CHE517 Advanced Process Control Prof. Shi-Shang Jang Chemical Engineering Department National...

CHE517CHE517Advanced Advanced

Process ControlProcess ControlProf. Shi-Shang JangProf. Shi-Shang Jang

Chemical Engineering DepartmentChemical Engineering DepartmentNational Tsing-Hua UniversityNational Tsing-Hua University

Hsin Chu, TaiwanHsin Chu, Taiwan

Course DescriptionCourse Description• Course: CHE517 Advanced Process Control• Instructor: Professor Shi-Shang Jang• Text: Seborg, D.E., Process Dynamics and

Control, 2nd Ed., Wiley, USA, 2003.• Course Objective: To study the application of

advanced control methods to chemical and electronic manufacturing processes

• Course Policies: One Exam(40%), a final project (30%) and biweekly homework(30%)

Course OutlineCourse Outline1. Review of Feedback Control System2. Dynamic Simulation Using MATLAB

and Simu-link3. Feedforward Control and Cascade

Control4. Selective Control System5. Time Delay Compensation6. Multivariable Control

Course Outline - ContinuedCourse Outline - Continued7. Computer Process Control8. Model Predictive Control9. R2R Process Control

Chapter 1 Review of Chapter 1 Review of Feedback Control SystemsFeedback Control Systems

• Feedback Control• Terminology• Modeling• Transfer Functions• P, PI, PID Controllers• Block Diagram Analysis• Stability• Frequency Response• Stability in Frequency Domain

Feedback ControlFeedback Control

Examples:• Room temperature control • Automatic cruise control • Steering an automobile • Supply and demand of chemical engineers

Controller

Transmitter

Set point

stream

Tempsensor

Heat loss

condensate

Feedback Control-block Feedback Control-block diagramdiagram

Terminology:• Set point • Manipulated variable (MV) • Controlled variable (CV)• Disturbance or load (DV)• Process • controller

Σ Controller process

Sensor +transmitter

+

-Set point

Measured value

error

Manipulated variable

Controlled variable

disturbance

InstrumentationInstrumentation

• Signal Transmission: Pneumatic 3-15psig, safe longer time lags, reliable• Electronic 4-20mA, current, fast, easy to interface with computers, may be

sensitive to magnetic and/or electric fields• Transducers: to transform the signals between two types of signals, I/P: current to

pneumatic, P/I, pneumatic to current

Controller

Transmitter

Set point

stream

Tempsensor

Heat loss

condensate

ModelingModeling

Rate of accumulation = Input – output + generation – consumption

At steady state : let T = TS and Q = QS 0 = QS – UA(TS - T0S)

Deviation variables : let T = TS+Td , Q = QS+Qd , T0 = T0s+T0d

Then :

If system is at steady state initially Td(0) = 0

)()( 0TTUAQTMCdt

dP

Mass M Cp T

QQ=UA(T-T0)

)()( 0ddddP TTUAQTdt

dMC

Transfer FunctionsTransfer FunctionsLaplace Transforming:Laplace Transforming:

M Cp S Td(S) = qd(S) - U A (Td(S) – Tod(S))

Or

UASMC

SUAT

UASMC

SqST

p

od

p

dd

UAMsC

UA

p

UAMsCp 1

∑Td(S)

+

+

qd(S)

Tod(S)

Non-isothermal CSTR Non-isothermal CSTR

• Total mass balance:

• Mass balance :

• Energy balance :

• Initial conditions : V(t=0) = Vi , T(t=0) = Ti , CA(t=0) = CAi

• Input variables : F0 , CA0 , T0 ,F

FFVdt

d 0)(

condensate

T V ρ CA CB

F0

ρ0

CA0

T0 FρCA

T

steam

A BrA = - KCA mol/ft3

K = αe-E/RT

VKCCFCFCVdt

dAAAA )()( 00

)())(()( 00 TTsUAKCHrTCFTCFTCVdt

dAPPP

Linearization of a FunctionLinearization of a Function

X0X0 -△ X0+△

- 0 △ △

F(X)

X

aX+b

LinearizationLinearization

0 0

0 0

0 0

0 0

,

( , )

0

Laplace Transform

( )

1

x x x xu u u u

dd d

d d d

d

d

dx f ff x u x x u u

dt x u

f x u

dxax bu

dt

sx s ax s bu s

or

x s b K

u s s a s

Linearization of Non-Linearization of Non-isothermal CSTR isothermal CSTR

12

12

11 0,

,21 22 , 11 0,

31 32 , 33 31 0, 32 , 33

11 12

, 21 22 , 21 22 23

31 32 33 31 32 3

. .

