chapter 2 507
TRANSCRIPT
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Chapter 2
Traditional Advanced Control
Approaches Feedforward, Cascadeand Selected Control
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2-1 Feed Forward Control (FFC)
Block Diagram
Design of FFC controllers
Examples
Applications
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Why Feedforward ?
Advantages of Feedback Control Corrective action is independent of sources of
disturbances
No knowledge of process (process model) is required
Versatile and robust
Disadvantages No corrective action until disturbance has affected the
output. Perfect control is impossible.
Nothing can be done about known process disturbance
If disturbances occur at a frequency comparable to thesettling time of the process. Then process may neversettle down.
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Feedforward Control
FeedforwardController Disturbance
Process
Output
ManipulatedVariable
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Feedforward Control
Advantages Corrective action is taken as soon as disturbances
arrives.
Controlled variable need not be measured.
Does not affect the stability of the processes
Disadvantages Load variable must be measured
A process model is required
Errors in modeling can result in poor control
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LI
FFC
steam
FI
LI
steam
FI
FFCFB
Feedback control
Feedforward control
Combined feedforward-feedback control
LI
FB
BoilerFeed control
steam
FI
EXAMPLES
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Design Procedures (Block diagram
Method)
GL(s)
Load transfer function
GF(s)
Load
ManipulatedVariable
Gp(s)
Process
X2
C
Output
L
M
FF
Controller
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Derivation
functiontransferprocess,(2)
functiontransferload),()1(:need
or
0)(Hence).(allfor0)(wantWe
)()(
)()()()()(
)()()()()(
sG
sG
sG
sGsG
sGsGsG
sLsC
sLsGsGsG
sLsGsGsLsG
sMsGsLsGsC
P
L
P
L
F
FPL
FPL
FPL
PL
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Examples
Example 1 Let Gp(s)=Kp/ps+1, GL(s)=KL/Ls+1
Then, GF(s)=-(KL/Kp)(ps+1)/(Ls+1)
Therefore,feedforward controller is a lead-lag unit.
Example 2 Let Gp(s)=Kpe
-Dps/ps+1, GL(s)=KLe-DLs/Ls+1
Then, GF(s)=-(KL/Kp)(ps+1)/(Ls+1)e(-DL+DP)s
If -DL+DPis positive, then this controller is unrealizable.However, an approximation would be to neglect thedelay terms, and readjusting the time constants. In thiscase, perfect FF compensation is impossible.
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Tuning feedforward controllers
Let
This has three adjustable constants, K, 1, 2
Tuning K, K is selected so that for a persistent
disturbance, there is no steady state error inoutput.
Adjusting1, 2 can be obtained from transferfunctions. Fine tune 1, 2 such that for a step
disturbance, the response is somewhatsymmetrical about the set point.
11)(
2
1
ssKsGF
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Example: A simulated disturbed plant
Waste watertreatment
Disturbedflow rate
BOD
Chemicals
(CV)
DV
MV
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Simulated Block Disgram
1
1 2 3s s s Disturbed flowrate
+
3
1
1s
Chemicals
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Feedforward v.s. Feedback Control
0 5 10 15 20 25 30-0.2
-0.15
-0.1
-0.05
0
0.05
0.1
0.15
FB
FF
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Example: Distillation Column
Example: Distillation Column Mass Balance: F=D+B
Fz=Dy+Bx
D=F(z-x)/(y-x)
In practice
For example: If light key increase in feed, increasedistillate rate.
setset
set
xy
xzFD
)(
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Design of Feedforward Control Using
Material and Energy Balances Consider the hear exchanger
Energy Balance yields Q=WC(T2-T1)=Ws
Where =hear of vaporization Ws=WC(T2-T1)/
This equation tells us the current stream demand based on (1)current flow rate, W, (2) current inlet temperature, T1, (3) desiredvalue of outlet temperature T2.
Steam
T2w, T1
Condensate
Ws
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Control Law and Design
Implementation:
Note no dynamics are incorporated
K
X
measured Tset
Gain
measured Ws
w
T1- +
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When to use Feedforward ?
Feedback control is unsatisfactory
Disturbance can be measured andcompensated for
Frequency of disturbance variations arecomparable to frequency of oscillation of thesystem
Output variable cannot be measured. There are large time delays in the system
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2-2 Cascade Control
Block Diagram
Design Considerations
Applications
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Illustrative Example : Steam Jacket
PCPT
TC
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Illustrative Example: Steam Jacket -
Continued
Energy Balance of the Tank:
Energy Balance of Jacket:
Material Balance of the Jacket
LossHeat TThAdt
dTV J
dt
nPd
R
V
Rn
VP
dt
d
dt
dT sJJ
s
JJJ /
condensatendt
dnin
s
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Illustrative Example: Steam Jacket -
Continued
Assume:
Where X=valve position
21110
113130
1
)(
sssX
sPsssP
sT
J
J
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Block Diagram
Feed backController
SteamValve
StirredTank
Tset
Steamsupply
pressure
Valve
position
Jacket
steampressure primary
Tank
Temp.
secondary
PrimaryController
JacketPressureController
SteamValve
StirredTank
Tset secondary
Jacketpressure
supplypressure
TankTemp.
