ChE / MET 433
Linearity, Windup, & PID
11 Apr 12
Process Linearity,
Integral Windup,
PID Controllers
1
ChE / MET 433
Quiz Solutions
2
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  C S
• Calculate the process gain K
3
K = -1.09
K = -0.69
K = 0.-45
K = -0.15
K = -0.26
K = -0.33
Adaptive Control ?
4
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)
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Integral (Reset) Windup
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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
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ChE / MET 433
Advanced control schemes
11 Apr 12
Cascade Control: Ch 9
8
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
9
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:
G P  ??
Split into 2 “processes” that can measure intermediate variable?
R s 
E s 
+
-
Gc
C s 
+
A
GC 2
GP2
G P1
KT2
Gp2 must be quicker responding than GP1.
• Inner (2nd-dary) loop faster than
primary loop
• Outer loop is primary loop
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Material Dryer Example
MT
% moisture
sp
MC
steam
Heat Exchger
T
R s 
E s 
+
air
blower
C s 
% moisture
Gc
GV
GP
KT
-
11
Separate Gp into 2 blocks
MT
% moisture
sp
MC
sp
TC
steam
TT
Heat Exchger
air
blower
T
R s 
E s 
+
-
G c1
C s 
+
A
GC 2
G PT
GV
G PM
K TM
K TT
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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
13
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
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Temperature Control of a Well-Mixed Reactor (CSTR)
Ti
Responds quicker to Ti
changes than coolant
temperature changes.
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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
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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
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Distillation Columns
Cascade
Control
Examples
Objective:
Regulate
temperature
(composition) at
top and bottom
of column
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Furnace
TP out
Objective:
Keep TP out
at the
set point
Heat Exchanger
T2 out
Objective:
Keep T2 out
at the
set point
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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)
20
In-Class Exercise: Cascade Control System Design
Does this design reject P changes
in the overhead vapor space?
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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
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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
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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?
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ChE / MET 433
Advanced control schemes
13 Apr 12
Ratio Control: Ch 10
25
Ratio Control
•Special type of feed forward control
A
B
• Blending/Reaction/Flocculation
• A and B must be in certain
ratio to each other
26
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.
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Ratio Control
Possible cascade control systems:
“wild” stream
A
sp Desired Ratio
FT

A
FY
B
A
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
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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
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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
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ChE / MET 433
Advanced control schemes
16 Apr 12
Feed Forward Control:
Ch 11
31
Feed Forward Control
steam
Suppose qi is primary disturbance
TC
TT
q i (t )
T i (t )
Heat Exchanger
? What is a drawback to this feedback control loop?
? Is there a potentially better way?
?
What if Ti changes?
FF
FT
q i (t )
steam
TT
Heat Exchanger
T i (t )
FF must be done with FB control!
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Feed Forward and Feedback Control
M (t )
?
FF
M
FF

(t )
steam
I
P
M  (t )
TY
TC
TY
FT
TT
q i (t )
T i (t )
Heat Exchanger
M  ( t )  M ( t )  M
Q i s 
+
-
FF

GL
G FF  FFC
G FF
E s 
(t )  M
K TD
Block diagram:
R s 
FF
GC
+
M
+
M
FF
M
GV
G PT
+
+
T s 

K TT
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Q i s 
Feed Forward
Control
GL
TD
K TD
G FF
R s 
E s 
+
-
GC
+
+
M
M
FF
M
GV
GP
+
+
T s 
TP

K TT
q i t 
TD
MFF
Response to MFF
TP
No change; perfect compensation!
T t 
0
t
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Q i s 
Feed Forward
Control
GD
K TD
FFC
R s 
E s 
+
GC
+
Q i s 
M
FF
M
-
Examine FFC T.F.
+
+
GM
+
C s 

C  s   G D  Q i ( s )  G M  FFC  K T D  Q i ( s )
gpm
For “perfect” FF control: C  s   0
GD
K TD
% TO
0  G D  Q i ( s )  G M  FFC  K T D  Q i ( s )
% TO D
FFC
M
FF
% CO FF
GM
+
% TO
+
C s 
FFC  
GD
K TD  G M
% TO
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Feed Forward Control: FFC Identification
Set by traditional means:
FFC  
% TO D
K TD
GD
K TD  G M
gpm
Model fit G D & G M to FOPDT equation:
GD 
K De
 to s
D
 Ds 1
% TO
GM 
gpm
 K
D
FFC   
K K
 TD M
FF Gain
{ FFC ss }
steady state FF
control
KMe
 to
M
s
M s 1
% TO
% CO
  s 1
s
 t  t
  M
  e o D oM
   s 1 

  D

Lead/lag
unit

Dead time
compensator
Accounts for time
differences in 2 legs

Often ignored; if t o D  t oM
set term to 1

{ FFC dyn }
dynamic FF control
36
ChE / MET 433
37
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 
+
-
G cT
T s 
P s 
+
GCP
G PP
GV
G PT
K TP
K TT
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