Solutions Manual  Ch12
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Smith & Corripio, 3rd edition Problem 121. Loop interaction for the processes of Fig. 121.1. (a) Blending tank, Fig. 121.1(a). Manipulated variables: flows of each of the two inlet streams. Controlled variables: product flow and composition. Increasing the flow of either inlet stream increases the product flow. Increasing the flow of the more concentrated stream increases the product composition, while increasing the flow of the more dilute stream decreases the product composition. This means the interaction is positive (the two loops help each other): when the flow of the stream controlling the product composition increases, the flow controller decreases the flow of the other stream; as the two inlet streams have opposite effects on the product composition, the change caused by the product flow controller changes the composition in the same direction.
(b) Chemical reactor, Fig. 121.1(b). Manipulated variables: coolant flow and reactants flow. Assume the product flow is manipulated to control the level in the reactor. Controlled variables: flow and reactant composition in the product stream. As the reactor is cooled, the reaction must be exothermic. • •
The coolant flow has a negative effect on the temperature and no direct effect on the composition. The reactants flow has a positive effect on the reactant composition and, if the feed is at a lower temperature than the reactor, a negative effect on temperature.
So, at first glance it appears that there is no intercation between the loops, but the interaction comes through the effect or reactant composition and temperature on each other through the reaction rate. When the reactants composition is not controlled, an increase in coolant flow causes a decrease in reactor temperature; this decreases the reaction rate and increases the reactants concentration, resulting in a higher reaction rate and temperature than if the reactants composition were kept constant. So, controlling the composition constant results in a smaller change in reactor temperature when the coolant flow is changed. Similarly, if the temperature is not controlled, an increase in reactants flow increases in reactants concentration and the reaction rate resulting in an increase in temperature, so that the increase in reaction rate is higher than if the temperature is kept constant. So, controlling the temperature constant results in a larger increase in composition when the temperature is allowed vto increase. So, for this particular case, it appears that the interaction is negative with respect to the temperature loop, and positive with respect to the composition loop. This is unusual.
(c) Evaporator, Fig. 121.1(c). Manipulated variables: valve on feed line, steam flow, and product flow. Controlled variables: evaporator level, product composition, and througput. • • •
Opening the valve on the feed line increases the throughput and the level in the evaporator, and decreases the product composition by diluting the contents of the evaporator. Increasing the steam flow increases the vaporization rate, decreasing the level and increasing the product composition by removing the solvent. Increasing the product flow decreases the level and increases the throughput, but it has no direct effect on the product composition.
As the level must be controlled, the effect of each manipulated variable on the controlled variables depends on which manipulated variable is used to control the level. Let us assu,e that the feed valve is used to control the level, since the feed is the largest of the three streams and the level must be controlled tightly in an evaporator. Then, an increase in steam causes the feed flow to increaseto maintain the level constant; this increases the throughput and also the product composition, since there is a net increase in the flow of solute into the eveporator. An increase in product flow increases the feed flow to maintain the level constant; this increases the throughput and decreases the product composition, since there is a net increase in the rate of solvent into the evaporator. The interaction is positive: three positive effects and one negative. When the troughput is maintained constant the steam and product flows must be chnaged in opposite directions causing the product composition to change more than if the throughput is allowed to vary. Similarly, when the product composition is maintained constant, the steam and product flows must change in the same direction causing the throughput to change more than if the composition is allowed to vary.
(d) Paperdrying machine, Fig. 121.1(d). Manipulated variables: stock feed flow and steam flow. Controlled variables: moisture content and drybasis weight (fibers per unit area) of the product. Assume that the water is removed at a constant rate by mechanical means (filtration) in the first part of the machine. • •
Increasing the stock feed rate increases the moisture content and the drybasis weight. increasing the steam rate decreases the moisture content and has no direct effect on the drybasis weight.
It appears that there is no interaction.
(e) Distillation column, Fig. 121.1(e). Manipulated variables: Coolant flow to the condenser, steam flow to the reboiler, reflux flow, distillate product flow, and bottoms product flow. Controlled variables: distillate and bottoms product compositions, condenser accumulator and
column bottom levels, and column pressure. • • • • •
Coolant rate has a negative effect on the column pressure and a positive effect on the accumulator level. Steam flow to the reboiler has positive effects on the bottoms product purity and the column pressure, and negative effects on the distillate product purity and the column bottom level. Reflux flow has positive effects on distillate product purity and column bottom level, and negative effects on bottoms product purity and condehnser accumulator level. Distillate product rate has a negative effect on the condenser accumulator level and no direct effect on any of the other variables. Bottoms product rate has a negative effect on column bottom level and no direct effect on any of the other variables.
