The system with a transfer function

G(s) = l-s + 1

is usually known as first-order lead element. Construct its Bode-

diagram and try to rationalize the word lead in its name. [Hint:

Contrast it to the familiar first-order lag element].

6 . Construct the Bode diagram and Nyquist plot of a first-order system with

dead time, having a transfer function

-T sKp*e

d

G(s) =ups+1

7 . Does the Nyquist plot have a meaning for the frequencies -0 5 w I O?

Show that the Nyquist plot for this range of frequencies is the mirror

image of the familiar Nyquist plot for the frequencies o_<ws+,=.

8 . Construct qualitatively the Nyquist plot of a sixth-order system with a

transfer function

KG(s) = P

(Tls+l) (r2s+l) (T3S+l.) (TqS+l) (TgS+l) (T6S+l)

where '1?29 T39 r49 T59 =6 are all real and positive.

9 . For a system like the one in item 8 above, we claim that the slope of

the overall asymptotes in the 1ogAR vs.- logw plot of its Bode diagram can

be given from the algebraic sum of the slopes of the asymptotes for the

individual subsystems,

1 1 1___ . . . ___T s+l1 ' 2T s+l ' ' rgs+l +

Explain why. Alos, construct qualitatively the Bode diagram indicating

the slopes of the asymptotes for the overall system. Draw a qualitative

Nyquist plot for the above system.

10. The Bode plots for a PI controller show that as w-t0 the AR-. This

is not physically realizable. Therefore, the transfer function

GC= Kc(l +$-)

I

11.

12.

represents the behavior of an ideal PI controller. How should we

modify the above transfer function so that it represents the behavior

of an actual PI controller? The transfer function of the actual PI

controller must be such that as W-+0 then AR-+finite value. [Note:

Consult Ref. and 1.

The Bode plots for a PD controller show that as w- the AR-. This

is, again, physically unrealizable. How should we modify the transfer

function of a PD controller so that as w- the AR+finite value?

[Note: Consult Ref. and to develop the transfer function of an

actual PD controller.

Based on the responses in Items 10 and 11 above, develop the transfer

function of an actual PID controller, which has the following

characteristics:

as w-4 , AR -+ c1 = finite

and as w- , AR -f 8 = finite.

,

q35

- - - _

- - -_

AR I

1 /Yjwe 178)

. .

I ’

A

I I

I~ i

-3 ,’

,I I

c

1 I

. , -41

1 III, ’ I

.-- - _ -’ 1

-, -I,- ’I

@)

-1 Iti

--_ C>c __- - ^ -I49-3

1 -^ --__ *_ -_

06

CHAPTER 18

DESIGN OF FEEDBACK CONTROL SYSTEMS USING FREQUENCY RESPONSE TECHNIQUES

In the previous chapter we studied the frequency response analysis and

its application to various dynamic systems. The question that may have been

raised in the mind of the reader, i.e. what do we do with it, will find its

answer in this chapter.

The frequency response analysis is a useful tool for designing feedback

controllers. It helps the designer,

- first, to study the stability characteristics of a closed-loop using the

Bode or Nyquist diagrams of the open-loop transfer function, and

- second, to select the most appropriate values for the adjusted parameters

of a controller.

18.1 THE BODE STABILITY CRITERION

Consider the closed-loop system shown in Figure 18.1. The open-loop

transfer function is given by (see Section 15.2, Remark 2)

y,(s)K ,e-o.ls

GOL =Y,,(s)

= oc5s + 1 (18.1)

The Bode diagram for GOL(s) can be constructed easily (see Example 17.7) and

is shown in Figure 18.2. We notice that,

when w = 17.0 rad/min $ = -180"

The frequency where the phase lag is equal to 180' is called crossover fre-

quency and it is denoted by wco. At this frequency the amplitude ratio is

found from the Bode diagram to be

AR =”

= 0.12 (18.2)

Consequently, if Kc = l/O.12 = 8.56 the amplitude ratio becomes equal to 1.

Now, let us consider the "opened" loop shown in Figure 18.3a with Kc =

8.56. Here, the measurement signal has been disconnected from the comparator

of the feedback controller. If the set point changes in a sinusoidal manner

with frequency w = 17.0 rad/min, and an amplitude equal to 1, i.e.

Y$) = sin(l7.0t)

then the ultimate open-loop response, y,(t), is given by

y,(t) = sin(l-/.Ot - 180") = -sin(l7.0t)

At some instant of time the set point yd is set to zero, while at the

same time we "close" the loop (Figure 18.3b). Under these conditions the

comparator inverts the sign of the y,, which now plays the same role played

by the set point in the "open" loop. Notice that the error E remains the

same. Theoretically, the response of the system will continue to oscillate

with constant amplitude, since AR = 1, despite the fact that both the load

and the set point do not change.

Let us examine the following cases:

a. If Kc > 8.56 then AR>1 when = -180". Therefore, the sustained

oscillation of the "closed" loop of Figure 18.3b will exhibit an ever

increasing amplitude leading to an unstable system.

b. On the contrary, if Kc < 8.56 then AR<1 when 4 = -180". Con-

sequently, the oscillating response of the "closed'! loop of Figure

18.3b will exhibit a continuously decreasing amplitude leading to an

eventual dying out of the oscillation.

The conclusion drawn from the above observations is the following:

"A feedback control system is unstable if the AR of the.-corresponding open-loop transfer function is larger than1 at the crossover frequency."

This is known as the Bode Stability Criterion.

Example

Using the Bode Criterion

(a) First-Order Open-Loop Response. Consider a control system with the

following dynamic components

KProcess: G = -$&

P P

Measuring sensor: Gm = Km

Controller: GC

= Kc, i.e. proportional

Valve (final control element): Gf = Kf

The open-loop transfer function is:

K K K KG cfpm = KOL = GGGG =c f p m rps+l

rPs+l

We know (see Section 17.1) that the phase lag for a first-order system

is between 0' and 90". Therefore, according to the Bode stability

criterion the above system is always stable since there is no cross-

over frequency.

(b) First-Order with Dead-Time Open-Loop Response. Consider again the

dynamic components of the loop in the case (a) above with the following

change.

G = K .-Os5’m m

Then, the open-loop transfer function becomes,

GK .-0.5s

OL = 'rps+l

The phase lag for this system is

$I = tan -1GTpW) + (-0.5w)

The last equation shows that 0 I $ < m. Consequently, there is a cross-

over frequency wco' where $ = -180" and according to the Bode

criterion the system may become unstable for a large Kc which leads

to AR>1 at this frequency. This example demonstrates a very important

characteristic for the stability of chemical processes:

"Dead-time is a principal source of destabilizing effectsin chemical process control systems."

Since most of the chemical processes exhibit an open-loop response

which can be approximated by a first-order system with dead-time, it

is clear that the possibility for closed-loop instability will be,

almost always, present. Therefore, the tuning of the feedback con-

troller becomes a crucial task.

(c) Higher Order Open-Loop Responses: Consider again the control system

for case (a) above with the following changes:

KmGm = ~-rms+l

The open-loop transfer function becomes,

G KOL = (rps+l)(rms+l)

and the phase lag becomes -180' when W=m . Therefore, according to the

Bode criterion such system is always stable since as w the AR-4.

If we consider

KmGm = ___r-,s+l and Kf

Gf = rfs+l

then the open-loop transfer function becomes

GOL =K

(rpS+l)(rms+l)(rfS+1)

and the phase lag is 0" 2 C$ 5 -270". Therefore, there is a crossover

frequency wCo where 4 = -180" and the system may become unstable for

large enough Kc. This leads to the second important observation about

the stability of chemical process control systems:

"In the absence of dead-time a closed-loop system may becomeunstable if its open-loop transfer function is of third-order or higher."

Remarks: (1) All systems in Example 18.1 have an important common feature;

the AR and $J of the corresponding open-loop transfer

functions decrease continuously as w increases. This

is also true for the large majority of chemical processing

systems. For such systems the Bode stability criterion

leads to rigorous conclusions. Thus it constitutes a very

useful tool for the stability analysis of most control sys-

tems of interest to a chemical engineer.

(2) It is possible though that the AR or 4 of an open-loop

transfer function may not be decreasing continuously with

w. In Figure 18.4 we see the Bode plots of an open-loop

transfer function where AR and 4 increase in a certain

range of frequencies. For such systems the Bode criterion

may lead to erroneous conclusions and we need the more general

Nyquist criterion which will be discussed in Section 18. .

Fortunately, systems with AR or 4 like those of Figure

18.4 are very few and consequently the Bode criterion will

be applicable in most of the cases.

(3) In order to use the Bode criterion, we need the Bode plots

for the open-loop transfer function of the controlled system.

These can be constructed in two ways; (a) numerically, if

the transfer functions of the process, measuring device,

controller and final control element are known and (b)

experimentally, if all or some of the transfer functions are

unknown. In the second case the system is disturbed with a

sinusoidal input at various frequencies, and the amplitude

and phase lag of the open-loop response are recorded. From

these data we can construct the Bode plots.

18.2 GAIN AND PHASE MARGINS

The Bode stability criterion indicates how we can establish a rational

method for tuning the feedback controllers in order to avoid unstable

behavior by the closed-loop response of a process.

Consider the Bode plots for the open-loop transfer function of a feed-

back system (Figure 18.5). The two important features of these plots are;

- the crossover frequency%o ' where Q! = -180" and

- the point where AR = 1.

Let M be the amplitude ratio at the corssover frequency (see Figure 18.5).

According to the Bode criterion

- if 14~1 the closed-loop system is stable and

- if M>l it is unstable.

Define,

Gain Margin = l/M (18.3)

Then, for a stable system M<l and

Gain Margin > 1

We can make the following observations on the practical significance of the

gain margin:

- It constitutes a measure of how far the system is from the brink of

instability.

- The higher the gain margin is above the value of 1, the more robust the

closed-loop behavior will be and thus the safer the operation of the

controlled process. In other words the higher the gain margin the higher

the safety factor we use for controller tuning.

- Typically, a control designer synthesizes a feedback system with gain

margin larger than 1.7. This means that the AR can increase 1.7 times

above the design value before the system becomes unstable.

Let us now study the above observations using an example.

Example 18.2 - Gain Margin and the Tuning of a Controller

Consider the closed-loop system of Figure 18.1. The crossover frequency

is %o = 17 rad/min, and the amplitude ratio at this frequency is (see eqn.

(18.2))

AR = C

&T&z

= 0.12 KC

2.

Therefore the gain margin is equal to

Gain Margin = o l12K.C

If we require a gain margin of 1.7 we find

10.12 K

= 1.7C

Let us assume now that the dead-time for the process has not been estimated

accurately and that its "true" value is 0.15 instead of 0.1. Then, the open-

loop transfer function is given by

K e-0.15s

GC

OL = 0.5s+l(18.4)

and not by the assumed eqn. (18.1). For the open-loop transfer function of

eqn. (18.4) we find that the crossover frequency is %o = 11.6 rad/min. At

this frequency the amplitude ratio is

AR = d& = J,,.~.- = o.83P

and the system is still stable despite the error by 50% we made in estimating

the dead-time of the process. Notice though that the amplitude ratio has

moved closer to the value 1, i.e. the system has noved closer to instability.

The present example demonstrates the practical significant of the gain

margin in tuning feedback controllers. Two points are worth emphasizing:

- Since process parameters like dead-times, static gains, time constant are

almost never known exactly, a gain margin larger than one, e.g. 1.7, is a

safety factor for stable operation.

- If the various parameters are known very well, only small safety factors

are needed, i.e. gain margins in the range 1.4-1.7. For systems with

parameters poorly known the safety factor must increase and the recommended

values for gain margins are in the range 1.7-3.0.

Besides the gain margin there is another safety factor which is used for

the design of a feedback control system; the phase margin. Consider again

Figure 18.5. Let $(,) be the phase lag at the frequency for which AR = 1.

The phase margin is defined as follows:

Phase Margin = 180' - 91)

i.e. it is the additional phase lag needed to destablize the system. It is

,._ .*. .

clear therefore that the higher the phase margin the larger the safety factor

used for designing a controller. Typical phase margins used by the designers

are larger than 30".

Example 18.3 - Phase Margin and the Tuning of a Controller

Consider again the closed-loop system of Figure 18.1. We know that

AR = C

6-k

and 4 = tan-'(-0.5o) + (-0.1~)2. w

Let us tune the controller using a phase margin equal to 30". Then, we have

Kc = m and 30"=180" -tan-'(-0.5~) + (-0.1~)

From the second euation we find w = 12.5 rad/min. Then, the first equations

gives Kc = 6.33.

(i> Assume now that the dead-time has been estimated wrong and that its

"true" value is 0.15. Then, the phase lag at the frequency h) =

12.5, where AR = 1, is given by,

$ = tan-'(-0.5w) + (-0.1501) = tan-1(-0.5*12.5)+(-0.15~12.5)=188"

We notice that the system has become unstable, i.e. a phase margin

of 30' was not enough to provide a safety factor for a 50% error in

dead-time.

(ii) The reader can easily show that a phase margin of 45" is enough to

tune the controller in case (i) above and provide the necessary

safety factor for absorbing a 50% error in the dead-time. The value

of the proportional gain Kc for a 45" phase margin is found to be

Kc = 5.05. Assume that there is an error in the time constant which

has a "true" value of 0.25 instead of the assumed 0.5. Then, the

crossover frequency is found from the equation

180" = tan-l(-0.25w) + (-0.1~)

and it is equal to %o = 17.9. At this frequency

AR = 5.05 = 1.1

J/(o.25*17.9)2+1

and we notice again that the system is unstable. Therefore,

although a phase margin of 45" was satisfactory for tuning the

controller in the presence of a 50% error in dead-time, it is

not enough for absorbing an error of up to 50% in the time constant.

Larger phase margin is needed.

18.3 THE ZIEGLER-NICHOLS TUNING TECHNIQUE

In Section 16.4 we discuss a tuning method based on the process reaction

curve. The method is primarily experimental and uses real process data from

the system's response. In this section we will discuss an alternate method

developed by Ziegler and Nichols which is based on the frequency response

analysis, discussed in earlier sections.

Unlike the process reaction curve method which uses data from the open-

loop response of a system, the Ziegler-Nichols tuning technique is a closed-

loop procedure. It goes through the following steps:

(i> Bring the system to the desired operational level (design condition).

(ii) Using proportional control only and with the feedback loop closed,

introduce sinusoidal set point changes with low amplitude and varying

frequencies until the system oscillates continuously. The frequency

of continuous oscillation is the crossover frequency, wCo. Let M

be the AR for the system's response at the crossover frequency.

(iii) Compute the following two quantities:

Ultimate Gain = Ku = l/M

Ultimate Period of Sustained Cycling = P = 27i/wCO (min/cycle).U

(iv> Using the values of Ku and Pu, Ziegler and Nichols recommended

the following settings for feedback controllers:

KCrI(min.) ~D(min.)

-

Proportional (P) KU/2

Proportional-Integral (PI) Ku/2.2 Pull.2

Proportional-Integral-Derivative (PID) Ku/l.7 PJ2 Pu/ 8

The above settings reveal the rationale of the Z-N methodology.

-For proportional control alone, use a gain margin equal to 2.

-For PI control use a lower proportional gain because the presence of the

integral control mode introduces additional phase lag in all frequencies

(see Figure 17.8b) with destabilizing effects on the system. The refore,

lower Kc maintains approximately the same gain margin. Similar

arguments were used in the process reaction curve tuning technique (see

Section 16.4).

- The presence of the derivative control mode introduces phase lead with

strong stabilizing effects in the closed-loop response. Consequently,

the proportional gain Kc for a PID controller can be increased without

threatening the stability of the system.

Example 18.4 - Controller Tuning by the Ziegler-Nichols and Cohen-Coon Methods

Consider the multicapacity process in case (ii) of Example 16.4. We

have;

G = (5s+&2s+l) '1

IP Gm = 10s + 1 and Gf = 1.0

I

The controller settings according to the process reaction curve method were

found to be:

- For proportional controller; Kc = 8.0

- For PI controller; Kc = 7.2 and TI = 8.15

- For PID controller; Kc = 10.7, rI = 5.85 and -cD = 0.89

Let us now find the Ziegler-Nichols settings and compare them to the

above.

Using proportional control only, the crossover frequency can be found

from the equation,

-180" = tan-'(-5wC0) + tan-'(-2wC0) + tan-l(-10wCC)

which yields %o = 0.415 rad/min. The amplitude ratio at the crossover

frequency is found from the following equation,

log(AR) = log

&l2 + log & + log &

and it is equal to 0.055. Therefore the ultimate gain is

KU

= l/O.055 = 18.22

Also, the ultimate period is found to be

Pu = -z..?L=Yzo

15.14 minutes/cycle

Then, the Ziegler-Nichols recommended settings are:

- For a proportional controller; Kc = 18.22/2 = 9.11

- For a PI controller; Kc = 18.22i2.2 = 8.28 and -cI = 15.1411.2 = 12.62

- For a PID controller; KC = 18.2211.7 = 10.72, rI = 15.1412 = 7.57 and

rD = 15.14/8 = 1.89

Comparing the Z-N to the C-C settings we observe that;

- the proportional gains are a little larger for the Z-N settings,

- the reset and rate time constants are also higher for the Z-N.

Figures 18.6a and 18.6b indicate the responses of the closed-loop system to

step changes in the set point and load respectively, using a PID controller

with Z-N and C-C settings. We notice that the responses with Z-N tuning are

better than those with the more conservative C-C settings. It must be

emphasized though, that no general conclusions can be drawn as to the

relative superiority of one method over the other. The only conclusion we

draw is that both methods provide very good first guesses for the values of

the controllers' adjusted parameters.

18.4 THE NYQUIST STABILITY CRITERION

As we pointed out in Section 18.1, the Bode stability criterion is valid

for systems with AR and $I monotonically decreasing with w. For feedback

systems with open-loop Bode plots like those of Figure 18.4 the more general

Nyquist criterion is employed. In this section we will present a simple out-

line of this criterion and its usage. For more details on the theoretical

background of the methodology the reader can consult Appendix 18.A at the end

of this chapter.

The Nyquist stability criterion states that:

"If the open-loop Nyquist plot of a feedback systemencircles the point (-1,0) as the frequency w takes anyvalue from -00 to +", then the closed-loop response isunstable."

To understand the concept of encirclement and therefore the correct usage of

the Nyquist criterion, let us study the following examples.

Example 18.5 - The Stability Characteristics of a Third-Order System Using the-~-Nyquist Stability Criterion

Consider the following open-loop transfer function:

G KcOL = (s+1)(2s+1)(4s+l)

Figure 18.7 shows the Nyquist plots for GOL when Kc = 1 (curve A) and

Kc = 50 (curve B). For each Nyquist plot the solid line covers the frequency

range 0 I w < -+ while the dotted part the frequencies from -a to 0. The

dotted segment of the Nyquist plot is the mirror image of the solid line

segment with respect to the real axis.

Figure 18.7 shows that curve A does not encircle the point (-l,O), while

curve B does. Thus, according to the Nyquist criterion the feedback system

with open-loop Nyquist plot the curve A is stable, while curve B indicates

an unstable closed-loop system. This in turn implies that for Kc = 1 the

system is stable while for KC= 50 is unstable.

Example 18.6 - Conditional Stability and the Nyquist Criterion

Consider the Nyquist plots shown in Figures 18.8a, 18.8b and 18.8~. All

correspond to the same open-loop transfer function with different values for

the proportional gain Kc. The plots in Figures 18.8a and 18.8~ do not

encircle the point (-1,0) while the Nyquist plot of Figure 18.8b does.

Therefore, the feedback systems corresponding to the first and third Nyquist

plots have stable closed-loop responses, while that of the second is unstable.

From the above plots it is clear that the closed-loop response becomes

unstable for a range of values KC

such that the point (-1,0) is between A

and B of the resulting Nyquist plot. When point (-1,0) is to the left of A

(Figure 18.8~) or to the right of B (Figure 18.8a), it is not encircled by

byquist plot and the corresponding closed-loop response is stable.

Remark: For fast conclusions on the encirclement or not of the point

(-1,0) by the open-loop Nyquist plot, the reader can use the

following practical method:

"Place a pencil at the point (-1,O). Attach one end of a thread

at the pencil and with the other end trace the whole length of

the Nyquist plot. If the thread has wrapped around the pencil

then we can say that the point (-k,O) is encircled by the

Nyquist plot."

The gain margin and phase margin of an open-loop response can be also

computed from a Nyquist plot. This must be expected since Bode and Nyquist

plots of a system contain exactly the same information.

Consider the Nyquist plot of Figure 18.9. At the frequency of the

point A the Nyquist plot intersects the unit circle around the origin.

Therefore, since the distance of point A from the origin is the AR at this

frequency, we conclude the angle $,M represents the phase margin.

Furthermore, at the frequency of point B the phase lag is equal to 180".

The amplitude ratio at this point is the distance between B and the origin,

i.e. AR = M. Consequently, the gain margin is easily found as l/M.


In this chapter we have demonstrated that the frequency response analysis

is a useful tool for designing feedback ocntrollers. In particular:

d The open-loop Bode and Nyquist plots of feedback systems reveal if the

closed-loop response will be stable or not.

r

,1 .-:., -

I

- The gain and phase margins can be computed easily and constitute the

safety factors incorporated in the tuning of feedback controllers to

account for uncertainties in the values of time constants, dead times and

static gains. Tuning with gain margins larger than 1.7 and phase margins

larger than 30” is the most typical. It should be emphasized though that

the size of the margins used, is dictated by the magnitude of the uncer-

tainty in the values of the process parameters.

- The Ziegler-Nichols settings stem from frequency response considerations

and provide a closed-loop tuning methodology, unlike the Cohen-Coon

settings.

Wtih this chapter we close Part IV which dealt with the analysis and

design of simple feedback systems.

In the following five chapters of Part V we will extend our attention

to more complex control configurations which are encountered quite often in

chemical plants. Thus, we will study;

- special compensatory control far processes with large dead time or inverse

response,

- systems with selective control loops,

- feedforward and ratio control configurations and

- adaptive control.

Finally, we will discuss the need for the experimental identification of

process dynamics and present some techniques used for this purpose.


1.

2 .

3.

4.

5 .

6 .

7 .

8.

(

9 .

1

t

Explain in your own words that by "opening" a feedback loop we place the

controller in a "manual" operation, while by "closing" it we place the

controller in the "automatic" mode.

What is the basis of the Bode criterion? Why

rigorous?

it is not general-lY

Do you think that the following modified statement of the Bode criterion

is generally rigorous? Explain.

"A feedback control system is unstable, if the AR of the

corresponding open-loop transfer function is larger than 1

at any crossover frequency."

Construct an open-loop transfer function whose AR or 4, or both,

are not continuously decreasing functions of the frequency w. Draw

its Bode and Nyquist plots.

Identify the two major sources of instability in closed-loop responses.

Elaborate on these two factors.

Using the Nyquist stability criterion show that feedback systems with

first and second-order open-loop responses are always stable.

Define the phase and gain margins and show how you can compute them

from Bode or Nyquist plots.

Explain in your own words what we mean when we say that phase and gain

margins constitute safety margins (safety factors) in tuning a feedback

controller. Why do we need a safety margin in tuning a feedback

controller?

Describe the Ziegler-Nichols tuning methodology. This procedure is

often called "continuous cycling" tuning method. Why?

10. The Ziegler-Nichols settings result from closed-loop considerations,

while the Cohen-Coon settings are determined from the open-loop

response of the control system. Would you choose one over the other

because it uses open or closed-loop data? Explain.

11. The experimental determination of the Ziegler-Nichols settings brings

the chemical process at the threshold between stable and unstable

operation. Can you tolerate this in an industrial environment?

12. State the Nyquist stability criterion and give some examples of stable

and unstable feedback control systems, different than those in this

chapter. Explain the concept of encirclement of the point (-1,O) by

the Nyquist plot, which is so central for the Nyquist criterion.

13. Answer the following questions and justify your answers:

- Larger gain margin implies smaller or larger allowable controller

gain?

- Larger gain margin makes closed-loop response of a process faster

or slower?

- Larger phase margin implies faster or slower closed-loop response?

- Larger phase margin implies smaller or larger allowable controller

gain?

14. The discussion in Section 18.2 and Examples 18.2 and 18.3 have indi-

cated that we could use very large phase and gain margins to guarantee

closed-loop stability under model inaccuracies. Why would you try not

to use larger margins than those needed?

15. Larger uncertainty in the parameters of a model (static gain, time

constant, dead time) require tuning of the controller's parameters

based on larger or smaller gain or phase margin?

IIIIIIIIIII1

.!iF:c.: :4 ,o

I

9\

Y

\\

\\\

4 .“. ,,, _ . _”

,

,

I’

,’

-2

I

,-0

.c

REFERENCES

Chapter 13: There is a variety of references that the reader can consult for

more information on the constructional and operational details of; measuring

devices, feedback controllers, transmission lines, transducers, and final

control elements. The following are some typical sources:

(1) Process Instruments and Controls Handbook, by D. M. Considine, McGraw-Hill Book Co., New York (1957).

(2) Handbook of Applied Instrumentation, by D. M. Considine and S. D. ROSS,McGraw-Hill Book Co., New York (1964).

(3) Instrument Engineers Handbook: Vol. 1. Process Measurement, by B.Liptak, Chilton Book Co., Philadelphia (1970).

For measuring devices Chapter 7 of Ref. 4 can be found very useful.

(4) Measurements and Control Applications for Practicing Engineers by J. 0...-Hougen, Cahners Books, Boston (1972).

For the dynamics of some typical sensors the reader can consult the article,

(5) "Process Dynamics. Part 2: Process Control Loops," by J. L. Guy, Chem.Engng., Aug. 24, p. 111 (1981).

While for the dynamics of thermocouples, valves, pumps, piping, et al. the

book by Guthrie [Ref. 61 contains useful information

(6) Techniques of Process Control, by P. S. Buckley, J. Wiley and Sons,New York (1964).

The selection of the appropriate control valve is discussed in Reference 6 and

in the book by Luyben [Ref. 71.

(7) Process Modeling, Simulation and Control for Chemical Engineers, byW. L. Luyben, McGraw-Hill Book Co., New York (1973).

The distributed character of the pneumatic transmission lines dynamics is

discussed in Ref. 6 and in the book by Weber, [Ref. 81.

(8) An Introduction to Process Dynamics and Control, by T. W. Weber, J.Wiley and Sons, New York (1973).

II) ,. . ___~ . .

Chapter 15: The mathematical proof the Routh-Hurwitz tests can be found in

the classic book,

(9) Dynamics of a System of Rigid Bodies, 3rd edition, by E. J. Routh,McMillan, London (1877)

wuile for an extensive discussion the reader can consult,

(10) Stability Theory of Dynamical Systems, by J. L. Willelms, Nelson,(1970) .

(11) Mathematical Methods in Chemical Engineering, by V. G. Jenson, and G. V.Jeffreys, Academic Press, London (1963).

The books by Willelms [Ref. lo] and Douglas [Ref. 121 can also be used for

studying alternative definitions of stability and more advanced treatment on

the subject.

(12) Process Dynamics and Control, Vol. 2., by J. M. Douglas, Prentice-Hall,Englewood Cliffs (1972).

The construction rules for the root locus of a closed-loop system can be

found in the books by Douglas [Ref. 121, Luyben [Ref. 71 and in the following

two classic texts:

(13) Proces Systems Analysis and Control, by D. R. Doughanowr and L. B.Koppel, McGraw-Hill Book Co., New York (1965).

(14) Modern Control Engineering, by K. Ogata, Prentice-Hall, EnglewoodCliffs (1970).

There is a variety of references on the use of root locus for the design of

closed-loop systems. The texts by Luyben [Ref. 71, Douglas [Ref. 121,

Coughanowr and Koppel [Ref. 131 and Ogata [Ref. 141 offer an excellent

treatment of the subject with a large number of examples.

Chapter 16: Two excellent references on the practical problems of controller

design are the books by Buckley [Ref. 61 and Shinskey [Ref. 151

(15) Process Control Systems, 2nd edition, by F. G. Shinskey, McGraw-HillBook Co., New York (1979).