Secondary loop
Primary loop
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Principal Advantages and Disadvantages
Advantages Disturbances in the secondary loop are corrected by
secondary controllers
Response of the secondary loop is improved, thusincreasing the speed of response of the primary loop
Gain variations in secondary loop are compensated bysecondary loop
Disadvantages Increased cost of instrumentation
Need to tune two loops instead of one Secondary variable must be measured
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Design Considerations
Secondary loop must be fast responding otherwisesystem will not settle
Time constant in the secondary loop must be smaller than primaryloop
Since secondary loop is fast, proportional actionalone is sufficient, offset is not a problem insecondary loop
Only disturbances within the secondary loop are
compensated by the secondary loop. Hence,cascading improves the response to thesedisturbances
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Applications: 1. Valve Position Control
Valve motion is affected by friction and pressuredrop in the line. Friction causes dead band. High
pressure drop also causes hysteresis in the valveresponse
Useful in most loops except flow and pressure
ControlValveMotor
Desired
position
Secondaryloop
Valveposition
Air Pressure to
Valve Motor
A li i 2 C d Fl L
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Application 2. Cascade Flow LoopOutput From
Primary Controller no cascade
Output From
Primary ControllerFC
FT
Fset
DP
cascade
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GC2 GC1 GP1 GP2
Primarycontroller
Secondarycontroller
Secondaryprocess
primaryprocess
cm2e2mset
Secondary loop
Primary loop
cset
2
11
112
1C
PC
PC
set
GGG
GGmm
GC2 GCL GP2 mset m2
ccset
22
22
1 PCLC
PCLC
set GGG
GGG
c
c
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Gc 12)110()1(
12 SS )13)(130(
1
SS
+
-
Primary
+
-
GC2
Secondary
G2(S) G3(S)
2
32
121
12
G
GG
c
For a cascade system
(open-loop)
Without cascade control
32GG
c
c
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Illustrative Example: Steam Jacket
Continued Cascade Case
Wu = 0.53
Mag = 20*log10(AR) = -30 (dB)
AR = 0.0316
6228.311
AR
Ku
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Illustrative Example: Steam Jacket
Continued No Cascade Case
Wu = 0.25
Mag = 20*log10(AR) = 0 (dB)
AR = 1
11
AR
Ku
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Illustrative Example: Steam Jacket
Continued No Cascade Case
Ku = 1;wu = 0.25;Pu = 2*Pi / wu = 25.1327
Kc = Ku/1.7 = 0.5882
Taui = Pu / 2 = 12.5664 Taud = Pu /8 = 3.1416
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Illustrative Example: Steam Jacket
Continued Cascade Case
Ku = 20;wu = 0.53;Pu = 2*Pi / wu = 12
Kc = Ku/1.7 = 11.8
Taui = Pu / 2 = 6
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2-3 Selective Control Systems
Override Control
Auctioneering Control
Ratio Control
Change from one controlled (CV) or manipulated variables
(MV) to another
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LT LC LSS
PC
Normalloop
water
steam
1. Override Control Example Boiler Control
LSS: Low Selective Switch Output a lower of two inputsPrevents: 1. Level from going too low, 2. Pressure fromexceeding limit (lower)
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motor
SC
HSS
PC FC
Gas out
Gas in
Normal loop
Example: Compressor Surge Control
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High Pressure Line
Low Pressure Line
PC
PC
HSS
Example: Steam Distribution System
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Length of reactor
Tem
perate
T1
T2
Hot spot
2. Auctioneering Control Systems
Temperature profiles in a tubular reactor
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TT TTTTTT
HSS
Cooling flow
Auctioneering Control Systems
TC
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Temperature Control
Steam
T2
Bypass
Exchanger
TC
Split Range Control: More than one manipulated variable is
adjusted by the controller
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Boiler 2 Steam Header
Boiler 2
Boiler 2
PC
Example: Steam Header: Pressure Control
3 Ratio Control Type of feedforward control
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FT
Driver
FT
A Wild stream
Controlled StreamB
FB
Gc
Desired
Ratio
FA
3. Ratio Control Type of feedforward control
However, one stream in proportion to another. Use if the ratio
must be measured and displayed
Disadvantage:Ratio may goTo erratic
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A Wild stream
FT
Desired Ratio
Multiplier
FC
FT
Controlled streamFB
FA
Another implementation of Ratio Control
+
-