There are positive ad negative interactions between the variables. As the two levels must be controlled, the interactions depend on which variables are manipulated to control the levels.
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mass% := %
Smith & Corripio, 3rd edition
klb := 1000lb
Problem 122. Control of caustic dilution process. Problem data:
SP
klb
wP := 40 hr
AC 3
SP
x P := 30mass%
FC
x C := 50mass%
SP FC
1
3
FT
Caustic
Assumptions: • Perfect mixing • Constant mass
AT
SP
FC
xP
2
FT
wP
FT
Manipulated variables: Flows of water and caustic wC
and
w Water
ww
Controlled variables: Product flow and mass fraction:
Steady state model:
wP
and
Total mass balance:
xP wC + ww = wP
Caustic balance:
wC⋅ x C = wP⋅ x P
Solve for x P:
xP =
wC⋅ x C wP
=
(pure water diluent) wC⋅ x C wC + ww
Openloop gain matrix: wC
ww wP
δwP δwC KOL = δxP δw C
Relative gain matrix:
δwP
1 1 1 1 w ⋅ x − w ⋅ x −w ⋅ x w ⋅ x −w ⋅ x C C C C= w C C C = P C δx P 2 2 2 w 2 wP wP wP P δww
δww
wC
ww
xP
−ww⋅ x C ww −wC⋅ x C wC −wC⋅ xC − ww⋅ xC −wC⋅ xC − ww⋅ xC wC + ww wC + ww µ = = −ww⋅ x C −wC⋅ x C wC ww −wC⋅ xC − ww⋅ xC −wC⋅ xC − ww⋅ xC wC + ww wC + ww
wP
xP
So, the result is as for the regular blender: the pairing that minimizes the interaction is the one in which the largest of the two inlet flows is used to control the product flow, and the smaller inlet flow is used to control the product mass fraction. The interaction is positive (the loops help each other) as the relative gains are positive.
Numerical results:
wC :=
wP⋅ x P
klb ww = 16 hr
ww := wP − wC
xC
Open loop gains:
wC
w ⋅x − w ⋅ x C C klb klb KOL := w C 2 hr⋅ mass% w 2 hr⋅ mass% wP P 1
Relative gains:
1
wC wP µ := ww wP
wC wP
wP
ww
1 1 KOL = 0.5 −0.75
wC
ww µ =
klb wC = 24 hr
wP xP
ww
0.6 0.4 0.4 0.6
wP xP
So, in this case, to minimize interaction, • the caustic stream flow, FC1, must be manipulated to control the product flow • the water inlet flow, FC2, must be used to control the product composition.
When the flow control loop is closed, the gain of the composition loop increases by 1 µ 1, 1
− 1 = 66.7 %
The gain is then:
KOL 1, 1 µ 1, 1
= −1.25
mass%⋅ hr klb
Notice that the gain of the composition loop is negative, so a direct acting controller is required for the composition loop. Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
degF := R
Smith & Corripio, 3rd edition
gal
gpm :=
min
Problem 123. Automatic control of a household shower. Problem data: fo := 3gpm T1 := 170degF ρ := 8.33
lb gal
SP
To := 110degF
SP
T2 := 80degF
TC TE
cp := 1
fo To
FE
FC
BTU lb⋅ degF
Assume • Negligible time delay • Constant properties • Reference temperature of 0ºF
S
S T1 f1 Hot water
T2 f2 Cold water
(a) Develop the model of the process Total mass balance:
f1 ⋅ ρ + f2 ⋅ ρ = fo ⋅ ρ
fo = f1 + f2
Enthalpy balance:
f1 ⋅ ρ ⋅ cp ⋅ T1 + f2 ⋅ ρ ⋅ cp ⋅ T2 = fo ⋅ ρ ⋅ cp ⋅ To To =
f1 ⋅ T1 + f2 ⋅ T2 f1 + f2
(b) Required flows of hot and cold water and the openloop steadystate gains. Combine balance equations:
(
)
f1 ⋅ T1 + fo − f1 T2 = fo ⋅ To To − T2 f1 := fo ⋅ T1 − T2
Openloop gains:
f2 := fo − f1
f1 = 1 gpm
f2 = 2 gpm
δfo δf1 KOL = δTo δf 1
δfo
1 f + f ⋅T − f ⋅T − f ⋅T = ( 1 2) 1 1 1 2 2 δTo 2 f1 + f2 ) ( δf1 δf2
(
f1 + f2 ⋅ T2 − f1 ⋅ T1 − f2 ⋅ T2
)
(f1 + f2)
f1
1 1 f ⋅ T − T 2 ) gpm −f1 ⋅ ( T1 − T2 ) gpm KOL := 2 ( 1 2 2 degF degF fo fo
1
2
f2
1 1 KOL = 20 −10
fo To
The first row in dimensionless and the second has units of ºF/gpm.