__ ..-. -___ __I, ..x”_ . -- .” .),....

In these two texts the reader will find useful practical guidelines in

selecting the most appropriate type of feedback controller for a particular

application. In addition, one can find alternate tuning techniques employed

by the industrial practice.

For an extensive discussion on the various types of performance criteria,

their advantages and shortcomings in designing feedback controllers, the

reader can consult the following reference,

(16) "Optimization of Closed-Loop Responses" by G. Stephanopoulos, ModuleNo. 12, in Process Dynamics and Control, an AIChE publication.

In Ref. 16, the reader will also find various techniques for solving the con-

troller design problems, which use time integral performance criteria. For

additional reading on this subject the following sources are also recommended:

(17) Linear Control System Analysis and Design, by J. J. D'Azzo and C. H.Houpis, McGraw-Hill Book Co., New York (1975).

(18) Digital Computer Process Control, by C. L. Smith, Intex Educ. Publish.,New York (1972).

(19) Analytical Design of Linear Feedback Controls, by G. C. Newton, Jr.,L. A. Gould and J. F. Kaiser, John Wiley and Sons, New York (1957).

For additional reading on the process reaction curve method and the Cohen and

Coon settings, the reader can consult References 8, 12, 13 and 15. The

details on the development of the Cohen and Coon settings can be found in the

original work by Cohen and Coon

(20) "Theoretical Considerations of Retarded Control," by G. H. Cohen andG. A. Coon, Trans. ASME, 75, p. 827 (1953).

Chapters 17 and 18: The books by Buckley [Ref. 61 and Caldwell, Coon and Zoss

[Ref. 211 are two very good sources for an in depth study of the frequency

response analysis and its ramifications in controller design.

(21) Frequency Response for Process Control, by W. I. Caldwell, G. A. Coonand L. M. Zoss, McGraw-Hill Book Co., New York (1959).

For systems with transfer functions very difficult to factor and consequently

very hard to complete the frequency response analysis, Luyben [Ref. 71 dis-

cusses various numerical solution techniques. He has also included a computer

program in FORTRAN which uses the "stepping" technique to develop the Bode

and Nyquist plots for a distillation column. More details on the philosophy

of the Ziegler-Nichols tuning method can be found in the original work,

(22) "Optimum Settings for Automatic Controllers," by J. G. Ziegler and N. B.Nichols, Trans. ASME, 64, p. 759 (1942).

In References 6, 7, 13 and 15 the reader can find a large number of examples

demonstrating the application of frequency response arguments in the design of

feedback controllers. In particular, References 6 and 15 analyze the frequency

response characteristics of flow, pressure, temperature, concentration, et al.

control systems, and draw some useful general inferences according to the

control system.

APPENDIX 18.A. COMPLEX MAPPING AND THE NYQUIST CRITERION FOR STABILITY.

Consider the function G(s) = l/(s+l) of the complex variables.

Let A, B, C and D be four points in the complex plane with coordinates

as shown in Figure 18.A-la. The value of G(s) at the point A can be

found easily as follows:

G(s) = (~) = (-2+jlo) + 1 = -l+ j-0A

If we consider the G-plane, i.e. a space with coordinates Re[G(s)] and

Im[G(s)], then, the value of G(s) at the point A is given by the point

A' (Figure 18.A-lb). We say that point A maps onto the point A'

through the function G(s). In the same way we find the maps of points

B, C, and D which are shown in Figure 18.A-lb (points B', C', and D').

Let us proceed a step further. Consider the line S in the com-

plex plane (s-plane) given by (Figure 18.A-2a)

s: s=a+ j (?a) with OLCt

Curve S' is the map of line S in the space of Re[G(s)] and Im[G(s)]

(G-plane) and is shown in Figure 18.A-2b. Curve S' is given by:

G(s) = A = -=[a+j (2:) ] + 1

for 0 Ia.

After having introduced the concept of complex mapping let us now

state Cauchy's Principle of the Argument, which constitutes the basis

of the Nyquist criterion for stability.

"Let G(s) be a function of the complex variable s. Let also

C be a closed contour in the s-plane, on and within which

the G(s) is analytic. Let the contour C encircle n points

at which G(s) takes a value G0 ’

Then, the complex map of C

in the G-plane encircles point Go n times."

To appreciate the importance of the above principle let us con-

sider the closed-loop characteristic equation.

1 + GPGmGcGf = O

Recall that G(s) 3 Gp(s)~Gm(s)*Gc(s)~Gf(s) is the open-loop transfer

function. Then the characteristic equation becomes:

G(s) = -1 (18.A-1)

Define a contour C (Figure 18.A-3a) which encloses all the right half

of the complex plane (s-plane). If there are k roots of equation

(18.A-1) in the right half of the s-plane, then according to Cauchy's

principle stated above, the map of C in the G-plane encircles the

point G = -1, k times.

Let us see what is the map of C in the G-plane. Contour C is

composed of the imaginary axis and a semicircle of radius R, where

R -f 00. For the imaginary axis we have

s = j-w with --m < w < +c=J

and its map in the G-plane will give us the Nyquist plot (see Section

17.4). For the large semicircle we have:

s = Rejw (18.A-2)

Q,(s)G(s) = p

n

where, Q,(s) is an m-th order polynomial and P,(s) is an n-th order one,

with n > m. Substittue s in G(s) by its equal from equation (18.A-2) and

take:

G[RejW] = Qm[RejW]/Pn[RejW]

Since the order of Pn is higher than that of Q,, then it is clear that

IGIRejW] 1 + 0

and the large semicircle maps in the origin of the G-plane. Also,

dG[ReJW] takes all values from -90' to +90°. Thus we conclude that;

"the map of the contour C (Figure 18.A-3) in the G-plane,

where G(s) 5 open-loop transfer function, is the Nyquist

plot for G(s)."

Consequently, we can state the following criterion for the stability of

a closed-loop response.

"If the Nyquist diagram of the open-loop transfer function

G(s) encircles the point (-l,O), k times, then according to

Cauchy's Principle of the Arguement, ther are k roots of the

closed-loop characteristic equation in the right half of the

--., . . .1

46’1

s-plane. If k = 0 then the closed-loop response is stable,

and if k > 0 the closed-loop response is unstable."

The above criterion is known as the Nyquist Criterion of closed-loop

stability.

n

I,

b

c

-..

4

47f

PART V

ANALYSIS AND DESIGN OF ADVANCED CONTROL SYSTEMS

Although feedback control is the most commonly encountered in chemical

processes, it is not the only one. There exist situations where feedback con-

trol action is insufficient to produce the desired response of a given process.

In such cases other control configurations are used like feedforward, ratio,

multivariable, cascade, override, split range, adaptive control, etc.

In the following four chapters of Part V we will study the static and

dynamic characteristics as well as methods for the design of the following.

advanced control systems:

- Compensatory control for processes with large dead time or inverse response.

- Multiple loop control (cascade; selective; split range).

- Feedforward and ratio control.

- Adaptive and inferential control.

Finally, in Chapter 23 we will study the need for experimental modeling of

processing systems and will examine several techniques which are commonly

used in the chemical plants for this purpose.

CHAPTER 19

FEEDBACK CONTROL OF SYSTEMS WITH LARGE DEAD TIME

OR INVERSE RESPONSE

All the previous chapters of Part IV have been devoted to the analysis

and design of feedback control systems for rather simple processes. In this

chpater we will be concerned with the feedback control of two special types

of systems; with large dead times and inverse responses. We will see in the

subsequent paragraphs of this chapter that for such systems conventional P,

PI OR PID controllers may not be sufficient to yield the desired response.

19.1 PROCESSES WITH LARGE DEAD TIME

Consider the general feedback control system of Figure 14.1. All the

dynamic components of the loop may exhibit significant time delays in their

response. Thus,

- the main process may involve transportation of fluids over long distances

or include phenomena with long incumbation periods;

- the measuring device may require long periods of time for completing the

sampling and the analysis of the measured output (a gas chromatograph is

such a device);

- the final control element may need some time to develop the actuating

signal;

- a human controller (decision maker) may need significant time to think and

take the proper control action.

In all of the above situations a conventional feedback controller would provide

quite unsatisfactory closed-loop response for the following reasons:

- A disturbance entering the process will not be detected until after a sig-

nificant period of time.

- The control action which will be taken on the basis of the last measurement

will be inadequate because it attempts to regulate a situation (eliminate

an error) which was originated while back in time.

- The control action also will take some time to make its effect felt by

the process.

- As a result of all the above significant dead time is the main source of

instability for closed-loop responses.

Example 19.1 - Dead Time as a Main Source of Closed-Loop Instability

Consider the following open-loop transfer function,. .

-ds

GKc-e

OL = 0.5s+l

(i> If td = 0.01 minutes, i.e. very small, then crossover frequency =

160 rad/min, and ultimate gain = 80.01.

(ii) Suppose that the dead time increased to td = 0.1. Then, the

crossover frequency = 17 rad/min and ultimate gain = 8.56. We

notice that the increase of the dead time has introduced significant

additional phase lag which reduces the crossover frequency and the

maximum allowable gain. In other words, the increase of dead time

has made the closed-loop response more sensitive to periodic dis-

turbances and has brought the system closer to the brink of

instability.

(iii) Further increase in dead time, i.e. td = 1.0 yields crossover

frequency = 2.3 rad/min and ultimate gain = 1.52. And we see the

same trends as above.

The results of Example 19.1 lead to the following general statements:

"As the dead time of an open-loop transfer function increases, the following

two undesirable effects take place:

- The crossover frequency decreases. This implies that the closed-loop

response will be sensitive even to lower frequency periodic disturbances

entering the system.

- The ultimate gain decreases. Therefore, in order to avoid the insta-

bilities of the closed-loop response, we must reduce the value of the

proportional gain Kc which leads to sluggish response."

Figure 19.1 depicts graphically the above results.

The discussion above indicates that a control'system different than the

typical feedback loop is needed to compensate for dead time effects.

19.2 DEAD-TIME COMPENSATION

In the previous section we identified the critical need for more effective

control of processes with significant dead time. In this section we will

discuss a modification of the classical feedback control system, which was

proposed by 0. J. M. Smith for the compensation of dead-time effects. It is

known as Smith Predictor or Dead-Time Compensator.

To understand the nature of the dead-time compensation proposed by

Smith, consider the simple feedback loop with set point changes only shown

in Figure 19.2a. We have assumed that all the dead time is caused by the

process, i.e.

Gp(4 = G(s)e-tds

and that for simplicity, G,(s) = Gf(s) = 1. The feedback measurement signal

used by the control mechanism is equal to,

P(s) = Gc(s> l ☯G(s) -e

-tas1 Y,,(s) (19.1)

i.e. it is delayed by td minutes.

In order to eliminate the undesired effects of a delayed information as

the y(s), we would like to have a measurement signal which carries current

and not delayed information, like the

Y,(s) = Gc(s)*~(d*~,,(s) (19.2)

This is possible if we add to the signal y(s) the following quantity

p'(s) = (1 - e-tds

Kc(s) -G(s) l Y,,W (19.3)

It is easy to verify that

Y(s) + Y'(s) = Y,(s) ;

The implication of adding y(s) to the signal f(s) is shown in Figure 19.2b.

There we notice that the signal y'(s) can be taken by a simple local loop

around the controller, which is called, dead-time compensator or Smith pre-

dictor. The simplified loop of Figure 19.2c is completely equivalent to'

that of Figure 19.2b and indicates the real effect of the dead-time compen-

sator, i.e.

"moving the effect of dead time outside the loop."

Remarks: (1) In the block diagram of Figure 19.2c it is not correct to

think that we take a measurement signal after the block

G(s) because such signal is not measurable in a real process

with dead time. The only measurable signals are the process

output, Y(s) , and the manipulated variable. Therefore, the

block diagram of Figure 19.2c is meant to give only a

schematic representation of what is the effect of the dead-

time compensator and not to depict physical reality.

(2) The dead-time compensator predicts the delayed effect that

the manipulated variable will have on the process output.

This prediction lent the name to the term Smith Predictor

and it is possible only if we have a model for the

dynamics of the process (transfer function, dead time).

(3) In most of the process control problems the model of the

process is not perfectly known, i.e. G(s) and td are

only approximately known. Therefore, if G(s) and td

represent the "true" characteristics of the process while

Gmodel(s) and td(model) represent their approximations

which we use for dead-time compensation then the control

system with the Smith predictor is shobm in Figure 19.3.

In this case we have:

7mw = y(s) + F'(s)

= &G e-tds

+(l-e-td(model)'s

jGc Gmodell*?'Sp(s)

or

Y,(S) = [GcGmodel+(G e-tds

-Gmodel e-td(model)'s

) 1 -Y,,W(19.4)

The above equation indicates some important features of the

dead-time compensators:

- Only for perfectly known ,processes we will have perfect

compensation, i.e. for G 2 Gmodel and td = td(model)'

- The larger the modeling error, i.e. the larger the dif-

ferences (G - Gmodel) and (td - td(model)), the less

effective is the compensation.

- The error is estimating the dead time is more detrimental

for effective dead-time compensation, i.e. (td - t d(mode1))

is more crucial than (G - Gmodel), because of the

exponential function.

(4) The dead time in a chemical process is usually caused by

material flows. Since the flowrate is not normally constant

but shows variations during the operation of a plant, the

value of the dead-time changes. Therefore, if the dead-time

478

compensator is designed for a certain value oftd'

when it takes a new value5 the compensation will not

be as effective.

Example 19.2 - Dead-Time Compensation and the Effect of Modeling Error

Consider the feedback loop shown in Figure 19.3a. Let the controller

be simple proportional and the "true" transfer function of the process be

Gp(s) = ’-1s

OJs+l' e

It is easy to recognize that : .?

G(s) = & and. td = 1

(a) Suppose that we use simple feedback control. For this system it was

found in Example 19.1 that the open-loop transfer function has

crossover frequency: %o = 2.3 radlmin and ultimate gain: Kc = 1.52.

The fact that the ultimate gain is 1.52 forced us to use Kc = 1.5 <

1.52. Nevertheless, the system is very close to the brink of

instability and has a rather unacceptable offset (see Section 14.2)

offset = 1 1l+KK = 1 + 1.1.5 = 0.4

PCCurve A (Figure 19.4) shows the response of the system to a unit

step-change in the set point.

:5> Let us introduce "perfect" dead-time compensation. This is possible

if the "true" transfer function of the process is known. Then, the

control system is given by the block diagram of Figure 19.3b. The

open-loop transfer function is (see eqn. (19.2))

7,w Kci&(s) = GcG = 0.5s+l

which has no crossover frequency. Consequently, we can use arbitrarily

large proportional gain to reduce the offset without endangering the

stability of the system. Curve B (Figure 19.4) shows the response

of the closed-loop system with Kc = 50. The offset has been greatly

reduced, i.e.

offset = 1 1l+KK = 1 + 1a5o = 0.0196

PC

(c) Suppose that the process gain and time constant are perfectly known

but not the dead time. In such case, Gmoael = G = l/(O.Ss+l). The

dead time of the process can only be approximated. Let ta(moael) =

0.8.

Let us examine a common error in the design of a process control system

which we must avoid. Were we to consider the value 0.8 as the "true" value

of the process dead time, we coda design a dead-time compensator as in case

(b) above. Since there would be no crossover frequency we could use an

arbitrarily large Kc in order to reduce the offset. Let Kc = 100..

Since the "true" value of the process dead time is not 0.8 but 1.0, the

compensation is not perfect. There is a dead time equal to 1.0 - 0.8 = 0.2

which has not been compensated by the dead-time compensator. Thus uncompensated

dead time gives rise to additional phase lag and leads eventually to a cross-

over frequency. If the ultimate gain is smaller than 100 then the system with

3c.'C = 100 is unstable. Indeed, for the present example the crossover frequency

is %o = 9 radlmin and the ultimate gain is 4.6. Therefore, if we are not

certain on the value of dead time we must be conservative in selecting the

------A,value of even with partial dead-time compensation.

19.3 CONTROL OF SYSTEMS WITH INVERSE RESPONSE

In Section 12.3 we analyzed the behavior of a special class of systems

with inverse response. There we saw that the net result of two opposing

effects may produce an initial response which is in the opposite direction to

where it will eventually end up (see Figures 12.4b and 12.5b).

The most common case of a process with inverse response is that resulting

from the conflict of two first-order systems with opposing effects (Figure

12.5). In this section we will limit our attention to the regulation of such

processes. Extensions to more complex systems like those of Table 12.1 are

easy and straightforward. !'*

Two are the most popular ways to control systems with inverse response;

the first uses PID feedback controller with Ziegler-Nichols tuning and the

second an inverse response compensator.

A. Simple PID Control

From all types of feedback controllers only PID can be used effectively

for the following simple reason. The derivative control mode by its nature

will anticipate the "wrong" direction of the system's response and will pro-

vide the proper corrective action to limit (never eliminate) the inverse

shoot. Wailer and Nyg?irdas [Ref. 261 have demonstrated numerically that the

Ziegler-Nichols classical tuning of a PID controller yields very good control

;uf systems with inverse response.

23 . Inverse Response Compensator

In Section 19.2 we discussed how we can develop a Smith predictor (dead-

time compensator) which cancels the effect of dead time. The same general

concept of the predictor (compensator) can be used to cope with the inverse

response of a process and was developed by Iinoya and Altpeter [Ref. 251. *

7* ’

Consider the feedback system of Figure 19.5a. The controlled process

exhibits inverse response when (see Example 12.4)

rp1K

-> J&l

=p2Kp2

The open-loop response of the system is:

(K T -K 'I )s+(K -K >

Y(s) = Gc(s>-21 p2 p2 Pl Pl p2

(T s+l)(r s+l) *f,,(s)

Pl p2

and has a positive zero at the point (see also Example 19.4)

(19.5)

K -Ki fPl p2

Z = -

KPlTP2-K T

> '0

: p2 p1

To eliminate the inverse response it is enough to eliminate the positive zero

of the above open-loop transfer function. This is possible if in the open-

loop response y(s) we add the quantity y'(s) given by,

T’(s) = Gc(s>*k( ls+l -rp2

i+$ ‘YspWTpl

Then, from eqns. (19.5) and (19.6) we can easily find that

(19.6)

Tm(s) = Y(s) +Y’(s

.and for

[(K T

> = Gc(s> l

Pl P2-KP2TPl)+k(~Pl-TP2)1S+(KPl-KP2)

(T s+l)(-rPl p2

s+l)

.y

SP

K 'c -K ^c

k Lp2 Pl Pl p2

TP1 - =P2(19.7)

we find that the zero of the resulting open-loop transfer function is non-

positive, i.e.

,(s)

(K -K >

z = - Pl p2(K T

0

Pl p2- K T )+k(T

p2 Pl-T ) '

Pl p2

Adding the signal Y'(S) to the main feedback signal y(s) it means the

creation of the local loop around the controller as it is shown in Figure

19.5b. The system in this local loop is the modified Smith predictor and

the actual compensator of the inverse response. Its transfer function as it

can be,seen from eqn. (19.6) is

(19.8)

where k must satisfy condition (19.7).

Remarks: (1) The inverse response compensator predicts the inverse behavior

of the process and provides a corrective signal to eliminate

it. The prediction is based on a model for the process.

The ideal prediction comes if the transfer function of the

process is completely known. In such case the compensator

is given by

K K

GCOMPENSATOR (s)p2 p2

= T s+1 - s+1P? TP,

L I

Therefore, the compensator given by eqn. (19.8) is only an

approximation to the process' transfer function.

(2) Modeling inaccuracies in terms of ~~ and1 *p2

will cause

increased inverse shoots and slower responses.

(3) For the controller, PI is the most common choice.


Chemical processes with significant dead times or inverse response present

a special challenge to the control designer. For the case of processes with

dead time there is the danger of instability even with low controller gains,

while for processes with inverse response the elimination or at least sig-

nificant reduction of the inverse shoot is of paramount importance.

both problems can be solved with the introduction of the so-called Smith

predictor or feedback compensation. Thus, in the case of dead time we intro-

duce a local feedback loop around the controller which anticipates the time

delay of the measurement signal and makes the proper compensation. If the

dead time is perfectly known the compensation is perfect and the control

feedback signal carries current not delayed process information. If the dead

time is not perfectly known (which is usually the case) or changing with time.~

(very common feature of chemical processes), the dead-time compensation is

only partial.

In the case of inverse response the added compensator modifies the

location of the zero of the open-loop transfer function, i.e. shifts it from

positive to nonpositive (negative or zero). This eliminates the inverse

behavior. It should be noted that simple PID control of inverse responses

gives rather satisfactory results.

With this chapter we close the analysis and design of feedback control

-systems. In Part V we will focus our attention to the analysis and design

of various other control configurations like feedforward, ratio, cascade,

split-range, override, multivariable, etc. which appear very often in chemical

processes.


1.

2.

3.

4.

5.

What is the effect of dead time in the response of simple feedback con-

trol loops? Explain in physical terms.

Why is the controller design of processes with dead time a particularly

sensitive and difficult problem? Demonstrate using a practical example.

Describe in physical terms the concept of dead-time compensation. Why

such a system is also called predictor?

Show that the dead time and inverse response compensators are based on

the same logic. What are their implementational difficulties?-

Consider the following feedback loop with load changes only. Construct

a dead-time compensator assuming G(s) and td perfectly known. Is it

: 1’-- --_

6.

7.

8.

the same as the dead-time compensator constructed for set point changes?

What is the effect of model inaccuracies on the effectiveness of dead-

time compensators?

Is the dead time of a process constant or it varies with time? If it

varies, give three relevant physical examples. What is the effect

of changing dead time on the design of a dead-time compensator?

What is our goal when designing a controller for a system with inverse

response? Describe what an inverse response compensator does.

10. Consider the system with inverse response described in Section 19.3.

Identify the transfer function of the compensator. Notice that it is

a function of the parameter k which must satisfy condition (19.7).

Do you have any ideas on how k would affect the quality of the con-

trolled response? [see also numerical example in Ref. 251.

I1 j,p I- - - --T----

cc>

1.0

0.5

0.4

II---------_---------_- ---- - _- -

-_---__

t

CONTROLLER MEOtAN’Sti-----------------, I

CHAPTER 20

CONTROL SYSTEMS WITH MULTIPLE LOOPS

The feedback control configuration involves one measurement (output) and

one manipulated variable in a single loop. There are though other simple

control configurations which may use,

- more than one measurement and one manipulation, or

- one measurement and more than one manipulated variables.

In such cases control systems with multiple loops may arise. Typical examples

of such configurations which we will study in the present chapter are the

following:

(a) Cascade control.

(b) Various types of selective control.

(c) Split range control.

Before proceeding we should emphasize that these control systems involve

loops which are not separate but share either the single manipulated, variable

or the only measurement. In this respect the multiple loop control systems of

this chapter are generically different from those we will study in Chapter 22.

20.1 CASCADE CONTROL

In a cascade control configuration we have one manipulated variable and

snore than one measurement. It is clear that with a single manipulation we can

-mtrol only one output. Let us now examine the motivation behind the cascade

control and its typical characteristics using an example from the experience

in the chemical processes.

Example 20.1 - Cascade Control for a Jacketed CSTR

Consider the CSTR shown in Figure 1.7. The reaction is exothermic and

the generated heat is removed by the coolant which flows in the jacket around

the tank. The control objective is to keep the temperature of the reacting

mixture, T, constant at a desired value. The possible disturbances to the

reactor are; the feed temperature Ti and the coolant temperature TC’

The

only manipulated variable is the coolant flowrate FC’

Simple Feedback Control

If we use simple feedback, we will take the control configuration shown-.

in Figure 20.la, i.e. measure temperature T and manipulate coolant flowrate

Fc’ It is clear that T will respond much faster to changes in Ti than to

changes in TC’

Therefore, the simple feedback control of Figure 20.la will

be very effective in compensating for changes in Ti and less effective in

compensating for changes in TC’

Cascade Control

We can improve the response of the simple feedback control to changes in

'the coolant temperature by measuring Tc and taking control action before

its effect has been felt by the reacting mixture. Thus, if Tc goes up,

increase the flowrate of the coolant in order to remove the same amount of

heat. Decrease the coolant flowrate when Tc decreases.

We notice, therefore, that we can have two control loops using two dif-

f-erent measurements, T and Tc, but sharing a common manipulated variable,

Fc' How these loops are related is shown in Figure 20.lb. There we notice

that,

- the loop that measures T (controlled variable) is the dominant, or primary,

or master control loop and uses a set point supplied by the operator, while

- the loop that measures TC

uses the output of the primary controller as

its set point and it is called secondary or slave loop.

The control configuration with the above two loops is known as cascade control

and is very common in chemical processes.

Let us generalize the above discussion. Consider a process consisting of

two parts as shown in Figure 20.2a; Process I and Process II. Process I

(primary) has as its output the variable we want to control. Process II

(secondary) has an output which we are not interested to control but which

affects the output we want to control. For the CSTR system of Example 20.1,

Process I is the reaction in the tank and the controlled"output is the tem-

perature T. Process II is the jacket and its output Tc affects Process I

(reactor) and consequently T.

Figure 20.2b shows the typical simple feedback control system, while

Figure 20.2~ indicates the general form of the cascade control. The last

figure demonstrates very clearly the major benefit to be gained by cascade

control, i.e.

"disturbances arising within the secondary loop are correctedby the secondary controller before they can affect the valueof the primary controlled output."

This important benefit has lead to an extensive usage of cascade control in

.&emical processes.

Example 20.2 - Cascade Control for Various Processes

Let us describe the use of cascade control in various typical processing

systems.

(a) Heat exchangers. The typical configuration is shown in Figure 20.3a.

The control objective is to keep the exit temperature of stream 2 at

a desired value. The secondary loop is used to compensate for changes

in the flowrate of stream 1.

(b) Distillation columns. Cascade control is usually employed to regulate

the temperature (and consequently the concentration) at the top or.

bottom of a distillation column. Figures 20.3b and 20.3~.show two such

typical cascade control systems. In both cases the secondary loop is

used to compensate flowrate changes.

(c) Furnaces. Cascade control can be used to regulate the temperature of a

process stream (e.g. feed to a reactor) exiting from a furnace.

Figure 20.3d shows the resulting cascade configuration. Again, the

secondary loop is used to compensate for flowrate changes (fuel flowrate).

The reader should notice that in ali the cascade configurations of

Example 20.2, the secondary loop is used to compensate for flowrate changes.

This observation is quite common in chemical processes and someone could state

that;

"in chemical processes, flowrate control loops are almostalways cascaded with other control loops."

Let us now turn our attention to the closed-loop behavior of cascade

control systems. Consider the block diagram of a general cascade system

shown in Figure 20.4a. To simplify the presentation we have assumed that the

transfer functions of the measuring devices are both equal to 1.

The closed-loop response of the primary loop is influenced by the dynamics

of the secondary loop, whose open-loop transfer function is equal to

GSECONDARY = Gc,II Gp,II (20.1)

The stability of the secondary loop is determined by the roots of the

characteristic equation,

l + Gc II Gp,II = 0 (20.2),

Figure 20.4b shows a simplified form of the general block diagram (Figure 20.4a),

where the secondary loop has been considered as a dynamic element.

For the primary loop the overall open-loop transfer function is

G

GPRIMARY = Gc,I (c,II Gp,II

l+ Gc II G ) Gp,I, P,II

and consequently the characteristic equation whose roots determine the-_

stability of the primary loop is the following,

G1 + Gc I 1 +‘;‘I

G

, (“I1

c,II Gp,II) G

P,I =0

(20.3)

(20.4) .

Remarks: (1) The two controllers of a cascade control system are standard

feedback controller, i.e. P, PI, PID.. Generally, proportional

controller is used for the'secondary loop although PI controller

with small integral action is not unusual. Any offset caused

by P control in the secondary loop is not important since

we are not interested to control the output of the secondary

process.

(2) The dynamics of the secondary loop is much faster than that

of the primary loop. Consequently, the phase lag of the

closed secondary loop will be less than that of the primary

loop. This feature leads to the following important

result which constitutes the rationale behind the use of

cascade control: "The crossover frequency for the

secondary loop is higher than that for the primary loop.

