CHE517 CHE517 Advanced Advanced Process Control Process Control Prof. Shi-Shang Jang Prof. Shi-Shang Jang Chemical Engineering Department Chemical Engineering Department National Tsing-Hua University National Tsing-Hua University Hsin Chu, Taiwan Hsin Chu, Taiwan
CHE517 Advanced Process Control. Prof. Shi-Shang Jang Chemical Engineering Department National Tsing-Hua University Hsin Chu, Taiwan. Course Description. Course: CHE517 Advanced Process Control Instructor: Professor Shi-Shang Jang - PowerPoint PPT Presentation
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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
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
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
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
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