(c) Relative gains and pairing that minimize interaction. f1
f2
−f2 ⋅ ( T1 − T2 ) f2 f1 −f1⋅ ( T1 − T2 ) f −( f1 + f2) ( T1 − T2) −( f1 + f2) ( T1 − T2) f1 + f2 f1 + f2 o µ = = −f1 ⋅ ( T1 − T2 ) f1 −f2⋅ ( T1 − T2 ) f2 T o −( f1 + f2) ( T1 − T2) −( f1 + f2) ( T1 − T2) f1 + f2 f1 + f2 The pairing that minimizes interaction is the total flow with the larger of the two inlet streams (in this case the cold water, f2 ) and the tempearture with the smaller stream (in this case the hot water, f 1 ).
f1 fo µ := f2 fo
fo f1 f2
fo
f1 µ =
f2
0.333 0.667 0.667 0.333
fo To
The interaction is positive (the loops help each other), as the relative gains are positive. When one loop is closed, the gain of the other loop increases by: 1 µ 1, 0
− 1 = 50 %
The closedloop gains are:
KOL 0, 1 µ 0, 1
= 1.5
KOL 1, 0 µ 1, 0
= 30
degF gpm
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
Smith & Corripio, 3rd edition Problem 124. Control of an evaporator. Problem data: Manipulated variables: Product and steam valve signals. mP
and
wv
mS mS
Controlled variables: Feed flow (throughput) Product composition wF
and
Vapors
wS
SP FC
Steam
SP
xP
LC
Assume the evaporator economy E is constant.
wF
SP
LT
AC
FT
Feed
mP
xP
xF
AT
wP
Product
(a) Steadystate model of the evaporator. Total mass balance:
wF = wP + wv
Solute balance:
wF⋅ x F = wP⋅ x P
Combine and rearrange:
wF = wP + E⋅ wS
Control valve gains:
wP = KvP⋅ mP
wv = E⋅ wS
Economy:
x P = x F⋅
(wP + E⋅ wS) wP
E⋅ wS = x F⋅ 1 + wP
wS = KvS⋅ mS
(b) Steadystate openloop gains and relative gains.
mP
mS
Openloop gains:
δwF δmP KOL = δx P δm P
δwF
KvS⋅ E KvP − E ⋅ w E S = KvS⋅ x F⋅ δx P KvP⋅ x F⋅ wP w 2 P δmS
δmS
wF
xP
Relative gains:
2 KvP⋅ KvS⋅ x F⋅ E ⋅ wS KvP⋅ KvS⋅ x F⋅ E E⋅ wS E⋅ wS 1 2 wP⋅ KvP⋅ Kvs⋅ xF⋅ E⋅ 1 + + wP ⋅ KvP⋅ Kvs ⋅ x F⋅ E⋅ wP w 2 wP w 2 P P µ = 2 KvP⋅ KvS⋅ x F⋅ E ⋅ wS KvP⋅ KvS⋅ x F⋅ E 2 E⋅ wS E⋅ wS 1 1 wP⋅ KvP⋅ Kvs ⋅ x F⋅ E⋅ wP ⋅ KvP⋅ Kvs⋅ x F⋅ E⋅ + + wP w 2 wP w 2 P P
mP
mS
wP E⋅ wS wP + E⋅ wS wP + E⋅ wS µ = wP E⋅ wS wP + E⋅ wS wP + E⋅ wS
wF
xP
The interaction is positive (the loops help each other), as the realative gains are positive.
(c) General pairing strategy. The pairing the minimizes interaction is: • if the product flow is larger than the vapor flow (E*W S), control throughput with the signal to the valve on the product line and the composition with the signal to the steam valve • if the product flow is smaller than the vapor flow (E*WS), control throughput with the signal to the steam valve and the composition with the signal to the product valve. Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
Smith & Corripio, 3rd edition Problem 125. Multivariable control of evaporator of Problem 124. x F := 30mass%
Problem data:
x P := 50mass%
klb wF := 80 hr
E := 0.9
klb klb
Combine balance equations to solve for design conditions: xF wF − wP klb klb wP := wF⋅ wS := wP = 48 wS = 35.556 xP E hr hr From the solution to Problem 124, ignoring the valve gains: Openloop gains: 1 E −x ⋅ E⋅ w x F⋅ E S klb klb KOL := F 2 w hr ⋅ mass% hr ⋅ mass% w P P Relative gains: wP E⋅ wS wP + E⋅ wS wP + E⋅ wS µ := wP E⋅ wS wP + E⋅ wS wP + E⋅ wS
wP
wS wF xP
0.9 1 KOL = −0.417 0.563
The first row is in klb/klb, and the second is in mass%/(klb/hr). wP µ =
wS
0.6 0.4 wF 0.4 0.6 xP
Pairing that minimizes interaction is • control throughput with the product rate • control composition with steam rate
When the feed flow loop is closed, the product composition loop gain changes by 1 µ 1, 1
− 1 = 66.7 %
The closedloop gain of the composition loop is: KOL 1, 1 µ 1, 1
mass%⋅ hr = 0.938
klb
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Smith & Corripio, 3rd edition Problem 126. Distillation product composition control. Manipulated variables: Steam and reflux flows
AT
Controlled variables: Bottoms and distillate compositions
PC
The pressure and level loops are arranged as shown in the figure.
