Solutions Manual - Ch12

(1)

Smith & Corripio, 3rd edition

Problem 12-1. Loop interaction for the processes of Fig. 12-1.1.

(a) Blending tank, Fig. 12-1.1(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. 12-1.1(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.

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(c) Evaporator, Fig. 12-1.1(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) Paper-drying machine, Fig. 12-1.1(d).

Manipulated variables: stock feed flow and steam flow.

Controlled variables: moisture content and dry-basis 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 dry-basis weight. •

increasing the steam rate decreases the moisture content and has no direct effect on the •

dry-basis weight.

It appears that there is no interaction.

(e) Distillation column, Fig. 12-1.1(e).

Manipulated variables: Coolant flow to the condenser, steam flow to the reboiler, reflux flow, distillate product flow, and bottoms product flow.

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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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Steady state model:

Total mass balance: _{wC ww}+ =_wP

Caustic balance: _{wC xC}⋅ =_{wP xP}⋅ (pure water diluent)

Solve for x_P: _xP wC xC⋅

wP

= wC xC⋅

wC ww+

=

Open-loop gain matrix:

wC ww wP KOL δwP δwC δxP δwC δwP δww δxP δww













= 1 wP xC⋅ − _{wC xC}⋅ wP2 1 wC − ⋅_xC wP2













= 1 ww xC⋅ wP2 1 wC − ⋅_xC wP2













= xP

Relative gain matrix:

_wC _ww

mass%:= % klb:= 1000lb

Smith & Corripio, 3rd edition

Problem 12-2. Control of caustic dilution process.

3 SP AT FC SP FT

w

_P

x

_P w FC FT FC FT SP SP Caustic Water AC 3 1 2 Problem data: wP 40:= klb_hr _{xP 30mass%}:= xC 50mass%:= Assumptions: Perfect mixing • Constant mass • Manipulated variables: Flows of water and caustic

wC and _ww

Controlled variables:

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Notice that the gain of the composition loop is negative, so a direct acting controller is required for the composition loop.

mass% hr⋅ klb KOL_{1 1}_,

µ_{1 1}_, = −1.25 The gain is then:

1

µ_{1 1}_, − 1=66.7 %

When the flow control loop is closed, the gain of the composition loop increases by

So, in this case, to minimize interaction,

the caustic stream flow, FC-1, must be manipulated to control the product flow •

the water inlet flow, FC-2, must be used to control the product composition. • xP µ 0.6 0.4 0.4 0.6











= µ wC wP ww wP ww wP wC wP













:= wP ww wC Relative gains: xP KOL=





_

_0.51 ₋_0.751



_

KOL 1 ww xC⋅ wP2 klb hr mass%⋅ 1 wC − ⋅_xC wP2 klb hr mass%⋅













:= wP ww wC Open loop gains:

wC 24= klb_hr ww 16= klb_hr ww wP wC:= − wC wP xP⋅ xC :=

Numerical results:

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. xP µ wC − ⋅_xC wC − ⋅_xC− _{ww xC}⋅ ww − ⋅_xC wC − ⋅_xC− _{ww xC}⋅ ww − ⋅_xC wC − ⋅_xC− _{ww xC}⋅ wC − ⋅_xC wC − ⋅_xC− _{ww xC}⋅













= wC wC ww+ ww wC ww+ ww wC ww+ wC wC ww+













= wP

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Open-loop gains: f2 2 gpm= f1 1 gpm= f2:= fo f1− f1 fo To T2− T1 T2− ⋅ := f1 T1⋅ +

(

_{fo f1}−

)

_T2 =_{fo To}⋅ Combine balance equations:

(b) Required flows of hot and cold water and the open-loop steady-state gains.