This allows us to use higher gains in the secondary

controller in order to regulate faster the effect of a

disturbance occurring in the secondary loop without

endangering the stability of the system."

(3) The tuning of the two controllers of a cascade control

system proceeds in two steps:

- First, we determine the settings for the secondary con-

troller using one of the methods we studied in Chapters

16 and 18, i.e. Cohen and Coon or Ziegler-Nichols or

others using time-integral criteria or phase and gain

margin considerations; The open-loop transfer function

we can use for tuning is given by eqn. (20.1).

- Second, from the Bode plots of the overall system we

determine the crossover frequency using the settings

for the secondary loop we found above. Then, using

the frequency response techniques we described in

Chapter 18, we choose the settings for the primary

controller. The open-loop transfer function needed

for the construction of the Bode plots is given by

eqn. (20.3).

,JXxample 20.3 - Dynamic Characteristics of a Cascade Control System

Consider a process with the following transfer functions for its primary

.and secondary elements,

G 1P,T = (5s+1~010s+1) and G

P,TT = 0.5s+l

The secondary process is faster than the primary as can be seen from the

corresponding time constants.

Were we to use simple feedback control, the open-loop transfer function

with PI control would be,

Gc,I'Gp,II'Gp,I = KcI, (1 +$*(O.sls+l) l

1 0 0(5s+l)(los+l)

The crossover frequency can be found from the equation that sets the total

phase lag equal to -180", i.e.

and it is equal to

%o = 4.9 -rad/min

Also, the overall amplitude ratio is given by,

tan-'(-l/wCO) + tan-l(-0.5wCO) + tan-l(-5wCO)+tan-l(-10wCO) ~-180"

The ultimate value of the gain Kc I can be found from the condition,

AR =l at w = %Ci

Thus,

and we find

Kc,I = 32.25

merefore, when the disturbance dII (of the secondary process) changes, the

simple feedback controller can use a gain up to 32.25 before the system becomes

unstable. Also, given the fact that the overall provess is 3rd order, we

expect that the closed-loop response of y(t) to changes in dII will be

rather slow.

Consider now a cascade control system similar to that of Figure 20.4a.

The open-loop transfer function for the secondary loop is given by eqn. (20.1)

and assuming simple proportional controller we find

Gc,II Gp,II1

= Kc,11 0.5s+1

There is no crossover frequency for the secondary control loop. Therefore we

can use large values for the gain Kc,II' which produce a very fast closed-

loop response to compensate for any changes in the disturbance dII, arising-

within the secondary process.

-.

Once we have selected the value of -Kc II for the secondary loop, wei .I ,

can find the crossover frequency for t!he overall open-loop transfer function

given by eqn. (20.3). Thenj we can select the value of Kc I for the primary,

controller, using the Ziegler-Nichols methodology. Quite often we will not

select arbitrarily a very large Kc II but in coordination with the resulting,

values of Kc,I'

20.2 SELECTIVE CONTROL SYSTEMS

These are control systems which involve one manipulated variable and

several controlled outputs. Since with one manipulated variable we can con-

trol only one output, the selective control systems transfer control action

from one controlled output to another according to the need. There are

.exral types of selective control systems and in this section we will discuss

only the following two:

(a) Override control for the protection of process equipment.

(b) Auctioneering control.

A. Override Control

During the normal operation of a plant or during its start-up or shut-down

it is possible that dangerous situations may arise, which may lead to destruction

of equipment and operating personnel. In such cases it is necessary to change

from the normal control action and attempt to prevent a process variable from

exceeding an allowable upper or lower limit. This can be achieved through

the use of special types of switches. The high selector switch (HSS) is used

whenever a variable should not exceed an upper limit, while the low selector

switch (LSS) is employed to prevent a process variable to exceed a lower limit.

Example 20.4 - Examples of Override Control

(a) Protection of a boiler system. Usually, the steam pressure in a boiler

is controlled through the use of a pressure control loop on the dis-

charge line. At the same time the water level in the boiler should

not fall below a lower limit wkich is necessary to keep the heating

coil immersed in water and thus prevent its burning out. Figure 20.5

shows the override control system using a low switch selector (LSS).

According to this system, whenever the liquid level falls below the

allowable limit, the LSS switches control action from pressure control

to level control and closes the valve on the discharge line.

Protection of a compressor system. The discharge of a compressor is

controlled with a flow control system. To prevent the discharge

pressure from exceeding an upper limit, an override control with a

high switch selector (HSS) is introduced. It transfers control

action from the flow control to the pressure control loop whenever

the discharge pressure exceeds the upper limit (Figure 20.6).

Notice that flow control or pressure control is actually cascaded

to the speed control of the compressor's motor.

(c) Protection of a steam distribution system. In any chemical process there

is a network distributing steam at various pressure levels to the pro-

cessing units. High pressure steam is "let-down" to lower pressure

levels at the let-down stations. The amount of steam "let-down" at

such stations is controlled by the demand on the low pressure steam

line (Loop 1 in Figure 20.7): To protect the high pressure line from

excessive pressures, we can install an override control system with a

HSS, which transfers control action from Loop 1 to Loop 2 when the

pressure in the high pressure line exceeds an upper limit.

B. Auctioneering Control Systems

Such control configurations select among several similar measurements

the one with the highest value and feed it to the controller. Thus, it is a

selective controller with several measured outputs and one manipulated input.

Example 20.5 - Examples of Auctioneering Control

(a) Catalytic tubular reactors with highly exothermic reactions. Several

highly exothermic reactions take place in tubular reactors filled with

a catalyst bed. Typical examples are the hydrocarbon oxidation

reactions like the oxidation of o-xylene or naphthalene to produce

phthalic anhydride. Figure 20.8 shows the temperature profile along

the length of the tubular reactor. The highest temperature is called

hot spot. The location of the hot spot moves along the length of the

reactor depending on the feed conditions (temperature, concentration,

flowrate) and the catalyst activity (Figure 20.8). The value of the

hot spot temperature depends also on the above factors and the tem-

perature and flowrate of the coolant. The control of such systems

is a real challenge for a chemical engineer.

The primary control objective is to keep the hot spot temperature

below an upper limit. Therefore, we need a control system that can

identify the location of the hot spot and provide the proper control

action. This can be achieved through; -

- the placement of several thermocouples along the length of the

reactor and

- the use of an auctioneering system to select the highest temperature,

which will be used to control the lflowrate of the coolant (Figure 20.9).

(b) Regeneration of catalytic reactors. The catalyst in catalytic reactors

undergoes deactivation as the reaction proceeds, due to carbonaceous

deposits on it. It can be regenerated by burning off these deposits

with air or oxygen. To avoid destruction of the catalyst, due to

excessive temperatures during the combustion of the deposits, we can

use an auctioneering system which;

- takes the temperature measurements from various thermocouples along

the length of the reactor,

- selects the highest which corresponds to the combustion front as

it moves through the bed, and

- controls appropriately the amount of air.

23-3 SPLIT-RANGE CONTROL

Unlike the cascade and selective control schemes examined in Sections

20.1 and 20.2, the split range control configuration has one measurement only

(controlled output) and more than one manipulated variables.

Since there is only one controlled output, we need only one control

signal which is thus split into several parts, each affecting one of the

available manipulations. In other words, we can control a single.process

output by coordinating the actions of several manipulated variables, all of

which have the same effect on the controlled output. Such systems are not

very common in chemical processes but provide added safety and optimality

operational whenever necessary as the following examples demonstrate.

Example 20.6 - Split-Range Control of a Chemical Reactor

Consider the reactor shown in Figure 20.10a where a gas phase reaction

takes place. Two control valves manipulate the flow fo the feed and the

reaction product. It is clear that in order to control the pressure in the

reactor the two valves cannot act independently but should be coordinated.

Thus, when valve Vl opens, valve V2 closes and vice versa. Figure 20.10b

indicates the coordination of the two valves' actions as a function of the

controller's output signal (see also Table 20.1).

Let the controller's output signal, corresponding to the desired

operation of the reactor by 6 psig. From Figure 20.10b we see that valve V2

is partly open while valve Vl is completely open. When for.various reasons

tie pressure in the reactor increases the controller's.output signal increases

too. Then, it is split into two parts and affects the two valves simul-

taneously. Thus:

-As the controller output increases from 6 psig to 9 psig, valve V2 opens

continuously while Vl remains completely open. Both actions lead to a

reduction in the pressure.

- For large increases in the reactor's pressure, the control output may

exceed 9 psig. In such case, as we can see from Figure 20.10b, the valve

V2 is completely open while Vl starts closing. Both actions again lead

to a reduction in pressure until the reactor has returned to the desired

operation.

Example 20.7 - Split-Range Control of the Pressure in a Steam Header

Let us consider another example of split-range control, which is

encountered very often in chemical plants. Several paraller boilers discharge

steam in a common steam header and from there to the process needs (Figure

20.11). The control objective is to maintain constant pressure in the steam.f

header when the steam demand at the various processing units changes. There

are several manipulated variables (steam flow from every boiler) which can be

used simultaneously. Figure 20.11 shows also the structure of the resulting

control system. It should be noted that instead of controlling the steam flow

from each boiler, we could control the firing rate and thus the steam pro-

duction rate at each boiler.

Similar structures can be developed for the pressure control of a common

discharge or suction header for N paraller compressors.

Table 20.1

Controller'soutputSignal

3 psig

Valve Vl Valve V2Stem Stem

Position Position

Open Closed

9 p s i g Open Open

15 psig Closed Open


Control systems with multiple loops arise when we have one manipulated

variable and several available measurements. Among them cascade and

selective control systems are the most often encountered in chemical

processes.

Cascade control is used to compensate for the effects of disturbances

which arise within the secondary part of a process. The response, when com-

pared to simple feedback, is much faster and quite robust. It should be

emphasized that cascade control is possible only if we can measure the output

of the secondary process. Cascade control is very popular in chemical pro-

cesses and almost always flow control systems will be cascaded with other

control loops.

Selective control systems are used whenever we want to be able to select

the measured output that we will connect to the single manipulated variable.

Override and auctioneering control are two typical examples of selective

control. The first is used to protect the operating personnel and process

equipment from excessive excursions of process variables beyond the acceptable

operating limits. The second is used whenever we need to identify the highest

value among similar measurements and use it for control purposes. Other types

of selective control systems are the variable structuring control and the use

,~E;redundant measurements, but will not be discussed in this text.

Split-range is a control configuration that uses a single measurement,

.produces a single control action which in turn is "Split" to activate more

than one manipulated variables, whose actions should be coordinated.


1. Consider a process with one manipulated input and two measured outputs.

Can you keep both outputs at the desired values, using only the single

manipulated variable? If not, explain why.

2. Starting from the premise that tht answer to item 1 (above) is negative,

explain how is it possible to have,

- a cascade control system or

- a selective control system

both of which have a single manipulation and two measured outputs.

3. Discuss the rationale of a' cascade control system and demonstrate why

it provides better response than simple feedback.

4. In Section 20.1 we assumed that the secondary process (PROCESS II,

Figure 20.2) in a cascade control system is faster than the primary

process (PROCESS I, Figure 20.2). Is this necessary to justify the

use of a cascade control configuration? In other words; would you

still recommend cascade control for a process (like that of Figure

20.2a) with a secondary process much slower than the primary?

5. What are the main advantages and disadvantages of cascade control?

For what kind of processes can you employ cascade control?

6. In chemical processes, flowrate control loops are almost always cas-

caded with other control loops. Why does this happen? [Note: Take

into account the following two facts: (a) The flowrate itself is

subject to changes and is regulated by the flow control loop, and

(b) flowrates are the most common manipulated variables in chemical

processes].

7. What types of controllers would you use for the two controllers of a

cascade system? How would you tune them? Discuss a methodology to

select the adjusted parameters of the two controllers.

8. Are the stability characteristics of the closed-loop response of a

cascade system better than those of a simple feedback? Elaborate on

your answer.

9. What is meant by selective control systems? How many different types

of selective control systems are available? Discuss their characteristics.

10. Discuss the sationale\,behind an override control system. Why is it very

useful, and what situations is called upon to control?

11. Describe 2-3 situations (different than those discussed in Example 20.4)

where you should use override control systems. *:

12. What is an auctioneering control system, and where would you use it?

Describe a situation (different than those of Example 20.5) where you

could use auctioneering control.

13. Consid%er a process with one controlled output and two active manipulated

variables. Under what conditions could you use both manipulated

variables to control the single output?

14. What is split-range control? In Example 20.6 we have a situation with

split-range control. To control the pressure in the reactor we could

use valve Vl or valve V2 with simple control configurations or both

valves in a split-range control configuration. Which of the three is

better? Why?

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CHAPTER 21

FEEDFORWARD AND RATIO CONTROL

Feedback control loops can never achieve perfect control of a chemical

process, i.e. keep the output of the process continuously at the desired

set point value in the presence of load or set point changes. The reason is

simple, a feedback controller reacts only after it has detected a deviation

in the value of the output from the desired set point.

Unlike the feedback systems, a feedforward control configuration

measures the disturbance (load) directly and takes control action to

eliminate its impact

have the theoretical

In this section

systems and describe

last section we will

ratio control.

I 6

on the process output. Therefore, feedforward controllers

potential for perfect control.

we will study the characteristics of feedforward control

the techniques which are used for their design. In the

examine a' special case of feedforward control, the

21.1 THE LOGIC OF FEEDFORWARD CONTROL

Consider the stirred tank heater shown in Figure 21.la. The control

objective is to keep the temperature of the liquid in the tank at a desired

value (set point) despite any changes in the temperature of the inlet

stream. Figure 21.lb shows the conventional feedback loop which measures the

llrsperature in the tank and after comparing it with the desired value

increases or decreases the steam pressure, thus providing more or less heat

-5nto the liquid. A feedforward control system uses a different approach. It

msures the temperature of the inlet stream (disturbance) and adjusts

appropriately the steam pressure (manipulated variable). Thus, it increases

the steam pressure if the inlet temperature decreases and decreases the steam

pressure when the inlet temperature goes up. Figure 21.1~ shows the feed-

forward control configuration.

In Figure 21.2a we can see the general form of a feedforward control

system. It measures the disturbances directly and then it anticipates the

effect that it will have on the process output. Subsequently, it changes the

manipulated variables by such amount as to eliminate completely the impact of

the disturbances on the process output (controlled variable). Control actionj

starts immediately after a change in the disturbance(s) has been detected.

In Figure 21.2b we have repeated the schematic of a typical feedback loop so

that the reader can contrast directly:the two control systems. It is clear

that feedback acts after the fact in a compensatory manner while feedforward

acts beforehand in an anticipatory manner.

Let us now look at some common feedforward control systems used in

chemical processes.

Example 21.1 - Feedforward Control of Various Processing Units

(a) Feedforward control of a heat exchanger, shown in Figure 21.3a. The

objective is to keep the exit temperature of the liquid constant by

manipulating the steam pressure. Two are the main disturbances

(loads) measured for feedforward control, liquid flowrate and liquid

inlet temperature.

,cb) Feedforward control of a drum boiler, shown in Figure 21.3b. Here,

the objective is to keep the liquid level in the drum constant. The

two disturbances are the steam flow from the boiler which is dictated

by varying demand elsewhere in the plant and the flow of the feed-

water. The last is also the main manipulation.

(c) Feedforward control of a distillation column, shown in Figure 21.3~.

The two disturbances here are the feed flowrate (F) and composition (C).

The available manipulated variables are the steam pressure in the reboiler,

and the reflux ratio. The composition of overhead or bottom product

is the control objective. Feedforward control is particularly useful

for a distillation column, because its response time can be measured

in hours leading to large amounts of off-spec products. - -

(d) Feedforward control of a CSTR, shown in Figure 20.3d. Inlet concentration

and temperature are the two disturbances while the product withdrawal

flow-rate and,'the coolant flowrate are the two manipulations. The con-

trol objectives are two, maintain constant temperature and composition

within the CSTR.

Remarks: (1) As the examples above have indicated, feedforward control

systems can be developed for more than one disturbance.

The controller acts according to which disturbance changed

value. Therefore, the schematic of Figure 21.2a with

several disturbances represents the general case of feed-

forward control with a single controlled variable.

(2) The feedforward control of a CSTR, in Example 21.1, indi-

cates that the extension to systems with multiple controlled

variables should be rather straightforward.

(3) With the exception of the controller, all the other hard-

ware elements in a feedforward loop are the same as for

a feedback loop (sensors, transducers, transmitters, final

control elements).

21.2 THE PROBLEM OF DESIGNING FEEDFORWARD CONTROLLERS

The question that arises is: How do we design feedforward controllers?

The reader may have suspected already that conventional P, PI or PID con-

trollers will not be appropriate. Let us start with an example; the design

of feedforward controllers for a stirred tank heater.

Example 21.2 - Designing Feedforward Controllers for a Stirred Tank Heater

In Example 4.4 we developed the dynamic mass and energy balances for

the stirred tank heater of Figure 21.la. They are given by eqns. (4.4a) and

(4.5b).''/i

A dh =dt

Fi

_ - F

Ah =dt = Fi(Ti - T) + Q/PC, (4.5b)

Assume that Fi does not change and that Fi = F. Then dh/dt = 0 and we

have only the heat balance, eqn. (4.5b). The inlet temperature Ti is the

disturbance and the amount of heat supplied by steam, Q, is the manipulated

variable. The control objective is to keep the liquid temperature, T, at the

desired set point value, TSP.

(a) Steady-state feedforward controller. The simplest form of feedforward

controller can be developed if we consider the steady state heat

balance, i.e.

0 = Fi(Ti - T) + Q/PC,

or

T = Ti++i Ocp

(21.1)

From eqn. (21.1) we find that in order to keep T = TSp, the manipulated

variable Q should change according to the equation

Q = Fi pc (TP SP - Ti> (21.2)

Equation (21.2) is the design equation for the steady state feedforward

controller. It shows how Q should change in the presence of dis-

turbance or set point changes. Figure 21.4a depicts the resulting

control system.

The steady state feedforward controller will always achieve the

desired steady

steady state.

sient.

state performance of the heater, i.e. T = TSP at

This will not be true, in general, during the tran-.,

(b) Dynamic feedforward controller. :To improve the response during the

transient we will design a feedforward controller using the dynamic

heat balance and not its equivalent steady state, as above. Equation

(4.5b) can be written as follows:

Fi FiFT=TT~+&-

d

where V = ah = liquid volume in the tank. Put eqn. (21.3) into a

form with deviation variables and take:

dT' Fi Fidt + VT' = ?T; + Q'vpc

P

Take the Laplace transforms of eqn. (21.3a):

T;(s)

T’(s) = x + 1

Fi PCP

(21.3)

(21.3a)

(21.4)

where T = V/Fi = retention time of liquid in the tank. The feedforward

controller should make sure that T'(s) = T;,(s) = set point, despite any

changes in the disturbance T!I' or set point THPS Therefore, from

eqn. (21.4) we find that Q should be given by,

S’(s) = Fi p~~[(rs+l)T;~(s) - T;(s)] (21.5)

Equation (21.5) is the design equation for the dynamic feedforward

controller and Figure 21.4b depicts the resulting control mechanism.

As it can be seen from Figures 21.4a and 21.4b, the only difference

between the steady state and dynamic feedforward controllers for the

tank heater is the transfer function (rs+l) multiplying the set

point. Therefore, we expect that for load (disturbance) changes the

two controllers will be equivalent. On the contrary, dynamic feed-

forward control will be better for set point changes. Figures 21.5a

and 21.5b verify this point.

Example 21.2 has pointed out a very essential characteristic in feedforward

control:

"The design of a feedforward controller comes out directly from themodel of a process."

Thus, the steady state design came out from the steady state heat balance,

and the dynamic controller from the dynamic heat balance. It is obvious

that,

"the better a model represents the behavior of a process, the betterthe resulting feedforward controller will be."

Let us now generalize the design procedure outlined in Example 21.2.

Consider the block diagram of an uncontrolled process (Figure 21.6a).

Zbc process output is given by,

Y(s) = Gp(s)iii(s) + Gd(s)&s) (21.6)

Let y,,(s) be the desired set point for the process output. Then, eqn. (21.6)

fm y(s) = B,,(s) yields:

7+) = Gp(s)iii(s) + Gd(S);i(S) (21.7)

We can solve eqn. (21.7) with respect to G(s) and find the value that the

manipulated variable should have in order to keep y(s) = y,,(s), in the

presence of disturbance or set point changes. Then, we take;

E(s) = (21.8)

Equation (21.8) determines the form that the feedforward control system should

have and which is shown in Figure 21.6b. It also determines the two transfer

functions, G, and GSP, which complete the design of the control mechanism,

i.e.

Gc(s)=Gd(s)/Go(s) -transfer function ofthemain feedforward controller

’ GSp(4 =

Remarks: (1)

(2)

(3)

r

(21.9)

l/Gd(s) f transfer function of the set point element (21.10)

From Figure 21.6a we notice that the feedforward loop retains

all the external characteristics of a feedback loop. Thus,

it has a main measurement which is compared to a set point

signal and the result of the comparison is the actuating

signal for the main controller. In substance though, the

two control systems differ significantly as it was pointed

out in Section 21.1.

From the design eqns. (21.9) and (21.10) it is clear that

a feedforward controller cannot be a conventional feedback

controller (P, PI, PID). Instead, it should be viewed as

a special purpose computation machine. This is the reason

that sometimes it is referred to as feedforward computer.

The design eqns. (21.9) and (21.10) demonstrate again that

feedforward control depends heavily on the good knowledge

of the process model (G p, Gd>* Perfect control necessitates

perfect knowledge of G andP

Gd, which is not practically

possible. This is the main drawback of feedforward con-

trol.

(4) In the control system of Figure 21.6b we left out the sen-

sor which measures the disturbance and the final control

element. The inclusion of these two elements alters the

design of the transfer functions, Gc(s) and GSP(s).

Consider the more general feedforward control system of

Figure 21.6c, including the measuring sensor and the final

control element. We can easily show that,

TX GpGVGcGs’;~~SP $ [Gd - GpGvGcGm]i (21.11)

The design transfer functions Gc and GSP can now be

identified by the following two requirements:

(c) Disturbance rejection. The controller should be capable to eliminate

completely the impact of a disturbance change on the process output.

This implies that the coefficient of 8 in eqn. (21.11) should be

zero, i.e.

Gd - G G G G =Op v c m

or

GC = Gd/GpGvGm (21.12)

(d) Set-point tracking. The control mechanism should be capable to make

the process output track exactly any changes in the set point, i.e.

keep 7 = ysP. This implies that the coefficient of y,P in eqn.

(21.11) should be equal to 1, i.e.

GGGG = 1pvcSP

GpGv(Gd/GpGvGm)Gsp = 1

5-17

and finally,

GSP = Gm/Gd (21.13)

Equations (21.12) and (21.13) are more general than (21.9) and (21.10),

with the latter resulting from the former for Gm = Gv = 1.

21.3 PRACTICAL ASPECTS ON THE DESIGN OF FEEDFORWARD CONTROLLERS

The design eqns. (21.9) and (21.10) or their more general counterparts

(21.12) and (21.13) indicate that the feedforward controller will be a special

purpose computational machine. Its practical implementation is rather easy if

we use a digital computer as the controller (see Part VII), but for analogi

controllers it is rather difficult and expensive to build these special purpose

machines. In this section we will examine some simplifications which lead to

practical implementations of the feedforward control concept.

To simplify the presentation, let us first assume that Gm = Gv = 1.

Then, eqns. (21.9) and (21.10) will be the basis of the controller design.

Each of the two process transfer functions, Gp(s) and Gd(s), ahs two elements;

(i) the static element whit corresponds to the static gain and (ii) the

purely dynamic element which is a function of s. Thus,

GpW = Kp*G;(s) and G&d = Kd*G;(s)

For instance, in Example 21.2, for the stirred tank heater we

.identify the static and dynamic parts of the process transfer

"sp. (21. 4) ) ;

can easily

functions (see

GpW = F ‘,c1

‘Gi indicates that K = 1

i P P Fi PCand G;(s) = -&-Ts+l

P

Gd(d = & indicates that Kd = 1 and G;(s) = & .

A. Design of Steady State Feedforward Controllers

The simplest feedforward controller and the easiest to implement is the

steady state. As it was demonstrated in Example 21.2, we use simple steady

state balances for design. How does this modify the design eqns, (21.9) and

(21.10)?

At steady state, we retain only the static elements of the process

transfer functions, since Gl;(s=O.) = Gi(s=O) = 0 (see Section 1. Thus,

GP = K and

P Gd = Kd

Then, the design transfer functions, Gc -and GSp, are given by,

GC = Kd/KP (21.14a)

and

GSP = l/Kd (21.14b)

i.e. they are simple constant. Therefore, the elements Gc and GSP can be

constructed easily in the same way as a proportional controller, which has

only the proportional gain. This the reason that the design elements given

by eqns. ((21.14a) and (21.14b)) are called gain-only elements.

B, Design of Simple Dynamic Feedforward Controllers

Instead of using the exact transfer functions, Gp(s) and Gd(s), it is

possible to use approximations to them and still obtain very good results.

athough they are approximations, they are expected to give improved results

:uv.er the steady state feedforward controller.

Consider that Gp(s) and Gd(s) are approximated by first-order lags.

Then,

GcW =Gd(d

GpW=

1as+1

1Bs+l

= Bs+las+1

(21.15a)

and

1GSp(s) = Go = as + 1 (21.15b)

The controller given by eqn. (21.15a) is called lag-lead element because

(Bs+l) intorduces phase lead and the l/as+1 adds phase lag. a and B are

adjustable parameters for the controller. For the set point element GSP(s)

eqn. (21.15b) indicates that we should use a lead element.

The lag-lead element is the most commonly used in dynamic feedforward

control. It is quite versatile because the two adjustable parameters a,B

allow it to be used as lead element, when“a is very small, or lag element-

when B is very small. Finally, lag-lead elements can be bought easily and

they are not expensive like special purpose analog computational devices.

Example 21.3 - Designing Feedforward Controller for a CSTR

Consider the CSTR system described in Example 4.10. In Example 9.2 we

developed the transfer functions for the linearized model of the system;

Z$s) =bl(s+a22)

cd, (s)a12bl a12b2

P(s) i- p(s> q(s) - p(s> T;(s) (9.15a)

T’ (s) a21bl -,= p(s> 'Ai +

bl(s+all)q(s) +

b2(s+all)

P(s) P(s) Tp) (9.15b)

..where

P(s) z s2 + (a11 + a22)S + (alla22 - a12a21)

For the definition of the constant parameters; all, a12, a21, a22 and bl, b2,

see Example 9.2. All variables are in deviation form.

Let US examine two different control problems and develop the necessary

feedforward control systems:

Problem 1. Control the concentration CA in the presence of changes in the

inlet concentration and temperature. The temperature of the coolant, Tc, is

the manipulated variable. Since we have two disturbances, we need two distinct

feedforward controllers. To develop the design equations for the two con-

trollers put in eqn. (9.15a) Ei(s) = 0. Then, we take:

blP(s) = - bla12b2

(s + a22)Eii(s) - 5 Tf(s) (21.16)

Equation (21.16) indicates that the first controller is a lead element while

the second is a gain-only element. The resulting feedforward system is shown4

in Figure 21.7a. 8

Problem 2. Control the temperature T considering CA and Ti as the two

disturbances, and Tc the manipulated variable. SettiZg T'(s) = 0, eqn.

(9.15b) yields

a21blF(s) = - . 1

b2- C's+all Ai (s)

' bl*- F T;(s)

2(21.17)

Equation (21.17) shows that the first controller is a lag element while the

second is a gain-only element. The resulting feedforward system is shown in

Figure 21.7b.

21.4 FEEDFORWARD-FEEDBACK CONTROL

Feedforward control has the potential for perfect control but it also

suffers from several inherent weaknesses. In particular;

--it requires the identification of all possible disturbances and their direct

-%easurement, something which may not be possible for many processes;

- any changes in the parameters of a process (e.g. deactivation of a catalyst

with time, reduction of a heat transfer coefficient due to fouling, etc.)

cannot be compensated by a feedforward controller because their impact

cannot be detected;

- feedforward control requires a very good model for the process which for

many systems in chemical industry is not possible.