LC 1 Condenser
LT
PT AC 1
Feed
Reflux
Distillate
AC 2 LC 2
LT
Steam Bottoms
AT
Results of tests performed on a smlation of the column: Base Test 1 Test 2
Case
wS := ( 24.0 25.0 24.0 )
Steam flow, klb/hr Reflux flow, lbmole/hr
wR := ( 70.0 70.0 75.0 )
Butane mole %
x B := ( 6.22 3.08 8.77 ) y D := ( 93.50 91.79 96.88 )
Openloop steady state gains: x B0 , 1 − wS − 0, 1 KOL := yD − 0, 1 wS − 0, 1 Relative gains:
xB
0, 0
xB
0, 2
− xB 0, 0
wS 0, 0
wR − wR 0, 2 0, 0
yD
yD
wS 0, 0
wR − wR 0, 2 0, 0
0, 0
0, 2
− yD 0, 0
wS
wR
−3.14 0.51 KOL = −1.71 0.676
xB yD
The first column is in mole%/(klb/hr) and the second in mole%/(lbmole/hr).
− KOL ⋅ KOL KOL0 , 0⋅ KOL1 , 1 0, 1 1 , 0 K K OL OL µ := KOL ⋅ KOL − KOL0 , 1⋅ KOL1 , 0 0, 0 1, 1 KOL KOL
wS µ =
wR
1.697 −0.697 xB −0.697 1.697 yD
The interaction is negative (loops fight each other): relative gains are either negative or greater than unity. Pairing the minimizes interaction: • control distillate composition with reflux rate • control bottoms composition with steam rate
Closedloop gain of distillate control loop:
KOL 1, 1 µ 1, 1
= 0.398
mole%⋅ hr lbmole
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Smith & Corripio, 3rd edition
kbl := 42000gal
Problem 127. Control of gasoline blender of example 122.5. Problem data:
Octane
1. Alkylate 2. Straight run x := 3. Reformate
RVP
97.0 80.0 y := 92.0
5.00 11.00 3.00
Gasoline x p := 89.0
SP
f1
FT
SP
SP
FC
AC
AC
1
2
3
AT
AT
FT
x
y
f
SP
Alkylate
y p := 7.00
FC
kbl fp := 60.0 day
f2
FT
SP
Straight run
Assume • gasoline properties are a linear combination of the feed properties weighted by volume • Constant standar density
FC f3
FT
Reformate
Required inlet flows: xP =
x0 x1 x2 f := y y y 0 1 2 1 1 1
SP
FC
−1
f1 ⋅ x 1 + f2 ⋅ x 2 + f3 ⋅ x 3 f1 + f2 + f3
xp fp ⋅ day yp kbl 1
yp =
22.5 f = 24.375 13.125
f1 ⋅ y 1 + f2 ⋅ y 2 + f3 ⋅ y 3 f1 + f2 + f3
kbl day
Steadystate openloop gains: f 1⋅ ( x0 − x 1) + f 2⋅ ( x0 − x 2) f 0⋅ ( x 1 − x0) + f 2⋅ ( x 1 − x2) f 0⋅ ( x2 − x 0) + f 1⋅ ( x2 − x 1) 2 2 2 2 day 2 day 2 day fp ⋅ fp ⋅ fp ⋅ kbl kbl kbl KOL := f ⋅ ( y − y ) + f ⋅ ( y − y ) f ⋅ ( y − y ) + f ⋅ ( y − y ) f ⋅ ( y − y ) + f ⋅ ( y − y ) 1 0 1 2 0 2 0 1 0 2 1 2 0 2 0 1 2 1 2 2 2 day day day 2 2 2 fp ⋅ fp ⋅ fp ⋅ kbl kbl kbl 1 1 1
Alk St run f1 f2
Ref f3
0.133 −0.15 0.05 KOL = −0.033 0.067 −0.067 1 1 1
day kbl
xp
Oct
yp
RVP
fp
Flow
Relative gains: 11.25 0.375 7.5 kbl B = −1.875 4.688 0.406 day −5.625 −15.938 0.219
−1
B := KOL
Inverse of the openloop gains:
KOL0 , 0⋅ B0 , 0 KOL0 , 1⋅ B1 , 0 KOL0 , 2⋅ B2 , 0 µ := KOL1 , 0⋅ B0 , 1 KOL1 , 1⋅ B1 , 1 KOL1 , 2⋅ B2 , 1 KOL ⋅ B ⋅B ⋅B K K 2 , 0 0 , 2 OL2 , 1 1 , 2 OL2 , 2 2 , 2
Alk f1
St run f2
Ref f3
0.281 −0.281 1 µ = −0.375 0.313 1.062 0.375 0.406 0.219
x p Oct y p RVP fp Flow
Pairing to minimize interaction: • Use alkylate (stream 1) to control the gasoline octane • Use the reformate (stream 3) to control the gasoline Reed vapor pressure • Use the straight run (stream 2) to control the product flow. These are the same pairings as in Example 122.5.