To f1 T1⋅ +f2 T2⋅ f1 f2+ = f1 ρ⋅ cp⋅ ⋅_T1 +_{f2 ρ}_{⋅ cp}⋅ ⋅_T2=_{fo ρ}_{⋅ cp}⋅ ⋅_To Enthalpy balance: fo f1 f2= + f1 ρ⋅ +_{f2 ρ}⋅ =_{fo ρ}⋅

Total mass balance:

(a) Develop the model of the process

Assume

Negligible time delay • Constant properties • Reference temperature of 0ºF • cp 1 BTU lb degF⋅ := ρ 8.33 lb gal := T2:= 80degF T1:= 170degF To:= 110degF fo:= 3gpm Problem data:

TE

TC

Hot

water

Cold

water

S

SP

S

FE

FC

SP

f

1

f

2

f

o

T

2

T

1

T

_o

Problem 12-3. Automatic control of a household shower.

(7)

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit 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 gpm KOL_{1 0}_, µ_{1 0}_, = 30 KOL_{0 1}_, µ_{0 1}_, = 1.5 The closed-loop gains are:

1

µ_{1 0}_, − 1=50 %

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:

To µ 0.333 0.667 0.667 0.333











= µ f1 fo f2 fo f2 fo f1 fo













:= fo f2 f1

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, f₁).

To µ f1 − ⋅

(

_{T1 T2}−

)

f1 f2+

(

)

−

(

_{T1 T2}−

)

f2 − ⋅

(

_{T1 T2}−

)

f1 f2+

(

)

−

(

_{T1 T2}−

)

f2 − ⋅

(

_{T1 T2}−

)

f1 f2+

(

)

−

(

_{T1 T2}−

)

f1 − ⋅

(

_{T1 T2}−

)

f1 f2+

(

)

−

(

_{T1 T2}−

)













= f1 f1 f2+ f2 f1 f2+ f2 f1 f2+ f1 f1 f2+













= fo f2 f1

(c) Relative gains and pairing that minimize interaction.

The first row in dimensionless and the second has units of ºF/gpm.

To KOL





₂₀1 ₋1₁₀







= KOL 1 f2 T1 T2⋅

(

−

)

fo2 gpm degF 1 f1 − ⋅

(

_{T1 T2}−

)

fo2 gpm degF













:= fo f2 f1 KOL δfo δf1 δTo δf1 δfo δf2 δTo δf1













= 1 f1 f2+

(

)

⋅T1− _{f1 T1}⋅ −_{f2 T2}⋅ f1 f2+

(

)

2 1 f1 f2+

(

)

⋅T2 −_{f1 T1}⋅ − _{f2 T2}⋅ f1 f2+

(

)

2













=

(8)

Economy: _{wv E wS}= ⋅

Solute balance: _{wF xF}⋅ =_{wP xP}⋅

Combine and rearrange: wF wP E wS= + ⋅ _{xP xF}

(

wP E wS+ ⋅

)

wP ⋅ = _{xF 1} E wS⋅ wP +











⋅ =

Control valve gains: _{wP KvP mP}= ⋅ _{wS KvS mS}= ⋅

(b) Steady-state open-loop gains and relative gains.

mP mS Open-loop gains: wF KOL δwF δmP δxP δmP δwF δmS δxP δmS













= KvP KvP xF⋅ − wSE⋅ wP2











⋅ KvS E⋅ KvS xF⋅ E wP ⋅













= xP

Smith & Corripio, 3rd edition

Problem 12-4. Control of an evaporator.

SP SP SP FC FT

LC

LT _AC AT Feed Product Vapors Steam m_S m_P

w

F

x

_P

w

P

w

S

w

v

x

_F Problem data: Manipulated variables:

Product and steam valve signals.

mP and _mS

Controlled variables: Feed flow (throughput) Product composition

wF and _xP

Assume the evaporator economy E is constant.

(a) Steady-state model of the evaporator.

