chapter 2 traditional advanced control approaches – feedforward, cascade and selected control
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Chapter 2
Traditional Advanced Control Approaches – Feedforward, Cascade and Selected Control
2-1 Feed Forward Control (FFC) Block Diagram Design of FFC controllers Examples Applications
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 the
settling time of the process. Then process may never settle down.
Feedforward Control
Feedforward Controller Disturbance
ProcessOutput
ManipulatedVariable
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
LI
FFC
steam
FI
LI
steam
FI
FFCFB
Σ
Feedback control
Feedforward control
Combined feedforward-feedback control
LI
FB
BoilerFeed control
steam
FI
EXAMPLES
Design Procedures (Block diagram Method)
GL(s)
Load transfer function
GF(s)
Load
Manipulated Variable
Gp(s)
Process
X2
C
Output
L
M∑
FF
Controller
Derivation
function transfer process ,(2)
function transfer load),()1(:need
or
0)(
Hence ).( allfor 0)( want We
)()(
)()()()()(
)()()()()(
sG
sG
sG
sGsG
sGsGsG
sLsC
sLsGsGsG
sLsGsGsLsG
sMsGsLsGsC
P
L
P
LF
FPL
FPL
FPL
PL
Examples
Example 1 Let Gp(s)=Kp/τps+1, GL(s)=KL/τLs+1
Then, GF(s)=-(KL/Kp)(τps+1)/(τLs+1)
Therefore, feedforward controller is a “lead-lag” unit.
Example 2 Let Gp(s)=Kpe-Dps/τps+1, GL(s)=KLe-DLs/τLs+1
Then, GF(s)=-(KL/Kp)(τps+1)/(τLs+1)e(-DL+DP)s
If -DL+DP is positive, then this controller is unrealizable. However, an approximation would be to neglect the delay terms, and readjusting the time constants. In this case, perfect FF compensation is impossible.
Tuning feedforward controllers Let This has three adjustable constants, K, τ1, τ2
Tuning K, K is selected so that for a persistent disturbance, there is no steady state error in output.
Adjustingτ1, τ2 can be obtained from transfer functions. Fine tune τ1, τ2 such that for a step disturbance, the response is somewhat symmetrical about the set point.
1
1)(
2
1
s
sKsGF
Example: A simulated disturbed plant
Waste water treatment
Disturbed flow rate
BOD
Chemicals
(CV)
DV
MV
Simulated Block Disgram
1
1 2 3s s s Disturbed flow rate
+
3
1
1s Chemicals
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
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, increase distillate rate.
setset
set
xy
xzFD
)(
Design of Feedforward Control Using Material and Energy Balances Consider the hear exchanger
Energy Balance yields Q=WC(T2-T1)=Wsλ
Where λ=hear of vaporization Ws=WC(T2-T1)/λ
This equation tells us the current stream demand based on (1) current flow rate, W, (2) current inlet temperature, T1, (3) desired value of outlet temperature T2.
Steam
T2w, T1
Condensate
Ws
Control Law and Design
Implementation:
Note no dynamics are incorporated
Σ
K
X
measured Tset
Gain
measured Ws
w
T1- +
When to use Feedforward ?
Feedback control is unsatisfactory Disturbance can be measured and
compensated for Frequency of disturbance variations are
comparable to frequency of oscillation of the system
Output variable cannot be measured. There are large time delays in the system
2-2 Cascade Control
Block Diagram Design Considerations Applications
Illustrative Example : Steam Jacket
PCPT
TC
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
Illustrative Example: Steam Jacket - Continued Assume:
Where X=valve position
21110
1
13130
1
)(
sssX
sP
sssP
sT
J
J
Block Diagram
Feed back Controller
Steam Valve
Stirred Tank
Tset
Steam supply
pressureValve
position
Jacket
steam pressure primary
Tank
Temp.secondary
Primary Controller
Jacket Pressure Controller
Steam Valve
Stirred Tank
Tset secondary
Jacket
pressure
supply pressure
Tank
Temp.
Secondary loop
Primary loop
Principal Advantages and Disadvantages Advantages
Disturbances in the secondary loop are corrected by secondary controllers
Response of the secondary loop is improved, thus increasing the speed of response of the primary loop
Gain variations in secondary loop are compensated by secondary loop
Disadvantages Increased cost of instrumentation Need to tune two loops instead of one Secondary variable must be measured
Design Considerations
Secondary loop must be fast responding otherwise system will not settle
Time constant in the secondary loop must be smaller than primary loop
Since secondary loop is fast, proportional action alone is sufficient, offset is not a problem in secondary loop
Only disturbances within the secondary loop are compensated by the secondary loop. Hence, cascading improves the response to these disturbances
Applications: 1. Valve Position Control
Valve motion is affected by friction and pressure drop in the line. Friction causes dead band. High pressure drop also causes hysteresis in the valve response
Useful in most loops except flow and pressure
ControlValve Motor
Desired
position
Secondary loop
Valve position
Air Pressure to Valve Motor
Application 2. Cascade Flow LoopOutput From
Primary Controller “ no cascade “
Output From
Primary ControllerFC
FT
Fset
DP
“ cascade “
GC2 GC1 GP1 GP2Σ
Primarycontroller
Secondarycontroller
Secondaryprocess
primaryprocess
cm2e2
mset
Secondary loop
Primary loop
cset
211
112
1 CPC
PC
set
GGG
GG
m
m
GC2 GCL GP2Σmset m2
ccset
22
22
1 PCLC
PCLC
set GGG
GGG
c
c
Gc 12ΣΣ)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
32GGc
θcθ
Illustrative Example: Steam Jacket – Continued – Cascade Case
Wu = 0.53
Mag = 20*log10(AR) = -30 (dB)
AR = 0.0316
6228.311
AR
Ku
Illustrative Example: Steam Jacket – Continued – No Cascade Case
Wu = 0.25
Mag = 20*log10(AR) = 0 (dB)
AR = 1
11
AR
Ku
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
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
2-3 Selective Control Systems Override Control Auctioneering Control Ratio Control
Change from one controlled (CV) or manipulated variables (MV) to another
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)
motor
SC
HSS
PC FC
Gas out
Gas in
Normal loop
Example: Compressor Surge Control
High Pressure Line
Low Pressure Line
PC
PC
HSS
Example: Steam Distribution System
Length of reactor
Tem
perate
T1
T2
Hot spot
2. Auctioneering Control Systems
Temperature profiles in a tubular reactor
TT TTTTTT
HSS
Cooling flow
Auctioneering Control Systems
TC
Temperature Control
Steam
T2
Bypass
Exchanger
TC
Split Range Control: More than one manipulated variable is adjusted by the controller
Boiler 2 Steam Header
Boiler 2
Boiler 2
PC
Example: Steam Header: Pressure Control
FT
Driver
FT
A Wild stream
Controlled StreamB
FB
ε Gc
Desired
Ratio
FA
3. Ratio Control – Type of feedforward control
However, one stream in proportion to another. Use if the ratiomust be measured and displayed
Disadvantage:Ratio may go To erratic
A Wild stream
FT
Desired Ratio
Multiplier
ε FC
FT
Controlled streamFB
FA
Another implementation of Ratio Control
+
-
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