Design of static decoupler, as in Example 123.3. To use the recommended pairing with the decoupler, we must swap columns 2 and 3 in the iopenloop gain matrix, and recalculate the inverse matrix. The new inverse matrix is matrix B with the second and third rows swapped. The decoupler is then, from Eq. 123.11: Oct
B0 , 0 B0 , 0 B2 , 0 D := B0 , 0 B 1, 0 B0 , 0
B B B B B B
0, 1 2, 1 2, 1 2, 1 1, 1 2, 1
0, 2 B 1, 2 B 2, 2 B 1, 2 B 1, 2
RVP
Flow
m2
m3
B
B
1, 2
m1
1 −0.706 0.923 D = −0.75 1 0.538 −0.25 −0.294 1
f1 f3
Alkylate
f2
Straight run
Reformate
The instrumentation diagram is the same as Fig. 123.6.
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
Smith & Corripio, 3rd edition Problem 128. Control of a wet grinding circuit. Manipulated variables: SF = solids flow MW = mill water flow SW = slurry tank water Controlled variables: TOR = mill torque FML = flow from mill DCF = cyclone feed density
Cyclone
TOR
XT
XC FC FML
SW
FT
Mill
SF
AC AT DCF
MW
Models from openloop tests by Hubert and Woodburn (1983): SF, kg/s Time is in sec
MW, kg/s
SW, kg/s
153 −21 119 337 ⋅ s + 1 10⋅ s + 1 217 ⋅ s + 1 0.37 0.767 −0.050 G( s) = 500 ⋅ s + 1 33⋅ s + 1 10⋅ s + 1 − 320s 0.930 −0.667 e −1.033 500 ⋅ s + 1 166s + 1 47s + 1
TOR, Nm FML, liters/s DCF, kg/liter
(a) Relative gains and pairings that minimize interaction. Obtain the openloop steadystate gains by setting s = 0 in the transfer functions: −21 119 153 KOL := 0.37 0.767 −0.050 0.930 −0.667 −1.033
−1
B := KOL
Obtain the relative gains by Eq. 122.13, page 422:
0.03089 −6.43731 −0.31641 B = −0.01256 3.8685 0.06809 0.03592 −8.29331 −1.29688
KOL0 , 0⋅ B0 , 0 KOL0 , 1⋅ B1 , 0 KOL0 , 2⋅ B2 , 0 µ := KOL1 , 0⋅ B0 , 1 KOL1 , 1⋅ B1 , 1 KOL1 , 2⋅ B2 , 1 KOL ⋅ B KOL ⋅ B KOL ⋅ B 2, 0 0, 2 2, 1 1, 2 2, 2 2, 2
SF kg/s
MW kg/s
SW kg/s
3.676 −1.922 −0.754 TOR, Nm FML, liters/s µ = −2.382 2.967 0.415 −0.294 −0.045 1.34 DCF, kg/liter
Pairing to minimize interaction: • Control the mill torque TOR with the solids flow SF • Control the flow from the mill FML with the water flow to the mill MW • Control the density of the cyclone feed DCF with the slurry water flow SW
(b) Design a decoupler and draw the block diagram and the instrumentation diagram of the decoupled system. Because of the complexity of the transfer functions, and the fact that five of the six interaction terms are negative, a full dynamic decoupler will probably create unstable poles. So, a static decoupler is designed. If dynamic compensation is required, leadlag units may be added and tuned online. The gain matrix is correct for the recommended pairing. Obtain the decoupler from Eq. 123.11, page 432:
1 B1 , 0 D := B0 , 0 B 2, 0 B0 , 0
B B
0, 1 1, 1
1 B B
2, 1
B 2, 2 B 1, 2 B 2, 2 1 B
0, 2
1, 1
Decoupled system gains:
mTOR mFML mDCF SF −1.664 0.244 1 D = −0.407 1 −0.053 1 1.163 −2.144
set
MW SW
set
set
All variables in kg/sec
mTOR mFML mDCF kg/sec kg/sec kg/sec