On the other hand, feedback control is rather insensitive to all three of the

above drawbacks but it has poor performance for a number of systems (multi-

capacity, dead time, tee.), and raises questions of closed-loop stability.

Table 21.1 summarizes the relative advantages and disadvantages of the two

control systems.

We would expect that a combined feedforward-feedback control system will

retain the superior performance of the first and the insensitivity of the

second to uncertainties and inaccuraci'es in disturbance identification and

measurement parameter changes, and exactness in the process model. Indeed,

any deviations caused by the various weaknesses of the feedforward control

will be corrected by the feedback controller. This is possible because a

feedback control loop monitors directly the behavior of the controlled pro-

cess (measures process output). Figure 21.8 shows the configuration of a

combined feedforward-feedback control system.

Let us now develop an equation for the closed-loop response of feed-

forward-feedback system of Figure 21.8. First of all recall that (we have

dropped the argument s to simplify the presentation)

7 = GPiii + Gda

,Zhe value of the manipulated variable is given by

iii = cvc = Cv(E1 + E2) = Cv Gc+ CvGc 2 =2

5 = Cv Gc (71 sp

- G - G ii>ml

j?> + Cv Gc (G2 SP 'SP m2

(21.6)

(21.18) .

Replace iii in eqn. (21.6) by its equal given from eqn. (21.18) and after

algebraic rearrangements take:

GpGV(G +G G

ji = c1 c2 SP ) Gd - G G G GP v c2 m2

l+GGG GP v cl ml

%P + l+GGG GP v cl ml

5.22

ii (21.19)

A close examination of eqn. (21.19), which yields the closed-loop process out-

put under feedforward-feedback control, reveals the following characteristics:

(1) The stability of the closed-loop response is determined by the roots of

the characteristic equation

l+GGGG = 0P v cl ml

which depends on the transfer functions of the feedback loop only.

Therefore,

"the stability characteristics of a feedback system will notchange with the addition of a feedforward loop."

(2) The transfer functions of the feedforward loop, G and Gc2

sp, will

be given by the design eqns. (21.12) and (21.13), i.e.

= G/GGG 'Gd pvm2

andc2 GSP = G /Gd

m2

If GP' Gd' Go' Gm2

are known exactly, then the feedforward loop

compensates completely for disturbance or set point changes and the

feedback action remains idel since El = 0.

(3) If any of the Gp, Gd, Gv, G, is known only approximately, then

Gd - G G G G #OP v c2 m2

and/or GpGvGc2GSP # 1

In such case the feedforward loop does not provide perfect control,

i.e. 'i + Ysp. Then, El # 0 and the feedback loop is activated and

offers the necessary compensation.

Example 21.4 - Feedforward-Feedback Control of the Tank Heater

Consider again the tank heater of Example 21.2. Under feedforward con-

trol only we have the configuration shown in Figure 21.4b. The design

transfer functions are:

Gc = FipcP

and Gsp = -rs+l

Assume that the density p or the heat capacity cP

are not known exactly.

Then, the feedforward loop does not provide for perfect control. Figure 21.9a

shows the temperature in the tank after a step change in the inlet temperature.

Notice the remaining deviation.

Introduce now in the system a feedback loop with PI controller (Figure

21.9b). In Figure 21.9a we have plotted again the temperature of the liquid

in the tank, for the same step change in the inlet temperature. Notice the

deviation has disappeared.

21.5 RATIO CONTROL

Ratio control is a special type of feedforward control where two dis-

turbances (loads) are measured and held in a constant ratio to each other. It

is mostly used to control the ratio of flowrates of two streams. Both flow-

rates are measured but only one can be controlled. The stream whose flowrate

is not under control is usually referred to as "wild" stream.

Figures 21.10a and 21.10b show two different ratio control configurations

for two streams, Stream A is the “wild" stream.

- ;En configuration 1 (Figure 21.10a) we measure both flowrates and take

their ratio. This ratio is compared to the desired ratio (set point) and

the deviation (error) between the measured and desired ratios constitutes

the actuating signal for the ratio controller.

- In configuration 2 (Figure 21.10b) we measure the flowrate of the "wild"

stream A and multiply it by the desired ratio. The result is the flow-

rate that the stream B should have and constitutes the set point value

which is compared to the measured flowrate of stream B. The deviation

constitutes the actuating signal for the controller, which adjusts

appropriately the flow of B.

Ratio control is used extensively in chemical processes with the fol-

lowing as the most commonly encountered examples:

(1) Keeping a constant ratio between the feed flowrate and the steam in the

reboiler of a distillation column (Figure 21.11a).

(2) Holding constant the reflux ratio in a distillation column (Figure 21.11b).

(3) Controlling the ratio of two reactants, entering a reactor, at a desired

level (Figure 21.11~). : 6

(4) Holding the ratio of two blended streams constant, in order to maintain

the composition of the‘blend at the desired value.

(5) Holding the ratio of a purge stream to the recycle stream constant

(Figure 21.11d).

'(6) Keeping the ratio of fuel/air in a burner at its optimum value (most

efficient combustion).

(7) Maintaining the ratio of the liquid flowrate to vapor flowrate in an

absorber constant, in order to achieve the desired composition in the

exit vapor stream. Figure 21.11e shows such a ratio control system in

a more reliable cascade control configuration. The secondary loop of

the cascade improves the response of the system by regulating better

the flow of the liquid feed.

,SuMMARY AND CONCLUDING REMARKS

A feedforward control system measures the value of a disturbance and

aicipates what its effect will be on the controlled output of a process.

Then, the controller changes the manipulated variable by such amount as to

eliminate the impact that the disturbance would have on the output. Such an

approach is substantially different from that of a feedback system. Its major

advantage is the ability to act beforehand, while its most serious drawback is

its heavy dependence on the process model, which is almost always incompletely

known. Additional weaknesses are; (i) the requirement to know and measure all

possible disturbances and (ii) the inability to cope with changing process

parameters. For all these reasons feedforward control is used with simple and

well known processes and almost always with feedback compensation. Processes

which benefit the most from feedforward control are those with very slow- - ._.

response (usually multicapacity) and/or significant dead time. .

Feedforward control, when appended with a feedback loop, offers sig-

nificant improvements, the response and robustness over pure feedforward or

feedback control. The combination of the two will be encountered quite often

in chemical processes.

Ratio control is a special purpose f.eedforward control system. Primarily,

it is used to control the ratio of the flowrate of a "wild" stream to the

floarate of a controllable stream. It is extensively used in chemical

processes.


1.

2.

3 .

4 .

5 .

6 .

7 .

8 .

9 .

18,

114

Define the concept of feedforward control on physical grounds.

Is driving a car mostly feedforward or feedback control? What about

riding a bicycle?

Discuss the relative advantages and disadvantages of feedforward and

feedback control systems. Why the addition of feedback control

improves the performance of a feedforward system? ~. .::I : .i ._..

What kinds of processes stand to benefit the most from feedforward

control? Why? .:

Do the stability characteristics of-a process change.with feedforward

control, as they do with feedback? Elaborate on your answer.

In Section 21.4 it was claimed that the stability characteristics of a

feedforward-feedback control system are affected only by the feedback

loop. Explain why.

Draw three different feedforward control configurations for the mixing

process of Example 4.11.

Under what conditions a steady state feedforward control system will

yield the same performance as a dynamic feedforward controller in

rejecting the effect of a disturbance?

What is a +ag-lead element and why is it considered to be a versatile

component for feedforward control?

Consider the feedforward control of a distillation column? What kind

of dynamic feedforward element will be needed: lag-lead, lag only,

lead only, gain only? Give a rather qualitative explanation.

What is ratio control and why is it useful in process control? Give

three specific examples.

12. How do you select the desired value of the ratio in a ratio control

system?

13. Draw the feedforward and feedback control systems which regulate the

flow through a pipe. Do you expect one of them to be significantly

better than the other in maintaining the desired flow or not?

14. In Figures 21.10a and 21.10b we see two different ratio control

configurations. Which one would you prefer and why? (Hint: Examine

the static gain of the control loop in Figure 21.10a and consult

Reference for details).

Table 21.1. Relative Advantages and Disadvantagesof Feedforward and Feedback Control

FEEDFORWARD

Advantages

1. Acts before the effect of a

disturbance has been felt by

the system.

2. Is good for slow systems

(multicapacity) or with sig-

nificant dead time.

1.

2.

/; 3.;

3. It does not introduce instability

in the closed-loop response. 4.

FEEDBACK

Advantages

1. It does not require identifi-

cation and measurement of any

disturbance.

2. It is insensitive to modeling

errors.

3h.. . It is insensitive to parameter

changes.

Disadvantages

Requires identification of all

possible disturbances and their

direct measurement.

Cannot cope with unmeasured

disturbances.

Insensitive to process parameter

variations.

Requires good knowledge of the

process model.

Disadvantages

It waits until the effect of the

disturbances has been felt by the

system, before control action is

taken.

It is unsatisfactory for slow

processes or with significant

dead time.

It may create instability in the

closed-loop response.

cc>

FEEDFO2WARD

CcNTRCLLEI:

IF‘.yure 21.3 1

53/

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533

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CHAPTER 22

ADAPTIVE AND INFERENTIAL CONTROL SYSTEW

In this chapter we will examine two additional types of control systems;

(a) adaptive, and (b) inferential. Although their basic objectives and

functions can be easily described in a qualitative manner, their practical

implementation is rather complicated, involves extensive computations, and it

is accomplished through the use of digital computers. For this reason we will

postpone and quantitative discussion on these two control systems for Part VII,

where we will study the use of digital computers for process control. There-i

fore, in this chapter we will only make a qualitative presentation of ‘each

control system, discussing its basic logic and giving examples of its practical

application.

22.1 ADAPTIVE CONTROL

Adaptive is called a control system,, which can adjust its parameters

automatically in such a way as to compensate for variations in the character-

istics of the process it controls. The various types of adaptive control

systems differ only in the way the parameters of the controller are adjusted.

But, why are adaptive controllers needed in chemical processes? Two are

the main reasons:

- Most of the chemical processes are nonlinear. Therefore, their linearized

*-:-~.;dels which are used to design linear controllers depend on the particular

steady state (around which the process is linearized). It is clear then

that as the desired steady state operation of a process changes, the "best"

values of the controller's parameters change. This implies the need for

controller adaptation. Example 10.5 demonstrates how the time constant

and process gain of a simple liquid storage tank depend on the value of

the steady state liquid level.

536

- Most of the chemical processes are nonstationary, i.e. their characteristics

change with time. Typical examples are; the decay of the catalyst activity

in a reactor and the decrease of the overall heat transfer coefficient in a

heat exchanger, due to fouling (Example 10.6). This change leads again to

a deterioration of the linear controller, which was designed using some

nominal values for the process parameters, thus, requiring adaptation of

the controller parameters.

What is the objective'of the adaptation procedure? Clearly, it is not to

keep the controlled variable at the specified set-point. This will be

accomplished by the control loop, however badly. We need an additional

criterion, an objective function which will guide the adaptation mechanism to

the "best" adjustment of the controller parameters. To phrase it differently,

we need a criterion to guide the adaptive tuning of the controller. Any of

the performance criteria we discussed in Chapters 16 and 18 could be used,

i.e.

- one-quarter decay ratio,

- integral of the square error,

- gain or phase margins, etc.

There are two different mechanisms for the adaptation of the controller

parameters.

A. Programmed or Scheduled Adaptive Control

Suppose that the process is well known and that an adequate mathematical

model for it is available. If there is an auxiliary process variable which

correlates well with the changes in process dynamics, then we can relate ahead

of time the "best" values of the controller parameters to the value of the

auxiliary process variable. Consequently, by measuring the value of the

auxiliary variable we can schedule or program the adaptation of the controller

parameters. Figure 22.1 shows the block diagram of a programmed adaptive

control system. We notice that it is composed of two loops. The inner loop

is an ordinary feedback control loop. The outer loop includes the parameter

adjustment (adaptation) mechanism and it is comparable to feedforward com-

pensation, with no feedback to compensate for incorrect adaptation. A typical

example is the so-called gain scheduling adaptive control.

Example 22.1 - Gain Scheduling Adaptive Control

In a normal feedback control loop (Figure 22.2a) the control valve or._. . i. .'. .L _ _ q .::.another of its components may exhibit?a nonlinear character.

6.. _ . . . i

In such case

the gain of the nonlinear component will depend on the current steady state.

Suppose that we want to keep the total gain of the overall system constant.

From Figure 22.2a we find easily that the overall system gain is given by

KOVERALL = K K K K = constantpmcf.

It is clear then that as the gain Kf of the nonlinear valve changes, the

gain of the controller, Kc, should change as follows

KC= constant/(KpKmKf) (22.1)

We assume that the gains KP

and Km are known exactly. Furthermore, if

the characteristics of the control valve are known well, then, its gain, Kf,

*can be calculated from the stem position. Therefore, by measuring the stem

;etion (auxiliary measurement) we can compute the Kf. Then, eqn. (22.1)

yields the adaptation mechanism of this simple gain scheduling adaptive

controller. Figure 22.2b shows the resulting control structure.

Remark: Notice that the gain scheduling is comparable to feedforward compen-

sation. There is no feedback to compensate for incorrect

adaptation.

Example 22.2 - Programmed Adapative Control of a Combustion System

Consider a burner where the fuel/air ratio is kept at its optimal value

to achieve the highest efficiency of combustion. Excess of fuel or air will

reduce the efficiency. The optimal fuel/air ratio is maintained through a

ratio control mechanism (section 21.5). The control system is shown in

. Figure 22.3a.

The otpimal value of the fuel/air ratio which maximizes the combustion

efficiency depends on the conditions prevailing within the process, e.g.

temperature of air. Consequently, as the temperature of air changes, so does-.

the optimal value of the fuel/air ratio.

From previous experimental data we know how the optimal fuel/air ratio

changes with air temperature for maximum efficiency. Therefore, in order to

maintain the ratio continuously at its optimal value despite any changes in

the air temperature, we can use a programmed adaptive control system. Such

a system is shown in Figure 22.3b. It measures the temperature of the air

(auxiliary measurement) and adjusts the value of the fuel/air ratio. Notice

again that the ratio adjustment mechanism is like feedforward compensation.

B. Self-Adaptive Control

If the process is not known well, we need to evaluate the objective

:finction on-line (while the process is operating) using the values of the

nntrolled output. Then, the adaptation mechanism will change the controller

parameters in such a way as to optimize (maximize or minimize) the value of

ttie objective function (criterion). In the following two examples we will

examine the logic of two special self-adaptive control systems; model

reference adaptive control (MRAC) and self-tuning regulator (STR).

Example 22.3 - Model-Reference Adaptive Control (MRAC)

Figure 22.4 illustrates a different way to adjust the parameters of the

controller. We postulate a reference-model which tells us how the controlled

process output ideally should respond to the command signal (set-point). The

model output is compared to the actual process output. The difference (error,

Ed) between the two outputs is used through a computer to adjust the parameters

of the controller in such a way as to minimize,the integral square error, i.e.

t

Minimize ISE =I

[em( dt0

i ./, 4The model chosen by the control designer for reference purposes is to a certain

extent arbitrary. Most often a rather simple linear model is used.

We notice that the model-reference adaptive control is composed of two

loops. The inner loop is an ordinary feedback control loop. The outer loop

includes the adaptation mechanism and looks like a feedback loop too. The

model output plays the role of the set-point while the process output is the

actual measurement. There is a comparator whose output (error, cm) is the

input of the adjustment mechanism. The key problem is to design the adaptation

mechanism in such a way as to provide a stable system, i.e. bring the error

Elll to zero. This is not a trivial problem and we will be concerned with it

in Chapter 32, i.e. after we have studied the use of digital computers for

fdTp->cess control.

Example 22.4 - Self-Tuning Regulator (STR)

Consider the block diagram of Figure 22.5. It represents the structure

of a self-tuning regulator, which constitutes another way for adjusting the

parameters of a controller.

The STR is composed, again, of two loops. The inner loop consists of the

process and an ordinary linear feedback controller. The outer loop is used to

adjust the parameters of the feedback controller and is composed of, (a) a

recursive parameter

troller parameters.

estimator and (b) an adjustment mechanism for the con-

The parameter estimator assumes a simple linear model for the process,

e.g.

Kpe-tds

Ts+l

Then, using measured values for the manipulated variable, m, and the controlled

output, Y, it estimates the values of the parameters K, T and td, employing

a least-squares estimation technique. Once the values of the process parameters

a and b are known, the adjustment mechanism can find the "best" values for

the controller parameters using various design criteria like,

- phase or gain margins,

- integral of the squared error, etc.

Both the parameter estimator and the adjustment mechanism require

involved computations. For this reason the STR can be implemented only

through the use of digital computers.

Adaptive control systems have been applied in chemical processes. The

range of their applicability has expanded with the introduction of digital

computers for process control. Several theoretical and experimental studies

have appeared in the chemical engineering literature, while the number of

industrial adaptive control mechanisms increases continuously. Most of the

adaptive control systems require extensive computations for parameter esti-

mation and optimal adjustment of controller parameters which can be performed

on-line only by digital computers. Therefore, we will delay any discussion on

the quantitative design of such systems until Chapter 32, i.e. after we have

studied the use of digital computers for control.

22.2 INFERENTIAL CONTROL

Quite often, the controlled output of a processing unit cannot be

measured directly. Consequently, we cannot use feedback control or any other

configuration which necessitates the direct measurement of the controlled

variable. If the disturbances which create the control problems can be

measured and an adequate process model is available, then we could use feed-

forward control to keep the unmeasured output at its desired value (see

Chapter 21).

What happens though if the disturbances cannot be measured? None of the

control configurations studied so far can be used to control an unmeasured

process output in the presence of unmeasured disturbances. This is the type

of control problems where inferential control is the only solution. Let us

now examine the structure of an inferential control system.

Consider the block diagram of the process shown in Figure 22.6a, with

one unmeasured controlled output (y) and one secondary measured output (2).

The manipulated variable m and the disturbance d affect both outputs. The

Wurbance is considered to be unmeasured. The transfer functions in the

&ock diagram indicate the relationships between the various inputs and out-

puts, and they are considered to be perfectly known.

From Figure 22.6a we can easily derive the following input-output

reliationships;

7 = G -iii +p1

Gd l d� (22.2)1

? = G l fi + Gd l iip2

(22.3)2

From eqn. (22.3) we can solve with respect to d and find the following

estimate of the unmeasured disturbance,

(22.4)

Substitute the above estimate into eqn. (22.2) and find the following

relationship,

-f. ., .-:y ..__ Gd

7 = bpl-2GpJ; + <;

-ii.-. :: ,:: 8 i:,l;

(22.5)

Equation (22.5) provides the needed estimator which relates the unmeasured

controlled output to measured quantities like m and z. Figure 22.6b

shows the structure of the resulting inferential control system. Notice that

the estimated value of the unmeasured output plays the same role as a regular

measured output, i.e. it is compared to the desired set-point and the dif-

ference is the actuating signal for the controller. Figure 22.7 shows a

simplified diagram of a typical inferential control system.

Remarks: (1) It is important to notice that the success of an inferential

control scheme depends heavily on the availability of a

good estimator, which in turn depends on how well we know

the process. Thus, if the process transfer functions, Gp1'

G' Gd and

p2 1 Gd2are perfectly known, then a perfect

estimator can be constructed and consequently we will have

perfect control. When the process transfer functions are only

approximately known (which is usually the case), then the

inferential scheme provides control of varying quality

depending on how well the process is known.

(2) In chemical process control the variable which is most com-

monly inferred from secondary measurements is composition.

This is due to the lack of reliable, rapid and economical

measuring devices for a wide spectrum of chemical systems.

Thus, inferential control may be used for the control of

chemical reactors, distillation columns and other mass

transfer operations like driers, absorbers, etc. Temperature

is the most common secondary measurement, used to infer the

unmeasured composition.

‘.-, ., -. ,; ,.

Example 22.5 - Inferential Control of a Distillation Column

Consider a distillation column with 16 trays, which separates a mixture

of propane-butane into two products. The feed composition is the unmeasured

disturbance and the control objective is to maintain the overhead product

molar composition 95% in propane. The reflux ratio is the manipulated

variable.

Since the feed and overhead compositions are considered unmeasured, we

can only use inferential control. The secondary measurement employed to

infer the overhead composition is the temperature at the top tray. Let us

now examine how we can develop and design the inferential control mechanism.

The process as defined above has two inputs and the two putputs, i.e.

- inputs; feed composition (disturbance), reflux ratio (manipulation),

- outputs; overhead propane composition (unmeasured controlled output) and

temperature of top tray (secondary measurement).

How can we identify the four process transfer functions? In Example 4.13 we

saw that a rigorous approach leads to an overwhelming mathematical model.

The process reaction curve method, which was ciscussed in Section 16.4, is a

simpler approach and yields the transfer functions between the various inputs

and outputs. Following this methodology, we developed the input-output

relationships (see also Figure 22.8a):

y(s) 2 0.90*e-2S ,;ics) + 1.20.e-lS ...(s)70s + 1 30s + 1

= 0.20 eB2’ ;i. +Z(s) - 60s + 1

Having developed the four process transfer functions it is easy to design

the inferential control system (Figure 22.8b).

Remarks: (1) The temperature of the top tray was selected arbitrarily to

be the secondary measurement. But why did we not select

the temperature of the second or third, etc. tray from the

top? The answer is rather complex and will be given in

Chapter 32.

(2) Were we to control the purity of the bottoms product, a

different temperature would be needed. Most likely would be

close to the bottom of the column.

(3) Recall that the effectiveness of an inferential control

scheme depends heavily on the goodness of the estimator,

which in turn depends on the model which is available for

the process. Assume that the overhead composition can be

measured intermittently, either by taking samples manually

and analyzing them or even better using on-line a gas

chromatograph. From the composition measurements we can

take the useful information needed to judge how effective

has been the inferential control. Thus, if the measured

steady state value of the overhead composition deviates

significantly from the desired set-point value, we can use

the deviation (error) through an adaptive mechanism to

correct the estimator. The resulting system in shown in

Figure 22.3. .-


Adaptive and inferential control schemes are gaining importance and wider

acceptability in the chemical process industry. Both depend on good process

models and for their implementation require extensive computations which are

possible only through the use of modern digital computers. Therefore, in

Chapter 32 we will return to these two schemes for further design developments.

Adaptive control is dictated by the nonlinearity and nonstationarity of

chemical processes. Its objective is to adjust the controller parameters in

such a way as to compensate for variations in the process itself. Depending

on how the controller parameters are adjusted we may have programmed or self-

adaptive mechanisms. The former is used in processes which are known well,

while the latter for processes which are poorly known. Gain scheduling is

the most characteristic example of the programmed adaptive control, while

model-reference adaptive control and self-tuning regulators are the most

tppical configurations of self-adaptive control schemes. With the use of

digital computers the range for adaptive control applications has expanded.

Inferential control is the only course of action if we want to regulate

.an unmeasured output in the presence of unmeasured disturbances. A secondary

process output is measured and from it the value of the unmeasured controlled

output is inferred. The effectiveness of inferential control depends very

heavily on the availability of good process models. If the "unmeasured"

controlled output can be measured intermittently, then an adapative scheme can

be constructed which corrects the estimator of the inferential loop. Compo-

sition is the most commonly inferred process variable and temperature the

usual secondary measurement.

With this chapter we close the series of advanced control systems of

Part V. It should be noted that all control configurations studied in the

last four chapters possess a single manipulation for the regulation of a

single controlled output. Thus, we have single-input, single-output (SISO)

systems. But, most of the processing units in a chemical plant have more than

one controlledoutputs,requiring more than one manipulated variables. This

leads to systems with multiple-inputs and multiple-outputs (MIMO). How do we

design control systems for such processes? This will be the subject of the

chapters in Part VI.


1. What is adaptive control and why is it needed in chemical process control?

2. Give two examples of adaptive control for processing units, different

than those described in Section 22.1. Describe qualitatively the

functions of the adaptive control schemes you proposed.

3. What is programmed adaptive and what is self-adaptive control? Give

one example of each, different than those in Section 22.1. When would

you recommend the programmed and when the self-adaptive scheme?

4 . What is gain scheduling control and why can you use it in chemical

process control? It was claimed in Example 22.1 that it resembles to

feedforward compensation. Explain why. What are its advantages and

disadvantages?

5. Discuss the logic of model-reference adapative control and self-tuning

regulator. Find the similarities and differences between the two

configurations.

547

6. Show qualitatively that the structure of a self-tuning regulator can be

derived from that of a model-reference adapative control, if the

parameter estimation is done by updating the reference model.

7. The outer loops in the MRAC and STR configurations, are they of feed-

forward or feedback nature?

8. Consider the neutralization with a caustic of an acidic effluent waste

from a chemical plant. The titration curve of the waste being neu--.

-'tralized is nonlinear and changes with time due to unmeasured_.

disturbances. Develop a qualitative self-adaptive control scheme and

describe the functions of its components..: (You can consult Reference

.>

9.’ As it was discussed in Section 22.1, the purpose of an adaptive con-

troller is not to keep the controlled output at its desired set point.

This is accomplished by the regular feedback loop. What is then the

criterion or the objective function ,that guides the parameter

adjustment of an adaptation mechanism? How is this objective evaluated

by the programmed or self-adaptive schemes.

10. As it was discussed in Chapter 21, the effectiveness of a feedforward

control loop depends heavily on the quality of the model which is

available for the process. Develop an adaptive control mechanism for

a feedforward controller which will compensate for any process

variations.

11. Discuss the logic of an inferential control scheme, Why is this control

scheme needed? Describe two examples of inferential control different

than those of Section 22.2.

12, What do we mean when we say that a process variable is "unmeasured"?

13. Consider two processes; one (process A) slow with time constant 5 hours

and another (process B) faster with time constant 1 hour. The compo-

sition of the output streams from the two processes is measured every

2-3 hours. Which of the two process outputs can be controlled by con-

ventional feedback and which one will require inferential control?

14. Show that the inferential control employed for process A or B in item

12 (above) can be improved through an adaptive mechanism which uses the

direct composition measurement every 2-3 hours. (Consult Example 22.5.)

15. Develop an inferential control scheme which can be used to control the

drying of solids with warm air. Discuss how would you develop the

estimator of the inferential structure. (Consult a reference on solids

drying with air and Reference .)

16. If, in addition to the unmeasured disturbance there are measured dis-

turbances in a system, we can develop a combined inferential-feedforward

configuration. Develop such a configuration for a system of your choice.

1ME,+SVRlNG SENSoR

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PART VI

DESIGN OF CONTROL SYSTEMS FOR MDLTIVARIABLE

PROCESSES. INTRODUCTION TO PLANT CONTROL.

The control configurations we have examined so far were confined to pro-

cesses with a single controlled output, requiring a single manipulated input.

Such single-input, single-output (SISO) systems are very simple and not the

typical processing units encountered by a chemical engineer. Chemical pro-

cesses usually have two-or more controlled outputs, requiring two or more.

manipulated variables. The design of control systems for such multiple-input,

multiple-output (MIMO) processes will be the subject of the three chapters in

Part VI of this text. In particular;

- Chapter 23 will discuss the new questions which must be answered for the

controller design of MIMO systems. It will also present a methodology

for the development of alternative control configurations for such systems,

based on their degrees of freedom.

- In Chapter 24 we will examine the selection of the appropriate measurements

rind manipulations in order to "close the loops". Furthermore, we will

study the design of "decoupled loops" for MIMO systems.

- Finally, in Chapter 25 we will present an introduction to the design of

control systems for complete plants, which constitute the most complex MIMO

systems to be encountered by a chemical engineer.