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Relative gains: µ KvP KvS⋅ ⋅_xF⋅E wP KvP⋅ ⋅_Kvs⋅_xF⋅E 1 wP E wS⋅ wP2 +











⋅ KvP KvS⋅ ⋅_xF⋅E2⋅_wS wP2⋅KvP⋅_Kvs⋅_xF⋅E 1 wP E wS⋅ wP2 +











⋅ KvP KvS⋅ ⋅_xF⋅E2⋅_wS wP2⋅KvP⋅_Kvs⋅_xF⋅E 1 wP E wS⋅ wP2 +











⋅ KvP KvS⋅ ⋅_xF⋅E wP KvP⋅ ⋅_Kvs⋅_xF⋅E 1 wP E wS⋅ wP2 +











⋅













= mP mS wF µ wP wP E wS+ ⋅ E wS⋅ wP E wS+ ⋅ E wS⋅ wP E wS+ ⋅ wP wP E wS+ ⋅













= 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*W_S), control throughput with the signal to •

the steam valve and the composition with the signal to the product valve.

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beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

(10)

KOL





₋_0.4171 _0.5630.9







= xP

The first row is in klb/klb, and the second is in mass%/(klb/hr). Relative gains: wP wS wF µ wP wP E wS+ ⋅ E wS⋅ wP E wS+ ⋅ E wS⋅ wP E wS+ ⋅ wP wP E wS+ ⋅













:= µ 0.6 0.4 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 closed-loop gain of the composition loop is:

mass% hr⋅ klb KOL_{1 1}_,

µ_{1 1}_, = 0.938

Smith & Corripio, 3rd edition

Problem 12-5. Multivariable control of evaporator of Problem 12-4.

Problem data: _{xF 30mass%}:= _{xP 50mass%}:= _{wF 80}klb

hr

:= E 0.9klb

klb := Combine balance equations to solve for design conditions:

wP wF xF xP ⋅ := _wS wF wP− E := _{wP 48}klb hr = _{wS 35.556}klb hr =

From the solution to Problem 12-4, ignoring the valve gains: Open-loop gains: wP wS wF KOL 1 xF − _{⋅ wS}E⋅ wP2 klb hr mass%⋅ E xF E⋅ wP klb hr mass%⋅













:=

(11)

Relative gains:

The first column is in

mole%/(klb/hr) and the second in mole%/(lbmole/hr). yD KOL





−₋3.14_1.71 _0.6760.51







= KOL xB_{0 1}_, − xB_{0 0}_, wS_{0 1}_, − wS_{0 0}_, yD_{0 1}_, − yD_{0 0}_, wS_{0 1}_, − wS_{0 0}_, xB_{0 2}_, −xB_{0 0}_, wR_{0 2}_, −wR_{0 0}_, yD_{0 2}_, −yD_{0 0}_, wR_{0 2}_, −wR_{0 0}_,













:= xB wR wS

Open-loop steady state gains:

yD:= (93.50 91.79 96.88) xB:= (6.22 3.08 8.77) Butane mole %

wR:= (70.0 70.0 75.0) Reflux flow, lbmole/hr

wS:= (24.0 25.0 24.0) Steam flow, klb/hr

Case Base Test 1 Test 2

Results of tests performed on a smlation of the column: Manipulated variables:

Steam and reflux flows Controlled variables:

Bottoms and distillate compositions The pressure and level loops are arranged as shown in the figure.

Steam Bottoms Distillate Feed Condenser LC PC LC AC AC 1 2 1 2 LT AT LT AT PT Reflux

Problem 12-6. Distillation product composition control.

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wS wR xB µ KOL_{0 0}_, ⋅KOL_{1 1}_, KOL KOL_{0 1}_, ⋅KOL_{1 0}_,









− KOL KOL_{0 1}_, ⋅KOL_{1 0}_,









− KOL KOL_{0 0}_, ⋅KOL_{1 1}_, KOL













:= µ 1.697 0.697 − 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 •

Closed-loop gain of distillate control loop:

KOL1 1,

µ_{1 1}_, = 0.398

mole% hr⋅ lbmole

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KOL f₁⋅

(

x₀− x₁

)

+ f₂⋅

(

x₀−x₂

)

fp2



_

day_kbl



_

2 ⋅ f₁⋅

(

y₀− y₁

)

