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
• 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 Outline
1. Review of Feedback Control System
2. Dynamic Simulation Using MATLAB
and Simu-link
3. Feedforward Control and Cascade
Control
4. Selective Control System
5. Time Delay Compensation
6. Multivariable Control
Course Outline - Continued
7. Computer Process Control
8. Model Predictive Control
9. R2R Process Control
Chapter 1 Review of
Feedback Control Systems
•
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Feedback Control
Terminology
Modeling
Transfer Functions
P, PI, PID Controllers
Block Diagram Analysis
Stability
Frequency Response
Stability in Frequency Domain
Feedback Control
Transmitter
Controller
Set point
Temp
sensor
Heat loss
stream
condensate
Examples:
• Room temperature control
• Automatic cruise control
• Steering an automobile
• Supply and demand of chemical engineers
Feedback Control-block
diagram
Manipulated
variable
error
+ Σ
Set point
Controller
disturbance
process
Controlled variable
-
Measured value
Sensor +
transmitter
Terminology:
• Set point
• Manipulated variable (MV)
• Controlled variable (CV)
• Disturbance or load (DV)
• Process
• controller
Instrumentation
Transmitter
Controller
Set point
Temp
sensor
Heat loss
stream
condensate
•
•
•
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
Modeling
Q
Mass M
Cp
T
Q=UA(T-T0)
Rate of accumulation = Input – output + generation –
consumption
d
( MC PT )  Q  UA(T  T0 )
dt
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 :
d
MC P
dt
(Td )  Qd  UA(Td  T0d )
If system is at steady state initially Td(0) = 0
Transfer Functions
Laplace Transforming:
M Cp S Td(S) = qd(S) - U A (Td(S) – Tod(S))
Or
q d S
UATod S
Td S 

SMC p  UA SMC p  UA
qd(S)
UA
MsC p  UA
+
∑
+
1
Tod(S)
MsC p  UA
Td(S)
Non-isothermal CSTR
F0
ρ0
CA0
T0
AB
rA = - KCA mol/ft3
T V ρ CA CB
steam
F
ρ
CA
T
K = αe-E/RT
condensate
• Total mass balance: d (V  )  F0   F 
dt
• Mass balance :
d
(V C A )  F0 C A0  F C A  ( KC A )V
dt
d
• Energy balance : (V CPT )  F0CPT0  F C PT  (Hr )( KC A )  UA(Ts  T )
dt
• Initial conditions : V(t=0) = Vi , T(t=0) = Ti , CA(t=0) =
CAi
• Input variables : F0 , CA0 , T0 ,F
Linearization of a Function
F(X)
aX+b
X0 -△ X0
-△
0
X0+△
△
X
Linearization
f
dx
 f  x, u  
x
dt
x  x0
u  u0
 x  x0  
 f ( x0 , u0 )
dxd
 axd  bud  0
dt
Laplace Transform
sxd  s   axd  s   bud  s 
or
xd ( s )
K
b


ud  s  s  a  s  1
f
u
x  x0
u  u0
 u  u0 
Linearization of Nonisothermal CSTR
dVd
 b11 F0, d  b12 Fd
dt
dC A,d
 a21Vd  a22C A,d  b11 F0, d  b12 Fd
dt
dTd
 a31Vd  a32C A,d  a33Td  b31 F0, d  b32C A, d  b33Tsd
dt
i.e.
 Vd   0
d 
  a
C
A
,
d
  21
dt 
 Td   a31
0
a22
a32
0   Vd   b11 b12
0  C A,d   b21 b22
a33   Td  b31 b32
0   F0,d 


b23   Fd   AX ( s )  BU ( s )
b33   Ts ,d 
 F0,d 
 Vd 


y   0 0 1 C A,d    0 0 0   Fd   CX  DU
 Ts ,d 
 Td 


Td  s   C ( sI  A) 1 B   D U ( s )
 G p  s  Tsd  s   GL  s  F0,d ( s )  GL '  s  Fd  s 
Common Transfer Functions
K=Gain; τ=time constant;
ζ=damping factor; D=delay
• First Order System
• Second Order System
CV ( s)
K

MV s  s  1
CV ( s)
K
 2 2
MV s   s  2s  1
• First Order Plus Time Delay
CV ( s)
K  Ds

e
MV s  s  1
• Second Order Plus Time Delay
CV ( s)
K
 2 2
e  Ds
MV s   s  2s  1
Transfer Functions of
Controllers
• Proportional Control (P)
m(s) = Kc[ e(s) ]
e(s)
m(s)
Kc
e = Tspt - T
• Proportional Integral Control (PI)

1
m( t )  K c e( t ) 
I


0 e(t)dt 
t


1
m(s)  K c e(s) 
e(s)
Is


e(s)
K c (1 
1
)
 Is
m(s)
• Proportional-Integral-Derivative Control (PID)

1
m( t )  K c e( t ) 
I

m(s)  K c
de 
0 e(t)dt  D dt 
t


1
e(s) 1   Ds
 Is

e(s)
K c (1 
1
  D s)
 Is
m(s)
The Stability of a Linear
System
• 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
Im
Exponential Decay
with oscillatory
Purdy oscillatory
Exponential growth
with oscillatory
Fast Exponential
growth
Exponential Decay
Fast Decay
Slow Decay
Slow growth
Purdy oscillatory
Stable (LHP)
Unstable (RHP)
Re
Partial 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
X(S)= G1(S)*U1(S) + G2(S)*U2(S)
X1(S)
U1(S)
X2(S)
G1(S)
X1(S)
+
Σ
G2(S)
U2(S)
X2(S)
+
X (S)
Transfer function Block
diagram
Tset
+
Kc
Σ
-
QS
control
1
MC P S  UA
process
Measuring device
1
Proportional control
No measurement lags
Td
Tset
1
MC P S  UA