CHAPTER 23

SYNTHESIS OF ALTERNATIVE CONTROL CONFIGURATIONS

FOR MULTIPLE-INPUT, MULTIPLE-OUTPUT PROCESSES

The presence of multiple controlled outputs and multiple manipulated

inputs creates a situation which we have not confronted so far, namely; there

are more than one possible control configurations for a MIMO process. In this

chapter we will develop a concise methodology for the development of all

feasible control systems for a single processing unit or processes composed

of more than one interacting unit.

23.1 DESIGN QUESTIONS FOR MIMO CONTROL SYSTEMS

Consider a general process with several inputs and outputs (Figure 2.1).

There are several questions which must be answered before we attempt the design

of a control system for such a process.

(a) What are the control objectives? In other words, how many and which

ones of all possible variables should be controlled at desired values?

This seemingly simple question is quite critical for the design of

efficient control systems.

(b) What outputs should be measured? Once the control objectives have been

identified we need to select the necessary measurements in order to

monitor the operation of the process. We can classify the measured

outputs into two categories:

- Primary measurements; these are the controlled outputs through which

we can determine directly if the control objectives are satisfied

or not.

- Secondary measurements; these are not used to monitor directly the

control objectives but are auxiliary measurements employed for

cascade, adaptive or inferential control (see Figures 20.2,

22.1, 22.7).

(c). What inputs can be measured? We assume that all of the manipulated

variables are measurable and therefore can be employed for adaptive

(model-reference or self-tuning regulator) and inferential control

(see Figures 22.5, 22.7). With respect to the disturbances only a

few can be measured easily, rapidly and reliably. These measurable

disturbances can,be used to construct feedforward (Figure 21.3),

feedforward-feedback (Figure 21.8) and ratio control configurations

(Figure 21.10).

(d) What manipulated variables should be used? A multiple input,

multiple-output system possesses several manipulated variables which

can be used for the design of a control system. The selection of

the most appropriate manipulations is a very critical problem and

should be approached with care. Some manipulations have a direct,

fast and strong effect on the controlled outputs, some others do not.

Furthermore, some variables are easy to manipulate in real life

(e.g. liquid flows), some others are not (e.g. flow of solids,

slurries, etc.).

(e) What is the configuration of the control loops? Once all the possible

measurements and manipulations have been identified, we need to decide

how they are going to be interconnected through the control loops. In

other words, what measurement will actuate a given manipulated variable

or what manipulation will be used to regulate a given controlled output

at its desired value?

For MIMO systems there is a large number of alternative control configurations.

The selection of the most appropriate is the central and critical question to

be resolved.

Let us now examine the above design questions in more detail and develop

systematic approaches to answer them.

23.2 DEGREES OF FREEDOM AND THE NUMBER OF CONTROLLED AND MANIPULATED VARIABLES

We have defined the degrees of freedom for a given process (see Section

5.2) as the independent variables which must be specified in order to define

the process completely. The number of degrees of freedom was also found to be

given by the following equation,

f = V - E (23.1)

where V = number of independent variables describing a process and

E = number of independent equations physically relating the V

variables.

It is clear that in order to have a completely determined process the

number of its degrees of freedom should be zero. There are two sources which

provide the additional equations needed to reduce the number of degrees of

freedom to zero.

(a) The external world which specifies the values of certain input variables.

As external world we mean everything outside the process like;

- the general surroundings influencing the operating conditions, or

- a previous unit which feed the process, or

- a following unit when the outflow of the process is a manipulated inflow

.~dor the following unit.

(b) The control system which imposes certain relationships between the con-

trolled outputs and the manipulated inputs (feedback) or between the measured

disturbances and the manipulated inputs (feedforward). Thus, we can state

easily that,

"the maximum number of independent controlled variables in aprocessing system is equal to the number of degrees of freedomminus the externally specified variables," i.e.

(number of control objectives) = f -(number of externally specified inputs)(23.2)

This relationship was used in Examples 5.7 and 5.8 in order to determine the

number of controlled outputs in a binary distillation and a mixing process,

respectively.

Having determined the number of independent controlled outputs, the

following question arises: How many independent manipulated inputs do we need

in order to keep the controlled outputs at their desired values (set points)?

To answer this question, let us consider a process with the following

specifications:

- N controlled outputs (yl,y2,***,yN),

- M independent manipulations (ml,m2,***,mM) with MzN, and

- L disturbances externally specified (dl,d2,***,dL).

Let the following N equations represent the relationships between the con-

trolled outputs, the manipulations and disturbances,

y1 = flbl,m2,***,y,+ dl,d2,***,dL)

y2 = f2bl,m29**‘9~; dl,d29***,dL)

---------me--

(23.3)

YN = fN(ml~m29”‘9mM; dl,d2,*-,dL)

As the values of the disturbances change (specified by the external world),

the values of the controlled outputs must remain the same. This is possible

S N of the M manipulated variables are free to change so as to satisfy

the system of eqns. (23.3). Therefore',

"for the design of a control system the number of the requiredindependent manipulated variables is equal to the number ofindependent controlled variables," i.e.

(number of independent manipulated variables) = (number of controlled variables)

= f - (number of externally specified inputs) (23.4)

Remarks: (1)

(2)

(3)

Let k be the number of controlled variables given by

eqn. (23.2). Then, if the actually controlled variables

are fewer than k, say Il<k, then there are (k-k) process

variables which change "wildly" in an uncontrolled manner

and may cause problems to the operation of the process.

But, if the effects of these "uncontrolled" variables on

the operation of the process are acceptable, then it is

perfectly legitimate to have fewer controlled variables

than the number dictated by eqn. (23.2).

It is impossible to design a control systemi

which can

regulate more controlled variables than the number given

by eqn. (23.2).

The degrees of freedom of a process at dynamic state are

equal in number of more than those at steady state. This

is due to the fact that the, dynamic balance equations

contain the accumulation terms, while for steady state

balances the accumulation is zero. An incorrect estimate

of the number of degrees of freedom can have a profound

effect on the design of the appropriate controller. Con-

sider the simple, liquid holding tank of Example 10.1.

The dynamic mass balance yields

A dh = Fdt i

- F0

Here we have three independent variables (h, Fi, Fo) and

one equation. The cross-sectional area, A, is a parameter

with given value. Therefore, we have two degrees of

freedom. Since F, is specified by the external world,

we can have only one controlled variable. This suggests

the conventional feedback loop between h and Fo. Had

we examined the steady state balance, where dh/dt = 0, we

would have concluded (erroneously) that there is only one

degree of freedom and consequently no controlled output.

(4) Recall Examples 20.6 and 20.7 on split-range control. Notice

that the number of manipulated variables used for control

is larger than the number of controlled outputs. Therefore,

eqn. (23.4) determined the minimum number of required

manipulations. i '!1

Example 23.1 - Determining the Number of Controlled and Manipulated Variablesfor a Flash Drum

Consider the flash drum shown in Figure 23.la. The feed is composed of

N components with molar fractions 'i9 2 = 1,2,***,N. As the liquid feed is

"flashed" from the high pressure pf to the lower pressure p of the drum,

vapor is produced and reaches equilibrium with the remaining liquid. Steam

flowing through a coil supplies'the necessary heat for maintaining the desired

temperature in the drum, despite any variations in the operating conditions.

For this process we would like to identify, (a) the controlled variables,

ib) the manipulated inputs and (c) generate all feasible loop configurations.

Let us first determine the degrees of freedom for the flash drum. The

modeling equations are:

- Total mass balance (assuming constant molar density and insignificant vapor

-boldup)

dh41 dt = Ff - (FV + FL>

- Component balances

d (hxi)Ap dt = Ffzi - (FvYi + FLXi) i = 1,2,***,N-1

- Heat balance

cP,LA d(W--- = cp,fFfTf - (cp,VFVT - c~,~F~T) + UAS(TS Y T)dt

- Vapor-liquid equilibrium relationships

yi = Ki(T,p)exi .’ .- i = 1;2,i .a ,N >’

- Consistency constraints

Nii1 xi = 1 and f

y i=l yi = 13

All the above relationships constiute a system of 2N+3 equations with

4N + 14 variables. These variables are classified as follows:

Constants (N+6); A, AS, zd, cP,f' cP,v' cP,L and KI(T,p) for i = 1,2,***,N

Externally specified (N+l); Tf, TE and zi for i = 1,2,***,N-1

Unspecified (2N+7); Ff, Fv, FL, p, T, h, kg and xiY Yi for i = 1,2,***,N

Therefore, the number of controlled variables is equal to

(2N+7) - (2N+3) = 4

But, which four of the (2N+7) unspecified variables will be selected as con-

trolled outputs? The operating requirements dictate that T and p should

be kept constant in order to achieve the desired separation. Furthermore,

%nr constant production the flowrate of the liquid feed should be maintained

* ST fie desired value. Finally, the liquid level should remain within certain

bounds. Thus, T, P, Ff and h are the controlled variables. All four con-

XGoiled variables can be measured directly, using simple and reliable sensors

(thermocouples, differential pressure cells, etc.) with fast responses.

Therefore, the measured variables for the control system are; T, p, Ff and h.

From the set of (2N+7) unspecified variables we can select the required four

manipulated variables. Clearly these are; Ff, FV' FL and WS.

23.3 GENERATION OF ALTERNATIVE LOOP CONFIGURATIONS

After the identification of the controlled and manipulated variables we

need to determine the control configuration, i.e. specify the manipulated

variable which will control a given controlled variable. In other words,

determine the configuration of the control loops.

For a system with N controlled and N manipulated variables there are

N! different loop configurations. Figure 23.2 shows the two possible loop

configurations for a process with 2 manipulations and 2 controlled outputs.

As the number N increases, the number of different loop configurations

increases very rapidly, e.g.

for N=3 there are 3! = 6 different configurations



etc.

The selection of the "best" among all possible loop configurations is a

difficult problem. Various criteria can be used to couple every controlled

triable with the "best" manipulation, like;

- choose the manipulation which has a direct and fast effect on a controlled

variable;

- choose the couplings so that there is a small dead time between every

manipulation and the corresponding controlled variable;

- select the couplings so that the interaction of the control loops is

minimal, etc.

In subsequent chapters we will develop more precise quantitative criteria for

the selection of the loops.

Example 23.2 - Alternative Loop Configurations for the Flash Drum

In Example 23.1 we identified the controlled and manipulated variables

for the flash drum. But, how are these interconnected to form the control

loops? Table 23.1 shows the 24(=4!) possible loop configurations resulting

from all possible combinations among the controlled and manipulated variables.

The "best" among the 24 can be found using the following qualitative ,._

arguments:

- The effects of Ff, FV and FL on the temperature T are indirect and

rather slow, while that of WS is direct and faster. Therefore, from

the 24 loop configurations of Table 23.1, only the No. 1, 3, 7, 9, 14 and

18 look promising for efficient temperature control.

- The effects of W8 and FL on the pressure p are also indirect and

slow. Therefore, Ff and FV are better manipulated variables for con-

trolling p and from the previously selected loop configurations only the

NO. 3, 7, 9 and 14 remain valid candidates.

- Among the No. 3, 7, 9 and 14, the loop configuration No. 3 seems to be the

best because it uses FL to achieve fast level control and manipulates

Yf directly. This loop configuration is shown in Figure 23.lb.

Remarks: (1) To select the most promising control configuration for the

flash drum we employed qualitative arguments. In subsequent

sections we will study quantitative techniques for selecting

the optimal coupling between controlled and manipulated

variables.

(2) It should be emphasized that the four loops of the control

configuration in Figure 23.lb interact with each other.

Thus, increasing the steam flow-rate to control the tem-

perature will affect and thus decontrol the pressure. The

interaction among the control loops is an important design

consideration. In Chapter 24 we will examine the relative

gain array method which determines how the manipulated

variables should be coupled with the controlled variables

in such a way as to minimize the interaction among the

resulting loops.

23.4 EXTENSIONS TO SYSTEMS WITH INTERACTING UNITS

In Sections 23.2 and 23.3 we studied the determination of the necessary

controlled and manipulated variables, as well as the generation of all

feasible loop configuration, for single processing units. In the present

section we will extend these results to systems composed of several inter-

acting processing units, since such are the systems encountered in a chemical

plant.

Consider a process composed of N units which interact with each other

through material or energy flows. To determine all feasible control con-

figurations for the overall process, we can adopt the following systematic

procedure:

Step 1. Divide the process into separate blocks. Every block may contain a

single processing unit or a small number of processing units with an

inherent, common operational goal. For example, the block containing

a distillation column should also contain the condenser and reboiler

attached to the column; two neighboring heat exchangers in series or

in parallel should be contained in the same block; a reactor and its

feed preheater could be in the same block, etc.

Step 2. Determine the degrees of freedom and the number of controlled and

manipulated variables for each block. To do this, follow the procedure

described in Section 23.2.

Step 3. Determine all feasible loop configurations for each block. Having

specified the controlled and manipulated variables for each block, it

is easy to generate all possible configuration, following the approach

described in Section 23.3. Using qualitative or quantitative

arguments, retain a small number of the "best" loop configurations

for each block.

Step 4. Recombine the blocks with their loop configurations. It is clear

that the number of the generated loop configurations for the overall

process is equal to the product of the retained configurations for all

blocks.

Step 5. Eliminate conflicts among the control systems of the various blocks.

The control configurations resulting in Step 4 usually lead to an over-

specification of the overall controlled process. This can be explained

as follows: Consider two units connected by a common flow (Figure

23.3a). When we design the loops for each unit separately, it is

possible to select the interconnecting flow as a controlled variable

for both units but in different loops (Figure 23.3b). Also, it is

possible to have the corm-non interconnecting flow as the manipulated

variable in two different control loops (Figure 23.3~). Both situations

correspond to overspecified systems and lead to conflicts among the

control systems. Such conflicts must be erased before we can have a

feasible control configuration for the overall process.

Let us now demonstrate the above procedure on two specific processing

systems composed of several interacting units.

Example 23.3 - Generate the Control Loop Configuration for a Simple ChemicalProcess

The heart of the process shown in Figure 23.4 is the continuous stirred

tank reactor (CSTR) where the simple, exothermic reaction, A-tB, takes place.

The reactor feed is preheated, first by the hot reactor effluent and then by

steam. Coolant, flowing through a jacket around the reactor, removes the

heat generated by the reaction,lin order to maintain the temperature of the

reacting mixture at the maximum allowable (for highest conversion). The: ;

coolant is provided with two branches one of which is cooled while the other

is heated. The rates of cooling and heating, i.e. Qc and Q, are constant.

With this configuration we can fine tune the temperature of the coolant

(increase or decrease it) before it enters the jacket of CSTR. The reactor

effluent is first cooled by the feed in the feed-effluent heat exchanger and

subsequently it is "flashed" in a flash drum. There it is separated into two

streams, a vapor and a liquid, which are further processed in separate units.

Cooling water is provided to regulate the temperature in the drum. We would

like to develop alternative loop configurations for this process which satisfy

.&he following operating objectives;

-.= &zp the conversion in the reactor at its highest permissible value,

- maintain a constant production rate and

- achieve constant composition in the liquid product of the flash drum.

l.32p1 . Divide the process into four blocks (Figure 23.4); coolant system,

feeti preheating, reactor, and flash drum with its feed cooler.

Steps 2 and 3. Determine the degrees of freedom as well as the controlled

and manipulated variables for each blocl. Also, generate all possible loop

configurations for each block and retain the "best".

Coolant System (Figure 23.5a)

Table 23.2 summarizes all the characteristics of the coolant system. There

are two controlled variables requiring two manipulations. From the operating

requirements we can easily determine that: . __.

- Fc and TCO are the two controlled variables.

- Fc and TCO are also the two measured variables.

- The two manipulated variables, can be selected from the set,I.

Fc' Fcl' Fc2, Fcl + F Y Fcl'Fc2

Table 23.2 also indicates a few of the possible loop configurations. Con-

figuration No. 1 seems to be the simplest and is selected for the control of

the coolant system (Figure 23.5a).

Feed Preheating System (Figure 23.5d)

This block requires one controlled variable (see Table 23.3) which is the tem-

perature Ti. The only available manipulated variable is the steam flowrate

Ws, thus yielding only one loop configuration (Figure 23.5b)

Reactor (Figure 23.5~)

Table 23.4 shows that there should be two controlled variables for the

'reactor which are easily identified as the temperature, Tr' and concentration,

Q9 of the reactor effluent stream. Available manipulations are,

Fi 3 Tco (01: Fc>

Table 23.4 shows the three possible loop configurations. The No. 3 corresponds

to cascade temperature control and, as we have seen in Section 20.1 and Example

20.1, it provides fast compensation. Thus, configuration No. 3 is selected for

the reactor.

543

Flash Drum (Figure 23.5d)

This is similar to the flash drum system analyzed in Examples 23.1 and 23.2

with one difference; instead of the steam heating (see Figure 23.la) there is

a water cooling system (Figure 23.5d). Therefore, following the same pro-

cedure as in Example 23.1 we conclude that there should be,

- four controlled variables [Fi, pf, Tf, h], and

- four manipulated variables [Fi, FV, FL, FW].

We can generate 24 possible loop configurations, similar to those tabulated in

Table 23.1. The configuration shown in Figure 23.5d is selected as the "best"

because it provides direct and fast regulation of all controlled variables.i

4 .Step Recombine the four blocks with their control configurations. Con-,"

sidering that the four blocks (coolant system, feed preheating, reactor,

flash drum) possess 6, 1, 2 and 24 possible loop configuration, we can

generate in principle 288 (=6xlx2~24) control configurations for the

overall process. Not all of them need to be examined for consistency

because some are obviously bad. Figure 23.6a shows the resulting con-

trol system if the "best" loop configurations are selected for each

block.

5 .Step Eliminate confliects among the control loops of the various blocks.

Consider the control system for the overall process shown in Figure

23.6a. We notice quickly two overspecifications which create conflicts

among the control loops.

(a) The coolant flowrate is used as manipulated variable by two

different loops; the temperature cascade loop of the reactor

and the feedforward flow control loop of the coolant system.

(b) The flowrate Fi (feed to the reactor, and reactor effluent) is

controlled by two different loops; the feedback concentration

control loop in the reactor and the feedforward flow control

loop in the flash drum.

To eliminate conflict (a) we can delete the feedforward flow control

loop in the coolant system. To erase conflict (b) we delete the flow

control loop in the flash drum. Thus, the final control configuration

for the overall process is shown in Figure 23.6b. It has no conflicts

among the loops and the process is exactly specified.

Example 23.4 - Generate the Control Loop Configuration for an IntegratedChemical Plant : i'

:Consider the process shown in Figure 23.7. An exothermic reaction

:;A+B+C takes place in the gas phase. The product C is taken from the

top of a distillation column. The unreacted raw materials A and B are

both recycled to the reactor from the flash drum and the bottom of the dis-

stillation column, respectively. Compressors (C-l, C-2) are used to increase

the pressure of the feed and recycled gas A. The liquid B is vaporized in

a series of two heat exchanger (E-l, E-2). The reactor is a tubes-and shell

heat exchanger with the reaction taking place in the tubes and the coolant

flowing in the shell around the tubes.

The basic control objective is; maintain the desired steady state pro-

dllirtion rate and quality of product C for a long period.

$?%Pi?* Divide the processinto seven blocks as shown in Figure 23.7:

(a) Compressor for the fresh feed gas A,

(b) Compressor for gas A recycled from the flash drum,

(c) .Mixing drum for the fresh feed B and the recycled from the-

bottom of the distillation column,

(d) Feed vaporizing and preheating,

(e) Reactor,

(f) Flash drum with its feed cooler, and

(g) Distillation column with its condenser and reboiler.

Steps 2 and 3. To simplify the presentation of this example, we have omitted

the details of modeling, accounting for variables, determination of

controlled and manipulated variables, etc. Figure 23.8a through 23.88

show the selected "best" loop configurations for each block.

4 .Step Figure 23.9 presents the control structure for the overall plant,

resulting from the particular loop configurations shown in Figure 23.8.

D Step 5. Close observation of the control-configuration in Figure 23.9 reveals

the following conflicts among the various control loops:

(i> The flowrate of the recycles stream B from the bottom of the

distillation column is controlled by two loops; the feedback

level control at the bottom of the distillation column (stream

22) and the feedforward flow,control in the mixing drum

(stream 3). To erase this conflict eliminate the feedforward

loop in the mixing drum because the level controller is

absolutely necessary for the good operation of the distillation

column.

(ii) The flowrate of the reactor feed is controlled by three loops;

the flow control loop in the mixing drum (stream 6), the flow

control loop in the feed preheating block (stream 8) and the

flow control loop in the reactor block (stream 9). Eliminate

the conflict by retaining the flow control on stream 6 only.

(iii) Delete the flow control on the feed of the flash drum (stream

11) because its flow is determined by the flow of stream 6.

(iv> The flow-rate of the feed to the distillation column is controlled

by two loops; the level control of the flash drum (stream 15)

and the flow control on stream 16. Retain only the second loop

and eliminate the first.

6 .Step Improve the control configuration generated in Step 5. After the

elmination of the four conflicts among the control loops, which we des-

cribed above, we can make two additional modifications which improve

the quality of the resulting control.

(CL) The pressure control of both gaseous streams 1 and 13 may be

excessive. Since the pressure of stream 5 is the one of prac-

tical importance, we can replace the two pressure control loops

by one, which measures the pressure of stream 5 and manipulates

the bypass flow around compressor C-l.

(B) For the pressure control in the flash drum we use the flowrate

of the vapor (stream 13) as the manipulated. But, the variations

in stream 13 are fed back to the main process and may cause

additional disruptions in the operation. For better pressure

control introduce a purge stream (stream 23) and manipulate its

flowrate. Figure 23.10 shows the final configuration of the

control loops after eliminating any conflicts (Step 5) and making

the two modifications described in Step 6.

"SUMMARY AND CONCLUDING REMARKS

The typical processes in a chemical plant involve more than one inputs

and outputs and necessitate the design of multivariable control systems.

Several questions need to be answered for either single MIMO units or pro-

cesses with several interacting units; (a) how many and which ones are the

controlled variables, (b) how many measurements and manipulated variables

are needed, (c) what is the configuration of the control loops, etc.

The notion of degrees of freedom in a system dictates that:

- the number of controlled variables is equal to the total number of degrees

of freedom minus the number of externally specified variables, also,

- the number of required manipulations is at least equal to the number of

controlled variables.

But, besides knowing the necessary number of controlled and manipulated

variables, how does one select them among several process variables? One can

use the following general guidelines: _I

- Identify as controlled the variables which are directly dictated by the

operating objectives for a process. Any remaining freedom should be used

to control hold-ups or flows (in a feedforward manner).

- Consider that all controlled variables are measurable and constitute the

set of required measurements. If any controlled variable is unmeasurable,

examine the possibility for inferential control.

- Select the necessary number of controlled variables among all those

available so that they provide direct, easy, and fast regulation of the

controlled variables.

For a process with N controlled and N manipulated variables there are N!

possible control configurations, composed of single loops. Which configuration

3.~: -ihe best? In this chapter we have not addressed directly this question,

and all evaluations were made at a qualitative level.

For complex processes composed of several interacting units we can use

the following generalized procedure for the synthesis of feasible control

configurations:

573

- Divide the process into blocks and generate alternative loop configurations

for each block.

- Recombine the blocks with their loop configurations and eliminate any

overspecification which creates conflicts among the loops.

This chapter has been concerned with the generation of feasible control con-

figurations and not their precise evaluation. Some general and fallible

qualitative arguments were used to select the "best" loop configuration. In

Chapter 24 we will study the most popular method for the quantitative

evaluation of alternative loop configuration, i.e. the relative-gain array.- . .

This method indicates what manipulations should be coupled. with every con-

trolled variable, so that the interaction among the control loops is minimal.

Furthermore, we will study the design of special decoupling control systems

in case that strong interaction among loops persists.

THINGS TO THINK &OUT

1. What is a MIMO process and in what sense is the design of a control sys-

tem for a MIMO process different than that for a SISO process?

2. Discuss the design questions related to a MIMO control system.

3, Why do we assume that all manipulated variables are measurable? Is

this assumption correct? How would you use the values of manipulated

variables in a control system?

-4. Why do we claim that there is a large number of control configurations

for a MIMO process? Find the number of alternative loop configurations

for a process with N controlled variables and M manipulations, where

M>N.

5. "Prove" equations (23.2) and (23.3) which determine the number of

necessary controlled and manipulated variables.

6. Construct a physical example where we can have fewer controlled variables

than dictated by equation (23.2) and the operation of the process is

acceptable.

7. Equation (23.3) determines the minimum number of required manipulations

for a process. Why is it minimum? Could you use more manipulations and

how? Construct a physical example with more manipulations than dictated

by equation (23.3).

8. Is it sufficient to control the temperature and pressure in a flash

drum in order to have vapor and liquid products of desired composition?

Elaborate on your answer and explain why yes or no.

9. Consider the stirred tank heater example. Show that there are two dis-

tinct loop configurations. One of them is unacceptable because it

renders an uncontrollable system. Which one is this configuration and

why is it unacceptable? (Hint: Consider the effects of manipulations

on the controlled variables.)

10. Extend the observation made for the uncontrollability of the stirred

tank heater to other systems. State qualitatively a test for rejecting

loop configurations leading to uncontrollable systems.

11. Assuming that all variables can be measured, how many measurements do

you need for the design of a control system with N controlled

variables?

':12 * Consider the process examined in Example 23.3. Are there more degrees

of freedom when the processing units are considered together in an

integrated whole, or when the various units are considered separately

detached from each other? Explain why.

13.. Why is the overall process overspecified when the various blocks with

their corresponding loop configuration are recombined to yield the

control configuration for the overall process (see Steps 4 and 5 in

Section 23.4)? How does the overspecification manifest itself in the

configuration of the control loops? Discuss how you can eliminate

it.

14. Determine the number of controlled and manipulated variables for the

flash drum (Example 23.1) assuming steady state operation. Why are the

results different than those of Example 23.1? State the danger involved

when we consider steady state models to design a MIMO control system.

15. What are the controlled variables which remain unidentified when we

use steady state models to'determine controlled and manipulated;

variables? How can you overcome this drawback and still use steady

state models?

/

__-- L_

-_.-

_

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I

[Fig uce23. Z]

,

r--- ---- -___-me

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FEED f’u.WfAT~N~t- - - - - ----me;

FLASH DRUM

* . I

. :* :

cc

1 Figure 23.5)

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II

CHAPTER 24

INTERACTION AND DECOUPLING OF CONTROL LOOPS

From the discussion in Chapter 23, two characteristics should be clear

concerning the design of control systems for processes with multiple inputs

and multiple outputs:

- First; a control system is composed of several interacting control loops.

- Second; there is a large number of feasible, alternative configurations

of control loops.

For example, to control the operation of a flash drum we need a configuration

of four loops, which must be selected from 24 possible such loop configurations

(Example 23.2).

These two characteristics dictate the content of this chapter. In par-

ticular we will study;

- the interaction among the control loops of a MI130 process,

- the relative-gain array method which determines how the controlled and

manipulated variables should be coupled to yield control loops with minimal

interaction, and finally

- the design of special control systems with non-interacting loops.

24.1 INTERACTION OF CONTROL LOOPS

Consider a

@Zgzzre 24.la).

Y,(S)

Y2W

Huts), H12W,

the two outputs

(24.2) indicate

outputs.

process with two controlled outputs and two manipulated inputs

The input-output relationships are given by

= Hll(d iiil(s) + H12(~) fii2W (24.1)

= H21W =l(s> + H2+4 i5,W (24.2)

H21(s) and H22(~) are the four transfer functions relating

to the two inputs (see Section 9.2). Equations (24.1) and

that a change in ml or m2 will affect both controlled

Let US form two control loops by coupling ml with yl and m2 with

y2 as shown in Figure 24.lb. To simplify the presentation, we have assumed

that the transfer functions of the measuring devices and final control

elements in both loops are equal to 1. If Gel(s) and Gc2(s) are the trans-

fer functions of the two controllers, then the values of the manipulations are

given by

y(s) = Gcl[~l,SpW - Y1l (24.3)

q4 = Gc2[Y2,sp(S) - T21 (24.4)

To understand the nature of interaction between two control loops and how it

arises, we will study the affects of input changes on the outputs when; (a)

one loop is closed and the other open and (b) both loops are closed.

