CBE 491 / CBE 433
Linearity, Windup, & PID
26 Nov 12
Process Linearity,
Integral Windup,
PID Controllers
1
Process Linearity
Test the Heat Exchanger process linearity by:
• Starting Loop Pro trainer
• Set %CO to 80%
• Make steps down (say 10% down) to the %CO
• Measure the response CS
• Calculate the process gain K
2
K = -1.09
K = -0.69
K = 0.-45
K = -0.15
K = -0.26
K = -0.33
Adaptive Control ?
3
Integral (Reset) Windup
• “Windup” can occur if integral action present
• Most modern controllers have anti-windup protection
• If doesn’t have windup protection, set to manual when reach point of
saturation, then switch back to auto, when drops below sat. level
•
•
•
•
•
IE: LoopPro Trainer, select Heat Exchanger
Set %CO to 90%; SP to 126; Kc to 1 %/deg C; Tau I to 1.0 min
Set Integral with Anti-Reset Windup ON
Change Set Point to 120 deg. C. (~10 min); then change back to 126 deg. C
Repeat with controller at ON: (Integral with Windup)
4
Integral (Reset) Windup
5
In-Class PID Controller Exercise
Tune the Heat Exchanger for a PID Controller:
• Use the built in IMC, and choose Moderately Aggressive
• Start Loop Pro trainer
• Tune at the initial %CO and exit temperature
• Compare PI with PID
• Compare PID with PID with filter
6
CBE 491 / CBE 433
Advanced control schemes
26 Nov 12
Cascade Control: Ch 9
7
Improve Feedback Control
Feedback control:
• Disturbance must be measured before action taken
• ~ 80% of control strategies are simple FB control
• Reacts to disturbances that were not expected
We’ll look at:
• Cascade Control (Master – Slave)
• Ratio Control
• Feed Forward
8
Cascade Control
•
•
Control w/ multiple loops
Used to better reject specific disturbances
Rs  +
Es 
C s 
M (s )
Gc
GP
Take slow process:
GP  ??
Split into 2 “processes” that can measure intermediate variable?
Rs  +
Es 
-
Gc
+
A
C s 
GC 2
GP 2
GP1
KT 2
Gp2 must be quicker responding than GP1.
• Inner (2nd-dary) loop faster than
primary loop
• Outer loop is primary loop
9
Material Dryer Example
MT
% moisture
sp
MC
steam
Heat Exchger
T
Rs  +
Es 
Gc
GV
air
blower
%moisture
GP
KT
C s 
-
10
Separate Gp into 2 blocks
MT
% moisture
sp
MC
sp
TC
steam
TT
Heat Exchger
air
blower
T
Rs  +
Es 
-
Gc1
+
A
GC 2
GV
GPT
GPM
K TM
C s 
K TT
11
cascade control can improve
rejection of this disturbance
but can not help rejection
of this disturbance
disturbance
variable II
disturbance
variable I
primary
set point
+–
Primary
Controller
secondary
set point
+–
Secondary
Controller
Final
Control
Element
Disturbance
Process II
Disturbance
Process I
Secondary
Process
secondary
process
variable
Primary
++
Process
++
primary
process
variable
secondary process variable
primary process variable
12
Problem Solving Exercise:
sp
Heat Exchanger
Single feedback loop.
Suppose known there will be steam
pressure fluctuations…
TC
steam
TT
Hot
water
Heat Exchger
T
Design cascade system that measures (uses) the steam pressure in the HX shell.
PT
Heat Exchger
steam
TT
Hot
water
T
13
Temperature Control of a Well-Mixed Reactor (CSTR)
Ti
Responds quicker to Ti
changes than coolant
temperature changes.
14
Temperature Control of a Well-Mixed Reactor (CSTR)
Use Cascade Control to
improve control.
Ti
If Tout (jacket) changes it is
sensed and controlled
before “seen” by primary T
sensor.
Secondary Loop
• Measures Tout (jacket)
• Faster loop
• SP by output primary loop
Primary Loop:
• Measures controlled var.
• SP by operator
15
Cascade Control
Benefits:
• Disturbances in secondary loop corrected by 2ndary loop controller
• Flowrate loops are frequently cascaded with another control loop
• Improves regulatory control, but doesn’t affect set point tracking
• Can address different disturbances, as long as they impact the
secondary loop before it significantly impacts the primary (outer loop).
Challenges:
• Secondary loop must be faster than primary loop
• Bit more complex to tune
• Requires additional sensor and controller
16
Distillation Columns
Cascade
Control
Examples
Objective:
Regulate
temperature
(composition) at
top and bottom
of column
17
Furnace
TP out
Objective:
Keep TP out
at the
set point
Heat Exchanger
T2 out
Objective:
Keep T2 out
at the
set point
18
In-Class Exercise: Cascade Control System Design
Design a cascade system to control level
(note overhead P can’t be controlled)
What affects flowrate?
• Valve position
• Height of liquid
• P (delta P across valve)
19
In-Class Exercise: Cascade Control System Design
Does this design reject P changes
in the overhead vapor space?
20
Tuning a Cascade System
• Both controllers in manual
• Secondary controller set as P-only (could be PI, but this might slow sys)
• Tune secondary controller for set point tracking
• Check secondary loop for satisfactory set point tracking performance
• Leave secondary controller in Auto
• Tune primary controller for disturbance rejection (PI or PID)
• Both controllers in Auto now
• Verify acceptable performance
21
In-Class Exercise: Tuning Cascade Controllers
•
•
•
•
Select Jacketed Reactor
Set T cooling inlet at 46 oC (normal operation temperature; sometimes it drops to 40 oC)
Set output of controller at 50%.
Desired Tout set point is 86 oC (this is steady state temperature)
•
•
•
•
Tune the single loop PI control
Criteria: IMC aggressive tuning
Use doublet test with +/- 5 %CO
Test your tuning with disturbance from 46 oC to 40 oC
22
In-Class Exercise: Tuning Cascade Controllers
•
•
•
•
•
•
•
•
•
•
Select Cascade Jacketed Reactor
•
Set T cooling inlet at 46 oC (again)
•
Set output of controller (secondary) at 50%.
•
Desired Tout set point is 86 oC (as before)
•
Note the secondary outlet temperature (69
oC) is the SP of the secondary controller
Tune the secondary loop; use 5 %CO doublet open loop
Criteria: ITAE for set point tracking (P only)
Use doublet test with +/- 5 %CO
Test your tuning with 3 oC setpoint changes
Tune the primary loop for PI control; make 3 oC set point changes (2nd-dary controller)
Note: MV = sp signal; and PV = T out of reactor
Criteria: IAE for aggressive tuning (PI)
Implement and with both controllers in Auto… change disturbance from 46 to 40 oC.
How does response compare to single PI feedback loop?
23
CBE 491 / CBE 433
Advanced control schemes
26 Nov 12
Ratio Control: Ch 10
24
Ratio Control
•Special type of feed forward control
A
B
• Blending/Reaction/Flocculation
• A and B must be in certain
ratio to each other
25
Ratio Control
Possible control system:
sp
sp
FC
FY
FC
FT
A
FY
FT
B
• What if one stream could
not be controlled?
• i.e., suppose stream A was
“wild”; or it came from an
upstream process and
couldn’t be controlled.
26
Ratio Control
Possible cascade control systems:
“wild” stream
A
sp Desired Ratio B A
FT