+ f₂⋅

(

y₀−y₂

)

fp2



_

day_kbl



_

2 ⋅ 1 f₀⋅

(

x₁−x₀

)

+ f₂⋅

(

x₁− x₂

)

fp2



_

day_kbl



_

2 ⋅ f₀⋅

(

y₁−y₀

)

+ f₂⋅

(

y₁− y₂

)

fp2



_

day_kbl



_

2 ⋅ 1 f₀⋅

(

x₂−x₀

)

+f₁⋅

(

x₂−x₁

)

fp2



_

day_kbl



_

2 ⋅ f₀⋅

(

y₂−y₀

)

+f₁⋅

(

y₂−y₁

)

fp2



_

day_kbl



_

2 ⋅ 1













:=

Steady-state open-loop gains:

kbl day f 22.5 24.375 13.125











= f x₀ y₀ 1 x₁ y₁ 1 x₂ y₂ 1













1 − xp yp 1













fp day⋅ kbl := yp f1 y1⋅ +f2 y2⋅ + f3 y3⋅ f1 f2+ + _f3 = xP f1 x1⋅ +f2 x2⋅ + f3 x3⋅ f1 f2+ + _f3 =

Required inlet flows:

Assume

gasoline properties are a linear •

combination of the feed properties weighted by volume Constant standar density • fp 60.0 kbl day := yp:= 7.00 xp:= 89.0 Gasoline y 5.00 11.00 3.00











:= x 97.0 80.0 92.0











:= 1. Alkylate 2. Straight run 3. Reformate RVP Octane Problem data:

FC

FT

FC

FT

FC

FT SP AC AT 1 SP AC AT 2 SP FC FT 3 SP SP SP x y f f f f 1 2 3 Alkylate Straight run Reformate

Problem 12-7. Control of gasoline blender of example 12-2.5.

kbl:= 42000gal

Smith & Corripio, 3rd edition

(14)

Oct µ 1 0.375 − 0.375 0.281 0.313 0.406 0.281 − 1.062 0.219











= _{yp 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 12-2.5.

Design of static decoupler, as in Example 12-3.3.

To use the recommended pairing with the decoupler, we must swap columns 2 and 3 in the iopen-loop 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. 12-3.11:

Oct RVP Flow m1 m2 m3 f1 Alkylate D B_{0 0}_, B_{0 0}_, B_{2 0}_, B_{0 0}_, B_{1 0}_, B_{0 0}_, B_{0 1}_, B_{2 1}_, B_{2 1}_, B_{2 1}_, B_{1 1}_, B_{2 1}_, B_{0 2}_, B_{1 2}_, B_{2 2}_, B_{1 2}_, B_{1 2}_, B_{1 2}_,













:= D 1 0.75 − 0.25 − 0.706 − 1 0.294 − 0.923 0.538 1











= _f3 Reformate f2 Straight run The instrumentation diagram is the same as Fig. 12-3.6.

Alk St run Ref

f1 f2 f3 xp Oct day kbl yp RVP KOL 0.133 0.033 − 1 0.15 − 0.067 1 0.05 0.067 − 1











= fp Flow

Relative gains:

Inverse of the open-loop gains: B:= _KOL−1 B

7.5 1.875 − 5.625 − 11.25 4.688 15.938 − 0.375 0.406 0.219











= kbl day

Alk St run Ref

f1 f2 f3 µ KOL_{0 0}_, ⋅B_{0 0}_, KOL_{1 0}_, ⋅B_{0 1}_, KOL_{2 0}_, ⋅B_{0 2}_, KOL_{0 1}_, ⋅B_{1 0}_, KOL_{1 1}_, ⋅B_{1 1}_, KOL_{2 1}_, ⋅B_{1 2}_, KOL_{0 2}_, ⋅B_{2 0}_, KOL_{1 2}_, ⋅B_{2 1}_, KOL_{2 2}_, ⋅B_{2 2}_,













:= xp

(15)

Obtain the relative gains by Eq. 12-2.13, page 422: B 0.03089 0.01256 − 0.03592 6.43731 − 3.8685 8.29331 − 0.31641 − 0.06809 1.29688 −











= B:= _KOL−1 KOL 119 0.37 0.930 153 0.767 0.667 − 21 − 0.050 − 1.033 −











:=

Obtain the open-loop steady-state gains by setting s = 0 in the transfer functions:

(a) Relative gains and pairings that minimize interaction.