(a) One Loop Closed. Assume that Loop 1 is closed and Loop 2 is open (Figure

24.2a). Assume also that m2 = constnat, i.e. E,(s), and make a change in the

set point yl,SPm After substituting eqn. (24.3) into eqns. (24.1) and

(24.2) we take:

Tl =HllGcl

1 + HllGcl 7l,SP (24.5)

1,SP (24.6)

Lt is clear then that any change in the set point yl,sp will not only affect

k& behavior of the controlled output yl, but also the uncontrolled output

3. The dotted lines in Figure 24.2a indicate schematically the routes

through which yl sp affects the two outputs. Similar conclusions are drawn,

if we consider Loop 1 open and Loop 2 closed. The situation becomes more

complex when both loops are closed.

(b) Both Loops Closed (Figure 24.2b). Initially the process is at steady

state with both outputs at their desired values. Consider a change in the

set point yl,SP only and keep the set point of Loop 2 the same, i.e.

72,SP = O. Then, the following things happen:

- The controller of Loop 1 will change the value of ml in such a way as to

bring the output yl to the new set-point value. This is the direct

effect of m1 on Yl through Loop 1, and is shown schematically by the

dashed line in Figure 24.2b.

- The control action of ml will not only attempt to bring yl to the new

set point, but will also disturb y2, from its steady state value. Then,

the controller of Loop 2 attempts to compensate for the variations in y2

by changing appropriately-the value of the manipulated variable m2. But a

change in m2 affects in turn output yl. This is an indirect effect of

ml on yly through Loop 2, and is shown schematically by the dotted line

of Figure 24.2b. It constitutes the essence of interaction between two

control loops.

Analogous conslusion on the loop interaction will be drawn if we consider a

change in y2,Sp, while keeping yl,Sp the same. In summary, we can make

the following statement which described the interaction between two control

loops:

"The regulatory action of a control loop deregulates the out-put of another loop (in the same process), which in turntakes control action to compensate for the variations in itscontrolled output, disturbing at the same time the output ofthe first loop."

Having completed a qualitative presentation, let us now examine the

quantitative ramifications of the interaction between two control loops. Sub-

stitute eqns. (24.3) and (24.4) into eqns. (24.1) and (24.2) respectively

and take:

(1 + HllGcl)Yl+ (H12Gc2)~2=H11Gcl*~l SP+H12Gc2’?2 sp (24.7), ,

(H21Gcl)71 + (1 + H22Gc2)~2=H21Gcl’~l,~~ + H22Gc2*55,~~ (24.8)

Solve eqns. (24.7) and (24.8) with respect to the controlled outputs yl and

y2 and take the following closed-loop input-output relationships,

3 = Pll(S) Qsp + p12w Y2,sp (24.9)

y2 = p21w iysp + p&s) 72,sp (24.10)

where

Pll(S) = {HllGcl + GclGc2(HllH22 -yH12H21)~/Q(s)

Ih

p&d = H12Gc2/Q(s> f

p21w = H21Gc2/Q(s>

p&J = (H22Gc2 + GclGc2(HllH22 - H12H21))/Q(d

and

Q(s) = (l + HllGcl)(l + H22Gc2) - %2H21GclGc2(24.11)

Remarks: (1) Equations (24.9) and (24.11) describe the response of outputs

y1 and y2 when both loops are closed, i.e. they have

accounted for the interaction between the two loops.

(2) When H12 = H21 = 0, there is no interaction between the

two control loops. The closed-loop outputs are given by

the following equations,

Tl =HllGcl

7 7, =H22Gc2

1 + HllGcl l,SP 1 + H22Gc2 '2,SP

The closed-loop stability of the two noninteracting loops

depends on the roots of their characteristic equations.

Thus, if the roots of the following two equations

1 + HllGcl = 0 1 + H22Gc2 = 0 (24.12)

have negative real parts, the two noninteracting loops are

stable.

(3) The stability of the closed-loop outputs of two inter-

acting loops is determined by the roots of the character-

istic equation

Q(s) 5 (1+H11Gcl(l+H22Gc2) -H12H21GclGc2 = 0 (24.13)

Thus, if the roots of eqn. (24.13) have negative real

parts, the two interacting loops are stable.

(4) Suppose that the two feedback controllers Gel and Gc2-..

are tuned separat'ely, ite. keeping the loop under tuning

closed and the other open. Then, we cannot guarantee

stability for the overall control system, where both loops

are closed. The reason is simple: Tuning each loop

separately we force the roots of the characteristic eqns.

(24.12) for the individual'loops to acquire negative real

parts. But the roots of these equations are different

from the roots of the characteristic eqn. (24.13) which

determines the stability of the overall system with both

loops closed.

(5) Normally, we tune the two controllers in such a way that

the roots of all eqns. (24.12) and (24.13) have negative

real parts. Such tuning guarantees stability when both

loops are closed (roots of eqn. (24.13)), or only one is

closed while the other is open due to a hardware failure

(roots of eqns. (24.12)).

(6) The previous discussion indicates that the interaction

between control loops is a significant factor and affects

in a very profound manner the "goodness" of a control

system. For this reason, a control designer attempts to

couple the manipulated variables with the controlled

outputs in such a way as to minimize the interaction

of the resulting control loops. If strong interactions

persist for any possible pairing, then he will design a

sp,ecial,control system which eliminates the interaction

(decoupling the loops).

Example 24.1 - Interaction of Contro14Loops in‘s Stirred Tank Heater

: Consider once more the.stirred tank heater (Example 4.4). Figure 24.3

shows the two control loops; Loop 1 controls the liquid level by manipulating

the effluent flowrate, and Loop 2 regulates the temperature by manipulating

the steam flowrate. Let us see how the two loops interact:

- When the inlet flowrate (load) or the desired value of liquid level

(set point) change, Loop 1 attempts to compensate for the changes by

manipulating the value of the effluent flowrate. This in turn will dis-

turb the temperature of the liquid in tank and Loop 2 will compensate by

adjusting appropriately the value of steam flowrate.

- If on the other hand, the temperature of the inlet stream (load) or the

%! &sired value of the temperature (set point) change, Loop 2 will adjust

.- ,-be steam flowrate to compensate for the changes. This will leave the

liquid level undisturbed.

.1y.5;.rz;-zs9 we notice that Loop 1 affects Loop 2 but not vice versa. In other words,

thsinteraction is in a single direction.

Example 24.2 - Interaction of Control Loops in a Stirred Tank Reactor

In the CSTR of Figure 24.4, the temperature is controlled by the flow of

coolant in the jacket while the effluent concentration is controlled by the

inlet flowrate. Assume that initially both effluent concentration and tem-

perature are at their desired values.

. - Consider a change in the inlet concentration (load) or the desired effluent

concentration (set point). Loop 1 will compensate for these changes by

manipulating the feed flowrate. However, this change in the feed rate also

disturbs the reactor temperature away from the desired value. The, Loop 2

attempts to compensate for the change in temperature by varying the coolant

flowrate, which in turn affects the effluent concentration.

- On the other hand, when to compensate for changes in feed temperature (load)

or the desired set point of reactor temperature, it also causes the effluent

concentration to vary. Then, Loop 1 attempts to compensate for the change

in effluent concentration by varying the feed rate, which in turn disturbs

the reactor temperature.

It is clear from the above that Loop 1 interacts with Loop 2 in both directions

(unlike the loops of the stirred tank heater which interac in a single

direction).

&ample 24.3 - Tuning the Controllers of Two Interacting LOOPS

Assume that the input-output relationships of a process with two controlled

outputs and two manipulated variables are given by,

7, 1=O.ls+l %

+ 5 -O.ls+l m2

Y2 1= ml + 20.5s+l 0.5s+l :2

Form two loops by coupling ml with yl and m2 with y2. The closed-

loop input-output relationships are given by eqns. (24.9) and (24.10) where

Hll =1

H12 =5 1 2

O.ls+l O.ls+l H21 = 0.5s+l H22 = 0.5s+l

Let the two controllers be simple

Gcl = Kcl and

(a) Tuning each loop separately:

when Loop 2 is open is given

proportional controllers with

Gc2 = Kc2

The charactewristic equation of Loop 1

by,

KC-l1 + Hll~cl = 1 + o.ls+l=O

and yields the closed-loop pole

s = -lO(l + Kcl) < 0

Therefore, when Loop 2 is open, Loop 1 is stable for any value of gain

&Cl. Similarly, the closed-loop pole for Loop 2 when Loop 1 is open

is given by

S = -2(1 + Kc2) < 0

and consequently, Loop 2 is stable for any value of Ka2, when Loop 1

is open.

(b) Tuning with both loops closed: When both loops are closed, the charac-

teristic equation is given by eqn. (24.13) and for this example takes

the following form:

(1 + oK;;+l)(l +.2Kc2, - 51

O.ls+1 l 0.5s+l'K,1'Kc2 = O

+(0.6+0.5Kcl+0.1Kc2)s+(1+KclfKc2-9KclKc2) = 0 (24.14)

According to the first test of the Routh-Hurwitz criterion for stability (see

Section 15.3), eqn. (24.14) has at least one root with positive real part if

any of its coefficients is negative. Thus, the closed-loop behavior of the

process is unstable if the following inequality is satisfied,

1 + Kc1 + Kc2 - 9KclKc2 < Cl (24.15)

Inequality (24.15) places restrictions on the values that Kc1 and Kc2 can

take, to render a stable performance when both loops are closed. This is in

direct contrast to our earlier result (see (a) above), whereby all values of

KC l

and Kc2 were acceptable if each loop were tuned separately. [Note: The

allowable range of values for n-Kcl--:and K--..g

which render stable responses

when both loops are closed can be found by applying the second test of the

Routh-Hurwitz criterion.]

24.2 THE RELATIVE-GAIN ARRAY AND THE SELECTION OF LOOPS

In the previous chapter we recognized that for a process with N con-

trolled outputs and N manipulated variables there are N! different ways

to form the control loops. Which one is the best? One way to answer this

question is to consider the interactions among the loops for all N! loop

configurations and select the one where the interactions are minimal. The.-

Relative-Gain Array is exactly such methodology whereby we select pairs of

input and output variables in order to minimize the amount of interaction

- among the resulting loops. It was first proposed by Bristol and today is a

very popular tool for the selection of control loops. Let us now study the-logic of the mehtod and present some examples describing its usage.

A. Definitions

Consider a process with two outputs and two inputs (Figure 24.la). Then,

do the following two experiments:

(a) Assume that m 2 remains constant (Figure 24.5a). Introduce a step

change in the input ml of magnitude Am1 and record the new steady

state value of output yl. Let Ayl be the change from the previous

steady state. It is clear that it has been caused only by the change

in ml' The open-loop static gain between yl and ml when m2 is

kept constant is given by (see Section 10.4 and eqn. (10.20))

(AyllAml)m2

(b) In addition to the static gain computed above, there is another open-loop

gain between yl and ml, when m2varies by a feedback loop con-

trolling the other output, y2 (Figure 24.5b). Thus, introducing a-step

change Am1 we record a change Ayl in the steady state value of

yl*Ayl will be, in general, different than Ayl for the following

reason: The input change Am1 does not only affect yl but also y2.

Then, the control loop attempts to keep y2 constant by varying m2,

which in turn affects the steady state value of yl. Therefore, Ay;

is the compound result of the effects from ml and m2' Let the new

open-loop gain between Yl and ml when y2 is kept constant by the

control loop, be given by

(AYi/AmlIy -2

The ratio of the two open-loop gains computed above defines the relative gain,

k, between output yl and input ml, i.e..^

(Ayl/Aml)m

x11 = (Ayljoml)y22

(24.16)

The relative gain provides a useful measure of interaction, In particular:

- If Xl1 = 0, then yl does not respond to ml and ml should not be used

to control Yl'

- If Xl1 = 1, then ~2 does not affect yl and the control loop between

Yl and ml does not interact with the loop of y2 and m2. In this case

we have completely decoupled loops.

- If 0 < Xl1 < 1, then an interaction exists and as m2 varies it affects

the steady state value of yl. The smaller the value of xl1' the larger

the interaction becomes.

- If Xl1 < 0, then m2 causes a strong effect on yl and in the opposite

direction than that caused by ml. In this case, the interaction effect is

very dangerous.

In a similar manner as above we can define the following three remaining

relative gains between the 2 inputs and 2 outputs of the process we have been

considering, i.e.

52 = (Ayl/Am2)m /(Ayl/Am2)y1 2

: relative gain between yl and m2

x21= (Ay2/Aml)m /(Ay2/Aml)y

2 1: relative gain between Y2 and ml

x22 = (Ay2/Am2)m /(Ay,/Am,)1 Yl

: relative gain between y2 and m2

The values of these gians can also be used as measured of interaction for the

corresponding cases in a similar way as it was done above for xll*

B. Selection of Loops

For a process .with two inputs and two outputs there are two different

..loop configurations, shown in Figure 23.2. Let us see how we can use the

relative gains to select the configuration with minimum interaction between

thf! loops.

Arrange the four relative gains, All, Xl2, X21 and X22 into a matrix

form, which is known as the relative-gain array, i.e.

A ==

ml

rXl11x21

m2

x12 y11x22 y2-1

It can be shown that the sum of the relative gians in any row or column of the

array is equal to 1. Thus,

x11 + Xl2 = 1

x21 + x22 = 1

x11 + A21 = 1

and _. /., :;* . - (2$*1!)%2 + x22 = '1 I

Therefore, we need to know only one of the four relative gains while the other

three can be easily computed. For example, if All = 0.75 then Xl2 = A21 =

0.25 and X22 = 0.75.

Depending on the value of Xll, we can distinguish the following dif-

ferent situations:

- All = 1. Then, the relative-gain array is,

and it is obvious that we can have two noninteracting loops formed by; ml

coupled with yl and m2 coupled with y2 (Figure 23.2a).

-31 = O.The, the relative-gain array is given by,

a = 711 I1 0

.Ihe 1 in the off-diagonal elements indicate that we can form two non-

interacting control loops by coupling ml with y2 and m2 with yl

(Figure 23.2b).

- 51 = 0.5. Then,

B =3.5 0.5

L I0.5 0.5

and the amount of interaction between the two loops is the same in both

configurations of Figure 23.2. In other words, it does not matter how we

.couple inputs and outputs.

- 0 < Al1 < 0.5, say Al1 = 0.25. Then,

The two larger number, i.e. 0.75, indicate the recommended coupling with

the samller amount of interaction. Thus, we couple ml with y2 and m2

with yl (Figure 23.2b).

- 0.5 -C xl1 < 1, say Xl1 = 0.8. Then,

and the recommended coupling is the opposite of the previous case, i.e.

Couple ml with yl and m2 with y2 (Figure 23.2a).

- All> 1. Then, A22 = Xl1 > 1 and Al2 = h21 = 1 - Xl1 < 0. Situations

with relative gains outside the range 0 to 1 create difficult control

voblems. Let us see why.

51) Suppose that you couple yl with ml and y2 with m2. The

corresponding relative gains, X11 and x22' are larger than 1.

Then, from the definition of the relative gains we conclude that,

(Ayl/Aml> > (~yl/A~l~y2 and (Ay2/Am2) >m2 ml

(Ay2/Am >2 y1

In other words the response of the outputs is held back by the

interaction from the other loop and the larger the values of the

relative gains above unity, the larger the "holding back" effect

will be. Thus, we need larger values for the controller gains.

(2) If you couple yl with m2 and y2 with ml, the corresponding

gains Xl2 and A21 are negative. In this case, the interaction

will take the controlled outputs in the opposite direction than

that desired by the control effort and control will be lost

altogether. Therefore, never form loops by coupling inputs to

outputs with negative relative gains.

We can summarize all the above observations with the following rule for

selecting the control loops:

"Select the control loops by pairing the controlled outputs

Yi with the manipulated variables m.J in such a way thatthe relative gains Aij are positive and as close aspossible to unity."

Remarks: (1) The relative gains provide a measure of interaction based on

steady state considerations. Therefore, the rule given above

for the selection of loops does not guarantee that the dynamic

interaction between the loops will be also minimal.

(2) The relative-gain array is a square matrix, which implies

that the number of manipulated variables is equal to the

number of controlled outputs. Now, suppose that we have a

process with two outputs and three possible manipulations,

ml' m2 and m3' There are three possible pairs of mani-

pulated variables; (ml,m,), (m,,m,) and (m3,ml). Therefore,

we can form three different relative-gain arrays,

ml m2 m2 m3 m3 ml

and we need to examine all of them before we can select the

set of two loops with minimal interaction. [Note: In

general, hil # All, xi2 # A12, etc.1

(3) There are two ways of obtaining the relative gains of a

process; a computational approach using a steady state input-. .

output model for the process and an experimental approach.

2When a steady state model is available, then we can obtain

the numerator and denominator of the relative gain (see eqn.

(24.16)) by simple differentiation. This way we can express

the relative gains in terms of the controlled and manipulated

variables themselves, which enables us to evaluate the

. interaction across a range of operating conditions.

(4) For an existing process we can evaluate the relative gains

experimentally, by performing the following two experiments:

Experiment 1 (all loops open). Keeping all loops open make

a small step change Am1 in ml, keeping m2 constant.

Record the changes in the steady state values of yl and

y2, i.e. Ayl and Ay2. Then, compute

(AYl/Aml)m2

and (Ay2/Aml)m2

Return the system to the initial steady state and repeat the

same experiment by varying m2 by Am2. Record the changes

Ayl and Ay2 and compute,

(AYl/Am2)m1

and (Ay2/Am2>"1

Experiment 2 (one loop closed). Make a small change Am1

in m1, while keeping y2 constant by feedback control using

m2' Record the change Ayl in the steady state of yl and

compute the gain

(aYl/Aml~y2

Repeat the same experiment, but now keep yl constant

through a control loop with m2. Record the change Ay2.

and compute the gain

(Ay2/Amlly1

Similarly, we can compute the following two gains,

(Ayl/Am2> andy2

(Ay2/Am2)y1

Taking the ratios of the corresponding gains in Experiments

1 and 2, we can compute the relative gains Xll, X12, X21,

x22* -[Note: Remember that you do not need to compute all

relative gains, since they are related by eqns. (24.17)].

(5) The definition of the relative gains and their use in

selecting the control loops are not limited to systems with

two inputs and two outputs. The extension to general pro-

cesses is straightforward. Thus, the relative gain between

an output y.1 and a manipulation m.J

is defined by

x ij = (Ayi/Amj)m/(Ayi/Amj)y

The subscript m denotes constant values for all manipula-

tions except m., i.e. all loops open, while subscript yJ

indicates all outputs except yi are kept constant by the

control loops, i.e. all loops closed. Similarly, the

relative-gain array is given by

ml m2 %

'12 � � l '1N '1

A == 1'22 l l l '2N y2

The entries of a satisfy the following two properties;

ill ' ij= 1 for j=1,2,***,N summation by columns. .

Nj&l ‘ij = 1 for: i=1,2,***,N summation by rows

The loop selection rule remains the same.

Example 24.4 - Select the Loops Using the Relative-Gain Array

Consider a process with the following input-output relationships:

Yl =l- 1

-iTi ml + O.ls+l %2

Y2 =-0.2 - 0.8

0.5s+l ml + s+1 m2

(24.18)

(24.19)

Let us compute the relative gains:

-Bake a unit step change in ml, i.e. iii1 = l/s, while keeping m2 constant,

3.e. ii i2 = 0. Then, from eqn. (24.18) we take,

Recall the final-value theorem (Section 7.5) and find the resulting new

steady state in yl, i.e.

Yl,ss = lim [s y,(s)] = lim [l/(s+l)] = 1S-to S-4

Therefore, (Ayl/Aml)m = l/l = 1.2

- Keep y2 constant under control by varying rn2' Intorduce a unit step in

ml' Since y2 must remain constant, i.e. 72 = 0, eqn. (24.19) will tell

us by how much m2 should change;

ii) = 0.2 s+l2 0.8’ 0. Ss+l

ti? \

Substitute this value in eqn. (24.18) and find

71 1 s+l= s+l 1 ml - .-. 0.2+ O.ls+l 0.8 0.5s+l

Then, the resulting new steady state for y1 is given by,

1 1yl SS=lim [s Yl]=lim s* -a-, S-+O s+o Ilr

s+l s1 0.2 s+l 1

+ O.ls+1'0.8' 0.5s+l'S = 1.25

Therefore, (Ayl/Am )IL Y2

= 1.2511 = 1.25 and

ii 1

x11 = (Ayl/Am,)m2/(Ayl~~ml)y2 = l/l.25 = 0.8

Using eqns..(24.17) we find A12 = x21 = 0.2 and X22 = 0.8. It is easy now

to conclude that we should pair ml with yl and m2 with y2 to form two

1~op.s with minimum interaction. It should be noted that had we selected the

loops differently, i.e. couple ml with y2 and m2 with yl, the interaction

of the loops would have been four times larger (i.e. 0.8/0.2 = 4).

Example 24.5 - Selecting the Loops in a Mixing Process

Two streams with flowrates Fl and F2 and compositions (moles per cent)

"1 = 80% and x2 = 20% in a chemical A, are mixed in a vessel (Figure 24.6a).

We would like to form two control loops to regulate the product composition, x,

and flowrate, F. Let F z yl and x z y2 be the two controlled outputs,

while Fl E ml and F2 Z m2 are the two available manipulated variables. Two

are the possible control configurations with different pairings between the

inputs and outputs, and they are shown in Figures 24.6b and 24.6~. Which one

should we prefer?

The steady state mass balances yield:

F = F1 + F2

Fx = FIXl + F2X2

(24.20)

(24.21)

[Note: We have neglected the energy balance because the temperature of the

product stream is not in our operating reqyirements.] The desired steady

state for operational purposes is,

F = 200 moles/hr and x = 60% (by moles)

With these values we find the following steady state solution of eqns. (24.20)

and (24.21)

Fl = 133.4 and F2 = 66.6

To compute the relative gain between F and Fl, do the following:

- Change Fl by one unit, i.e. F1 = 134.4, while holding F2 = 66.6 the

same. Solve eqns. (24.20) and (24.21) for F and x and find the

following new steady states

F = 201 X = 0.6012

Therefore,

(AF/AFl)F = l/l = 1 (Ax/AF~)~ = 0.6012/1 = 0.60122 2

- Change Fl by one unit, i.e. F1 = 134.4, while holding x = 60% constant.

Solve eqns. (24.20) and (24.21) and find:

F = 201.67 F2 = 67.27

Therefore,

(AF/AFl)x = 1.6711 = 1.67

Consequently, the relative gain between F and F1 is:

x11 = (AF/AFl),2/(AF/AFl)x = l/l.67 = 0.6

It follows easily that the complete relative-gain array is

F1 F2

-0.6 0.4 FA =I

1 I0.4 0.6 x

Two are the main conclusions we can draw:

First; the two loops with minimum interaction are formed when we couple F

with F1 and x with F2 (Figure 24.6b).

Second; although the interaction between the two selected groups is smaller

than that of the other alternative configuration (Figure 24.6c), it is still

significant. Thus, any control action to regulate F will seriously disturb

x and vice versa.

24.3 DESIGN OF NONINTERACTING CONTROL LOOPS

The relative-gain array indicates how the inputs should be coupled with

the outputs to form loops with the smaller amount of interaction. But, the

persisting interaction, although it is the smaller possible, may not be small

enough. Example 24.5 demonstrated this aspect clearly. In such case, the two

control loops still affect each other's operation very seriously, and the

overall control system is characterized unacceptable.

When the designer is confronted with two strongly interacting loops, he

introduces in the control system special new elements called decouplers. The

purpose of the decouplers is to cancel the interaction effects between the

two loops and thus render two noninteracting control loops. Let us now study

how we can design the decouplers for a process with two strongly interacting

loops.

667

consider the process whose input-output relationships are given by eqns.

(24.1) and (24.2). Form two interacting loops by coupling ml with yl and

m2 with y2 (see Figure 24.lb).

Assume that initially both outputs are at their desired set point values.

Suppose that a disturbance or a set point change cause the controller of

Loop 2 to vary the value of m2. This will create an undesired disturbance

for Loop 1 and will cause yl to deviate from its desired value. However, we

could change ml by such an amount as to cancel the interaction effect from

m2' But, the question arises; how much should we change m2?

From eqn. (24.1) we find that in order to keep yl constant, i.e.

% = 0, m2 should change by the following amount,

{ =- H12(s) -1 HllW m2

(24.22)

Equation (24.22) implies that we can introduce a dynamic element with a

transfer function,

I+) = -H12W

Hll(d(24.23)

which uses the value of m2 as input and provides as output the amount by

which we should change ml, in order to cancel the effect of m2 on yl'

This dynamic element is called decoupler and when is installed in the control

system (Figure 24.7a) it cancels any effect that Loop 2 might have on Loop 1

but not vice versa.

To eliminate the interaction from Loop 1 to Loop 2, we can follow the

same reasoning as above and we find that the transfer function of the second

decoupler is given by,

D2(s) = -H21(d

H22(~)(24.24)

The block diagram of the process with two feedback control loops and two

decouplers is given in Figure 24.7b.

From the block diagram of Figure 24.7b it is easy to develop the fol-

lowing two closed-loop input-output relationships:

Yl =Gcl[H1l - H12H21'H223

l + Gcl[Hll - H12H21'H221-71,SP

(24.25)

Y2 =Gc2[H22 - H12H21'H223

l + Gc2[H22 - H22H21'H211 l Y2,SP (24.26)

The last two equations demonstrate the complete decoupling of the two loops

since the controlled variable of each loop depends only on its own set point

and not on the set point of the other loop. Figure 24.7~ shows the net block

diagram of the two noninteracting loops described by eqns. (24.25) and (24.26).

It is completely equivalent to that of Figure 24.7b.

Remarks: (1) Two interacting control loops are perfectly decoupled only

when the process is perfectly known, because only in this

case the transfer functions Hll, H12, H21 and H22 are

known exactly. Since this requirement is rarely satisfied

in practice, the decouplers offer only partial decoupling

with some weak interaction still persisting between the

two loops.

(2) As we have mentioned repeatedly, chemical processes are mostly

nonlinear and nonstationary (i.e. the values of their

parameters change). Therefore, even if the decoupling is

initially perfect, as the desired operating conditions

change the decoupling deteriorates. One solution to this

problem is to use adaptive decouplers. Thus, as the pro-

cess changes we estimate the new transfer functions H1ly

H12' H21 and H22 and compute new decouplers. How to

design adaptive decouplers is quite complex and goes

beyond the scope of the present text.

(3) Perfect or very good decoupling allows the independent

tuning of each controller without risking the stability

of the overall system.

(4) A close examination of Figure 24.7b reveals that for all

practical purposes the decouplers are essentially feed-

forward control elements. Thus, decoupler Dl measures

the changes in m2 and takes appropriate action to can-

cel the effect that m2 would have on yl before it has

been felt by yl.

(5) If the decouplers are designed using steady state models

for the process, then we talk about steady state or static

decoupling. Equations (24.23) and (24.24) provide the

design of dynamic decouplers. It should be emphasized

that for severely interacting loops static decoupling is

better than no decoupling at all.

(6) For a general process with two inputs and two outputs we

need two decouplers to produce n&interacting loops.

Whenever we use only one decoupler, despite the fact that

two are needed, we, talk about partial or one-way decoupling.

Such systems allow the interaction to travel in one

direction. Figure 24.7a shows a partial decoupling of

the loops. Thus, disturbances entering Loop 2 cannot

enter Loop 1 due to the decoupler Dl' On the other hand,

disturbances originating in Loop 1 may enter Loop 2 but

cannot be returned.

Example 24.7 - Partial Decoupling;

Let us return to the mixing process we studied in Example 24.5. Suppose

that the operating requirements allow small variations in the product flowrate

F, while dictating very tight control on the concentration x of the product.

Then, we can use partial or one-way decoupling to cancel any effects that

interaction might have on x, leaving the simple feedback loop to regulate

the value of the product flowrate, F.