A
FY
B
A
FC
B
FT
B
A
“wild” stream
FT
This unit multiplies A by
the desired ratio; so
output = A B A
A
FY
Desired Ratio
BA
sp B
FC
B
FT
B
27
Ratio Control Uses:
• Constant ratio between feed flowrate and steam in reboiler of
distillation column
• Constant reflux ratio
• Ratio of reactants entering reactor
• Ratio for blending two streams
• Flocculent addition dependent on feed stream
• Purge stream ratio
• Fuel/air ratio in burner
• Neutralization/pH
28
In-Class Exercise: Furnace Air/Fuel Ratio
•
•
•
•
Furnace Air/Fuel Ratio model
disturbance: liquid flowrate
“wild” stream: air flowrate
ratioed stream: fuel flowrate
•
•
•
Minimum Air/Fuel Ratio 10/1
Fuel-rich undesired (enviro, econ, safety)
If air fails; fuel is shut down
Check TC tuning to disturbance & SP changes.
Desired 2 – 5% excess O2
PV
Disturbance var.
TC
Dependent MV
Ratio set point
TC output
Independent MV
29
CBE 491 / CBE 433
Advanced control schemes
26 Nov 12
Feed Forward Control:
Ch 11
30
Feed Forward Control
steam
Suppose qi is primary disturbance
TC
TT
qi (t )
Ti (t )
Heat Exchanger
? What is a drawback to this feedback control loop?
? Is there a potentially better way?
?
What if Ti changes?
FF
FT
qi (t )
steam
TT
Heat Exchanger
Ti (t )
FF must be done with FB control!
31
Feed Forward and Feedback Control
?
FF

