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This lecture was delivered by Dr. Sakal Japendu for Process Control course at Ambedkar University, Delhi. It includes: Feedback, Performance, Evaluating, Control, Simulation, Frequency, Response, Dead, Time, Constant, Disturbance
Typology: Slides
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When I complete this chapter, I want to be
able to do the following.
Outline of the lesson.
TC
v
v
∫
−
−
=
−
=
−
=
∆
=
=
≈
−
−
∆
−
−
=
−
−
=
−
−
t
m
SP
I
m
SP
c
valve
m
m
sensor
c
v
c
in
cout
cin
A
RT
E
rxn
p
P
A
RT
E
A
A
A
dt
T T T T T K
MV
MV
v
dt dv
T
T
dt
dT
P
C
v
aF
h
U
T
T
T
UA
C
e
Vk
H
T
T
C
F
dt
dT
C
V
C
e
Vk
C
C
F
dt
dC
V
0
6
0
0
0
0
0
1
'
)
(
)
(
)
(
)
(
)
(
))
(
(
)
(
)
(
)
(
max
.
/
/
τ τ
ρ
ρ
ρ
Many numerical
methods; Euler,
Runge-Kutta,
or other.
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0
50
100
150
0.9880.9860.9840.
IAE = 0.22254 ISE = 0.
Time (min)
XD (mol frac)
0
50
100
150
0.05 0.04 0.03 0.02 0.
IAE = 0.8863 ISE = 0.
Time (min)
XB (mol frac)
0
50
100
150
8550 8500 8450 8400 8350
Time (min)
R (mol/min)
0
50
100
150
1.38 1.36 1.
x 10
4
Time (min)
V (mol/min)
F
R
F
V
x
B
x
D
AC
AC
DISTIL: Results of detailed, non-linear, tray-by-tray dynamic model with PID feedbackcontrollers. Simulation in MATLAB.
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PRESENT VALUES
1) Total simulation time
100.
2) Time step for simulation
0.
3) Set point change
0.
4) Disturbance change
0.
5) Process reaction curve MV input
0.
6) Select continuous/digital controller, currently continuous
(Controller executed every simulation time step)
7) Execute dynamic simulation8) Return to main menu
0
20
40
60
80
100
120
0
0.4 0.3 0.2 0.
S-LOOP plots deviation variables (IAE = 7.5136)
Time
Controlled Variable
0
20
40
60
80
100
120
-10 -15 -20 -
0
Time
Manipulated Variable
)
1
2
)(
1
)(
1
(
)
1
(
3
2
2 3
2
1
s
−
s
s
s
s
s
e
K
lead
p
ξτ
τ
τ
τ
τ
θ
s
T
s
T
K
d
I
c
1
1
SP(s)
MV(s)
D(s)
CV(s)
Disturbance model
Process feedback modelPID controller model
Controlledvariable
Setpoint
Disturbance
Manipulatedvariable
)
1
2
)(
1
)(
1
(
)
1
(
3
2
2
3
2
1
_
s
−
s
s
s
s
s
e
K
d
d
d
d
d
d
lead
d
d
τ
ξ
τ
τ
τ
τ
θ
Simulation of single-loop linear systems is easily achieved using theS_LOOP program in MATLAB. Cases are possible for systems withand without control for step inputs
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No!
No!
No! Yes!
Yes!
Are you sure
of this answer?
Process dynamics for
disturbance T
in
to T
d
θθθθ
τ τ
τ τ
min
Process dynamics for
MV v
2
to T
p
θθθθ
ττττ
= 5 min
Three cases withamplitude 1 K anddifferent T
in
sine
periods, P.
P = 5000 minP = 50 minP = 0.05 min
For each case,what is the outputamplitude?
Let’s do a
thought
experiment,
without
calculating!
T
v
v
T
in
Process dynamics for disturbance T
in
to T
d
θθθθ
ττττ
= 5 min
Process dynamics for MV v
2
to T
p
θθθθ
= 5 min ;
τ τ
τ τ
= 5 min
T
v
v
T
in
FREQUENCY =0.12629 rad/time & AMP RATIO =1.
0
100
200
300
400
-0.
0
1
CV
0
100
200
300
400
-0.
0
1
Disturb., magnitude = 1
P = 50 min
Process dynamics for disturbance T
in
to T
d
θθθθ
ττττ
= 5 min
Process dynamics for MV
v
2
to T
p
θθθθ
= 5 min ;
ττττ
= 5 min
T
v
v
T
in
P = .05 min
FREQUENCY =126.2939 rad/time & AMP RATIO =0.
0
-0.
0
1
CV
0
-0.
0
1
Disturb., magnitude = 1
Process dynamics for
disturbance T
in
to T
d
θ θ
θ θ
= 0 min ;
ττττ
5 min
Process dynamics for
v
2
to T
p
θθθθ
ττττ
= 5 min
Three cases withamplitude 1 K anddifferent T
in
sine
periods, P.
P = 5000 minP = 5 minP = 0.005 min
For each case,what is the outputamplitude?
Let’s do a
thought
experiment,
without
calculating!
TC v
v
T
in
Process dynamics for disturbance T
in
to T
d
θθθθ
= 0 min ;
τ τ
τ τ
= 5 min
Process dynamics for MV
v
2
to T
p
θθθθ
τ τ
τ τ
= 5 min
TC v
v
T
in
FREQUENCY =0.0012496 rad/time & AMP RATIO =0.
0
1
2
3
4
5
x 10
4
-0.
0
1
CV
0
1
2
3
4
5
x 10
4
-0.
0
1
Disturb., magnitude = 1
P = 5000 min
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Process dynamics for disturbance T
in
to T
d
θθθθ
= 0 min ;
τ τ
τ τ
= 5 min
Process dynamics for MV
v
2
to T
p
θθθθ
ττττ
= 5 min
TC v
v
T
in
P = 50 min
FREQUENCY =0.12629 rad/time & AMP RATIO =1.
0
100
200
300
400
-0.
0
1
CV
0
100
200
300
400
-0.
0
1
Disturb., magnitude = 1
10
10
10
10
0
10
1
10
2
10
3
10
10
10
10
0
10
1
Frequency, w (rad/time)
Amplitude Ratio, |CV| / |D|
Please
discuss