0 0 0 0

0

dd d

A dd A d d d

dd A d d d A d d

d d

A d A d

d d

dVb F b F

dtdC

a V a C b F b Fdt

dTa V a C a T b F b C b Ts

dti e

V V b bd

C a a C b b bdt

T a a a T b b b

0,

3 ,

0,

,

,

1

0,

( ) ( )

0 0 1 0 0 0

( ) ( )

( ) '

d

d

s d

d d

A d d

d s d

d

p d L d L d

F

F AX s BU s

T

V F

y C F CX DU

T T

T s C sI A B D U s

G s Ts s G s F s G s F s

Common Transfer FunctionsCommon Transfer FunctionsK=Gain; τ=time constant;K=Gain; τ=time constant;

ζ=damping factor; D=delay ζ=damping factor; D=delay

• First Order System

• Second Order System

• First Order Plus Time Delay

• Second Order Plus Time Delay

1

)(

s

K

sMV

sCV

Dse

ss

K

sMV

sCV

12

)(22

Dse

s

K

sMV

sCV

1

)(

12

)(22

ss

K

sMV

sCV

Transfer Functions of Transfer Functions of ControllersControllers

• Proportional Control (P)

• Proportional Integral Control (PI)

• Proportional-Integral-Derivative Control (PID)

m(s) = Kc[ e(s) ]

e = Tspt - TKc

e(s) m(s)

t

0I

c dt)t(e1

)t(eK)t(m

)s(es

1)s(eK)s(m

Ic

e(s) m(s))s

11(K

Ic

e(s) m(s))ss

11(K D

Ic

t

0 DI

c dt

dedt)t(e

1)t(eK)t(m

s

s

11)s(eK)s(m D

Ic

The Stability of a Linear The Stability of a Linear SystemSystem

• Given a linear system y(s)/u(s)=G(s)=N(s)/D(s) where N, D are

polynomials• A linear system is stable if and only if

all the roots of D(s) is at LHS, i.e., the real parts of the roots of D(s) are negative.

Stability in a Complex Plane

Re

Im

Purdy oscillatory

Purdy oscillatory

Fast Decay Slow Decay

Exponential Decay

Exponential Decay with oscillatory

Slow growth

Fast Exponential growth

Exponential growthwith oscillatory

Stable (LHP) Unstable (RHP)

Partial Proof of the TheoryPartial Proof of the Theory• For example: y(s)/u(s)=K/(τs+1)• The root of D(s)=-1/τ• In time domain: τy’+y=ku(t)• The solution of this ODE can be

derived by y(t)=e-t/τ [∫e1/τku(t)dt+c]

• It is clear that if τ<0, limt→∞y →∞.

Transfer functions in parallel Transfer functions in parallel

X(S)= G1(S)*U1(S) + G2(S)*U2(S)

Σ

U1(S)

U2(S)

G1(S)

G2(S)

X1(S)

X2(S)

+

+

X (S)

X1(S) X2(S)

Transfer function Block Transfer function Block diagramdiagram

Σ Kc+

-

Tset

control

QS

process

1

Measuring device

Td

UASMCP 1

UASMCK

UASMCK

T

T

PC

PC

set

d

11

1

Proportional control Proportional control No measurement lagsNo measurement lags

Block Diagram AnalysisBlock Diagram Analysis

e = Xs – Xm

m = Gc (S) e(s) = Gc e

X1 = Gp m = Gp Gc e

X = GL L + X1 = GL L + Gp Gc e

Xm = Gm X = Gm GL L + Gp Gc e

X = GL L + Gp Gc[Xs – Xm]