Assuming that x is kept at the desired value of 0.6, then eqn. (24.21)

yield

(Fl + F2)0.6 = 0.8Fl + 0.2F2

or

F2 = Fl/2

The last equation describes the necessary steady state decoupler which cancels

any effects that the flow control loop might have on the composition control

loop.

Example 24.8 - Physically Unrealizable Decouplers- - -

Consider a process whose input-output relationships are given by

y1 =0.5e-1'5S ml + e-Os5' _

s+l 2s+l m2

Y2 =2e-l.0s 10.5s+l ml + s+l iii2

Form the two control loops by coupling yl with ml and y2 with m2. Then,

the transfer functions of the two decouplers are given by eqns. (24.23) and

(24.24), i.e.

Dl(s) = - R12(s) = s+l +1.0sHll(s) 22s+l' e

D2(s) = -H21(S) = s+l -1.0sH22(4 2 0.5s+l e

Decoupler Dl +1.0sis physically unrealizable due to term e . On the other

hand D2 is realizable. The reader should elaborate more on the reason that

makes D1 unrealizable.


Interaction among the control loops creates several undesirable effects

on the operation of a process. The loops disturb each other, i.e. load or

set point changes entering a control loop are propagated to other loops. But

what is even more serious, the disturbances propagating from one loop to the.I .

others return back to the original loop and may destabilize the process

altogether. These hidden feedback actions deteriorate the performance of

the control system and constitute the main "headache" for the control

designer.

Bristol's relative-gain array provides a simple and useful tool to select

the couplings between the various inputs and outputs in such a way that the

interaction among the resulting loops is minimal. The foundations of the

method are heuristic in nature and there have been instances where the

relative-gain array gave the wrong recommendations. But its simplicity and

practicality are two very attractive features which made it very popular.

Whenever severe interaction still persists even for the best configuration

of control loops, the addition of decouplers in the control system is

recommended. The main function of a decoupler is to provide a control action

equal in size but of opposite sign to the propagating loop interaction. Thus,

the interaction effect is cancelled and the loops behave as if they were

completely independent. It must be noted that perfect decoupling is possible

only if the process is completely known. Since this is hardly ever the case

with chemical processes, the decoupling is only partial, but even so some

form of decoupling should be preferred over no decoupling at all.

There is still large body of material concerning the design of control

systems for processes with multiple inputs and multiple outputs. In the last

two chapters we have only scratched the surface of the problem. Several

theories and design techniques are available, either in the time or s domain.

Due to their mathematical complexity, they are beyond the scope of this text.

The interested reacer can find several relevant references on these methods at

the end of Part VI.


1.

2.

3.

4.

Explain in your own words the interaction among the control loops of a

flash drum (Figure 23.lb). Do the same for the loops of a distillation

column (Figure 5.6).

Can you tune separately two interacting loops and retain the stability

of the overall process? Explain why yes or no.

Consider the process of Figure 24.la. Couple yl with m2 and y2

with m1, to form the two loops. Draw the corresponding block diagram.

Develop the resulting closed-loop input-output relationships, similar

to those given by eqns. (24.9) and (24.10). Has the closed-loop

characteristic equation changed or not?

Define the two open-loop gains used in the definition of the relative

gain X12. Give two different ways for computing X12. Why is Xl2 a

good measure of loop interaction? Can you compute All, A21 and X22

when you only know X12? If yes show how, if not explain why.

5 . Repeat item 4. for the relative gain. A.. of a general process with1J

N inputs and N outputs. What do the subscripts i and j denote?

6. Define the relative-gain array for a process with 2 inputs and 2 out-

puts. Extend the d-finition to a process with N inputs and N

outputs.

7. Consider a process with the following transfer functions; H12(s) =

H21W = 0 and Hll(s), H22(~) # 0. Show that the relative-gain array

is given by

1 0I I0 1'

8. What are the properties of a relative-gain array? How many relative

gains do you need to compute in order to specify completely the relative-

gain array of a process with; (a) 3 inputs and 3 outputs, (b) PJ inputs

and N outputs?

9. Explain how you can use the relative-gain array to select the loops

with minimum interaction. Why would you avoid coupling an output yi

with a manipulated variablemj

if h ij < O? Does Xij < 0 imply

that another relative gain is larger than 1 or not? Explain.

10. In Example 24.5 let xl = 0.3 and x2 = 0.7 and select the control

loops. Have they remained the same or not? Explain your result on

physical grounds. Has the interaction between the two loops increased

or decreased? Explain why.

11. Define an interaction index as follows,

(1 - Xij)/Xij

Consider the following relative-gain array

A ==

and take the interaction index array (using the above definition)

- 41 1/4I =

= 1/4 4 1-Which of the two arrays i and f shows more clearly the amount of

relative interaction between the corresponding loops?

12. What do we mean by the term decoupling two control loops? Do the two

loops of the process in item 7 need decoupling? Why, yes or no?

13. Consider the process of Figure 24.la. Form the two loops by coupling

y1 with m2 and y2 with ml. Find the transfer functions of the

required two decouplers.

14. Find the steady state decouplers for the two control loops selected in

the process of Example 24.4.

15. What is one-way decoupling of two control loops and why could it be

acceptable?

16. Explain in your own words the feedforward control nature of a decoupler.

When do you have perfect decoupling and when not?

17. After introducing the necessary decouplers, can you tune the controllers

of two loops separately so that the stability of the overall process is

guaranteed or not? [Hint: Examine'closely the closed-loop character-

istic equations of two decoupled loops.]

18. What do we mean when we say that a decoupler is physically unrealizable?

Explain why decoupler D1 in Example 24.8 is physically unrealizable.

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PRDCE55 I

CHAPTER 25

DIGITAL COMPUTER CONTROL LOOPS

A typical control loop consists of the following components (Figure

26.1); (a) process, (b) measuring sensor and the accompanying transducer (if

necessary), (c) controller, (d) final control element with the associated

electropneumatic converter (if necessary) and (e) transmission lines for

either the process measurement or the control command signal. As long as

the controller is an analog device (pneumatic or electronic) it can, (i) pro-

cess continuously the analog signals generated by the sensors and/or the

corresponding transducers, and (ii) produce continuous, analog command.

signals for the final control element. It is obvious that in such case

all transmission lines carry continuous, analog signals.

The above picture has been the basis for all control systems we have

examined so far. The introduction though of a digital computer in the place

of an analog controller creates the need for new hardware elements and new

control design problems. Before we examine what they are, let us briefly

review the characteristics of a digital computer and how it is interfaced to

the external world.

26.1 THE DIGITAL COMPUTER

Despite the differences in capacity, speed, architecture, all digital

computers designed for process control have much the same functions. Figure

26.2 indicates the basic components of such a typical digital computer and

the associated peripherals. Let us briefly describe their basic features.

A. Central Processing Unit

Usually it is designated and referred to as CPU for short. CPU is the

heart of a computer system and maintains control over all its functions.*

Thus, it is the CPU that,

- determined what is the next operation to be performed by the computer,

- executes the various arithmetic or logic operations,

- retrieves or stores information from or to the memory,

- directs various other operations.like data transfer between memory and

peripheral devices.

The smallest unit of information that the computer uses for communication

or computations is the bit (from binary digir), which can assume values 0 or-

1. A specified number of bits together form a word. In addition to the bit

and word, an intermediate unit called byte is also used to characterize infor-

mation of 8-bit long (i.e. 1 byte - 8 bits).' The byte is a useful unit

because all alphabet letters, numbers, other typing characters, control

characters, etc. can be fully specified by one byte according to the ASCI

Code (industry standards). The CPU stores information in the memory or

verifies information from it in terms of words. The length of a word varies

among the various computers with 8-, 16-, or 32-bit words being the most

common. Thus, consider the following 16-bit word; 1001101110010110. A usual

16-bit word computer arranges the binary digits in an octal system (composed

of three binary digits); i.e.

1 001 101 110 010 110

The first digit is reserved for the sign (e.g. 0 = +, 1 = -> and the other 15

form five groups of three digits each and are used to represent an integer

number. Thus, the above binary is equal to the following octal integer,

-156268 = -{l.84 + 5.83 + 6.82 + 2.8l + 6.81 E -705210

Therefore, the range of integers in a l&bit word computer is,

from 1 111 111 111 111 111 = -777778 - 1 = 32,767lO

to 0 111 111 l-l 111 111 = +777778 = +32,76810

The CPU is equipped with a set of hardware instruction to perform some

very basic operations like;

- addition, subtraction and logical comparisons between integer numbers,

- storing data in and recalling data from the memory, and

- transferring data between CPU and the various peripheral devices.

More complicated operations can be performed using combinations of the above

basic hardware instructions.

'To expedite and facilitate operations, modern CPU's are equipped with

various hardware options. Among all possible options the following are of

great value to process control computers.

(1) Hardware Floating-Point Processor. Performs with very high speed

floating-point arithmetic operations and expands tremendously the computational

speed of the machine.

(2) Real-Time Clock. Every digital computer used for process control must

have a real-time clock. This is the device that keeps track of the real

world's time and allows the computer to schedule its functions at time

intervals, in coordination with the various needs of the real world. Thus,

it is the real-time clock that determines when the computer should take data

from measuring sensors or change the values of manipulated variables.

'(3) Power Fail-Safe/Automatic Restart. In the event that power to the com-

puter is lost, this option senses the power failure and executes a prespecified

set of instructions before the machine becomes inoperable. These instructions

may transfer the control of the process from the digital computer to another

back-up control system and/or save information necessary for an orderly and

automatic restart of the control programs, when electrical power has been

restored to the computer. This option enhances the safety of computer con-

trolled processes.

(4) Watchdog Timer. This is another valuable option for process control

computers. It allows the computer to determine if the control program is

being executed smoothly or if the program is "hung-up" in a never ending

loop. In the second case an alarm alerts the operator that the computer has

lost control of the process, due to software problems.

B. Memory

This is the place where the computer stores, (a) the instructions of

the program it executes and (b) the values of the initial data, intermediate

and final results from computations. The smallest unit of storage is the bit,

but the memory is organized in terms of words. Thus, 4K words memory is com-

posed of 4,000 16-bit words (for 16-bit word computers). Each memory word is

characterized by a unique address and during the execution of a program the

CPU keeps track of the memory address which contains the data or the instruction

under execution.

There are two general types of memory: The Random Access Memory (RAM)

allows data and instructions to be "written" and "read" at any location

Jaddress) in the memory. On the contrary, the Read Only Memory (ROM), as its

name implies, does not allow alterations of its content, i.e. a program can

"read" information from the locations of the ROM but cannot "write" in it.

The RAN is used for the storage of any size general purpose programs, while

ROM is employed for the execution of highly specific and small in size pro-

grams. Most of the ROM is "programmed" in the factory and it is used to store

basic instructions for starting up the computer or basic input/output

commands, etc. Recently field-programmable ROM's have been introduced.

The cycle time of a computer is the time required by the CPU to read the

content of one word from memory and restore its content. According to the

value of the cycle time, we can distinguish the following types of RAM:

- Core memory, with typical cycle time -1 us (slow) and low cost, it is con-

structed with ferrite rings which retain the stored information when power

fails.

- Metal-Oxide Silicon Nemory (MOS), with typical cycle -500 ns, is faster and

cheaper than core memory and is based on simple semiconductor device.

- Bipolar Transistor Memory, with typical cycle -300 ns is still faster but,

more expensive. It is constructed from complex integrated circuits.

The cycle time is not the only factor

is. Various additional determinants, like

instructions in CPU, the number of general

the speed with which a computer executes a

C. Mass Storage Devices

that determines how fast a computer

the number and type of basic

purpose registers, etc., affect

program.

Are used to store large amounts of data and/or instructions. Various

types of mass storage devices are available with different, (a) capacity for

storage, (b) purchase cost and (c) speed for accessing and retrieving infor-

mation. The most common units are:

- Disks, with very large capacity for storage (1 f 100 million 16-bit words),

low access time (5 -:- 100 us) and high cost. The disks are distinguished

into fixed-head and moving head disks. The first have capacities in the

range 1-10 million 16-bit words and access time -5 us, while the second

have longer access times (50 :- 75 US) but higher capacities (up to 100

million words). Floppy disks are low cost, small capacity devices and are

the most common mass storage facilities for microcomputers.

- Magnetic Tapes. These are slow speed mass storage devices with significant

capacity (10 f 20 million words). They are seldomly found on process con-

trol computers and they are used to store off-line large programs and large

amounts of data.

D. Communication Peripherals

These equipments are used for communication between the operator and the

computer and include; typewriter terminals, line printers, video display units,

storage scope graphics terminals, card readers, X-Y plotters, etc. Uith such

devices the computer can display data describing the current state in the

operation of the process it controls, or inform the operator about the current

control actions taken by the computer. Furthermore, the communication

peripherals allow the operator to intervene and change set points, gains and

other characteristic parameters of a control loop, or switch control from the

computer to manual or other back-up control systems. The communication

peripherals must be supported by easy to use,

organized software. If this is not the case,

frustrated or lose confidence in the computer

it useless.

highly informative and well

then the operator may become

control system, thus rendering

E. Input/Output (I/O) Interface

This is the device which allows the communication between the computer and

the process to be controlled. In particular, the I/O performs the following

functions:

(a) It receives the signals from the measuring sensors and transducers

associated with the various measured process variables. These signals

may be continuous, alalog electrical voltages (thermocouple output,

flow or pressure transducer signals), or simple digital information (on-

off), e.g. various relays, on-off valves, etc.

(b) It sends out command signals to the various manipulated variables,

either in analog or digital form.

(c) It allows communication with other computers, which are used either as

process controllers or number "crunchers". This feature permits the

use of several computers for the "distributed digital control systems".

A digital computer without an input/output interface cannot function as a

process controller. But, what are the features of such interface, how does it

operate and how one selects the appropriate interface for a given process

control application? These questions are quite central in the design of a

computer based control system and will be covered in the next section with

some details.

Remark: It is common practice to characterize the digital computers as large

or maxi-computers, mini-computers or micro-computers. The stan-

dards for such classification are often obscure. Generally though

a large computer has words of 32 or 64 bits, memory larger than

128,000 words and a large number of associated peripherals. It is

primarily used for scientific or business purposes and physically it

occupies a large number of cabinets (10 to 15). Its cost is nor-

mally larger than $500,000. A minicomputer is a 12-, 16-, or 18-bit

machine (16-bit the most popular) with 4,000 up to 128,000 words of

memory. It has several peripherals and its cost may go up to

$200,000, depending on the size of memory and associated peripherals.

Microcomputers are normally characterized machines with 8-bit

words, although one may see 16-bit processors characterized as

micros. They possess from 1,000 up to 32,000 words of memory and

a few peripheral devices. The cost of the basic CPU is less than

$1,000 and goes up depending on the memory size and peripherals.

1laxicomputers are very costly for process control purposes and

are not used. Minicomputers are well suited to control a large

number of control loops. But, the future in process control

applications belongs to micros due to their low cost and

tremendous abilities.

26.2 COMPUTER-PROCESS INTERFACE FOR DATA ACQUISITION AND CONTROL

Return to Figure 26.1, which shows all necessary hardware elements in a

loop with analog controller. Replace the controller by a digital computer.

Then, the control functions (e.g. feedback P, PI, PID laws) will be performed

by an executable program (in BASIC, FORT&W, assembly language, etc.), which

resides in the memory of the computer. It is obvious that such control

program requires as data (input) the values of the measured outputs and

produces as results (output) the values that the manipulated variables should

have in order to keep the controlled variables a't the desired set points.

For a digital computer both input (data) and output (results) are in digital

form and correspond to discrete-time values. Here is where problems of

incompatibility arise and dictate the necessary hardware elements for an

input/output interface between a digital computer and the controlled process.

A. Samplers

The process measurement data (flow rates pressures, liquid levels,

temperatures, etc.) are provided continuously in time by the various measuring

sensors and transducers. The computer though can handle information on a

discreteltime basis, i.e. at given time instants for the following reason:

the time taken by the computer to "read" the measured value, calculate the

error, and make a control correction is finite. If during this period the

measured value changed, this is not recognized by the computer. Then, the

computer "reads" in effect at discrete time intervals. This is denoted

through the use of a sampler, which is simply a switch closing at specified

time intervals. In other words, a sampler takes in values of a continuous

signal and produces a sequence of sampled values at particular time instants

(Figure 26.3a).

B. Hold Elements

On the ohter hand, most of the final elements (pneumatic valves in

particular) are actuated by continuous in time signals (e.g. compressed air).

Therefore, the control commands produced by the computer program should be

converted from discrete time to continuous in time signals. This is accom-

plished by the hold elements. Figure 26.3b shows schematically the conversion

of a sequen-e of discrete in time signals to a stair-step like continuous

signal. Here, the hold element keeps the value of a discrete in time signal

constant for all the period until the next signal comes along.

C. Analog to Digital Converters (ADC)

The measurement data are not only provided continuously in time, but they

are also analog electrical signals in nature. They cannot be used directly by

the control program which requires data in a digital form (e.g. information

coded in l&bit words, for a 16-bit word machine). Therefore, the input

interface should contain an analog to digital converter (ADC or A/D converter).

The analog signals coming from measuring devices and sensors are modified

so that they fall within a prespecified voltage range, e.g. 0-lOV, 0-5V, tlOV,

or +5V, etc. The digital signal produced by an A/D converter is expressed by

an integer number in a binary form. The resolution of the conversion depends

on the number of bits used by the converter to encode an analog value in

digital form. The most common converters use 8-bit or 12-bit resolution, with

the second providing smaller error and being more costly.

Consider a voltage range 0-1V and an n-bit converter. The n bits

define 2n integer numbers (including zero), which in turn define 2n-l

voltage intervals between 0 and 1. Thus, the accuracy of the conversion

expressed by the value of resolution is given by, *

1Resolution = -2"-1

(26.1)

For a 12-bit converter the resolution is about 0.05 per cent, i.e. when two

voltage values differ by more than 0.05% of the prespecified voltage range,

the converter will distinguish the two signals and assign two different

integers for them. For an 8-bit converter the resolution is smaller, about

0.4%. Usually, both 8-bit and 12-bit converter's are satisfactory for process

control purposes. Converters with more than 12 bits are used only when

extremely high precision is required and are quite costly.

The conversion speed is very high and typical A/'D converters used for

process control allow 20,000-lOO,O-0 conversions per second. Higher rates

can be achieved by high-performance converters and are useful only for very

special problems. ,

To avoid the need for a large number of A/D converters handling the con-

version of a large number of different analog signals, it is usual practice

to use a multiplexer. This is an electronic switch with several ports, which

can serve sequentially several lines carrying analog signals (Figure 26.4).

D. D i g i t a l (DBC)

The control commands produced by the control program are in digital form,

but most of the final control elements, pneumatic valves in particular, are

actuated by analog signals (e.g. compressed air). To erase this incompati-

bility, the output interface should include a ) (DAC

or D/A converter).

D/A converters function in the reverse manner than A/D converters. Thus,

for a 12-bit converter we can have 212 = 4096 integer numbers defining 4095.

intervals of the prespecified voltage range, say 0-10 volts. Then, the integer

number 516 causes an analog output of

$$lO = 1.26 volts.

E. Digital I/O

A digital computer control system may be required to handle digital

inputs or outputs for a variety of reasons. Typical examples are:

(a) Information concerning

l the status of relays turning pumps, valves, lights and other devices

o n o r o f f ,

l the status of multiplexers,

l the settings of various switches,

a the status of communication peripherals and various digital logic

devices.

(b) Control commands to

l relays, switches, solenoids, digital logic devices,

0 stepping motors.

(c) Communication between

0 several computers

l a computer and its peripherals, etc.

Therefore, the I/O interface is not designed to receive analog digital input

and output signals.

The digital signals are fully compatible with a computer so that no

special converters are needed in the I/O interface. The transmission (input

or output) of digital signals by the I/O interface can be done either in

parallel (two-way, in and out, simultaneous transmission) or in series (one-

way, in or out transmission). The length of a digital information transmitted

in or out is one word (i.e. 16 bits for a 16-bit machine). The transmission

rates vary from very low to very high, and are expressed in terms of band rates,

where one band = lOx(number of bytes transmitted/second).

Remark: When the prespecified range of voltages involves negative and

positive values then the first bit of an A/D or D/A converter is

used to denote the sign. Thus, consider the range k5V. For a

12-bit converter we have 2(12-1) = 2048 positive integer numbers

(including zero) to represent voltage values in the positive

range 0 5 volts 2 5. Also, there are 2047 negative integer num-

bers (excluding zero) covering the range -5 5 volts < 0.

26.3 COMPUTER CONTROL LOOPS

In the previous two sections we gave a brief description of a digital

computer and its characteristics, of the associated peripheral and of the

I/O interface required for data acquisition and control. In this section we

will examine the various types of control loops, which result when a digital

computer is used as the main controller, as well as the necessary hardware

components.

A. Single-Loop Control

Figure 26.1 shows the hardware elements of a single-loop control system,

using an analog controller. When we replace the analog controller by a

digital computer, the following changes take place:

(a) The measurement signal from the sensor or transducer is sampled at

prespecified intervals of time, using a simple sampler. Thus, it is

converted from continuous to discrete-time signal. This in turn is

converted from analog to digital by an A/D converter and enters the

computer.

(b) The hardwired analog logic of an analog controller is replaced by the

software of the control program which resides in the memory and is

executed by the computer whenever it is called.

(c) The control commands produced by the control program are digital and

discrete-time signals. They are first converted to analog by a D/A

converter and then to continuous in time signals by simple hold elements

before they actuate the final control elements.

Figure 26.5 summarizes the above changes and indicates all hardware components

present in a single computer control loop. We observe that both continuous

and discrete-time signals are present in the loop. They are denoted by c:

and d: respectively. It should also,be noted that the set-point values,

as well as the values of the adjustable control parameters (e.g. gains, reset

- or rate time constants, etc.), are now introduced by the operator through a

typewriter terminal.

B. Multiple-Loop Control

A digital computer can be used to control simultaneously several outputs

and not only one as discussed above. We will still need an interface between

the computer and the process, but now is somewhat different. Thus:

- Instead of using one A/D converter for every measured variable, we employ a

single A/D converter which serves all measured variables sequentially

through a multiplexer.

- A multiplexer can also be used to obtain several outputs from a single D/A

converter.

- The control program is now composed of several subprograms, each one used

to control a different loop. Furthermore, the control program should be

able to coordinate the execution of the various subprograms so that each

loop functions properly.

Figure 26.6 shows the use of a single computer (CPU) to control two outputs.

When a digital computer has assumed all control actions of a conventional

controller, then we talk about direct digital control (DDC). Both systems in

Figures 26.5 and 26.6 are examples of direct digital control.

C. Supervisory Control

Unlike the situation of direct digital control, we may use the computer

to change only the set points or the values of the adjustable control

parameters of the local controllers. The resulting system is known as

supervisory control (Figure 26.7a).

The local controllers may be conventional, analog devices or digital

computers implementing direct digital control. An I/O interface is needed to

inform the supervising computer about the state of the local control loops and

for the computer to provide the set point or other changes to the local con-

trollers. When the local controllers are digital computers, then the I/O

interface carries only digital signals, allowing the communication between the

supervising computer and the local DDC's.

Supervisory control has been applied extensively in chemical processes,

to optimize their operation (minimize operating cost, maximize efficiency in

energy or raw materials utilization, maximize production profit, etc.).

Thus, one computer supervises and coordinates the operating of several control

loops, deciding what are the best set-point values for the various loops.

Figure 26.7b shows schematically the architecture of a supervisory control

system for a chemical plant.

Remarks: (1) In some cases, the voltage signal produced by the measuring

sensor or transducer is very low and easily corrupted by

noise. Typical example is the ou.tput of a thermocouple which

is of the order of milivolts. Low voltage signals are

normally amplified to the prespecified voltage range for

A/D conversion. If several signals need amplification, then

we may use a multiplexer first, followed by a single common

amplifier.

(2) The computer is physically located at some distance from

the controlled process. We can put the computer-process

I/O interface, (a) close to the computer or (b) close to

the controlled process.

In the first case we have analog signals transmitted over some distance

between the process and the interface. This approach yields satisfactory

results if the analog signals are transmitted over short distances, i.e. no

longer than 200-300 ft. For longer distances there is significant deterio-

ration in the transmitted signal due to voltage losses and cable capacitance.

Furthermore, external noise may seriously corrupt the transmitted analog

signal.

Alternative (b) is preferred when the transmission distance is long and

there are strong sources of external noise. In such case the proximity

between process and interface allows conversion of "clean" and "strong"

analog signals to ditital, which can then be transmitted to the computer.

Digital signals are less susceptible to external noise and can be transmitted

over long distances by telephone lines.

(3) Microcomputers are normally used for local direct digital

control of several loops (5 to 10). Minicomputers are

usually employed as supervising computers in a super-

visory control architecture.

26.4 NEW CONTROL DESIGN PROBLEMS

The introduction of a digital computer for process control raises some

new design questions, which were not covered by earlier chapters.

(a>

(b)

(4

Cd)

(e>

The digital computer uses and produces information in discrete-time

form. Therefore, the continuous process models which we have used for

the design of analog controllers are not appropriate. We need to

develop a mechanism which will convert the differential equations

describing the process to difference equations, which are convenient

for discrete-time representation.

How fast should we sample a measured variable to produce its discrete-

time equivalent? Does the sampling rate affect the quality of control?

How should we reconstruct a continuous signal from its discrete-time

equivalent, so that we can actuate the final control elements? How

does the type of reconstruction (i.e. type of hold element) affect

the quality of control?

The Laplace transforms allowed us to develop simple input-output

relationships for a process and provided the framework for easy

analysis and design of loops with continuous, analog controllers.

For discrete-time systems we need to introduce new analytical tools.

This will be provided by the z-transforms.

Does the design of a control loop change when we use digital computer

control? What about the stability conditions and the tuning of a

loop?

(f) How can we use the tremendous computational power of a computer to

implement some advanced notions of process control like; feedforward,

adaptive, inferential, optimizing, etc.?

In the following chapters we will address all the above and other questions

related to the design of computer control system.

Before closing this section, let us make a few remarks on the software

required to implement the control laws, and which constitutes a neiJ control

design problem introduced by the use of digital computers.

There are two classes of software programs needed for computer process

control applications; the computer system nrograms and the application-

programs.

(1) Computer System Programs. Are supplied by the manufacturer of process

control computers or specialized software houses. They include:

- Operating Systems, which deal with the real-time operation of the computer

control system. They supervise the execution of the control programs and

to this end organize the various operations of the hardware components in an

orderlymanner- providingefficientuse of the CPU, memory, communication

peripherals and I/O interface.

- Utility Programs, such as assemblers, editors, debuggers, compilers, etc.

which support the development of the application programs written by the

users.

(2) Application Programs. These are written by the user and perform the

specific functions required by the control problem such as;

- monitoring the measured process variables at specified time intervals,

- executing the algorithms of the control laws,

- coordinating the control actions to the various final control elements,

- computing and changing set points,

- computing and changing the values of the adjustable controller parameters,

- calling alarms if process variables exceed preset limits, etc.

The application programs may be written in high level languages such as

FORTRAN or low level such as machine language. High level languages are

easily understood by the programmer and allow an easy statement for the

solution procedure. They require though increased memory and slow down the

execution because the FORTRAN statements for example must be translated into

the machine language before they can be executed. Normally, one writes the

complex part of a control program'in a high level language because it is an

overwhelming task for machine language programming, which is only used to

encode those functions performed at high speeds and repeatedly (data

acquisition, implementation of control commands, etc.).

SLWMARY AND CONCLUDING REMARKS

Digital computers in chemical process control are already a successful

reality and offer exciting possibilities for the future. The high compu-

tational speed coupled with large capacities to store information make the

digital computers very "intelligent" process controllers. It must be noted

though that the advanced and continuously advancing technology of the micro-

processors has not been fully exploited by the process control designer, and

that a tremendous potential exists. This is the challenge for the new

chemical engineers.

A computer control system is composed basically of (a) the central pro-

cessing unit with the associated peripherals (memory, mass storage devices,

communication devices) and (b) the I/O interface between the computer and the

process it controls. From all specific hardware features we should single out

the real-time clock which allows the computer operations to be synchronized

with the real-time needs of the controlled process. It is not easy to have

computer control without a real-time clock.