DCF, kg/liter FML, liters/s G s( ) 119 217 s⋅ +1 0.37 500 s⋅ + 1 0.930 500 s⋅ + 1 153 337 s⋅ +1 0.767 33 s⋅ + 1 0.667 − e−320s 166s+ 1 21 − 10 s⋅ + 1 0.050 − 10 s⋅ + 1 1.033 − 47s+ 1













= TOR, N-m Time is in sec SW, kg/s MW, kg/s SF, kg/s

Models from open-loop tests by Hubert and Woodburn (1983): 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

_AT

XT

FT

AC

XC FC

DCF

TOR

FML

Mill

Cyclone

SW

SF

MW

Problem 12-8. Control of a wet grinding circuit.

(16)

Block diagram of decoupled system: DCF, kg/liter FML, liters/sec KOL D⋅ 32.372 0 0 0 0.258 0 0 − 0 0.771 −











= TOR, N-m

Decoupled system gains: kg/sec kg/sec kg/sec

mDCF mFML

mTOR

All variables in kg/sec SWset MWset SF kg/s MW kg/s SW kg/s µ KOL_{0 0}_, ⋅B_{0 0}_, KOL_{1 0}_, ⋅B_{0 1}_, KOL_{2 0}_, ⋅B_{0 2}_, KOL_{0 1}_, ⋅B_{1 0}_, KOL_{1 1}_, ⋅B_{1 1}_, KOL_{2 1}_, ⋅B_{1 2}_, KOL_{0 2}_, ⋅B_{2 0}_, KOL_{1 2}_, ⋅B_{2 1}_, KOL_{2 2}_, ⋅B_{2 2}_,













:= TOR, N-m µ 3.676 2.382 − 0.294 − 1.922 − 2.967 0.045 − 0.754 − 0.415 1.34











= FML, liters/s 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, lead-lag units may be added and tuned on-line. The gain matrix is correct for the recommended pairing.

Obtain the decoupler from Eq. 12-3.11, page 432:

mTOR mFML mDCF SFset D 1 B_{1 0}_, B_{0 0}_, B_{2 0}_, B_{0 0}_, B_{0 1}_, B_{1 1}_, 1 B_{2 1}_, B_{1 1}_, B_{0 2}_, B_{2 2}_, B_{1 2}_, B_{2 2}_, 1













:= D 1 0.407 − 1.163 1.664 − 1 2.144 − 0.244 0.053 − 1











=

(17)

R

1

(s)

R

2

(s)

R

3

(s)

M

1

(s)

M

₂

(s)

M

₃

(s)

SF

MW

SW

TOR

FML

DCF

G

C1

(s)

G

C2

(s)

G

C3

(s)

G

11

(s)

G

₁₂

(s)

G

₁₃

(s)

G

21

(s)

G

22

(s)

G

23

(s)

G

₃₁

(s)

G

32

(s)

G

33

(s)

1.664 -0.244

0.407

0.053

1.163

2.144 -+

+

(18)

AT

XT

FT

AC

XC

FC

DCF

TOR

FML

Mill

SW

SF

MW

XY

FY

AY

-1.664 0..244 -0.407 _-0.053 1.163 -2.144

TOR

SP

_FML

SP

_DCF

SP

3 ₃

₃

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.