= GL L + Gp Gc [Xs] – Gp Gc [Xm]

=GL L + Gp Gc Xs – Gp Gc Gm X

smcp

cp

mcp

L XGGG1

GGL

GGG1

GX

∑ X(S)++

GL(S)

GP(S)

Gm(S)

L(S)

mGc(S)∑+

-Xs

Xm

X1

e

Stability of a Closed Loop Stability of a Closed Loop SystemSystem

• A closed loop system is stable if and only of the roots of its characteristic equation :

1+Gc(s)Gp(s)Gm(s)=0

are all in LHP

Level SystemLevel System

11/

/11

Laplacing

2

,point reference aGiven

.

,

0

,00

00

s

K

saA

a

aAssF

sh

or

sFsah(s)Ash

hh

kFhh

h

fFF

F

f

dt

dhA

hF

hkFFFdt

dhA

din

d

dindd

ddinininin

d

in

inoutin

The jacketed CSTRThe jacketed CSTR

TRC

FC

Tc

T, Ca

W

Set Point

Wc

2A B

A Nonisothermal Jacketed CSTRA Nonisothermal Jacketed CSTR• (i) Material balance of species A

• (ii) Energy balance of the jacket

• (iii) Energy balance for the reactor

• (iv) Dependence of the rate constant on temperature

2)(

A

AAA kCV

CCW

dt

dC f

P

A

P

cf

C

HkC

VC

TTA

V

TTW

dt

dT

2)()(

c

wcc

Pc

cc

M

TTW

CM

TTA

dt

dT )(

'

)(

)273

exp(0 T

QAk

Linearization of Linearization of Nonisothermal CSTRNonisothermal CSTR

• CV=T(t)

• MV=Wc(t)

• It can be shown that

123

,

csbsas

K

sW

sT

dc

d

A Practical Example A Practical Example ––Temperature Temperature Control of a CSTRControl of a CSTR

Method of Reaction CurveMethod of Reaction Curve

τ DDead time

Maximum slope

△C

Process output

Time constant time

Ziegler-Nichols Reaction Curve Ziegler-Nichols Reaction Curve Tuning RuleTuning Rule

P only PI PID

Kc /DKp 0.9/DKp 1.2/DKp

I n.a. D/0.3 D/0.5

D n.a. n.a. 0.5D

△C

τ

D

△m

D= 1τ =13k = -0.0202

Kc= -579.2079τi =3.33

setpoint

Ziegler-Nichols Ultimate Gain Ziegler-Nichols Ultimate Gain TuningTuning Find the ultimate gain of the process Find the ultimate gain of the process Ku. The period of the oscillation is Ku. The period of the oscillation is called ultimate period Pucalled ultimate period Pu

P only PI PID

Kc Ku/2 Ku/2.2 Ku/1.7

I n.a. Pu/1.2 Pu/2

D n.a. n.a. Pu/8

Measuring Controller Measuring Controller PerformancePerformance

00

0

2

0

2

00

dttetdtyytITAE

dttedttyyISE

dttedttyyIAE

s

s

s

Upper Limit of Designed Upper Limit of Designed Controller Parameters of PID Controller Parameters of PID

ControllersControllers• Q: Given a plant with a transfer

function G(s), one implements a PID controller for closed loop control, what is the upper limit of its parameters?

• A: The upper limit of a controller should be bounded at its closed loop stability.