A digital computer is a machine that handles information in digital and

discrete-time form, which is incompatible with the largely analog and con-

tinuous in time signals of a chemical process. This incompatibility is

erased by the computer process I/O interface with its samplers, hold elements,

multiplexers, A/D and D/A converters.

Two are the main modes of computer control; the direct digital (DDC) and

the supervisory control. The first is implemented by local microcomputers

which can handle from 5 to 10 loops each, thus replacing as many analog con-

trollers. The economics of DDC are favorable and their use expands continuously.

The second does not implement all control actions but it is limited to the

supervision of local controllers and the change of their set points and

adjustable parameters. Supervisory control is employed to optimize the

performance of a chemical plant by optimally adjusting the set points of the

local controllers, which may analog or digital.

The use of a digital computer as process controller creates a new set of

design questions such as; mathematical tools to handle discrete-time signals

and process models, process analysis and controller design techniques for

discrete-time systems, how to use the computational power for advanced control

techniques, etc. These questions define also the scope of the chapters to

follow.


1. What is a digital and what is an analog signal (information)? Identify

their differences and discuss how one can be converted to the other.

2. Define the units of bit, tyte, word. Why are all needed to encode

information in a digital system?

637

3.

4.

5.

6.

7.

8.

9.

Describe the structure of a conventional digital computer, and identify

the characteristics of each hardware component in this structure.

What are the basic and what are the optional features of a central

processing unit? Are all of them needed for a process control

computer?

What is the real-time clock needed for? How does it function? Why is

it very difficult to use a computer for process control without a

real-time clock? Do you have any suggestions on how you can count

time elapsed without a real-time clock?

Identify the functions and hardware components of a computer-process

I/O interface.

Explain in simple physical terms how you can convert a continuous to

a discrete in time signal and vice versa. Why are these two operations

necessary in a computer-process I/O interface?

Describe in physical terms the conversion of a signal from analog to

ditital and vice versa. Why are these conversions needed in the I/O

interface?

What is easier and less costly to do; (a) multiplex N analog signals

first and then use a common A/D converter or (b) convert the N analog

signals to digital first and then multiplex them to enter the computer

through a single, common word of storage?

10. Repeat question 9., but now use N digital signals, a multiplexer and

D/A converters.

11. For a prespecified voltage range of +lOV, find the resolution of a

12-bit A/D converter. What are the integer numbers representing

voltages of -2V and +5V? What is the possible conversion error in

volts?

12.

13.

14.

15.

16.

17.

18.

19.

Find the voltages which are represented by the integer numbers -712 and

+1514, within a prespecified range +lOV. What is the possible conversion

error in volts?

Find the number of bits needed for a D/A converter to yield an error

less than 0.0001 volts, for a prespecified range of voltages 0 to 5V.

Define the direct digital and supervisory control. Which one is used

for regulatory control actions and which for servo operations? In a

supervisory control mode, what are better as local controllers,

analog or digital devices? Discuss relative advantages and disadvantages.

What size computers would you use for DDC and supervisory control?

Why?, How do the local DDC's communicate with the supervising computer?

Identify all components of a DDC system suing one microprocessor to

handle the four loops of a flash drum (see Example 23.2, Figure 23.lb).

Describe supervisory control system for the plant of Example 23.4

(Figure 23.10) using as local controllers; (a) analog devices con-

trolling one loop each, and (b) microcomputers which are capable of

handling four loops each. How would you select the loops to be con-

trolled by each local microprocessor?

What is high level and what a low level language for computer control

applications? Which one would you use and why?

Discuss the new design problems raised by the use of a digital computer

for process control.

!TRAPJSDUCER

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[ANALOG O R PIGI-T~L)

CHAPTER 26

FROM CONTINUOUS TO DISCRETE-TIME SYSTEMSr

When we use.continuous, analog controllers, all signals in a loop are

continuous in time. Then, the dynamic behavior of each component in the loop

(process, measuring device, controller, final control element). as well as the

response of the overall control system can be effectively analyzed by continuous

models (differential equations in the time domain or transfer functions in the

Laplace domain).

Theintroduction of a digital computer in a process control loop changes

the above picture because a computer can handle information on a discrete-time

basis only, i.e. at particular time instants. As we can see from Figure 26.5,

in a computer control loop we have both continuous and discrete-time signals

present. The implication of this feature is twofold:

(a) Continuous signals must be converted to discrete-time before they can

be "read" by the computer and the discrete-time control commands pro- -

duced by the computer must be converted to continuous signals before

they can actuate the final control elements.

(b) The continuous models (e.g. differential equations in the time-domain,

or input-output models in the Laplace domain) are not convenient to

analyze the dynamic behavior of loops with computer control and discrete-

time models are needed.

Therefore, before proceeding with the development of design techniques for

computer control systems, we should study how to convert continuous signals

and models to discrete-time equivalents and vice versa. This is the subject

of Chapter 27.

27.1 SAMPLING CONTINUOUS SIGNALS

Consider a line carrying a measurement signal y, which varies con-

tinuously in time as shown in Figure 27.la. The line is interrupted by a

switch, called sampler, which closes every Ts seconds and remains closed

for an infinitesimally short period of time (theoretically a time point).

The x's of Figure 27.lb show the value y, of the signal at the other end

of the line, when TS

= 1 second. We notice that the resulting signal has

values only when the time is multiple of T5, i.e. at time points

t = nTS

n=0,1,2,***

and it is zero for any other times. The signal of Figure 27.lb is called

discrete-time or sampled representation of the continuous signal in Figure

27.la, with a sampling period of 1 second. Figure 27.1~ shows another sampled

representation of the same continuous signal but with sampling period of

three seconds. Two observations are easy to make:

- As the sampling period tends to zero, the sampled representation comes

closer to the continuous signal but requires an increasingly large number

of sampled values.

- On the other hand, as

are required, but the

deteriorates, and the

or impossible.

the sampling period increases, fewer sampled values

sampled representation of a continuous signal

reconstruction of the original signal becomes poor

Thus, the obvious question arises:

"How does one select the best sampling period so that thesampled representation of a continuous signal is satis-factory, without requiring excessively large number ofsampled values?"

There exists a mathematical answer to the above question whose development is

quite complex and goes beyond the scope of this text. We will try to give a

practical answer based on the typical dynamic responses encountered in

chemical processes.

Example 27.1 - Sampling the Response of First-Order Systems

Consider a first-order linear system subject to an input step change.

Figure 10.4 shows the response of the system with time. In Section 10.4 we

found that the response reaches the 63.2% of its final value when the time

elapsed is equal to one time constant rP'

Also, when t = 2rP

the response

has reached 86.5% of the final value, at t = 3rP

the 95%, etc. Therefore,

if the sampled represenntation of the response is going to be of any value,

the sampling period must be smaller than one time constant. How much smaller?

Practical experience suggests that a sampling period between 0.1 and 0.2 of

one time constant yields satisfactory results.

Suppose that the first-order system possesses dead time. Then its

response to a step input is given by Figure 12.3a. If the dead time, td, is

of the same order of magnitude as the time constant, TP'

then select the

sampling period equal to 0.1 ~~ or 0.1 TP'

whichever is smaller. If td

is much smaller thanrP

then neglect the dead time and take Ts = 0.1 T .P

The above rational can be extended.

2Exam le 27.2 - Sam line

The rule developed in Example 27.1 for the sampling rate of a first-

order response can be extended to cover a large class of overdamped systems.

Figure 16.12a shows the experimental response of an overdamped process to an

input step change. The S-shaped response of Figure 16.12a can be approximated

by the response of a first-order plus dead-time system, as described in

Section 16.5. Thus, we can identify the dominant time constant rP

and the

dead time td. The sampling period should be 0.1 -cp or p.1 td, whichever

is smaller, or 0.1 T ifP td is much smaller than

rP'.

Example 27.3 - Sampling the Oscillating Response of a System

Oscillatory behavior is exhibited by underdamped open or closed-loop

systems (see Chapter 11) and by the steady state response of linear systems

in general subject to periodic, sinusoidal input changes (Chapter 17).

To develop a good sampled representation of an oscillating signal follow

the rule:

"Sample an oscillating signal more than two times per cycleof oscillation, otherwise it is impossible to reconstructthe original signal from its sampled values."

To domonstrate the above rule consider the sinusoidal signal of Figure 27.2a

sampled once per cycle. The sampled values are shown in Figure 27.2b and

Figure 27.2~ shows clearly that we cannot reconstruct uniquely the original

sinusoid because there exist several waves passing through the sampled

values. Therefore, sampling with a period equal to the period of oscillation

renders useless sampled values.

The example of Figure 27.3 demonstrates a serious error which can be

committed by an improper selection of the sampling period. The sinusoid

of Figure 27.3a is sampled with a period equal to 314 of the period of

oscillation (i.e. 413 samples per period or better expressed 4 samples per 3

cycles of oscillation). The sampled values are shown in Figure 27.3b. When

we attempt to reconstruct the sinusoid going through these sampled values, we

take the signal of Figure 27.3~ which is clearly different than the original.

Let us now develop a quantitative description for the sampling operation

and the resulting sampled values of a continuous signal.

The sampler is a physical switch which stays closed for a very small but

finite period of time, At, around the sampling instant. During this time the

sampler output takes the value of the continuous signal and has the form

shown in Figure 27.4a, or approximately the pulse form of Figure 27.4b. To

develop a concise mathematical descripti.on we assume that the sampler acts

instantly, i.e. At-@. To retain the same area under the pulse as At+0 the

height of the pulse goes to infinity and at the limit we take an impulse of

infinite magnitude, zero duration and an area ("strength") under the impulse

equal to the magnitude of the continuous signal at a sampling instant. Thus,

the impulse at the sampling point t = n Ts; n=0,1,2,*** is given by

Y* (nTs) = y(nT,)*b(t - nTs) (27.1)

where &(t - nTs) is the. unit impulse or Dirac function at t = nT, (see

Section 7.2 and Figure 7.3). A sampler with the idealized output given by

eqn. (27.1) is known as ideal impulse sampler.

We can extend eqn. (27.1) to apply for any time. Thus, the sequence of

impulses y*(t) coming out of an impulse sampler is expressed by the following

equation in the time-domain

y"(t) = y*(OTs) + y*(lT,) + y*(2Ts) + l . ..

= y(0)6(t) + y(Ts)6(t - Ts) + y(2Ts)6(t - 2Ts) + l *a

and finally,

y"(t) = nzo y(nTs>S(t - nTs) (27.2)

Equation (27.2) is compatible with the idealized physical picture we have

considered, i.e.

- at the sampling instants the "strength" of the impulses is equal to the

value of the continuous signal, and

- between the sampling instants the "strength" is zero, that is, no output

value.

Remark: Take the Laplace transforms of both sides of eqn. (27.2)

y*(s) = ngo y(nTs,dh-nT

- nT,)] = nY& Y(nTs) e s c&s(t) 1

From the last equation we find an expression for the sequence of

impulses coming out of the impulse sampler in the Laplace domain

(i.e. s-domain),

T”(s) = nzo y(nTs) e-TS (27.2a)

27.2 RECONSTRUCTION OF CONTINUOUS SIGNALS FROM THEIR DISCRETE-TIXE VALUES

The discrete-time nature of a digital computer implies that when a com-

puter is used to control a process the control commands are given periodically

as impulses at particular time instants and not continuously in time. Such

sequence of control impulses cannot maintain a final control element con-

tinuously in operation. Thus, a valve opens when a control impulse from the

computer reaches the valve, but then it closes until the next control impulse

arrives at the valve. Such control action is undesirable and the question is;

how can we construct a continuous signal from its discrete-time values?

Consider a control signal produced intermittently every T seconds by a

computer and expressed by a series of impulses (discrete-time values) shown

in Figure 27.5a,

m*(O) = m(O)d(t), m*(T) = m(T)G(t-T), m*(2T) = m(2T)6(t-2T),**-

The simplest way to convert a sequence of discrete-time values into a con-

tinuous signal is to keep the discrete-time value of the signal at t = nT,

m(t) = m(nT) for nT L t s (n+l)T '(27.3)

and n=0,1,2,***

In particular;

- for O<t<T m(t) = m(O)

- for T5t< 2T m(t) = m(T)

- for 2T -< t < 3T m(t) = m(2T), etc.

The resulting stair-step continuous signal is shown in Figure 27.5b. The

conversion represented by eqn. (27.3) is known as zero-order hold. It does

not represent the only way to construct a continuous signal from its discrete-

time values.

n=0,1,2,*** constant, until the.next one comes along.. Thus, if m(t) is

the resulting continuous signal, we have,

Consider two successive discrete-time values, say m[(n-l)T] and m(nT).

We assume that for the next period nT 5 t -< (n+l)T the continuous signal can

be given by a linear extrapolation of the'previous two values, i.e.

m(t) = m(nT) + m(nT) - m[(n-1)Tl .(tT - nT)

for nT I t < (n+l)T

and n=2,3,4,***

Equation (27.4) yields the so-called first-order hold and the continuous

signal it produces is shown in Figure 27.5~. Notice that the first-order hold

element needs at least two values to start the construction of the continuous

signal while the zero-order hbld need only one.

It is possible to develop second-, third-, or higher-order hold elements.

They need 3, 4, or more initial discrete-time values before they can start the

construction of a continuous. As the order of a hold element increases, the

computational load increases and becomes more complex, with marginal improvements

in the quality of reconstructed signal. Therefore, for most of the process

control applications the zero-order hold element provides satisfactory results

with low computational load and it is normally used. To improve the quality of

a reconstructed signal is better to decrease the period between two successive

discrete-time values, rather than increase the order of the hold element.

Example 27.4 - Comparing the Results of Zero- and First-Order Hold Elements

We will consider two distinct cases of discrete-time signals; (a) slowly

varying with time (Figure 27.6a) and (b) rapidly changing with time (Figure

27.7a):

- For the slowly varying signal the superviority of the first-order hold

is obvious (see Figures 27.6b and 27.6~). This is due to the almost

constant slope of the changing signal over large periods of time which

permits a successful linear extrapolation. Nevertheless, the performance

of the zero-order hold is also considered satisfactory.

- For the rapidly changing signal both reconstructions are rather poor (see

Figures 27.7b and 27.7~). This is mainly due to the long period of

occurrence of discrete-time values and any improvement should come from

shortening this period, i.e. have more discrete-time values of the signal

per unit of time. Nevertheless, Figures 27.7b and 27.7~ indicate some very

useful features:

l The zero-order hold element by this nature it never generates

"extreme" values, outside the range of the discrete-time values. In

other words, the zero-order hold produces a "conservative" continuous

signal which is satisfactory during periods of slow change and unsatis-

factory during periods of fast change in the values of the discrete-time

signal.

l The first-order hold produces significant excursions beyond the range

of the discrete-time values. This could produce undesirably large

control actions which may endanger the stability of the controlled-

process.

Remarks: (1) The mathematical basis behind the construction of a hold

element, independently of its order, is the following.

Consider the continuous signal, m(t), which must be con-

structed from discrete-time values m(T), m(wT), m(3T),

etc. The Taylor series of m(t) around a sampled value

m(nT) is given by

2m(t) = m(nT) + ($)t=nT(t-nT)++ (d)dt2 t=nT(t-nT)2+***

If we retain only the zero-order terms (i.e. constant) we

take the zero-order hold element (eqn. (27.3)),

m(t) = m(nT) nT < t < (n+l)T (27.3)

If we retain the zero- and first-order terms we take,

m(t) = m(nT) + (g)t=nT(t-nT)

The derivative (dm/dt)t=nT can be approximated by,

(chn) t=nT = m(nT) - ;[ (n-l)Tl

Thus, we take the first-order hold element (eqn. (27.4)),

m(t) = m(nT) + m(nU - m[(n-1)TlT (t-nT) (27.4)

Similarly, by retaining additional terms, 2nd, 3rd, etc.

order we can develop higher-order hold elements. All

necessary derivatives will be numerically approximated

as above, but they will require an increasing number of

discrete-time values.

(2) The output of a zero-order hold element is like pulse,

having a constant height equal to m(nT) and duration

T. After recalling that the Laplace transform of a unit

pulse is given by eqn. (7.12), then from eqn. (27.3) we

find that the Laplace transform of a zero-order hold out-

put is given by

iids> = m(nT)1 - emST

S

The last equation implies that the transfer function of a

zero-order hold element is given by

-STHo(s) = 1 - e

S(27.5)

(3) In a similar manner we can find the transfer functions

of a first-order 'hold element

HIW = T1 + ST (1 - ewST 2

S> (27.6)

27.3 CONVERSION OF CONTINUOUS TO DISCRETE-TIME MODELS

We will start by recalling the typical computer loop for direct digital

control shown in Figure 26.5. For presentational purposes only we simplify

the loop to that shown in Figure 27.8 by retaining its basic four components;

process, A/D converter with the associated sampler, digital controller, and

D/A converter with teh associated hold element. We notice that both con-

tinuous and discrete-time signals are *present in the loop. Pius:

- The process has continuous input and output signals and consequently it

can be described by continuous models (differential equations in the time

domain, transfer functions in the Laplace domain).

- The discrete-time output of the A/D converter can be modeled as a function

of the continuous input, by eqn. (27.2) in the time domain or by eqn.

(27.2a) in the Laplace domain.

- The hold elements can be represented by the corresponding transfer

functions (eqns. (27.5) and (27.6)).

- The digital controller has both input and output signals discrete in time.

So far we have not studied any techniques to model such systems which from

now on we will call discrete.

Let us go a step further. If the main controller were a continuous feed-

back PID device, the output of the controller would be given by eqn. (13.6) in

Section 13.2,

c(t)KC= K$t) +T

ic(t)dt + Kc rD g + cs (13.6)

I

A continuous model for the control action such as that of eqn. (13.6) is

inconvenient for a digital controller which uses error values at particular .

time instants, i.e.

E(O), c(TS), E(2TS) ,*-*,E(nTS) ,***

and produces control commands at discrete time points, i.e.

c(T), c(2T),***,c(nT),***

But, how can we convert a continuous model to an equivalent discrete-time one?

This is the question to resolve in this section.

Example 27.5 - The Discrete-Time Model of a Digital PID Controller

Start with the continuous analog of a PID control action, given by eqn.

(13.6). We will examine each term (proportiaonal, integral, derivative)

separately:

- Every sampling period a sampled value of the process output enters the

computer. Let y(nTs) be the sampled value at the n-th sampling instant..

y(nTs) is compared to the set-point value at the same instant and yields

the value of the discrete-time error,

E(nT,) = ySp(nTs) - y(nTs)

Then, the discrete-time control action produced by the proportional mode

is,

Kc E(nTs)

- The control action produced by the integral mode is based on the inte-

gration of errors over a time period. Since the values of the errors are

available on a discrete-time basis, the integral /s(t)dt can only be

approximated by numerical integration. Figure 27.9a shows the numerical

evaluation of an integral, using rectangular integration. It is easy to

see that

i’c(t>dt 2 Ts k;. 4kTs)0

Therefore, the integral model control action is given by

Kc Ts n- $0 ENS)3

- For the derivative mode action we need a numerical evaluation of the

derivative deldt. Figure 27.9b shows a first-order difference approxi-

mation of the derivative. Therefore,

K ds rDc rD dt = Kc < {E(nTs) - c[(n-l)Tsl)

Consequently, the control action of a digital PID controller is determined

by the follwoing discrete-time model,

c(nTs) = K;E KcTs n,(nTs) + -~~ kz0KCTD

dkTs) + -y-S

{s(nTs) - s[(n-l)Tsll+cs

(27.6)

which is nothing else but a numerical approximation of its continuous

counterpart. Due to the use of finite differences for the approxi-

mation of integrals and derivatives, eqn. (27.6) is known as the

difference equation.

Example (27.5) is very instructive on how to develop a discrete-time

model from its equivalent continuous one. The procedure can be generalized

for any continuous dynamic model as follows:

(a) Start with the differential equations describing a continuous model in

the time domain.

(b) Approximate the derivatives of any order by finite differences.

(c) Approximate any integral terms in the model by a scheme of numerical

integration.

(d) The values of any simple terms are equated to the corresponding

discrete-time values at the sampling instants. The discrete-time

modeling equation(s) resulting from the above procedure is known

as the difference equation(s) in contrast to the term differential

equation(s) used for continuous model.

From courses in numerical analysis we know that there exists a variety

of techniques to approximate derivatives and integral terms. A detailed

exposition of such methods goes beyond the scope of the present text and

the interested reader can consult the various references on numberical analysis

cited in the References section at the end of Part VII.

Let us close this section with more examples on the time discretization

of continuous models.

(27.7)

Example 27.6 - Discrete-Time Model of a First-Order Process

A nonlinear first-order process is described by,

AYdt = f(y,d

Approximate the derivative by first-order difference, i.e.

LiY yn+l - yndt = T

Then, at a given time instant t = nT eqn. (27.7) yields,

Yn+l = yn + T-f(yn,mn> (27.8)

Equation (27.8) is the discrete-time dynamic model of a first-order process

and shows what the output of the process will be at the next time instant,

using current values of the input, mn, and output, y,.

For a linear first-order system we have (see eqn. (10.2))

ayrP = + y = KPm

and using the above we can easily derive the difference equation which is the

discrete-time model,

Yn+l = (1 KPT--;T-)y, + - m (27.9)P rP n

[Note: In all the above expressions we have used the following simplifying

notation, y(nT) 5 Y, and m(nT) z mn. T is the time period between two

successive discrete-time values].

Example 27.7 - Discrete-Time Model of a Second-Order Process

A linear second-order system is described by eqn. (11.2),

p & aydt2

+ 251 dt + y = Kpm (11.2)

We have already seen how to approximate the first-order derivative, i.e.

h57

4.Y Yn+1 - YIldt = T

For the second-order derivative we have:

= $ (Y,+2 - 2Yn+1 + Y,)

Replace the derivatives in eqn. (11.2) by their approximations and take the

following difference equation,

12 (Yn+2T2

- 2yni-l + Y,) + 25 I (Y,+~ - Y,) + Y, = Kp mn

or

2 2Yn-i-2

= 2(1 - 5 +)Y~+~ - (T - 25 $ + l)y, + Kp G mT2

n (27.10)T

Equation (27.10) represents the discrete-time model of a second-order process.

Notice that in order to compute the next value (Y~+~) of y, we need its

previous two values (~,+~,y,). For 3rd, 4th, and higher-order systems we will

need the previous 3, 4, and more values of y to approximate all derivatives.

Example 27.8 - Discrete-Time Model of a Multivariable Process

Consider the following process with two inputs and two outputs:

dyldt + allyl + a12Y2 = bllml + b12m2

dy2- + a2gl +. a22Y2 = b2pl + b22m2dt

With the first-order difference approximation fot the derivatives we take the

following difference equations which represent the discrete-time model of the

multivariable process,

y1,*+1 - y1,nT + allyl,n + a12y2,n = b11m1,n + b12m2,n -

y2,n+l - y2,n .T + a21Yl,n + a22Y2,n = b21ml,n + b22m2,n

O K

yl,n+l = (l--Tall)yl,n - Ta12y2,n + Ubllml,n + b12m2,J (27.11)

y2,n+l = -T a21yl,n + (1 - Ta22)y2,n + T(b21ml,n + b22m2,n) (27.12)

Remarks: (1) Numerical differentiation of process measurement data can

cause serious problems when there is appreciable process

noise, i.e. random effects appearing during the operation

but not included in the assumed model. To overcome this

difficulty we can use digital filters which filter out any

' noise and yield "smooth" measurement data. In a sub-

sequent chapter we will study the development of such

filters.

(2) The discretizationof continuous models with dead time is

rather straightforward. For example, consider a first-

order process with dead time, td, between the input, m(t),

and the output, y(t), i.e.

dv + yrp z = Kp m(t - td>

Let td = k T, i.e. the dead time is an integer multiple

of the period, T. Then the discrete-time model is easily

found to be

Yn+l = (1 -+-)y, + y mp (n-k'0

(27.13)P

(3) The quality of an approximate discrete-time model improves

as the value of the discretization time interval, T,

decreases. Why?

(4) The conversion of a discrete to continuous model in the

time domain is possible but not simple. In the next

chapter we will see that such conversions are more easily

done in the Laplace domain.

SUM?URY AND CONCLLDING REMARKS

D The presence of a digital computer in a control loop implies the

coexistence of continuous and discrete-time signals and systems in the loop.

Therefore, we should have the mathematical facility for converting continuous

signals and systems to discrete-time ones and vice versa.

The assumption of an ideal impulse sampler allows an easy quantification- -

of discrete-time signals produced from their continuous counterpoarts. On

the other hand it is easy to construct and describe hold elements of various

orders, with the zero-order being the most popular. Laplace transforms for

both components (sampler, hold element) are possible and yield an analytical

description of the two conversions (continuous to discrete and discrete to

continuous).

The discrete nature of a digital computer imposes the need for a

discrete-time representation of the various components in the loop. Such

discrete models for the process, controllers, etc. can be easily derived by

numerical approximation of the corresponding continuous models. There exist

various forms for approximating numerically derivatives of any order,

integrals, etc., with varying complexity and computational requirements. The

preferred approximation depends on the charactewristics of the continuous

model and the desired quality of the discrete model.

In the present chapter all conversions have been confined in the time

domain. With the introduction of z-transforms in the next chapter, we will

develop the mathematical framework which allows easy analyses of discrete-

time process dynamic and design of discrete-time computer control systems.


1.

2.

3.

4.

5.

6.

7.

8.

Explain in your own words why we need to convert continuous signals to

discrete-time ones and vice versa. Give a physical example for this

need using as reference the computer control of a stirred tank heater.

Using the same computer control system as in 1. above, explain why we

need to convert continuous to discrete-time models and vice versa.

What is a sampled signal? Sketch one and indicate how it is related to

its continuous counterpart.

Define the ideal impulse sampler. How does it differ from a real

sampler? Draw two sketches indicating the outputs of an ideal impulse

and a real sampler.

Develop two mathematical expressions describing the output of an ideal

impulse smapler; one in the time domain and the other in the Laplace

domain.

How would you select the sampling rate for, (a) the response of a general

underdamped open-loop system and (b) the oscillating response of a

closed-loop system?

Is it satisfactory sampling rate to take three samples every two

cycles of an oscillating sinusoidal signal? Demonstrate graphically

why yes or no.

Discuss the mathematical basis for the construction of various orders of

hold elements. Develop the time domain expressions for zero- and

first-order hold elements. Describe their functions in physical terms.

Can you construct simple electrical circuits which function as zero-.

and first-order hold elements?

9. Consider the discrete-time signals shown by the two sequences of

impulses in Figures 27.Q-la and 27.Q-lb. What type of hold element

would you select to construct the corresponding continuous signals?

Elaborate on your answer.

10. Describe the different ways which can be used to improve the quality

of a reconstructed continuous signal from its discrete-time values.

Outline relative advantages and disadvantages of the two methods.

11. Describe the general procedure for converting a continuous model to a

discrete-time one.

12. Why is a discrete-time model an approximation of its continuous counter-

part?

13. Discretize in time the continuous mode& of a stirred tank heater.

14. Consider the following two first-order systems

0.01 $f + y = m and

With a discretization time interval T=l second, which one of the

two systems will have a better discrete-time representation? Explain

Why. Also, show how you can improve the quality of the other (worst)

discrete-time model.

15. How many sampled output values do you need to construct the discrete-

time model for a third-order process? Explain why.

Ii--iqc;tre 27.3 i

(ai

::x

,

,

(a) FIRST- ORDER HOLD

ii >

(b)

Tm k) I Fi j ure .2T.6 I

I Stt--Po;nt---- ----- _____ (

\. # r ----.---- ---------- --

I

. :; jp=&p&q t)oLD i-&z-yL!L-, fL --- -_----- ___-_--1.L- iv-----------!

I -- - _--------- ----I 4 -7

I :

I,

( I / ’

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Chemical Process Control and Chemical Modeling-Stephanopulos