(19)

xB µ 1.217 0.217 − 0.217 − 1.217











= yD mS mR µ KOL_{0 0}_, ⋅KOL_{1 1}_, KOL KOL −

(

)

_{0 1}_, ⋅KOL_{1 0}_, KOL KOL −

(

)

_{0 1}_, ⋅KOL_{1 0}_, KOL KOL_{0 0}_, ⋅KOL_{1 1}_, KOL













:= xB KOL





−_2.53.0 0.75₋_3.5







%TO %CO := yD

Relative gains, from Eq. 12-2.7, page 418:: mS

mR

Obtain the open-loop steady-state gains by setting s = 0 in the transfer functions: %CO:= %

(a) Relative gains and pairing that minimizes interaction.

%TO:= %

XB s( ) 2.5 1+ 0.35sMR s( ) 3.5 1 +0.25s 1+0.35s









MS s( ) − = YD s( ) −3.0 1− 0.11s 1+ 0.35s









MR s( ) 0.75 1 +0.35sMS s( ) + =

Transfer functions from open-loop step tests (time parameters in hours): = composition of the light

key in the bottoms product xB

= composition of heavy key in the distillate yD xB yD mS mR Manipulated variables: Reflux flow

Steam flow to reboiler Controlled variables: Distillate purity

Bottoms product purity

Steam Bottoms Distillate Feed Condenser LC PC LC AC AC 1 2 1 2 LT AT LT AT PT Reflux

Problem 12-9. Decoupler design for distillation product composition control.

(20)

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.

YD s( ) −3.0 1− 0.11 s⋅ 1+ 0.35 s⋅









⋅ 0.75 1+ 0.35 s⋅ ⋅D1 s( ) +









⋅MS s( ) = =0 _{D1 s}( ) 3.0 0.75(1−0.11s) = XB s( ) 2.5 1+ 0.35 s⋅ ⋅D2 s( ) 3.5 1+ 0.25 s⋅ 1+ 0.35 s⋅









⋅ −









⋅MR s( ) = =0 D2 s( ) 3.5 2.5(1+ 0.25s) =

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. D1 s( ) 4.0%CO %CO = _{D2 s}( ) 1.4%CO %CO 1+ 0.28s 1+ 0.03s









=

The second decoupler introduces a net lead of 0.25 hour (15 minutes). Block diagram of decoupled system:

G

11

(s)

G

12

(s)

G

₂₁

(s)

G

22

(s)

D

1

(s)

D

₂

(s)

G

c1

(s)

G

_c2

(s)

R

1

(s)

R

₂

(s)

M

R

(s)

M

_S

(s)

Y

D

(s)

X

B

(s)

+

-Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit 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

(21)

D2 s( ) GP2 2, − ( )s GP_{2 1}_, ( )s = −0.9 1.1 1.5s+1 2.0s+1 e 1 0.3− ( )s − = GP_{2 2}_, ( )s _GP 2 1, ( ) D2 ss⋅ ( ) +









M2 s( ) =0 D1 s( ) GP1 1, − ( )s GP_{1 2}_, ( )s = −0.81 1.2 2.4s+1 1.4s+1 e 0.6 1.1− ( )s − = GP_{1 2}_, _{( ) D1 s}s ⋅ ( ) _GP 1 1, ( )s +









M1 s( ) =0

(b) Design decoupler and show in the block diagram.

Pairing that minimizes interaction: Control c₁ with m₂

•

Control c₂ with m₁ •

The interaction is negative (loops fight each other): relative gains are either negative or greater than unity. c2 µ −1.234 2.234 2.234 1.234 −











= µ KOL_{0 0}_, ⋅KOL_{1 1}_, KOL KOL_{0 1}_, − _KOL 1 0, ⋅ KOL KOL_{0 1}_, − _KOL 1 0, ⋅ KOL KOL_{0 0}_, ⋅KOL_{1 1}_, KOL













:= c1 m2 m1 Relative gains by Eq. 12-2.7, page 418:

KOL





0.81_1.1 1.2_0.9







:=

Obtain open-loop steady-state gains by setting s = 0 in the transfer functions:

(a) Relative gains and pairing that minimizes interaction.