−0 TOR, Nm 32.372 0 FML, liters/sec KOL⋅ D = 0 0.258 0 0 −0.771 DCF, kg/liter 0 Block diagram of decoupled system:
R1(s)

+
GC1(s)
M1(s)
SF
+ 
+
1.664
R2(s)

+
GC2(s)
G13(s)
0.407
G21(s)
M2(s)
+ 
MW


GC3(s)
+ + G22(s) + FML G31(s) G32(s)
2.144  + M3(s) +
Instrumentation diagram of decoupler:
+
TOR
G23(s)
1.163
+
+ +
G12(s)
0.244
0.053
R3(s)
G11(s)
SW
+ G33(s) +
+
DCF
FMLSP
TORSP
XC
1.664
3
XY
DCFSP
FC
0.407
0..244
3 FY
AC
2.144 0.053
1.163
AY 3
TOR
XT
FML
SW
FT SF MW
Mill
AT DCF
The summer devices, XY, FY, and AY, are the decouplers. Each signal into the summer is multiplied by the factor shown by it (unity if no factor is shown). The factors are not corrected for the valve gains. Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
Smith & Corripio, 3rd edition Problem 129. Decoupler design for distillation product composition control. Manipulated variables: Reflux flow Steam flow to reboiler Controlled variables: Distillate purity Bottoms product purity
mR mS yD
AT
PC
Condenser
xB
xB
Reflux
AC 1
Distillate
AC 2
= composition of heavy key in the distillate = composition of the light key in the bottoms product
LT
PT
Feed yD
LC 1
LC 2
LT
Steam Bottoms
AT
Transfer functions from openloop step tests (time parameters in hours): 0.75 1 − 0.11s YD( s) = −3.0 M R( s) + M ( s) 1 + 0.35s S 1 + 0.35s 2.5 1 + 0.25s XB( s) = M R( s) − 3.5 M S( s) 1 + 0.35s 1 + 0.35s %TO := %
(a) Relative gains and pairing that minimizes interaction.
%CO := %
Obtain the openloop steadystate gains by setting s = 0 in the transfer functions: mR KOL :=
mS
−3.0 0.75 %TO yD 2.5 −3.5 %CO xB
Relative gains, from Eq. 122.7, page 418::
KOL0 , 0⋅ KOL1 , 1 KOL µ := ( −KOL) 0 , 1⋅ KOL1 , 0 KOL
(−KOL)0 , 1⋅ KOL1 , 0 KOL KOL ⋅ KOL 0, 0 1, 1 KOL mR
µ =
mS
1.217 −0.217 yD −0.217 1.217 xB
The interaction is negative (loops fight each other): relative gains are negative or greater than one. Pairing that minimizes interaction: • Control distillate purity with reflux flow • Control bottoms purity with steam flow
(b) Design the block diagram and draw the block diagram for this system. 1 − 0.11⋅ s 0.75 YD( s) = −3.0⋅ + ⋅ D1 ( s) ⋅ M S( s) = 0 1 0.35 s 1 + 0.35 ⋅ s + ⋅ XB( s) =
2.5 ⋅ D ( s) − 3.5⋅ 1 + 0.25⋅ s ⋅ M ( s) = 0 1 + 0.35⋅ s 2 1 + 0.35⋅ s R
3.0 D1 ( s) = ( 1 − 0.11s) 0.75 3.5 D2 ( s) = ( 1 + 0.25s) 2.5
The first decoupler required a negative lead that will produce an undersirable result in the reflux flow, so we will use a simple gain. The second decoupler can be implemented, but a small lag must be included. %CO D1 ( s) = 4.0 %CO
%CO 1 + 0.28s D2 ( s) = 1.4 %CO 1 + 0.03s
The second decoupler introduces a net lead of 0.25 hour (15 minutes). Block diagram of decoupled system:
R1(s)

+
MR(s) D1(s)
G11(s) + + YD(s) G12(s)
D2(s)
G21(s)
+ +
R2(s) +
Gc1(s)

Gc2(s)
MS(s)
+ +
G22(s)
+ XB(s) +
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Smith & Corripio, 3rd edition Problem 1210. Decoupler design for a 2x2 process. C( s) = GP( s) M ( s) + GPU ( s) ⋅ U( s)
