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Learning Objectives
Understand when, why & how to apply
cascade control
Understand benefits of cascade control
loops
Draw a cascade block diagram and
analyse response
Be able to simplify cascade control loops
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What about the extra disturbance
to the heat exchanger?
Steam supply
pressure may
vary
Steam flow
will change
without valve
movement
This is a
disturbance
to a
manipulated
variable
Cold fluid
T change
Hot fluid
Steam
TC
101
TT
101
SP
Steam supply pressure Ps
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Cascade Control: Controllingfor Manipulated Variable
DisturbancesCold fluidT change
Hot fluid
SteamTT
101
TC
101
TSP
FC
101
FSP
Steam pressure disturbance
T control is the OUTER loop
F control is the INNER loop
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Heat exchanger control: Feedback
control
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Cascade control heat exchanger:
Option 1 Option 2Smith & Corropio
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Cascade control example
Smith & Corropio
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Cascade Control to jacketed reactor
Cooling water
Variablesupply P
Outer
Temp
Controller
InnerFlow
controller
TC101
SP
FC101
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Simplified Block Diagram
Simplify inner loopreplace with single
transfer function
GpYFs
1Tm
GD
++
SP E
+-
Kc(1+1/Tis) GVU
Kc2+
-
D2
+
Outer loopInner loop
KC2Gv1+KC2Gv
D11
1+KC2Gv
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Analysis of a Cascade Loop:
1 - Inside the Inner Loop
0.5s1
0.5
c2K1
0.5s1
0.5
c2K
F
sF
SP
FSP
IF Kc2= 10
0.083s1
0.83
F
sF
SP
0.5
1 +0.5s
UKc2
+ -
4e-2s
1 + 15s
TFs
1Tm
1e-s
1+5s
TD
++
Ts E
+-
Kc(1+1/Tis)
Ps
+
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Analysis of a Cascade Loop:
1 - Equivalent Block for Inner Loop
FSP
0.083s1
0.83
F
sF
SP
4e-2s
1 + 15s
TFs
1Tm
1e-s
1+5s
TD
++
Ts E
+-
Kc(1+1/Tis)0.83
1 + 0.083s
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Analysis of a Cascade Loop:
2Impact of second disturbance
1
sP
sF
If Kc2= 10
Fset0.5
1 +0.5s
UKc2
+ -
4e-2s
1 + 15s
TFs
1Tm
1e-s
1+5s
TD
++
Ts E
+-
Kc(1+1/Tis)
Ps
+
No cascade:
Unit change
in Ps gives unit
change in Fs
With cascade:
Unit change
in Ps gives 1/6change in Fs
(final value theorem)
Open loop TF:
0.5s1
0.5
c2K1
1
sP
sF
Closed loop TF:
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Reactor with preheater:
feedback controlMaintain constant temperature
in the reactor (controlled variable)
By manipulating fuel to the
preheater furnace
Feed temperature
Is a disturbance
Smith & Corropio
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Reactor with preheater:
Cascade control
Smith & Corropio
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FEEDBACK
CONTROL
Smith & Corropio
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CASCADE
CONTROLInner loop
= Preheater loop
Smith & Corropio
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Tuning Cascade loops
Inner loop is a normal feedback loop Tune by ZN ultimate frequency, ZN quarter decay, or
CohenCoon method
Then start tuning the outer loop : Set inner loop to automatic mode
Tune by ZN ultimate frequency
A one-step method (Austin and Lopez) also
existssee Smith Ch 9 and table 9.31 Find gain, timeconstant and deadtime of both loops Set parameters via table
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2005 EXAM: Question 2Solution slides by Prof Brisk
You have recently been assigned the task of examining acascade control system on a furnace. The previous engineer incharge of this system has told you that the transfer functionrelating gas flow to the burner, F, to the signal from the slavecontroller is first order with a gain of 0.2 and a time constant of
10 s. The transfer function relating the change in furnacetemperature , T, to the gas flow, F, is first order + dead timewith a gain of 10, a time constant of 100 s, and a dead time of1 s. In the absence of control it was found that when the gaspressure, P, increased by one unit, the gas flowrate reached
63% of its final value in 10 s, and the temperature increasedby 100oC at steady state. The master controller is PI and theslave controller is P.
The topic areas for this question are Block Diagrams,Cascade Control and Controller Tuning
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Solution to Q2a
Draw a block diagram of the system
10e-s
1 + 100s
TTs E
+-
Kc(1+1/Tis)
1ry loop
2ry loop
Kc2FU 0.2
1 +10s+-
Fs +
10
1+10sP
Product ofthese = 100
PI control: Kc(1+1/[Tis]) Note block diagrammust have signs shownfor feedback, setpoint
and disturbance
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Solution to Q2b / Learning item:
Closed Loop Transfer Eqns & Fns.Determine the closed-loop transfer functions F/FSand F/P asfunctions of the secondary controller gain Kc2, where FSis theinner controller set point
YKcGc U
Km
Gm
Ym
KdGdD
++
Ys E+
-
KvGv KpGp
D
mG
mKp
Gp
Kv
Gv
Kc
Gc
K1d
Gd
K
sY
mG
mK
pG
pKv
Gv
Kc
Gc
K1
pG
pKv
Gv
Kc
Gc
KY
Forward1+Forward*Feedback
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Solution to Q2b - continued
Kc2
FU 0.2
1 +10s+-
Fs
)s101/(K2.01
)s101/(K2.0
F
F
2c
2c
s
2c
2c
K2.0s101K2.0
s)]K2.01/(10[1
)K2.01/(K2.0
2c
2c2c
Do NOT leave in this form
Ts1
K
This standard form allowsclear physical interpretation
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Solution to Q2b - continued
Kc2
FU 0.2
1 +10s+-
Fs
)s101/(K2.01
)s101/(10
P
F
2c
2cK2.0s10110
s)]K2.01/(10[1
)K2.01/(10
2c
2c
+
10
1+10sP
Kv= 0.2 here
Ts1
K
We see immediately that theeffect of the disturbance is
reduced by (1 +KvKc2)
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Solution to Q2c
Draw a simplified block diagram of the cascade control system
+-
UKc2
0.2
1 +10s
10
1+10s
P
F+
FsTs E
+-
Kc(1+1/Tis)10e-s
1 + 100s
T
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Solution to Q2c
Draw a simplified block diagram of the cascade control system
Fs F+
Ts E
+-
Kc(1+1/Tis)10e-s
1 + 100s
T
s)]K2.01/(10[1
)K2.01/(K2.0
2c
2c2c
s)]K2.01/(10[1
)K2.01/(10
2c
2c
P
Equivalent TF for inner loop
EquivalentTF for inner loop