GU s( ) 0.5 2.2s+ 1 1.5 − 1.8s+ 1













= GP s( ) 0.81e−0.6s 1.4s+1 1.1e−0.3s 1.5s+1 1.2e−1.1s 2.4s+ 1 0.9e−s 2s+1













= C s( )=_{GP s}( ) M s( ) +_{GPU s}( ) U s⋅ ( )

Problem 12-10. Decoupler design for a 2x2 process.

(22)

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:

D1 s( ) −0.675 2.9s+1 1.4s+1 = D2 s( ) −0.818 1.5s+1 2.0s+1 e 0.7s − =

Block diagram of the complete system:

G

P12

(s)

D

1

(s)

D

2

(s)

G

c1

(s)

G

c2

(s)

R

1

(s)

R

2

(s)

M

2

(s)

M

1

(s)

C

1

(s)

C

₂

(s)

+

-G

P11

(s)

G

U1

(s)

G

_U2

(s)

G

P22

(s)

G

_P21

(s)

+

U(s)

(23)

Gv1 s( ) G21 s⋅ ( ) −Gv1 s( )⋅G21 s( ) Gv2 s( ) G22 s⋅ ( ) ⋅Gv2 s( )⋅G22 s( ) +











⋅M1 s( ) + C2 s( ) _{Gv2 s}_{( ) G22 s}⋅ ( ) −Gv2 s( )⋅G12 s( ) Gv1 s( ) G11 s⋅ ( ) ⋅Gv1 s( )⋅G21 s( ) +











⋅M2 s( ) =

If the decouoplers can be implemeted exactly, then:

Gv1 s( ) G21 s⋅ ( )+ _{D21 s}_{( ) Gv2 s}⋅ ( )⋅_{G22 s}( )

(

)

⋅M1 s( )

+

C2 s( )=

(

_{Gv2 s}_{( ) G22 s}⋅ ( )+ _{D12 s}_{( ) Gv1 s}⋅ ( )⋅_{G21 s}( )

)

⋅_{M2 s}( ) From the block diagram, the transfer functions are:

Similarly for the other variable:

C1 s( ) _{Gv1 s}_{( ) G11 s}( ) G21 s( ) G22 s( )⋅G12 s( ) −











⋅ ⋅_{M1 s}( )

= So, the system would be _decoupled.

Simplify: Gv2 s( ) G12 s⋅ ( ) −Gv2 s( )⋅G12 s( ) Gv1 s( ) G11 s⋅ ( ) ⋅Gv1 s( )⋅G11 s( ) +











⋅M2 s( ) + C1 s( ) _{Gv1 s}_{( ) G11 s}⋅ ( ) −Gv1 s( )⋅G21 s( ) Gv2 s( ) G22 s⋅ ( ) ⋅Gv2 s( )⋅G12 s( ) +











⋅M1 s( ) =

If the decouoplers can be implemeted exactly, then:

Gv2 s( ) G12 s⋅ ( )+ _{D12 s}_{( ) Gv1 s}⋅ ( )⋅_{G11 s}( )

(

)

⋅M2 s( )

+

C1 s( )=

(

_{Gv1 s}_{( ) G11 s}⋅ ( )+ _{D21 s}_{( ) Gv2 s}⋅ ( )⋅_{G12 s}( )

)

⋅_{M1 s}( ) From the block diagram, the transfer functions are:

D21 s( ) −Gv1 s( )⋅G21 s( ) Gv2 s( ) G22 s⋅ ( ) = D12 s( ) −Gv2 s( )⋅G12 s( ) Gv1 s( ) G11 s⋅ ( ) =

The decouplers are given in Eq. 12-3.2, page 426:

Problem 12-11. Decoupling of 2x2 process of Fig. 12-3.2--show decoupled

closed-loop transfer functions

(24)

Simplify:

So, the system would be decoupled. C2 s( ) _{Gv2 s}_{( ) G22 s}( ) G12 s( ) G11 s( )⋅G21 s( ) −