0.81e− 0.6s 1.2e− 1.1s 1.4s + 1 2.4s + 1 GP( s) = − 0.3s −s 0.9e 1.1e 2s + 1 1.5s + 1
0.5 2.2s + 1 GU( s) = −1.5 1.8s + 1
(a) Relative gains and pairing that minimizes interaction. Obtain openloop steadystate gains by setting s = 0 in the transfer functions:
KOL :=
0.81 1.2 1.1 0.9
Relative gains by Eq. 122.7, page 418:
KOL0 , 0⋅ KOL1 , 1 −KOL0 , 1⋅ KOL1 , 0 KOL KOL µ := −KOL0 , 1⋅ KOL1 , 0 KOL0 , 0⋅ KOL1 , 1 KOL KOL
m1
µ =
m2
−1.234 2.234 c1 2.234 −1.234 c2
The interaction is negative (loops fight each other): relative gains are either negative or greater than unity. Pairing that minimizes interaction: • Control c1 with m2 •
Control c2 with m1
(b) Design decoupler and show in the block diagram. GP1 , 2( s) ⋅ D1 ( s) + GP1 , 1( s) M1 ( s) = 0 GP2 , 2( s) + GP2 , 1( s) ⋅ D2( s) M2 ( s) = 0
−GP ( s) 1, 1
D1 ( s) = = GP ( s) 1, 2
−GP ( s) 2, 2
−0.81 2.4s + 1 − ( 0.6− 1.1)s e 1.2 1.4s + 1
−0.9 1.5s + 1 − ( 1− 0.3)s D2 ( s) = = e 1.1 2.0s + 1 GP ( s) 2, 1
The first decoupler term cannot be implemented because it requires a negative delay which would require knowledge of the signal in the future. So, we will add the negative delay of 0.5 min to the lead in the lead lag unit, making the new lead: ( 2.4 − 0.6 + 1.1)min = 2.9 min
The decouplers are then: 2.9s + 1 D1 ( s) = −0.675 1.4s + 1 1.5s + 1 − 0.7s D2 ( s) = −0.818 e 2.0s + 1
Block diagram of the complete system:
R1(s) +

Gc1(s)
M2(s)
+
GP12(s)
+ GP11(s)
D1(s) U(s)
+ + C1(s) +
GU1(s) GU2(s)
D2(s) R2(s) +

Gc2(s)
M1(s)
GP22(s) + +
GP21(s)
+ +
+C2(s)
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Smith & Corripio, 3rd edition Problem 1211. Decoupling of 2x2 process of Fig. 123.2show decoupled closedloop transfer functions The decouplers are given in Eq. 123.2, page 426: −Gv2( s) ⋅ G12( s) D12( s) = Gv1( s) ⋅ G11( s)
−Gv1( s) ⋅ G21( s) D21( s) = Gv2( s) ⋅ G22( s)
From the block diagram, the transfer functions are:
(
)
C1 ( s) = Gv1( s) ⋅ G11( s) + D21( s) ⋅ Gv2( s) ⋅ G12( s) ⋅ M 1 ( s)
(
)
+ Gv2( s) ⋅ G12( s) + D12( s) ⋅ Gv1( s) ⋅ G11( s) ⋅ M 2 ( s) If the decouoplers can be implemeted exactly, then: −Gv1( s) ⋅ G21( s) C1 ( s) = Gv1( s) ⋅ G11( s) + ⋅ Gv2( s) ⋅ G12( s) ⋅ M 1 ( s) Gv2( s) ⋅ G22( s) −Gv2( s) ⋅ G12( s) + Gv2( s) ⋅ G12( s) + ⋅ Gv1( s) ⋅ G11( s) ⋅ M 2 ( s) Gv1( s) ⋅ G11( s) Simplify:
G21( s) C1 ( s) = Gv1( s) ⋅ G11( s) − ⋅ G12( s) ⋅ M 1 ( s) G22( s)
So, the system would be decoupled.
Similarly for the other variable: From the block diagram, the transfer functions are:
(
)
C2 ( s) = Gv2( s) ⋅ G22( s) + D12( s) ⋅ Gv1( s) ⋅ G21( s) ⋅ M 2 ( s)
(
)
+ Gv1( s) ⋅ G21( s) + D21( s) ⋅ Gv2( s) ⋅ G22( s) ⋅ M 1 ( s) If the decouoplers can be implemeted exactly, then: −Gv2( s) ⋅ G12( s) C2 ( s) = Gv2( s) ⋅ G22( s) + ⋅ Gv1( s) ⋅ G21( s) ⋅ M 2 ( s) Gv1( s) ⋅ G11( s) −Gv1( s) ⋅ G21( s) + Gv1( s) ⋅ G21( s) + ⋅ Gv2( s) ⋅ G22( s) ⋅ M 1 ( s) Gv2( s) ⋅ G22( s)
Simplify:
G12( s) C2 ( s) = Gv2( s) ⋅ G22( s) − ⋅ G21( s) ⋅ M 2 ( s) G11( s)
So, the system would be decoupled.