ApproachesApproaches• Direct Substitution for Kc• Root Locus method for Kc• Frequency Analysis for all

parameters

An ExampleAn Example

)3)(2)(1(

1

sssKc

○

－

＋

1. Stability Limit by Direct 1. Stability Limit by Direct SubstitutionSubstitution

• At the stability limit (maximum value of Kc permissible), roots cross over to the RHP. Hence when Kc=Ku, there are two roots on the imaginary axis s=±iω

• (s+1)(s+2)(s+3)+Ku=0, and set s= ±iω, we have (iω+1)(iω+2)(iω+3)+Ku= 0, i.e. (6+Ku-6ω2)+i(11ω-ω3)=0. This can be true only if both real and imaginary parts vanishes: 11ω-ω3=0→ ω= ±√11 ; 6+Ku-6×11=0 →Ku=60

2. Method of Root Locus2. Method of Root Locus

Rlocus (sys,k)

k(12) ans =69.6706

3. Frequency Domain 3. Frequency Domain AnalysisAnalysis

• Definitions: Given a transfer function G(s)=y(s)/x(s); Given x(t)=Asinωt; we have y(t) →Bsin(ωt+ψ)

• We denote Amplitude Ratio=AR(ω) =B/A; Phase Angle=ψ(ω)

• Both AR and ψ are function of frequency ω; we hence define AR and ψ is the frequency response of system G(s)

An ExampleAn Example

321

1

sssA sin(t) B = sin(t+)

Frequency Response of a Frequency Response of a first order systemfirst order system

1

22

1

22

22

22

tan1

tan);sin(1

)(

1)(

)(sin)(;1

)()()(

KAR

tKA

ty

sK

sA

sy

sA

sxtAtxsK

sGsxsy

Basic TheoremBasic Theorem• Given a process with transfer function

G(s);• AR(ω)= ︳ G(iω) ︳• φ(ω)=∠ G(iω)• Basically, G(iω)=a+ib

ab

baAR

/tan 1

22

Example: First Order SystemExample: First Order System

90lim

0lim

thatNote

tantan

1

11

)(

1

1

1

)(1

1

11

1)(

11

22

22

222222

AR

a

b

baAR

ibaii

iiG

ssG

CorollaryCorollary

• If G(s)=G1(s)G2(s)G3(s)

• Then AR(G)=AR(G1) AR(G2) AR(G3)

• φ(G)=φ (G1) +φ (G2)+φ (G3)

• Proof: Omitted

ExampleExample

2

11

121

222

2

221

121

212

2

1

1

tantan

11

11)(

KKARARAR

sGsGKK

sG

Bode Plot: An exampleBode Plot: An exampleG(s)=1/(s+1)(s+2)(s+3)G(s)=1/(s+1)(s+2)(s+3)where db=20logwhere db=20log1010(AR)(AR)

Nyquist PlotNyquist Plotsys=tf(num,den)sys=tf(num,den)

NYQUIST(sys,{wmin,wmax}))NYQUIST(sys,{wmin,wmax}))

Nyquist Stability CriteriaNyquist Stability Criteria• Given G(iω), assume that at a

frequency ωu, such that φ=-180° and one has AR(ωu), the sufficient and necessary condition of the stability of the closed loop of G(s) is such that: AR(ωu) ≦1

The Extension of Nyquist The Extension of Nyquist Stability CriteriaStability Criteria

• Given plant open loop transfer function G(s), such that at a frequency ωu, the phase angle φ(ωu)=-180°. At that point, the amplitude ratio AR= ｜ G (ωu) ｜ , then the ultimate gain of the closed loop system is Ku=1/AR, ultimate period Pu=2π/ ωu.

Simulink ExampleSimulink Example

time

Resp

on

s

e

D1.4 3.7-1.4=2.3

sP e

sG 5.0

15.2

165.0

Simulink Example - Simulink Example - ContinuedContinued

>> sys=tf(1,[1 6 11 6])

Transfer function:

1

----------------------

s^3 + 6 s^2 + 11 s + 6

>> bode(sys)

u=3.5ARu=-38db=10-38/20

=0.0162

Ku=1/ARu=80

Simulink Example - Simulink Example - ContinuedContinued

1. Reaction Curve Approach: KC=1.2/DKp=1.2*2.5/(0.5*0.165)=36; I=D/0.5=1;D=D*0.5=0.25

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Simulink Example - Simulink Example - ContinuedContinued

1. Ultimate properties Approach: Ku/1.7=80/1.7=47;I=Pu/2= 2* / 2U =0.9;D=Pu/8=0.22