⋅ ⋅_{M2 s}( ) =

(25)

w1 50= _minlb _{w2 50} lb 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. w=_{w1 w2}+ =_{Kv1 m1}⋅ +_{Kv2 m2}⋅ _{∆w Kv1 ∆m1}= ⋅ +_{Kv2 ∆m2}⋅ =0 _∆m1 −Kv2 Kv1 ∆m2 = x w1 x1⋅ +w2 x2⋅ w1 w2+ = ∆x

(

w1 w2+

)

x1 w1 x1− ⋅ −w2 x2⋅ w1 w2+

(

)

2 Kv1 ∆m1⋅ w1 w2+

(

)

x2 w1 x1− ⋅ −_{w2 x2}⋅ w1 w2+

(

)

2 Kv2 ∆m2⋅ + = =0 w2 x1 x2⋅

(

−

)

⋅_Kv1⋅_∆m1+ _{w1 x2 x1}⋅

(

−

)

⋅_Kv2⋅_∆m2=0 _∆m2 Kv1 w2⋅ Kv2 w1⋅ ∆m1 =

Kv1:= _100%COwmax _Kv2 wmax

100%CO := _{Kv1 1.5} lb min %CO⋅ = _{Kv2 1.5} lb min %CO⋅ =

The gains of the decouplers are: ₋_Kv1 Kv2 = −1

Kv1 w2⋅ Kv2 w1⋅ =1

Smith & Corripio, 3rd edition

Problem 12-12. Decoupler design for blending tank of Example 12-5.1.

AC SP AT SP w x₁1

w

x w x₂2

m

1 2 V FC FT Design conditions: x1:= 10mass% _x2:= 30mass% w 100 lb min := x:= 20mass%

Valves are linear with constant pressure drop.

wmax 150:= _minlb Analyzer transmitter AT:

xmin 5mass%:= _{xmax 35mass%}:=

At the design conditions: _w1 w x −x1 x2 x1− ⋅

(26)

+

SP AT SP w x₁1

w

x w x₂2

m

1

m

₂ V FC FT FY AY 3

+

-AC 3 Instrumentation diagram:

FY: summer, keeps flow constant when the analyzer controller varies m₂

AY: summer, keeps composition constant when the flow controller varies m₁

Both require a bias of: w1

Kv1 = 33.33 %CO w2

Kv2 = 33.33 %CO

(27)

Ratio xF xPset

=

where, at design conditions: wP Ratio wF= ⋅

This is a ratio controller with the ratio adjustable by the output of the composition controller AC:

wP xF xPset wF ⋅ = xPset xF wF wP ⋅ =

Want the output of the analyzer controller AC to be the composition: mS





_{100%CO E}wFmax_{⋅ KvS}_⋅







mFC KvP E KvS⋅











mP − =

wFset= _100%COwFmax

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:

mS wF set E KvS⋅ KvP E KvS⋅ mP − = wS wF set wP − E = So wFset=wP E wS+ ⋅

Want the output of the feed flow controller FC to be the sum of the vapors and the product:

Decoupler design:

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. wS KvS mS= ⋅ wP KvP mP= ⋅ xP xF wF⋅ wP = wF wP E wS= + ⋅

From the solution of Problem 12-4:

SP SP SP FC FT

LC

LT _AC AT Feed Product Vapors Steam m_S m_P

w

F

x

_P

w

P

w

S

w

v

x

_F

Problem 12-13. Non-linear static decoupler for evaporator of Problem 12-4.

(28)

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:

Ratio xF xPset wFmax 100%CO KvP⋅ = mP Ratio mFC= ⋅ where Instrumentation diagram: SP SP SP FC FT

LC

AC LT AT Feed Product Vapors Steam mS m_P

w

F

x

P

w

P

w

S

x

F FY AY 3 X

m

FC

Ratio

FY: summer, subtracts the

scaled signal to the product valve from the signal to the seam valve to maintain the feed flow constant. 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.