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Smith & Corripio, 3rd edition Problem 1212. Decoupler design for blending tank of Example 125.1. Design conditions: x 1 := 10mass% w := 100
SP
AC
x 2 := 30mass%
lb
SP
FC
x := 20mass%
min
m1
Valves are linear with constant pressure drop. lb wmax := 150 min
w1 x1
V
m
2
AT
FT
x
w
Analyzer transmitter AT: x min := 5mass%
x max := 35mass%
At the design conditions:
w2 x2
x − x1 w1 := w⋅ x2 − x1
w2 := w − w1
lb w1 = 50 min
lb w2 = 50 min
So, there is no best pairing and the relative gains are 0.5 for both pairings. As there are no flow transmitters on the inlet flows, ratio control cannot be implemented. We design then a linear decoupler. −Kv2 w = w1 + w2 = Kv1⋅ m1 + Kv2⋅ m2 ∆w = Kv1⋅ ∆m1 + Kv2⋅ ∆m2 = 0 ∆m1 = ∆m2 Kv1 w1 ⋅ x 1 + w2 ⋅ x 2 x= w1 + w2
(w1 + w2)x1 − w1⋅ x1 − w2⋅ x2 K ⋅ ∆m + (w1 + w2)x2 − w1⋅ x1 − w2⋅ x2 K ⋅ ∆m = 0 v1 1 v2 2 (w1 + w2)2 (w1 + w2)2
∆x =
(
)
(
)
w2 ⋅ x 1 − x 2 ⋅ Kv1⋅ ∆m1 + w1 ⋅ x 2 − x 1 ⋅ Kv2⋅ ∆m2 = 0 wmax Kv1 := 100%CO
The gains of the decouplers are:
wmax Kv2 := 100%CO
−Kv1 Kv2
= −1
∆m2 =
lb Kv1 = 1.5 min⋅ %CO Kv1⋅ w2 Kv2⋅ w1
=1
Kv1⋅ w2 Kv2⋅ w1
∆m1
lb Kv2 = 1.5 min⋅ %CO
Instrumentation diagram: FY: summer, keeps flow constant when the analyzer controller varies m2 AY: summer, keeps composition constant when the flow controller varies m1

Kv1 w2 Kv2
= 33.33 %CO
AC SP
m1
3 FY +
FC
+
3 AY
w1 x1
Both require a bias of: w1
SP
m2
+ V
AT
FT
x
w
w2 x2
= 33.33 %CO
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Smith & Corripio, 3rd edition Problem 1213. Nonlinear static decoupler for evaporator of Problem 124. From the solution of Problem 124:
wv
wF = wP + E⋅ wS
xP =
Vapors
x F⋅ wF
mS
wP
wS
SP FC
wP = KvP⋅ mP
SP
LC
wS = KvS⋅ mS
Assume the vapor rate is greater than the product rate, so that the steam is used to control the feed rate and the product to control the composition.
wF
Steam SP
LT
AC
FT
Feed
mP
xP
xF
AT
wP
Product
Decoupler design: Want the output of the feed flow controller FC to be the sum of the vapors and the product: set
= wP + E⋅ wS
wF
So
set
wS =
wF
− wP
E
set
KvP mS = − m E⋅ KvS P E⋅ KvS wF
Now, the feed flow setpoint is obtained using a scale factor on the controller output in %CO. This scale factor is obtained from the maximum expected feed rate over 100%CO: set
wF
=
wFmax 100%CO
mS =
wFmax 100%CO⋅ E⋅ K mFC − vS
KvP E⋅ K mP vS
Want the output of the analyzer controller AC to be the composition: set
xP
wF = x F⋅ wP
wP =
xF set
xP
⋅ wF
This is a ratio controller with the ratio adjustable by the output of the composition controller AC: xF where, at design conditions: wP = Ratio⋅ wF Ratio = set xP
This would work best if a flow controller is installed on the product line, but this is probably too much of an expense for an evaporator. So, the ratio of the signals are used:
mP = Ratio⋅ mFC
Ratio =
where
xF
wFmax
set 100%CO⋅ KvP
xP
Instrumentation diagram: FY: summer, subtracts the scaled signal to the product valve from the signal to the seam valve to maintain the feed flow constant.
SP AC SP FC
Ratio
Vapors
mFC
AY
AY: multiplier, makes the signal to the product valve proportional to the total signal from the feed controller, scaled, to make the product flow proportional to the feed flow and mintain the product com position constant.
X
FY 3 mS
SP
LC wF
mP
wS Steam
LT
FT
Feed
xF
xP wP
AT Product
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
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