M FF (t )
FT
I
M  (t )
TY
TC
steam
P
TY
TT
qi (t )
Ti (t )
M (t )
Heat Exchanger
M  (t )  M (t )  M FF (t )  M FF 
Qi s 
GL
K TD
Block diagram:
GFF  FFC
GFF
Rs  +
Es 
-
GC
+
M
+
M FF
M
GV
GPT
+
+
T s 
K TT
32
Qi s 
Feed Forward
Control
GL
TD
K TD
GFF
Rs  +
Es 
-
GC
+
+
M
M FF
M
GV
GP
+
+
T s 
TP
K TT
qi t 
TD
MFF
Response to MFF
TP
No change; perfect compensation!
T t 
0
t
33
Qi s 
Feed Forward
Control
GD
K TD
FFC
Rs  +
Es 
GC
+
Qi s 
+
C s 
For “perfect” FF control: C s   0
GD
K TD
%TO
0  GD  Qi ( s )  GM  FFC  K TD  Qi ( s )
%TOD
FFC
%COFF
GM
+
C s   GD  Qi ( s )  GM  FFC  K TD  Qi ( s )
gpm
M FF
M FF
M
-
Examine FFC T.F.
+
GM
+
%TO
+
C s 
%TO
FFC  
GD
K TD  GM
34
Feed Forward Control: FFC Identification
Set by traditional means:
FFC  
%TOD
gpm
K TD
GD
K TD  GM
Model fit GD & GM to FOPDT equation:
t
s
K D e oD
GD 
 Ds 1
t
%TO
gpm
 KD
FFC  
 KT K M
 D
FF Gain
{ FFC ss }
steady state FF
control
s
K M e oM
GM 
 M s 1

%TO
%CO
   M s  1  to toM s
D



e
   D s  1 

Lead/lag
unit

Dead time
compensator
Accounts for time
differences in 2 legs
Often ignored; if
set term to 1
t
oD
 toM

{ FFC dyn }
dynamic FF control
35
Feed Forward Control: FFC Identification
Qi s 
How to determine FOPDT
models GD & GM :
With Gc disconnected:
• Step change COFB, say 5%
• Fit C(s) response to FOPDT
 t oM s
K e
GM  M
 M s 1
%TO
%CO
Still in open loop:
• Step change Q, say 5 gpm
• Fit C(s) response to FOPDT
t
gpm
K TD
s
K D e oD
GD 
 Ds 1
%TO
gpm
GD
%TOD
FFC
M FF
GM
%COFB
FFC  
+
+
C s 
%TO
GD
K TD  GM
 KD
FFC 
K TD  K M
  Ld s  1 


  s 1
 Lg

 m   Ld lead time
 D   Lg lag time
36
Lead/Lag or Dynamic Compensator
Look at effect of these two to step change in input
 Ld
 Lg
 ld/ lg = 2
cff
Output or
response
 ld/ lg = 1
c(t )
 ld/ lg = ½
Time
Final Change from:
• Magnitude of step change,
• Initial response by the lead/lag,
• Exponential decay from lag,  Lg
  Ld


 Lg




37
Feed Forward Control
Rule of Thumb: if 0.65 
 Ld
 1.3 lead-lag won’t help much; use FFCss
 Lg
(p 389)
In text: pp 393-395, useful comments if implementing FFC
+
-
1. Compensates for disturbances
before they affect the process
1. Requires measurement or
estimation of the disturbance
2. Can improve the reliability of the
feedback controller by reducing
the deviation from set point
2. Does not compensate for
unmeasured disturbances
3. Offers advantages for slow
processes or processes with
large deadtime.
3. Linear based correction; only as
good as the models; performance
decreases with nonlinear
processes.
No improvement using FFC with set point changes.
38
In-Class PS Exercise: Feed Forward Control
What is the Gm, and what is
the GD?
Determine FCC
Tune PI controller to
aggressive IMC
For disturbance: Tjacket in
50oC – 60oC – 50oC
• Test PI Controller
• Test PI + FFCss only
• Test PI + FFC full
39
In-Class PS Exercise: Feed Forward Control
PI only
PI + FFCss only
PI + full FFC
40
CBE 491 / CBE 433
41
Problem Solving Exercise: Heat Exchanger
sp
TC
PC
PT
steam
TT
Hot
water
Heat Exchger
T
Draw the block diagram: what is the primary and what is the secondary loop?
Rs  +
Es 
-
GcT
+
GC P
GV
GPP
Ps 
GPT
T s 
K TP
K TT
42