1
1  KC
MC P S  UA
KC
Td
Block Diagram Analysis
Xs
+
e
∑
-
Gc(S)
L(S)
GL(S)
m
GP(S)
Xm
X1
+
+
∑
Gm(S)
e = X s – 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
G pG c
GL
X
L 
Xs
1  G pG cG m
1  G pG cG m
X(S)
Stability of a Closed Loop
System
• 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 System
dh
 Fin  Fout  Fin  k h
dt
Given a reference point Fin0 , h0
A
A
dhd
f
Fin  Fin0   f h  h0   Fin,d  k hd

dt
Fin
h
2 h0
Laplacing
Ashd (s)  ahd s   Fin,d s 
or
hd s 
1
1/ a
K



Fin.d s  As  a  A / a s  1 s  1
The jacketed CSTR
W
Set Point
TRC
FC
2A  B
Tc
Wc
T, Ca
A Nonisothermal Jacketed CSTR
• (i) Material balance of species A
dC A W (C A f  C A )
2

 kCA
dt
V
• (ii) Energy balance of the jacket
dTc A(T  Tc ) Wc (Tc  Tw )


dt
M cCP '
Mc
• (iii) Energy balance for the reactor
2
dT W (T f  T ) A(T  Tc ) HkC A



dt
V
C PV
CP
• (iv) Dependence of the rate constant on
temperature
Q
k  A0 exp( 
)
T  273
Linearization of
Nonisothermal CSTR
• CV=T(t)
• MV=Wc(t)
• It can be shown that
Td s 
K
 3
Wc,d s  as  bs 2  cs  1
A Practical Example –Temperature
Control of a CSTR
Method of Reaction Curve
Process
output
Maximum slope
△C
D
Dead time
τ
Time constant
time
Ziegler-Nichols Reaction Curve
Tuning 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
D
τ
△m
D= 1
τ =13
k = -0.0202
Kc= -579.2079
τi =3.33
setpoint
Ziegler-Nichols Ultimate Gain
Tuning
Find the ultimate gain of the process
Ku. The period of the oscillation is
called 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
Performance


0
0
IAE   y s  y t  dt   et  dt


ISE    y s  y t  dt   et  dt
2
2
0
0


0
0
ITAE   t y s  y dt   t et  dt
Upper Limit of Designed
Controller Parameters of
PID Controllers
• 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.
Approaches
• Direct Substitution for Kc
• Root Locus method for Kc
• Frequency Analysis for all
parameters
An Example
＋
－
○
Kc
1
( s  1)( s  2)( s  3)
1. Stability Limit by Direct
Substitution
• 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 Locus
Rlocus (sys,k)
k(12) ans =69.6706
3. Frequency Domain
Analysis
• 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 Example
A sin(wt)
1
s  1s  2s  3
B = sin(wt+f)
Frequency Response of a
first order system
y(s)
K
A
 G ( s) 
; x(t )  A sin wt  x( s ) 
x( s )
s  1
s w
A
K
y(s) 
s  w s  1
KA
y (t ) 
sin(wt  f );f  tan  w 
1 w 
K
AR 
1 w 
f   tan w
2
2
2
1
2
2
1
2
2
2
Basic Theorem
• Given a process with transfer function
G(s);
• AR(ω)=︳G(iω)︳
• φ(ω)=∠ G(iω)
• Basically, G(iω)=a+ib
AR  a 2  b 2
f  tan
1
b / a 
Example: First Order
System
1
G (s) 
s  1
1
1  i (w )
1
 (w )
G iw  


i
 a  ib
2 2
2 2
2 2
iw  1 1   w
1 w
1 w
1
AR  a 2  b 2 
1   2w 2
b
f  tan 1    tan 1  w 
a
Note that
lim AR  0
w 
lim f  90
w 
Corollary
• 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
Example
K1
K2
G ( s) 
 G1 s G2 s 
 1w  1  2w  1
AR  AR1 AR2 

K1
 
1 w 1
2
2
K2
2

w 2 1
2
f  f1  f2   tan 1 w 1   tan 1 w 2 
Bode Plot: An example
G(s)=1/(s+1)(s+2)(s+3)
where db=20log10(AR)
Nyquist Plot
sys=tf(num,den)
NYQUIST(sys,{wmin,wmax}))
Nyquist 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
Stability 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 Example
Response
0.165 0.5 s
GP 
e
2.5s  1
D1.4
3.7-1.4=2.3
time
Simulink Example Continued
>> sys=tf(1,[1 6 11 6])
Transfer function:
1
---------------------s^3 + 6 s^2 + 11 s + 6
>> bode(sys)
wu=3.5
ARu=-38db
=10-38/20
=0.0162
Ku=1/ARu=80
Simulink Example Continued
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
2
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0
1
2
3
4
5
6
7
8
9
10
Simulink Example Continued
1. Ultimate properties Approach:
Ku/1.7=80/1.7=47;I=Pu/2= 2* / 2wU =0.9;D=Pu/8=0.22